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Unformatted text preview: Actuarial Mathematics
for
Life Contingent Risks
David C M Dickson,
Mary R Hardy,
Howard R Waters
February 2, 2009 2 To Carolann,
Vivien
and Phelim Contents
1 Introduction to Life Insurance 19 1.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.3 Life insurance and annuity contracts . . . . . . . . . . . . . . . . . . . . . 22
1.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 1.3.2 Traditional insurance contracts . . . . . . . . . . . . . . . . . . . . 23 1.3.3 Modern insurance contracts . . . . . . . . . . . . . . . . . . . . . . 26 1.3.4 Distribution methods . . . . . . . . . . . . . . . . . . . . . . . . . . 28 1.3.5 Underwriting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.3.6 Premiums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 1.3.7 Life annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 1.4 Other insurance contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 1.5 Pension beneﬁts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
1.5.1
1.5.2 1.6 Deﬁned beneﬁt and deﬁned contribution pensions . . . . . . . . . . 35
Deﬁned beneﬁt pension design . . . . . . . . . . . . . . . . . . . . . 35 Mutual and proprietary insurers . . . . . . . . . . . . . . . . . . . . . . . . 36
3 4 CONTENTS
1.7 Typical problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 1.8 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 1.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 2 Survival Models 41 2.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.2 The future lifetime random variable . . . . . . . . . . . . . . . . . . . . . . 41 2.3 The force of mortality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.4 Actuarial notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 2.5 Mean and standard deviation of Tx . . . . . . . . . . . . . . . . . . . . . . 57 2.6 Curtate future lifetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
2.6.1 Kx and ex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 2.6.2 The complete and curtate expected future lifetimes, ex and ex . . . 64 ◦ 2.7 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 2.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 3 Life Tables and Selection 75 3.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 3.2 Life tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 3.3 Fractional age assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . 79 3.3.1 Uniform distribution of deaths . . . . . . . . . . . . . . . . . . . . . 80 3.3.2 Constant force of mortality . . . . . . . . . . . . . . . . . . . . . . 85 3.4 National life tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 3.5 Survival models for life insurance policyholders . . . . . . . . . . . . . . . 91 CONTENTS 5 3.6 Life insurance underwriting . . . . . . . . . . . . . . . . . . . . . . . . . . 93 3.7 Select and ultimate survival models . . . . . . . . . . . . . . . . . . . . . . 96 3.8 Notation and formulae for select survival models . . . . . . . . . . . . . . . 99 3.9 Select life tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 3.10 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
3.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
4 Insurance Beneﬁts 121 4.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 4.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 4.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 4.4 Valuation of insurance beneﬁts . . . . . . . . . . . . . . . . . . . . . . . . . 124
4.4.1
4.4.2 Whole life insurance: the annual case, Ax 4.4.3 Whole life insurance: the 1/mthly case, Ax 4.4.4 Recursions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 4.4.5 Term insurance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 4.4.6 Pure endowment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 4.4.7 Endowment insurance . . . . . . . . . . . . . . . . . . . . . . . . . 142 4.4.8
4.5 ¯
Whole life insurance: the continuous case, Ax . . . . . . . . . . . . 124 Deferred insurance beneﬁts . . . . . . . . . . . . . . . . . . . . . . . 145 . . . . . . . . . . . . . . 128
(m) . . . . . . . . . . . . 130 (m)
¯
Relating Ax , Ax and Ax
. . . . . . . . . . . . . . . . . . . . . . . . . . . 148 4.5.1 Using the uniform distribution of deaths assumption . . . . . . . . 148 4.5.2 Using the claims acceleration approach . . . . . . . . . . . . . . . . 150 6 CONTENTS
4.6 Variable insurance beneﬁts . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 4.7 Functions for select lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 4.8 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 4.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 5 Annuities 169 5.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 5.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 5.3 Review of annuitiescertain 5.4 Annual life annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 . . . . . . . . . . . . . . . . . . . . . . . . . . 170 5.4.1
5.4.2 Term annuitydue . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 5.4.3 Whole life immediate annuity . . . . . . . . . . . . . . . . . . . . . 177 5.4.4
5.5 Whole life annuitydue . . . . . . . . . . . . . . . . . . . . . . . . . 172 Term immediate annuity . . . . . . . . . . . . . . . . . . . . . . . . 179 Annuities payable continuously . . . . . . . . . . . . . . . . . . . . . . . . 180
5.5.1
5.5.2 5.6 Whole life continuous annuity . . . . . . . . . . . . . . . . . . . . . 180
Term continuous annuity . . . . . . . . . . . . . . . . . . . . . . . . 183 Annuities payable m times per year . . . . . . . . . . . . . . . . . . . . . . 184
5.6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 5.6.2 Life annuities payable m times a year . . . . . . . . . . . . . . . . . 185 5.6.3 Term annuities payable m times a year . . . . . . . . . . . . . . . . 186 5.7 Comparison of annuities by payment frequency . . . . . . . . . . . . . . . . 188 5.8 Deferred annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 CONTENTS
5.9 7 Guaranteed annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 5.10 Increasing annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
5.10.1 Arithmetically increasing annuities . . . . . . . . . . . . . . . . . . 197
5.10.2 Geometrically increasing annuities . . . . . . . . . . . . . . . . . . . 199
5.11 Evaluating annuity functions . . . . . . . . . . . . . . . . . . . . . . . . . . 201
5.11.1 Recursions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
5.11.2 Applying the UDD assumption . . . . . . . . . . . . . . . . . . . . 202
5.11.3 Woolhouse’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . 204
5.12 Numerical illustrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
5.13 Functions for select lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
5.14 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
5.15 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
6 Premium Calculation 219 6.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 6.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 6.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 6.4 The present value of future loss random variable . . . . . . . . . . . . . . . 222 6.5 The equivalence principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
6.5.1 Net premiums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 6.6 Gross premium calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 6.7 Proﬁt 6.8 The portfolio percentile premium principle . . . . . . . . . . . . . . . . . . 245 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 8 CONTENTS
6.9 Extra risks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249
6.9.1 Age rating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 6.9.2 Constant addition to µx . . . . . . . . . . . . . . . . . . . . . . . . 250 6.9.3 Constant multiple of mortality rates . . . . . . . . . . . . . . . . . 252 6.10 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 255
6.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
7 Policy Values 265 7.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 7.2 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 7.3 Policies with annual cash ﬂows . . . . . . . . . . . . . . . . . . . . . . . . . 266
7.3.1
7.3.2 Policy values for policies with annual cash ﬂows . . . . . . . . . . . 274 7.3.3 Recursive formulae for policy values . . . . . . . . . . . . . . . . . . 283 7.3.4 Annual proﬁt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 7.3.5
7.4 The future loss random variable . . . . . . . . . . . . . . . . . . . . 266 Asset shares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297 Policy values with 1/mthly cash ﬂows . . . . . . . . . . . . . . . . . . . . . 301
7.4.1
7.4.2 7.5 Recursions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
Valuation between premium dates . . . . . . . . . . . . . . . . . . . 303 Continuous cash ﬂows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306
7.5.1
7.5.2 7.6 Thiele’s diﬀerential equation . . . . . . . . . . . . . . . . . . . . . . 306
Numerical solution of Thiele’s diﬀerential equation . . . . . . . . . 311 Policy alterations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 CONTENTS 9 7.7 Retrospective policy value . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 7.8 Negative policy values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 7.9 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 7.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
8 Multiple State Models 341 8.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 8.2 Examples of multiple state models . . . . . . . . . . . . . . . . . . . . . . . 342
8.2.1 The alive–dead model . . . . . . . . . . . . . . . . . . . . . . . . . 342 8.2.2 Term insurance with increased beneﬁt on accidental death . . . . . 343 8.2.3 The permanent disability model . . . . . . . . . . . . . . . . . . . . 345 8.2.4 The disability income insurance model . . . . . . . . . . . . . . . . 345 8.2.5 The joint life and last survivor model . . . . . . . . . . . . . . . . . 346 8.3 Assumptions and notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 8.4 Formulae for probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . 353
8.4.1 Kolmogorov’s forward equations . . . . . . . . . . . . . . . . . . . 358 8.5 Numerical evaluation of probabilities . . . . . . . . . . . . . . . . . . . . . 359 8.6 Premiums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364 8.7 Policy values and Thiele’s diﬀerential equation . . . . . . . . . . . . . . . . 369
8.7.1 The disability income model . . . . . . . . . . . . . . . . . . . . . . 370 8.7.2 Thiele’s diﬀerential equation – the general case . . . . . . . . . . . 375 8.8 Multiple decrement models . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 8.9 Joint life and last survivor beneﬁts . . . . . . . . . . . . . . . . . . . . . . 383 10 CONTENTS
8.9.1 The model and assumptions . . . . . . . . . . . . . . . . . . . . . . 383 8.9.2 Joint life and last survivor probabilities . . . . . . . . . . . . . . . . 384 8.9.3 Joint life annuity and insurance functions . . . . . . . . . . . . . . . 386 8.9.4 An important special case: independent survival models . . . . . . 394 8.10 Transitions at speciﬁed ages . . . . . . . . . . . . . . . . . . . . . . . . . . 399
8.11 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 405
8.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407
9 Pension Mathematics 425 9.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 9.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 9.3 The salary scale function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427 9.4 Setting the DC contribution . . . . . . . . . . . . . . . . . . . . . . . . . . 430 9.5 The service table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 9.6 Valuation of beneﬁts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447
9.6.1 Final salary plans . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 9.6.2 Career average earnings plans . . . . . . . . . . . . . . . . . . . . . 454 9.7 Funding plans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 9.8 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 9.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 10 Interest Rate Risk 477 10.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477
10.2 The yield curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 CONTENTS 11 10.3 Valuation of insurances and life annuities . . . . . . . . . . . . . . . . . . . 482
10.3.1 Replicating nonpar cash ﬂows . . . . . . . . . . . . . . . . . . . . . 484
10.4 Diversiﬁable and nondiversiﬁable risk . . . . . . . . . . . . . . . . . . . . . 487
10.4.1 Diversiﬁable mortality risk . . . . . . . . . . . . . . . . . . . . . . . 489
10.4.2 Nondiversiﬁable risk . . . . . . . . . . . . . . . . . . . . . . . . . 491 10.5 Monte Carlo simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498
10.6 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 506
10.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507
11 Emerging Costs for Traditional Life Insurance 515 11.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515
11.2 Proﬁt testing for traditional life insurance . . . . . . . . . . . . . . . . . . 516
11.2.1 The net cash ﬂows for a policy . . . . . . . . . . . . . . . . . . . . . 516
11.2.2 Reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519
11.3 Proﬁt measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523
11.4 A further example of a proﬁt test . . . . . . . . . . . . . . . . . . . . . . . 525
11.5 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 537
11.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538
12 Emerging Costs for EquityLinked Insurance 547 12.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547
12.2 Equitylinked insurance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548
12.3 Deterministic proﬁt testing . . . . . . . . . . . . . . . . . . . . . . . . . . . 549
12.4 Stochastic proﬁt testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561 12 CONTENTS
12.5 Stochastic pricing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567
12.6 Stochastic reserving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569
12.6.1 Reserving for policies with nondiversiﬁable risk . . . . . . . . . . . 569
12.6.2 Quantile reserving . . . . . . . . . . . . . . . . . . . . . . . . . . . 570
12.6.3 CTE reserving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573
12.6.4 Comments on reserving . . . . . . . . . . . . . . . . . . . . . . . . . 574
12.7 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 575
12.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577 13 Option Pricing 587 13.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587
13.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587
13.3 The ‘no arbitrage’ assumption . . . . . . . . . . . . . . . . . . . . . . . . . 588
13.4 Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 590
13.5 The binomial option pricing model . . . . . . . . . . . . . . . . . . . . . . 592
13.5.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592
13.5.2 Pricing over a single time period . . . . . . . . . . . . . . . . . . . . 593
13.5.3 Pricing over two time periods . . . . . . . . . . . . . . . . . . . . . 599
13.5.4 Summary of the binomial model option pricing technique . . . . . . 603
13.6 The BlackScholesMerton model . . . . . . . . . . . . . . . . . . . . . . . 604
13.6.1 The model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
13.6.2 The BlackScholesMerton option pricing formula . . . . . . . . . . 606
13.7 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 621 CONTENTS 13 13.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623
14 Embedded Options 629 14.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629
14.2 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629
14.3 Guaranteed minimum maturity beneﬁts . . . . . . . . . . . . . . . . . . . . 632
14.3.1 Pricing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632 14.3.2 Reserving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 636
14.4 Guaranteed minimum death beneﬁt . . . . . . . . . . . . . . . . . . . . . . 639
14.4.1 Pricing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 639 14.4.2 Reserving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643
14.5 Pricing methods for embedded options . . . . . . . . . . . . . . . . . . . . 647
14.6 Risk management . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651
14.7 Emerging costs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653
14.8 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 666
14.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668
A Probability theory 677 A.1 Probability distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677
A.1.1 Binomial distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 677
A.1.2 Uniform distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 678
A.1.3 Normal distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 678
A.1.4 Lognormal distribution . . . . . . . . . . . . . . . . . . . . . . . . . 680
A.2 The central limit theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 683 14 CONTENTS
A.3 Functions of a random variable . . . . . . . . . . . . . . . . . . . . . . . . 684
A.3.1 Discrete random variables . . . . . . . . . . . . . . . . . . . . . . . 684
A.3.2 Continuous random variables . . . . . . . . . . . . . . . . . . . . . 685
A.3.3 Mixed random variables . . . . . . . . . . . . . . . . . . . . . . . . 686
A.4 Conditional expectation and conditional variance . . . . . . . . . . . . . . 687
A.5 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 689 B Numerical techniques 691 B.1 Numerical integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691
B.1.1 The trapezium rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 692
B.1.2 Repeated Simpson’s rule . . . . . . . . . . . . . . . . . . . . . . . . 693
B.1.3 Integrals over an inﬁnite interval . . . . . . . . . . . . . . . . . . . 694
B.2 Woolhouse’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 696
B.3 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 697
C Simulation 699 C.1 The inverse transform method . . . . . . . . . . . . . . . . . . . . . . . . . 699
C.2 Simulation from a normal distribution . . . . . . . . . . . . . . . . . . . . 701
C.2.1 The BoxMuller method . . . . . . . . . . . . . . . . . . . . . . . . 701
C.2.2 The polar method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702
C.3 Notes and further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 703 Preface
Life insurance has undergone enormous change in the last two to three decades. New and
innovative products have been developed at the same time as we have seen vast increases
in computational power. In addition, the ﬁeld of ﬁnance has experienced a revolution
in the development of a mathematical theory of options and ﬁnancial guarantees, ﬁrst
pioneered in the work of Black, Scholes and Merton, and actuaries have come to realize
the importance of that work to risk management in actuarial contexts.
Given the changes occurring in the interconnected worlds of ﬁnance and life insurance,
we believe that this is a good time to recast the mathematics of life contingent risk to be
better adapted to the products, science and technology that are relevant to current and
future actuaries.
In this book we have developed the theory to measure and manage risks that are contingent
on demographic experience as well as on ﬁnancial variables. The material is presented with
a certain level of mathematical rigour; we intend for readers to understand the principles
involved, rather than to memorize methods or formulas. The reason is that a rigorous
approach will prove more useful in the long run than a short term utilitarian outlook,
as theory can be adapted to changing products and technology in ways that techniques,
without scientiﬁc support, cannot.
We start from a traditional approach, and then develop a more contemporary perspective.
The ﬁrst seven chapters set the context for the material, and cover traditional actuarial
models and theory of life contingencies, with modern computational techniques integrated 15 16 CONTENTS throughout, and with an emphasis on the practical context for the survival models and
valuation methods presented. Through the focus on realistic contracts and assumptions,
we aim to foster a general business awareness in the life insurance context, at the same
time as we develop the mathematical tools for risk management in that context.
In Chapter 8 we introduce multiple state models, which generalize the lifedeath contingency structure of previous chapters. Using multiple state models allows a single framework for a wide range of insurance, including beneﬁts which depend on health status, on
cause of death beneﬁts, or on two or more lives.
In Chapter 9 we apply the theory developed in the earlier chapters to problems involving
pension beneﬁts. Pension mathematics has some specialized concepts, particularly in
funding principles, but in general this chapter is an application of the theory in the
preceding chapters.
In Chapter 10 we move to a more sophisticated view of interest rate models and interest
rate risk. In this chapter we explore the crucially important diﬀerence between diversiﬁable and nondiversiﬁable risk. Investment risk represents a source of nondiversiﬁable
risk, and in this chapter we show how we can reduce the risk by matching cash ﬂows from
assets and liabilities.
In Chapter 11 we continue the cash ﬂow approach, developing the emerging cash ﬂows for
traditional insurance products. One of the liberating aspects of the computer revolution
for actuaries is that we are no longer required to summarize complex beneﬁts in a single
actuarial value; we can go much further in projecting the cash ﬂows to see how and when
surplus will emerge. This is much richer information that the actuary can use to assess
proﬁtability and to better manage portfolio assets and liabilities.
In Chapter 12 we repeat the emerging cash ﬂow approach, but here we look at equity linked
contracts, where a ﬁnancial guarantee is commonly part of the contingent beneﬁt. The
real risks for such products can only be assessed taking the random variation in potential
outcomes into consideration, and we demonstrate this with Monte Carlo simulation of the
emerging cash ﬂows. CONTENTS 17 The products that are explored in Chapter 12 contain ﬁnancial guarantees embedded in
the life contingent beneﬁts. Option theory is the mathematics of valuation and risk management of ﬁnancial guarantees. In Chapter 13 we introduce the fundamental assumptions
and results of option theory.
In Chapter 14 we apply option theory to the embedded options to the ﬁnancial guarantees
in insurance products. The theory can be used for pricing and for determining appropriate
reserves, as well as for assessing proﬁtability.
The material in this book is designed for undergraduate and graduate programs in actuarial science, and for those selfstudying for professional actuarial exams. Students should
have suﬃcient background in probability to be able to calculate moments of functions
of one or two random variables, and to handle conditional expectations and variances.
We also assume familiarity with binomial, uniform, exponential, normal and lognormal
distributions. Some of the more important results are reviewed in Appendix A. We also
assume that readers have completed an introductory level course in the mathematics of
ﬁnance, and are aware of the actuarial notation for annuitiescertain.
Throughout, we have opted to use examples that liberally call on spreadsheet style software. Spreadsheets are a ubiquitous tool in actuarial practice, and it is natural to use
them throughout, allowing us to use more realistic examples, rather than having to simplify for the sake of mathematical tractability. Other software could be used equally
eﬀectively, but spreadsheets represent a fairly universal language that is easily accessible.
To keep the computation requirements reasonable, we have ensured that every example
and exercise can be completed in Microsoft Excel, without needing any VBA code or
macros. Readers who have suﬃcient familiarity to write their own code may ﬁnd more
eﬃcient solutions than those that we have presented, but our principle was that no reader
should need to know more than the basic Excel functions and applications. It will be
very useful for anyone working through the material of this book to construct their own
spreadsheet tables as they work through the ﬁrst seven chapters, to generate mortality
and actuarial functions for a range of mortality models and interest rates. In the worked
examples in the text, we have worked with greater accuracy than we record, so there will 18 CONTENTS be some diﬀerences from rounding when working with intermediate ﬁgures.
One of the advantages of spreadsheets is the ease of implementation of numerical integration algorithms. We assume that students are aware of the principles of numerical
integration, and we give some of the most useful algorithms in Appendix B.
The material in this book is appropriate for two onesemester courses. The ﬁrst seven
chapters form a fairly traditional basis, and would reasonably constitute a ﬁrst course.
Chapters 8 to 14 introduce more contemporary material. Chapter 13 may be omitted by
readers who have studied an introductory course covering pricing and delta hedging in
a BlackScholesMerton model. Chapter 9, on pension mathematics, is not required for
subsequent chapters, and could be omitted if a single focus on life insurance is preferred. Acknowledgements
Many of our students and colleagues have made valuable comments on earlier drafts of
parts of the book. Particular thanks go to Carole Bernard, Phelim Boyle, Johnny Li, Ana
Maria Mera, Kok Keng Siaw and Matthew Till.
The authors gratefully acknowledge the contribution of the Departments of Statistics
and Actuarial Science, University of Waterloo, and Actuarial Mathematics and Statistics,
HeriotWatt University, in welcoming the nonresident authors for short visits to work on
this book. These visits signiﬁcantly shortened the time it has taken to write the book (to
only one year beyond the original deadline). David Dickson
University of Melbourne
Mary Hardy
University of Waterloo
Howard Waters
HeriotWatt University. Chapter 1
Introduction to Life Insurance
1.1 Summary Actuaries apply scientiﬁc principles and techniques from a range of other disciplines to
problems involving risk, uncertainty and ﬁnance. In this chapter we set the context
for the mathematics of later chapters, by describing some of the background to modern
actuarial practice in life insurance, followed by a brief description of the major types of
life insurance products that are sold in developed insurance markets. Because pension
liabilities are similar in many ways to life insurance liabilities, we also describe some
common pension beneﬁts. We give examples of the actuarial questions arising from the
risk management of these contracts. How to answer such questions, and solve the resulting
problems, is the subject of the following chapters. 1.2 Background The ﬁrst actuaries were employed by life insurance companies in the early 18th century to
provide a scientiﬁc basis for managing the companies’ assets and liabilities. The liabilities
19 20 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE depended on the number of deaths occurring amongst the insured lives each year. The
modeling of mortality became a topic of both commercial and general scientiﬁc interest,
and it attracted many signiﬁcant scientists and mathematicians to actuarial problems,
with the result that much of the early work in the ﬁeld of probability was closely connected
with the development of solutions to actuarial problems.
The earliest life insurance policies provided that the policyholder would pay an amount,
called the premium, to the insurer. If the named life insured died during the year that
the contract was in force, the insurer would pay a predetermined lump sum, the sum
insured, to the policyholder or his or her estate. So, the ﬁrst life insurance contracts
were annual contracts. Each year the premium would increase as the probability of death
increased. If the insured life became very ill at the renewal date, the insurance might not
be renewed, in which case no beneﬁt would be paid on the life’s subsequent death. Over a
large number of contracts, the premium income each year should approximately match the
claims outgo. This method of matching income and outgo annually, with no attempt to
smooth or balance the premiums over the years, is called assessmentism. This method
is still used for group life insurance, where an employer purchases life insurance cover for
its employees on a year to year basis.
The radical development in the later 18th century was the level premium contract. The
problem with assessmentism was that the annual increases in premiums discouraged policyholders from renewing their contracts. The level premium policy oﬀered the policyholder the option to lockin a regular premium, payable perhaps weekly, monthly, quarterly or annually, for a number of years. This was much more popular with policyholders,
as they would not be priced out of the insurance contract just when it might be most
needed. For the insurer, the attraction of the longer contract was a greater likelihood of
the policyholder paying premiums for a longer period. However, a problem for the insurer
was that the longer contracts were more complex to model, and oﬀered more ﬁnancial
risk. For these contracts then, actuarial techniques had to develop beyond the year to
year modeling of mortality probabilities. In particular, it became necessary to incorporate
ﬁnancial considerations into the modeling of income and outgo. Over a 1 year contract, 1.2. BACKGROUND 21 the time value of money is not a critical aspect. Over, say, a 30 year contract, it becomes
a very important part of the modeling and management of risk.
Another development in the life insurance in the 19th century was the concept of insurable interest. This was a requirement in law that the person contracting to pay the life
insurance premiums should face a ﬁnancial loss on the death of the insured life that was
no less than the sum insured under the policy. The insurable interest requirement disallowed the use of insurance as a form of gambling on the lives of public ﬁgures, but more
importantly, removed the incentive for a policyholder to hasten the death of the named
insured life. Subsequently, insurance policies tended to be purchased by the insured life,
and in the rest of this book we use the convention that the policyholder who pays the
premiums is also the life insured, whose survival or death triggers the payment of the sum
insured under the conditions of the contract.
The earliest studies of mortality include life tables constructed by John Graunt and Edmund Halley. A life table summarizes a survival model by specifying the proportion of
lives that are expected to survive to each age. Using London mortality data from the early
17th century, Graunt proposed, for example, that each new life had a probability of 40%
of surviving to age 16, and a probability of 1% of surviving to age 76. Edmund Halley,
famous for his astronomical calculations, used mortality data from the city of Breslau in
the late 17th century as the basis for his life table, which, like Graunt’s, was constructed
by proposing the average (‘medium’ in Halley’s phrase) proportion of survivors to each
age from an arbitrary number of births. Halley took the work two steps further. First, he
used the table to draw inference about the conditional survival probabilities at intermediate ages. That is, given the probability that a newborn life survives to each subsequent
age, it is possible to infer the probability that a life aged, say, 20, will survive to each
subsequent age, using the condition that a life aged zero survives to age 20. The second
major innovation was that Halley combined the mortality data with an assumption about
interest rates to ﬁnd the value of a whole life annuity at diﬀerent ages. A whole life
annuity is a contract paying a level sum at regular intervals while the named life (the
annuitant) is still alive. The calculations in Halley’s paper bear a remarkable similarity 22 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE to some of the work still used by actuaries in pensions and life insurance.
This book continues in the tradition of combining models of mortality with models in
ﬁnance to develop a framework for pricing and risk management of long term policies in
life insurance. Many of the same techniques are relevant also in pensions mathematics.
However, there have been many changes since the ﬁrst long term policies of the late 18th
century. 1.3
1.3.1 Life insurance and annuity contracts
Introduction The life insurance and annuity contracts that were the object of study of the early actuaries
were very similar to the contracts written up to the 1980s in all the developed insurance
markets. Recently, however, the design of life insurance products has radically changed,
and the techniques needed to manage these more modern contracts are more complex
than ever. The reasons for the changes include:
• Increased interest by the insurers in oﬀering combined savings and insurance products. The original life insurance products oﬀered a payment to indemnify (or oﬀset)
the hardship caused by the death of the policyholder. Many modern contracts
combine the indemnity concept with an opportunity to invest.
• More powerful computational facilities allow more complex products to be modeled.
• Policyholders have become more sophisticated investors, and require more options
in their contracts, allowing them to vary premiums or sums insured, for example.
• More competition has led to insurers creating increasingly complex products in order
to attract more business. 1.3. LIFE INSURANCE AND ANNUITY CONTRACTS 23 • The risk management techniques in ﬁnancial products have also become increasingly
complex, and insurers have oﬀered some beneﬁts, particularly ﬁnancial guarantees,
that require modern sophisticated techniques from ﬁnancial engineering to measure
and manage the risk.
In the remainder of this section we describe some of the most important modern insurance
contracts, which will later be used as examples in the book. Diﬀerent countries have
diﬀerent names and types of contracts; we have tried to cover the major contract types
in North America, the United Kingdom and Australia.
The basic transaction of life insurance is an exchange; the policyholder pays premiums in
return for a later payment from the insurer which is life contingent, by which we mean
that it depends on the death or survival or possibly the state of health of the policyholder.
We usually use the term ‘insurance’ when the beneﬁt is paid as a single lump sum, either
on the death of the policyholder or on survival to a predetermined maturity date. (In
the UK it is common to use the term ‘assurance’ for insurance contracts involving lives,
and insurance for contracts involving property.) An annuity is a beneﬁt in the form of a
regular series of payments, usually conditional on the survival of the policyholder. 1.3.2 Traditional insurance contracts Term, whole life and endowment insurance are the traditional products, providing cash
beneﬁts on death or maturity, usually with predetermined premium and beneﬁt amounts.
We describe each in a little more detail here.
Term Insurance pays a lump sum beneﬁt on the death of the policyholder, provided
death occurs before the end of a speciﬁed term. Term insurance allows a policyholder
to provide a ﬁxed sum for his or her dependents in the event of the policyholder’s
death.
Level term insurance indicates a level sum insured and regular, level premiums. 24 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE
Decreasing term insurance indicates that the sum insured and (usually) premiums
decrease over the term of the contract. Decreasing term insurance is popular in the
UK where it is used in conjunction with a home mortgage; if the policyholder dies,
the remaining mortgage is paid from the term insurance proceeds.
Renewable term insurance oﬀers the policyholder the option of renewing the policy
at the end of the original term, without further evidence of the policyholder’s health
status. In North America, Yearly Renewable Term (YRT) insurance is common,
under which insurability is guaranteed for some ﬁxed period, though the contract is
only written for one year at a time.
Convertible term insurance oﬀers the policyholder the option to convert to a whole
life or endowment insurance at the end of the original term, without further evidence
of the policyholder’s health status. Whole Life Insurance pays a lump sum beneﬁt on the death of the policyholder whenever it occurs. For regular premium contracts, the premium is often payable only
up to some maximum age, such as 80. This avoids the problem that older lives may
be less able to pay the premiums.
Endowment Insurance oﬀers a lump sum beneﬁt paid either on the death of the policyholder or at the end of a speciﬁed term, whichever occurs ﬁrst. This is a mixture
of a term insurance beneﬁt and a savings element. If the policyholder dies, the sum
insured is paid just as under term insurance; if the policyholder survives, the sum
insured is treated as a maturing investment. Endowment insurance is obsolete in
many jurisdictions. Traditional endowment insurance policies are not currently sold
in the UK, but there are large portfolios of policies on the books of UK insurers,
because until the late 1990s, endowment insurance policies were often used to repay
home mortgages. The policyholder (who is the home owner) paid interest on the
mortgage loan, and the principal was paid from the proceeds on the endowment insurance, either on the death of the policyholder or at the ﬁnal mortgage repayment
date. 1.3. LIFE INSURANCE AND ANNUITY CONTRACTS 25 Endowment insurance policies are becoming popular in developing nations, particularly for ‘microinsurance’ where the amounts involved are small. It is hard for small
investors to achieve good rates of return on investments, because of heavy expense
charges. By pooling the death and survival beneﬁts under the endowment contract,
the policyholder gains on the investment side from the resulting economies of scale,
and from the investment expertise of the insurer.
With proﬁts insurance
Also part of the traditional design of insurance is the division of business into ‘with
proﬁt’, (also known, especially in North America, as ‘participating’, or ‘par’ business),
and ‘without proﬁt’ (also known as ‘nonparticipating’ or ‘nonpar’). Under with proﬁt
arrangements, the proﬁts earned on the invested premiums are shared with the policyholders. In North America, the with proﬁt arrangement often takes the form of cash
dividends or reduced premiums. In the UK and in Australia the traditional approach
is to use the proﬁts to increase the sum insured, through bonuses called ‘reversionary
bonuses’ and ‘terminal bonuses’. Reversionary bonuses are awarded during the term
of the contract; once a reversionary bonus is awarded it is guaranteed. Terminal bonuses
are awarded when the policy matures, either through the death of the insured, or when an
endowment policy reaches the end of the term. Reversionary bonuses may be expressed
as a percentage of the total of the previous sum insured plus bonus, or as a percentage of
the original sum insured plus a diﬀerent percentage of the previously declared bonuses.
Reversionary and terminal bonuses are determined by the insurer based on the investment
performance of the invested premiums.
For example, suppose an insurance is issued with sum insured $100 000. At the end of the
ﬁrst year of the contract a bonus of 2% on the sum insured and 5% on previous bonuses is
declared; in the following two years, the rates are 2.5% and 6%. Then the total guaranteed
sum insured increases each year as follows:
If the policyholder dies, the total death beneﬁt payable would be the original sum in 26 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE
Bonus on Original
Year
Sum Insured
Bonus on Bonus
1
2%
5%
2
2.5%
6%
3
2.5%
6%
.
.
.
.
.
.
.
.
. Total Bonus
2 000.0
4 620.0
7 397.2
.
.
. sured plus reversionary bonuses already declared, increased by a terminal bonus if the
investment returns earned on the premiums have been suﬃcient.
With proﬁts contracts may be used to oﬀer policyholders a savings element with their life
insurance. However, the traditional with proﬁts contract is designed primarily for the life
insurance cover, with the savings aspect a secondary feature. 1.3.3 Modern insurance contracts In recent years insurers have provided more ﬂexible products that combine the death
beneﬁt coverage with a signiﬁcant investment element, as a way of competing for policyholders’ savings with other institutions, for example, banks, openended investment
companies (e.g. mutual funds in North America, or unit trusts in the UK). Additional
ﬂexibility also allows policyholders to purchase less insurance when their ﬁnances are
tight, and then increase the insurance coverage when they have more money available.
In this section we describe some examples of modern, ﬂexible insurance contracts.
Universal Life Insurance combines investment and life insurance. The policyholder
determines a premium and a level of life insurance cover. Some of the premium
is used to fund the life insurance; the remainder is paid into an investment fund.
Premiums are ﬂexible, as long as they are suﬃcient to pay for the designated sum
insured under the term insurance part of the contract. Under variable universal life, 1.3. LIFE INSURANCE AND ANNUITY CONTRACTS 27 there is a range of funds available for the policyholder to select from. Universal life
is a common insurance contract in North America.
Unitised With Proﬁt is a UK insurance contract; it is an evolution from the conventional withproﬁt policy, designed to be more transparent than the original. Premiums are used to purchase units (shares) of an investment fund, called the with
proﬁts fund. As the fund earns investment return, the shares increase in value (or
more shares are issued), increasing the beneﬁt entitlement as reversionary bonus.
The shares will not decrease in value. On death or maturity, a further terminal
bonus may be payable depending on the performance of the with proﬁts fund.
After some poor publicity surrounding with proﬁts business, and, by association,
unitised with proﬁts business, these product designs were withdrawn from the UK
and Australian markets by the early 2000s. However, they will remain important for
many years as many companies carry very large portfolios of with proﬁt (traditional
and unitised) policies issued during the second half of the 20th century.
EquityLinked Insurance has a beneﬁt linked to the performance of an investment
fund. There are two diﬀerent forms. The ﬁrst is where the policyholder’s premiums
are invested in an openended investment company style account; at maturity, the
beneﬁt is the accumulated value of the premiums. There is a guaranteed minimum
death beneﬁt payable if the policyholder dies before the contract matures. In some
cases, there is also a guaranteed minimum maturity beneﬁt payable. In the UK
and most of Europe, these are called unitlinked policies, and they rarely carry
a guaranteed maturity beneﬁt. In Canada they are known as segregated fund
policies and always carry a maturity guarantee. In the USA these contracts are
called variable annuity contracts; maturity guarantees are increasingly common
for these policies. (The use of the term ‘annuity’ for these contracts is very misleading. The beneﬁts are designed with a single lump sum payout, though there may
be an option to convert the lump sum to an annuity.)
The second form of equitylinked insurance is the EquityIndexed Annuity (EIA) in 28 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE
the USA. Under an EIA the policyholder is guaranteed a minimum return on their
premium (minus an initial expense charge). At maturity, the policyholder receives
a proportion of the return on a speciﬁed stock index, if that is greater than the
guaranteed minimum return.
EIAs are generally rather shorter in term than unitlinked products, with seven year
policies being typical; variable annuity contracts commonly have terms of twenty
years or more. EIAs are much less popular with consumers than variable annuities. 1.3.4 Distribution methods Most people ﬁnd insurance dauntingly complex. Brokers who connect individuals to an
appropriate insurance product have, since the earliest times, played an important role in
the market. There is an old saying amongst actuaries that “insurance is sold, not bought”,
which means that the role of an intermediary in persuading potential policyholders to take
out an insurance policy is crucial in maintaining an adequate volume of new business.
Brokers, or other ﬁnancial advisors, are often remunerated through a commission system. The commission would be speciﬁed as a percentage of the premium paid. Typically,
there is a higher percentage paid on the ﬁrst premium than on subsequent premiums. This
is referred to as a front end load. Some advisors may be remunerated on a ﬁxed fee
basis, or may be employed by one or more insurance companies on a salary basis.
An alternative to the broker method of selling insurance is direct marketing. Insurers
may use television advertising or other telemarketing methods to sell direct to the public.
The nature of the business sold by direct marketing methods tends to diﬀer from the
broker sold business. For example, often the sum insured is smaller. The policy may
be aimed at a niche market, such as older lives concerned with insurance to cover their
own funeral expenses (called preneed insurance in the USA). Another mass marketed
insurance contract is loan or credit insurance, where an insurer might cover loan or credit
card payments in the event of the borrower’s death, disability or unemployment. 1.3. LIFE INSURANCE AND ANNUITY CONTRACTS 1.3.5 29 Underwriting It is important in modeling life insurance liabilities to consider what happens when a life
insurance policy is purchased. Selling life insurance policies is a competitive business and
life insurance companies (also known as life oﬃces) are constantly considering ways in
which to change their procedures so that they can improve the service to their customers
and gain a commercial advantage over their competitors. The account given below of how
policies are sold covers some essential points but is necessarily a simpliﬁed version of what
actually happens.
For a given type of policy, say a 10 year term insurance, the life oﬃce will have a schedule
of premium rates. These rates will depend on the size of the policy and some other factors
known as rating factors. An applicant’s risk level is assessed by asking them to complete
a proposal form giving information on relevant rating factors, generally including their
age, gender, smoking habits, occupation, any dangerous hobbies, and personal and family
health history. The life oﬃce may ask for permission to contact the applicant’s doctor
to enquire about their medical history. In some cases, particularly for very large sums
insured, the life oﬃce may require that the applicant’s health be checked by a doctor
employed by the oﬃce.
The process of collecting and evaluating this information is called underwriting. The
purpose of underwriting is, ﬁrst, to classify potential policyholders into broadly homogeneous risk categories, and secondly to assess what additional premium would be appropriate for applicants whose risk factors indicate that standard premium rates would be
too low.
On the basis of the application and supporting medical information, potential life insurance policyholders will generally be categorized into one of the following groups:
• Preferred lives have very low mortality risk based on the standard information.
The preferred applicant would have no recent record of smoking; no evidence of drug
or alcohol abuse; no high risk hobbies or occupations; no family history of disease 30 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE
known to have a strong genetic component; no adverse medical indicators such as
high blood pressure or cholesterol level or body mass index.
The preferred life category is common in North America, but has not yet caught on
elsewhere. In other areas there is no separation of preferred and normal lives.
• Normal lives may have some higher rated risk factors than preferred lives (where
this category exists), but are still insurable at standard rates. Most applicants fall
in this category.
• Rated lives have one or more risk factors at raised levels and so are not acceptable
at standard premium rates. However, they can be insured for a higher premium.
An example might be someone having a family history of heart disease. These
lives might be individually assessed for the appropriate additional premium to be
charged. This category would also include lives with hazardous jobs or hobbies
which put them at increased risk.
• Uninsurable lives have such signiﬁcant risk that the insurer will not enter an
insurance contract at any price. Within the ﬁrst three groups, applicants would be further categorized according to the
relative values of the various risk factors, with the most fundamental being age, gender
and smoking status. Most applicants (around 95% for traditional life insurance) will be
accepted at preferred or standard rates for the relevant risk category. Another 23% may
be accepted at nonstandard rates because of an impairment, or a dangerous occupation,
leaving around 23% who will be refused insurance.
The rigour of the underwriting process will depend on the type of insurance being purchased, on the sum insured and on the distribution process of the insurance company.
Term insurance is generally more strictly underwritten than whole life insurance, as the
risk taken by the insurer is greater. Under whole life insurance, the payment of the sum
insured is certain, the uncertainty is in the timing. Under, say, 10 year term insurance, it
is assumed that the majority of contracts will expire with no death beneﬁt paid. If the 1.3. LIFE INSURANCE AND ANNUITY CONTRACTS 31 underwriting is not strict there is a risk of adverse selection by policyholders – that is,
that very high risk individuals will buy insurance in disproportionate numbers, leading to
excessive losses. Since high sum insured contracts carry more risk than low sum insured,
high sums insured would generally trigger more rigorous underwriting.
The marketing method also aﬀects the level of underwriting. Often, direct marketed
contracts are sold with relatively low beneﬁt levels, and with the attraction that no
medical evidence will be sought beyond a standard questionnaire. The insurer may assume
relatively heavy mortality for these lives to compensate for potential adverse selection.
By keeping the underwriting relatively light, the expenses of writing new business can be
kept low, which is an attraction for high volume low sum insured contracts.
It is interesting to note that with no third party medical evidence the insurer is placing
a lot of weight on the veracity of the policyholder. Insurers have a phrase for this – that
both insurer and policyholder may assume ‘utmost good faith’ or ‘uberrima ﬁdes’ on the
part of the other side of the contract. In practice, in the event of the death of the insured
life, the insurer may investigate whether any pertinent information was withheld from the
application. If it appears that the policyholder held back information, or submitted false
or misleading information, the insurer may not pay the full sum insured. 1.3.6 Premiums A life insurance policy may involve a single premium, payable at the outset of the contract,
or a regular series of premiums payable provided the policyholder survives, perhaps with
a ﬁxed end date. In traditional contracts the regular premium is generally a level amount
throughout the term of the contract; in more modern contracts the premium might be
variable, at the policyholder’s discretion for investment products such as equitylinked
insurance, or at the insurer’s discretion for certain types of term insurance.
Regular premiums may be paid annually, semiannually, quarterly, monthly or weekly.
Monthly premiums are common as it is convenient for policyholders to have their outgoings 32 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE payable with approximately the same frequency as their income.
An important feature of all premiums is that they are paid at the start of each period.
Suppose a policyholder contracts to pay annual premiums for a 10 year insurance contract.
The premiums will be paid at the start of the contract, and then at the start of each
subsequent year provided the policyholder is alive. So, if we count time in years from
t = 0 at the start of the contract, the ﬁrst premium is paid at t = 0, the second is paid
at t = 1 and so on, to the tenth premium paid at t = 9. Similarly, if the premiums are
monthly, then the ﬁrst monthly installment will be paid at t = 0, and the ﬁnal premium
will be paid at the start of the ﬁnal month at t = 9 11 years. (Throughout this book we
12
1
assume that all months are equal in length, at 12 years.) 1.3.7 Life annuities Annuity contracts oﬀer a regular series of payments. When an annuity depends on the
survival of the recipient, it is called a ‘life annuity’. The recipient is called an annuitant. If
the annuity continues until the death of the annuitant, it is called a whole life annuity.
If the annuity is paid for some maximum period, provided the annuitant survives that
period, it is called a term life annuity.
Annuities are often purchased by older lives to provide income in retirement. Buying a
whole life annuity guarantees that the income will not run out before the annuitant dies.
Single Premium Deferred Annuity (SPDA) Under an SPDA contract, the policyholder pays a single premium in return for an annuity which commences payment
at some future, speciﬁed date. The annuity is ‘life contingent’, by which we mean
the annuity is paid only if the policyholder survives to the payment dates. If the
policyholder dies before the annuity commences, there may be a death beneﬁt due.
If the policyholder dies soon after the annuity commences, there may be some minimum payment period, called the guarantee period, and the balance would be paid
to the policyholder’s estate. 1.4. OTHER INSURANCE CONTRACTS 33 Single Premium Immediate Annuity (SPIA) This contract is the same as the
SPDA, except that the annuity commences as soon as the contract is eﬀected. This
might, for example, be used to convert a lump sum retirement beneﬁt into a life
annuity to supplement a pension. As with the SPDA, there may be a guarantee
period applying in the event of the early death of the annuitant.
Regular Premium Deferred Annuity (RPDA) The RPDA oﬀers a deferred life annuity with premiums paid through the deferred period. It is otherwise the same as
the SPDA.
Joint life annuity A joint life annuity is issued on two lives, typically a married couple.
The annuity (which may be single premium or regular premium, immediate or
deferred) continues while both lives survive, and ceases on the ﬁrst death of the
couple.
Last survivor annuity A last survivor annuity is similar to the joint life annuity, except
that payment continues while at least one of the lives survives, and ceases on the
second death of the couple.
Reversionary annuity A reversionary annuity is contingent on two lives, usually a
couple. One is designated as the annuitant, and one the insured. No annuity
beneﬁt is paid while the insured life survives. On the death of the insured life, if
the annuitant is still alive, the annuitant receives an annuity for the remainder of
his or her life. 1.4 Other insurance contracts The insurance and annuity contracts described above are all contingent on death or survival. There are other life contingent risks, in particular involving short or long term
disability. These are known as morbidity risks. 34 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE Income Protection Insurance When a person becomes sick and cannot work, their
income will, eventually, be aﬀected. For someone in regular employment, the employer may cover salary for a period, but if the sickness continues the salary will
be decreased, and ultimately will stop being paid at all. For someone who is self
employed, the eﬀects of sickness on income will be immediate. Income protection
policies replace at least some income during periods of sickness. They usually cease
at retirement age.
Critical Illness Insurance Some serious illnesses can cause signiﬁcant expense at the
onset of the illness. The patient may have to leave employment, or alter their
home, or incur severe medical expenses. Critical illness insurance pays a beneﬁt on
diagnosis of one of a number of severe conditions, such as certain cancers or heart
disease. The beneﬁt is usually in the form of a lump sum.
Long Term Care Insurance This is purchased to cover the costs of care in old age,
when the insured life is unable to continue living independently. The beneﬁt would
be in the form of the long term care costs, so is an annuity beneﬁt. 1.5 Pension beneﬁts Many actuaries work in the area of pension plan design, valuation and risk management.
The pension plan is usually sponsored by an employer. Pension plans typically oﬀer employees (also called pension plan members) either lump sums or annuity beneﬁts or both
on retirement, or deferred lump sum or annuity beneﬁts (or both) on earlier withdrawal.
Some oﬀer a lump sum beneﬁt if the employee dies while still employed. The beneﬁts
therefore depend on the survival and employment status of the member, and are quite similar in nature to life insurance beneﬁts – that is, they involve investment of contributions
long into the future to pay for future life contingent beneﬁts. 1.5. PENSION BENEFITS 1.5.1 35 Deﬁned beneﬁt and deﬁned contribution pensions Deﬁned beneﬁt (DB) pensions oﬀer retirement income based on service and salary with
an employer, using a deﬁned formula to determine the pension. For example, suppose an
employee reaches retirement age with n years of service (i.e. membership of the pension
plan), and with pensionable salary averaging S in, say, the ﬁnal three years of employment.
A typical ﬁnal salary plan might oﬀer an annual pension at retirement of B = Snα,
where α is called the accrual rate, and is usually around 1%2%. The formula may be
interpreted as a pension beneﬁt of, say, 2% of the ﬁnal average salary for each year of
service.
The deﬁned beneﬁt is funded by contributions paid by the employer and (usually) the
employee over the working lifetime of the employee. The contributions are invested, and
the accumulated contributions must be enough, on average, to pay the pensions when
they become due.
Deﬁned contribution (DC) pensions work more like a bank account. The employee and
employer pay a predetermined contribution (usually a ﬁxed percentage of salary) into a
fund, and the fund earns interest. When the employee leaves or retires, the proceeds
are available to provide income throughout retirement. In the UK most of the proceeds
must be converted to an annuity. In the USA and Canada there are more options – the
pensioner may draw funds to live on without necessarily purchasing an annuity from an
insurance company. 1.5.2 Deﬁned beneﬁt pension design The age retirement pension described in the section above deﬁnes the pension payable
from retirement in a standard ﬁnal salary plan. Career average salary plans are also
common in some jurisdictions, where the beneﬁt formula is the same as the ﬁnal salary
formula above, except that the average salary over the employee’s entire career is used in
place of the ﬁnal salary. 36 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE Many employees leave their jobs before they retire. A typical withdrawal beneﬁt would
be a pension based on the same formula as the age retirement beneﬁt, but with the start
date deferred until the employee reaches the normal retirement age. Employees may have
the option of taking a lump sum with the same value as the deferred pension, which can
be invested in the pension plan of the new employer.
Some pension plans also oﬀer deathinservice beneﬁts, for employees who die during
their period of employment. Such beneﬁts might include a lump sum, often based on
salary and sometimes service, as well as a pension for the employee’s spouse. 1.6 Mutual and proprietary insurers A mutual insurance company is one that has no shareholders. The insurer is owned by
the withproﬁt policyholders. All proﬁts are distributed to the withproﬁt policyholders
through dividends or bonuses.
A proprietary insurance company has shareholders, and usually has withproﬁt policyholders as well. The participating policyholders are not owners, but have a speciﬁed right
to some of the proﬁts. Thus, in a proprietary insurer, the proﬁts must be shared in some
predetermined proportion, between the shareholders and the withproﬁt policyholders.
Many early life insurance companies were formed as mutual companies. More recently,
in the UK, Canada and the USA, there has been a trend towards demutualization, which
means the transition of a mutual company to a proprietary company, through issuing
shares (or cash) to the withproﬁt policyholders. Although it would appear that a mutual
insurer would have marketing advantages, as participating policyholders receive all the
proﬁts and other beneﬁts of ownership, the advantages cited by companies who have
demutualized include increased ability to raise capital, clearer corporate structure and
improved eﬃciency. 1.7. TYPICAL PROBLEMS 1.7 37 Typical problems We are concerned in this book with developing the mathematical models and techniques
used by actuaries working in life insurance and pensions. The primary responsibility of
the life insurance actuary is to maintain the solvency and proﬁtability of the insurer.
Premiums must be suﬃcient to pay beneﬁts; the assets held must be suﬃcient to pay the
contingent liabilities; bonuses to policyholders should be fair.
Consider, for example, a whole life insurance contract issued to a life aged 50. The sum
insured may not be paid for thirty years or more. The premiums paid over the period
will be invested by the insurer to earn signiﬁcant interest; the accumulated premiums
must be suﬃcient to pay the beneﬁts, on average. To ensure this, the actuary needs to
model the survival probabilities of the policyholder, the investment returns likely to be
earned and the expenses likely to be incurred in maintaining the policy. The actuary
may take into consideration the probability that the policyholder decides to terminate
the contract early. The actuary may also consider the proﬁtability requirements for the
contract. Then, when all of these factors have been modeled, they must be combined to
set a premium.
Each year or so, the actuary must determine how much money the insurer or pension
plan should hold to ensure that future liabilities will be covered with adequately high
probability. This is called the valuation process. For withproﬁt insurance, the actuary
must determine a suitable level of bonus.
The problems are rather more complex if the insurance also covers morbidity risk, or
involves several lives. All of these topics are covered in the following chapters.
The actuary may also be involved in decisions about how the premiums are invested. It
is vitally important that the insurer remains solvent, as the contracts are very long term
and insurers are responsible for protecting the ﬁnancial security of the general public. The
way the underlying investments are selected can increase or mitigate the risk of insolvency.
The precise selection of investments to manage the risk is particularly important where 38 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE the contracts involve ﬁnancial guarantees.
The pensions actuary working with deﬁned beneﬁt pensions must determine appropriate
contribution rates to meet the beneﬁts promised, using models that allow for the working
patterns of the employees. Sometimes, the employer may want to change the beneﬁt
structure, and the actuary is responsible for assessing the cost and impact. When one
company with a pension plan takes over another, the actuary must assist with determining
the best way to allocate the assets from the two plans, and perhaps how to merge the
beneﬁts. 1.8 Notes and further reading A number of essays describing actuarial practice can be found in Renn (ed.) (1998). This
book also provides both historical and more contemporary context for life contingencies.
The original papers of De Witt and Halley are available online (and any search engine
will ﬁnd them). Anyone interested in the history of probability and actuarial science will
ﬁnd these interesting, and remarkably modern. 1.9. EXERCISES 1.9 39 Exercises Exercise 1.1 Why do insurers generally require evidence of health from a person applying for life insurance but not for an annuity? Exercise 1.2 Explain why an insurer might demand more rigorous evidence of a prospective policyholder’s health status for a term insurance than for a whole life insurance. Exercise 1.3 Explain why premiums are payable in advance, so that the ﬁrst premium
is due now rather than in one year’s time. Exercise 1.4 Lenders oﬀering mortgages to home owners may require the borrower to
purchase life insurance to cover the outstanding loan on the death of the borrower, even
though the mortgaged property is the loan collateral.
(a) Explain why the lender might require term insurance in this circumstance.
(b) Describe how this term insurance might diﬀer from the standard term insurance
described in Section 1.3.2.
(c) Can you see any problems with lenders demanding term insurance from borrowers? 40 CHAPTER 1. INTRODUCTION TO LIFE INSURANCE Exercise 1.5 Describe the diﬀerence between a cash bonus and a reversionary bonus for
withproﬁt whole life insurance. What are the advantages and disadvantages of each for
(a) the insurer and (b) the policyholder? Exercise 1.6 It is common for insurers to design whole life contracts with premiums
payable only up to age 80. Why? Exercise 1.7 Andrew is retired. He has no pension, but has capital of $500 000. He is
considering the following options for using the money:
(a) Purchase an annuity from an insurance company that will pay a level amount for
the rest of his life.
(b) Purchase an annuity from an insurance company that will pay an amount that
increases with the cost of living for the rest of his life.
(c) Purchase a 20 year annuity certain.
(d) Invest the capital and live on the interest income.
(e) Invest the capital and draw $40 000 per year to live on.
What are the advantages and disadvantages of each option? Chapter 2
Survival Models
2.1 Summary In this chapter we represent the future lifetime of an individual as a random variable,
and show how probabilities of death or survival can be calculated under this framework.
We then deﬁne an important quantity known as the force of mortality, introduce some
actuarial notation, and discuss some properties of the distribution of future lifetime. We
introduce the curtate future lifetime random variable. This is a function of the future
lifetime random variable which represents the number of complete years of future life. We
explain why this function is useful and derive its probability function. 2.2 The future lifetime random variable In Chapter 1 we saw that many insurance policies provide a beneﬁt on the death of the
policyholder. When an insurance company issues such a policy, the policyholder’s date of
death is unknown, so the insurer does not know exactly when the death beneﬁt will be
payable. In order to estimate the time at which a death beneﬁt is payable, the insurer
41 42 CHAPTER 2. SURVIVAL MODELS needs a model of human mortality, from which probabilities of death at particular ages
can be calculated, and this is the topic of this chapter.
We start with some notation. Let (x) denote a life aged x, where x ≥ 0. The death of
(x) can occur at any age greater than x, and we model the future lifetime of (x) by a
continuous random variable which we denote by Tx . This means that x + Tx represents
the ageatdeath random variable for (x).
Let Fx be the distribution function of Tx , so that
Fx (t) = Pr[Tx ≤ t].
Then Fx (t) represents the probability that (x) does not survive beyond age x + t, and we
refer to Fx as the lifetime distribution from age x. In many life insurance problems we
are interested in the probability of survival rather than death, and so we deﬁne Sx as
Sx (t) = 1 − Fx (t) = Pr[Tx > t].
Thus, Sx (t) represents the probability that (x) survives for at least t years, and Sx is
known as the survival function.
Given our interpretation of the collection of random variables {Tx }x≥0 as the future lifetimes of individuals, we need a connection between any pair of them. To see this, consider
T0 and Tx for a particular individual who is now aged x. The random variable T0 represented the future lifetime at birth for this individual, so that, at birth, the individual’s
age at death would have been represented by T0 . This individual could have died before
reaching age x – the probability of this was Pr[T0 < x] – but has survived. Now that the
individual has survived to age x, so that T0 > x, his or her future lifetime is represented
by Tx and the age at death is now x + Tx . If the individual dies within t years from
now, then Tx ≤ t and T0 ≤ x + t. Loosely speaking, we require the events [Tx ≤ t] and
[T0 ≤ x + t] to be equivalent, given that the individual survives to age x. We achieve this
by making the following assumption for all x ≥ 0 and for all t > 0
Pr[Tx ≤ t] = Pr[T0 ≤ x + tT0 > x]. (2.1) 2.2. THE FUTURE LIFETIME RANDOM VARIABLE 43 This is an important relationship.
Now, recall from probability theory that for two events A and B
Pr[AB ] = Pr[A and B ]
,
Pr[B ] so, interpreting [T0 ≤ x + t] as event A, and [T0 > x] as event B , we can rearrange the
right hand side of (2.1) to give
Pr[Tx ≤ t] = Pr[x < T0 ≤ x + t]
.
Pr[T0 > x] That is,
Fx (t) = F0 (x + t) − F0 (x)
.
S0 (x) (2.2) Also, using Sx (t) = 1 − Fx (t),
Sx (t) = S0 (x + t)
,
S0 (x) (2.3) which can be written as
S0 (x + t) = S0 (x) Sx (t). (2.4) This is a very important result. It shows that we can interpret the probability of
survival from age x to age x + t as the product of
(1) the probability of survival to age x from birth, and,
(2) the probability, having survived to age x, of further surviving to age x + t.
Note that Sx (t) can be thought of as the probability that (0) survives to at least age
x + t given that (0) survives to age x, so this result can be derived from the standard
probability relationship
Pr[A and B ] = Pr[AB ] Pr[B ] 44 CHAPTER 2. SURVIVAL MODELS where the events here are A = [T0 > x + t] and B = [T0 > x], so that
Pr[AB ] = Pr[T0 > x + tT0 > x],
which we know from (2.1) is equal to Pr[Tx > t].
Similarly, any survival probability for (x), for, say, t + u years can be split into the
probability of surviving the ﬁrst t years, and then, given survival to age x + t, subsequently
surviving another u years. That is,
Sx (t + u) = ⇒ Sx (t + u) = S0 (x + t + u)
S0 (x)
S0 (x + t) S0 (x + t + u)
S0 (x)
S0 (x + t) ⇒ Sx (t + u) = Sx (t)Sx+t (u). (2.5) We have already seen that if we know survival probabilities from birth, then, using formula
(2.4), we also know survival probabilities for our individual from any future age x. Formula
(2.5) takes this a stage further. It shows that if we know survival probabilities from any
age x (≥ 0), then we also know survival probabilities from any future age x + t (≥ x).
Any survival function for a lifetime distribution must satisfy the following conditions to
be valid.
Condition 1. Sx (0) = 1; that is, the probability that a life currently aged x
survives 0 years is 1.
Condition 2. limt→∞ Sx (t) = 0; that is, all lives eventually die.
Condition 3. The survival function must be a non–increasing function of t; it
cannot be more likely that (x) survives, say 10.5 years than 10 years, because in
order to survive 10.5 years, (x) must ﬁrst survive 10 years. 2.2. THE FUTURE LIFETIME RANDOM VARIABLE 45 These conditions are both necessary and suﬃcient, so that any function Sx which satisﬁes
these three conditions as a function of t (≥ 0), for a ﬁxed x (≥ 0), deﬁnes a lifetime
distribution from age x, and, using formula (2.5), for all ages greater than x.
For all the distributions used in this book, we make three additional assumptions:
Assumption 1. Sx (t) is diﬀerentiable for all t > 0. Note that together with
d
Condition 3 above, this means that dt Sx (t) ≤ 0 for all t > 0.
Assumption 2. limt→∞ t Sx (t) = 0.
Assumption 3. limt→∞ t2 Sx (t) = 0.
These last two assumptions ensure that the mean and variance of the distribution of Tx
exist. These are not particularly restrictive constraints – we do not need to worry about
distributions with inﬁnite mean or variance in the context of individuals’ future lifetimes.
These three extra assumptions are valid for all distributions that are feasible for human
lifetime modeling.
Example 2.1 Let
F0 (t) = 1 − (1 − t/120)1/6 for 0 ≤ t ≤ 120. Calculate the probability that
(a) a newborn life survives beyond age 30,
(b) a life aged 30 dies before age 50, and
(c) a life aged 40 survives beyond age 65.
Solution 2.1 (a) The required probability is
S0 (30) = 1 − F0 (30) = (1 − 30/120)1/6 = 0.9532. 46 CHAPTER 2. SURVIVAL MODELS (b) From formula (2.2), the required probability is
F30 (20) = F0 (50) − F0 (30)
= 0.0410.
1 − F0 (30) (c) From formula (2.3), the required probability is
S40 (25) = S0 (65)
= 0.9395.
S0 (40) We remark that in the above example, S0 (120) = 0, which means that under this model,
survival beyond age 120 is not possible. In this case we refer to 120 as the limiting age of
the model. In general, if there is a limiting age, we use the Greek letter ω to denote it.
In models where there is no limiting age, it is often practical to introduce a limiting age
in calculations, as we will see later in this chapter. 2.3 The force of mortality The force of mortality is an important and fundamental concept in modeling future lifetime.
We denote the force of mortality at age x by µx and deﬁne it as
µx = lim+
dx→0 1
Pr[T0 ≤ x + dx  T0 > x].
dx (2.6) ¿From equation (2.1) we see that an equivalent way of deﬁning µx is
µx = lim+
dx→0 1
Pr[Tx ≤ dx],
dx which can be written in tems of the survival function Sx as
µx = lim+
dx→0 1
(1 − Sx (dx)) .
dx (2.7) 2.3. THE FORCE OF MORTALITY 47 Note that the force of mortality depends, numerically, on the unit of time; if we are
measuring time in years, then µx is measured per year.
The force of mortality is best understood by noting that for very small dx, formula (2.6)
gives the approximation
µx dx ≈ Pr[T0 ≤ x + dx  T0 > x]. (2.8) Thus, for very small dx, we can interpret µx dx as the probability that a life who has
attained age x dies before attaining age x + dx. For example, suppose we have a male
aged exactly 50 and that the force of mortality at age 50 is 0.0044 per year. A small value
of dx might be a single day, or 0.00274 years. Then the approximate probability that (50)
dies on his birthday is 0.0044 × 0.00274 = 1.2 × 10−5 .
We can relate the force of mortality to the survival function from birth, S0 . As
Sx (dx) = S0 (x + dx)
,
S0 (x) formula (2.7) gives
µx = = 1
S0 (x) − S0 (x + dx)
lim+
S0 (x) dx→0
dx
1
S0 (x) − d
S0 (x) .
dx Thus, µx = −1 d
S0 (x).
S0 (x) dx (2.9) ¿From standard results in probability theory, we know that the probability density function for the random variable Tx , which we denote fx , is related to the distribution function 48 CHAPTER 2. SURVIVAL MODELS Fx and the survival function Sx by
d
d
Fx (t) = − Sx (t).
dt
dt fx (t) = So, it follows from equation (2.9) that
µx = f0 (x)
.
S0 (x) We can also relate the force of mortality function at any age x + t, t > 0, to the lifetime
distribution of Tx . Assume x is ﬁxed and t is variable. Then d(x + t) = dt and so
µ x+ t = − 1
d
S0 (x + t)
S0 (x + t) d(x + t) =− 1
d
S0 (x + t)
S0 (x + t) dt =− 1
d
S0 (x)Sx (t)
S0 (x + t) dt =− S0 (x) d
Sx (t)
S0 (x + t) dt = −1 d
Sx (t).
Sx (t) dt Hence
µ x+ t = fx (t)
.
Sx (t) (2.10) 2.3. THE FORCE OF MORTALITY 49 This relationship gives a way of ﬁnding µx+t given Sx (t). We can also use equation (2.9)
to develop a formula for Sx (t) in terms of the force of mortality function. We use the fact
that for a function h whose derivative exists,
d
1d
log h(x) =
h(x),
dx
h(x) dx
so from equation (2.9) we have
d
log S0 (x),
dx
and integrating this identity over (0, y ) yields
µx = −
y
0 µx dx = − (log S0 (y ) − log S0 (0)) . As log S0 (0) = log Pr[T0 > 0] = log 1 = 0, we obtain
y S0 (y ) = exp − µx dx ,
0 from which it follows that
Sx (t) = S0 (x + t)
= exp −
S0 (x) x+t t µr dr
x = exp − µx+s ds . (2.11) 0 This means that if we know µx for all x ≥ 0, then we can calculate all the survival
probabilities Sx (t), for any x and t. In other words, the force of mortality function fully
describes the lifetime distribution, just as the function S0 does. In fact, it is often more
convenient to describe the lifetime distribution using the force of mortality function than
the survival function.
Example 2.2 As in Example 2.1, let
F0 (x) = 1 − (1 − x/120)1/6
for 0 ≤ x ≤ 120. Derive an expression for µx . 50 CHAPTER 2. SURVIVAL MODELS Solution 2.2 As S0 (x) = (1 − x/120)1/6 , it follows that
d
S0 (x) = 1 (1 − x/120)−5/6
6
dx −1
120 , and so
µx = −1 d
S0 (x) =
S0 (x) dx 1
(1
720 − x/120)−1 = 1
.
720 − 6x As an alternative, we could use the relationship
µx = − d
d
log S0 (x) = −
dx
dx 1
log(1 − x/120)
6 = 1
1
=
.
720(1 − x/120)
720 − 6x Example 2.3 Let µx = Bcx , x > 0, where B and c are constants such that 0 < B < 1
and c > 1. This model is called Gompertz’ law of mortality. Derive an expression for
Sx (t).
Solution 2.3 From equation (2.11),
x+ t Sx (t) = exp − Bcr dr .
x Writing cr as exp{r log c},
x+ t x+ t
r Bc dr = B
x x = = exp{r log c}dr B
exp{r log c}
log c
B
c x+ t − c x ,
log c x+ t
x 2.3. THE FORCE OF MORTALITY 51 giving
Sx (t) = exp −B x t
c (c − 1) .
log c The force of mortality under Gompertz’ law increases exponentially with age. At ﬁrst sight
this seems reasonable, but as we will see in the next chapter, the force of mortality for most
populations is not an increasing function of age over the entire age range. Nevertheless,
the Gompertz model does provide a fairly good ﬁt to mortality data over some age ranges,
particularly from middle age to early old age.
Example 2.4 Calculate the survival function and probability density function for Tx
using Gompertz’ law of mortality, with B = 0.0003 and c = 1.07, for x = 20, x = 50 and
x = 80. Plot the results and comment on the features of the graphs.
Solution 2.4 For x = 20, the force of mortality is µ20+t = Bc20+t and the survival
function is
S20 (t) = exp −B 20 t
c (c − 1) .
log c The probability density function is found from (2.10):
µ20+t = f20 (t)
⇒ f20 (t) = µ20+t S20 (t) = Bc20+t exp
S20 (t) −B 20 t
c (c − 1) .
log c Figure 2.1 shows the survival functions for ages 20, 50 and 80, and Figure 2.2 shows the
corresponding probability density functions. These ﬁgures illustrate some general points
about lifetime distributions.
First, we see an eﬀective limiting age, even though, in principle there is no age to which
the survival probability is exactly zero. Looking at Figure 2.1, we see that although
Sx (t) > 0 for all combinations of x and t, survival beyond age 120 is very unlikely. 52 CHAPTER 2. SURVIVAL MODELS 1 0.9 0.8 Survival probability 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
0 10 20 30 40 50 60 70 80 90 100 Time, t Figure 2.1: Sx (t) for x = 20 (bold), 50 (solid) and 80 (dotted). 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0
0 10 20 30 40 50 60 70 80 90 Time, t Figure 2.2: fx (t) for x = 20 (bold), 50 (solid) and 80 (dotted). 100 2.4. ACTUARIAL NOTATION 53 Second, we note that the survival functions are ordered according to age, with the probability of survival for any given value of t being highest for age 20 and lowest for age 80.
For survival functions that give a more realistic representation of human mortality, this
ordering can be violated, but it usually holds at ages of interest to insurers. An example
of the violation of this ordering is that S0 (1) may be smaller than Sx (1) for x ≥ 1, as a
result of perinatal mortality.
Looking at Figure 2.2, we see that the densities for ages 20 and 50 have similar shapes,
but the density for age 80 has a quite diﬀerent shape. For ages 20 and 50, the densities
have their respective maximums at (approximately) t = 60 and t = 30, indicating that
death is most likely to occur around age 80. The decreasing form of the density for age
80 also indicates that death is more likely to occur at age 80 than at any other age for
a life now age 80. A further point to note about these density functions is that although
each density function is deﬁned on (0, ∞), the spread of values of fx (t) is much greater
for x = 20 than for x = 50, which, as we will see in Table 2.1, results in a greater variance
of future lifetime for x = 20 than for x = 50. 2.4 Actuarial notation The notation used in the previous sections, Sx (t), Fx (t) and fx (t) is standard in statistics.
Actuarial science has developed its own notation, International Actuarial Notation, that
encapsulates the probabilities and functions of greatest interest and usefulness to actuaries. The force of mortality notation, µx , comes from International Actuarial Notation.
We summarize the relevant actuarial notation in this section, and rewrite the important
results developed so far in this chapter in terms of actuarial functions. The actuarial
notation for survival and mortality probabilities is
t px = Pr[Tx > t] = Sx (t), (2.12) 54 CHAPTER 2. SURVIVAL MODELS t qx = Pr[Tx ≤ t] = 1 − Sx (t) = Fx (t), u t qx = Pr[u < Tx ≤ u + t] = Sx (u) − Sx (u + t). (2.13)
(2.14) So
t px is the probability that (x) survives to at least age x + t, t qx is the probability that (x) dies before age x + t, u t qx is the probability that (x) survives u years, and then dies in the subsequent t
years, that is, between ages x + u and x + u + t. This is called a deferred mortality
probability, because it is the probability that death occurs in some interval following
a deferred period.
We may drop the subscript t if its value is 1, so that px represents the probability that
(x) survives to at least age x + 1. Similarly, qx is the probability that (x) dies before age
x + 1. In actuarial terminology qx is called the mortality rate at age x.
The relationships below follow immediately from the deﬁnitions above and the previous
results in this chapter:
t px + t qx = 1, u t qx
t+u px = u p x − u+ t p x ,
= t p x u p x+ t µx = − 1d
x p0
x p0 dx from (2.5),
from (2.9). (2.15)
(2.16) Similarly,
µ x+ t = − 1d
d
t px ⇒
t px = −t px µx+t ,
dt
t px dt (2.17) 2.4. ACTUARIAL NOTATION 55 fx (t)
⇒ fx (t) = t px µx+t
Sx (t) µ x+ t = from (2.10), (2.18) t
t px = exp − µx+s ds from (2.11). (2.19) 0 As Fx is a distribution function and fx is its density function, it follows that
t Fx (t) = fx (s)ds,
0 which can be written in actuarial notation as
t
t qx = s px µx+s ds. (2.20) 0 This is an important formula, which can be interpreted as follows. Consider time s, where
0 ≤ s < t. The probability that (x) is alive at time s is s px , and the probability that (x)
dies between ages x + s and x + s + ds, having survived to age x + s, is (loosely) µx+s ds,
provided that ds is very small. Thus s px µx+s ds can be interpreted as the probability that
(x) dies between ages x + s and x + s + ds. Now, we can sum over all the possible death
intervals s to s + ds – which requires integrating because these are inﬁnitesimal intervals
– to obtain the probability of death before age x + t.
We can illustrate this event sequence using a time line diagram: Time Age s 0 x Event Probability (x) survives s years
s px s+ds x+s x+s+ds
(x)
dies
µx+s ds t x+t 56 CHAPTER 2. SURVIVAL MODELS This type of interpretation is important as it can be applied to more complicated situations, and we will employ the time line again in later chapters.
In the special case when t = 1, formula (2.20) becomes
1 qx = s px µx+s ds. 0 When qx is small, it follows that px is close to 1, and hence s px is close to 1 for 0 ≤ s < 1.
Thus
1 qx ≈ 0 µx+s ds ≈ µx+1/2 , where the second relationship follows by the midpoint rule for numerical integration.
Example 2.5 As in Example 2.1, let
F0 (x) = 1 − (1 − x/120)1/6
for 0 ≤ x ≤ 120. Calculate both qx and µx+1/2 for x = 20 and for x = 110, and comment
on these values.
Solution 2.5 We have
S0 (x + 1)
px =
=
S0 (x) 1
1−
120 − x 1 /6 , giving q20 = 0.00167 and q110 = 0.01741, and from the solution to Example 2.2, µ20 1 =
2
0.00168 and µ110 1 = 0.01754. We see that µx+1/2 is a good approximation to qx when the
2
mortality rate is small, but is not such a good approximation, at least in absolute terms,
when the mortality rate is not close to 0. 2.5. MEAN AND STANDARD DEVIATION OF TX 2.5 57 Mean and standard deviation of Tx Next, we consider the expected future lifetime of (x), E[Tx ], denoted in actuarial notation
◦
by ex . We also call this the complete expectation of life.
◦ In order to evaluate ex , we note from formulae (2.17) and (2.18) that
fx (t) = t px µx+t = − d
t px .
dt (2.21) From the deﬁnition of an expected value, we have
∞ ◦ ex = tfx (t)dt 0
∞ = tt px µx+t dt. 0 We can now use (2.21) to evaluate this integral using integration by parts as
∞ ◦ ex = − d
t px dt
dt t 0 = − t t px ∞
0 − ∞ t px dt . 0 In Section 2.2 we stated the assumption that limt→∞ tt px = 0, which gives
∞ ◦ ex = t px dt. 0 Similarly, for the variance, V[Tx ], we have
2
E[Tx ] = ∞
0 t2 t px µx+t dt (2.22) 58 CHAPTER 2. SURVIVAL MODELS =− ∞ d
t px dt
dt t2 0 ∞
0 t2 t px =− ∞ =2 − ∞ t px 2t dt 0 t t px dt. (2.23) 0
2
So we have integral expressions for E[Tx ] and E[Tx ]. For some lifetime distributions we are
able to integrate directly. In other cases we have to use numerical integration techniques
to evaluate the integrals in (2.22) and (2.23). The variance of Tx can then be calculated
as
2 ◦ 2
V [Tx ] = E Tx − ex . Example 2.6 As in Example 2.1, let
F0 (x) = 1 − (1 − x/120)1/6
◦ for 0 ≤ x ≤ 120. Calculate ex and V[Tx ] for (a) x = 30 and (b) x = 80.
Solution 2.6 As S0 (x) = (1 − x/120)1/6 , we have
t px = S0 (x + t)
=
S0 (x) 1− t
120 − x 1/6 . Now recall that this formula is valid for 0 ≤ t ≤ 120 − x, since under this model survival
beyond age 120 is impossible. Technically, we have
t px = 1−
0 1/6
t
120−x for x + t ≤ 120,
for x + t > 120. 2.5. MEAN AND STANDARD DEVIATION OF TX
So the upper limit of integration in equation (2.22) is 120 − x, and
120−x ◦ ex =
0 1− 1/6 t
120 − x dt. We make the substitution y = 1 − t/(120 − x), so that t = (120 − x)(1 − y ), giving
1 ◦ y 1/6 dy ex = (120 − x)
= 6
(120
7 0 − x). ◦ ◦ Then e30 = 77.143 and e80 = 34.286.
Under this model the expectation of life at any age x is 6/7 of the time to age 120.
2
For the variance we require E[Tx ]. Using equation (2.23) we have 2
E Tx 120−x =2 t t px dt 0
120−x =2 t
1−
120 − x t 0 1/6 dt. Again, we substitute y = 1 − t/(120 − x) giving
1 E 2
Tx 2 = 2(120 − x) 0 (y 1/6 − y 7/6 ) dy 6
6
−
7 13 = 2(120 − x)2 . Then
◦ 2
V[Tx ] = E[Tx ] − ex 2 = (120 − x)2 2(6/7 − 6/13) − (6/7)2 59 60 CHAPTER 2. SURVIVAL MODELS
= (120 − x)2 (0.056515) = ((120 − x) (0.23773))2 . So V[T30 ] = 21.3962 and V[T80 ] = 9.5092 .
Since we know under this model that all lives will die before age 120, it makes sense that
the uncertainty in the future lifetime should be greater for younger lives than for older
lives. A feature of the model used in Example 2.6 is that we can obtain formulae for quantities
◦
of interest such as ex , but for many models this is not possible. For example, when we
◦
model mortality using Gompertz’ law, there is no explicit formula for ex and we must use
numerical integration to calculate moments of Tx . In Appendix B we describe in detail
how to do this.
◦ Table 2.1 shows values of ex and the standard deviation of Tx (denoted SD[Tx ]) for a range
of values of x using Gompertz’ law, µx = Bcx , where B = 0.0003 and c = 1.07. For this
survival model, 130 p0 = 1.9 × 10−13 , so that using 130 as the maximum attainable age in
our numerical integration is accurate enough for practical purposes.
◦ We see that ex is a decreasing function of x, as it was in Example 2.6. In that example
◦
ex was a linear function of x, but we see that this is not true in Table 2.1. 2.6
2.6.1 Curtate future lifetime
Kx and ex In many insurance applications we are interested not only in the future lifetime of an
individual, but also in what is known as the individual’s curtate future lifetime. The
curtate future lifetime random variable is deﬁned as the integer part of future lifetime, 2.6. CURTATE FUTURE LIFETIME
x
0
10
20
30
40
50
60
70
80
90
100 ◦ ex
SD[Tx ]
71.938 18.074
62.223 17.579
52.703 16.857
43.492 15.841
34.252 14.477
26.691 12.746
19.550 10.693
13.555 8.449
8.848
6.224
5.433
4.246
3.152
2.682 61
◦ x + ex
71.938
72.223
72.703
73.492
74.752
76.691
79.550
83.555
88.848
95.433
103.152 ◦ Table 2.1: Values of ex , SD[Tx ] and expected age at death for the Gompertz model with
B = 0.0003 and c = 1.07.
and is denoted by Kx for a life aged x. If we let denote the ﬂoor function, we have Kx = Tx .
We can think of curtate future lifetime as the number of whole years lived in the future
by an individual. As an illustration of the importance of curtate future lifetime, consider
the situation where a life aged x at time 0 is entitled to payments of 1 at times 1, 2, 3, . . .
provided that (x) is alive at these times. Then the number of payments made equals the
number of complete years lived after time 0 by (x). This is the curtate future lifetime.
We can ﬁnd the probability function of Kx by noting that for k = 0, 1, 2, . . ., Kx = k if
and only if (x) dies between the ages of x + k and x + k + 1. Thus for k = 0, 1, 2, . . .
Pr[Kx = k ] = Pr[k ≤ Tx < k + 1]
= k qx 62 CHAPTER 2. SURVIVAL MODELS
=
=
= k px
k px − k+1 px − k p x p x+ k k p x q x+ k . The expected value of Kx is denoted by ex , so that ex = E[Kx ], and is referred to as the
curtate expectation of life (even though it represents the expected curtate lifetime).
So
E[Kx ] = ex
= ∞ k Pr[Kx = k ] k=0 = ∞
k=0 k ( k px − k+1 px ) = (1 px − 2 px ) + 2(2 px − 3 px ) + 3(3 px − 4 px ) + . . .
= ∞ k px . (2.24) k=1 Note that the lower limit of summation is k = 1.
Similarly,
2
E[Kx ] = ∞
k=0 k 2 ( k px − k+1 px ) = (1 px − 2 px ) + 4(2 px − 3 px ) + 9(3 px − 4 px ) + 16(4 px − 5 px ) + . . . 2.6. CURTATE FUTURE LIFETIME
x
0
10
20
30
40
50
60
70
80
90
100 63
◦ ex
ex
71.438 71.938
61.723 62.223
52.203 52.703
42.992 43.492
34.252 34.752
26.192 26.691
19.052 19.550
13.058 13.555
8.354 8.848
4.944 5.433
2.673 3.152 ◦ Table 2.2: Values of ex and ex for Gompertz’ law with B = 0.0003 and c = 1.07. =2 ∞
k=1 =2 ∞
k=1 k k px − ∞ k px k=1 k k px − ex . As with the complete expectation of life, there are a few lifetime distributions that allow
2
E[Kx ] and E[Kx ] to be calculated analytically. For more realistic models, such as Gompertz’, we can calculate the values easily using Excel or other suitable software. Although
in principle we have to evaluate an inﬁnite sum, at some age the survival probability will
be suﬃciently small that we can treat it as an eﬀective limiting age. 64 CHAPTER 2. SURVIVAL MODELS
◦ 2.6.2 The complete and curtate expected future lifetimes, ex
and ex As the curtate future lifetime is the integer part of future lifetime, it is natural to ask if
◦
there is a simple relationship between ex and ex . We can obtain an approximate relationship by writing
∞ ◦ ex = t px dt = 0 ∞
j =0 j +1
t px dt. j If we approximate each integral using the trapezoidal rule for numerical integration, we
obtain
j +1
t px
j 1
dt ≈ 2 (j px + j +1 px ) , and hence
◦ ex ≈ ∞ 1
(p
2jx j =0 + j +1 px ) = 1
2 + ∞ j px . j =1 Thus, we have an approximation that is frequently applied in practice, namely
◦ ex ≈ ex + 1 .
2 (2.25) In Chapter 5 we will meet a reﬁned version of this approximation. Table 2.2 shows values
◦
of ex and ex for a range of values of x when the survival model is Gompertz’ law with
B = 0.0003 and c = 1.07. Values of ex were calculated by applying formula (2.24) with
◦
a ﬁnite upper limit of summation of 130 − x, and values of ex are as in Table 2.1. This
table illustrates that formula (2.25) is a very good approximation in this particular case
for younger ages, but is less accurate at very old ages. This observation is true for most
realistic survival models. 2.7. NOTES AND FURTHER READING 2.7 65 Notes and further reading Although laws of mortality such as Gompertz’ law are appealing due to their simplicity,
they rarely represent mortality over the whole span of human ages. A simple extension of
Gompertz’ law is Makeham’s law (Makeham, 1860), which models the force of mortality
as
µx = A + Bcx . (2.26) This is very similar to Gompertz’ law, but adds a ﬁxed term that is not age related, that
allows better for accidental deaths. The extra term tends to improve the ﬁt of the model
to mortality data at younger ages.
In recent times, the GompertzMakeham approach has been generalized further to give
the GM(r, s) (GompertzMakeham) formula, µx = h1 (x) + exp{h2 (x)},
r
s
where h1 and h2 are polynomials in x of degree r and s respectively. A discussion of this
r
s
formula can be found in Forfar et al (1988). Both Gompertz’ law and Makeham’s law are
special cases of the GM formula.
In Section 2.3, we noted the importance of the force of mortality. A further signiﬁcant
point is that when mortality data are analyzed, the force of mortality is a natural quantity
to estimate, whereas the lifetime distribution is not. The analysis of mortality data is
a huge topic and is beyond the scope of this book. An excellent summary article on
this topic is Macdonald (1996). For more general distributions, the quantity f0 (x)/S0 (x),
which actuaries call the force of mortality at age x, is known as the hazard rate in
survival analysis and the failure rate in reliability theory. 66 2.8 CHAPTER 2. SURVIVAL MODELS Exercises Exercise 2.1 Let F0 (t) = 1 − (1 − t/105)1/5 for 0 ≤ t ≤ 105. Calculate
(a) the probability that a newborn life dies before age 60,
(b) the probability that a life aged 30 survives to at least age 70,
(c) the probability that a life aged 20 dies between ages 90 and 100,
(d) the force of mortality at age 50,
(e) the median future lifetime at age 50,
(f) the complete expectation of life at age 50,
(g) the curtate expectation of life at age 50. Exercise 2.2 The function
G(x) = 18000 − 110x − x2
18000 has been proposed as the survival function S0 (x) for a mortality model.
(a) What is the implied limiting age ω ?
(b) Verify that the function G satisﬁes the criteria for a survival function.
(c) Calculate 20 p0 . (d) Determine the survival function for a life aged 20.
(e) Calculate the probability that a life aged 20 will die between ages 30 and 40. 2.8. EXERCISES 67 (f) Calculate the force of mortality at age 50. Exercise 2.3 Calculate the probability that a life aged 0 will die between ages 19 and
36, given the survival function
S0 (x) = 1√
100 − x ,
10 0 ≤ x ≤ 100 (= ω ). Exercise 2.4 Let
1
C
C
S0 (x) = exp − Ax + Bx2 +
Dx −
2
log D
log D
where A, B, C and D are all positive.
(a) Show that the function S0 is a survival function.
(b) Derive a formula for Sx (t).
(c) Derive a formula for µx .
(d) Now suppose that
A = 0.00005, B = 0.0000005, C = 0.0003, (i) Calculate t p30 for t = 1, 5, 10, 20, 50, 90.
(ii) Calculate t q40 for t = 1, 10, 20.
(iii) Calculate t 10 q30 for t = 1, 10, 20.
(iv) Calculate ex for x = 70, 71, 72, 73, 74, 75. D = 1.07. 68 CHAPTER 2. SURVIVAL MODELS
◦ (v) Calculate ex for x = 70, 71, 72, 73, 74, 75, using numerical integration. Exercise 2.5 Let F0 (t) = 1 − e−λt , where λ > 0.
(a) Show that Sx (t) = e−λt .
(b) Show that µx = λ.
(c) Show that ex = (eλ − 1)−1 .
(d) What conclusions do you draw about using this lifetime distribution to model human
mortality? Exercise 2.6 Given that px = 0.99, px+1 = 0.985, 3 px+1 = 0.95 and qx+3 = 0.02, ﬁnd
(a) px+3 ,
(b) 2 px ,
(c) 2 px+1 ,
(d) 3 px ,
(e) 1 2 qx . 2.8. EXERCISES 69 Exercise 2.7 Given that
F0 (x) = 1 − 1
1+x for x ≥ 0, ﬁnd expressions for, simplifying as far as possible,
(a) S0 (x),
(b) f0 (x),
(c) Sx (t), and calculate:
(d) p20 , and
(e) 10 5 q30 . Exercise 2.8 Given that
S0 (x) = e−0.001 x 2 for x ≥ 0, ﬁnd expressions for, simplifying as far as possible,
(a) f0 (x), and
(b) µx . Exercise 2.9 Show that
d
t px = t px (µx − µx+t ) .
dx 70 CHAPTER 2. SURVIVAL MODELS Exercise 2.10 Suppose that Gompertz’ law applies with µ30 = 0.000130 and µ50 =
0.000344. Calculate 10 p40 . Exercise 2.11 A survival model follows Makeham’s law, so that
µx = A + Bcx for x ≥ 0. (a) Show that under Makeham’s law
t px = st g c x (ct −1) , (2.27) where s = e−A and g = exp{−B/ log c}.
(b) Suppose you are given the values of
c= log( 10 p70 ) − log( 10 p60 )
log( 10 p60 ) − log( 10 p50 ) 10 p50 , 10 p60 and 10 p70 . Show that 0 .1 . Exercise 2.12 (a) Construct a table of px for Makeham’s model with parameters A =
0.0001, B = 0.00035 and c = 1.075, for integer x from age 0 to age 130, using Excel
or other appropriate computer software. You should set the parameters so that
they can be easily changed, and you should keep the table, as many exercises and
examples in future chapters will use it.
(b) Use the table to determine the age last birthday at which a life currently age 70 is
most likely to die.
(c) Use the table to calculate e70 .
◦ (d) Using a numerical approach, calculate e70 . 2.8. EXERCISES 71 Exercise 2.13 A life insurer assumes that the force of mortality of smokers at all ages
is twice the force of mortality of nonsmokers.
(a) Show that, if * represents smokers’ mortality, and the ‘unstarred’ function represents
nonsmokers’ mortality, then
∗
t px = (t px )2 . (b) Calculate the diﬀerence between the life expectancy of smokers and nonsmokers
age 50, assuming that nonsmokers mortality follows the Gompertz model, with
B = 0.0005 and c = 1.07.
(c) Calculate the variance of the future lifetime for a nonsmoker age 50 and for a
smoker age 50 under the Gompertz model.
Hint: You will need to use numerical integration for parts (b) and (c). Exercise 2.14
◦ (a) Show that
◦ ex ≤ ex+1 + 1.
(b) Show that
◦ ex ≥ ex .
(c) Explain (in words) why
1
◦
ex ≈ ex + .
2
◦ (d) Is ex always a nonincreasing function of x? 72 CHAPTER 2. SURVIVAL MODELS Exercise 2.15 (a) Show that o ex = 1
S0 (x) ∞ S0 (t)dt, x where S0 (t) = 1 − F0 (t), and hence, or otherwise, prove that
do
o
ex = µx ex − 1
dx
Hint: x d
dx g (t)dt
a = g (x). What about d
dx a g (t)dt ?
x (b) Deduce that
o x + ex
is an increasing function of x, and explain this result intuitively. 2.8. EXERCISES 73 Answers to selected exercises
2.1 (a) 0.1559
(b) 0.8586
(c) 0.1394
(d) 0.0036
(e) 53.28
(f) 45.83
(g) 45.18
2.2 (a) 90
(c) 0.8556
(d) 1 − 3x/308 − x2 /15 400
(e) 0.1169
(f) 0.021
2.3 0.1
0.9064, 0.3812, 3.5 × 10−7 12.001, 11.499, 11.009, 10.533 12.498, 11.995, 11.505, 11.029 2.4 (d) (i) 0.9976, 0.9862, 0.9672, (ii) 0.0047, 0.0629, 0.1747 (iii) 0.0349, 0.0608, 0.1082 (iv) 13.046, 12.517, (v) 13.544, 13.014, 2.6 (a) 0.98
(b) 0.97515
(c) 0.96939 74 CHAPTER 2. SURVIVAL MODELS
(d) 0.95969
(e) 0.03031 2.7 (d) 0.95455
(e) 0.08218
2.10 0.9973
2.12 (b) 73
(c) 9.339
(d) 9.834
2.13 (b) 6.432
(c) 125.89 (nonsmokers), 80.11 (smokers) Chapter 3
Life Tables and Selection
3.1 Summary In this chapter we deﬁne a life table. For a life table tabulated at integer ages only, we
show, using fractional age assumptions, how to calculate survival probabilities for all ages
and durations.
We discuss some features of national life tables from Australia, England & Wales and the
United States.
We then consider life tables appropriate to individuals who have purchased particular
types of life insurance policy and discuss why the survival probabilities diﬀer from those
in the corresponding national life table. We consider the eﬀect of ‘selection’ of lives for
insurance policies, for example through medical underwriting. We deﬁne a select survival
model and we derive some formulae for such a model.
75 76 CHAPTER 3. LIFE TABLES AND SELECTION 3.2 Life tables Given a survival model, with survival probabilities t px , we can construct the life table
for the model from some initial age x0 to a maximum age ω . We deﬁne a function {lx }
for x0 ≤ x ≤ ω as follows. Let lx0 be an arbitrary positive number (called the radix of
the table) and, for 0 ≤ t ≤ ω − x0 , deﬁne
lx0 +t = lx0 t px0 .
From this deﬁnition we see that for x0 ≤ x ≤ x + t ≤ ω ,
lx+t = lx0 x+t−x0 px0
= lx0 x−x0 px0 t px
= lx t px ,
so that
t px = lx+t /lx . (3.1) For any x ≥ x0 , we can interpret lx+t as the expected number of survivors to age x + t
out of lx independent individuals aged x. This interpretation is more natural if lx is an
integer, and follows because the number of survivors to age x + t is a random variable
with a binomial distribution with parameters lx and t px . That is, suppose we have lx
independent lives aged x, and each life has a probability t px of surviving to age x + t.
Then the number of survivors to age x + t is a binomial random variable, Lt , say, with
parameters lx and t px . The expected value of the number of survivors is then
E[Lt ] = lx t px = lx+t . 3.2. LIFE TABLES 77 We always use the table in the form ly /lx which is why the radix of the table is arbitrary
– it would make no diﬀerence to the survival model if all the lx values were multiplied by
100, for example.
From (3.1) we can use the lx function to calculate survival probabilities. We can also
calculate mortality probabilities. For example,
q30 = 1 − l31
l30 − l31
=
l30
l30 (3.2) and
15 30 q40 = 15 p40 30 q55 = l55
l40 1− l85
l55 = l55 − l85
.
l40 (3.3) In principle, a life table is deﬁned for all x from the initial age, x0 , to the limiting age,
ω . In practice, it is very common for a life table to be presented, and in some cases even
deﬁned, at integer ages only. In this form, the life table is a useful way of summarizing
a lifetime distribution since, with a single column of numbers, it allows us to calculate
probabilities of surviving or dying over integer numbers of years starting from an integer
age.
It is usual for a life table, tabulated at integer ages, to show the values of dx , where
dx = lx − lx+1 , (3.4) in addition to lx , as these are used to compute qx . From (3.4) we have
dx = lx 1 − lx+1
lx = lx (1 − px ) = lx qx . We can also arrive at this relationship if we interpret dx as the expected number of deaths
in the year of age x to x + 1 out of lx lives aged exactly x, so that, using the binomial
distribution again
dx = lx qx . (3.5) 78 CHAPTER 3. LIFE TABLES AND SELECTION
x
30
31
32
33
34
35
36
37
38
39 lx
10 000.00
9 965.22
9 927.12
9 885.35
9 839.55
9 789.29
9 734.12
9 673.56
9 607.07
9 534.08 dx
34.78
38.10
41.76
45.81
50.26
55.17
60.56
66.49
72.99
80.11 Table 3.1: Extract from a life table.
Example 3.1 Table 3.1 gives an extract from a life table. Calculate
(a) l40 ,
(b) 10 p30 , (c) q35 ,
(d) 5 q30 , and
(e) the probability that a life currently aged exactly 30 dies between ages 35 and 36.
Solution 3.1 (a) From equation (3.4), l40 = l39 − d39 = 9 453.97.
(b) From equation (3.1),
10 p30 = l40
9 453.97
=
= 0.94540.
l30
10 000 3.3. FRACTIONAL AGE ASSUMPTIONS 79 (c) From equation (3.5),
q35 = d35
55.17
=
= 0.00564.
l35
9 789.29 (d) Following equation (3.2),
5 q30 = l30 − l35
= 0.02107.
l30 (e) This probability is 5  q30 . Following equation (3.3),
5  q30 3.3 = l35 − l36
d35
=
= 0.00552.
l30
l30 Fractional age assumptions A life table {lx }x≥x0 provides exactly the same information as the corresponding survival
distribution, Sx0 . However, a life table tabulated at integer ages only does not contain
all the information in the corresponding survival model, since values of lx at integer ages
x are not suﬃcient to be able to calculate probabilities involving noninteger ages, such
as 0.75 p30.5 . Given values of lx at integer ages only, we need an additional assumption
or some further information to calculate probabilities for noninteger ages or durations.
Speciﬁcally, we need to make some assumption about the probability distribution for the
future lifetime random variable between integer ages.
We use the term fractional age assumption to describe such an assumption. It may
be speciﬁed in terms of the force of mortality function or the survival or mortality probabilities.
In this section we assume that a life table is speciﬁed at integer ages only and we describe
the two most useful fractional age assumptions. 80 CHAPTER 3. LIFE TABLES AND SELECTION 3.3.1 Uniform distribution of deaths The uniform distribution of deaths (UDD) assumption is the most common fractional age
assumption. It can be formulated in two diﬀerent, but equivalent, ways as follows.
UDD1
For integer x, and for 0 ≤ s < 1, assume that
s qx = sqx . (3.6) UDD2
Recall from Chapter 2 that Kx is the integer part of Tx , and deﬁne a new random
variable Rx such that
Tx = Kx + Rx .
The UDD2 assumption is that, for integer x, Rx ∼ U(0, 1), and Rx is independent
of Kx .
The equivalence of these two assumptions is demonstrated as follows. First, assume that
UDD1 is true. Then for integer x, and for 0 ≤ s < 1,
Pr[Rx ≤ s] = = ∞
k=0
∞
k=0 = ∞ Pr[Rx ≤ s and Kx = k ] Pr[k ≤ Tx ≤ k + s] k p x s q x+ k k=0 = ∞
k=0 k px s (qx+k ) using UDD1 3.3. FRACTIONAL AGE ASSUMPTIONS
∞ =s 81 k p x q x+ k k=0
∞ =s Pr[Kx = k ] k=0 = s.
This proves that Rx ∼ U(0, 1). To prove the independence of Rx and Kx , note that
Pr[Rx ≤ s and Kx = k ] = Pr[k ≤ Tx ≤ k + s]
= k p x s q x+ k = s k p x q x+ k
= Pr[Rx ≤ s] Pr[Kx = k ]
since Rx ∼ U(0, 1). This proves that UDD1 implies UDD2.
To prove the reverse implication, assume that UDD2 is true. Then for integer x, and for
0 ≤ s < 1,
s qx = Pr[Tx ≤ s]
= Pr[Kx = 0 and Rx ≤ s]
= Pr[Rx ≤ s] Pr[Kx = 0] as Kx and Rx are assumed independent. Thus,
s qx = s qx . (3.7) 82 CHAPTER 3. LIFE TABLES AND SELECTION Formulation UDD2 explains why this assumption is called the Uniform Distribution of
Deaths, but in practical applications of this assumption, formulation UDD1 is the more
useful of the two.
An immediate consequence is that
lx+s = lx − sdx (3.8) for 0 ≤ s < 1. This follows because
s qx =1− lx+s
lx and substituting s qx for s qx gives
s dx
lx − lx+s
=
.
lx
lx Hence
lx+s = lx − s dx
for 0 ≤ s ≤ 1. Thus, we assume that lx+s is a linearly decreasing function of s.
Diﬀerentiating equation (3.6) with respect to s, we obtain
d
s qx = qx ,
ds 0≤s≤1 and we know that the left hand side is the probability density function for Tx at s, because
we are diﬀerentiating the distribution function. The pdf for Tx at s is s px µx+s so that
under UDD
q x = s p x µ x+ s
for 0 ≤ s < 1. (3.9) 3.3. FRACTIONAL AGE ASSUMPTIONS 83 The left hand side does not depend on s, which means that the density function is a
constant for 0 ≤ s < 1, which also follows from the uniform distribution assumption for
Rx .
Since qx is constant with respect to x, and s px is a decreasing function of s, we can
see that µx+s is an increasing function of s, which is appropriate for ages of interest
to insurers. However, if we apply the approximation over successive ages, we obtain a
discontinuous function for the force of mortality, with discontinuities occurring at integer
ages, as illustrated in Example 3.4. Although this is undesirable, it is not a serious
drawback. Example 3.2 Given that p40 = 0.999473, calculate
uniform distribution of deaths. 0.4 q40.2 under the assumption of a Solution 3.2 We note that the fundamental result in equation (3.7), that for fractional
age s, s qx = s qx requires x to be an integer. We can manipulate the required probability
0.4 q40.2 to involve only probabilities from integer ages as follows
0.4 q40.2 = 1−
= 1− 0.4 p40.2 0.6 p40
0.2 p40 =1− =1− l40.6
l40.2 1 − 0.6q40
1 − 0.2q40 = 2.108 × 10−4 . Example 3.3 Use the life table in Example 3.1 above, with the UDD assumption, to
calculate (a) 1.7 q33 and (b) 1.7 q33.5 . 84 CHAPTER 3. LIFE TABLES AND SELECTION Solution 3.3
1.7 q33 (a) We note ﬁrst that
= 1 − 1.7 p33 = 1 − (p33 ) (0.7 p34 ) . We can calculate p33 directly from the life table as l34 /l33 = 0.995367 and
1 − 0.7 q34 = 0.996424 under UDD, so that 1.7 q33 = 0.008192.
(b) To calculate
1.7 q33.5 1.7 q33.5 0.7 p34 = using UDD, we express this as = 1 − 1.7 p33.5
= 1− l35.2
l33.5 = 1− l35 − 0.2d35
l33 − 0.5d33 = 0.008537. Example 3.4 Under the assumption of a uniform distribution of deaths, calculate
lim µ40+t using p40 = 0.999473, and calculate lim µ41+t using p41 = 0.999429.
+
−
t→0 t→1 Solution 3.4 From formula (3.9), we have µx+t = qx /t px . Setting x = 40 yields
lim µ40+t = q40 /p40 = 5.273 × 10−4 , t→1− while setting x = 41 yields
lim µ41+t = q41 = 5.71 × 10−4 . t→0+ 3.3. FRACTIONAL AGE ASSUMPTIONS 85 Example 3.5 Given that q70 = 0.010413 and q71 = 0.011670, calculate
a uniform distribution of deaths. 0.7 q70.6 assuming Solution 3.5 As deaths are assumed to be uniformly distributed between ages 70 to 71
and ages 71 to 72, we write
0.7 q70.6 = 0.4 q70.6 + (1 − 0.4 q70.6 ) 0.3 q71 . Following the same arguments as in Solution 3.3, we obtain
0.4 q70.6 and as 3.3.2 0.3 q71 =1− 1 − q70
= 4.191 × 10−3 ,
1 − 0.6q70 = 0.3q71 = 3.501 × 10−3 , we obtain 0.7 q70.6 = 7.678 × 10−3 . Constant force of mortality A second fractional age assumption is that the force of mortality is constant between
integer ages. Thus, for integer x and 0 ≤ s < 1, we assume that µx+s does not depend on
s, and we denote it µ∗ . We can obtain the value of µ∗ by using the fact that
x
x
1 px = exp − µx+s ds .
0
∗ Hence the assumption that µx+s = µ∗ for 0 ≤ s < 1 gives px = e−µx or µ∗ = − log px .
x
x
Further, under the assumption of a constant force of mortality, for 0 ≤ s < 1 we obtain
s
s px = exp − 0 µ∗ du
x ∗ = e−µx s = (px )s . Similarly, for t, s > 0 and t + s < 1,
s
s p x+ t = exp − 0 µ∗ du
x = (px )s . 86 CHAPTER 3. LIFE TABLES AND SELECTION Thus, under the constant force assumption, the probability of surviving for a period of
s < 1 years from age x + t is independent of t provided that s + t < 1.
The assumption of a constant force of mortality leads to a step function for the force of
mortality over successive years of age. By its nature, the assumption produces a constant
force of mortality over the year of age x to x + 1, whereas we would expect the force
of mortality to increase for most ages. However, if the true force of mortality increases
slowly over the year of age, the constant force of mortality assumption is reasonable.
Example 3.6 Given that p40 = 0.999473, calculate
constant force of mortality.
Solution 3.6 We have 0.4 q40.2 =1− 0.4 p40.2 0.4 q40.2 under the assumption of a = 1 − (p40 )0.4 = 2.108 × 10−4 . Example 3.7 Given that q70 = 0.010413 and q71 = 0.011670, calculate
assumption of a constant force of mortality. 0.7 q70.6 under the Solution 3.7 As in Solution 3.5 we write
0.7 q70.6 = 0.4 q70.6 + (1 − 0.4 q70.6 ) 0.3 q71 , where 0.4 q70.6 = 1 − (p70 )0.4 = 4.178 × 10−3 and
−3
0.7 q70.6 = 7.679 × 10 . 0.3 q71 = 1 − (p71 )0.3 = 3.515 × 10−3 , giving Note that in Examples 3.2 and 3.5 and in Examples 3.6 and 3.7 we have used two diﬀerent
methods to solve the same problems, and the solutions agree to ﬁve decimal places. It
is generally true that the assumptions of a uniform distribution of deaths and a constant
force of mortality produce very similar solutions to problems. The reason for this can be 3.4. NATIONAL LIFE TABLES 87 seen from the following approximations. Under the constant force of mortality assumption
∗ qx = 1 − e−µ ≈ µ∗
provided that µ∗ is small, and for 0 < t < 1,
t qx ∗ = 1 − e−µ t ≈ µ∗ t. In other words, the approximation to t qx is t times the approximation to qx , which is what
we obtain under the uniform distribution of deaths assumption. 3.4 National life tables Life tables based on the mortality experience of the whole population of a country are
regularly produced for many countries in the world. Separate life tables are usually
produced for males and for females and possibly for some other groups of individuals, for
example on the basis of smoking habits.
Table 3.2 shows values of qx × 105 , where qx is the probability of dying within one year,
for selected ages x, separately for males and females, for the populations of Australia,
England & Wales and the United States. These tables are constructed using records of
deaths in a particular year, or a small number of consecutive years, and estimates of
the population in the middle of that period. The relevant years are indicated in the
column headings for each of the three life tables in Table 3.2. Data at the oldest ages are
notoriously unreliable. For this reason, the United States Life Tables do not show values
of qx for ages 100 and higher.
For all three national life tables and for both males and females, the values of qx follow
exactly the same pattern as a function of age, x. Figure 3.1 shows the US 2002 mortality
rates for males and females; the graphs for England & Wales and for Australia are similar.
(Note that we have plotted these on a logarithmic scale in order to highlight the main features. Also, although the information plotted consists of values of qx for x = 0, 1, . . . , 99, 88 CHAPTER 3. LIFE TABLES AND SELECTION Australian Life Tables English Life Table 15 US Life Tables
200002
199092
2002
x
Males
Females Males
Females Males Females
0
567
466
814
632
764
627
1
44
43
62
55
53
42
2
31
19
38
30
37
28
10
13
8
18
13
18
13
20
96
36
84
31
139
45
30
119
45
91
43
141
63
40
159
88
172
107
266
149
50
315
202
464
294
570
319
60
848
510
1392
830
1210
758
70
2 337
1 308 3 930
2 190 2 922
1 899
80
6 399
4 036 9 616
5 961 7 028
4 930
90 15 934
12 579 20 465
15 550 16 805
13 328
100 24 479
23 863 38 705
32 489
−
−
Table 3.2: Values of qx × 105 from some national life tables. 3.4. NATIONAL LIFE TABLES 89 1 Mortality rates 0.1 0.01 0.001 0.0001
0 10 20 30 40 50 60 70 80 90 100 Age Figure 3.1: US 2002 Mortality rates, Male (dotted) and Female (solid).
we have plotted a continuous line as this gives a clearer representation.) We note the
following points from Table 3.2 and Figure 3.1.
• The value of q0 is relatively high. Mortality rates immediately following birth,
perinatal mortality, are high due to complications arising from the later stages of
pregnancy and from the birth process itself. The value of qx does not reach this
level again until about age 55. This can be seen from Figure 3.1.
• The rate of mortality is much lower after the ﬁrst year, less than 10% of its level in
the ﬁrst year, and declines until around age 10.
• In Figure 3.1 we see that the pattern of male and female mortality in the late teenage
years diverges signiﬁcantly, with a steeper incline in male mortality. Not only is this
feature of mortality for young adult males common for diﬀerent populations around
the world, it is also a feature of historical populations in countries such as the UK 90 CHAPTER 3. LIFE TABLES AND SELECTION 0.35 0.30 Mortality rates 0.25 0.20 0.15 0.10 0.05 0.00
50 60 70 80 90 100 Age Figure 3.2: US 2002 Male mortality rates (solid), with ﬁtted Gompertz mortality rates
(dotted).
where mortality data has been collected for some time. It is sometimes called the
accident hump, as many of the deaths causing the ‘hump’ are accidental.
• Mortality rates increase from age 10, with the accident hump creating a relatively
large increase between ages 10 and 20 for males, a more modest increase from ages
20 to 40, and then steady increases from age 40.
• For each age, all six values of qx are broadly comparable, with, for each country, the
rate for a female almost always less than the rate for a male of the same age. The one
exception is the Australian Life Table, where q100 is slightly higher for a female than
for a male. According to the Australian Government Actuary, Australian mortality
data indicate that males are subject to lower mortality rates than females at very
high ages, although there is some uncertainty as to where the crossover occurs due
to small amounts of data at very old ages. 3.5. SURVIVAL MODELS FOR LIFE INSURANCE POLICYHOLDERS 91 • The Gompertz model introduced in Chapter 2 is relatively simple, in that it requires
only two parameters and has a force of mortality with a simple functional form,
µx = Bcx . We stated in Chapter 2 that this model does not provide a good ﬁt
across all ages. We can see from Figure 3.1 that the model cannot ﬁt the perinatal
mortality, nor the accident hump. However, the mortality rates at later ages are
rather better behaved, and the Gompertz model often proves useful over older age
ranges. Figure 3.2 shows the older ages US 2002 Males mortality rate curve, along
with a Gompertz curve ﬁtted to the US 2002 table mortality rates. The Gompertz
curve provides a pretty close ﬁt – which is a particularly impressive feat, considering
that Gompertz proposed the model in 1825.
A ﬁnal point about Table 3.2 is that we have compared three national life tables using
values of the probability of dying within one year, qx , rather than the force of mortality,
µx . This is because values of µx are not published for any ages for the US Life Tables.
Also, values of µx are not published for age 0 for the other two life tables – there are
technical diﬃculties in the estimation of µx within a year in which the force of mortality
is changing rapidly, as it does between ages 0 and 1. 3.5 Survival models for life insurance policyholders Suppose we have to choose a survival model appropriate for a man, currently aged 50
and living in the UK, who has just purchased a 10 year term insurance policy. We
could use a national life table, say English Life Table 15, so that, for example, we could
assume that the probability this man dies before age 51 is 0.00464, as shown in Table
3.2. However, in the UK, as in some other countries with welldeveloped life insurance
markets, the mortality experience of people who purchase life insurance policies tends
to be diﬀerent from the population as a whole. The mortality of diﬀerent types of life
insurance policyholders is investigated separately, and life tables appropriate for these
groups are published. 92 CHAPTER 3. LIFE TABLES AND SELECTION Table 3.3 shows values of the force of mortality (×105 ) at two year intervals from age
50 to age 60 taken from English Life Table 15, Males, (ELTM 15) and from a life table
prepared from data relating to term insurance policyholders in the UK in 19992002 and
which assumes the policyholders purchased their policies at age 50. This second set of
mortality rates come from Table A14 of a 2006 working paper of the Continuous Mortality
Investigation in the UK. Hereafter we refer to this working paper as CMI, and further
details are given at the end of this chapter. The values of the force of mortality for ELTM
15 correspond to the values of qx shown in Table 3.2.
x ELTM 15
50
440
52
549
54
679
56
845
58
1 057
60
1 323 CMI
78
152
240
360
454
573 Table 3.3: Values of the force of mortality ×105 from English Life Table 15 and CMI
(Table A14) for UK males who purchase a term insurance policy at age 50.
The striking feature of Table 3.3 is the diﬀerence between the two sets of values. The
values from CMI are very much lower than those from ELTM 15, by a factor of more than
5 at age 50 and by a factor of more than two at age 60. There are at least three reasons
for this diﬀerence. Two of these are discussed below, the third is discussed in the next
section.
(a) The data on which the two life tables are based relate to diﬀerent calendar years;
199092 in the case of ELTM 15 and 19992002 in the case of CMI. Mortality rates
in the UK, as in many other countries, have been decreasing for some years so we
might expect rates based on more recent data to be lower. However, this explains 3.6. LIFE INSURANCE UNDERWRITING 93 only a small part of the diﬀerences in Table 3.3. An interim life table for England
and Wales based on population data from 20022004, gives the following values for
males: µ50 = 391 × 10−5 and µ60 = 1008 × 10−5 . Clearly, mortality in England
and Wales has improved over the twelve year period, but not to the extent that it
matches the CMI values shown in Table 3.3. Other explanations for the diﬀerences
in Table 3.3 are needed.
(b) A major reason for the diﬀerence between the values in Table 3.3 is that ELTM 15
is a life table based on the whole male population of England and Wales, whereas
CMI (Table A14) is based on the experience of males who are term insurance policyholders. Within any large group, there are likely to be variations in mortality
rates between subgroups. This is true in the case of the population of England and
Wales, where social class, deﬁned in terms of occupation, has a signiﬁcant eﬀect on
mortality. Put simply, the better your job, and hence the wealthier you are likely to
be, the lower your mortality rates. Given that people who purchase term insurance
policies are likely to be among the better paid people in the population, we have an
explanation for a large part of the diﬀerence between the values in Table 3.3.
CMI (Table A2) shows values of the force of mortality based on data from males in the
UK who purchased whole life or endowment insurance policies. These are similar to
those shown in Table 3.3 for term insurance policyholders and hence much lower than the
values for the whole population. People who purchase whole life or endowment policies,
like those who purchase term insurance policies, tend to be among the wealthier people
in the population. 3.6 Life insurance underwriting The values of the force of mortality in Table 3.3 taken from CMI are values based on
data for males who purchased term insurance at age 50. CMI (Table A14) gives values 94 CHAPTER 3. LIFE TABLES AND SELECTION for diﬀerent ages at the purchase of the policy ranging from 17 to 90. Values for ages at
purchase 50, 52, 54 and 56 are shown in Table 3.4. x
50
52
54
56
58
60
62
64
66 Age at purchase of policy
50
52
54
56
78
−
−
−
152
94
−
−
240
186
113
−
360
295
227
136
454
454
364
278
573
573
573
448
725
725
725
725
917
917
917
917
1 159 1 159 1 159 1 159 Table 3.4: Values of the force of mortality ×105 from CMI (Table A14) for diﬀerent ages
at purchase of a term insurance policy.
There are two signiﬁcant features of the values in Table 3.4, which can be seen by considering the rows of values for ages 56 and 62.
(a) Consider the row of values for age 56. Each of the four values in this row is the force
of mortality at age 56 based on data from the UK over the period 19992002 for males
who are term insurance policyholders. The only diﬀerence is that they purchased
their policies at diﬀerent ages. The more recently the policy was purchased, the
lower the force of mortality. For example, for a male who purchased his policy at
age 56, the value is 0.00136, whereas for someone of the same age who purchased
his policy at age 50, the value of is 0.00360.
(b) Now consider the row of values for age 62. These values, all equal to 0.00725, do
not depend on whether the policy was purchased at age 52, 54 or 56. 3.6. LIFE INSURANCE UNDERWRITING 95 These features are due to life insurance underwriting, which we described in Chapter 1.
Recall that the life insurance underwriting process evaluates medical and lifestyle information to assess whether the policyholder is in normal health.
The important point for this discussion is that the mortality rates in CMI are based on
individuals accepted for insurance at normal premium rates, i.e. individuals who have
passed the health checks outlined above. This means, for example, that a man aged 50
who has just purchased a term insurance at the normal premium rate is in good health,
assuming the health checks are eﬀective, and so is likely to be much healthier, and hence
have a lower mortality rate, than a man of age 50 picked randomly from the population
of England and Wales. This explains a major part of the diﬀerence between the mortality
rates in Table 3.3. When this man reaches age 56, we can no longer be certain he is in
good health – all we know is that he was in good health six years ago. Hence, his mortality
rate at age 56 is higher than that of a man of the same age who has just passed the health
checks and been permitted to buy a term insurance policy at normal rates. This explains
the diﬀerences between the values of the force of mortality at age 56 in Table 3.4.
The eﬀect of passing the health checks at an earlier age eventually wears oﬀ, so that at
age 62, the force of mortality does not depend on whether the policy was purchased at
age 52, 54 or 56. This is point (b) above. However, note that these rates, 0.00725, are
still much lower than µ62 (= 0.01664) from ELTM 15. This is because people who buy
insurance tend, at least in the UK, to have lower mortality than the general population.
In fact the population is made up of many heterogeneous lives, and the eﬀect of initial
selection is only one area where actuaries have tried to manage the heterogeneity. In the
US, there has been a lot of activity recently developing tables for ‘preferred lives’, who are
assumed to be even healthier than the standard insured population. These preferred lives
tend to be from higher socioeconomic groups. Mortality and wealth are closely linked. 96 3.7 CHAPTER 3. LIFE TABLES AND SELECTION Select and ultimate survival models A feature of the survival models studied in Chapter 2 is that probabilities of future survival
depend only on the individual’s current age. For example, for a given survival model and
a given term t, t px , the probability that an individual currently aged x will survive to
age x + t, depends only on the current age x. Such survival models are called aggregate
survival models, because lives are all aggregated together.
The diﬀerence between an aggregate survival model and the survival model for term
insurance policyholders discussed in Section 3.6 is that in the latter case, probabilities of
future survival depend not only on current age but also on how long ago the individual
entered the group of policyholders, i.e. when the policy was purchased.
This leads us to the following deﬁnition. The mortality of a group of individuals is described by a select and ultimate survival model, usually shortened to select survival
model, if the following statements are true.
(a) Future survival probabilities for an individual in the group depend on the individual’s current age and on the age at which the individual joined the group.
(b) There is a positive number (generally an integer), which we denote by d, such that
if an individual joined the group more than d years ago, future survival probabilities
depend only on current age. The initial selection eﬀect is assumed to have worn oﬀ
after d years.
We use the following terminology for a select survival model. An individual who enters
the group at, say, age x, is said to be selected, or just select, at age x. The period d
after which the age at selection has no eﬀect on future survival probabilities is called the
select period for the model. The mortality that applies to lives after the select period is
complete is called the ultimate mortality, so that the complete model comprises a select
period followed by the ultimate period. 3.7. SELECT AND ULTIMATE SURVIVAL MODELS 97 Going back to the term insurance policyholders in Section 3.6, we can identify the ‘group’
as male term insurance policyholders in the UK. A select survival model is appropriate
in this case because passing the health checks at age x indicates that the individual is in
good health and so has lower mortality rates than someone of the same age who passed
these checks some years ago. There are indications in Table 3.4 that the select period, d,
for this group is less than or equal to six years. See point (b) in Section 3.6. In fact, the
select period is 5 years for this particular model. Select periods typically range from 1
year to 15 years for life insurance mortality models.
For the term insurance policyholders in Section 3.6, being selected at age x meant that
the mortality rate for the individual was lower than that of a term insurance policyholder
of the same age who had been selected some years earlier. Selection can occur in many
diﬀerent ways and does not always lead to lower mortality rates, as Example 3.8 shows.
Example 3.8 Consider men who need to undergo surgery because they are suﬀering
from a particular disease. The surgery is complicated and there is a probability of only
50% that they will survive for a year following surgery. If they do survive for a year, then
they are fully cured and their future mortality follows the Australian Life Tables 200002,
Males, from which you are given the following values:
l60 = 89 777, l61 = 89 015, l70 = 77 946. Calculate
(a) the probability that a man aged 60 who is just about to have surgery will be alive
at age 70,
(b) the probability that a man aged 60 who had surgery at age 59 will be alive at age
70, and
(c) the probability that a man aged 60 who had surgery at age 58 will be alive at age
70. 98 CHAPTER 3. LIFE TABLES AND SELECTION Solution 3.8 In this example, the ‘group’ is all men who have had the operation. Being
selected at age x means having surgery at age x. The select period of the survival model
for this group is one year, since if they survive for one year after being ‘selected’, their
future mortality depends only on their current age.
(a) The probability of surviving to age 61 is 0.5. Given that he survives to age 61, the
probability of surviving to age 70 is
l70 /l61 = 77 946/89 015 = 0.8757.
Hence, the probability that this individual survives from age 60 to age 70 is
0.5 × 0.8757 = 0.4378.
(b) Since this individual has already survived for one year following surgery, his mortality follows the Australian Life Tables 200002, Males. Hence, his probability of
surviving to age 70 is
l70 /l60 = 77 946/89 777 = 0.8682.
(c) Since this individual’s surgery was more than one year ago, his future mortality
is exactly the same, probabilistically, as the individual in part (b). Hence, his
probability of surviving to age 70 is 0.8682. Selection is not a feature of national life tables since, ignoring immigration, an individual
can enter the population only at age zero. It is an important feature of many survival
models based on data from, and hence appropriate to, life insurance policyholders. We
can see from Tables 3.3 and 3.4 that its eﬀect on the force of mortality can be considerable.
For these reasons, select survival models are important in life insurance mathematics. 3.8. NOTATION AND FORMULAE FOR SELECT SURVIVAL MODELS 99 The select period may be diﬀerent for diﬀerent survival models. For CMI (Table A14),
which relates to term insurance policyholders, it is ﬁve years, as noted above; for CMI
(Table A2), which relates to whole life and endowment policyholders, the select period is
two years.
In the next section we introduce notation and develop some formulae for select survival
models. 3.8 Notation and formulae for select survival models A select survival model represents an extension of the ultimate survival model studied in
Chapter 2. In Chapter 2, survival probabilities depended only on the current age of the
individual. For a select survival model, probabilities of survival depend on current age
and (within the select period) age at selection, i.e. age at joining the group. However,
the survival model for those individuals all selected at the same age, say x, depends only
on their current age and so ﬁts the assumptions of Chapter 2. This means that, provided
we ﬁx and specify the age at selection, we can adapt the notation and formulae developed
in Chapter 2 to a select survival model. This leads to the following deﬁnitions:
S[x]+s(t) = Pr[a life currently aged x + s who was select at age x survives to age x + s + t]
t q[x]+s = Pr[a life currently aged x + s who was select at age x dies before age x + s + t] t p[x]+s = 1 − t q[x]+s ≡ S[x]+s (t) µ[x]+s is the force of mortality at age x + s for an individual who was select at age x
1 − S[x]+s (h)
h→0
h
From these deﬁnitions we can derive the following formula
µ[x]+s = lim+ t
t p[x]+s = exp − µ[x]+s+u du .
0 This formula is derived precisely as in Chapter 2. It is only the notation which has 100 CHAPTER 3. LIFE TABLES AND SELECTION changed.
For a select survival model with a select period d and for t ≥ d, that is, for durations at
or beyond the select period, the values of µ[x−t]+t , s p[x−t]+t and us q[x−t]+t do not depend
on t, they depend only on the current age x. So, for t ≥ d we drop the more detailed
notation, µ[x−t]+t , s p[x−t]+t and us q[x−t]+t , and write µx , s px and us qx . For values of t < d,
we refer to, for example, µ[x−t]+t as being in the select part of the survival model and for
t ≥ d we refer to µ[x−t]+t (≡ µx ) as being in the ultimate part of the survival model. 3.9 Select life tables For an ultimate survival model, as discussed in Chapter 2, the life table {lx } is useful
since it can be used to calculate probabilities such as tu qx for nonnegative values of t, u
and x. We can construct a select life table in a similar way but we need the table to
reﬂect duration as well as age, during the select period. Suppose we wish to construct this
table for a select survival model for ages at selection from, say, x0 (≥ 0). Let d denote
the select period, assumed to be an integer number of years.
The construction in this section is for a select life table speciﬁed at all ages and not just
at integer ages. However, select life tables are usually presented at integer ages only, as
is the case for ultimate life tables.
First we consider the survival probabilities of those individuals who were selected at least
d years ago and hence are now subject to the ultimate part of the model. The minimum
age of these people is x0 + d. For these people, future survival probabilities depend only
on their current age and so, as in Chapter 2, we can construct an ultimate life table, {ly },
for them from which we can calculate probabilities of surviving to any future age.
Let lx0 +d be an arbitrary positive number. For y ≥ x0 + d we deﬁne
ly =
Note that (y −x0 −d) px0 +d lx0 +d .
(y −x0 −d) px0 +d = (y −x0 −d) p[x0 ]+d (3.10)
since, given that the life was select at least d 3.9. SELECT LIFE TABLES 101 years ago, the probability of future survival depends only on the current age, x0 + d.
From this deﬁnition we can show that for y > x ≥ x0 + d
ly = y −x px lx . (3.11) This follows because
ly = (y −x0 −d) px0 +d = y −x p[x0 ]+x−x0 = (y−x px ) = lx0 +d
(x−x0 −d) p[x0 ]+d (x−x0 −d) px0 +d lx0 +d lx0 +d y −x px lx . This shows that within the ultimate part of the model we can interpret ly as the expected
number of survivors to age y out of lx lives currently age x (< y ), who were select at least
d years ago.
Formula (3.10) deﬁnes the life table within the ultimate part of the model. Next, we need
to deﬁne the life table within the select period. We do this for a life select at age x by
‘working backwards’ from the value of lx+d . For x ≥ x0 and for 0 ≤ t ≤ d, we deﬁne
l[x]+t = lx+d / d−t p[x]+t (3.12) which means that if we had l[x]+t lives age x + t, selected t years ago, then the expected
number of survivors to age x + d is lx+d . This deﬁnes the select part of the life table.
Example 3.9 For y ≥ x + d > x + s > x + t ≥ x ≥ x0 , show that
y −x−t p[x]+t = ly
l[x]+t (3.13) 102 CHAPTER 3. LIFE TABLES AND SELECTION and
s−t p[x]+t = l[x]+s
.
l[x]+t (3.14) Solution 3.9 First,
y −x−t p[x]+t = y −x−d p[x]+d d−t p[x]+t = y −x−d px+d d−t p[x]+t ly = lx+d
ly = l[x]+t lx+d
l[x]+t
, which proves (3.13). Second,
s−t p[x]+t = d−t p[x]+t / d−s p[x]+s = lx+d l[x]+s
l[x]+t lx+d = l[x]+s
,
l[x]+t which proves (3.14). This example, together with formula (3.11), shows that our construction preserves the
interpretation of the ls as expected numbers of survivors within both the ultimate and
the select parts of the model. For example, suppose we have l[x]+t individuals currently 3.9. SELECT LIFE TABLES 103 aged x + t who were select at age x. Then, since y−x−t p[x]+t is the probability that any
one of them survives to age y , we can see from formula (3.13) that ly is the expected
number of survivors to age y . For 0 ≤ t ≤ s ≤ d, formula (3.14) shows that l[x]+s can be
interpreted as the expected number of survivors to age x + s out of l[x]+t lives currently
aged x + t who were select at age x.
Example 3.10 Write an expression for 2 6 q[30]+2 in terms of l[x]+t and ly for appropriate
x, t, and y , assuming a select period of 5 years.
Solution 3.10 Note that 2 6 q[30]+2 is the probability that a life currently age 32, who
was select at age 30, will die between ages 34 and 40. We can write this probability as
the product of the probabilities of the following events:
– a life aged 32, who was select at age 30, will survive to age 34, and,
– a life aged 34, who was select at age 30, will die before age 40.
Hence,
2 6 q[30]+2 = 2 p[30]+2 6 q[30]+4 = l[30]+4
l[30]+2 = l[30]+4 − l40
.
l[30]+2 1− l[30]+10
l[30]+4 Note that l[30]+10 ≡ l40 since ten years is longer than the select period for this survival
model. Example 3.11 A select survival model has a select period of 3 years. Its ultimate mor 104 CHAPTER 3. LIFE TABLES AND SELECTION tality is equivalent to the US Life Tables, 2002, Females. Some lx values for this table are
shown in Table 3.5.
x
70
71
72
73
74
75 lx
80 556
79 026
77 410
75 666
73 802
71 800 Table 3.5: An extract from the US Life Tables, 2002, Females.
You are given that for all ages x ≥ 65,
p[x] = 0.999, p[x−1]+1 = 0.998, p[x−2]+2 = 0.997. Calculate the probability that a woman currently aged 70 will survive to age 75 given
that
(a) she was select at age 67,
(b) she was select at age 68,
(c) she was select at age 69, and
(d) she is select at age 70.
Solution 3.11 (a) Since the woman was select 3 years ago and the select period for
this model is 3 years, she is now subject to the ultimate part of the survival model.
Hence the probability she survives to age 75 is l75 /l70 , where the ls are taken from
US Life Tables, 2002, Females. The required probability is
l75 /l70 = 71 800/80 556 = 0.8913. 3.9. SELECT LIFE TABLES 105 (b) In this case, the required probability is
5 p[68]+2 = l[68]+2+5 /l[68]+2 = l75 /l[68]+2 = 71 800/l[68]+2 . We can calculate l[68]+2 by noting that
l[68]+2 p[68]+2 = l[68]+3 = l71 = 79 026.
We are given that p[68]+2 = 0.997. Hence, l[68]+2 = 79 264 and so
5 p[68]+2 = 71 800/79 264 = 0.9058. (c) In this case, the required probability is
5 p[69]+1 = l[69]+1+5 /l[69]+1 = l75 /l[69]+1 = 71 800/l[69]+1 . We can calculate l[69]+1 by noting that
l[69]+1 p[69]+1 p[69]+2 = l[69]+3 = l72 = 77 410.
We are given that p[69]+1 = 0.998 and p[69]+2 = 0.997. Hence, l[69]+1 = 77 799 and so
5 p[69]+1 = 71 800/77 799 = 0.9229. (d) In this case, the required probability is
5 p[70] = l[70]+5 /l[70] = l75 /l[70] = 71 800/l[70] . Proceeding as in parts (b) and (c), we have
l[70] p[70] p[70]+1 p[70]+2 = l[70]+3 = l73 = 75 666,
giving
l[70] = 75 666/(0.997 × 0.998 × 0.999) = 76 122.
Hence
5 p[70] = 71 800/76 122 = 0.9432. 106 CHAPTER 3. LIFE TABLES AND SELECTION Example 3.12 CMI (Table A5) is based on UK data from 19992002 for male non–
smokers who are whole life or endowment insurance policyholders. It has a select period
of two years. An extract from this table, showing values of q[x−t]+t , is given in Table 3.6.
Use this survival model to calculate the following probabilities:
(a) 4 p[70] ,
(b) 3 q[60]+1 , and
(c) 2 q73 . Age, x
60
61
62
63
...
70
71
72
73
74
75 Duration 0
q[x]
0.003469
0.003856
0.004291
0.004779
...
0.010519
0.011858
0.013401
0.015184
0.017253
0.019664 Duration 1
q[x−1]+1
0.004539
0.005059
0.005644
0.006304
...
0.014068
0.015868
0.017931
0.020302
0.023034
0.026196 Duration 2+
qx
0.004760
0.005351
0.006021
0.006781
...
0.015786
0.017832
0.020145
0.022759
0.025712
0.029048 Table 3.6: CMI (Table A5) extract: mortality rates for male non–smokers who have whole
life or endowment policies. Solution 3.12 Note that CMI (Table A5) gives values of q[x−t]+t for t = 0 and t = 1
and also for t ≥ 2. Since the select period is 2 years q[x−t]+t ≡ qx for t ≥ 2. Note also 3.9. SELECT LIFE TABLES 107 that each row of the table relates to a man currently age x, where x is given in the ﬁrst
column. Select life tables, tabulated at integer ages, can be set out in diﬀerent ways –
for example, each row could relate to a ﬁxed age at selection – so care needs to be taken
when using such tables.
(a) We calculate 4 p[70] as
4 p[70] = p[70] p[70]+1 p[70]+2 p[70]+3
= p[70] p[70]+1 p72 p73
= (1 − q[70] ) (1 − q[70]+1 ) (1 − q72 ) (1 − q73 )
= 0.989481 × 0.984132 × 0.979855 × 0.977241
= 0.932447. (b) We calculate 3 q[60]+1 as
3 q[60]+1 = q[60]+1 + p[60]+1 q62 + p[60]+1 p62 q63
= q[60]+1 + (1 − q[60]+1 ) q62 + (1 − q[60]+1 ) (1 − q62 ) q63
= 0.005059 + 0.994941 × 0.006021 + 0.994941 × 0.993979 × 0.006781
= 0.017756. (c) We calculate 2 q73 as
2 q73 = 2 p73 q75 = (1 − q73 ) (1 − q74 ) q75 108 CHAPTER 3. LIFE TABLES AND SELECTION
= 0.977241 × 0.974288 × 0.029048
= 0.027657. Example 3.13 A select survival model has a 2 year select period and is speciﬁed as
follows. The ultimate part of the model follows Makeham’s Law, so that
µx = A + Bcx
where A = 0.00022, B = 2.7 × 10−6 and c = 1.124. The select part of the model is such
that for 0 ≤ s ≤ 2,
µ[x]+s = 0.92−s µx+s .
Starting with l20 = 100 000, calculate values of
(a) lx for x = 21, 22, . . . , 82,
(b) l[x]+1 for x = 20, 21, . . . , 80, and,
(c) l[x] for x = 20, 21, . . . , 80.
Solution 3.13 First, note that
t px = exp −At − B xt
c (c − 1)
log c and for 0 ≤ t ≤ 2,
t
t p[x] = exp − µ[x]+s ds
0 = exp 0.92−t 1 − 0.9t
ct − 0.9t
A+
Bcx
log(0.9)
log(0.9/c) . (3.15) 3.10. NOTES AND FURTHER READING 109 (a) Values of lx can be calculated recursively from
lx = px−1 lx−1 for x = 21, 22, . . . , 82. (b) Values of l[x]+1 can be calculated from
l[x]+1 = lx+2 /p[x]+1 for x = 20, 21, . . . , 80. (c) Values of l[x] can be calculated from
l[x] = lx+2 /2 p[x] for x = 20, 21, . . . , 80. The values are shown in Table 3.7. 3.10 Notes and further reading The mortality rates in Section 3.4 are drawn from the following sources:
• Australian Life Tables 20002002 were produced by the Australian Government
Actuary (2004).
• English Life Table 15 was prepared by the UK Government Actuary and published
by the Oﬃce for National Statistics (1997).
• US Life Tables 2002 were prepared in the Division of Vital Statistics of the National
Center for Health Statistics in the US – see Arias (2004).
The Continuous Mortality Investigation in the UK has been ongoing for many years.
Findings on mortality and morbidity experience of UK policyholders are published via a
series of formal reports and working papers. In this chapter we have drawn on CMI (2006). 110
x 20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49 CHAPTER 3. LIFE TABLES AND SELECTION
l [x] 99 995.08
99 970.04
99 944.63
99 918.81
99 892.52
99 865.69
99 838.28
99 810.20
99 781.36
99 751.69
99 721.06
99 689.36
99 656.47
99 622.23
99 586.47
99 549.01
99 509.64
99 468.12
99 424.18
99 377.52
99 327.82
99 274.69
99 217.72
99 156.42
99 090.27
99 018.67
98 940.96
98 856.38
98 764.09
98 663.15 l[x]+1 lx+2 x+2 x l [x] l[x]+1 lx+2 x+2 99 973.75
99 948.40
99 922.65
99 896.43
99 869.70
99 842.38
99 814.41
99 785.70
99 756.17
99 725.70
99 694.18
99 661.48
99 627.47
99 591.96
99 554.78
99 515.73
99 474.56
99 431.02
99 384.82
99 335.62
99 283.06
99 226.72
99 166.14
99 100.80
99 030.10
98 953.40
98 869.96
98 778.94
98 679.44
98 570.40 100 000.00
99 975.04
99 949.71
99 923.98
99 897.79
99 871.08
99 843.80
99 815.86
99 787.20
99 757.71
99 727.29
99 695.83
99 663.20
99 629.26
99 593.83
99 556.75
99 517.80
99 476.75
99 433.34
99 387.29
99 338.26
99 285.88
99 229.76
99 169.41
99 104.33
99 033.94
98 957.57
98 874.50
98 783.91
98 684.88
98 576.37
98 457.24 20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51 50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80 98 552.51
98 430.98
98 297.24
98 149.81
97 987.03
97 807.07
97 607.84
97 387.05
97 142.13
96 870.22
96 568.13
96 232.34
95 858.91
95 443.51
94 981.34
94 467.11
93 895.00
93 258.63
92 551.02
91 764.58
90 891.07
89 921.62
88 846.72
87 656.25
86 339.55
84 885.49
83 282.61
81 519.30
79 584.04
77 465.70
75 153.97 98 450.67
98 318.95
98 173.79
98 013.56
97 836.44
97 640.40
97 423.18
97 182.25
96 914.80
96 617.70
96 287.48
95 920.27
95 511.80
95 057.36
94 551.72
93 989.16
93 363.38
92 667.50
91 894.03
91 034.84
90 081.15
89 023.56
87 852.03
86 555.99
85 124.37
83 545.75
81 808.54
79 901.17
77 812.44
75 531.88
73 050.22 98 326.19
98 181.77
98 022.38
97 846.20
97 651.21
97 435.17
97 195.56
96 929.59
96 634.14
96 305.75
95 940.60
95 534.43
95 082.53
94 579.73
94 020.33
93 398.05
92 706.06
91 936.88
91 082.43
90 133.96
89 082.09
87 916.84
86 627.64
85 203.46
83 632.89
81 904.34
80 006.23
77 927.35
75 657.16
73 186.31
70 507.19 52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82 Table 3.7: Select life table with a 2 year select period. 3.10. NOTES AND FURTHER READING 111 In Section 3.5 we noted that there can be considerable variability in the mortality experience of diﬀerent groups in a national population. Coleman and Salt (1992) give a very
good account of this variability in the UK population.
The paper by Gompertz (1825), who was the Actuary of the Alliance Insurance Company
of London, introduced the force of mortality concept. 112 CHAPTER 3. LIFE TABLES AND SELECTION 3.11 Exercises Exercise 3.1 Sketch the following as functions of age x for a typical (human) population,
and comment on the major features.
(a) µx ,
(b) lx , and
(c) dx . Exercise 3.2 You are given the following life table extract.
Age, x
52
53
54
55
56
57
58
59
60 lx
89 948
89 089
88 176
87 208
86 181
85 093
83 940
82 719
81 429 Calculate
(a) 0.2 q52.4 assuming UDD (fractional age assumption), (b) 0.2 q52.4 assuming constant force of mortality (fractional age assumption), 3.11. EXERCISES
(c) 5.7 p52.4 assuming UDD, (d) 5.7 p52.4 assuming constant force of mortality, (e) 3.2 2.5 q52.4 assuming UDD, and (f) 3.2 2.5 q52.4 113 assuming constant force of mortality. Exercise 3.3 Table 3.8 is an extract from a (hypothetical) select life table with a select
period of 2 years. Note carefully the layout – each row relates to a ﬁxed age at selection.
Use this table to calculate
(a) the probability that a life currently aged 75 who has just been selected will survive
to age 85,
(b) the probability that a life currently aged 76 who was selected one year ago will die
between ages 85 and 87, and
(c) 4 2 q[77]+1 . Exercise 3.4 CMI (Table A23) is based on UK data from 19992002 for female nonsmokers who are term insurance policyholders. It has a select period of 5 years. An
extract from this table, showing values of q[x−t]+t , is given in Table 3.9.
Use this survival model to calculate
(a) 2 p[72] ,
(b) 3 q[73]+2 , 114 CHAPTER 3. LIFE TABLES AND SELECTION
x l [x] l[x]+1 lx+2 x+2 75
76
77
.
.
. 15 930
15 508
15 050
.
.
. 15 668
15 224
14 744
.
.
. 15 286
14 816
14 310
.
.
. 77
78
79
.
.
. 12 576
11 928
11 250
10 542
9 812
9 064 82
83
84
85
86
87 80
81
82
83
84
85 Table 3.8: Extract from a (hypothetical) select life table. Age, x
69
70
71
72
73
74
75
76
77 Duration 0
q [ x]
0.003974
0.004285
0.004704
0.005236
0.005870
0.006582
0.007381
0.008277
0.009281 Duration 1
q[x−1]+1
0.004979
0.005411
0.005967
0.006651
0.007456
0.008361
0.009376
0.010514
0.011790 Duration 2
q[x−2]+2
0.005984
0.006537
0.007229
0.008066
0.009043
0.010140
0.011370
0.012751
0.014299 Duration 3
q[x−3]+3
0.006989
0.007663
0.008491
0.009481
0.010629
0.011919
0.013365
0.014988
0.016807 Duration 4
q[x−4]+4
0.007994
0.008790
0.009754
0.010896
0.012216
0.013698
0.015360
0.017225
0.019316 Duration 5+
qx
0.009458
0.010599
0.011880
0.013318
0.014931
0.016742
0.018774
0.021053
0.023609 Table 3.9: Mortality rates for female non–smokers who have term insurance policies. 3.11. EXERCISES 115 (c) 1 q[65]+4 , and
(d) 7 p[70] . Exercise 3.5 CMI (Table A21) is based on UK data from 19992002 for female smokers
who are term insurance policyholders. It has a select period of 5 years. An extract from
this table, showing values of q[x−t]+t , is given in Table 3.10.
Age x
70
71
72
73
74
75
76
77 Duration 0
q [x]
0.010373
0.011298
0.012458
0.013818
0.015308
0.016937
0.018714
0.020649 Duration 1
q[x−1]+1
0.013099
0.014330
0.015825
0.017553
0.019446
0.021514
0.023772
0.026230 Duration 2
q[x−2]+2
0.015826
0.017362
0.019192
0.021288
0.023584
0.026092
0.028830
0.031812 Duration 3
q[x−3]+3
0.018552
0.020393
0.022559
0.025023
0.027721
0.030670
0.033888
0.037393 Duration 4
q[x−4]+4
0.021279
0.023425
0.025926
0.028758
0.031859
0.035248
0.038946
0.042974 Duration 5+
qx
0.026019
0.028932
0.032133
0.035643
0.039486
0.043686
0.048270
0.053262 Table 3.10: Mortality rates for female smokers who have term insurance policies.
Calculate
(a) 7 p[70] ,
(b) 1 2 q[70]+2 , and
(c) 3.8 q[70]+0.2 assuming UDD. 116 CHAPTER 3. LIFE TABLES AND SELECTION Exercise 3.6 A select survival model has a select period of 3 years. Calculate 3 p53 , given
that
q[50] = 0.01601, 2 p[50] = 0.96411, 2 q[50] = 0.02410, 2 3 q[50]+1 = 0.09272. Exercise 3.7 When posted overseas to country A at age x, the employees of a large
company are subject to a force of mortality such that, at exact duration t years after
arrival overseas (t = 0, 1, 2, 3, 4),
q[A]+t = (6 − t)qx+t
x
where qx+t is on the basis of US Life Tables, 2002, Females. For those who have lived in
country A for at least 5 years the force of mortality at each age is 50% greater than that
of US Life Tables, 2002, Females, at the same age. Some lx values for this table are shown
in Table 3.11.
Age, x
30
31
32
33
34
35
...
40 lx
98 424
98 362
98 296
98 225
98 148
98 064
...
97 500 Table 3.11: An extract from the United States Life Tables, 2002, Females. Calculate the probability that an employee posted to country A at age 30 will survive to
age 40 if she remains in that country. 3.11. EXERCISES 117 Exercise 3.8 A special survival model has a select period of 3 years. Functions for this
model are denoted by an asterisk, ∗ . Functions without an asterisk are taken from the
Canada Life Tables 200002, Males. You are given that, for all values of x,
p∗x] = 4 px−5
[ ; , p∗x]+1 = 3 px−1
[ ; , p∗x]+2 = 2 px+2
[ ;, and p∗ = px+1 .
x A life table, tabulated at integer ages, is constructed on the basis of the special survival
∗
model and the value of l25 is taken as 98 363 (i.e. l26 for Canada Life Tables 200002,
Males). Some lx values for this table are shown in Table 3.12.
Age, x
15
16
17
18
19
20
21
22
23
24
25
26
.
.
.
62
63
64
65 lx
99 180
99 135
99 079
99 014
98 942
98 866
98 785
98 700
98 615
98 529
98 444
98 363
.
.
.
87
86
85
84 503
455
313
074 Table 3.12: An extract from the Canada Life Tables 200002, Males. 118 CHAPTER 3. LIFE TABLES AND SELECTION ∗
(a) Construct the l[∗x] , l[∗x]+1 , l[∗x]+2 , and lx+3 columns for x = 20, 21, 22.
∗
(b) Calculate 2 38 q[21]+1 , ∗
40 p[22] , ∗
40 p[21]+1 , ∗
40 p[20]+2 , and ∗
40 p22 . Exercise 3.9 (a) Show that a constant force of mortality between integer ages implies
that the distribution of Rx , the fractional part of the future life time, conditional on
Kx = k , has the following truncated exponential function for integer x, for 0 ≤ s < 1
and for k = 0, 1, . . .
Pr[Rx ≤ s  Kx = k ] = 1 − exp{−µ∗ +k s}
x
1 − exp{−µ∗ +k }
x (3.16) where µ∗ +k = − log px+k .
x
(b) Show that if formula (3.16) holds for k = 0, 1, 2, . . . then the force of mortality is
constant between integer ages. Exercise 3.10 Verify formula (3.15). 3.11. EXERCISES Answers to selected exercises
3.2 (a) 0.001917
(b) 0.001917
(c) 0.935422
(d) 0.935423
(e) 0.030957
(f) 0.030950
3.3 (a) 0.66177
(b) 0.09433
(c) 0.08993
3.4 (a) 0.987347
(b) 0.044998
(c) 0.010514
(d) 0.920271
3.5 (a) 0.821929
(b) 0.055008
(c) 0.065276
3.6 0.90294
3.7 0.977497 119 120 CHAPTER 3. LIFE TABLES AND SELECTION 3.8 (a) The values are as follows:
x l[∗x] l[∗x]+1 l[∗x]+2 lx+3 20
21
22 99 180
99 135
99 079 98 942
98 866
98 785 98 700
98 615
98 529 98 529
98 444
98 363 (b) 0.121265, 0.872587, 0.874466, 0.875937, 0.876692. Chapter 4
Insurance Beneﬁts
4.1 Summary In this chapter we develop formulae for the valuation of traditional insurance beneﬁts.
In particular, we consider whole life, term and endowment insurance. For each of these
beneﬁts we identify the random variables representing the present values of the beneﬁts
and we derive expressions for moments of these random variables. The functions we
develop for traditional beneﬁts will also be useful when we move to modern variable
contracts.
We develop valuation functions for beneﬁts based on the continuous future lifetime random
variable, Tx , and the curtate future lifetime random variable, Kx from Chapter 2. We
(m)
introduce a new random variable, Kx , which we use to value beneﬁts which depend
on the number of complete periods of length 1/m years lived by a life (x). We explore
relationships between the present values of payment streams with diﬀerent frequencies.
We also introduce the actuarial notation for the expected values of the present value of
beneﬁts.
121 122 4.2 CHAPTER 4. INSURANCE BENEFITS Introduction In the previous two chapters, we have looked at models for future lifetime. The main
reason that we need these models is to apply them to the valuation of payments which
are dependent on the death or survival of a policyholder or pension plan member. Because
of the dependence on death or survival, the timing and possibly the amount of the beneﬁt
are uncertain, so the present value of the beneﬁt can be modelled as a random variable.
In this chapter we combine survival models with time value of money functions to derive
the distribution of the present value of an uncertain, life contingent future beneﬁt.
We generally assume in this chapter (and in the following three chapters) that the interest
rate is constant and ﬁxed. This is appropriate, for example, if the premiums for an
insurance policy are invested in risk free bonds, all yielding the same interest rate, so that
the term structure is ﬂat. In Chapter 10 we introduce more realistic term structures, and
consider some models of interest that allow for uncertainty.
For the development of present value functions, it is generally easier, mathematically, to
work in continuous time. In the case of a death beneﬁt, working in continuous time means
that we assume that the death payment is paid at the exact time of death. In the case
of an annuity, a continuous beneﬁt of, say, $1 per year would be paid in inﬁnitesimal
units of $dt in every interval (t, t + dt). Clearly both assumptions are impractical; it will
take time to process a payment after death, and annuities will be paid at most weekly,
not every moment (though the valuation of weekly payments is usually treated as if the
payments were continuous, as the diﬀerence is very small). In practice, insurers and
pension plan actuaries work in discrete time, often with cash ﬂow projections that are,
perhaps, monthly or quarterly. In addition, when the survival model being used is in
the form of a life table with annual increments (that is, lx for integer x), it is simplest
to use annuity and insurance present value functions that assume payments are made
at integer durations only. We work in continuous time in the ﬁrst place because the
mathematical development is more transparent, more complete and more ﬂexible. It is
then straightforward to adapt the results from continuous time analysis to discrete time 4.3. ASSUMPTIONS 123 problems. 4.3 Assumptions To perform calculations in this chapter, we require assumptions about mortality and
interest. We use the term basis to denote a set of assumptions used in life insurance or
pension calculations, and we will meet further examples of bases when we discuss premium
calculation in Chapter 6, policy values in Chapter 7 and pension liability valuation in
Chapter 9.
Throughout this chapter we illustrate the results with examples using the same mortality
model, which we call the Standard Ultimate Survival Model: Mortality Makeham’s law with A = 0.00022
B = 2.7 × 10−6
c = 1.124 We call this an ultimate basis to diﬀerentiate it from the standard select basis that we
will use in Chapter 6. This model is the ultimate part of the model used in Example 3.13.
As discussed above, this interest assumption is equivalent to assuming a ﬂat yield curve,
and can be criticized as unrealistic. However, it is a convenient assumption from a pedagogical point of view, is often accurate enough for practical purposes (but not always)
and we relax the assumption in later chapters.
It is also convenient to work with other interest theory functions that are in common
actuarial and ﬁnancial use. We review some of these here.
We use
v= 1
1+i 124 CHAPTER 4. INSURANCE BENEFITS as a shorthand for discounting. The present value of a payment of S which is to be paid
in t years’ time is Sv t . The force of interest per year is denoted δ where
δ = log(1 + i), 1 + i = eδ , and v = e−δ ; δ is also known as the continuously compounded rate of interest. In ﬁnancial mathematics
and corporate ﬁnance contexts, and in particular if the rate of interest is assumed risk
free, the common notation for the continuously compounded rate of interest is r.
The nominal rate of interest compounded p times per year is denoted i(p) where
i(p) = p (1 + i)1/p − 1 ⇔ 1 + i = 1 + i(p) /p p . The eﬀective rate of discount per year is d where
d = 1 − v = iv = 1 − e−δ ,
and the nominal rate of discount compounded p times per year is d(p) where
d(p) = p 1 − v 1/p ⇔ (1 − d(p) /p)p = v. 4.4
4.4.1 Valuation of insurance beneﬁts
¯
Whole life insurance: the continuous case, Ax For a whole life insurance policy, the time at which the beneﬁt will be paid is unknown
until the policyholder actually dies and the policy becomes a claim. Since the present
value of a future payment depends on the payment date, the present value of the beneﬁt
payment is a function of the time of death, and is therefore modeled as a random variable.
Given a survival model and an interest rate we can derive the distribution of the present
value random variable for a life contingent beneﬁt, and can therefore compute quantities
such as the mean and variance of the present value. 4.4. VALUATION OF INSURANCE BENEFITS 125 We start by considering the value of a beneﬁt of amount $1 payable following the death of
a life currently aged x. Using a beneﬁt of $1 allows us to develop valuation functions per
unit of sum insured, then we can multiply these by the actual sum insured for diﬀerent
beneﬁt amounts.
We ﬁrst assume that the beneﬁt is payable immediately on the death of (x). This is known
as the continuous case since we work with the continuous future lifetime random variable
Tx . Although in practice there would normally be a short delay between the date of a
person’s death and the time at which an insurance company would actually pay a death
beneﬁt (due to notiﬁcation of death to the insurance company and legal formalities) the
eﬀect is slight and we will ignore that delay here.
For our life (x), the present value of a beneﬁt of $1 payable immediately on death is a
random variable, Z , say, where
Z = v Tx = e−δ Tx .
We are generally most interested in the expected value of the present value random
variable for some future payment. We refer to this as the Expected Present Value or
EPV. It is also commonly referred to as the Actuarial Value.
The EPV of the whole life insurance beneﬁt payment with sum insured $1 is E[e−δ Tx ]. In
¯
actuarial notation, we denote this expected value by Ax , where the bar above A denotes
that the beneﬁt is payable immediately on death.
As Tx has probability density function fx (t) = t px µx+t , we have
∞ ¯
Ax = E[e−δ Tx ] = e−δ t t px µx+t dt. (4.1) 0 It is worth looking at the intuition behind this formula. We use the timeline format that
was introduced in Section 2.4 in Figure 4.1.
Consider time s, where x ≤ x + s < ω . The probability that (x) is alive at time s is s px ,
and the probability that (x) dies between ages x + s and x + s + ds, having survived to 126 CHAPTER 4. INSURANCE BENEFITS age x + s, is, loosely, µx+s ds, provided that ds is very small. In this case, the present
value of the death beneﬁt of $1 is e−δs .
Now we can integrate (that is, sum the inﬁnitesimal components of) the product of present
value and probability over all the possible death intervals s to s + ds to obtain the EPV
of the death beneﬁt that will be paid in exactly one of these intervals. Time 0 Age (x) survives s years x s (x) s+ds
dies x+s x+s+ds Probability
s px ω µx+s ds
e−δs Present value Figure 4.1: Timeline diagram for continuous whole life insurance. Similarly, the second moment (about zero) of the present value of the death beneﬁt is
E[Z 2 ] = E[(e−δ Tx )2 ] = E[e−2δ Tx ]
∞ = e−2δt t px µx+t dt 0 = 2 ¯
Ax (4.2) where the superscript 2 indicates that calculation is at force of interest 2δ , or, equivalently,
at rate of interest j , where 1 + j = e2δ = (1 + i)2 . 4.4. VALUATION OF INSURANCE BENEFITS 127 The variance of the present value of a unit beneﬁt payable immediately on death is
¯
¯
V[Z ] = V[e−δ Tx ] = E [Z 2 ] − E[Z ]2 = 2 Ax − Ax 2 . (4.3) Now, if we introduce a more general sum insured, S , say, then the EPV of the death
beneﬁt is
¯
E[SZ ] = E[Se−δ Tx ] = S Ax
and the variance is
V[SZ ] = V[S e−δ Tx ] = S 2 2 ¯
¯
Ax − A2 .
x In fact we can calculate any probabilities associated with the random variable Z from the
probabilities associated with Tx . Suppose we are interested in the probability Pr[Z ≤ 0.5],
for example. We can rearrange this into a probability for Tx :
Pr[Z ≤ 0.5] = Pr[e−δ Tx ≤ 0.5]
= Pr[−δ Tx ≤ log(0.5)]
= Pr[δ Tx > − log(0.5)]
= Pr[δ Tx > log(2)]
= Pr[Tx > log(2)/δ ]
= u px where u = log(2)/δ . We note that low values of Z are associated with high values of Tx .
This makes sense because the beneﬁt is more expensive to the insurer if it is paid early,
as there has been little opportunity to earn interest. It is less expensive if it is paid later. 128 CHAPTER 4. INSURANCE BENEFITS 4.4.2 Whole life insurance: the annual case, Ax Suppose now that the beneﬁt of $1 is payable at the end of the year of death of (x), rather
than immediately on death. To value this we use the curtate future lifetime random
variable, Kx , introduced in Chapter 2. Recall that Kx measures the number of complete
years of future life of (x). The time to the end of the year of death of (x) is then Kx + 1.
For example, if (x) lived for 25.6 years from the issue of the insurance policy, the observed
value of Kx would be 25, and the death beneﬁt payable at the end of the year of death
would be payable 26 years from the policy’s issue.
We again use Z to denote the present value of the whole life insurance beneﬁt of $1, so
that Z is the random variable
Z = v Kx +1 .
The EPV of the beneﬁt, E[Z ], is denoted by Ax in actuarial notation.
In Chapter 2 we derived the probability function for Kx , Pr[Kx = k ] =
of the beneﬁt is Ax = E[v Kx +1
= ∞
k=0 v k+1 k qx = vqx + v 2 1 qx + v 3 2 qx + . . . . k qx , so the EPV (4.4) Each term on the right hand side of this equation represents the EPV of a death beneﬁt
of $1, payable at time k conditional on the death of (x) in (k − 1, k ].
In fact, we can always express the EPV of a life–contingent beneﬁt by considering each
time point at which the beneﬁt can be paid, and summing over all possible payment times
the product of
(1) the amount of the beneﬁt,
(2) the appropriate discount factor and 4.4. VALUATION OF INSURANCE BENEFITS 129 (3) the probability that the beneﬁt will be paid at that time.
We will justify this more rigorously in Section 4.6. Here, we will illustrate the process for
the whole life insurance example, in Figure 4.2. Time 0 1 2 3
...... Amount $1 $1 $1 Discount v v2 v3 Probability qx 1 qx 2 qx Figure 4.2: Timeline diagram for discrete whole life insurance. The second moment of the present value is
∞ v 2(k+1) k=0 k qx ∞ = k=0 (v 2 )(k+1) k qx = (v 2 )qx + (v 2 )2 1 qx + (v 2 )3 2 qx + . . . . Just as in the continuous case, we can calculate the second moment about zero of the
present value by an adjustment in the rate of interest from i to (1 + i)2 − 1. We deﬁne
2 Ax = ∞ k=0 v 2(k+1) k qx , (4.5) and so the variance of the present value of a beneﬁt of S payable at the end of the year
of death is
S2 2 Ax − (Ax )2 . (4.6) 130 CHAPTER 4. INSURANCE BENEFITS 4.4.3 (m) Whole life insurance: the 1/mthly case, Ax In Chapter 2 we introduced the random variable Kx , representing the curtate future
lifetime of (x), and we saw in Section 4.4.2 that the present value of an insurance beneﬁt
payable at the end of the year of death can be expressed in terms of Kx .
(m) We now deﬁne the random variable Kx , where m > 1 is an integer, to be the future
1
lifetime of (x) in years rounded to the lower m th of a year. The most common values
of m are 2, 4 and 12, corresponding to half years, quarter years and months. Thus, for
(4)
example, Kx represents the future lifetime of (x), rounded down to the lower 1/4.
Symbolically, if we let denote the integer part (or ﬂoor) function, then 1
mTx .
m (
Kxm) = (4.7) For example, suppose (x) lives exactly 23.675 years. Then
(2) (4) (12) Kx = 23, Kx = 23.5, Kx = 23.5, and Kx
(m) 8
= 23 12 = 23.6667.
(m) Note that Kx is a discrete random variable. Kx
12
the interval [k, k + 1/m), for (k = 0, m , m , . . .).
(m) The probability function for Kx
12
For k = 0, m , m , . . . , = k indicates that the life (x) dies in can be derived from the associated probabilities for Tx . (
Pr[Kxm) = k ] = Pr k ≤ Tx < k + 1
m = k  1 qx =
m k px − 1
k + m px . In Figure 4.3 we show the timeline for the mthly beneﬁt. At the end of each 1/m year
period, we show the amount of beneﬁt due, conditional on the death of the insured life in
the previous 1/m year interval, the probability that the insured life dies in the relevant
interval, and the appropriate discount factor.
Suppose, for example, that m = 12. A whole life insurance beneﬁt payable at the end of
the month of death has present value random variable Z where
(12) Z = v Kx +1/12 . 4.4. VALUATION OF INSURANCE BENEFITS Time 0 1/m 2/m 131 3/m
...... Amount $1 $1 $1 Discount v 1/m v 2/m v 3/m Probability 1
m qx 1
m  1 qx 2
m m  1 qx
m Figure 4.3: Timeline diagram for mthly whole life insurance. (12) We let Ax denote the EPV of this beneﬁt, so that E[Z ] = A(12) = v 1/12
x 1
12 qx + v 2/12 1
12  1 qx + v 3/12
12 2
12  1 qx + v 4/12
12 3
12  1 qx + . . . .
12 Similarly, for any m,
A(m) = v 1/m 1 qx + v 2/m 1  1 qx + v 3/m 2  1 qx + v 4/m 3  1 qx + . . .
x
m = ∞ v m k+1
m k=0 k
m m m m  1 qx m m (4.8)
(4.9) m As for the continuous and annual cases, we can derive the variance of the present value
of the mthly whole life beneﬁt by adjusting the interest rate for the ﬁrst term in the
variance. We have
(12) E [Z 2 ] = E[v 2(Kx +1/12) (12) = E[(v 2 )Kx +1/12 = 2 A(12) ,
x 132 CHAPTER 4. INSURANCE BENEFITS so the variance is
2 4.4.4 A(12) − (A(12) )2 .
x
x Recursions In practice, it would be very unusual for an insurance policy to provide the death beneﬁt
at the end of the year of death. Nevertheless, the annual insurance function Ax is still
useful. We are often required to work with annual life tables, such as those in Chapter
3, in which case we would start by calculating the annual function Ax , then adjust for a
more appropriate frequency using the relationships and assumptions we develop later in
this chapter.
Using the annual life table in a spreadsheet, we can calculate the values of Ax using
backwards recursion. To do this, we start from the highest age in the table, ω . We
assume all lives expire by age ω , so that qω−1 = 1. If the life table does not have a limiting
age, we choose a suitably high value for ω so that qω−1 is as close to 1 as we like. This
means that any life attaining age ω − 1 may be treated as certain to die before age ω , in
which case we know that Kω−1 = 0 and so
Aω−1 = E[v Kω−1 +1 ] = v.
Now, working from the summation formula for Ax we have
Ax = ω − x− 1 v k+1 k px qx+k k=0 = v qx + v 2 px qx+1 + v 3 2 px qx+2 + . . .
= v qx + v px v qx+1 + v 2 px+1 qx+2 + v 3 2 px+1 qx+3 + . . . , 4.4. VALUATION OF INSURANCE BENEFITS 133 giving the important recursion formula
Ax = v qx + v px Ax+1 . (4.10) This formula can be used in spreadsheet format to calculate Ax backwards from Aω−1
back to Ax0 , where x0 is the minimum age in the table.
The intuition for equation (4.10) is that we separate the EPV of the whole life insurance
into the value of the beneﬁt due in the ﬁrst year, followed by the value at age x + 1 of
all subsequent beneﬁts, multiplied by px to allow for the probability of surviving to age
x + 1, and by v to discount the value back from age x + 1 to age x.
(m) We can use the same approach for mthly beneﬁts; now the recursion will give Ax in
(m)
terms of Ax+ 1 . Again, we split the beneﬁt into the part payable in the ﬁrst period – now
m
of length 1/m years, followed by the EPV of the insurance beginning after 1/m years. We
have A(m) = v 1/m 1 qx + v 2/m 1 px 1 qx+ 1 + v 2/m 2 px 1 qx+ 2 + . . .
x
m m m m m m m = v 1/m 1 qx + v 1/m 1 px v 1/m 1 qx+ 1 + v 2/m 1 px+ 1
m m m m m 1
mm q x+ 2 + . . . ,
m giving the recursion formula
(m) A(m) = v 1/m 1 qx + v 1/m 1 px Ax+ 1 .
x
m m m Example 4.1 Using the Standard Ultimate Survival Model from Section 4.3, and an
interest rate of 5% per year eﬀective, construct a spreadsheet of values of Ax for x =
20, 21, . . . , 100. Assume that A129 = v .
Solution 4.1 The survival model for the Standard Ultimate Survival Model is the ulti 134 CHAPTER 4. INSURANCE BENEFITS mate part of the model used in Example 3.13 and so values of t px can be calculated as
explained in the solution to that example. The value of q129 is 0.99996, which is indeed
close to 1. We can use the formula
Ax = vqx + vpx Ax+1
to calculate recursively A128 , A127 , . . . , A20 , starting from A129 = v . Values for x =
20, 21, . . . , 100, are given in Table 4.1. Example 4.2 Using the Standard Ultimate Survival Model from Section 4.3, and an
(12)
interest rate of 5% per year eﬀective, develop a spreadsheet of values of Ax for x starting
at age 20, in steps of 1/12.
Solution 4.2 For this example, we follow exactly the same process as for the previous
example, except that we let the ages increase by 1/12 year in each row. We construct a
column of values of 1 px using
12 1
12 px = exp −A/12 − Bcx (c1/12 − 1)/ log(c) .
(12) We again use 130 as the limiting age of the table. Then set A129 11 = v 1/12 , and for all the
other values of (12)
Ax A(12) = v 1/12
x 1
12 12 use the recursion
qx + v 1/12 1
12 (12) p x Ax+ 1 .
12 The ﬁrst and last few lines of the spreadsheet are reproduced in Table 4.2. It is worth making a remark about the calculations in Examples 4.1 and 4.2. In Example
4.1 we saw that q129 = 0.99996, which is suﬃciently close to 1 to justify us starting
our recursive calculation by setting A129 = v . In Example 4.2, our recursive calculation 4.4. VALUATION OF INSURANCE BENEFITS
x
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46 Ax
0.04922
0.05144
0.05378
0.05622
0.05879
0.06147
0.06429
0.06725
0.07034
0.07359
0.07698
0.08054
0.08427
0.08817
0.09226
0.09653
0.10101
0.10569
0.11059
0.11571
0.12106
0.12665
0.13249
0.13859
0.14496
0.15161
0.15854 x
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73 Ax
x
0.16577 74
0.17330 75
0.18114 76
0.18931 77
0.19780 78
0.20664 79
0.21582 80
0.22535 81
0.23524 82
0.24550 83
0.25613 84
0.26714 85
0.27852 86
0.29028 87
0.30243 88
0.31495 89
0.32785 90
0.34113 91
0.35477 92
0.36878 93
0.38313 94
0.39783 95
0.41285 96
0.42818 97
0.44379 98
0.45968 99
0.47580 100 135
Ax
0.49215
0.50868
0.52536
0.54217
0.55906
0.57599
0.59293
0.60984
0.62666
0.64336
0.65990
0.67622
0.69229
0.70806
0.72349
0.73853
0.75317
0.76735
0.78104
0.79423
0.80688
0.81897
0.83049
0.84143
0.85177
0.86153
0.87068 Table 4.1: Spreadsheet results for Example 4.1, for the calculation of Ax using the Standard Ultimate Survival Model. 136 CHAPTER 4. INSURANCE BENEFITS
x 1
12 px 1
12 qx 20
1
20 12
2
20 12
3
20 12
.
.
. 0.999979
0.999979
0.999979
0.999979
.
.
. 50
1
50 12
.
.
. 0.999904 0.000096 0.19357
0.999903 0.000097 0.19429
.
.
.
.
.
.
.
.
.
0.413955 0.586045 0.99427
1 129 10
12
129 11
12 0.000021
0.000021
0.000021
0.000021
.
.
. (12) Ax 0.05033
0.05051
0.05070
0.05089
.
.
. (12) Table 4.2: Spreadsheet results for Example 4.2, for the calculation of Ax
Standard Ultimate Survival Model. using the started from A129 11 = v 1/12 . If we calculate q129 11 we ﬁnd its value is 0.58960, which is
12
12
certainly not close to 1.
What is happening in these calculations is that, for Example 4.1, we are replacing the
exact calculation
A129 = v (q129 + p129 A130 )
by A129 = v , which is justiﬁable because p129 is small and A130 is close to 1, meaning
that v (q129 + p129 A130 ) is very close to v . Similarly, for Example 4.2, we replace the exact
calculation
(12) A129 11 = v 1/12
12 by (12)
A129 11
12 1
12 q129 11 +
12 1
12 (12) p129 11 A130
12 (12) = v 1/12 . As the value of A130 is very close to 1, it follows that v 1/12 1
12 q129 11 +
12 1
12 (12) p129 A130 4.4. VALUATION OF INSURANCE BENEFITS Age, x
20
40
60
80
100 Continuous (a)
Mean St. Dev.
5 043
5 954
12 404
9 619
29 743 15 897
60 764 17 685
89 341
8 127 Monthly (b)
Mean St. Dev.
5 033
5 942
12 379
9 600
29 683 15 865
60 641 17 649
89 158
8 110 137
Annual (c)
Mean St. Dev.
4 922
5 810
12 106
9 389
29 028 15 517
59 293 17 255
87 068
7 860 Table 4.3: Mean and standard deviation of the present value of a whole life insurance
beneﬁt of $100 000, for Example 4.3.
can by approximated by v 1/12 . Example 4.3 Using the Standard Ultimate Survival Model speciﬁed in Section 4.3, and
an interest rate of 5% per year eﬀective, calculate the mean and standard deviation of the
present value of a beneﬁt of $100 000 payable (a) immediately on death, (b) at the end of
the month of death, and (c) at the end of the year of death for lives aged 20, 40, 60, 80
and 100, and comment on the results.
¯
Solution 4.3 For part (a), we must calculate 100 000Ax and
100 000 2A
¯x ¯
− (Ax )2 ¯
for x = 20, 40, 60 and 80, where 2 Ax is calculated at eﬀective rate of interest j = 10.25%,
(12)
¯
and for parts (b) and (c) we replace each Ax by Ax and Ax , respectively. The values
are shown in Table 4.3. The continuous beneﬁt values in the ﬁrst column are calculated
by numerical integration, and the annual and monthly beneﬁt values are calculated using
the spreadsheets from Examples 4.1 and 4.2.
We can make the following observations about these values. First, values for the continuous beneﬁt are larger than the monthly beneﬁt, which are larger than the annual 138 CHAPTER 4. INSURANCE BENEFITS beneﬁt. This is because the death beneﬁt is payable soonest under (a) and latest under
(c). Second, as x increases the mean increases for all three cases. This occurs because the
smaller the value of x, the longer the expected time until payment of the death beneﬁt
is. Third, as x increases, the standard deviation decreases relative to the mean, in all
three cases. And further, as we get to very old ages, the standard deviation decreases in
absolute terms, as the possible range of payout dates is reduced signiﬁcantly.
It is also interesting to note that the continuous and monthly versions of the whole life
beneﬁt are very close. That is to be expected, as the diﬀerence arises from the change in
the value of money in the period between the moment of death and the end of the month
of death, a relatively short period. 4.4.5 Term insurance ¯1
The continuous case, Ax:n
Under a term insurance policy, the death beneﬁt is payable only if the policyholder dies
within a ﬁxed term of, say, n years.
In the continuous case, the beneﬁt is payable immediately on death. The present value
of a beneﬁt of $1, which we again denote by Z , is
Z= v Tx = e−δ Tx
0 if Tx ≤ n,
if Tx > n. ¯1
The EPV of this beneﬁt is denoted Ax:n in actuarial notation. The bar above A again
denotes that the beneﬁt is payable immediately on death, and the 1 above x indicates
that the life (x) must die before the term of n years expires in order for the beneﬁt to be
payable.
Then 4.4. VALUATION OF INSURANCE BENEFITS n ¯1
Ax:n = e−δt t px µx+t dt 139 (4.11) 0 and similarly, the expected value of the square of the present value is
2 ¯1
Ax:n = n e−2δt t px µx+t dt 0 which, as with the whole life case, is calculated by a change in the rate of interest used.
1
The annual case, Ax:n Next, we consider the situation when a death beneﬁt of 1 is payable at the end of the year
of death, provided this occurs within n years. The present value random variable for the
beneﬁt is now
Z= v Kx +1
0 if Kx ≤ n − 1,
if Kx ≥ n. 1
The EPV of the beneﬁt is denoted Ax:n so that 1
Ax:n = n−1
k=0 v k+1 k qx . The mthly case, A (4.12) (m)1
x:n We now consider the situation when a death beneﬁt of 1 is payable at the end of the
1/mth year of death, provided this occurs within n years. The present value random
variable for the beneﬁt is
(m) Z= v Kx
0 1
+m (m) if Kx ≤ n −
(m)
if Kx ≥ n. 1
m , 140 CHAPTER 4. INSURANCE BENEFITS
(m)1
x:n The EPV of the beneﬁt is denoted A (m)1
A x: n = mn−1 v (k+1)/m k  1 qx .
m k=0 m so that (4.13) Example 4.4 Using the Standard Ultimate Survival Model as speciﬁed in Section 4.3,
(4)1
1
¯1
with interest at 5% per year eﬀective, calculate Ax:10 , A x:10 and Ax:10 for x = 20, 40,
60 and 80 and comment on the values.
¯1
Solution 4.4 We use formula (4.11) with n = 10 to calculate Ax:10 (using numerical
(4)1
x:10 integration), and formulae (4.13) and (4.12), with m = 4 and n = 10 to calculate A
1
and Ax:10 . The values are shown in Table 4.4, and we observe that values in each case increase as x
increases, reﬂecting the fact that the probability of death in a ten year period increases
with age for the survival model we are using. The ordering of values at each age is the
same as in Example 4.3, for the same reason – the ordering reﬂects the fact that any
payment under the continuous beneﬁt will be paid earlier than a payment under the
quarterly beneﬁt. The end year beneﬁt is paid later than the quarterly beneﬁt, except
when the death occurs in the ﬁnal quarter of the year, in which case the beneﬁt is paid
at the same time. 4.4.6 Pure endowment Pure endowment beneﬁts are conditional on the survival of the policyholder. For example,
a 10 year pure endowment with sum insured $10 000, issued to (x), will pay $10 000 in 4.4. VALUATION OF INSURANCE BENEFITS
x ¯1
Ax:10 20
40
60
80 0.00214
0.00587
0.04356
0.34550 A (4)1
x:10 0.00213
0.00584
0.04329
0.34341 141
1
Ax:10 0.00209
0.00573
0.04252
0.33722 Table 4.4: EPVs of term insurance beneﬁts.
10 years if (x) is still alive at that time, and will pay nothing if (x) dies before age
x + 10. Pure endowment beneﬁts are not sold as standalone policies, but may be sold
in conjunction with term insurance beneﬁts to create the endowment insurance beneﬁt
described in the following section. However, pure endowment valuation functions turn
out to be very useful.
The pure endowment beneﬁt of $1, issued to a life aged x, with a term of n years has
present value Z , say, where: Z= 0
vn if Tx < n,
if Tx ≥ n. There are two ways to denote the EPV of the pure endowment beneﬁt using actuarial
1
notation. It may be denoted Ax:n . The ‘1’ over the term subscript indicates that the term
must expire before the life does for the beneﬁt to be paid. This notation is consistent
with the term insurance notation, but it can be cumbersome, considering that this is a
function which is used very often in actuarial calculations. A more convenient standard
actuarial notation for the EPV of the pure endowment is n Ex .
If we rewrite the deﬁnition of Z above, we have Z= 0 with probability 1 − n px ,
v n with probability n px . (4.14) 142 CHAPTER 4. INSURANCE BENEFITS Then we can see that the EPV is
1
Ax:n = v n n px . (4.15) Note that because the pure endowment will be paid only at time n, assuming the life
survives, there is no need to specify continuous and discrete time versions; there is only a
discrete time version.
We will generally use the more direct notation v n n px or n Ex for the pure endowment
1
function, rather than the Ax:n notation. 4.4.7 Endowment insurance An endowment insurance provides a combination of a term insurance and a pure endowment. The sum insured is payable on the death of (x) should (x) die within a ﬁxed term,
say n years, but if (x) survives for n years, the sum insured is payable at the end of the
nth year.
Traditional endowment insurance policies were popular in Australia, North America and
the UK up to the 1990s, but are rarely sold these days in these markets. However, as
with the pure endowment, the valuation function turns out to be quite useful in other
contexts. Also, companies operating in these territories will be managing the ongoing liabilities under the policies already written for some time to come. Furthermore, traditional
endowment insurance is still relevant and popular in some other insurance markets.
We ﬁrst consider the case when the death beneﬁt (of amount 1) is payable immediately
on death. The present value of the beneﬁt is Z , say, where
Z= v Tx = e−δTx
vn if Tx < n,
if Tx ≥ n = v min(Tx ,n) = e−δ min(Tx ,n) . 4.4. VALUATION OF INSURANCE BENEFITS 143 Thus, the EPV of the beneﬁt is
n E [Z ] = −δ t e t px µx+t dt + e−δn t px µx+t dt n 0
n = ∞ e−δt t px µx+t dt + e−δn n px 0
1
¯1
= Ax:n + Ax:n and in actuarial notation we write
1
¯
¯
Ax:n = A1 :n + Ax:n .
x (4.16) Similarly, the expected value of the squared present value of the beneﬁt is
n e−2δt t px µx+t dt + e−2δn n px 0 ¯
which we denote 2 Ax:n .
In the situation when the death beneﬁt is payable at the end of the year of death, the
present value of the beneﬁt is
Z= v Kx +1
vn if Kx ≤ n − 1,
if Kx ≥ n = v min(Kx +1,n) .
The EPV of the beneﬁt is then
n−1
k=0 1
v k+1 k qx + v n P [Kx ≥ n] = Ax:n + v n n px , (4.17) 144 CHAPTER 4. INSURANCE BENEFITS and in actuarial notation we write
1
1
Ax:n = Ax:n + Ax:n . (4.18) Similarly, the expected value of the squared present value of the beneﬁt is
2 Ax:n = n−1
k=0 v 2(k+1) k qx + v 2n n px . Finally, when the death beneﬁt is payable at the end of the 1/mth year of death, the
present value of the beneﬁt is
(m) Z= v Kx
vn 1
+m (m) = v min(Kx 1
+ m ,n) (m) if Kx ≤ n −
(m)
if Kx ≥ n 1
m , . The EPV of the beneﬁt is
mn−1 v (k+1)/m k=0 k
m (m)1
x:n (
 1 qx + v n P [Kxm) ≥ n] = A
m + v n n px , and in actuarial notation we write
(m) (m)1
x:n Ax:n = A 1
+ Ax:n . (4.19) Example 4.5 Using the Standard Ultimate Survival Model as speciﬁed in Section 4.3,
(4)
¯
with interest at 5% per year eﬀective, calculate Ax:10 , Ax:10 and Ax:10 for x = 20, 40, 60
and 80 and comment on the values. 4.4. VALUATION OF INSURANCE BENEFITS 145 x ¯
Ax:10 Ax:10 (4) Ax:10 20
40
60
80 0.61438
0.61508
0.62220
0.68502 0.61437
0.61504
0.62194
0.68292 0.61433
0.61494
0.62116
0.67674 Table 4.5: EPVs of endowment insurance beneﬁts.
1
¯
Solution 4.5 We can obtain values of Ax:10 and Ax:10 by adding Ax:10 = v 10 10 px to the
1
¯1
values of Ax:10 and Ax:10 in Example 4.4. The values are shown in Table 4.5. The actuarial values of the 10 year endowment insurance functions do not vary greatly
with x, unlike the values of the 10 year term insurance functions. The reason for this is
that the probability of surviving 10 years is large (10 p20 = 0.9973, 10 p60 = 0.9425) and so
for each value of x, the beneﬁt is payable after 10 years with a high probability. Note that
v 10 = 0.6139, and as time 10 years is the latest possible payment date for the beneﬁt, the
(4)
¯
values of Ax:10 , Ax:10 and Ax:10 must be greater than this for any age x. 4.4.8 Deferred insurance beneﬁts Deferred insurance refers to insurance which does not begin to oﬀer death beneﬁt cover
until the end of a deferred period. Suppose a beneﬁt of $1 is payable immediately on the
death of (x) provided that (x) dies between ages x + u and x + u + n. The present value
random variable is
Z= 0
if Tx < u or Tx ≥ u + n,
−δ Tx
e
if u ≤ Tx < u + n. This random variable describes the present value of a deferred term insurance. We can,
similarly, develop random variables to value deferred whole life or endowment insurance. 146 CHAPTER 4. INSURANCE BENEFITS ¯1
The actuarial notation for the EPV of the deferred term insurance beneﬁt is u Ax:n . Thus
u+ n ¯1
u Ax:n = e−δt t px µx+t dt. (4.20) u Changing the integration variable to s = t − u gives
n ¯1
u Ax:n = e−δ(s+u) s+u px µx+s+u ds 0
n = e−δu u px e−δs s px+u µx+s+u ds 0 ¯1
¯1
¯1
= e−δu u px Ax+u:n = v u u px Ax+u:n = u Ex Ax+u:n . (4.21) ¯1
A further expression for u Ax:n is
¯1
u Ax:n ¯1
¯1
= Ax:u+n − Ax:u (4.22) which follows from formula (4.20) since
u+ n
u u+ n e−δt t px µx+t dt =
0 e−δt t px µx+t dt − u e−δt t px µx+t dt. 0 Thus, the EPV of a deferred term insurance beneﬁt can be found by diﬀerencing the
EPVs of term insurance beneﬁts for terms u + n and u.
Note the role of the pure endowment term u Ex = v u u px in equation (4.21). This acts
similarly to a discount function. If the life survives u years, to the end of the deferred
¯1
period, then the EPV at that time of the term insurance is Ax+u:n . Multiplying by v u u px
converts this to the EPV at the start of the deferred period.
Our main interest in this EPV is as a building block. We observe, for example, that an n
year term insurance can be decomposed as the sum of n deferred term insurance policies, 4.4. VALUATION OF INSURANCE BENEFITS 147 each with a term of one year, and we can write
¯1
Ax:n n = e−δt t px µx+t dt 0 = n−1
r=0 = n−1
r=0 r +1 e−δt t px µx+t dt r ¯1
r Ax:1 . (4.23) A similar decomposition applies to a whole life insurance policy and we can write
¯
Ax = ∞
r =0 ¯1
r Ax:1 . We can derive similar results for the deferred beneﬁt payable at the end of the year of
1
death, with EPV denoted u Ax:n .
In particular, it is useful to note that
1
Ax = Ax:n + n Ax (4.24) so that
1
Ax:n = Ax − n Ax
= Ax − v n n px Ax+n . 1
This relationship can be used to calculate Ax:n for integer x and n given a table of values
of Ax and lx . 148 4.5 CHAPTER 4. INSURANCE BENEFITS
(m)
¯
Relating Ax, Ax and Ax We mentioned in the introduction to this chapter that, even though insurance contracts
with death beneﬁts payable at the end of the year of death are very unusual, functions
(m)
¯
like Ax are still useful. The reason for this is that we can approximate Ax or Ax from
Ax , and we might wish to do this if the only information we had was a life table, with
integer age functions only, rather than a formula for the force of mortality that could be
applied for all ages.
(4) In Table 4.6 we show values of the ratios of Ax to Ax and Ax to Ax , using the Standard
Ultimate Survival Model from Section 4.3, with interest at 5% per year eﬀective.
(4) x Ax /Ax
20
1.0184
40
1.0184
60
1.0184
80
1.0186
100 1.0198
120 1.0296 Ax /Ax
1.0246
1.0246
1.0246
1.0248
1.0261
1.0368 (4) Table 4.6: Ratios of Ax to Ax and Ax to Ax , Standard Ultimate Survival Model.
(4) We see from Table 4.6 that, over a very wide range of ages, the ratios of Ax to Ax and
Ax to Ax are remarkably stable, giving the appearance of being independent of x. In the
(m)
following section we show how we can approximate values of Ax to Ax using values of
Ax . 4.5.1 Using the uniform distribution of deaths assumption ¯
The diﬀerence between Ax and Ax depends on the lifetime distribution between ages y
and y + 1 for all y ≥ x. If we do not have information about this, for example, because (M )
¯
4.5. RELATING AX , AX AND AX 149 we only have mortality information at integer ages, we can approximate the relationship
¯
between the continuous function Ax and the discrete function Ax using the fractional
age assumptions that we introduced in Section 3.3. The most convenient fractional age
assumption for this purpose is the uniform distribution of deaths assumption, or UDD.
Recall, from Equation (3.9), that under UDD, we have for 0 < s < 1, and for integer y ,
s py µy +s = qy . Using this assumption
∞ ¯
Ax = e−δt t px µx+t dt 0 = ∞ k+1
k k=0 = ∞ e−δt t px µx+t dt
1 k px v k+1
0 k=0 = ∞ 1
k p x q x+ k v k+1 e(1−s)δ ds using UDD 0 k=0 = Ax e(1−s)δ s px+k µx+k+s ds eδ − 1
.
δ Because eδ = 1 + i, under the assumption of UDD we have
¯
Ax = i
Ax .
δ (4.25) This exact result under the UDD assumption gives rise to the approximation
i
¯
Ax ≈ Ax .
δ (4.26) The same approximation applies to term insurance and deferred insurance, which we can 150 CHAPTER 4. INSURANCE BENEFITS show by changing the limits of integration in the proof above.
(m) We may also want to derive an mthly death beneﬁt EPV, such as Ax , from the annual
function Ax .
Under the UDD assumption we ﬁnd that
A(m) =
x i
i(m) Ax , (4.27)
(m) and the right hand side is used as an approximation to Ax . The proof of formula (4.27)
is left as an exercise for the reader.
We stress that these approximations apply only to death beneﬁts. The endowment insurance combines the death and survival beneﬁts, so we need to split oﬀ the death beneﬁt
before applying one of the approximations. That is, under the UDD approach
i
1
¯
Ax:n ≈ δ Ax:n + v n n px . 4.5.2 (4.28) Using the claims acceleration approach The claims acceleration approach is a more heuristic way of deriving an approximate
relationship between the annual death beneﬁt EPV, Ax , and the mthly or continuous
(m)
¯
EPVs, Ax and Ax .
The only diﬀerence between these beneﬁts is the timing of the payment. Consider, for
(4)
example, Ax and Ax . Since the insured life (x) dies in the year of age x + Kx to x + Kx +1,
under the end year of death beneﬁt (valued by Ax ), the sum insured is paid at time Kx +1.
(4)
Under the end of quarteryear of death beneﬁt (valued by Ax ), the beneﬁt will be paid
either at Kx + 1/4, Kx + 2/4, Kx + 3/4 or Kx + 1 depending on the quarter year in which
the death occurred. If the deaths occur evenly over the year (the same assumption as we
use in the UDD approach), then, on average, the beneﬁt is paid at time Kx + 5/8, which
is 3/8 years earlier than the end of year of death beneﬁt. (M )
¯
4.5. RELATING AX , AX AND AX 151 Similarly, suppose the beneﬁt is paid at the end of the month of death. Assuming deaths
occur uniformly over the year, then on average the beneﬁt is paid at Kx + 13/24, which
is 11/24 years earlier than the end year of death beneﬁt.
In general, for an mthly death beneﬁt, assuming deaths are uniformly distributed over
the year of age, the average time of payment of death beneﬁt is (m + 1)/2m in the year
of death.
So we have the resulting approximation
A(m) ≈ qx v
x
= ∞
k=0 m+1
2m + 1 qx v 1+ k qx v = (1 + i) m+1
2m + 2 qx v 2+ m+1
2m + ... +1
k+ mm
2 m−1
2m ∞
k=0 k qx v k+1 . That is
(m) Ax ≈ (1 + i) m−1
2m Ax . (4.29) ¯
For the continuous beneﬁt EPV, Ax , we let m → ∞ in equation (4.29), to give the
approximation
¯
Ax ≈ (1 + i)1/2 Ax . (4.30) This is explained by the fact that, if the beneﬁt is paid immediately on death, and lives
die uniformly through the year, then, on average, the beneﬁt is paid halfway through the
year of death, which is half a year earlier than the beneﬁt valued by Ax .
As with the UDD approach, these approximations apply only to death beneﬁts. Hence, 152 CHAPTER 4. INSURANCE BENEFITS for an endowment insurance using the claims acceleration approach we have
1
¯
Ax:n ≈ (1 + i)1/2 Ax:n + v n n px . (4.31)
(m) Note that both the UDD and the claims acceleration approaches give values for Ax
(m)
or Ax such that the ratios Ax /Ax = i/i(m) and Ax /Ax = i/δ are independent of x.
Note also that for i = 5%, i/i(4) = 1.0186 and i/δ = 1.0248, whilst (1 + i)3/8 = 1.0185
and (1 + i)1/2 = 1.0247. The values in Table 4.6 show that both approaches give good
approximations in these cases. 4.6 Variable insurance beneﬁts For all the insurance beneﬁts studied in this chapter the EPV of the beneﬁt can be
expressed as the sum over all the possible payment dates of the product of three terms:
• the amount of beneﬁt paid,
• the appropriate discount factor for the payment date, and
• the probability that the beneﬁt will be paid at that payment date.
This approach works for the EPV of any traditional beneﬁt – that is, where the future
lifetime is the sole source of uncertainty. It will not generate higher moments or probability
distributions.
The approach can be justiﬁed technically using indicator random variables. Consider
a life contingent event E – for example, E is the event that a life aged x dies in the
interval (k, k + 1]. The indicator random variable is
I (E ) = 1
0 if E is true,
if E is false. 4.6. VARIABLE INSURANCE BENEFITS 153 In this example, Pr[E is True ] = k qx , so the expected value of the indicator random
variable is
E [I (E )] = 1(k qx ) + 0(1 − k qx ) = k qx ,
and in general, the expected value of an indicator random variable is the probability of
the indicator event.
Consider for example an insurance that pays $1 000 after 10 years if (x) has died by that
time, and $2 000 after 20 years if (x) dies in the second 10 year period, with no beneﬁt
otherwise.
We can write the present value random variable as
1 000 I (Tx ≤ 10)v 10 + 2 000 I (10 < Tx ≤ 20)v 20
and the EPV is then
1 000 10 qx v 10 + 2 000 10 10 qx v 20 .
Indicator random variables can also be used for continuous beneﬁts. Here we consider
indicators of the form
I (t < Tx ≤ t + dt)
for inﬁnitesimal dt, with associated probability
E [I (t < Tx ≤ t + dt)] = Pr[t < Tx ≤ t + dt]
= Pr[Tx > t]Pr[Tx < t + dtTx > t]
≈ t px µx+t dt. Consider, for example, an increasing insurance policy with a death beneﬁt of Tx payable
at the moment of death. That is, the beneﬁt is exactly equal to the number of years lived 154 CHAPTER 4. INSURANCE BENEFITS by an insured life from age x to his or her death. This is a continuous whole life insurance
under which the beneﬁt is a linearly increasing function.
To ﬁnd the EPV of this beneﬁt, we note that the payment may be made at any time, so
we consider all the inﬁnitesimal intervals (t, t + dt), and we sum over all these intervals
by integrating from t = 0 to t = ∞.
First, we identify the amount, discount factor and probability for a beneﬁt payable in the
interval (t, t + dt). The amount is t, the discount factor is e−δt . The probability that the
beneﬁt is paid in the interval (t, t + dt) is the probability that the life survives from x to
x + t, and then dies in the inﬁnitesimal interval (t, t + dt), which gives an approximate
probability of t px µx+t dt.
So, we can write the EPV of this beneﬁt as
∞ t e−δt t px µx+t dt. (4.32) 0 ¯¯
In actuarial notation we write this as (I A)x . The I here stands for ‘increasing’ and the
bar over the I denotes that the increases are continuous.
An alternative approach to deriving Equation (4.32) is to identify the present value random variable for the beneﬁt, denoted by Z , say, in terms of the future lifetime random
variable,
Z = Tx e−δTx .
Then any moment of Z can be found from
∞ E[Z k ] = t e−δt k t px µx+t dt . 0 The advantage of the ﬁrst approach is that it is very ﬂexible and generally quick, even
for very complex beneﬁts.
If the policy term ceases after a ﬁxed term of n years, the EPV of the death beneﬁt is
¯¯ 1
(I A)x:n = n
0 t e−δt t px µx+t dt. 4.6. VARIABLE INSURANCE BENEFITS 155 There are a number of other increasing or decreasing beneﬁt patterns that are fairly
common. We present several in the following examples.
Example 4.6 Consider an n year term insurance policy issued to (x) under which the
death beneﬁt is k +1 if death occurs between ages x + k and x + k +1, for k = 0, 1, 2, . . . , n −
1.
(a) Derive a formula for the EPV of the beneﬁt using the ﬁrst approach described,
that is multiplying together the amount, the discount factor and the probability of
payment, and summing for each possible payment date.
(b) Derive a formula for the variance of the present value of the beneﬁt.
Solution 4.6 (a) Suppose that the beneﬁt is payable at time k +1, for k = 0, 1, . . . , n −
1. Then if the beneﬁt is paid at time k + 1, the beneﬁt amount is $k + 1. The
discount factor is v k+1 and the probability that the beneﬁt is paid at that date is
the probability that the policyholder died in the year (k, k + 1], which is k qx , so the
EPV of the death beneﬁt is
n−1
k=0 v k+1 (k + 1) k qx . 1
In actuarial notation the above EPV is denoted (IA)x:n . If the term n is inﬁnite, so that this is a whole life version of the increasing annual
policy, with beneﬁt k + 1 following death in the year k to k + 1, the EPV of the
death beneﬁt is denoted (IA)x where
(IA)x = ∞
k=0 v k+1 (k + 1) k qx . 156 CHAPTER 4. INSURANCE BENEFITS (b) We must go back to ﬁrst principles. First, we identify the random variable as
(Kx + 1)v Kx +1
0 Z= if Kx ≤ n,
if Kx > n. So
2 E[Z ] = n−1
k=0 v 2(k+1) (k + 1)2 k qx , and the variance is
V [Z ] = n−1
k=0 1
v 2(k+1) (k + 1)2 k qx − (IA)x:n 2 . Example 4.7 A whole life insurance policy oﬀers an increasing death beneﬁt payable at
the end of the quarter year of death. If (x) dies in the ﬁrst year of the contract, then the
beneﬁt is 1, in the second year it is 2 and so on. Derive an expression for the EPV of the
death beneﬁt.
Solution 4.7 First, we note that the possible payment dates are 1/4, 2/4, 3/4, . . . .
Next, if (x) dies in the ﬁrst year, then the beneﬁt payable is 1, if death occurs in the
second year the beneﬁt payable is 2, and so on. Third, corresponding to the possible
payment dates, the discount factors are v 1/4 , v 2/4 , . . . .
The probabilities associated with the payment dates are 1 qx , 1  1 qx , 2  1 qx , 3  1 qx , . . . .
4 4 4 (4) Hence, the EPV, which is denoted (IA )x , can be calculated as
1 1
4 2 3 qx v 4 + 1  1 qx v 4 + 2  1 qx v 4 + 3  1 qx v 1
4 +2 4 1  1 qx
4 4 4 1 4 4 2 3 v 1 4 + 1 1  1 qx v 1 4 + 1 2  1 qx v 1 4 + 1 3  1 qx v 2
4 4 4 4 4 4 4 4 4 4 4.6. VARIABLE INSURANCE BENEFITS +3 2  1 qx
4 1 157 2 3 v 2 4 + 2 1  1 qx v 2 4 + 2 2  1 qx v 2 4 + 2 3  1 q x v 3
4 4 4 4 4 4 + ...
(4)1
x:1 =A (4)1
x:1 + 2 1 A (4)1
x:1 + 3 2 A + ... . We now consider the case when the amount of the death beneﬁt increases in geometric
progression. This is important in practice because compound reversionary bonuses will
increase the sum insured as a geometric progression.
Example 4.8 Consider an n year term insurance issued to (x) under which the death
beneﬁt is paid at the end of the year of death. The beneﬁt is 1 if death occurs between
ages x and x + 1, 1 + j if death occurs between ages x + 1 and x + 2, (1 + j )2 if death
occurs between ages x + 2 and x + 3, and so on. Thus, if death occurs between ages x + k
and x + k + 1, the death beneﬁt is (1 + j )k for k = 0, 1, 2, . . . , n − 1. Derive a formula for
the EPV of this death beneﬁt.
Solution 4.8 The amount of beneﬁt is 1 if the beneﬁt is paid at time 1, (1 + j ) if the
beneﬁt is paid at time 2, (1 + j )2 if the beneﬁt is paid at time 3, and so on up to time n.
The EPV of the death beneﬁt is then
v qx + (1 + j )v 2 1 qx + (1 + j )2 v 3 2 qx + . . . + (1 + j )n−1 v n n−1 qx
= n−1
k=0 v k+1 (1 + j )k k qx 1
=
1+j n−1
k=0 v k+1 (1 + j )k+1 k qx 158 CHAPTER 4. INSURANCE BENEFITS
1
= Ax:n i∗ (4.33) where
i∗ = i−j
1+i
−1=
.
1+j
1+j 1
The notation Ax:n i∗ indicates that the EPV is calculated using the rate of interest i∗ ,
rather than i. In most practical situations, i > j so that i∗ > 0. Example 4.9 Consider an insurance policy issued to (x) under which the death beneﬁt
is (1 + j )t if death occurs at age x + t where 0 < t < n, with the death beneﬁt being
payable immediately on death.
(a) Derive an expression for the EPV of the death beneﬁt if the policy is an n year term
insurance.
(b) Derive an expression for the EPV of the death beneﬁt if the policy is a whole life
insurance.
Solution 4.9 (a) The present value of the death beneﬁt is (1 + j )Tx v Tx if Tx < n, and
is zero otherwise, so that the EPV of the death beneﬁt is
n (1 + j )t v t t px µx+t dt
0 ¯1
= Ax:n i∗
where
i∗ = 1+i
− 1.
1+j 4.7. FUNCTIONS FOR SELECT LIVES 159 (b) Similarly, if the policy is a whole life insurance rather than a term insurance, then
the EPV of the death beneﬁt would be
∞
0 ¯
(1 + j )t v t t px µx+t dt = Ax i∗ where
i∗ = 4.7 1+i
− 1.
1+j Functions for select lives Throughout this chapter we have developed results in terms of lives subject to ultimate
mortality. We have taken this approach simply for ease of presentation. All of the above
development equally applies to lives subject to select mortality.
¯
For example, A[x] denotes the EPV of a beneﬁt of 1 payable immediately on the death of
1
a select life [x]. Similarly, A[x]:n denotes the EPV of a beneﬁt of 1 payable at the end of
the year of death of a select life [x] should death occur within n years, or after n years if
[x] survives that period. 4.8 Notes and further reading The Standard Ultimate Survival Model incorporates Makeham’s law as its survival model.
A feature of Makeham’s law is that we can integrate the force of mortality analytically
and hence we can evaluate, for example, t px analytically, as in Exercise 2.11. This in turn
means that the EPV of an insurance beneﬁt payable immediately on death, for example
¯
Ax , can be written as an integral where the integrand can be evaluated directly, as follows
∞ ¯
Ax =
0 e−δt t px µx+t dt . 160 CHAPTER 4. INSURANCE BENEFITS This integral cannot be evaluated analytically but can be evaluated numerically. In many
practical situations, the force of mortality cannot be integrated analytically, for example if
µx is a GM(r, s) function with s ≥ 2, from Section 2.7. In such cases, t px can be evaluated
¯
numerically but not analytically. Functions such as Ax can still be evaluated numerically
but, since the integrand has to be evaluated numerically, the procedure may be a little
more complicated. See Exercise 4.18 for an example. The survival model in Exercise 4.18
has been derived from data for UK wholelife and endowment insurance policyholders
(nonsmokers), 1999  2002. See CMI (2006, Table 1) . 4.9. EXERCISES 4.9 161 Exercises Exercise 4.1 You are given the following table of values for lx and Ax , assuming an
eﬀective interest rate of 6% per year.
x
35
36
37
38
39
40 lx
100 000.00
99 737.15
99 455.91
99 154.72
98 831.91
98 485.68 Ax
0.151375
0.158245
0.165386
0.172804
0.180505
0.188492 Calculate
(a) 5 E35 ,
(b) A1 5 ,
35:
(c) 5 A35 , and
¯
(d) A35:5 assuming UDD. Exercise 4.2 Assuming a uniform distribution of deaths over each year of age, show that
(m)
Ax = (i/i(m) )Ax . Exercise 4.3 A withproﬁt whole life insurance policy issued to a life aged exactly 30
has a basic sum insured of $100 000. The insurer assumes compound reversionary bonuses 162 CHAPTER 4. INSURANCE BENEFITS at the rate of 3% will vest at the end of each policy year. Using the Standard Ultimate
Survival Model, with interest at 5% per year, calculate the EPV of this beneﬁt. Exercise 4.4 (a) Show that Ax:n = n−2
k=0 v k+1 k qx + v n n−1 px . (b) Compare this formula with formula (4.17) and comment on the diﬀerences. Exercise 4.5 Show that
(m) (m) (IA(m) )x = A(m) + vpx Ax+1 + 2 px v 2 Ax+2 + . . .
x
and explain this result intuitively. Exercise 4.6 (a) Derive the following recursion formula for an n year increasing term
insurance:
(IA)1 :n = vqx + vpx (IA)x1 n−1 + Ax1 n−1
x
+1:
+1: . (b) Give an intuitive explanation of the formula in part (a).
1
(c) You are given that (IA)50 = 4.99675, A50:1 = 0.00558, A51 = 0.24905 and i = 0.06. Calculate (IA)51 . 4.9. EXERCISES 163 Exercise 4.7 You are given that Ax = 0.25, Ax+20 = 0.40, Ax:20 = 0.55 and i = 0.03.
¯
Calculate 10 000Ax:20 using
(a) claims acceleration, and
(b) UDD. Exercise 4.8 Show that
n
1
(IA)x:n = (n + 1
1)Ax:n − 1
Ax:k
k=1 and explain this result intuitively. ¯
Exercise 4.9 Show that Ax is a decreasing function of i, and explain this result by
general reasoning. Exercise 4.10 Calculate A70 given that
A50:20 = 0.42247, 1
A50:20 = 0.14996, A50 = 0.31266. Exercise 4.11 Under an endowment insurance issued to a life aged x, let X denote the
present value of a unit sum insured, payable at the moment of death or at the end of the
n year term. 164 CHAPTER 4. INSURANCE BENEFITS Under a term insurance issued to a life aged x, let Y denote the present value of a unit
sum insured, payable at the moment of death within the n year term.
Given that
V[X ] = 0.0052, v n = 0.3, n px = 0.8, E(Y ) = 0.04, calculate V(Y ). Exercise 4.12 Show that if νy = − log py for y = x, x + 1, x + 2, . . ., then under the
assumption of a constant force of interest between integer ages, ¯
Ax = ∞
t=0 v t t px νx+t (1 − vpx+t )
.
δ + νx+t Exercise 4.13 Let Z1 denote the present value of an n year term insurance beneﬁt, issued
to (x). Let Z2 denote the present value of a whole of life insurance beneﬁt, issued to the
same life.
Express the covariance of Z1 and Z2 in actuarial functions, simpliﬁed as far as possible. Exercise 4.14 You are given the following excerpt from a select life table. 4.9. EXERCISES 165 [x ]
[40]
[41]
[42]
[43]
[44] l[x]
100 000
99 802
99 597
99 365
99 120 l[x]+1
l[x]+2
l[x]+3
lx+4
x+4
99 899 99 724 99 520 99 288
44
99 689 99 502 99 283 99 033
45
99 471 99 268 99 030 98 752
46
99 225 99 007 98 747 98 435
47
98 964 98 726 98 429 98 067
48 Assuming an interest rate of 6% per year, calculate
(a) A[40]+1:4 ,
(b) the standard deviation of the present value of a 4 year term insurance, deferred 1
year, issued to a newly selected life aged 40, with sum insured $100 000, payable at
the end of the year of death, and
(c) the probability that the present value of the beneﬁt described in (b) is less than or
equal to $85 000. Exercise 4.15 (a) Describe in words the insurance beneﬁts with the present values
given below. (i) (ii) Z1 = 20 v Tx
10 v Tx 0 Z2 =
10 v Tx 10 v 15 if Tx ≤ 15,
if Tx > 15. if Tx ≤ 5,
if 5 < Tx ≤ 15,
if Tx > 15. 166 CHAPTER 4. INSURANCE BENEFITS (b) Write down in integral form the formula for the expected value for (i) Z1 and (ii)
Z2 .
(c) Derive expressions in terms of standard actuarial functions for the expected values
of Z1 and Z2 .
(d) Derive an expression in terms of standard actuarial functions for the covariance of
Z1 and Z2 . Exercise 4.16 Suppose that Makeham’s law applies with A = 0.0001, B = 0.00035 and
c = 1.075. Assume also that the eﬀective rate of interest is 6% per year.
(a) Use Excel and backward recursion in parts (i) and (ii).
(i) Construct a table of values of Ax for integer ages, starting at x = 50.
(4) (ii) Construct a table of values of Ax for x = 50, 50.25, 50.5, . . . . (Do not use
UDD for this.)
(4) (4) (iii) Hence, write down the values of A50 , A100 , A50 and A100 .
(4) (4) (b) Use your values for A50 and A100 to estimate A50 and A100 using the UDD assumption.
(c) Compare your estimated values for the A(4) functions (from (b)) with your accurate
values (from (a)). Comment on the diﬀerences. Exercise 4.17 A life insurance policy issued to a life age 50 pays $2 000 at the end of
the quarter year of death before age 65 and $1 000 at the end of the quarter year of death 4.9. EXERCISES 167 after age 65. Use the Standard Ultimate Survival Model, with interest at 5% per year, in
the following.
(a) Calculate the EPV of the beneﬁt.
(b) Calculate the standard deviation of the present value of the beneﬁt.
(c) The insurer charges a single premium of $500. Assuming that the insurer invests
all funds at exactly 5% per year eﬀective, what is the probability that the policy
beneﬁt has greater value than the accumulation of the single premium? Exercise 4.18 The force of mortality for a survival model is given by
2 µx = A + BC x Dx ,
where
A = 3.5 × 10−4 , B = 5.5 × 10−4 , C = 1.00085, D = 1.0005.
(a) Calculate t p60 for t = 0, 1/40, 2/40, . . . , 2.
Hint: Use the repeated Simpson’s Rule.
¯
(b) Calculate A1 2 using an eﬀective rate of interest of 5% per year.
60:
Hint: Use the repeated Simpson’s Rule. 168 CHAPTER 4. INSURANCE BENEFITS Answers to selected exercises
4.1 (a) 0.073594
(b) 0.137503
(c) 0.013872
(d) 0.215182
4.3 $33 569.47
4.6 (c) 5.07307
4.7 (a) 5 507.44
(b) 5 507.46
4.10 0.59704
4.11 0.01
4.14 (a) 0.79267
(b) $7 519.71
(c) 0.99825
4.16 (a) (iii) 0.33587,
(b) 0.34333, 0.87508, 0.34330, 0.89647 0.89453 4.17 (a) $218.83
(b) $239.73
(c) 0.04054
4.18 (a) Selected values are
(b) 0.007725 1/4 p60 = 0.999031, p60 = 0.996049 and 2 p60 = 0.991903 Chapter 5
Annuities
5.1 Summary In this chapter we derive expressions for the valuation and analysis of life contingent
annuities. As we did for the insurance functions in Chapter 4, we consider beneﬁt valuation
for diﬀerent payment frequencies, and we relate the valuation of annuity beneﬁts to the
valuation of the related insurance beneﬁts.
We consider how to calculate annuity valuation functions. If full survival model information is available, then the calculation can be exact for beneﬁts payable at discrete time
points, and as exact as required, using numerical integration, for beneﬁts payable continuously. Where we are calculating beneﬁts payable more frequently than annual (monthly or
weekly, say) using only an integer age life table, a very common situation in practice, then
some approximation is required. We derive several commonly used approximations, using
the UDD assumption and Woolhouse’s formula, and explore their accuracy numerically.
169 170 5.2 CHAPTER 5. ANNUITIES Introduction We use the term life annuity to refer to a series of payments to (or from) an individual
as long as the individual is alive on the payment date. The payments are normally made
at regular intervals and the most common situation is that the payments are of the same
amount. The valuation of annuities is important as annuities appear in the calculation
of premiums (see Chapter 6), policy values (see Chapter 7) and pension beneﬁts (see
Chapter 9). The present value of a life annuity is a random variable, as it depends on
the future lifetime; however, we will use some results and notation from the valuation of
annuitiescertain, where there is no uncertainty in the term, so we start with a review of
these. 5.3 Review of annuitiescertain Recall that, for integer n,
1 − vn
(5.1)
d
denotes the present value of an annuitycertain of 1 payable annually in advance for n
years.
an = 1 + v + v 2 + . . . + v n−1 =
¨ Also
an = v + v 2 + v 3 + . . . + v n = an − 1 + v n
¨
denotes the present value of an annuitycertain of 1 payable annually in arrear for n years.
Thirdly, for any n > 0,
n v t dt = an =
¯
0 1 − vn
δ (5.2) denotes the present value of an annuitycertain payable continuously at rate 1 per year
for n years. 5.4. ANNUAL LIFE ANNUITIES 171 When payments of 1 per year are made every 1/m years in advance for n years, in
instalments of 1/m, the present value is
(m) an
¨ = 1
m 2 1 1 1 + v m + v m + . . . + v n− m = 1 − vn
d(m) and for payments made in arrear
(m) an = 1
m 1 2 v m + v m + . . . + vn = 1 − vn
(m)
= an −
¨
(m)
i 1
m (1 − v n ) . (5.3) In these equations for mthly annuities, we assume that n is an integer multiple of 1/m. 5.4 Annual life annuities The annual life annuity is paid once each year, conditional on the survival of a life (the
annuitant) to the payment date. If the annuity is to be paid throughout the annuitant’s
life, it is called a whole life annuity. If there is to be a speciﬁed maximum term, it is
called a term or temporary annuity.
Annual annuities are quite rare. We would more commonly see annuities payable monthly
or even weekly. However, the annual annuity is still important in the situation where we
do not have full information about mortality between integer ages, for example because
we are working with an integer age life table. Also, the development of the valuation
functions for the annual annuity is a good starting point before considering more complex
payment patterns.
As with the insurance functions, we are primarily interested in the EPV of a cash ﬂow,
and we also identify the present value random variables in terms of the future lifetime
(m)
random variables from Chapter 2, speciﬁcally, Tx , Kx , and Kx . 172 5.4.1 CHAPTER 5. ANNUITIES Whole life annuitydue Consider ﬁrst an annuity of 1 per year payable annually in advance throughout the
lifetime of an individual now aged x. The life annuity with payments in advance is known
as a whole life annuitydue. The ﬁrst payment occurs immediately, the second in one
year from now, provided that (x) is alive then, and payments follow at annual intervals
with each payment conditional on the survival of (x) to the payment date. In Figure 5.1
we show the payments and associated probabilities and discount functions in a timeline
diagram. Time 0 1 2 3
... Amount 1 1 1 1 Discount 1 v v2 v3 px 2 px 3 px Probability 1 Figure 5.1: Timeline diagram for whole life annuitydue. We note that if (x) were to die between ages x + k and x + k + 1, for some positive
integer k , then annuity payments would be made at times 0, 1, 2, . . . , k , for a total of
k + 1 payments. We deﬁned Kx such that the death of (x) occurs between x + Kx and
x + Kx + 1, so, the number of payments is Kx + 1, including the initial payment. This
means that , for k = 0, 1, 2, . . ., the present value of the annuity is ak+1 if Kx = k . Thus,
¨
using equation (5.1), the present value random variable for the annuity payment series, 5.4. ANNUAL LIFE ANNUITIES 173 Y , say, can be written as
Y = aKx +1 =
¨ 1 − v Kx +1
.
d There are three useful ways to derive formulae for calculating the expected value of this
present value random variable.
First, the mean and variance can be found from the mean and variance of v Kx +1 , which
were derived in Section 4.4.2. For the expected value of Y , which is denoted ax , we have
¨
1 − v Kx +1
d ax = E
¨ = 1 − E[v Kx +1 ]
.
d That is,
ax =
¨ 1 − Ax
.
d (5.4) This is a useful approach, as it also immediately gives us the variance of Y as
V[Y ] = V = 1
V[v Kx +1 ]
d2
2 = 1 − v Kx +1
d Ax − A2
x
.
d2 (5.5) Secondly, we may use the indicator random variable approach from Section 4.6. The
condition for the payment at k , say, is that (x) is alive at age x + k , that is, that Tx > k .
The present value random variable can be expressed as
Y = I(Tx > 0) + v I(Tx > 1) + v 2 I(Tx > 2) + v 3 I(Tx > 3) + . . . (5.6) 174 CHAPTER 5. ANNUITIES and the EPV of the annuity is the sum of the expected values of the individual terms.
Recall that E[I(Tx > t)] = Pr[Tx > t] = t px , so that
ax = 1 + v p x + v 2 2 p x + v 3 3 p x + . . . ,
¨
that is
∞ ax =
¨ v k k px . (5.7) k=0 This is a very useful equation for ax . However, this approach does not lead to useful
¨
expressions for the variance and higher moments of Y . This is because the individual
terms in expression (5.6) are dependent random variables.
Finally, we can work from the probability function for Kx , that is using Pr[Kx = k ] = k qx ,
so that
ax =
¨ ∞
k=0 ak+1 k qx .
¨ (5.8) This is less used in practice than equations (5.4) and (5.7). The diﬀerence between the
formulations for ax in equations (5.7) and (5.8) is that in equation (5.7) the summation
¨
is taken over the possible payment dates, and in equation (5.8) the summation is taken
over the possible years of death.
Example 5.1 Show that equations (5.7) and (5.8) are equivalent – that is, show that
∞
k=0 ak+1 k qx =
¨ ∞ v k k px . k=0 Solution 5.1 We can show this by using
k vt ak+1 =
¨
t=0 5.4. ANNUAL LIFE ANNUITIES 175 and
∞
k =t k qx = ∞
k =t (k px − k+1 px ) = t px . Then
∞
k=0 ∞ ak+1 k qx =
¨ k k=0 t=0 v t k qx = qx + (1 + v ) 1 1 qx + (1 + v + v 2 ) 2 1 qx + (1 + v + v 2 + v 3 ) 3 1 qx + ...
Changing the order of summation on the right hand side (that is, collect together terms
in powers of v ) gives
∞ k
t k=0 t=0 v k qx = = ∞ t=0 k=t
∞
t=0 = ∞ ∞ vt v t k qx
∞ k =t k qx v t t px t=0 as required. 5.4.2 Term annuitydue Now suppose we wish to value a term annuitydue of 1 per year. We assume the annuity
is payable annually to a life now aged x for a maximum of n years. Thus, payments are 176 CHAPTER 5. ANNUITIES made at times k = 0, 1, 2, . . . , n − 1, provided that (x) has survived to age x + k . The
present value of this annuity is Y , say, where
Y= aKx +1
¨
an
¨ if Kx = 0, 1, . . . , n − 1,
if Kx ≥ n. That is
Y = amin(Kx +1, n) =
¨ 1 − v min(Kx +1, n)
.
d The EPV of this annuity is denoted ax:n .
¨
We have seen the random variable v min(Kx +1, n) before, in Section 4.4.7, where the EPV
Ax:n is derived. Thus, the EPV of the annuity can be determined as ax:n = E[Y ] =
¨ 1 − E[v min(Kx +1, n) ]
d that is,
ax : n =
¨ 1 − Ax:n
.
d (5.9) The timeline for the term annuitydue cash ﬂow is shown in Figure 5.2. Notice that,
because the payments are made in advance, there is no payment due at time n, the end
of the annuity term.
Using Figure 5.2, and summing the EPVs of the individual payments, we have
ax:n = 1 + v px + v 2 2 px + v 3 3 px + . . . + v n−1 n−1 px
¨
that is
ax : n =
¨ n−1
t=0 v t t px . (5.10) 5.4. ANNUAL LIFE ANNUITIES Time 0 1 177 2 3 n1 n ......
Amount 1 1 1 1 1 Discount 1 v v2 v3 v n−1 px 2 px 3 px n− 1 p x Probability 1 Figure 5.2: Timeline diagram for term life annuitydue.
Also, we can write the EPV as
ax : n =
¨ n−1
k=0 ak+1 k qx + n px an
¨
¨ adapting equation (5.8) above. The second term here arises from the second term in the
deﬁnition of Y – that is, if the annuitant survives for the full term, then the payments
constitute an n year annuity. 5.4.3 Whole life immediate annuity Now consider a whole life annuity of 1 per year payable in arrear, conditional on the
survival of (x) to the payment dates. We use the term immediate annuity to refer to an
annuity under which payments are made at the end of the time periods, rather than at
the beginning. The actuarial notation for the EPV of this annuity is ax , and the timeline
for the annuity cash ﬂow is shown in Figure 5.3.
Let Y ∗ denote the present value random variable for the whole life immediate annuity. 178 CHAPTER 5. ANNUITIES Time 0 1 2 3
... Amount 1 1 1 Discount v v2 v3 Probability px 2 px 3 px Figure 5.3: Timeline diagram for whole life immediate annuity.
Using the indicator random variable approach we have
Y ∗ = v I(Tx > 1) + v 2 I(Tx > 2) + v 3 I(Tx > 3) + v 4 I(Tx > 4) + . . . .
We can see from this expression and from the time line, that the diﬀerence in present
value between the annuitydue and the immediate annuity payable in arrear is simply the
ﬁrst payment under the annuitydue, which, under the annuitydue, is assumed to be paid
at time t = 0 with no uncertainty.
So, if Y is the random variable for the present value of the whole life annuity payable in
advance, and Y ∗ is the random variable for the present value of the whole life annuity
payable in arrear, we have Y ∗ = Y − 1, so that E[Y ∗ ] = E[Y ] − 1, and hence
ax = ax − 1.
¨ (5.11) Also, from equation (5.5) and the fact that Y ∗ = Y − 1, we have
V [Y ∗ ] = V [Y ] = 2 Ax − A2
x
.
d2 5.4. ANNUAL LIFE ANNUITIES 5.4.4 179 Term immediate annuity The EPV of term immediate annuity of 1 per year is denoted ax:n . Under this annuity
payments of 1 are made at times k = 1, 2, . . . , n, conditional on the survival of the
annuitant.
The random variable for the present value is
Y = amin(Kx ,n) ,
and the timeline for the annuity cash ﬂow is given in Figure 5.4. Time 0 1 2 3 n1 n 1 ......
Amount 1 1 1 1 Discount v v2 v3 v n−1 Probability px 2 px 3 px n− 1 p x vn
n px Figure 5.4: Timeline diagram for term life immediate annuity.
Summing the EPVs of the individual payments, we have
n
2 3 ax : n = v p x + v 2 p x + v 3 p x + . . . + v n n px v t t px . = (5.12) t=1 The diﬀerence between the annuitydue EPV, ax:n , and the immediate annuity EPV, ax:n ,
¨
is found by diﬀerencing equations (5.10) and (5.12), to give
ax:n − ax:n = 1 − v n n px
¨ 180 CHAPTER 5. ANNUITIES so that
ax : n = ax : n − 1 + v n n p x .
¨ (5.13) The diﬀerence comes from the timing of the ﬁrst payment under the annuity due and the
last payment under the immediate annuity. 5.5 Annuities payable continuously 5.5.1 Whole life continuous annuity In practice annuities are payable at discrete time intervals, but if these intervals are
close together, for example weekly, it is convenient to treat payments as being made
continuously. Consider now the case when the annuity is payable continuously at a rate
of 1 per year as long as (x) survives. If the annuity is payable weekly (and we assume 52
weeks per year), then each week, the annuity payment is 1/52. If payments were daily, for
an annuity of 1 per year, the daily payment would be 1/365. Similarly, for an inﬁnitesimal
interval (t, t + dt) the payment under the annuity is dt provided (x) is alive through the
interval.
The EPV is denoted ax . The underlying random variable is Y , say, where
¯
Y = aTx .
¯
Analogously to the annual annuitydue, we can derive formulas for the EPV of the annuity
in three diﬀerent ways.
The ﬁrst approach is to use the annuity certain formula
1 − vn
an =
¯
δ
so that
1 − v Tx
Y=
δ 5.5. ANNUITIES PAYABLE CONTINUOUSLY 181 and
ax = E[Y ] =
¯ 1 − E[v Tx ]
.
δ That is, ax =
¯ ¯
1 − Ax
.
δ (5.14) Using this formulation for the random variable Y , we can also directly derive the variance
for the continuous annuity present value from the variance for the continuous insurance
beneﬁt
2¯
¯x
1 − v Tx
Ax − A2
=
.
δ
δ2 V[Y ] = V The second approach is to use the sum (here an integral) of the product of the amount
paid in each inﬁnitesimal interval (t, t + dt), the discount factor for the interval and the
probability that the payment is made. For each such interval, the amount is dt, the
discount factor is e−δt and the probability of payment is t px , giving
∞ ax =
¯ e−δt t px dt . (5.15) 0 We remark that this EPV can also be derived using indicator random variables by expressing the present value as
∞ Y= e−δt I(Tx > t) dt . 0 The development of formula (5.15) is illustrated in Figure 5.5; we show the contribution
to the integral from the contingent annuity payment made in an inﬁnitesimal interval of 182 CHAPTER 5. ANNUITIES Time 0 t t+dt
... Amount dt
e−δt
t px Discount
Probability Figure 5.5: Timeline diagram for continuous whole life annuity. time (t, t + dt). The interval is so small that payments can be treated as being made
exactly at t.
Finally, we can directly write down the EPV from the distribution of Tx as
∞ ax =
¯
0 at t px µx+t dt.
¯ We can evaluate this using integration by parts, noting that if we diﬀerentiate equation
(5.2) we get
d
a = v t = e−δt .
¯
dt t
Then
∞ ax =
¯
0 at
¯ d
(−t px )dt
dt = − at t px ∞ −
¯
0 ∞
0 t px e−δt dt 5.5. ANNUITIES PAYABLE CONTINUOUSLY
∞ = 183 e−δt t px dt . 0
◦ When δ = 0, we see that ax is equal to ex .
¯ 5.5.2 Term continuous annuity The term continuous life annuity present value random variable
amin(Tx ,n) =
¯ 1 − v min(Tx ,n)
δ has EPV denoted by ax:n . Analogously to the term annuitydue, we have three expressions
¯
for this EPV.
Using results for endowment insurance functions from Section 4.4.7, we have
ax : n =
¯ ¯
1 − Ax:n
.
δ (5.16) Using the indicator random variable approach we have
n ax : n =
¯ e−δt t px dt , (5.17) 0 and taking the expected value of the present value random variable we obtain
n ax : n =
¯
0 at t px µx+t dt + an n px .
¯
¯ One way to understand the diﬀerence between the second and third approaches is to see
that in the second approach we integrate over the possible payment dates, and in the third
approach we integrate over the possible dates of death. The third approach is generally
the least useful in practice. 184 CHAPTER 5. ANNUITIES 5.6 Annuities payable m times per year 5.6.1 Introduction For premiums, annuities and pension beneﬁts, the annual form of the annuity would
be unusual. Premiums are more commonly payable monthly, quarterly, or sometimes
weekly. Pension beneﬁts and purchased annuities are payable with similar frequency to
salary beneﬁts, which means that weekly and monthly annuities are common.
We can deﬁne the present value of an annuity payable m times per annum in terms of
(m)
(m)
the random variable Kx , which was introduced in Section 4.4.3. Recall that Kx is the
complete future lifetime rounded down to the lower 1/mth of a year.
We will also use the formula for the present value of an 1/m thly annuity certain. For
(m)
example, an is the present value of an annuity of 1 per year, payable in m instalments of
¨
1/m each year, for n years, with the ﬁrst payment at time t = 0 and the ﬁnal payment at
(m)
1
time n − m . It is important to remember that an is an annual factor, that is, it values
¨
a payment of 1 per year, and therefore for valuing annuities for other amounts, we need
(m)
to multiply the an factor by the annual rate of annuity payment.
¨
Suppose we are interested in valuing an annuity of $12 000 per year, payable monthly in
advance to a life aged 60. Each monthly payment is $1 000. The relevant future lifetime
(12)
(12)
random variable is K60 . If K60 = 0, then (60) died in the ﬁrst month, there was a
single payment made at t = 0 of $1 000, and the present value is
1
12 000 12 = 12 000 a
¨ (12)
1/12 . (12) If K60 = 1/12 then (60) died in the second month, there are two monthly annuity
payments, each of $1 000, and the relevant annuity factor is
12 000 1
12 + 1
1 12
v
12 = 12 000 a
¨ (12)
2/12 . Continuing, we see that the present value random variable for this annuity can be written 5.6. ANNUITIES PAYABLE M TIMES PER YEAR 185 as
12 000 5.6.2 1
12 + 1
1 12
v
12 2
1 12
v
12 + + ... + (12) 1 Kx
v
12 = 12 000 a
¨ (12)
(12) K60 +1/12 . Life annuities payable m times a year Consider ﬁrst an annuity of total amount 1 per year, payable in advance m times per year
1
throughout the lifetime of (x), with each payment being m . Figure 5.6 shows the whole
life 1/mthly annuity timeline cash ﬂow. Time 0 1/m 2/m 3/m 4/m
... Amount 1/m Discount 1/m 1 v 1/m Probability 1 1
m px 1/m
v 2/m
2
m px 1/m
v 3/m
3
m px 1/m
v 4/m
4
m px Figure 5.6: Timeline diagram for whole life 1/mthly annuitydue. The present value random variable for this annuity is
(m) a
¨ (m)
(m) 1
Kx + m 1 − v Kx
=
d(m) 1
+m .
(m) The EPV of this annuity is denoted by ax
¨
(m) a(m)
¨x 1 − E[v Kx
=
d(m) 1
+m and is given by 186 CHAPTER 5. ANNUITIES giving
(m) a(m)
¨x 1 − Ax
=
d(m) . (5.18) Using the indicator random variable approach we ﬁnd that a(m)
¨x = ∞
r =0 1 r/m
rp .
v
x
m
m (5.19) For annuities payable 1/mthly in arrear, we can use a comparison with the 1/mthly
annuitydue. Similar to the annual annuity case, the only diﬀerence in the whole life case
is the ﬁrst payment, of $1/m, so that the EPV of the 1/mthly immediate annuity is
a(m) = a(m) −
¨x
x 5.6.3 1
m . (5.20) Term annuities payable m times a year We can extend the above derivation to cover the term life annuity case, when the 1/mthly
annuity payment is limited to a maximum of n years. Consider now an annuity of total
amount 1 per year, payable in advance m times per year throughout the lifetime of (x) for a
1
maximum of n years, with each payment being m . The payments, associated probabilities
and discount factors for the 1/mthly term annuitydue are shown in the timeline diagram
in Figure 5.7.
The present value random variable for this annuity is
a
¨ (m)
(m) 1
min Kx + m ,n = 1−v (m) 1
min Kx + m , n d(m) . 5.6. ANNUITIES PAYABLE M TIMES PER YEAR Time 0 1/m 2/m 3/m 4/m 187 n1 n 1/m 0 ......
Amount 1/m Discount 1 1/m 1/m v 1/m Probability 1 1
m 1/m v 2/m px 2
m px v 3/m
3
m px 1/m
v 4/m
4
m px v n−1/m
1
n− m px Figure 5.7: Timeline diagram for term life 1/mthly annuitydue. (m) The EPV of this annuity is denoted by ax:n and is given by
¨
(m) (m)
ax : n
¨ 1 − E[v min(Kx
=
d(m) 1
+ m ,n) so that
(m) (m)
ax : n
¨ 1 − Ax:n
=
.
d(m) (5.21) Using the indicator random variable approach we ﬁnd that
(m) ax : n =
¨ mn−1
r =0 1 r/m
rp .
v
x
m
m (5.22) For the 1/mthly term immediate annuity, by comparison with the 1/mthly annuitydue,
the diﬀerence is the ﬁrst payment under the annuity due, with EPV 1/m, and the ﬁnal 188 CHAPTER 5. ANNUITIES payment under the immediate annuity, with EPV
(m) (m) ax : n = ax : n −
¨ 1
(1 − v n n px ) .
m 1n
v n px ,
m so that
(5.23) This is analogous to the result in equation (5.3) for the annuitycertain. Further, by
setting m = 1 in equations (5.19) and (5.22) we obtain equations (5.7) and (5.10) for ax
¨
and ax:n . Also, by letting m → ∞ in equations (5.19) and (5.22) we obtain equations
¨
(5.15) and (5.17) for continuous annuities, ax and ax:n .
¯
¯
We can derive expressions for the EPV of other types of annuity payable m times per
year, and indeed we can also ﬁnd higher moments of present values as we did for annuities
payable annually. 5.7 Comparison of annuities by payment frequency
(4) (4) In Table 5.1 we show values for ax , ax , ax , ax and ax for x = 20, 40, 60 and 80, using
¨¨
¯
the Standard Ultimate Survival Model from Section 4.3, with interest of 5% per year.
Using equations (5.11), (5.20), (5.15), (5.19) and (5.7), we obtain the values shown in
Table 5.1.
We observe that each set of values decreases with age, reﬂecting the shorter expected life
span as age increases. We also have, for each age, the ordering
ax < a(4) < ax < a(4) < ax .
¯
¨x
¨
x
There are two reasons for this ordering.
• While the life is alive, the payments in each year sum to $1 under both annuities,
but on average, the payments under the annuitydue are paid earlier than under the
continuous annuity. The time value of money means that the value of an annuity
with earlier payments will be higher than an annuity with later payments, for interest rates greater than zero, so the annuity values are in increasing order from the 5.7. COMPARISON OF ANNUITIES BY PAYMENT FREQUENCY
(4) ax
18.966
17.458
13.904
7.548 ax
19.338
17.829
14.275
7.917 (4) ax
¯
19.462
17.954
14.400
8.042 ax
¨
19.588
18.079
14.525
8.167 (4) x
20
40
60
80 189 ax
¨
19.966
18.458
14.904
8.548 (4) ¨
¯¨
Table 5.1: Values of ax , ax , ax , ax and ax
latest average payment date (ax payments are at each year end) to the earliest (¨x
a
payments are at the start of each year).
• In the year that (x) dies, the diﬀerent annuities pay diﬀerent amounts. Under the
annual annuitydue the full year’s payment of $1 is paid, as the life is alive at the
payment date at the start of the year. Under the annual immediate annuity, in the
year of death no payment is made as the life does not survive to the payment date
at the year end. For the mthly and continuous annuities, less than the full year’s
annuity may be paid in the year of death.
For example, suppose the life dies after 7 months. Under the annual annuity due,
the full annuity payment is made for that year, at the start of the year. Under the
quarterly annuity due, three payments are made, each of 1/4 of the total annual
amount, at times 0, 1/4 and 1/2. The ﬁrst year’s ﬁnal payment, due at time 3/4, is
not made, as the life does not survive to that date. Under the continuous annuity,
the life collects 7/12ths of the annual amount. Under the quarterly immediate
annuity, the life collects payments at times 1/4, 1/2, and misses the two payments
due at times 3/4 and 1. Under the annual immediate annuity, the life collects no
annuity payments at all, as the due date is the year end.
This second point explains why we cannot make a simple interest adjustment to relate the
annuitydue and the continuous annuity. The situation here is diﬀerent to the insurance
(4)
beneﬁts; Ax and Ax , for example, both value a payment of $1 in the year of death, 190 CHAPTER 5. ANNUITIES
(4) Ax at the end of the year, and Ax at the end of the quarter year of death. There is
no diﬀerence in the amount of the payment, only in the timing. For the annuities, the
(4)
diﬀerence between ax and ax arises from diﬀerences in both cashﬂow timing and beneﬁt
¨
¨
amount in the year of death.
We also note from Table 5.1 that the ax values are close to being halfway between ax and
¯
1
ax , suggesting the approximation ax ≈ ax + 2 . We will see in Section 5.11.3 that there
¨
¯
is indeed a way of calculating an approximation to ax from ax , but it involves an extra
¯
adjustment term to ax .
Example 5.2 Using the Standard Ultimate Survival Model, with 5% per year interest,
(4)
(4)
calculate values of ax:10 , ax:10 , ax:10 , ax:10 and ax:10 for x = 20, 40, 60 and 80, and
¨
¨
¯
comment.
Solution 5.2 Using equations (5.10), (5.12), (5.17), (5.23) and (5.22) with n = 10 we
obtain the values shown in Table 5.2.
x ax:10 ax:10 (4) ax:10
¯ ax:10
¨ (4) ax:10
¨ 20
40
60
80 7.711
7.696
7.534
6.128 7.855
7.841
7.691
6.373 7.904
7.889
7.743
6.456 7.952
7.938
7.796
6.539 8.099
8.086
7.956
6.789 (4) (4) Table 5.2: Values of ax:10 , ax:10 , ax:10 , ax:10 and ax:10
¯
¨
¨ We note that for a given annuity function, the values do not vary greatly with age, since
the probability of death in a 10 year period is small. That means, for example, that the
second term in equation (5.10) is much greater than the ﬁrst term. The present value of
an annuity certain provides an upper bound for each set of values. For example, for any
(4)
(4)
age x, ax:10 < a10 = 7.722 and ax:10 < a10 = 7.962.
¨
¨ 5.8. DEFERRED ANNUITIES 191 Due to the diﬀerences in timing of payments, and in amounts for lives who die during the
10 year annuity term, we have the same ordering of annuity values by payment frequency
for any age x:
(4) (4) ax:10 < ax:10 < ax:10 < ax:10 < ax:10 .
¯
¨
¨ 5.8 Deferred annuities A deferred annuity is an annuity under which the ﬁrst payment occurs at some speciﬁed
future time. Consider an annuity payable to an individual now aged x under which annual
payments of 1 will commence at age x + u, where u is an integer, and will continue until
the death of (x). This is an annuitydue deferred u years. In standard actuarial notation,
the EPV of this annuity is denoted by u ax . Note that we have used the format u ...
¨
to indicate deferment before, both for mortality probabilities (u t qx ) and for insurance
beneﬁts (u Ax ).
Figure 5.8 shows the timeline for a u year deferred annuitydue.
Combining Figure 5.8 with the time line for a u year term annuity, see Figure 5.2, we
can see that the combination of the payments under a u year temporary annuitydue and
a u year deferred annuitydue gives the same sequence of payments as under a lifetime
annuity in advance, so we obtain
ax:u + u ax = ax ,
¨
¨
¨ (5.24) or, equivalently,
¨
u ax = ax − ax : u .
¨
¨ (5.25) 192 CHAPTER 5. ANNUITIES Time 0 1 2 u1 u u+1 ...... ... Amount 0 0 0 0 1 Discount 1 v1 v2 v u− 1 vu v u+1 1 px 2 px u− 1 p x u px u+1 px Probability 1 1 Figure 5.8: Timeline diagram for deferred annual annuitydue. Similarly, the EPV of an annuity payable continuously at rate 1 per year to a life now
aged x, commencing at age x + u is denoted by u ax and given by
¯
¯
u ax = ax − ax : u .
¯
¯ Summing the EPVs of the individual payments for the deferred whole life annuitydue
gives
¨
u ax = v u u px + v u+1 u+1 px + v u+2 u+2 px + . . .
= v u u px 1 + v px+u + v 2 2 px+u + . . . so that
¨
u ax = v u u px ax+u = u Ex ax+u .
¨
¨ (5.26) We see again that the pure endowment function acts like a discount function. In fact, 5.8. DEFERRED ANNUITIES 193 we can use the u Ex function to ﬁnd the EPV of any deferred beneﬁt. For example, for a
deferred term immediate annuity,
u ax:n = u Ex ax+u:n , and for an annuitydue payable 1/mthly,
¨(m)
u ax (m) = u Ex ax+u .
¨ (5.27) This result can be helpful when working with tables. Suppose we have available a table
of whole life annuitydue values, say ax , along with the life table function lx , and we need
¨
the term annuity value ax:n . Then, using equations (5.24) and (5.26), we have
¨
ax : n = ax − v n n p x ax + n .
¨
¨
¨ (5.28) For 1/mthly payments, the corresponding formula is
(m) (m) ax:n = a(m) − v n n px ax+n .
¨
¨x
¨ (5.29) Example 5.3 Let Y1 , Y2 and Y3 denote present value random variables for a u year
deferred whole life annuitydue, a u year term annuitydue and a whole life annuitydue,
respectively. Show that Y3 = Y1 + Y2 . Assume annual payments.
Solution 5.3 The present value random variable for a u year deferred whole life annuitydue, with annual payments is
Y1 = = 0
v u aKx +1−u
¨
0
aKx +1 − au
¨
¨ if Kx ≤ u − 1,
if Kx ≥ u,
if Kx ≤ u − 1,
if Kx ≥ u. (5.30) 194 CHAPTER 5. ANNUITIES From Section 5.4.2 we have
Y2 = aKx +1
¨
au
¨ if Kx ≤ u − 1,
if Kx ≥ u. Hence
Y1 + Y2 = aKx +1
¨
aKx +1
¨ if Kx ≤ u − 1,
if Kx ≥ u, = aKx +1
¨ = Y3 , as required. We use deferred annuities as building blocks in later sections, noting that an n year
term annuity, with any payment frequency, can be decomposed as the sum of n deferred
annuities, each with term 1 year. So, for example,
ax : n =
¯ n−1
u=0 5.9 ¯
u ax:1 . (5.31) Guaranteed annuities A common feature of pension beneﬁts is that the pension annuity is guaranteed to be paid
for some period even if the life dies before the end of the period. For example, a pension
beneﬁt payable to a life aged 65, say, might be guaranteed for 5, 10 or even 15 years.
Suppose an annuitydue of 1 per year is paid annually to (x), and is guaranteed for a
period of n years. Then the payment due at k years is paid whether or not (x) is then
alive if k = 0, 1, . . . , n − 1, but is paid only if (x) is alive at age x + k for k = n, n + 1, . . . .
The present value random variable for this beneﬁt is
Y = an
¨
aKx +1
¨ if Kx ≤ n − 1,
if Kx ≥ n 5.9. GUARANTEED ANNUITIES
an
¨
an + aKx +1 − an
¨
¨
¨ = if Kx ≤ n − 1,
if Kx ≥ n 0
aKx +1 − an
¨
¨ 195 if Kx ≤ n − 1,
if Kx ≥ n = an +
¨ = an + Y 1 ,
¨ where Y1 denotes the present value of an n year deferred annuitydue of 1 per year, from
equation (5.30), and
E[Y1 ] = n ax = n Ex ax+n .
¨
¨
The expected present value of the unit n year guaranteed annuitydue is denoted ax:n , so
¨
ax:n = an + n Ex ax+n .
¨
¨
¨ (5.32) Figure 5.9 shows the timeline for an n year guaranteed unit whole life annuitydue. This
timeline looks like the regular whole life annuitydue timeline, except that the ﬁrst n
payments, from time t = 0 to time t = n − 1, are certain and not life contingent.
We can derive similar results for guaranteed beneﬁts payable 1/mthly; for example, a
monthly whole life annuitydue guaranteed for n years has EPV
(12) (12) ax : n = an
¨
¨ (12) + n Ex ax+n .
¨ Example 5.4 A pension plan member is entitled to a beneﬁt of $1 000 per month, in
advance, for life from age 65, with no guarantee. She can opt to take a lower beneﬁt, with
a 10 year guarantee. The revised beneﬁt is calculated to have equal EPV to the original
beneﬁt. Calculate the revised beneﬁt using the Standard Ultimate Survival Model, with
interest at 5% per year. 196 CHAPTER 5. ANNUITIES Time 0 1 2 n1 n n+1 ...... ... Amount 1 1 1 1 1 Discount 1 v1 v2 v n−1 vn v n+1 1 1 n px n+1 px Probability 1 1 1 Figure 5.9: Timeline diagram for guaranteed annual annuitydue. Solution 5.4 Let B denote the revised monthly beneﬁt. To determine B we must equate
the EPV of the original beneﬁt with that of the revised beneﬁt. The resulting equation
of EPVs is usually called an equation of value. Our equation of value is
(12) 12000¨65 = 12 B a
a
¨ (12) 65:10 (12) where a65 = 13.0870, and
¨
a
¨ (12)
65:10 (12) (12) = a10 + 10 p65 v 10 a75 = 13.3791.
¨
¨ Thus, the revised monthly beneﬁt is B = $978.17.
So the pension plan member can gain the security of the 10 year guarantee at a cost of a
reduction of $21.83 per month in her pension. 5.10. INCREASING ANNUITIES 5.10 197 Increasing annuities In the previous sections we have considered annuities with level payments. Some of the
annuities which arise in actuarial work are not level. For example, annuity payments may
increase over time. For these annuities, we are generally interested in determining the
EPV, and are rarely concerned with higher moments. To calculate higher moments it is
generally necessary to use ﬁrst principles, and a computer.
The best approach for calculating the EPV of nonlevel annuities is to use the indicator
random, or timeline, approach – that is, sum over all the payment dates the product of
the amount of the payment, the probability of payment (that is, the probability that the
life survives to the payment date) and the appropriate discount factor. 5.10.1 Arithmetically increasing annuities We ﬁrst consider annuities under which the amount of the annuity payment increases
arithmetically with time. Consider an increasing annuitydue where the amount of the
annuity is t + 1 at times t = 0, 1, 2, . . . , n − 1 provided that (x) is alive at time t. The
timeline is shown in Figure 5.10.
The EPV of the annuity is denoted by (I a)x in standard actuarial notation. From the
¨
diagram we see that (I a)x =
¨ ∞ v t (t + 1) t px . (5.33) t=0 Similarly, if the annuity is payable for a maximum of n payments rather than for the
whole life of (x), the EPV, denoted by (I a)x:n in standard actuarial notation, would be
¨ 198 CHAPTER 5. ANNUITIES Time 0 1 2 3 4
... Amount 1 2 3 4 5 Discount 1 v1 v2 v3 v4 1 px 2 px 3 px 4 px Probability 1 Figure 5.10: Timeline diagram for arithmetically increasing annual annuitydue. given by (I a)x:n =
¨ n−1 v t (t + 1) t px . (5.34) t=0 If the annuity is payable continuously, with the payments increasing by 1 at each year
end, so that the rate of payment in the tth year is constant and equal to t, for t =
1, 2, . . . , n, then we may consider the n year temporary annuity as a sum of one year
deferred annuities. By analogy with formula (5.31), the EPV of this annuity, denoted in
standard actuarial notation by (I a)x:n , is
¯
(I a)x:n =
¯ n−1
m=0 (m + 1) m ax:1 .
¯ We also have standard actuarial notation for the continuous annuity under which the rate
of payment at time t > 0 is t; that is, the rate of payment is changing continuously. The 5.10. INCREASING ANNUITIES Time 0 199 t t+dt
... Amount
Discount
Probability t dt
e−δt
t px Figure 5.11: Timeline diagram for increasing continuous whole life annuity. ¯¯
¯¯
notation for the EPV of this annuity is (I a)x if it is a whole life annuity, and (I a)x:n if
it is a term annuity. For every inﬁnitesimal interval, (t, t + dt), the amount of annuity
paid, if the life (x) is still alive, is t dt, the probability of payment is t px and the discount
function is e−δ t = v t . The timeline is shown in Figure 5.11.
To determine the EPV we integrate over all the possible intervals (t, t + dt), so that
n ¯¯
(I a)x:n = t e−δt t px dt . (5.35) 0 5.10.2 Geometrically increasing annuities An annuitant may be interested in purchasing an annuity that increases geometrically, to
oﬀset the eﬀect of inﬂation on the purchasing power of the income. The approach is similar
to the geometrically increasing insurance beneﬁt which was considered in Examples 4.8
and 4.9. 200 CHAPTER 5. ANNUITIES Example 5.5 Consider an annuitydue with annual payments where the amount of the
annuity is (1 + j )t at times t = 0, 1, 2, . . . , n − 1 provided that (x) is alive at time t. Derive
an expression for the EPV of this beneﬁt, and simplify as far as possible.
Solution 5.5 First, consider the timeline diagram in Figure 5.12. Time 0 1 2 3 4
...... Amount 1 Discount 1 Probability 1 (1 + j ) (1 + j )2 (1 + j )3 (1 + j )4 v1 v2 v3 v4 1 px 2 px 3 px 4 px Figure 5.12: Timeline diagram for geometrically increasing annual annuitydue. By summing the product of
• the amount of the payment at time t,
• the discount factor for time t, and
• the probability that the payment is made at time t,
over all possible values of t, we obtain the EPV as
n−1
t=0 (1 + j )t v t t px = ax:n i∗
¨ 5.11. EVALUATING ANNUITY FUNCTIONS 201 where ax:n i∗ is the EPV of a term annuitydue evaluated at interest rate i∗ where
¨
i∗ = 5.11 1+i
i−j
−1=
.
1+j
1+j Evaluating annuity functions If we have full information about the survival function for a life, then we can use summation or numerical integration to compute the EPV of any annuity.
Often, though, we have only integer age information, for example when the survival
function information is derived from a life table with integer age information only. In
this section we consider how to evaluate the EPV of 1/mthly and continuous annuities
given only the EPVs of annuities at integer ages. For example, we may have ax values
¨
for integer x. We present two methods that are commonly used, and we explore the
accuracy of these methods for a fairly typical (Makeham) mortality model. First we
consider recursive calculation of EPVs of annuities. 5.11.1 Recursions In a spreadsheet, with values for t px available, we may calculate ax using a backward
¨
recursion. We assume that there is an integer limiting age, ω , so that qω−1 = 1. First, we
set aω−1 = 1.0.
¨
The backward recursion for x = ω − 2, ω − 3, ω − 4, . . . is
ax = 1 + v px ax+1
¨
¨
since
ax = 1 + v p x + v 2 2 p x + v 3 3 p x + . . .
¨ (5.36) 202 CHAPTER 5. ANNUITIES
= 1 + v px 1 + v px+1 + v 2 2 px+1 + . . .
= 1 + v px ax+1 .
¨ Similarly, for the 1/mthly annuity due,
(m) aω−1/m =
¨ 1
m , 2
3
4
and the backward recursion for x = ω − m , ω − m , ω − m , . . . is a(m) =
¨x 1
m 1 + vm 1
m (m) p x ax + 1 .
¨ (5.37) m We can calculate EPVs for term annuities and deferred annuities from the whole life
annuity EPVs, using, for example, equations (5.24) and (5.26).
To ﬁnd the EPV of an annuity payable continuously we can use numerical integration.
Note, however, that Woolhouse’s formula, which is described in Section 5.11.3, gives an
excellent approximation to 1/mthly and continuous annuity EPVs. 5.11.2 Applying the UDD assumption
(m) We ﬁrst consider evaluation of ax:n under the assumption of a uniform distribution of
¨
deaths (UDD). The indication from Table 4.6 is that, in terms of EPVs for insurance beneﬁts, UDD oﬀers a reasonable approximation at younger ages, but may not be suﬃciently
accurate at older ages.
From Section 4.5.1 recall the results from equations (4.27) and (4.26) that, under the
UDD assumption,
i
i
¯
and Ax = Ax .
A(m) = (m) Ax
x
i
δ
We also know, from equations (5.4), (5.18) and (5.14) in this Chapter that for any survival
model
(m)
¯
1 − Ax
1 − Ax
1 − Ax
(m)
ax =
¨
, ax =
¨
and ax =
¯
.
d
d(m)
δ 5.11. EVALUATING ANNUITY FUNCTIONS 203 Now, putting these equations together we have
(m) a(m) =
¨x 1 − Ax
d(m) = i
1 − i(m) Ax
d(m) = i(m) − iAx
i(m) d(m) using UDD i(m) − i(1 − dax )
¨
=
(m) d(m)
i
= using (5.4) id
i − i(m)
ax − (m) (m)
¨
i(m) d(m)
id = α(m) ax − β (m)
¨
where
α(m) = id
i(m) d(m) and β (m) = i − i(m)
.
i(m) d(m) (5.38) For continuous annuities we can let m → ∞, so that
ax =
¯ id
i−δ
ax − 2 .
¨
δ2
δ For term annuities, starting from equation (5.29) we have,
(m) ax : n
¨ (m) = a(m) − v n n px ax+n
¨x
¨
= α(m)¨x − β (m) − v n n px (α(m)¨x+n − β (m))
a
a
= α(m) (¨x − v n n px ax+n ) − β (m) (1 − v n n px )
a
¨ using UDD 204 CHAPTER 5. ANNUITIES
= α(m) ax:n − β (m) (1 − v n n px ) .
¨ Note that the functions α(m) and β (m) depend only on the frequency of the payments,
not on the underlying survival model. It can be shown (see Exercise 5.12) that α(m) ≈ 1
and β (m) ≈ (m − 1)/2m, leading to the approximation
(m) ax : n ≈ ax : n −
¨
¨ 5.11.3 m−1
(1 − v n n px ) .
2m (5.39) Woolhouse’s formula Woolhouse’s formula is a method of calculating the EPV of annuities payable more frequently than annually that is not based on a fractional age assumption. It is based on
(m)
the EulerMaclaurin formula and expresses ax in terms of ax .
¨
¨
The EulerMaclaurin formula is a numerical integration method. It gives a series expansion for the integral of a function, assuming that the function is diﬀerentiable a certain
number of times. As discussed in Appendix B, in the case of a function g (t), where
limt→∞ g (t) = 0, the formula can be written in terms of a constant h > 0 as
∞ g (t)dt = h 0 ∞ h
h2
h4
g (kh) − g (0) + g (0) −
g (0) + . . . ,
2
12
720
k=0 (5.40) where we have omitted terms on the right hand side that involve higher derivatives of g .
To obtain our approximations we apply this formula twice to the function g (t) = v t t px ,
in each case ignoring second and higher order derivatives of g , which is reasonable as the
function is usually quite smooth. Note that g (0) = 1, limt→∞ g (t) = 0, and
g (t) = t px dt
d
v + v t t px
dt
dt = t px d −δ t
d
e
+ vt
t px
dt
dt 5.11. EVALUATING ANNUITY FUNCTIONS 205 = − t px δ e−δ t − v t t px µx+t ,
so g (0) = − (δ + µx ).
First, let h = 1. As we are ignoring second and higher order derivatives, the right hand
side of (5.40) becomes
∞ 1
1
g (k ) − + g (0) =
2 12
k=0 ∞
k=0 v k k px − = ax −
¨ 1
1
−
(δ + µx )
2 12 1
1
−
(δ + µx ) .
2 12 (5.41) Second, let h = 1/m. Again ignoring second and higher order derivatives, the right hand
side of (5.40) becomes
1
m ∞ 1
1
1
g (k/m) −
+
g (0) =
2
2m 12m
m
k=0 ∞ k v m k px −
m k=0 = a(m) −
¨x 1
1
−
(δ + µx )
2m 12m2 1
1
−
(δ + µx ) .
2m 12m2 (5.42) Since each of (5.41) and (5.42) approximates the same quantity, ax , we can obtain an
¯
(m)
approximation to ax by equating them, so that
¨
a(m) −
¨x 1
1
1
1
−
(δ + µx ) ≈ ax − − (δ + µx ) .
¨
2
2m 12m
2 12 Rearranging, we obtain the important formula
a(m) ≈ ax −
¨x
¨ m − 1 m2 − 1
−
(δ + µx ) .
2m
12m2 (5.43) The right hand side of equation (5.43) gives the ﬁrst three terms of Woolhouse’s formula,
and this is the basis of our actuarial approximations. 206 CHAPTER 5. ANNUITIES For term annuities, we again start from equation (5.29),
(m) (m) ax:n = a(m) − v n n px ax+n
¨
¨x
¨
and applying formula(5.43) gives
(m) ax : n
¨ ≈ ax −
¨ m − 1 m2 − 1
−
(δ + µx )
2m
12m2
−v n n px = ax:n −
¨
− ax + n −
¨ m − 1 m2 − 1
−
(δ + µx+n )
2m
12m2 m−1
(1 − v n n px )
2m
m2 − 1
δ + µx − v n n px (δ + µx+n ) .
12m2 (5.44) For continuous annuities, we can let m → ∞ in equations (5.43) and (5.44) (or just apply
equation (5.41)), so that
ax ≈ ax −
¯
¨ 1
1
− (δ + µx )
2 12 (5.45) and
1
1
ax:n ≈ ax:n − (1 − v n n px ) −
¯
¨
δ + µx − v n n px (δ + µx+n ) .
2
12
(m) An important diﬀerence between the approximation to ax:n based on Woolhouse’s for¨
mula and the UDD approximation is that we need extra information for the Woolhouse
approach, speciﬁcally values for the force of mortality. In practice, the third term in
equation (5.44) is often omitted (leading to the same approximation as equation (5.39)),
but as we shall see in the next section, this leads to poor approximations in some cases.
If the integer age information available does not include values of µx , then we may still 5.12. NUMERICAL ILLUSTRATIONS 207 use Woolhouse’s formula. As
x+1
2 p x− 1 = exp − µs ds
x− 1 ≈ exp{−2µx } , we can approximate µx as
1
µx ≈ − (log(px−1 ) + log(px )) ,
2 (5.46) and the results for the illustrations given in the next section are almost identical to where
the exact value of the force of mortality is used.
In fact, Woolhouse’s formula (with three terms) is so accurate that even if the full force
of mortality curve is known, it is often a more eﬃcient way to calculate annuity values
than the more direct formulae with comparable accuracy. Also, since we have a simple
relationship between annuity and insurance functions, we may use Woolhouse’s formula
(m)
also for calculating Ax , for example, using
A(m) = 1 − d(m) a(m)
¨x
x
In Section 2.6.2 we saw an approximate relationship between the complete expectation of
life and the curtate expectation of life, namely
◦ ex ≈ ex + 1 .
2
Setting the interest rate to 0 in equation (5.45) gives a reﬁnement of this approximation,
namely
◦ ex ≈ ex + 1 −
2 5.12 1
µ
12 x . Numerical illustrations In this section we give some numerical illustrations of the diﬀerent methods of computing
(m)
(12)
ax:n . Table 5.3 shows values of ax:10 for x = 20, 30, . . . , 100 when i = 0.1, while Table
¨
¨ 208 CHAPTER 5. ANNUITIES
x
20
30
40
50
60
70
80
90
100 Exact
6.4655
6.4630
6.4550
6.4295
6.3485
6.0991
5.4003
3.8975
2.0497 UDD
6.4655
6.4630
6.4550
6.4294
6.3482
6.0982
5.3989
3.8997
2.0699 W2
6.4704
6.4679
6.4599
6.4344
6.3535
6.1044
5.4073
3.9117
2.0842 W3
6.4655
6.4630
6.4550
6.4295
6.3485
6.0990
5.4003
3.8975
2.0497 W3*
6.4655
6.4630
6.4550
6.4295
6.3485
6.0990
5.4003
3.8975
2.0496 (12) Table 5.3: Values of ax:10 for i = 0.1.
¨
(2) 5.4 shows values of ax:25 when i = 0.05. The mortality basis for the calculations is the
¨
Standard Ultimate Survival Model, from Section 4.3.
The legend for each table is as follows:
Exact denotes the true EPV, calculated from formula (5.37);
UDD denotes the approximation to the EPV based on the uniform distribution of deaths
assumption;
W2 denotes the approximation to the EPV based on Woolhouse’s formula, using the
ﬁrst two terms only;
W3 denotes the approximation to the EPV based on Woolhouse’s formula, using all
three terms, including the exact force of mortality;
W3* denotes the approximation to the EPV based on Woolhouse’s formula, using all
three terms, but using the approximate force of mortality estimated from integer
age values of px . 5.12. NUMERICAL ILLUSTRATIONS
x
20
30
40
50
60
70
80
90
100 Exact
14.5770
14.5506
14.4663
14.2028
13.4275
11.5117
8.2889
4.9242
2.4425 UDD
14.5770
14.5505
14.4662
14.2024
13.4265
11.5104
8.2889
4.9281
2.4599 209
W2
14.5792
14.5527
14.4684
14.2048
13.4295
11.5144
8.2938
4.9335
2.4656 W3
14.5770
14.5506
14.4663
14.2028
13.4275
11.5117
8.2889
4.9242
2.4424 W3*
14.5770
14.5506
14.4663
14.2028
13.4275
11.5117
8.2889
4.9242
2.4424 (2) Table 5.4: Values of ax:25 for i = 0.05.
¨ From these tables we see that approximations based on Woolhouse’s formula with all
three terms yield excellent approximations, even where we have approximated the force
of mortality from integer age px values. Also, note that the inclusion of the third term is
important for accuracy; the two term Woolhouse formula is the worst approximation. We
also observe that the approximation based on the UDD assumption is good at younger
ages, with some deterioration for older ages. In this case approximations based on Woolhouse’s formula are superior, provided the three term version is used.
(12) It is also worth noting that calculating the exact value of, for example, a20 using a
¨
spreadsheet approach takes around 1,200 rows, one for each month from age 20 to the
limiting age ω . Using Woolhouse’s formula requires only the integer age table, of 100 rows,
and the accuracy all the way up to age 100 is excellent, using the exact or approximate
values for µx . Clearly, there can be signiﬁcant eﬃciency gains using Woolhouse. 210 CHAPTER 5. ANNUITIES 5.13 Functions for select lives Throughout this chapter we have assumed that lives are subject to an ultimate survival
model, just as we did in deriving insurance functions in Chapter 4. Just as in that chapter,
all the arguments in this chapter equally apply if we have a select survival model. Thus,
for example, the EPV of an n year term annuity payable continuously at rate 1 per year
to a life who is aged x + k and who was select at age x is a[x]+k:n , with
¯
¯
A[x]+k:n = 1 − δ a[x]+k:n .
¯
The approximations we have developed also hold for select survival models, so that, for
example
(m) a[x]+k = a[x]+k −
¨
¨ m − 1 m2 − 1
−
(δ + µ[x]+k )
2m
12m2 where
a[x]+k =
¨ ∞ v t t p[x]+k t=0 and
(m) a[x]+k =
¨ 5.14 1
m ∞ v t/m t
m p[x]+k . t=0 Notes and further reading Woolhouse (1869) presented the formula which bears his name in a paper to the Institute
of Actuaries in London. In this paper he also showed that his theory applied to jointlife annuities, a topic we discuss in Chapter 8. A more direct derivation of Woolhouse’s
formula comes from the EulerMaclaurin formula, and we give this in Appendix B. The
EulerMaclaurin formula was derived independently (about 130 years before Woolhouse’s 5.14. NOTES AND FURTHER READING 211 paper) by the famous Swiss mathematician Leonhard Euler and by the Scottish mathematician Colin Maclaurin. A proof of the EulerMaclaurin formula, and references to the
original works, can be found in Graham et al (1994). 212 CHAPTER 5. ANNUITIES 5.15 Exercises When a calculation is required in the following exercises, unless otherwise stated you
should assume that mortality follows the Standard Ultimate Survival Model as speciﬁed
in Section 4.3 and that interest is at 5% per year eﬀective.
Exercise 5.1 Describe in words the beneﬁts with the present values given and write
down an expression in terms of actuarial functions for the expected present value. (a) Y1 = aTx
¯
a15
¯ if Tx ≤ 15,
if Tx > 15. (b) Y2 = a15
aKx if 0 < Kx ≤ 15,
if Kx > 15. Exercise 5.2
able: (a) Describe the annuity with the following present value random vari Y= v Tx an−Tx
¯
0 if Tx ≤ n,
if Tx ≥ n. This is called a Family Income Beneﬁt.
(b) Show that E[Y ] = an − ax:n
¯
¯
(c) Explain the answer in (b) by general reasoning. 5.15. EXERCISES 213 Exercise 5.3 Given that a50:10 = 8.2066, a50:10 = 7.8277, and
¨
the eﬀective rate of interest per year? 10 p50 = 0.9195, what is Exercise 5.4 Given that a60 = 10.996, a61 = 10.756, a62 = 10.509 and i = 0.06, calculate
2 p60 . Exercise 5.5 You are given the following extract from a select life table.
[x ]
40
41
42
43
44 l[x]
33 519
33 467
33 407
33 340
33 265 l[x]+1
33 485
33 428
33 365
33 294
33 213 lx+2
x+2
33 440
42
33 378
43
33 309
44
33 231
45
33 143
46 Calculate the following, assuming an interest rate of 6% per year:
(a) a[40]:4 ,
¨
(b) a[40]+1:4 ,
(c) (Ia)[40]:4 ,
(d) (IA)[40]:4 ,
(e) the standard deviation of the present value of a 4 year term annuitydue, with annual
payment $1 000, payable to a select life age 41, and 214 CHAPTER 5. ANNUITIES (f) the probability that the present value of an annuitydue of 1 per year issued to a
select life aged 40 is less than 3.0. Exercise 5.6 The force of mortality for a certain population is exactly half the sum of
the forces of mortality in two standard mortality tables, denoted A and B . Thus
µx = (µA + µB )/2
x
x
for all x. A student has suggested the approximation
ax = (aA + aB )/2.
x
x
Will this approximation overstate or understate the true value of ax ? Exercise 5.7 Consider a life aged x. Obtain the formula
(IA)x = ax − d(I a)x
¨
¨
by writing down the present value random variables for
(a) an increasing annuitydue to (x) with payments of t + 1 at times t = 0, 1, 2, . . . , and
(b) a whole life insurance beneﬁt of amount t at time t, t = 1, 2, 3 . . . , if the death of
(x) occurs between ages x + t − 1 and x + t.
Hint: use the result
n tv t−1 = (I a)n =
¨
t=1 an − nv n
¨
.
d 5.15. EXERCISES 215 Exercise 5.8 Let H = min(K, n).
(a) Show that 2 Ax:n+1 − Ax:n+1
V[aH ] =
d2 2 . (b) An alternative form given for this variance is
1
1
1
(1 + i)2 [2 Ax:n − (Ax:n )2 ] − 2(1 + i)Ax:n v n n px + v 2n n px (1 − n px )
.
i2 Prove that this is equal to the expression in (a). Exercise 5.9 Consider the random variables Y = aTx and Z = v Tx .
¯
(a) Explain brieﬂy why the covariance of Y and Z will be negative, in general.
(b) Derive an expression for the covariance, in terms of standard actuarial functions.
(c) Show that the covariance is negative. Exercise 5.10 Find, and simplify where possible:
(a) d
a,
¨
dx x and (b) d
a
¨
dx x:n . 216 CHAPTER 5. ANNUITIES Exercise 5.11 Consider the following portfolio of annuitiesdue currently being paid
from the assets of a pension fund. Age
60
70
80 Number of
Annuitants
40
30
10 Each annuity has an annual payment of $10 000 as long as the annuitant survives. The
lives are assumed to be independent. Calculate
(a) the present value of the total outgo on annuities,
(b) the standard deviation of the present value of the total outgo on annuities, and
(c) the 95th percentile of the distribution of the present value of the total outgo on
annuities using a Normal approximation. Exercise 5.12 Consider the quantities α(m) and β (m) in formula (5.38). By expressing
i, i(m) , d and d(m) in terms of δ , show that
α(m) ≈ 1 and β (m) ≈ m−1
.
2m Exercise 5.13 Using a spreadsheet, calculate the mean and variance of the present value
of 5.15. EXERCISES 217 (a) an arithmetically increasing term annuitydue payable to a life aged 50 for at most
10 years under which the payment at time t is t + 1 for t = 0, 1, . . . , 9, and
(b) a geometrically increasing term annuitydue payable to a life aged 50 for at most 10
years under which the payment at time t is 1.03t for t = 0, 1, . . . , 9. Exercise 5.14 Using a spreadsheet, calculate the mean and variance of the present value
of
(a) a whole life annuitydue to a life aged 65, with annual payments of 1, and
(b) a whole life annuitydue to a life aged 65, with annual payments of 1 and a guarantee
period of 10 years.
Explain the ordering of the means and variances. Exercise 5.15 Jensen’s inequality states that for a function f , whose ﬁrst derivative is
positive and whose second derivative is negative, and a random variable X ,
E[f (X )] ≤ f (E[X ]) .
Use Jensen’s inequality to show that
ax ≤ aE[Tx ] .
¯
¯ 218 CHAPTER 5. ANNUITIES Answers to selected exercises
5.3 4.0014%
5.4 0.98220
5.5 (a) 3.66643
(b) 3.45057
(c) 8.37502
(d) 3.16305
(e) 119.14
(f) 0.00421
5.11 (a) 10 418 961
(b) 311 534
(c) 10 931 390
5.13 (a) 40.95, 11.057 (b) 9.121, 0.32965 5.14 (a) 13.550,
(b) 13.814, 12.497
8.380 Chapter 6
Premium Calculation
6.1 Summary In this chapter we discuss principles of premium calculation for insurance policies and
annuities. We start by reviewing what we mean by the terms ‘premium’, ‘net premium’
and ‘gross premium’. We next introduce the present value of future loss random variable.
We deﬁne the equivalence premium principle and we show how this premium principle
can be applied to calculate premiums for diﬀerent types of policy. We look at how we
can use the future loss random variable to determine when a contract moves from loss to
proﬁt or vice versa. We introduce a diﬀerent premium principle, the portfolio percentile
premium principle, and show how, using the mean and variance of the future loss random
variable, the portfolio percentile premium principle can be used to determine a premium.
The chapter concludes with a discussion of how a premium can be calculated when the
insured life is subject to some extra level of risk.
219 220 6.2 CHAPTER 6. PREMIUM CALCULATION Preliminaries An insurance policy is a ﬁnancial agreement between the insurance company and the policyholder. The insurance company agrees to pay some beneﬁts, for example a sum insured
on the death of the policyholder within the term of a term insurance, and the policyholder
agrees to pay premiums to the insurance company to secure these beneﬁts. The premiums
will also need to reimburse the insurance company for the expenses associated with the
policy.
The calculation of the premium may not explicitly allow for the insurance company’s
expenses. In this case we refer to a net premium (also, sometimes, a risk premium or
mathematical premium). If the calculation does explicitly allow for expenses, the premium
is called a gross premium or oﬃce premium.
The premium may be a single payment by the policyholder – a single premium – or
it maybe a regular series of payments, possibly annually, quarterly, monthly or weekly.
Monthly premiums are very common since many employed people receive their salaries
monthly and it is convenient to have payments made with the same frequency as income
is received.
It is common for regular premiums to be a level amount, but they do not have to be.
A key feature of any life insurance policy is that premiums are payable in advance, with
the ﬁrst premium payable when the policy is purchased. To see why this is necessary,
suppose it were possible to purchase a whole life insurance policy with annual premiums
where the ﬁrst premium were payable at the end of the year in which the policy was
purchased. In this case, a person could purchase the policy and then withdraw from the
contract at the end of the ﬁrst year before paying the premium then due. This person
would have had a year of insurance cover without paying anything for it.
Regular premiums for a policy on a single life cease to be payable on the death of the
policyholder. The premium paying term for a policy is the maximum length of time for
which premiums are payable. The premium paying term may be the same as the term of 6.3. ASSUMPTIONS 221 the policy, but it could be shorter. If we consider a whole life insurance policy, it would be
usual for the death beneﬁt to be secured by regular premiums and it would be common
for premium payment to cease at a certain age – perhaps at age 65 when the policyholder
is assumed to retire, or at age 80 when the policyholder’s real income may be diminishing.
As we discussed in Chapter 1, premiums are payable to secure annuity beneﬁts as well as
life insurance beneﬁts. Deferred annuities may be purchased using a single premium at
the start of the deferred period, or by regular premiums payable throughout the deferred
period. Immediate annuities are always purchased by a single premium. For example,
a person aged 45 might secure a retirement income by paying regular premiums over a
twenty year period to secure annuity payments from age 65. Or, a person aged 65 might
secure a monthly annuity from an insurance company by payment of a single premium.
For traditional policies, the benchmark principle for calculating both gross and net premiums is called the equivalence principle, and we discuss its application in detail in this
chapter. However, there are other methods of calculating premiums and we discuss one
of these, the portfolio percentile principle.
A more contemporary approach, which is commonly used for nontraditional policies, is
to consider the cash ﬂows from the contract, and to set the premium to satisfy a speciﬁed
proﬁt criterion. This approach is discussed in Chapter 11. 6.3 Assumptions As in Chapter 4, unless otherwise stated, we use a standard set of assumptions for mortality and interest in the numerical examples in this chapter. We use the select survival
model with a 2 year select period speciﬁed in Example 3.13 with an interest rate of 5%
per year eﬀective. Recall that the survival model is speciﬁed as follows:
µx = A + Bcx 222 CHAPTER 6. PREMIUM CALCULATION where A = 0.00022, B = 2.7 × 10−6 and c = 1.124. and
µ[x]+s = 0.92−s µx+s
for 0 ≤ s ≤ 2. The select and ultimate life table, at integer ages, for this model is shown
in Table 3.7 and values of Ax at an eﬀective rate of interest of 5% per year are shown in
Table 4.1. We refer to this model as the Standard Select Survival Model.
Example 6.1 Use the Standard Select Survival Model described above, with interest at 5% per year, to produce a table showing values of a[x] , a[x]+1 and ax+2 for
¨
¨
¨
x = 20, 21, . . . , 80. Assume that q131 = 1.
Solution 6.1 The calculation of survival probabilities p[x] , px]+1 and px for this survival
model was discussed in Example 3.13. Since we are assuming that q131 = 1, we have
a131 = 1. Annuity values can then be calculated recursively using
¨
ax = 1 + v px ax+1 ,
¨
¨
a[x]+1 = 1 + v p[x]+1 ax+2 ,
¨
¨
a[x] = 1 + v p[x] a[x]+1 .
¨
¨
Values are shown in Table 6.1. 6.4 The present value of future loss random variable The cash ﬂows for a traditional life insurance contract consist of the insurance or annuity
beneﬁt outgo (and associated expenses) and the premium income. Both are generally life
contingent, that is, the income and outgo cash ﬂows depend on the future lifetime of the 6.4. THE PRESENT VALUE OF FUTURE LOSS RANDOM VARIABLE
x
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50 a[x]
¨
19.96732
19.92062
19.87165
19.82030
19.76647
19.71003
19.65087
19.58887
19.52389
19.45581
19.38449
19.30979
19.23156
19.14965
19.06390
18.97415
18.88024
18.78201
18.67927
18.57184
18.45956
18.34224
18.21969
18.09172
17.95814
17.81876
17.67340
17.52187
17.36397
17.19952
17.02835 a[x]+1
¨
19.91993
19.87095
19.81959
19.76574
19.70929
19.65012
19.58810
19.52310
19.45500
19.38365
19.30892
19.23066
19.14871
19.06292
18.97313
18.87917
18.78088
18.67807
18.57058
18.45822
18.34081
18.21815
18.09007
17.95637
17.81686
17.67135
17.51965
17.36156
17.19691
17.02551
16.84718 ax+2
¨
x+2
19.87070
22
19.81934
23
19.76549
24
19.70903
25
19.64985
26
19.58783
27
19.52282
28
19.45471
29
19.38336
30
19.30862
31
19.23034
32
19.14838
33
19.06258
34
18.97277
35
18.87880
36
18.78049
37
18.67766
38
18.57014
39
18.45776
40
18.34031
41
18.21763
42
18.08951
43
17.95577
44
17.81621
45
17.67065
46
17.51889
47
17.36074
48
17.19602
49
17.02453
50
16.84612
51
16.66060
52 x
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80 a[ x ]
¨
16.85028
16.66514
16.47277
16.27303
16.06579
15.85091
15.62831
15.39789
15.15960
14.91340
14.65927
14.39724
14.12736
13.84972
13.56444
13.27169
12.97168
12.66467
12.35097
12.03093
11.70495
11.37350
11.03709
10.69629
10.35171
10.00402
9.65395
9.30225
8.94973
8.59722 a[x]+1
¨
16.66175
16.46908
16.26899
16.06137
15.84608
15.62302
15.39210
15.15325
14.90644
14.65165
14.38890
14.11822
13.83972
13.55351
13.25975
12.95864
12.65045
12.33547
12.01406
11.68661
11.35359
11.01550
10.67291
10.32644
9.97676
9.62458
9.27067
8.91584
8.56093
8.20681 223 ax+2
¨
x+2
16.46782
53
16.26762
54
16.05987
55
15.84443
56
15.62122
57
15.39012
58
15.15109
59
14.90407
60
14.64906
61
14.38606
62
14.11512
63
13.83632
64
13.54979
65
13.25568
66
12.95420
67
12.64561
68
12.33019
69
12.00830
70
11.68035
71
11.34678
72
11.00812
73
10.66491
74
10.31778
75
9.96740
76
9.61449
77
9.25981
78
8.90416
79
8.54841
80
8.19341
81
7.84008
82 Table 6.1: Annuity values using the Standard Select Survival Model. 224 CHAPTER 6. PREMIUM CALCULATION policyholder, unless the contract is purchased by a single premium, in which case there
is no uncertainty regarding the premium income. So we can model the future outgo less
future income with the random variable that represents the present value of the future
loss. When expenses are excluded we call this the net future loss, which we denote by
Ln . When expenses are included, then the premiums are the gross premiums, and the
0
random variable is referred to as the gross future loss, denoted Lg . In other words,
0
Ln = PV of beneﬁt outgo − PV of net premium income
0
Lg = PV of beneﬁt outgo + PV of expenses − PV of gross premium income.
0
In cases where the meaning is obvious from the context, we will drop the n or g superscript.
Example 6.2 An insurer issues a whole life insurance to [60], with sum insured S payable
immediately on death. Premiums are payable annually in advance, ceasing at age 80 or
on earlier death. The net annual premium is P .
Write down the net future loss random variable, Ln , for this contract in terms of lifetime
0
random variables for [60].
Solution 6.2 From Chapter 4, we know that the present value random variable for the
beneﬁt is Sv T[60] and from Chapter 5 we know that the present value random variable for
the premium income is P amin(K +1,20) , so
¨
[60] Ln = Sv T[60] − P amin(K
¨
0 [60] +1,20) . Since both terms of the random variable depend on the future lifetime of the same life,
[60], they are clearly dependent.
Note that since premiums are payable in advance, premiums payable annually in advance,
ceasing at age 80 or on earlier death means that the last possible premium is payable on
the policyholder’s 79th birthday. No premium is payable on reaching age 80. 6.5. THE EQUIVALENCE PRINCIPLE 225 Given an appropriate survival model together with assumptions about future interest
rates and, for gross premiums, expenses, the insurer can then determine a distribution
for the present value of the future loss. This distribution can be used to ﬁnd a suitable
premium for a given beneﬁt, or an appropriate beneﬁt for a speciﬁed premium. To do
this, the insurer needs to use a premium principle. This is a method of selecting an
appropriate premium using a given loss distribution. We discuss two premium principles
in this chapter. 6.5 The equivalence principle 6.5.1 Net premiums For net premiums, we take into consideration outgo on beneﬁt payments only. Thus,
expenses are not a part of net premium calculation. The beneﬁt may be a death beneﬁt
or a survival beneﬁt or a combination.
We start by stating the equivalence principle. Under the equivalence principle, the net
premium is set such that the expected value of the future loss is zero at the start of the
contract.
That means that
E[Ln ] = 0
0
which implies that
E[PV of beneﬁt outgo − PV of net premium income] = 0.
That is, under the equivalence premium principle,
EPV of beneﬁt outgo = EPV of net premium income (6.1) 226 CHAPTER 6. PREMIUM CALCULATION The equivalence principle is the most common premium principle in traditional life insurance, and will be our default principle – that is, if no other principle is speciﬁed, it is
assumed that the equivalence principle is to be used.
Example 6.3 Consider an endowment insurance with term n years and sum insured S
payable at the earlier of the end of the year of death or at maturity, issued to a select
life aged x. Premiums of amount P are payable annually throughout the term of the
insurance.
Derive expressions in terms of S , P and standard actuarial functions for
(a) the net future loss, Ln ,
0
(b) the mean of Ln ,
0
(c) the variance of Ln , and,
0
(d) the annual net premium for the contract.
Solution 6.3 (a) The the future loss random variable is Ln = Sv min(K[x] +1,n) − P amin(K
¨
0 [x] +1,n) . (b) The mean of Ln is
0
E[Ln ] = S E v min(K[x] +1,n) − P E amin(K
¨
0 [x] +1,n) = SA[x]:n − P a[x]:n .
¨
(c) Expanding the expression above for Ln gives
0
Ln
0 = Sv min(K[x] +1,n) 1 − v min(K[x] +1,n)
−P
d 6.5. THE EQUIVALENCE PRINCIPLE = S+ P
d v min(K[x] +1,n) − 227
P
,
d which isolates the random variable v min(K[x] +1,n) . So the variance is
V [Ln ] =
0 = P
d 2 P
S+
d 2 S+ V v min(K[x] +1,n) 2 A[x]:n − (A[x]:n )2 . (d) Setting the EPVs of the premiums and beneﬁts to be equal gives the net premium
as
P =S A[x]:n
.
a[x]:n
¨ (6.2) Furthermore, using formula (6.2) and recalling that
ax : n =
¨ 1 − Ax:n
,
d we see that the solution can be written as
P =S 1
a[x]:n
¨ −d so that the only actuarial function needed to calculate P for a given value of S is a[x]:n .
¨
Example 6.4 An insurer issues a regular premium deferred annuity contract to a select
life aged x. Premiums are payable monthly throughout the deferred period. The annuity
beneﬁt of X per year is payable monthly in advance from age x + n for the remainder of
the life of (x). 228 CHAPTER 6. PREMIUM CALCULATION (a) Write down the net future loss random variable in terms of lifetime random variables
for [x].
(b) Derive an expression for the monthly net premium.
(c) Assume now that, in addition, the contract oﬀers a death beneﬁt of S payable
immediately on death during the deferred period. Write down the net future loss
random variable for the contract, and derive an expression for the monthly net
premium.
Solution 6.4 (a) Let P denote the monthly net premium, so that the total premium
payable in a year is 12P . Then Ln =
0 0 − 12P a(12)
¨ (12) 1 K[x] + 12 if T[x] ≤ n, (12) X v n a(12)
¨ (12) 1
− 12P an
¨
if T[x] > n. K
+ −n
[x] 12 (b) The EPV of the annuity beneﬁt is
(12) Xv n n p[x] a[x]+n ,
¨
and the EPV of the premium income is
(12) 12P a[x]:n .
¨
By equating these EPVs we obtain the premium equation which gives
(12) P= Xv n n p[x] a[x]+n
¨
(12) 12¨[x]:n
a = n E[x] (12) X a[x]+n
¨
(12) 12¨[x]:n
a . 6.5. THE EQUIVALENCE PRINCIPLE 229 (c) We now have Sv T[x] − 12P a(12)
¨ (12) 1
if T[x] ≤ n, K[x] + 12 Ln =
0 (12) X v n a(12)
¨ (12) 1
− 12P an
¨
if T[x] > n. K
+ −n
[x] 12 The annuity beneﬁt has the same EPV as in part (b); the death beneﬁt during
¯
deferral is a term insurance beneﬁt with EPV S A1x]:n , so the premium equation
[
now gives
(12) P= ¯
S A1x]:n + Xv n n p[x] a[x]+n
¨
[
(12) 12¨[x]:n
a . Example 6.4 shows that the future loss random variable can be quite complicated to
write down. Usually, the premium calculation does not require the identiﬁcation of the
future loss random variable. We may go directly to the equivalence principle, and equate
the EPV of the beneﬁt outgo to the EPV of the net premium income to obtain the net
premium.
Example 6.5 Consider an endowment insurance with sum insured $100 000 issued to a
select life aged 45 with term 20 years under which the death beneﬁt is payable at the
end of the year of death. Using the Standard Select Survival Model, with interest at 5%
per year, calculate the total amount of net premium payable in a year if premiums are
payable (a) annually, (b) quarterly, and (c) monthly, and comment on these values.
Solution 6.5 Let P denote the total amount of premium payable in a year. Then the
(m)
EPV of premium income is P a[45]:20 (where m = 1, 4 or 12) and the EPV of beneﬁt outgo
¨ 230 CHAPTER 6. PREMIUM CALCULATION
m=4
Method
Exact
UDD
W3 (4)
a[45]:20
¨ 12.69859
12.69839
12.69859 m = 12
P (12)
a[45]:20
¨ 3 022.11 12.64512
3 022.16 12.64491
3 022.11 12.64512 P
3 034.89
3 034.94
3 034.89 Table 6.2: Annuity values and premiums. is 100 000A[45]:20 , giving
P= 100 000A[45]:20
(m) a[45]:20
¨ . Using Tables 3.7 and 6.1, we have
a[45]:20 = a[45] −
¨
¨ l65 20
v a[65] = 12.94092.
¨
l[45] From this we get
A[45]:20 = 1 − da[45]:20 = 0.383766.
¨
Hence, for m = 1 the net premium is P = $2 965.52.
(m) The values of a[45]:20 for m = 4 and 12 can either be calculated exactly or from a[45]:20
¨
¨
using one of the approximations in Section 5.11. Notice that the approximation labelled
W3∗ in that section is not available since p[x]−1 is meaningless and so we cannot estimate
µ[45] from the life table tabulated at integer ages. Table 6.2 shows values obtained using
the UDD assumption and Woolhouse’s formula with three terms. The ordering of these
premiums for m = 1, 4, 12 reﬂects the ordering of EPVs of 1/mthly annuities which
we observed in Chapter 5. In this example, Woolhouse’s formula provides a very good
approximation, whilst the UDD assumption gives a reasonably accurate premium. 6.6. GROSS PREMIUM CALCULATION 6.6 231 Gross premium calculation When we calculate a gross premium for an insurance policy or an annuity, we take account
of the expenses the insurer incurs. There are three main types of expense associated with
policies – initial expenses, renewal expenses and termination expenses.
Initial expenses are incurred by the insurer when a policy is issued, and when we calculate
a gross premium, it is conventional to assume that the insurer incurs these expenses
at exactly the same time as the ﬁrst premium is payable, although in practice these
expenses are usually incurred slightly ahead of this date. There are two major types of
initial expenses – commission to agents for selling a policy and underwriting expenses.
Commission is often paid to an agent in the form of a high percentage of the ﬁrst year’s
premiums plus a much lower percentage of subsequent premiums, payable as the premiums
are paid. Underwriting expenses may vary according to the amount of the death beneﬁt.
For example, an insurer is likely to require much more stringent medical tests on an
individual wanting a $10 million death beneﬁt compared with an individual wanting a
$10 000 death beneﬁt.
Renewal expenses are normally incurred by the insurer each time a premium is payable,
and in the case of an annuity, they are normally incurred when an annuity payment is
made. These costs arise in a variety of ways. The processing of renewal and annuity
payments involves staﬀ time and investment expenses. Renewal expenses also cover the
ongoing ﬁxed costs of the insurer such as staﬀ salaries and rent for the insurer’s premises,
as well as speciﬁc costs such as annual statements to policyholders about their policies.
Initial and renewal expenses may be proportional to premiums, proportional to beneﬁts
or may be ‘per policy’, meaning that the amount is ﬁxed for all policies, and is not related
to the size of the contract. Often, per policy renewal costs are assumed to be increasing
at a compound rate over the term of the policy, to approximate the eﬀect of inﬂation.
Termination expenses occur when a policy expires, typically on the death of a policyholder
(or annuitant) or on the maturity date of a term insurance or endowment insurance. 232 CHAPTER 6. PREMIUM CALCULATION Generally these expenses are small, and are largely associated with the paperwork required
to ﬁnalize and pay a claim. In calculating gross premiums, speciﬁc allowance is often not
made for termination expenses. Where allowance is made, it is usually proportional to
the beneﬁt amount.
In practice, allocating the diﬀerent expenses involved in running an insurance company
is a complicated task, and in the examples in this chapter we simply assume that all
expenses are known.
The equivalence principle applied to the gross premiums and beneﬁts states that the EPV
of the gross future loss random variable should be equal to zero. That means that
E[Lg ] = 0,
0
that is
EPV of beneﬁt outgo + EPV of expenses − EPV of gross premium income = 0.
In other words, under the equivalence premium principle, EPV of beneﬁts + EPV of expenses = EPV of gross premium income
(6.3)
We conclude this section with three examples in each of which we apply the equivalence
principle to calculate gross premiums.
Example 6.6 An insurer issues a 25 year annual premium endowment insurance with
sum insured $100 000 to a select life aged 30. The insurer incurs initial expenses of $2 000
plus 50% of the ﬁrst premium, and renewal expenses of 2.5% of each subsequent premium.
The death beneﬁt is payable immediately on death. 6.6. GROSS PREMIUM CALCULATION 233 (a) Write down the gross future loss random variable.
(b) Calculate the gross premium using the Standard Select Survival Model with 5% per
year interest.
Solution 6.6 (a) Let S = 100 000, x = 30, n = 25 and let P denote the annual gross
premium. Then
Lg = S v min(T[x] ,n) + 2000 + 0.475P + 0.025P amin(K
¨
0 [x] +1,n) = S v min(T[x] ,n) + 2000 + 0.475P − 0.975P amin(K
¨ − P amin(K
¨ [x] +1,n) [x] +1,n) . Note that the premium related expenses, of 50% of the ﬁrst premium plus 2.5%
of the second and subsequent premiums are more conveniently written as 2.5% of
all premiums, plus an additional 47.5% of the ﬁrst premium. By expressing the
premium expenses this way, we can simplify the gross future loss random variable,
and the subsequent premium calculation.
(b) We may look separately at the three parts of the gross premium equation of value.
The EPV of premium income is
P a[30]:25 = 14.73113 P.
¨
Note that a[30]:25 can be calculated from Tables 3.7 and 6.1.
¨
The EPV of all expenses is
2 000 + 0.475P + 0.025P a[30]:25
¨
= 2 000 + 0.475P + 0.025 × 14.73113P
= 2 000 + 0.843278P. 234 CHAPTER 6. PREMIUM CALCULATION
The EPV of the death beneﬁt can be found using numerical integration, and we
obtain
¯
100 000A[30]:25 = 100 000 × 0.298732 = 29 873.2 .
Thus, the equivalence principle gives
P= 29 873.2 + 2 000
= $2 295.04 .
14.73113 − 0.843278 Example 6.7 Calculate the monthly gross premium for a 10 year term insurance with
sum insured $50 000 payable immediately on death, issued to a select life aged 55, using
the following basis:
Mortality
Interest
Initial Expenses
Renewal Expenses Standard Select Survival Model
Assume UDD for fractional ages
5% per year
$500 +10% of each monthly premium in the ﬁrst year
1% of each monthly premium in the second and
subsequent policy years Solution 6.7 Let P denote the monthly premium. Then the EPV of premium income is
(12)
12P a[55]:10 . To ﬁnd the EPV of premium related expenses, we can apply the same idea as
¨
in the previous example, noting that initial expenses apply to each premium in the ﬁrst
year. Thus, we can write the EPV of all expenses as
(12) (12) 500 + 0.09 × 12P a[55]:1 + 0.01 × 12P a[55]:10
¨
¨
where the expenses for the ﬁrst year have been split as 9% plus 1%, so that we have 9%
¯1
in the ﬁrst year and 1% every year. The EPV of the insurance beneﬁt is 50 000A[55]:10 6.6. GROSS PREMIUM CALCULATION 235 and so the equivalence principle gives
(12) (12) 12P 0.99¨[55]:10 − 0.09¨[55]:1
a
a ¯1
= 500 + 50 000A[55]:10 . (12)
(12)
¯1
We ﬁnd that a[55]:10 = 7.8341, a[55]:1 = 0.9773 and A[55]:10 = 0.024954, giving P = $18.99
¨
¨
per month. Calculating all the EPVs exactly gives the same answer for the premium to four signiﬁcant
ﬁgures. Example 6.8 Calculate the gross single premium for a deferred annuity of $80 000 per
year payable monthly in advance, issued to a select life now aged 50 with the ﬁrst annuity
payment on the life’s 65th birthday. Allow for initial expenses of $1 000, and renewal
expenses on each anniversary of the issue date, provided that the policyholder is alive.
Assume that the renewal expense will be $20 on the ﬁrst anniversary of the issue date,
and that expenses will increase with inﬂation from that date at the compound rate of 1%
per year. Assume the standard survival model with interest at 5% per year.
Solution 6.8 The single premium is equal to the EPV of the deferred annuity plus the
EPV of expenses. The renewal expense on the tth policy anniversary is 20 (1.01t−1 ) for
t = 1, 2, 3, . . . so that the EPV of renewal expenses is
20 ∞ t−1 1.01 t v t p[50] t=1 20
=
1.01 ∞ t t 1.01 v t p[50] t=1 20
=
1.01 ∞ t
vj t p[50] = t=1 20
(¨[50]j − 1)
a
1.01 where the subscript j indicates that the calculation is at rate of interest j where
(12)
1.01v = 1/(1 + j ), that is j = 0.0396. The EPV of the deferred annuity is 80 000 15 a[50] ,
¨
so the single premium is
1 000 + 20
(¨[50]j − 1) + 80 000
a
1.01 As a[50]j = 19.4550 and
¨ (12)
¨
15 a[50] (12)
¨
15 a[50] . = 6.04129, the single premium is $484 669. 236 CHAPTER 6. PREMIUM CALCULATION We end this section with a comment on the premiums calculated in Examples 6.6 and 6.7.
In Example 6.6, the annual premium is $ 2 295.04 and the expenses at time 0 are $2 000
plus 50% of the ﬁrst premium, a total of $3 146.75, which exceeds the ﬁrst premium.
Similarly, in Example 6.7 the total premium in the ﬁrst year is $227.88 and the total
expenses in the ﬁrst year are $500 plus 10% of premiums in the ﬁrst year. In each
case, the premium income in the ﬁrst year is insuﬃcient to cover expenses in the ﬁrst
year. This situation is common in practice, especially when initial commission to agents
is high, and is referred to as new business strain. A consequence of new business
strain is that an insurer needs to have funds available in order to sell policies. From
time to time insurers get into ﬁnancial diﬃculties through pursuing an aggressive growth
strategy without suﬃcient capital to support the new business strain. Essentially, the
insurer borrows from shareholder (or participating policyholder) funds in order to write
new business. These early expenses are gradually paid oﬀ by the expense loadings in
future premiums. The part of the premiums that funds the initial expenses is called the
deferred acquisition cost. 6.7 Proﬁt The equivalence principle does not allow explicitly for a loading for proﬁt. Since writing
business generally involves a loan from shareholder or participating policyholder funds, it
is necessary for the business to be suﬃciently proﬁtable for the payment of a reasonable
rate of return – in other words, to make a proﬁt. In traditional insurance, we often load
for proﬁt implicitly, by margins in the valuation assumptions. For example, if we expect
to earn an interest rate of 6% per year on assets, we might assume only 5% per year in the
premium basis. The extra income from the invested premiums will contribute to proﬁt.
In participating business, much of the proﬁt will be distributed to the policyholders in
the form of cash dividends or bonus. Some will be paid as dividends to shareholders, if 6.7. PROFIT 237 the company is proprietary.
We may also use margins in the mortality assumptions. For a term insurance, we might
use a slightly higher overall mortality rate than we expect. For an annuity, we might use
a slightly lower rate.
More modern premium setting approaches, which use projected cash ﬂows, are presented
in Chapter 11, where more explicit allowance for proﬁt is incorporated in the methodology.
Each individual policy sold will generate a proﬁt or a loss. Although we calculate a
premium assuming a given survival model, for each individual policy the experienced
mortality rate in any year can take only the values 0 or 1. So, while the expected outcome
under the equivalence principle is zero proﬁt (assuming no margins), the actual outcome
for each individual policy will either be a proﬁt or a loss. For the actual proﬁt from a group
of policies to be reliably close to the expected proﬁt, we need to sell a large number of
individual contracts, whose future lifetimes can be regarded as statistically independent,
so that the losses and proﬁts from individual policies are combined.
As a simple illustration of this, consider a life who purchases a one year term insurance
with sum insured $1 000 payable at the end of the year of death. Let us suppose that the
life is subject to a mortality of rate of 0.01 over the year, that the insurer can earn interest
at 5% per year, and that there are no expenses. Then, using the equivalence principle,
the premium is
P = 1 000 × 0.01/1.05 = 9.52.
The future loss random variable is
Ln =
0 1 000v − P = 942.86 if Tx ≤ 1,
−P = −9.52
if Tx > 1, with probability 0.01,
with probability 0.99. The expected loss is 0.01 × 942.86 + 0.99 × (−9.52) = 0, as required by the equivalence
principle, but the probability of proﬁt is 0.99, and the probability of loss is 0.01. The
balance arises because the proﬁt, if the policyholder survives the year, is small, and the 238 CHAPTER 6. PREMIUM CALCULATION loss, if the policyholder dies, is large. Using the equivalence principle, so that the expected
future loss is zero, makes sense only if the insurer issues a large number of policies, so that
the overall proportion of policies becoming claims will be close to the assumed proportion
of 0.01.
Now suppose the insurer were to issue 100 such policies to independent lives. The insurer
would expect to make a (small) proﬁt on 99 of them. If the outcome from this portfolio
is that all lives survive for the year, then the insurer makes a proﬁt. If one life dies, there
is no proﬁt or loss. If more than one life dies, there will be a loss on the portfolio. Let D
denote the number of deaths in the portfolio, so that D ∼ B (100, 0.01). The probability
that the proﬁt on the whole portfolio is greater than or equal to zero is
Pr[D ≤ 1] = 0.73576
compared with 99% for the individual contract. In fact, as the number of policies issued
increases, the probability of proﬁt will tend, monotonically, to 0.5. On the other hand,
while the probability of loss is increasing with the portfolio size, the probability of very
large aggregate losses (relative, say, to total premiums) is much smaller for a large portfolio, since there is a balancing eﬀect from diversiﬁcation of the risk amongst the large
group of policies.
Let us now consider a whole life insurance policy with sum insured S payable at the end
of the year of death, initial expenses of I , renewal expenses of e associated with each
premium payment (including the ﬁrst) issued to a select life aged x by annual premiums
of P . For this policy
Lg = Sv K[x] +1 + I + e aK[x] +1 − P aK[x] +1 ,
¨
¨
0
where K[x] denotes the curtate future lifetime of [x].
If death occurs shortly after the policy is issued, so that only a few premiums are paid,
the insurer will make a loss, and, conversely, if the policyholder lives to a ripe old age,
we would expect that the insurer would make a proﬁt as the policyholder will have paid
a large number of premiums, and there will have been plenty of time for the premiums 6.7. PROFIT 239 to accumulate interest. We can use the future loss random variable to ﬁnd the minimum
future lifetime for the policyholder in order that the insurer makes a proﬁt on this policy.
The probability that the insurer makes a proﬁt on the policy, Pr[Lg < 0], is given by
0
¨
Pr[Lg < 0] = Pr S v K[x] +1 + I + e aK[x] +1 − P aK[x] +1 ≤ 0 .
¨
0
Rearranging and replacing aK[x] +1 with (1 − v K[x] +1 )/d, gives
¨
P −e
d Pr[Lg < 0] = Pr v K[x]+1 ≤
0 S+ = Pr K[x]+1 > 1
log
δ −I P −e
d P − e + Sd
P − e − Id . (6.4) Suppose we denote the right hand side term of the inequality in equation (6.4) by τ , so
that the contract generates a proﬁt for the insurer if K[x] + 1 ≥ τ . Generally, τ is not an
integer. Thus, if τ denotes the integer part of τ , then the insurer makes a proﬁt if the
life survives at least τ years, the probability of which is τ p[x] .
Let us continue this illustration by assuming that x = 30, S = 100 000, I = 1 000, and
e = 50. Then we ﬁnd that P = 498.45, and from equation (6.4) we ﬁnd that there is
a proﬁt if K[30] + 1 > 52.57. Thus, there is a proﬁt if the life survives for 52 years, the
probability of which is 52 p[30] = 0.70704.
Figure 6.1 shows the proﬁts that arise should death occur in a given year, in terms of values
at the end of that year. We see that large losses occur in the early years of the policy,
and even larger proﬁts occur if the policyholder dies at an advanced age. The probability
of realizing either a large loss or proﬁt is small. For example, if the policyholder dies in
the ﬁrst policy year, the loss to the insurer is 100 579, and the probability of this loss is
q[30] = 0.00027. Similarly, a proﬁt of 308 070 arises if the death beneﬁt is payable at time
80, and the probability of this is 79 q[30] = 0.00023. It is important to appreciate that the
premium has been calculated in such a way that the EPV of the proﬁt from the policy is 240 CHAPTER 6. PREMIUM CALCULATION 600000 500000 400000 Profit 300000 200000 100000 0
0 10 20 30 40 50 60 70 80 90 100000 200000
Year Figure 6.1: Proﬁt at yearend if death occurs in that year for the whole life insurance
described in Section 6.7.
zero
Example 6.9 A life oﬃce is about to issue a 25 year endowment insurance with a basic
sum insured of $250 000 to a select life aged exactly 30. Premiums are payable annually
throughout the term of the policy. Initial expenses are $1 200 plus 40% of the ﬁrst premium
and renewal expenses are 1% of the second and subsequent premiums. The oﬃce allows
for a compound reversionary bonus of 2.5% of the basic sum insured, vesting on each
policy anniversary (including the last). The death beneﬁt is payable at the end of the
year of death. Assume the Standard Select Survival Model with interest at 5% per year.
(a) Derive an expression for the future loss random variable, Lg , for this policy.
0
(b) Calculate the annual premium for this policy.
(c) Let L0 (k ) denote the present value of the loss on the policy given that K[30] = k for 6.7. PROFIT 241 k ≤ 24 and let L0 (25) denote the present value of the loss on the policy given that
the policyholder survives to age 55. Calculate L0 (k ) for k = 0, 1, . . . , 25.
(d) Calculate the probability that the insurer makes a proﬁt on this policy.
(e) Calculate V[Lg ].
0
Solution 6.9 (a) First, we note that if the policyholder’s curtate future lifetime, K[30] ,
is k years where k = 0, 1, 2, . . . , 24, then the number of bonus additions is k , the
death beneﬁt is payable k + 1 years from issue, and hence the present value of the
death beneﬁt is 250 000(1.025)K[30] v K[30] +1 . However, if the policyholder survives for
25 years, then 25 bonuses vest. Thus, if P denotes the annual premium,
Lg = 250 000(1.025min(K[30] , 25) )v min(K[30] +1, 25)
0
+1 200 + 0.39P − 0.99P amin(K
¨ [30] +1, 25) . (b) The EPV of the premiums, less premium expenses, is
0.99P a[30]:25 = 14.5838P.
¨
As the death beneﬁt is $250 000(1.025t ) if the policyholder dies in the tth policy
year, the EPV of the death beneﬁt is
24 250 000
t=0 v t+1 t q[30] (1.025t ) = 250 000 1
A1
1.025 [30]:25 j where 1 + j = (1 + i)/(1.025), so that j = 0.02439.
The EPV of the survival beneﬁt is
250 000v 25 25 p[30] 1.02525 = 134 295.43, and the EPV of the remaining expenses is
1 200 + 0.39P. = 3099.37 242 CHAPTER 6. PREMIUM CALCULATION
Value of K[30] ,
k
0
1
.
.
. PV of loss,
L0 (k )
233 437
218 561
.
.
. 23
24
≥ 25 1 737
−4 517
−1 179 Table 6.3: Values of the future loss random variable for Example 6.9. Hence, equating the EPV of premium income with the EPV of beneﬁts plus expenses
we ﬁnd that P = $9 764.44.
(c) Given that K[30] = k , where k = 0, 1, . . . , 24, the present value of the loss is the
present value of the death beneﬁt payable at time k + 1 less the present value of
k + 1 premiums plus the present value of expenses. Hence
L0 (k ) = 250 000(1.025k ) v k+1 + 1 200 + 0.39P − 0.99P ak+1 .
¨
If the policyholder survives to age 55, there is one extra bonus payment, and the
present value of the future loss is
L0 (25) = 250 000(1.02525 ) v 25 + 1 200 + 0.39P − 0.99P a25 .
¨
Some values of the present value of the future loss are shown in Table 6.3.
(d) The full set of values for the present value of the future loss shows that there is a
proﬁt if and only if the policyholder survives 24 years and pays the premium at the
start of the 25th policy year. Hence the probability of a proﬁt is 24 p[30] = 0.98297.
Note that this probability is based on the assumption that future expenses and
future interest rates are known and will be as in the premium basis. 6.7. PROFIT 243 (e) From the full set of values for L0 (k ) we can calculate
24 E[(Lg )2 ]
0 =
k=0 (L0 (k ))2 k q[30] + (L0 (25))2 25 p[30] = 12 115.552 which is equal to the variance as E[Lg ] = 0.
0 Generally speaking, for an insurance policy, the longer a life survives, the greater is
the proﬁt to the insurer, as illustrated in Figure 6.1. However, the converse is true for
annuities, as the following example illustrates.
Example 6.10 An insurance company is about to issue a single premium deferred annuity to a select life aged 55. The ﬁrst annuity payment will take place 10 years from
issue, and payments will be annual. The ﬁrst annuity payment will be $50 000, and each
subsequent payment will be 3% greater than the previous payment. Ignoring expenses,
and using the Standard Select Survival Model with interest at 5% per year, calculate
(a) the single premium,
(b) the probability the insurance company makes a proﬁt from this policy, and
(c) the probability that the present value of the loss exceeds $100 000.
Solution 6.10 (a) Let P denote the single premium. Then P = 50 000 ∞ v t (1.03t−10 ) t p[55] = $546 812. t=10 (b) Let L0 (k ) denote the present value of the loss given that K[55] = k , k = 0, 1, . . .
Then
L0 (k ) = −P
for k = 0, 1, . . . , 9,
10
−P + 50 000v ak−9 j for k = 10, 11, . . . ,
¨ (6.5) 244 CHAPTER 6. PREMIUM CALCULATION
where j = 1.05/1.03 − 1 = 0.019417.
Since ak−9 j is an increasing function of k , formula (6.5) shows that L0 (k ) is an
¨
increasing function of k for k ≥ 10. The present value of the proﬁt will be positive
if L0 (k ) < 0. Using formula (6.5), this condition can be expressed as
−P + 50 000 v 10 ak−9 j < 0,
¨
or, equivalently,
ak−9 j < 1.0510 P/50 000.
¨
Writing ak−9 j = (1 − v k−9 )/dj where dj = j/(1 + j ), this condition becomes
¨
k
vj −9 > 1 − dj 1.0510 P/50 000 , and as vj = exp{−δj } where δj = log(1 + j ) this gives
k − 9 < − log 1 − dj 1.0510 P/50 000 /δj .
Hence we ﬁnd that L0 (k ) < 0 if k < 30.55, and so there will be a proﬁt if the
policyholder dies before age 86. The probability of this is 1 − 31 p[55] = 0.41051.
(c) The present value of the loss will exceed 100 000 if
−P + 50 000v 10 ak−9 j > 100 000 ,
¨
and following through exactly the same arguments as in part (b) we ﬁnd that
L0 (k ) > 100 000 if k > 35.68. Hence the present value of the loss will be greater
than $100 000 if the policyholder survives to age 91, and the probability of this is
36 p[55] = 0.38462.
Figure 6.2 shows L0 (k ) for k = 1, 2, . . . , 50. We can see that the loss is constant for the
ﬁrst 10 years at −P and then increases due to annuity payments. In contrast to Figure
6.1, longevity results in large losses to the insurer. We can also clearly see from this ﬁgure
that the loss is negative if k takes a value less than 31, conﬁrming our answer to part (b). 6.8. THE PORTFOLIO PERCENTILE PREMIUM PRINCIPLE 245 400000 300000 200000 Present value of loss 100000 0
0
100000 5 10 15 20 25 30 35 40 45 50 Year of death, k 200000 300000 400000 500000 600000 Figure 6.2: Present value of loss from Example 6.10. 6.8 The portfolio percentile premium principle The portfolio percentile premium principle is an alternative to the equivalence premium
principle. We assume a large portfolio of identical and independent policies. By ‘identical’
we mean that the policies have the same premium, beneﬁts, term, and so on, and that
the policyholders are all subject to the same survival model. By ‘independent’ we mean
that the policyholders are independent of each other with respect to mortality.
Suppose we know the sum insured for these policies, and wish to ﬁnd an appropriate
premium. As the policies are identical, each policy has the same future loss random
variable. Let N denote the number of policies in the portfolio and let L0,i represent the
future loss random variable for the ith policy in the portfolio, i = 1, 2, 3, . . . , N . The total 246 CHAPTER 6. PREMIUM CALCULATION future loss in the portfolio is L, say, where
N L= N L0,i ; E[L] = i=1 N E[L0,i ] = N E[L0,1 ]; V[L] = i=1 V[L0,i ] = N V[L0,1 ].
i=1 (Note that as {L0,i }N are identically distributed, the mean and variance of each L0,i are
i=1
equal to the mean and variance of L0,1 .)
The portfolio percentile premium principle sets a premium so that there is a speciﬁed
probability, say α, that the total future loss is negative. That is, P is set such that
Pr[L < 0] = α.
Now, if N is suﬃciently large (say, greater than around 30), the central limit theorem
tells us that L is approximately normally distributed, with mean E[L] = N E[L0,1 ] and
variance V[L] = N V[L0,1 ]. In this case, the portfolio percentile principle premium can be
calculated from
Pr[L < 0] = Pr L − E[L]
V[L] < −E[L] V[L] =Φ −E[L] V[L] = α, which implies that
E[L]
V[L] = −Φ−1 (α) where Φ is the cumulative distribution function of the standardized normal distribution.
Our aim is to calculate P , but P does not appear explicitly in either of the last two
equations. However, as illustrated in the next example, both the mean and variance of L
are functions of P .
Example 6.11 An insurer issues whole life insurance policies to select lives aged 30.
The sum insured of $100 000 is paid at the end of the month of death and level monthly
premiums are payable throughout the term of the policy. Initial expenses, incurred at the 6.8. THE PORTFOLIO PERCENTILE PREMIUM PRINCIPLE 247 issue of the policy, are 15% of the total of the ﬁrst year’s premiums. Renewal expenses
are 4% of every premium, including those in the ﬁrst year.
Assume the Standard Select Survival Model with interest at 5% per year.
(a) Calculate the monthly premium using the equivalence principle.
(b) Calculate the monthly premium using the portfolio percentile principle, such that
the probability that the future loss on the portfolio is negative is 95%. Assume a
portfolio of 10 000 identical, independent policies.
Solution 6.11 (a) Let P be the monthly premium. Then the EPV of premiums is
(12) 12P a[30] = 227.065P.
¨
The EPV of beneﬁts is
(12) 100 000A[30] = 7 866.18,
and the EPV of expenses is
(12) 0.15 × 12P + 0.04 × 12P a[30] = 10.8826P.
¨
Equating the EPV of premiums with the EPVs of beneﬁts and expenses gives the
equivalence principle premium as $36.39 per month.
(b) The future loss random variable for the ith policy is
(12) 1 L0,i = 100 000v K[30] + 12 + 0.15 × 12P − 0.96 × 12P a
¨ (12)
(12) 1
K[30] + 12 . and its expected value can be calculated using the solution to part (a) as
E[L0,i ] = 7 866.14 − 216.18P. 248 CHAPTER 6. PREMIUM CALCULATION
To ﬁnd V[L0,i ] we can rewrite L0,i as
L0,i = 100 000 + 0.96 × 12P
d(12) (12) 1 v K[30] + 12 + 0.15 × 12P − 0.96 × 12P
d(12) so that
V[L0,i ] = 0.96 × 12P
100 000 +
d(12) 2
2 (12) (12) A[30] − (A[30] )2 = (100 000 + 236.59P )2 (0.0053515)
giving
V[L0,i ] = (100 000 + 236.59P ) (0.073154).
The future loss random variable for the portfolio of policies is L = 10 000
i=1 L0,i , so E[L] = 10 000(7866.18 − 216.18P )
and
V[L] = 10 000 (100 000 + 236.59P )2 (0.0053515).
Using the normal approximation to the distribution of L, we set P such that
Pr[L < 0] = Φ −E[L] V[L] =Φ 10 000(216.18P − 7 866.18)
100 (100 000 + 236.59P ) (0.073154) For the standardized normal distribution, Φ(1.645) = 0.95, so we set
100(216.18P − 7 866.18)
= 1.645
(100 000 + 236.59P ) (0.073154)
which gives P = $36.99. = 0.95. 6.9. EXTRA RISKS 249
n
P
1 000 38.31
2 000 37.74
5 000 37.24
10 000 36.99
20 000 36.81 Table 6.4: Premiums according to portfolio size. Note that the solution to part (b) above depends on the number of policies in the portfolio
(10 000) and the level of probability we set for the future loss being negative (0.95). If
the portfolio had n policies instead of 10 000, then the equation we would have to solve
for the premium, P , is
√ n(216.18P − 7 866.18)
= 1.645.
(100 000 + 236.59P ) (0.073154) (6.6) Table 6.4 shows some values of P for diﬀerent values of n. We note that P decreases as
n increases. In fact, as n → ∞, P → $36.39, which is the equivalence principle premium.
The reason for this is that as n → ∞ the insurer diversiﬁes the mortality risk. We discuss
diversiﬁcation of risk further in Chapter 10. 6.9 Extra risks As we discussed in Section 1.3.5, when an individual wishes to eﬀect a life insurance policy,
underwriting takes place. If underwriting determines that an individual should not be
oﬀered insurance at standard rates, the individual might still be oﬀered insurance, but
above standard rates. There are diﬀerent ways in which we can model the extra mortality
risk in a premium calculation. 250 CHAPTER 6. PREMIUM CALCULATION 6.9.1 Age rating One reason why an individual might not be oﬀered insurance at standard rates is that
the individual suﬀers from a medical condition. In such circumstances we refer to the
individual as an impaired life, and the insurer may compensate for this extra risk by
treating the individual as being older. For example, an impaired life aged 40 might be
asked to pay the same premium paid by a nonimpaired life aged 45. This approach
to modelling extra risk involves no new ideas in premium calculation – for example, we
could apply the equivalence principle in our calculation, and we would simply change the
policyholder’s age. This is referred to as age rating. 6.9.2 Constant addition to µx Individuals can also be deemed to be ineligible for standard rates if they regularly participate in hazardous pursuits, for example parachuting. For such individuals the extra
risk is largely independent of age, and so we could model this extra risk by adding a constant to the force of mortality – just as Makeham extended Gompertz’ law of mortality.
The application of this approach leads to some computational shortcuts for the following
reason. We are modelling the force of mortality as
µ[x]+s = µ[x]+s + φ
where functions with the superscript relate to the impaired life, functions without this
superscript relate to the standard survival model and φ is the constant addition to the
force of mortality. Then
t
t p [x] = exp − 0 t µ[x]+s ds = exp − µ[x]+s + φ ds = e−φt t p[x] . 0 This formula is useful for computing the EPV of a survival beneﬁt since
e−δt t p[x] = e−(δ+φ)t t p[x] , 6.9. EXTRA RISKS 251 so that, for example,
a[x]:n =
¨ n−1 e−δt t p[x] = t=0 n−1 e−(δ+φ)t t p[x] = a[x]:n j ,
¨ (6.7) t=0 where j denotes calculation at interest rate j = eφ+δ − 1. Note that a[x]:n j is calculated
¨
using rate of interest j and the standard survival model.
Now suppose that the impaired life has curtate future lifetime K[x] . We know that
a[x]:n = E amin(K
¨
¨ [x] +1,n) = 1 − A[x]:n
1 − E[v min(K[x] +1,n) ]
=
.
d
d So
A[x]:n = 1 − d a[x]:n = 1 − d a[x]:n j .
¨
¨ (6.8) It is important to note here that for the insurance beneﬁt we cannot just change the
interest rate. In formula (6.8), the annuity is evaluated at rate j , but the function d uses
the original rate of interest, that is d = i/(1 + i). Generally, when using the constant
addition to the force of mortality, it is simplest to calculate the annuity function ﬁrst,
using a simple adjustment of interest, then use formula (6.8) for any insurance factors.
Note that the standard discount function n Ex = v n n px is a survival beneﬁt value, and so
can be calculated for the extra risk by an interest adjustment, so that
n Ex n
= vj n px . Example 6.12 Calculate the annual premium for a 20 year endowment insurance with
sum insured $200 000 issued to a life aged 30 whose force of mortality at age 30 + s is
given by µ[30]+s + 0.01. Allow for initial expenses of $2 000 plus 40% of the ﬁrst premium,
and renewal expenses of 2% of the second and subsequent premiums. Use the Standard
Select Survival Model with interest at 5% per year. 252 CHAPTER 6. PREMIUM CALCULATION Solution 6.12 Let P denote the annual premium. Then by applying formula (6.7), the
EPV of premium income is
19 v t t p[30] = P a[30]:20 j
¨ P
t=0 where j = 1.05e0.01 − 1 = 0.06055. Similarly, the EPV of expenses is
2 000 + 0.38P + 0.02P a[30]:20 j .
¨
The EPV of the beneﬁt is 200 000A[30]:20 , where the dash denotes extra mortality and
the interest rate is i = 0.05. Using formula (6.8)
A[30]:20 = 1 − d a[30]:20 j .
¨
As a[30]:20 j = 12.072 and d = 0.05/1.05, we ﬁnd that A[30]:20 = 0.425158 and hence we
¨
ﬁnd that P = $7 600.84. 6.9.3 Constant multiple of mortality rates A third method of allowing for extra mortality is to assume that lives are subject to
mortality rates that are higher than the standard lives’ mortality rates. For example, we
might set q[x]+t = 1.1q[x]+t where the superscript again denotes extra mortality risk. With
such an approach we can calculate the probability of surviving one year from any integer
age, and hence we can calculate the probability of surviving an integer number of years.
A computational disadvantage of this approach is that we have to apply approximations
in calculating EPVs if payments are at other than annual intervals. Generally, this form
of extra risk would be handled by recalculating the required functions in a spreadsheet.
Example 6.13 Calculate the monthly premium for a 10 year term insurance with sum
insured $100 000 payable immediately on death, issued to a life aged 50. Assume that each
year throughout the 10 year term the life is subject to mortality rates that are 10% higher 6.9. EXTRA RISKS 253 than for a standard life of the same age. Allow for initial expenses of $1 000 plus 50%
of the ﬁrst monthly premium and renewal expenses of 3% of the second and subsequent
monthly premiums. Use the UDD assumption where appropriate, and use the Standard
Select Survival Model with interest at 5% per year.
Solution 6.13 Let P denote the total premium per year. Then the EPV of premium
(12)
(12)
income is P a50:10 and, assuming UDD, we compute a50:10 as
¨
¨
(12) a50:10 = α(12)¨50:10 − β (12) 1 − v 10
¨
a 10 p50 , where
α(12) = id
i(12) d(12) = 1.0002 and
β (12) = i − i(12)
= 0.4665.
i(12) d(12) As the initial expenses are 1 000 plus 50% of the ﬁrst premium, which is
write the EPV of expenses as
1 000 + 1
P,
12 we can 0.47P
(12)
+ 0.03P a50:10 .
¨
12 ¯1
Finally, the EPV of the death beneﬁt is 100 000(A50:10 ) and, using UDD, we can compute
this as
¯1
(A50:10 ) = i
1
(A50:10 )
δ = i
(A50:10 ) − v 10 10 p50
δ = i
1 − da50:10 − v 10 10 p50 .
¨
δ 254 CHAPTER 6. PREMIUM CALCULATION
(1)
t
0
1
2
3
4
5
6
7
8
9 (2)
t p50
1.0000
0.9989
0.9975
0.9959
0.9941
0.9921
0.9899
0.9875
0.9849
0.9819 (3)
t q50
0.0011
0.0014
0.0016
0.0018
0.0020
0.0022
0.0024
0.0027
0.0030
0.0033 (4)
vt
1.0000
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
0.7107
0.6768
0.6446 (5)
(6)
t+1
v
(2) × (4)
0.9524
1.0000
0.9070
0.9513
0.8638
0.9047
0.8227
0.8603
0.7835
0.8178
0.7462
0.7774
0.7107
0.7387
0.6768
0.7018
0.6446
0.6666
0.6139
0.6329
Total
8.0516 (7)
(3) × (5)
0.0011
0.0013
0.0014
0.0015
0.0015
0.0016
0.0017
0.0018
0.0019
0.0020
0.0158 Table 6.5: Spreadsheet calculations for Example 6.13. The formula for a50:10 is
¨
9 v t t p50 a50:10 =
¨
t=0 where
t p50 = t−1
r =0 (1 − 1.1q[50]+r ) . (6.9) (We have written q[50]+r in formula (6.9) as standard lives are subject to select mortality.)
(12)
¯1
Hence a50:10 = 8.0516, a50:10 = 7.8669 and (A50:10 ) = 0.01621, which give P = $345.18
¨
¨
and so the monthly premium is $28.76.
Table 6.5 shows how we could set out a spreadsheet to perform calculations. Column (2)
was created from the original mortality rates using formula (6.9), with column (3) being 6.10. NOTES AND FURTHER READING 255 calculated as
t q50 = t p50 (1 − 1.1q50+t ) . The total in column (6) gives a50:10 while the total in column (7) gives the value for
¨
1
¯1
(A50:10 ) . (Note that this must then by multiplied by i/δ to get (A50:10 ) .) 6.10 Notes and further reading The equivalence principle is the traditional approach to premium calculation, and we
apply it again in Chapter 7 when we consider the possibility that a policy may terminate
for reasons other than death. However, other approaches to premium calculation are
possible. We have seen one in Section 6.8, where we computed premiums by the portfolio
percentile principle.
A modiﬁcation of the equivalence principle which builds an element of proﬁt into a premium calculation is to select a proﬁt target amount for each policy, Π, say, and set the
premium to be the smallest possible such that E[L0 ] ≤ Π. Under this method of calculation we eﬀectively set a level for the expected present value of future proﬁt from the
policy and calculate the premium by treating this amount as an additional cost at the
issue date which will be met by future premium income.
Besides the premium principles discussed in this chapter, there is one further important
method of calculating premiums. This is proﬁt testing, which is the subject of Chapter
11.
The international actuarial notation for premiums may be found in Bowers et al (1997).
We have omitted it in this work because we ﬁnd it has no particular beneﬁt in practice. 256 6.11 CHAPTER 6. PREMIUM CALCULATION Exercises When a calculation is required in the following exercises, unless otherwise stated you
should assume that mortality follows the Standard Select Survival Model as speciﬁed in
Example 3.13 in Section 3.9, that interest is at 5% per annum eﬀective, and that the
equivalence principle is used for the calculation of premiums. Exercise 6.1 You are given the following extract from a select life table with 4 year
select period. A select individual aged 41 purchased a 3 year term insurance with a net
premium of $350 payable annually. The sum insured is paid at the end of the year of
death.
[x ]
[40]
[41]
[42] l[x]
100 000
99 802
99 597 l[x]+1
l[x]+2
l[x]+3
lx+4
x+4
99 899 99 724 99 520 99 288
44
99 689 99 502 99 283 99 033
45
99 471 99 628 99 030 98 752
46 Use an eﬀective rate of interest of 6% per year to calculate
(a) the sum insured, assuming the equivalence principle,
(b) the standard deviation of L0 , and
(c) Pr[L0 > 0]. Exercise 6.2 Consider a 10 year annual premium term insurance issued to a select life
aged 50, with sum insured $100 000 payable at the end of the year of death. 6.11. EXERCISES 257 (a) Write down an expression for the net lossatissue random variable.
(b) Calculate the net annual premium. Exercise 6.3 Consider a 20 year annual premium endowment insurance with sum insured
$100 000 issued to a select life aged 35. Assume initial expenses of 3% of the basic sum
insured and 20% of the ﬁrst premium, and renewal expenses of 3% of the second and
subsequent premiums. Assume that the death beneﬁt is payable at the end of the year of
death.
(a) Write down an expression for the gross lossatissue random variable.
(b) Calculate the gross annual premium.
(c) Calculate the standard deviation of the gross lossatissue random variable.
(d) Calculate the probability that the contract makes a proﬁt. Exercise 6.4 Consider an annual premium withproﬁt whole life insurance issued to a
select life aged exactly 40. The basic sum insured is $200 000 payable at the end of the
month of death, and the premium term is 25 years. Assume a compound reversionary
bonus of 1.5% per year, vesting on each policy anniversary, initial expenses of 60% of the
annual premium, renewal expenses of 2.5% of all premiums after the ﬁrst, plus per policy
expenses (incurred when a premium is payable) of $5 at the beginning of the ﬁrst year,
increasing by 6% per year compound at the beginning of each subsequent year.
Calculate the annual premium. 258 CHAPTER 6. PREMIUM CALCULATION Exercise 6.5 A select life aged exactly 40 has purchased a deferred annuity policy. Under
the terms of the policy, the annuity payments will commence 20 years from the issue date
and will be payable at annual intervals thereafter. The initial annuity payment will be
$50 000, and each subsequent payment will be 2% greater than the previous one. The
policy has monthly premiums, payable for at most 20 years. Calculate the gross monthly
premium allowing for initial expenses of 2.5% of the ﬁrst annuity payment and 20% of
the ﬁrst premium, renewal expenses of 5% of the second and subsequent premiums, and
terminal expenses, incurred at the end of the year of death, of $20 inﬂated from the issue
date assuming an inﬂation rate of 3% per year. Exercise 6.6 Find the annual premium for a 20 year term insurance with sum insured
$100 000 payable at the end of the year of death, issued to a select life aged 40 with
premiums payable for at most 10 years, with expenses, which are incurred at the beginning
of each policy year, as follows:
Year 1
% of premium Constant
Taxes
4%
0
Sales commission
25%
0
Policy maintenance
0%
10 Years 2+
% of premium Constant
4%
0
5%
0
0%
5 Exercise 6.7 A life insurer is about to issue a 30 year deferred annuitydue with annual
payments of $20 000 to a select life aged 35. The policy has a single premium which
is refunded without interest at the end of the year of death if death occurs during the
deferred period.
(a) Calculate the single premium for this annuity. 6.11. EXERCISES 259 (b) The insurer oﬀers an option that if the policyholder dies before the total annuity
payments exceed the single premium, then the balance will be paid as a death
beneﬁt, at the end of the year of death. Calculate the revised premium.
This is called a Cash Refund Payout Option. Exercise 6.8 A whole life insurance with unit sum insured payable at the end of the
year of death with a level annual premium is issued to (x). Let L0 be the net future loss
random variable with the premium determined by the equivalence principle. Let L∗ be
0
∗
the net future loss random variable if the premium is determined such that E [L0 ] = −0.5.
Given V[L0 ] = 0.75, calculate V[L∗ ].
0 Exercise 6.9 Calculate both the net and gross premiums for a whole life insurance issued
to a select life aged 40. The sum insured is $100 000 on death during the ﬁrst 20 years, and
$20 000 thereafter, and is payable immediately on death. Premiums are payable annually
in advance for a maximum of 20 years.
Use the following basis:
Mortality:
ultimate rates
select rates
Interest:
Premium expenses:
Other expenses: Makeham’s law with A = 0.0001, B = 0.00035, c = 1.075
2 year select period, q[x] = 0.75qx , q[x]+1 = 0.9qx+1
6% per year eﬀective
30% of the ﬁrst year’s premium
plus 3% of all premiums after the ﬁrst year
On each premium date an additional expense
starting at $10 and increasing at a compound rate of 3% per year 260 CHAPTER 6. PREMIUM CALCULATION Exercise 6.10 A life insurance company issues a 10 year term insurance policy to a
life aged 50, with sum insured $100 000. Level premiums are paid monthly in advance
throughout the term. Calculate the gross premium allowing for initial expenses of $100
plus 20% of each premium payment in the ﬁrst year, renewal expenses of 5% of all premiums after the ﬁrst year, and claim expenses of $250. Assume the sum insured and claim
expenses are paid 1 month after the date of death, and use claim acceleration. Exercise 6.11 For a special whole life insurance on (55), you are given:
• initial annual premiums are level for 10 years; thereafter annual premiums equal
onehalf of initial annual premiums,
• the death beneﬁt is $100 000 during the ﬁrst 10 years of the contract, is $50,000
thereafter, and is payable at the end of the year of death, and
• expenses are 25% of the ﬁrst year’s premium plus 3% of all subsequent premiums.
Calculate the initial annual gross premium. Exercise 6.12 For a whole life insurance with sum insured $150 000 paid at the end of
the year of death, issued to (x), you are given:
(i) 2 Ax = 0.0143,
(ii) Ax = 0.0653, and
(iii) the annual premium is determined using the equivalence principle. 6.11. EXERCISES 261 Calculate the standard deviation of Ln .
0 Exercise 6.13 A life is subject to extra risk that is modelled by a constant addition to
the force of mortality, so that, if the extra risk functions are denoted by , µx = µx + φ.
Show that at rate of interest i,
¯
¯
Ax = Axj + φaxj ,
¯
where j is a rate of interest that you should specify. Exercise 6.14 A life insurer is about to issue a 25 year annual premium endowment
insurance with a basic sum insured of $250 000 to a life aged exactly 30. Initial expenses
are $1 200 plus 40% of the ﬁrst premium and renewal expenses are 1% of the second and
subsequent premiums. The oﬃce allows for a compound reversionary bonus of 2.5% of
the basic sum insured, vesting on each policy anniversary (including the last). The death
beneﬁt is payable at the end of the year of death.
(a) Let L0 denote the gross future loss random variable for this policy. Show that
L0 = 250 000Z1 + 0.99P
P
Z2 + 1 200 + 0.39P − 0.99
d
d where P is the gross annual premium,
Z1 = v K[30] +1 (1.025)K[30] if K[30] ≤ 24,
v 25 (1.025)25
if K[30] ≥ 25, Z2 = v K[30] +1 if K[30] ≤ 24,
v 25
if K[30] ≥ 25. and 262 CHAPTER 6. PREMIUM CALCULATION (b) Using the equivalence principle, calculate P .
2
2
(c) Calculate E[Z1 ], E[Z1 ], E[Z2 ], E[Z2 ] and Cov[Z1 , Z2 ]. Hence calculate V[L0 ] using
the value of P from part (b). (d) Find the probability that the insurer makes a proﬁt on this policy.
Hint: recall the standard results from probability theory, that for random variables X
and Y and constants a, b and c, V[X + c] = V[X ], and
V[aX + bY ] = a2 V[X ] + b2 V[Y ] + 2abCov[X, Y ],
with Cov[X, Y ] = E[XY ] − E[X ]E[Y ]. Exercise 6.15 An insurer issues a 20 year endowment insurance policy to (40) with a
sum insured of $250 000, payable at the end of the year of death. Premiums are payable
annually in advance throughout the term of the contract.
(a) Calculate the premium using the equivalence principle.
(b) Find the mean and standard deviation of the future net loss at issue random variable
using the premium in (a).
(c) Assuming 10 000 identical, independent contracts, estimate the 99th percentile of
the future net loss at issue random variable using the premium in (a). 6.11. EXERCISES 263 Answers to selected exercises
6.1 (a) $216 326.38
(b) $13 731.03
(c) 0.0052
6.2 (b) $178.57
6.3 (b) $3 287.57
(c) $4 981.10
(d) 0.98466
6.4 $3 262.60
6.5 $2 377.75
6.6 $212.81
6.7 (a) $60 694.00
(b) $60 774.30
6.8 1.6875
6.9 $1 341.40 (net), $1 431.08 (gross) 6.10 $214.30
6.11 $1 131.13
6.12 $16 076.72
6.14 (b) $9 764.44
(c) $0.54958, 0.30251, $0.29852, 0.09020, 0.00071, 146 786 651. 264 CHAPTER 6. PREMIUM CALCULATION
(d) 0.98297 6.15 (a) $7 333.84
(b) 0,
(c) $33 696 $14 485 Chapter 7
Policy Values
7.1 Summary In this chapter we introduce the concept of a policy value for a life insurance policy.
Policy values are a fundamental tool in insurance risk management since they are used to
determine the economic or regulatory capital needed to remain solvent, and are also used
to determine the proﬁt or loss for the company over any time period.
We start by considering the case where all cash ﬂows take place at the start or end of a
year. We deﬁne the policy value and we show how to calculate it recursively from year to
year. We also show how to calculate the proﬁt from a policy in any year and we introduce
the asset share for a policy. Later in the chapter we consider policies where the cash
ﬂows are continuous and we derive Thiele’s diﬀerential equation for policy values – the
continuous time equivalent of the recursions for policies with annual cash ﬂows. We also
consider policy alterations.
265 266 7.2 CHAPTER 7. POLICY VALUES Assumptions In almost all the examples in this chapter we assume the standard select survival model
speciﬁed in Example 3.13 and used throughout Chapter 6. We assume, generally, that
lives are select at the time they purchase their policies.
The default rate of interest is 5% per year, though diﬀerent rates are used in some examples. This means that the life table in Table 3.7, the (ultimate) whole life insurance
values in Table 4.1 and the whole life annuity values in Table 6.1 may all be useful for
some calculations in this chapter. 7.3
7.3.1 Policies with annual cash ﬂows
The future loss random variable In Chapter 6 we introduced the future loss random variable, L0 . In this chapter we
are concerned with the estimation of future losses at intermediate times during the term
of a policy, not just at inception. We therefore extend the future loss random variable
deﬁnition, in net and gross versions. Consider a policy which is still in force t years after
it was issued. The present value of future net loss random variable is denoted Ln and the
t
g
present value of gross future loss random variable is denoted Lt , where
Ln
t = Present value, at time t, of future beneﬁts
− Present value, at time t, of future net premiums and
Lg
t = Present value, at time t, of future beneﬁts
+ Present value, at time t, of future expenses
− Present value, at time t, of future gross premiums 7.3. POLICIES WITH ANNUAL CASH FLOWS 267 We drop the n or g superscript where it is clear from the context which is meant. Note
that the future loss random variable Lt is deﬁned only if the contract is still in force t
years after issue.
The example below will help establish some ideas. The important features of this example
for our present purposes are that premiums are payable annually and the sum insured is
payable at the end of the year of death, so that all cash ﬂows are at the start or end of
each year.
Example 7.1 Consider a 20 year endowment policy purchased by a life aged 50. Level
premiums are payable annually throughout the term of the policy and the sum insured,
$500 000, is payable at the end of the year of death or at the end of the term, whichever
is sooner.
The basis used by the insurance company for all calculations is the standard select survival
model, 5% per year interest and no allowance for expenses.
(a) Show that the annual net premium, P , calculated using the equivalence principle,
is $15 114.33.
(b) Calculate E[Ln ] for t = 10 and t = 11, in both cases just before the premium due
t
at time t is paid.
Solution 7.1 (a) You should check that the following values are correct for this survival
model at 5% per year interest:
a[50]:20 = 12.8456 and A[50]:20 = 0.38830.
¨
The equation of value for P is
P a[50]:20 − 500 000 A[50]:20 = 0,
¨
giving
P= 500 000 A[50]:20
a[50]:20
¨ = $15 114.33. (7.1) 268 CHAPTER 7. POLICY VALUES (b) Ln is the present value of the future net loss 10 years after the policy was pur10
chased, assuming the policyholder is still alive at that time. The policyholder will
then be aged 60 and the select period for the survival model, two years, will have
expired eight years ago. The present value at that time of the future beneﬁts is
500 000 v min(K60 +1,10) and the present value of the future premiums is P amin(K60 +1,10) .
¨
n
n
Hence, the formulae for L10 and L11 are
Ln = 500 000 v min(K60 +1,10) − P amin(K60 +1,10)
¨
10
and
Ln = 500 000 v min(K61 +1,9) − P amin(K61 +1,9) .
¨
11
Taking expectations and using the annuity values
a60:10 = 7.9555 and a61:9 = 7.3282
¨
¨
we have
E[Ln ] = 500 000A60:10 − P a60:10 = $190 339
¨
10
and
E [Ln ] = 500 000A61:9 − P a61:9 = $214 757.
¨
11 We are now going to look at Example 7.1 in a little more detail. At the time when the
policy is issued, at t = 0, the future loss random variable, Ln , is given by
0
Ln = 500 000 v min(K[50] +1,20) − P amin(K
¨
0 [50] +1,20) . 7.3. POLICIES WITH ANNUAL CASH FLOWS 269 Since the premium is calculated using the equivalence principle, we know that E[Ln ] = 0,
0
which is equivalent to equation (7.1). That is, at the time the policy is issued, the expected
value of the present value of the loss on the contract is zero, so that, in expectation, the
future premiums (from time 0) are exactly suﬃcient to provide the future beneﬁts.
Consider the ﬁnancial position of the insurer at time 10 with respect to this policy. The
policyholder may have died before time 10. If so, the sum insured will have been paid and
no more premiums will be received. In this case the insurer no longer has any liability with
respect to this policy. Now suppose the policyholder is still alive at time 10. In this case
the calculation in part (b) shows that the future loss random variable, Ln , has a positive
10
expected value ($190 339) so that future premiums (from time 10) are not expected to
be suﬃcient to provide the future beneﬁts. For the insurer to be in a ﬁnancially sound
position at time 10, it should hold an amount of at least $190 339 in its assets so that,
together with future premiums from time 10, it can expect to provide the future beneﬁts.
Speaking generally, when a policy is issued the future premiums should be expected to
be suﬃcient to pay for the future beneﬁts and expenses. (If not, the premium should be
increased!) However, it is usually the case that for a policy which is still in force t years
after being issued, the future premiums (from time t) are not expected to be suﬃcient to
pay for the future beneﬁts and expenses. The amount needed to cover this shortfall is
called the policy value for the policy at time t.
The insurer should be able to build up its assets during the course of the policy because,
with a regular level premium and an increasing level of risk, the premium in each of the
early years is more than suﬃcient to pay the expected beneﬁts in that year, given that
the life has survived to the start of the year. For example, in the ﬁrst year the premium
of $15 114.33 is greater than the EPV of the beneﬁt the insurer will pay in that year,
500 000 v q[50] = $492.04. In fact, for the endowment insurance policy studied in Example
7.1, for each year except the last the premium exceeds the EPV of the beneﬁts, that is P > 500 000 v q[50]+t for t = 0, 1, . . . , 18. 270 CHAPTER 7. POLICY VALUES 20000
30000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 80000
130000 EPV 180000
230000
280000
330000
380000
430000
480000
Duration (years) Figure 7.1: EPV of premiums minus claims for each year of a 20 year endowment insurance, sum insured $500 000, issued to (50).
The ﬁnal year is diﬀerent because
P = 15 114.33 < 500 000 v = 476 190.
Note that if the policyholder is alive at the start of the ﬁnal year, the sum insured will
be paid at the end of the year whether or not the policyholder survives the year.
Figure 7.1 shows the excess of the premium over the EPV of the beneﬁt payable at the
end of the year for each year of this policy.
Figure 7.2 shows the corresponding values for a 20 year term insurance issued to (50).
The sum insured is $500 000, level annual premiums are payable throughout the term and
all calculations use the same basis as in Example 7.1. The pattern is similar in that there
is a positive surplus in the early years which can be used to build up the insurer’s assets.
These assets are needed in the later years when the premium is not suﬃcient to pay for
the expected beneﬁts. 7.3. POLICIES WITH ANNUAL CASH FLOWS 271 1200 600 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 EPV 600 1200 1800 2400 3000
Duration (years) Figure 7.2: EPV of premiums minus claims for each year of a 20 year term insurance,
sum insured $500 000, issued to (50).
The insurer will then, for a large portfolio, hold back some of the excess cash ﬂow from the
early years of the contract in order to meet the shortfall in the later years. This explains
the concept of a policy value – we need to hold capital during the term of a policy to
meet the liabilities in the period when outgo on beneﬁts exceeds income from premiums.
We give a formal deﬁnition of a policy value later in this section.
Before doing so, we return to Example 7.1. Suppose the insurer issues a large number,
say N , of policies identical to the one in Example 7.1, to independent lives all aged 50.
Suppose also that the experience of this group of policyholders is precisely as assumed
in the basis used by the insurer in its calculations. In other words, interest is earned on
investments at 5% every year, the mortality of the group of policyholders follows precisely
the standard select survival model and there are no expenses.
Consider the ﬁnancial situation of the insurer after these policies have been in force for
10 years. Some policyholders will have died, so that their sum insured of $500 000 will
have been paid at the end of the year in which they died, and some policyholders will still 272 CHAPTER 7. POLICY VALUES be alive. With our assumptions about the experience, precisely 10 p[50] N policyholders
will still be alive, q[50] N will have died in the ﬁrst year, 1 q[50] N will have died in the
second year, and so on, until the 10th year, when 9 q[50] N policyholders will have died.
The accumulation to time 10 at 5% interest of all premiums received (not including the
premiums due at time 10) minus all sums insured which have been paid is
N P (1.0510 + p[50] 1.059 + . . . + 9 p[50] 1.05)
− 500 000N (q[50] 1.059 + 1 q[50] 1.058 + . . . + 9 q[50] )
= 1.0510 N P (1 + p[50] 1.05−1 + . . . + 9 p[50] 1.05−9 )
− 1.0510 500 000N (q[50] 1.05−1 + 1 q[50] 1.05−2 + . . . + 9 q[50] 1.05−10 )
= 1.0510 N (P a[50]:10 − 500 000A1 10 )
¨
[50]:
= 186 634N.
(Note that, using the values in part (a) of Example 7.1, we have
a[50]:10
¨
1
A[50]:10 = a[50]:20 − v 10 10 p[50] a60:10 = 8.0566
¨
¨
= 1 − da[50]:10 − v 10 10 p[50] = 0.01439.)
¨ So, if the experience over the ﬁrst 10 years follows precisely the assumptions set out in
Example 7.1, the insurer will have built up a fund of $186 634N after 10 years. The
number of policyholders still alive at that time will be 10 p[50] N and so the share of this
fund for each surviving policyholder is 186 634N/(10 p[50] N ) = $190 339. This is precisely
the amount the insurer needs. This is not a coincidence! This happens in this example
because the premium was calculated using the equivalence principle, so that the EPV of
the proﬁt was zero when the policies were issued, and we have assumed the experience up 7.3. POLICIES WITH ANNUAL CASH FLOWS 273 to time 10 was exactly as in the calculation of the premium. Given these assumptions, it
should not be surprising that the insurer is in a ‘break even’ position at time 10. We can
prove that this is true in this case by manipulating the equation of value, equation (7.1),
as follows:
P a[50]:20 = 500 000 A50:20
¨
⇒ P (¨[50]:10 + v 10 10 p[50] a60:10 ) = 500 000 A1 10 + v 10 10 p[50] A60:10
a
¨
[50]:
⇒ P a[50]:10 − 500 000A1 10 = v 10 10 p[50] 500 000A60:10 − P a60:10
¨
¨
[50]:
⇒ 1.0510
P a[50]:10 − 500 000A1 10
¨
[50]:
10 p[50] = 500 000A60:10 − P a60:10 .
¨ (7.2) The left hand side of equation (7.2) is the share of the fund built up at time 10 for each
surviving policyholder; the right hand side is the expected value of the future net loss
random variable at time 10, E[Ln ], and so is the amount needed by the insurer at time
10
10 for each policy still in force.
For this example, the proof that the total amount needed by the insurer at time 10 for all
policies still in force is precisely equal to the amount of the fund built up, works because
(a) the premium was calculated using the equivalence principle,
(b) the expected value of the future loss random variable was calculated using the
premium basis, and
(c) we assumed the experience followed precisely the assumptions in the premium basis.
In practice, (a) and (b) may or may not apply; assumption (c) is extremely unlikely to
hold. 274 CHAPTER 7. POLICY VALUES 7.3.2 Policy values for policies with annual cash ﬂows In general terms, the policy value for a policy in force at duration t (≥ 0) years after it
was purchased is the expected value at that time of the future loss random variable. At
this stage we do not need to specify whether this is the gross or net future loss random
variable – we will be more precise later in this section.
The general notation for a policy value t years after a policy was issued is t V (the V comes
from ‘Policy Value’) and we use this notation in this book. There is a standard actuarial
notation associated with policy values for certain traditional contracts. This notation
is not particularly useful, and so we do not use it. (Interested readers can consult the
references in Section 7.9.)
Intuitively, the policy value at time t represents the amount the insurer should have in
its investments at that time in respect of a policy which is still in force, so that, together
with future premiums, the insurer can, in expectation, exactly pay future beneﬁts and
expenses. In general terms, we have the equation
tV + EPV at t of future premiums = EPV at t of future beneﬁts + expenses. An important element in the ﬁnancial control of an insurance company is the calculation
at regular intervals, usually at least annually, of the sum of the policy values for all policies
in force at that time and also the value of all the company’s investments. For the company
to be ﬁnancially sound, the investments should have a greater value than the total policy
value. This process is called a valuation of the company. In most countries, valuations
are required annually by the insurance supervisory authority.
In the literature, the terms reserve, prospective reserve and prospective policy
value are sometimes used in place of policy value. We use policy value to mean the
expected value of the future loss random variable, and restrict reserve to mean the actual
capital held in respect of a policy, which may be greater than or less than the policy value. 7.3. POLICIES WITH ANNUAL CASH FLOWS 275 The precise deﬁnitions of policy value are as follows.
Deﬁnition 7.1 The gross premium policy value for a policy in force at duration
t (≥ 0) years after it was purchased is the expected value at that time of the gross future
loss random variable on a speciﬁed basis. The premiums used in the calculation are the
actual premiums payable under the contract.
Deﬁnition 7.2 The net premium policy value for a policy in force at duration t (≥ 0)
years after it was purchased is the expected value at that time of the net future loss random
variable on a speciﬁed basis (which makes no allowance for expenses). The premiums
used in the calculation are the net premiums calculated on the policy value basis using the
equivalence principle, not the actual premiums payable.
We make the following comments about these deﬁnitions.
1. Throughout Section 7.3 we restrict ourselves to policies where the cash ﬂows occur
only at the start or end of a year since these policies have some simplifying features
in relation to policy values. However, Deﬁnitions 7.1 and 7.2 apply to more general
types of policy, as we show in later sections.
2. The numerical value of a gross or net premium policy value depends on the assumptions – survival model, interest, expenses, future bonuses – used in its calculation.
These assumptions, called the policy value basis, may diﬀer from the assumptions
used to calculate the premium, that is, the premium basis.
3. A net premium policy value can be regarded as a special case of a gross premium
policy value. The two are the same numerically if the actual premiums for the
contract are calculated using the equivalence principle and the policy value basis,
which does not include expenses.
4. When the policy value basis diﬀers from the premium basis, the net premium policy
value requires the recalculation of the premium. See Example 7.2 below. This is a 276 CHAPTER 7. POLICY VALUES
vestige of a time before modern computers, when easy calculation was a key issue –
using a net premium policy value allowed the use of computational short cuts. The
net premium policy value is becoming obsolete, but is still suﬃciently widely used
that it is helpful to understand the concept. We make more use of gross, rather than
net, premium policy values in this book. Where it is clear from the context which is
meant, or where the distinction is not important, we refer simply to a policy value. 5. If we are calculating a policy value at an integer duration, that is at the start/end of
a year, there may be premiums and/or expenses and/or beneﬁts payable at precisely
that time and we need to be careful about which cash ﬂows are included in our future
loss random variable. It is the usual practice to regard a premium and any premium–
related expenses due at that time as future payments and any insurance beneﬁts (i.e.
death or maturity claims) and related expenses as past payments. Under annuity
contracts, the annuity payments and related expenses may be treated either as future
payments or as past payments, so we need to be particularly careful to specify which
it is in such cases.
6. Note that if an insurance policy has a ﬁnite term, n years, for example for an
endowment insurance or a term insurance, then n V = 0 since the future loss random
variable on any basis is zero. Note also that if the premium is calculated using the
equivalence principle and the policy value basis is the same as the premium basis,
then 0 V = E[L0 ] = 0.
7. For an endowment insurance which is still in force at the maturity date, the policy
value at that time must be suﬃcient to pay the sum insured, S , say, so in this case
−
n− V = S and n V = 0, where n denotes the moment before time n.
8. In the discussion following Example 7.1 in Section 7.3.1 we saw how the insurer
built up the reserve for policies still in force by accumulating past premiums minus
claims for a group of similar policies. Broadly speaking, this is what would happen
in practice, though not with the artiﬁcial precision we saw in Section 7.3.1 that led
to the accumulated funds being precisely the amount required by the insurer. 7.3. POLICIES WITH ANNUAL CASH FLOWS 277 Example 7.2 An insurer issues a whole life insurance policy to a life aged 50. The sum
insured of $100 000 is payable at the end of the year of death. Level premiums of $1 300
are payable annually in advance throughout the term of the contract.
(a) Calculate the gross premium policy value 5 years after the inception of the contract,
assuming that the policy is still in force, using the following basis:
Mortality: the standard select survival model
Interest: 5% per year
Expenses: 12.5% of each premium
(b) Calculate the net premium policy value 5 years after the issue of the contract,
assuming that the policy is still in force, using the following basis:
Mortality: the standard select survival model
Interest: 4% per year
Solution 7.2 We assume that the life is select at age 50, when the policy is purchased.
At duration 5, the life is aged 55 and is no longer select since the select period for the
standard select survival model is only two years. Note that a premium due at age 55 is
regarded as a future premium in the calculation of a policy value.
(a) The gross future loss random variable at time 5 is
Lg = 100 000v K55 +1 − 0.875 × 1 300 aK55 +1 ,
¨
5
so
5V g = E[Lg ] = 100 000 A55 − 0.875 × 1 300 a55 = $5 256.35.
¨
5 (b) For the net premium policy value we calculate the net premium for the contract on
the net premium policy value basis. At 4% per year,
P = 100 000 A[50]
= $1 321.31.
a[50]
¨ 278 CHAPTER 7. POLICY VALUES
So, at 4% per year,
Ln = 100 000v K55 +1 − 1321.31¨K55 +1
a
5
and hence
5V n = 100 000 A55 − 1321.31¨55 = $6 704.75.
a Notice in this example that the net premium calculation ignores expenses, but uses
a lower interest rate, which provides a margin, implicitly allowing for expenses and
other contingencies. Example 7.3 A woman aged 60 purchases a 20 year endowment insurance with a sum
insured of $100 000 payable at the end of the year of death or on survival to age 80,
whichever occurs ﬁrst. An annual premium of $5 200 is payable for at most 10 years. The
insurer uses the following basis for the calculation of policy values:
Mortality: the standard select survival model
Interest: 5% per year
Expenses: 10% of the ﬁrst premium, 5% of subsequent premiums, and $200 on
payment of the sum insured
Calculate 0 V, 5 V, 6 V and 10 V , that is, the gross premium policy values for this policy at
times t = 0, 5, 6 and 10.
Solution 7.3 You should check the following values, which will be needed for the calculation of the policy values:
a[60]:10 = 7.9601,
¨
A[60]:20 = 0.41004, a65:5 = 4.4889,
¨ a66:4 = 3.6851,
¨ A65:15 = 0.51140, A66:14 = 0.53422, A70:10 = 0.63576. 7.3. POLICIES WITH ANNUAL CASH FLOWS 279 At time 0, when the policy is issued, the future loss random variable, allowing for expenses
as speciﬁed in the policy value basis, is
L0 = 100 200v min(K[60] +1,20) + 0.05 P − 0.95 P amin(K
¨ [60] +1,10) where P = $5 200. Hence
0V = E[L0 ] = 100 200A[60]:20 − (0.95 a[60]:10 − 0.05)P = $2 023.
¨ Similarly,
L5 = 100 200v min(K65 +1,15) − 0.95 P amin(K65 +1,5)
¨
so that
5V = E[L5 ] = 100 200A65:15 − 0.95 P a65:5 = $29 068,
¨ and
L6 = 100 200v min(K66 +1,14) − 0.95 P amin(K66 +1,4)
¨
so that
6V = E[L6 ] = 100 200A66:14 − 0.95 P a66:4 = $35 324.
¨ Finally, as no premiums are payable after time 9,
L10 = 100 200v min(K70 +1,10)
so that
10 V = E[L10 ] = 100 200A70:10 = $63 703. 280 CHAPTER 7. POLICY VALUES In Example 7.3, the initial policy value, 0 V is greater than zero. This means that from
the outset the insurer expects to make a loss on this policy. This sounds uncomfortable
but is not uncommon in practice. The explanation is that the policy value basis may
be more conservative than the premium basis. For example, the insurer may assume an
interest rate of 6% in the premium calculation, but, for policy value calculations, assumes
investments will earn only 5%. At 6% per year interest, and with a premium of $5 200,
this policy generates an EPV of proﬁt at issue of $2 869.
Example 7.4 A man aged 50 purchases a deferred annuity policy. The annuity will be
paid annually for life, with the ﬁrst payment on his 60th birthday. Each annuity payment
will be $10 000. Level premiums of $11 900 are payable annually for at most 10 years. On
death before age 60, all premiums paid will be returned, without interest, at the end of
the year of death. The insurer uses the following basis for the calculation of policy values:
Mortality: the standard select survival model
Interest: 5% per year
Expenses: 10% of the ﬁrst premium, 5% of subsequent premiums, $25 each time an
annuity payment is paid, and $100 when a death claim is paid
Calculate the gross premium policy values for this policy at the start of the policy, at the
end of the ﬁfth year, and at the end of the ﬁfteenth year, just before and just after the
annuity payment and expense due at that time.
Solution 7.4 We are going to need the following values, all of which you should check:
a[50]:10 = 8.0566,
¨
v 5 5 p55 = 0.77382,
1
A[50]:10 = 0.01439, a55:5 = 4.5268,
¨ a60 = 14.9041,
¨ a65 = 13.5498,
¨ v 10 10 p[50] = 0.60196,
1
(IA)[50]:10 = 0.08639, 1
A55:5 = 0.01062, 1
(IA)55:5 = 0.03302. 7.3. POLICIES WITH ANNUAL CASH FLOWS 281 Then, using the notation 15− V and 15+ V to denote the policy values at duration 15 years
just before and just after the annuity payment and expense due at that time, respectively,
and noting that P = 11 900, we can calculate the policy value at any time t as
EPV at t of future beneﬁts + expenses − EPV at t of future premiums.
At the inception of the contract, the EPV of the death beneﬁt is
1
P (IA)[50]:10 , the EPV of the death claim expenses is
1
100A[50]:10 , the EPV of the annuity beneﬁt and associated expenses is
10 025 v 10 10 p[50] a60 ,
¨
and the EPV of future premiums less associated expenses is
0.95P a[50]:10 − 0.05P ,
¨
so that
0V 1
1
= P (IA)[50]:10 + 100A[50]:10 + 10 025v 10 10 p[50] a60 − (0.95 a[50]:10 − 0.05)P
¨
¨ = $485.
At the 5th anniversary of the inception of the contract, assuming it is still in force, the
future death beneﬁt is 6P, 7P, . . . , 10P depending on whether the life dies in the 6th,
7th,. . .,10th years, respectively. We can write this beneﬁt as a level beneﬁt of 5P plus an
increasing beneﬁt of P, 2P, . . . , 5P . 282 CHAPTER 7. POLICY VALUES So at time 5, the EPV of the death beneﬁt is
P 1
1
(IA)55:5 + 5A55:5 , 1
the EPV of the death claim expenses is 100A55:5 , the EPV of the annuity beneﬁt and associated expenses is 10 025 v 5 5 p55 a60 ,
¨
and the EPV of future premiums less associated expenses is 0.95P a55:10 , so that
¨
5V 1
1
1
= P (IA)55:5 + 5 P A55:5 + 100A55:5 + 10 025 v 5 5 p55 a60 − 0.95P a55:5
¨
¨ = $65 470.
Once the premium payment period of 10 years is completed, there are no future premiums
to value, so the policy value is the EPV of the future annuity payments and associated
expenses. Thus,
15− V = 10 025 a65 = $135 837,
¨ 15+ V = 10 025 a65 = and
15− V − 10 025 = $125 812. We can make two comments about Example 7.4.
1. As in Example 7.3, 0 V > 0, which implies that the valuation basis is more conservative than the premium basis.
2. In Example 7.4 we saw that 15+ V = 15− V − 10 025. This makes sense if we regard
the policy value at any time as the amount of assets being held at that time in 7.3. POLICIES WITH ANNUAL CASH FLOWS 283 respect of a policy still in force. The policy value 15− V (= $135 837) represents the
assets being held at time 15 just before the payment of the annuity, $10 000, and the
associated expense, $25. Immediately after making these payments, the insurer’s
assets will have reduced by $10 025, and the new policy value is 15+ V .
We conclude this section by plotting policy values for the endowment insurance discussed
in Example 7.1 and for the term insurance with the same sum insured and term. For
these policies Figures 7.1 and 7.2 respectively show the EPV of premiums minus claims
for each year of the policy. Figures 7.3 and 7.4 respectively show the policy values. In
Figure 7.3 we see that the policy values build up over time to provide the sum insured on
maturity. By contrast, in Figure 7.4 the policy values increase then decrease. A further
contrast between these ﬁgures is the level of the policy values. In Figure 7.4 the largest
policy value occurs at time 13, with 13 V = $9 563.00, which is a small amount compared
with the sum insured of $500 000. The reason why small policy values occur for the term
insurance policy is simply that there is a small probability of the death beneﬁt being paid. 7.3.3 Recursive formulae for policy values In this section we show how to derive recursive formulae for policy values for policies
with discrete cash ﬂows. These formulae can be useful in the calculation of policy values
in some cases – we give an example at the end of this section to illustrate this point –
and they also provide an understanding of how the policy value builds up and how proﬁt
emerges while the policy is in force. We use Examples 7.1 and 7.4 to demonstrate the
principles involved.
Example 7.5 For Example 7.1 and for t = 0, 1, . . . , 19, show that
( t V + P )(1 + i) = 500 000 q[50]+t + p[50]+t t+1 V (7.3) where P = $15 114.33, i = 5% and the policy value is calculated on the basis speciﬁed in
Example 7.1. 284 CHAPTER 7. POLICY VALUES 500000 450000 400000 Policy value, tV 350000 300000 250000 200000 150000 100000 50000 0
0 2 4 6 8 10 12 14 16 18 20 Time, t Figure 7.3: Policy values for each year of a 20 year endowment insurance, sum insured
$500 000, issued to (50). 10000 9000 8000 Policy value, tV 7000 6000 5000 4000 3000 2000 1000 0
0 2 4 6 8 10 12 14 16 18 20 Time, t Figure 7.4: Policy values for each year of a 20 year term insurance, sum insured $500 000,
issued to (50). 7.3. POLICIES WITH ANNUAL CASH FLOWS 285 Solution 7.5 From the solution to Example 7.1 we know that for t = 0, 1, . . . , 19,
tV = 500 000 A[50]+t:20−t − P a[50]+t:20−t .
¨ Splitting oﬀ the terms for the ﬁrst year for both the endowment and the annuity functions,
we have
tV = 500 000 (vq[50]+t + vp[50]+t A[50]+t+1:19−t ) − P (1 + vp[50]+t a[50]+t+1:19−t )
¨
= v 500 000q[50]+t + p[50]+t (500 000A[50]+t+1:19−t − P a[50]+t+1:19−t ) − P.
¨ Rearranging, multiplying both sides by (1 + i) and recognising that
t+1 V = 500 000 A[50]+t+1:19−t − P a[50]+t+1:19−t
¨ gives equation (7.3). We comment on Example 7.5 after the next example.
Example 7.6 For Example 7.4 and for t = 1, 2, . . . , 9 show that
( t V + 0.95P )(1 + i) = ((t + 1)P + 100) q[50]+t + p[50]+t t+1 V (7.4) where P = $11 900, i = 5% and the policy value is calculated on the basis speciﬁed in
Example 7.4. Solution 7.6 For Example 7.4 and for t = 1, 2, . . . , 9, t V has the same form as 5 V , that
is
tV 1
1
= P (IA)[50]+t:10−t + (tP + 100)A[50]+t:10−t 286 CHAPTER 7. POLICY VALUES
+10 025v 10−t 10−t p[50]+t a60 − 0.95P a[50]+t:10−t
¨
¨ Recall that recurrence relations for insurance and annuity functions can be derived by
separating out the EPV of the ﬁrst year’s payments, so that
a[x]+t:n−t = 1 + vp[x]+t a[x]+t+1:n−t−1 ,
¨
¨
1
1
A[x]+t:n−t = vq[x]+t + vp[x]+t A[x]+t+1:n−t−1 and
1
1
1
(IA)[x]+t:n−t = vq[x]+t + vp[x]+t (IA)[x]+t+1:n−t−1 ) + A[x]+t+1:n−t−1 . Using these relations to split oﬀ the terms for the year t to t + 1 in the policy value
equation, we have, for t = 1, 2, ..., 9,
tV 1
1
= P v q[50]+t + vp[50]+t (IA)[50]+t+1:10−t−1 ) + A[50]+t+1:10−t−1
1
+(tP + 100) v q[50]+t + vp[50]+t A[50]+t+1:10−t−1 +10 025 vp[50]+t v 10−t−1 10−t−1 p[50]+t+1 a60
¨ −0.95P 1 + vp[50]+t a[50]+t+1:10−t−1
¨
⇒ tV = vq[50]+t ((t + 1)P + 100) − 0.95P
1
1
+vp[50]+t P (IA)[50]+t+1:10−t−1 + ((t + 1)P + 100) A[50]+t+1:10−t+1 +10 025 10−t−1 p[50]+t+1 v 10−t−1 a60 − 0.95P a[50]+t+1:10−t−1
¨
¨ . 7.3. POLICIES WITH ANNUAL CASH FLOWS
Notice that the expression in curly braces, {}, is t+1 V 287
, so, substituting and rearranging, ( t V + 0.95P ) (1 + i) = ((t + 1)P + 100) q[50]+t + p[50]+t t+1V , (7.5) as required.
Equations (7.3) and (7.4) are recursive formulae for policy values since they express t V in
terms of t+1V . Such formulae always exist but the precise form they take depends on the
details of the policy being considered. The method we used to derive formulae (7.3) and
(7.4) can be used for other policies: ﬁrst write down a formula for t V and then break up
the EPVs into EPVs of payments in the coming year, t to t + 1, and EPVs of payments
from t + 1 onwards. We can demonstrate this in a more general setting as follows.
Consider a policy issued to a life (x) where cash ﬂows – premiums, expenses and claims
– can occur only at the start or end of a year. Suppose this policy has been in force for t
years, where t is a nonnegative integer. Consider the (t + 1)st year, and let
Pt denote the premium payable at time t,
et denote the premiumrelated expense payable at time t,
St+1 denote the sum insured payable at time t + 1 if the policyholder dies in the year,
Et+1 denote the expense of paying the sum insured at time t + 1,
tV
t+1 V denote the gross premium policy value for a policy in force at time t, and,
denote the gross premium policy value for a policy in force at time t + 1. Let q[x]+t denote the probability that the policyholder, alive at time t, dies in the year and
let it denote the rate of interest assumed earned in the year. The quantities et , Et , q[x]+t
and it are all as assumed in the policy value basis.
Let Lt and Lt+1 denote the gross future loss random variables at times t and t + 1,
respectively, in both cases assuming the policyholder is alive at that time. Note that Lt 288 CHAPTER 7. POLICY VALUES involves present values at time t whereas Lt+1 involves present values at time t + 1. Then,
by considering what can happen in the year, we have
Lt = (1 + it )−1 (St+1 + Et+1 ) − Pt + et if K[x]+t = 0, with probability q[x]+t ,
(1 + it )−1 Lt+1 − Pt + et
if K[x]+t ≥ 1, with probability p[x]+t . Taking expected values, we have
tV = E[Lt ] = q[x]+t (1 + it )−1 (St+1 + Et+1 ) − (q[x]+t + p[x]+t )(Pt − et )
+p[x]+t (1 + it )−1 E[Lt+1 ] , which, after a little rearranging and recognizing that
equation t+1 V = E[Lt+1 ], gives the important ( t V + Pt − et )(1 + it ) = q[x]+t (St+1 + Et+1 ) + p[x]+t t+1V . (7.6) Equation (7.6) includes equations (7.3) and (7.4) as special cases and it is a little more
general than either of them since it allows the premium, the sum insured, the expenses
and the rate of interest all to be functions of t or of t + 1, so that they can vary from year
to year.
For policies with cash ﬂows only at the start/end of each year, the recursive formulae
always have the same general form. This form can be explained by considering equation
(7.6).
• Assume that at time t the insurer has assets of amount t V in respect of this policy.
Recall that t V is the expected value on the policy value basis of the future loss
random variable, assuming the policyholder is alive at time t. Hence we can interpret
t V as the value of the assets the insurer should have at time t (in respect of a policy
still in force) in order to expect to break even over the future course of the policy. 7.3. POLICIES WITH ANNUAL CASH FLOWS 289 • Now add to t V the net cash ﬂow received by the insurer at time t as assumed in
the policy value basis. In equation (7.6) this is Pt − et ; in Example 7.5 this was just
the premium, P = $15 114.33; in Example 7.6 this was the premium, P = $11 900,
less the expense assumed in the policy value basis, 0.05P . The new amount is the
amount of the insurer’s assets at time t just after these cash ﬂows. There are no
further cash ﬂows until the end of the year.
• These assets are rolled up to the end of the year with interest at the rate assumed
in the policy value basis, it (= 5% in the two examples). This gives the amount of
the insurer’s assets at the end of the year before any further cash ﬂows (assuming
everything is as speciﬁed in the policy value basis). This gives the left hand sides
of equations (7.6), (7.3) and (7.4).
• We assumed the policyholder was alive at the start of the year, time t; we do not
know whether the policyholder will be alive at the end of the year. With probability
p[x]+t the policyholder will be alive, and with probability q[x]+t the policyholder will
die in the year (where these probabilities are calculated on the policy value basis).
• If the policyholder is alive at time t + 1 the insurer needs to have assets of amount
t+1 V at that time; if the policyholder has died during the year, the insurer must pay
any death beneﬁt and related expenses. The expected amount the insurer needs
for the policy being considered above is given by the right hand side of equation
(7.6) (equations (7.3) and (7.4) for Examples 7.5 and 7.6). For the general policy
and both examples, this is precisely the amount the insurer will have (given our
assumptions). This happens because the policy value is deﬁned as the expected
value of the future loss random variable and because we assume cash ﬂows from t
to t + 1 are as speciﬁed in the policy value basis. We assumed that at time t the
insurer had suﬃcient assets to expect (on the policy value basis) to break even over
the future course of the policy. Since we have assumed that from t to t + 1 all cash
ﬂows are as speciﬁed in the policy value basis, it is not surprising that at time t + 1
the insurer still has suﬃcient assets to expect to break even. 290 CHAPTER 7. POLICY VALUES One further point needs to be made about equations (7.6), (7.3) and (7.4). We can rewrite
these three formulae as follows:
( t V + Pt − et )(1 + it ) = t+1 V + q[x]+t (St+1 + Et+1 − t+1 V ( t V + P )(1 + i) = t+1 V + q[x]+t (500 000 − ), ( t V + 0.95P )(1 + i) = t+1 V + q[x]+t ((t + 1)P q50+t − t+1 V ), t+1 V (7.7) ). The left hand sides of these formulae are unchanged – they still represent the amount of
assets the insurer is assumed to have at time t + 1 in respect of a policy which was in
force at time t. The right hand sides can now be interpreted slightly diﬀerently.
• For each policy in force at time t the insurer needs to provide the policy value,
at time t + 1, whether the life died during the year or not. t+1 V , • In addition, if the policyholder has died in the year (the probability of which is
q[x]+t ), the insurer must also provide the extra amount to increase the policy value
to the death beneﬁt payable plus any related expense: St+1 + Et+1 − t+1 V for the
general policy, 500 000 − t+1 V in Example 7.5 and (t + 1)P − t+1 V in Example 7.6.
This extra amount required to increase the policy value to the death beneﬁt is called
the death strain at risk (DSAR), or the Sum at Risk or the Net Amount
at Risk, at time t + 1. If the policy value basis does not explicitly allow for
claim expenses, the DSAR in the tth year, where the death beneﬁt payable is St , is
St − t V . This is an important measure of the insurer’s risk if mortality exceeds the
basis assumption, and is useful in determining risk management strategy, including
reinsurance – which is the insurance that an insurer buys to protect itself against
adverse experience.
In all the examples so far in this section it has been possible to calculate the policy value
directly as the EPV on the given basis of future beneﬁts plus future expenses minus future 7.3. POLICIES WITH ANNUAL CASH FLOWS 291 premiums. In more complicated examples, in particular where the beneﬁts are deﬁned in
terms of the policy value, this may not be possible. In these cases the recursive formula
for policy values, equation (7.6), can be very useful as the following example shows.
Example 7.7 Consider a 20 year endowment policy purchased by a life aged 50. Level
premiums of $23 500 per year are payable annually throughout the term of the policy. A
sum insured of $700 000 is payable at the end of the term if the life survives to age 70.
On death before age 70 a sum insured is payable at the end of the year of death equal to
the policy value at the start of the year in which the policyholder dies.
The policy value basis used by the insurance company is as follows:
Mortality: the standard select survival model
Interest: 3.5% per year
Expenses: nil
Calculate 15 V , the policy value for a policy in force at the start of the 16th year.
Solution 7.7 For this example, formula (7.6) becomes
( t V + P ) × 1.035 = q[50]+t St+1 + p[50]+t t+1 V for t = 0, 1, . . . , 19, where P = $23 500. For the ﬁnal year of this policy, the death beneﬁt payable at the end
of the year is 19 V and the survival beneﬁt is the sum insured, $700 000. Putting t = 19
in the above equation gives:
( 19 V + P ) × 1.035 = q69 19 V + p69 × 700 000.
Tidying this up and noting that St+1 = t V , we can work backwards as follows:
19 V = (p69 × 700 000 − 1.035P )/(1.035 − q69 ) = 652 401, 18 V = (p68 × 19 V − 1.035P )/(1.035 − q68 ) = 606 471, 292 CHAPTER 7. POLICY VALUES
17 V = (p67 × 18 V − 1.035P )/(1.035 − q67 ) = 562 145, 16 V = (p66 × 17 V − 1.035P )/(1.035 − q66 ) = 519 362, 15 V = (p65 × 16 V − 1.035P )/(1.035 − q65 ) = 478 063. Hence, the answer is $478 063. 7.3.4 Annual proﬁt Consider a group of identical policies issued at the same time. The recursive formulae for
policy values show that if all cash ﬂows between t and t + 1 are as speciﬁed in the policy
value basis, then the insurer will be in a break even position at time t + 1, given that it
was in a break even position at time t. These cash ﬂows depend on mortality, interest,
expenses and, for participating policies, bonus rates. In practice, it is very unlikely that
all the assumptions will be met in any one year. If the assumptions are not met, then the
value of the insurer’s assets at time t + 1 may be more than suﬃcient to pay any beneﬁts
due at that time and to provide a policy value of t+1 V for those policies still in force. In
this case, the insurer will have made a proﬁt in the year. If the insurer’s assets at time
t + 1 are not suﬃcient to pay any beneﬁts due at that time and to provide a policy value
of t+1 V for those policies still in force, the insurer will have made a loss in the year.
In general terms:
• Actual expenses less than the expenses assumed in the policy value basis will be a
source of proﬁt.
• Actual interest earned on investments less than the interest assumed in the policy
value basis will be a source of loss.
• Actual mortality less than the mortality assumed in the policy value basis can be a
source of either proﬁt or loss. For whole life, term and endowment policies it will 7.3. POLICIES WITH ANNUAL CASH FLOWS 293 be a source of proﬁt; for annuity policies it will be a source of loss.
• Actual bonus or dividend rates less than the rates assumed in the policy value basis
will be a source of proﬁt.
The following example demonstrates how to calculate annual proﬁt from a nonparticipating life insurance policy.
Example 7.8 An insurer issued a large number of policies identical to the policy in
Example 7.3 to women age 60. Five years after they were issued, a total of 100 of these
policies were still in force. In the following year,
• expenses of 6% of each premium paid were incurred,
• interest was earned at 6.5% on all assets,
• one policyholder died, and
• expenses of $250 were incurred on the payment of the sum insured for the policyholder who died.
(a) Calculate the proﬁt or loss on this group of policies for this year.
(b) Determine how much of this proﬁt/loss is attributable to proﬁt/loss from mortality,
from interest and from expenses.
Solution 7.8 (a) At duration t = 5 we assume the insurer held assets for the portfolio
with value exactly equal to the total of the policy values at that time for all the
policies still in force. From Example 7.3 we know the value of 5 V and so we assume
the insurer’s assets at time 5, in respect of these policies, amounted to 100 5 V . If
the insurer’s assets were worth less (resp. more) than this, then losses (resp. proﬁts)
have been made in previous years. These do not concern us – we are concerned only
with what happens in the 6th year. 294 CHAPTER 7. POLICY VALUES
Now consider the cash ﬂows in the 6th year. For each of the 100 policies still in
force at time 5 the insurer received a premium P (= $5 200) and paid an expense
of 0.06P at time 5. Hence, the total assets at time 5 after receiving premiums and
paying premiumrelated expenses were
100 5 V + 100 × 0.94 P = $3 395 551.
There were no further cash ﬂows until the end of the year, so this amount grew
for one year at the rate of interest actually earned, 6.5%, giving the value of the
insurer’s assets at time 6, before paying any death claims and expenses and setting
up policy values, as
(100 5 V + 100 × 0.94 P ) × 1.065 = $3 616 262.
The death claim plus related expenses at the end of the year was 100 250. A policy
value equal to 6 V (calculated in Example 7.3) is required at the end of the year for
each of the 99 policies still in force. Hence, the total amount the insurer requires at
the end of the year is
100 250 + 99 6 V = $3 597 342.
Hence the insurer has made a proﬁt in the sixth year of
(100 5 V + 100 × 0.94 P ) × 1.065 − (100 250 + 99 6 V ) = $18 919. (b) In this example the sources of proﬁt and loss in the sixth year are as follows.
(i) Interest: This is a source of proﬁt since the actual rate of interest earned, 6.5%, is
higher than the rate assumed in the policy value basis.
(ii) Expenses: These are a source of loss since the actual expenses, both premium related
(6% of premiums) and claim related ($250), are higher than assumed in the policy
value basis (5% of premiums and $200). 7.3. POLICIES WITH ANNUAL CASH FLOWS 295 (iii) Mortality: The probability of dying in the year for any of these policyholders is
q65 (= 0.0059). Hence, out of 100 policyholders alive at the start of the year, the
insurer expects 100 q65 (= 0.59) to die. In fact, one died. Each death reduces the
proﬁt since the amount required for a death, $100 250, is greater than the amount required on survival, 6 V (= $35 324), and so more than the expected deaths increases
the insurer’s loss.
Since the overall proﬁt is positive, (i) has had a greater eﬀect than (ii) and (iii) combined
in this year.
We can attribute the total proﬁt to the three sources as follows.
Interest: If expenses at the start the start of the year had been as assumed in the
policy value basis, 0.05 P per policy still in force, and interest had been earned at
5%, the total interest received in the year would have been
0.05 × (100 5 V + 100 × 0.95 P ) = $170 038.
The actual interest earned, before allowing for actual expenses, was
0.065 × (100 5 V + 100 × 0.95 P ) = $221 049.
Hence, there was a proﬁt of $51 011 attributable to interest.
Expenses: Now, we allow for the actual interest rate earned during the year (because
the diﬀerence between actual and expected interest has already been accounted for
in the interest proﬁt above) but using the expected mortality. That is, we look at
the loss arising from the expense experience given that the interest rate earned is
6.5%, but on the hypothesis that the number of deaths is 100 q65 .
The expected expenses on this basis, valued at the year end, are
100 × 0.05P × 1.065 + 100 q65 × 200 = $27 808. 296 CHAPTER 7. POLICY VALUES
The actual expenses, if deaths were as expected, are
100 × 0.06P × 1.065 + 100 q65 × 250 = $33 376.
The loss from expenses, allowing for the actual interest rate earned in the year but
allowing for the expected, rather than actual, mortality, was
33 376 − 27 808 = $5 568.
Mortality: Now, we use actual interest (6.5%) and actual expenses, and look at the
diﬀerence between the expected cost from mortality and the actual cost. For each
death, the cost to the insurer is the death strain at risk, in this case 100 000 + 250 −
6 V , so the mortality proﬁt is
(100 q65 − 1) × (100 000 + 250 − 6 V ) = −$26 524. This gives a total proﬁt of
51 011 − 5 568 − 26 524 = $18 919
which is the amount calculated earlier.
We have calculated the split in the order: Interest, Expenses, Mortality. At each step we
assume that factors not yet considered are as speciﬁed in the policy value basis, whereas
factors already considered are as actually occurred. This avoids ‘double counting’ and
gives the correct total.
However, we could follow the same principle, building from expected to actual, one basis
element at a time, but change the order of the calculation as follows.
Expenses: The loss from expenses, allowing for the assumed interest rate earned in
the year and allowing for the expected mortality, was
100 × (0.06 − 0.05) P × 1.05 + 100 q65 × (250 − 200) = $5 490. 7.3. POLICIES WITH ANNUAL CASH FLOWS 297 Interest: Allowing for the actual expenses at the start of the year, the proﬁt from
interest was
(0.065 − 0.05) × (100 5 V + 100 × 0.94 P ) = $50 933.
Mortality: The proﬁt from mortality, allowing for the actual expenses, was
(100q65 − 1) × (100 000 + 250 − 6 V ) = −$26 524.
This gives a total proﬁt of
−5 490 + 50 933 − 26 524 = $18 919
which is the same total as before, but with (slightly) diﬀerent amounts of proﬁt attributable to interest and to expenses.
This exercise of breaking down the proﬁt or loss into its component parts is called analysis
of surplus, and it is an important exercise after any valuation. The analysis of surplus
will indicate if any parts of the valuation basis are too conservative or too weak; it will
assist in assessing the performance of the various managers involved in the business, and
in determining the allocation of resources, and, for participating business it will help to
determine how much surplus should be distributed. 7.3.5 Asset shares In Section 7.3.1 we showed, using Example 7.1, that if the three conditions, (a), (b)
and (c), at the end of the section were fulﬁlled, then the accumulation of the premiums
received minus the claims paid for a group of identical policies issued simultaneously would
be precisely suﬃcient to provide the policy value required for the surviving policyholders
at each future duration. We noted that condition (c) in particular would be extremely
unlikely to hold in practice; that is, it is virtually impossible for the experience of a policy
or a portfolio of policies to follow exactly the assumptions in the premium basis. In 298 CHAPTER 7. POLICY VALUES practice, the invested premiums may have earned a greater or smaller rate of return than
that used in the premium basis, the expenses and mortality experience will diﬀer from
the premium basis. Each policy contributes to the total assets of the insurer through the
actual investment, expense and mortality experience.
It is of practical importance to calculate the share of the insurer’s assets attributable
to each policy in force at any given time. This amount is known as the asset share
of the policy at that time and it is calculated by assuming the policy being considered
is one of a large group of identical policies issued simultaneously. The premiums minus
claims and expenses for this notional group of policies are then accumulated using values
for expenses, interest, mortality and bonus rates based on the insurer’s experience for
similar policies over the period. At any given time, the accumulated fund divided by
the (notional) number of survivors gives the asset share at that time for each surviving
policyholder. If the insurer’s experience is close to the assumptions in the policy value
basis, then we would expect the asset share to be close to the policy value.
The reserve at duration t represents the amount the insurer needs to have at that time
in respect of each surviving policyholder; the asset share represents (an estimate of) the
amount the insurer actually does have.
Example 7.9 Consider a policy identical to the policy studied in Example 7.4 and suppose that this policy has now been in force for ﬁve years. Suppose that over the past ﬁve
years the insurer’s experience in respect of similar policies has been as follows.
• Annual interest earned on investments has been as shown in the following table.
Year
1
Interest % 4.8 2
5.6 3
5.2 4
4.9 5
4.7 • Expenses at the start of the year in which a policy was issued were 15% of the
premium. 7.3. POLICIES WITH ANNUAL CASH FLOWS 299 • Expenses at the start of each year after the year in which a policy was issued were
6% of the premium.
• The expense of paying a death claim was, on average, $120.
• The mortality rate, q[50]+t , for t = 0, 1, . . . , 4, has been approximately 0.0015.
Calculate the asset share for the policy at the start of each of the ﬁrst six years.
Solution 7.9 We assume that the policy we are considering is one of a large number, N ,
of identical policies issued simultaneously. As we will see, the value of N does not aﬀect
our ﬁnal answers.
Let ASt denote the asset share per policy surviving at time t = 0, 1, . . . , 5. We calculate
ASt by accumulating to time t the premiums received minus the claims and expenses paid
in respect of this notional group of policies using our estimates of the insurer’s actual
experience over this period and then dividing by the number of surviving policies. We
adopt the convention that ASt does not include the premium and related expense due at
time t. With this convention, AS0 is always 0 for any policy since no premiums will have
been received and no claims and expenses will have been paid before time 0. Note that
for our policy, using the policy value basis speciﬁed in Example 7.4, 0 V = $490.
The premiums minus expenses received at time 0 are
0.85 × 11 900 N = 10 115 N.
This amount accumulates to the end of the year with interest at 4.8%, giving
10 601 N.
A notional 0.0015 N policyholders die in the ﬁrst year so that death claims plus expenses
at the end of the year are
0.0015 × (11 900 + 120) N = 18 N 300 CHAPTER 7. POLICY VALUES Fund at
start of
Year, t
year
1
0
2
10 582 N
3
22 934 N
4
35 805 N
5
49 170 N Cash ﬂow Fund at end
at start of year before
of year
death claims
10 115 N
10 601 N
11 169 N
22 970 N
11 152 N
35 859 N
11 136 N
49 241 N
11 119 N
63 123 N Fund at
Death claims
end of
and expenses
year
18 N
10 582 N
36 N
22 934 N
54 N
35 805 N
71 N
49 170 N
89 N
63 034 N Survivors
0.9985 N
0.99852 N
0.99853 N
0.99854 N
0.99855 N ASt
10 598
23 003
35 967
49 466
63 509 Table 7.1: Asset share calculation for Example 7.9.
which leaves
10 601 N − 18 N = 10 582 N
at the end of the year. Since 0.9985 N policyholders are still surviving at the start of the
second year, AS1 , the asset share for a policy surviving at the start of the second year is
given by
AS1 = 10 582N/(0.9985 N ) = 10 598.
These calculations, and the calculations for the next four years, are summarized in Table
7.1. You should check all the entries in this table. For example, the death claims and
expenses in year 5 are calculated as
0.99854 × 0.0015 × (5 × 11 900 + 120) N = 89 N
since 0.99854 N policyholders are alive at the start of the ﬁfth year, a fraction 0.0015 of
these die in the coming year, the death beneﬁt is a return of the ﬁve premiums paid and
the expense is $120.
Note that the ﬁgures in Table 7.1, except the ‘Survivors’ column, have been rounded to
the nearest integer for presentation; the underlying calculations have been carried out
using far greater accuracy. 7.4. POLICY VALUES WITH 1/M THLY CASH FLOWS 301 We make the following comments about Example 7.9.
1. As predicted, the value of N does not aﬀect the values of the asset shares, ASt . The
only purpose of this notional group of N identical policies issued simultaneously is
to simplify the presentation.
2. The experience of the insurer over the ﬁve years has been close to the assumptions
in the policy value basis speciﬁed in Example 7.4. The actual interest rate has been
between 4.7% and 5.6%; the rate assumed in the policy value basis is 5%. The actual
expenses, both premiumrelated (15% initially and 6% thereafter) and claimrelated
($120), are a little higher than the expenses assumed in the policy value basis (10%,
5% and $100, respectively). The actual mortality rate is comparable to the rate in
the policy value basis, e.g. 0.99855 = 0.99252 is close to 5 p[50] = 0.99283.
As a result of this, the asset share, AS5 (= $63 509) is reasonably close to the policy
value, 5 V (= $65 470) in this example. 7.4 Policy values for policies with cash ﬂows at discrete intervals other than annually Throughout Section 7.3 we assumed all cash ﬂows for a policy occurred at the start or
end of each year. This simpliﬁed the presentation and the calculations in the examples.
In practice, this assumption does not often hold; for example, premiums are often payable
monthly and death beneﬁts are usually payable immediately following, or, more realistically, soon after, death. The deﬁnition of a policy value from Deﬁnitions 7.1 and 7.2 can
be directly applied to policies with more frequent cash ﬂows. The policy value at duration
t is still the expected value of the future loss random variable, assuming the policyholder
is still alive at that time – and our interpretation of a policy value is unchanged – it is
still the amount the insurer needs so that, with future premiums, it can expect (on the
policy value basis) to pay future beneﬁts and expenses. 302 CHAPTER 7. POLICY VALUES The following example illustrates these points.
Example 7.10 A life aged 50 purchases a 10 year term insurance with sum insured
$500 000 payable at the end of the month of death. Level quarterly premiums, each of
amount P = $460, are payable for at most ﬁve years.
Calculate the (gross premium) policy values at durations 2.75, 3 and 6.5 years using the
following basis.
Mortality: the standard select survival model
Interest: 5% per year
Expenses: 10% of each gross premium
Solution 7.10 To calculate 2.75 V we need the EPV of future beneﬁts and the EPV of
premiums less expenses at that time, assuming the policyholder is still alive. Note that
the premium and related expense due at time t = 2.75 are regarded as future cash ﬂows.
Note also that from duration 2.75 years the policyholder will be subject to the ultimate
part of the survival model since the select period is only two years.
Hence
2.75 V (12) 1
52.75: 7.25 = 500 000A (4) − 0.9 × 4 × P a52.75: 2.25
¨ = $3 091.02.
(12) 1
52.75: 7.25 where A = 0.01327 (4) a52.75: 2.25 = 2.14052
¨ Similarly
3V (12) 1
53: 7 = 500 000A = $3 357.94, (4) − 0.9 × 4 × P a53: 2
¨ 7.4. POLICY VALUES WITH 1/M THLY CASH FLOWS
(12) 1
53: 7 where A = 0.013057 303 (4) a53: 2 = 1.91446
¨ and
6 .5 V (12) 1
56.5: 3.5 = 500 000A = 500 000 × 0.008532 = $4 265.63. 7.4.1 Recursions We can derive recursive formulae for policy values for policies with cash ﬂows at discrete
times other than annually. We use the recursion to connect policy values at any possible
cash ﬂow date, premiums beneﬁts or expenses. Consider 2.75 V and 3 V in Example 7.10.
We need to be careful here because the premiums and beneﬁts are paid with diﬀerent
frequency. We can use a recurrence relationship to generate the policy value at each
month end, allowing for premiums only every third month. So, for example,
( 2.75 V + 460 − 0.1 × 460) (1.05)0.0833 = 500 000 0.0833 q52.75 + ( 0.0833 p52.75 ) 2.8333 V . and similarly
2.8333 V (1.05)0.0833 = 500 000 0.0833 q52.8333 + ( 0.0833 p52.8333 ) 2.9167 V (1.05)0.0833 = 500 000 0.0833 q52.9167 + ( 0.0833 p52.9167 ) 3 V 7.4.2 2.9167 V Valuation between premium dates All of the calculations in the sections above considered policy values at a premium date,
or after premiums have ceased. We often need to calculate policy values between premium 304 CHAPTER 7. POLICY VALUES dates; typically, we will value all policies on the same calendar date each year as part of
the insurer’s liability valuation process. The principle when valuing between premium
dates is the same as when valuing on premium dates, that is, the policy value is the
EPV of future beneﬁts minus premiums. The calculation may be a little more awkward.
We demonstrate in the following example, which uses the same contract as example 7.10
above.
Example 7.11 For the contract described in Example 7.10, calculate the policy value
after (a) 2 years and 10 months and (b) 2 years and 9.5 months, assuming the policy is
still in force at that time in each case.
Solution 7.11
SA (a) The EPV of future beneﬁts is
(12) 1
52.8333:7.1333 = S × 0.0132012 = 6 600.58
(m) (m) 1
Note that the functions A x:n and a x:n are deﬁned only if n is an integer multiple
¨
(12) 1
(4)
1
of m , so that A 52.8333:7.1333 is well deﬁned, but a 52.8333:7.1333 is not.
¨ The EPV of future premiums less premium expenses is
(4) (0.9)(4P )v 0.1667 0.1667 p52.8333 a53:2 = (0.9)(4P )(1.898466) = 3 143.86
¨
So the policy value is 2.83333 V = $3 456.72. (b) Now, the valuation is at neither a beneﬁt nor a premium date. We know that the
EPV of beneﬁts minus premiums at 2 years and 10 months is 2.8333 V . Onehalf of a
month earlier, we know that the life must either survive the time to the month end,
in which case the EPV of future beneﬁts less premiums is 2.8333 V v 0.0417 , or the life
will die, in which case the EPV of beneﬁts less premiums is S v 0.0417 . Allowing for
the appropriate probabilities of survival or death, the value at t = 2.7917 is
2.7917 V = 0.0417 q52.7917 S v 0.0417 + 0.0417 p52.7917 v 0.0417 2.83333 V = $3 480.99 7.4. POLICY VALUES WITH 1/M THLY CASH FLOWS 305 The principle here is that we have split the EPV into the part relating to cash ﬂows up
to the next premium date, plus the EPV of the policy value at the next premium date.
It is interesting to note here that it would not be appropriate to apply simple interpolation
to the two policy values corresponding to the premium dates before and after the valuation
date, as we have, for example,
2.75 V = $3 091.02, 2.7917 V = $3 480.99 and 3V = $3 357.94. The reason is that the curve of t V is not smooth if premiums are paid at discrete intervals,
since the policy value will jump immediately after each premium payment by the amount
of that payment. Before the premium payment, the premium immediately due is included
in the EPV of future premiums, which is deducted from the EPV of future beneﬁts to
give the policy value. Immediately after the premium payment, it is no longer included,
so the policy value increases by the amount of the premium.
In Figure 7.5 we show the policy values at all durations for the policy in Examples 7.10
and 7.11. The curve jumps at each premium date, and has an increasing trend until the
premiums cease. In the second half of the contract, after the premium payment term,
the policy value is run down. Other types of policy will have diﬀerent patterns for policy
values as we have seen in Figures 7.1 and 7.2.
A reasonable approximation to the policy value between premium dates can usually be
achieved by interpolating between the policy value just after the previous premium
and the policy value just before the next premium. That is, suppose the premium dates
are k years apart, then for s < k , we approximate t+k+s V by interpolating between
t+k V + Pt+k − Et+k and t+2k V ; more speciﬁcally,
t+k+s V ≈ (t+k V + Pt+k − Et+k ) 1 − s
s
+ (t+2k V )
.
k
k In the example above, this would give approximate values for 2.7917 V and 2.8333 V of
$3 480.51 and $3 455.99 respectively, compared with the accurate values of $3 480.99 and
$3 456.72 respectively. 306 CHAPTER 7. POLICY VALUES 6000 5000 Policy Value ($) 4000 3000 2000 1000 0
0 1 2 3 4 5 6 7 8 9 10 Duration (Years) Figure 7.5: Policy values for the limited premium term insurance contract, Example 7.11. 7.5 Policy values with continuous cash ﬂows 7.5.1 Thiele’s diﬀerential equation We have seen in Section 7.3 how to deﬁne policy values for policies with cash ﬂows at
discrete intervals and also how to derive recursive formulae linking reserves at successive
cash ﬂow time points. These ideas extend to policies where regular payments – premiums
and/or annuities – are payable continuously and sums insured are payable immediately
on death. In this case we can derive a diﬀerential equation, known as Thiele’s diﬀerential
equation. This is a continuous time version of the recursion equation (7.7), which we
derived in Section 7.3.3. Recall that for the discrete case
( t V + Pt − et )(1 + it ) = t+1 V + q[x]+t (St+1 + Et+1 − t+1 V ) (7.7) Our derivation of Thiele’s diﬀerential equation is somewhat diﬀerent to the derivation of
equation (7.7). However, once we have completed the derivation, we explain the link with 7.5. CONTINUOUS CASH FLOWS 307 this equation.
Consider a policy issued to a select life aged x under which premiums and premium–
related expenses are payable continuously and the sum insured, together with any related
expenses, is payable immediately on death. Suppose this policy has been in force for t
years, where t ≥ 0. Let
Pt denote the annual rate of premium payable at time t,
et denote the annual rate of premiumrelated expense payable at time t,
St denote the sum insured payable at time t if the policyholder dies at exact time t,
Et denote the expense of paying the sum insured at time t,
µ[x]+t denote the force of mortality at age [x] + t,
δt denote the force of interest per year assumed earned at time t, and,
tV denote the policy value for a policy in force at time t. We assume that Pt , et , St , µ[x]+t and δt are all continuous functions of t and that
et , Et , µ[x]+t and δt are all as assumed in the policy value basis.
Note that just as we allowed the rate of interest to vary from year to year in Section 7.3.3,
we are here letting the force of interest be a continuous function of time. Thus, if v (t)
denotes the present value of a payment of 1 at time t, we have
t v (t) = exp − δs ds . (7.8) 0 As t V represents the diﬀerence between the EPV of beneﬁts plus beneﬁt–related expenses
and the EPV of premiums less premium–related expenses, we have
tV ∞ =
0 v (t + s)
(St+s + Et+s ) s p[x]+t µ[x]+t+s ds
v (t) 308 CHAPTER 7. POLICY VALUES − ∞
0 v (t + s)
(Pt+s − et+s ) s p[x]+t ds.
v (t) Note that we are measuring time, represented by s in the integrals, from time t, so that if,
for example, the sum insured is payable at time s, the amount of the sum insured is St+s
and as we are discounting back to time t, the discount factor is v (t + s)/v (t). Changing
the variable of integration to r = t + s gives
tV ∞ =
t v (r)
(Sr + Er )
v (t) r −t p[x]+t µ[x]+r dr − ∞
t v (r)
(Pr − er )
v (t) r−t p[x]+t dr. (7.9) We could use formula (7.9) to calculate t V by numerical integration. However, we are
instead going to turn this identity into a diﬀerential equation. There are two main reasons
why we do this:
1. There exist numerical techniques to solve diﬀerential equations, one of which is
discussed in the next section. As we will see, an advantage of such an approach
over numerical integration is that we can easily calculate policy values at multiple
durations.
2. In Chapter 8 we consider more general types of insurance policy than we have so
far. For such policies it is usually the case that we are unable to calculate policy
values using numerical integration, and we must calculate policy values using a set
of diﬀerential equations. The following development of Thiele’s diﬀerential equation
sets the scene for the next chapter.
In order to turn equation (7.9) into a diﬀerential equation, we note that
r−t p[x]+t = r p [x]
t p [x] so that
1
tV =
v (t) t p[x] ∞
t v (r) (Sr + Er ) r p[x] µ[x]+r dr − ∞
t v (r) (Pr − er ) r p[x] dr , 7.5. CONTINUOUS CASH FLOWS 309 which we can write as
∞ v (t) t p[x] t V =
t v (r) (Sr + Er ) r p[x] µ[x]+r dr − ∞
t v (r) (Pr − er ) r p[x] dr. (7.10) Diﬀerentiation of equation (7.10) with respect to t leads to Thiele’s diﬀerential equation.
First, diﬀerentiation of the right hand side yields
−v (t) (St + Et ) t p[x] µ[x]+t + v (t) (Pt − et ) t p[x]
= v (t) t p[x] Pt − et − (St + Et ) µ[x]+t . (7.11) Diﬀerentiation of the left hand side is most easily done in two stages, applying the product
rule for diﬀerentiation at each stage. Treating v (t) t px as a single function of t we obtain
d
d
d
v (t) t p[x] t V = v (t) t p[x]
v (t) t p[x] .
tV + tV
dt
dt
dt
Next,
d
d
d
v (t) t p[x] = v (t) t p[x] + t p[x]
v (t).
dt
dt
dt
From the Chapter 2 we know that
d
t p[x] = − t p[x] µ[x]+t
dt
and from formula (7.8)
d
v (t) = −δt exp −
dt t δs ds
0 = −δt v (t). Thus, the derivative of the left hand side of equation (7.10) is
d
v (t) t p[x] t V
dt = v (t) t p[x]
= v (t) t p[x] d
t V − t V v (t) t p[x] µ[x]+t + t p[x] δt v (t)
dt
d
t V − t V µ[x]+t + δt
dt . 310 CHAPTER 7. POLICY VALUES Equating this to (7.11) yields Thiele’s diﬀerential equation, namely
d
t V = δt t V + Pt − et − (St + Et − t V ) µ[x]+t .
dt (7.12) Formula (7.12) can be interpreted as follows. The left hand side of the formula, d t V /dt,
is the rate of increase in the policy value at time t. We can derive a formula for this rate
of increase by considering the individual factors aﬀecting the value of t V :
• Interest is being earned on the current amount of the policy value. The amount of
interest earned in the time interval t to t + h is δt t V h (+o(h)), so that the rate of
increase at time t is δt t V .
• Premium income, minus premium–related expenses, is increasing the policy value
at rate Pt − et . If there were annuity payments at time t, this would decrease the
policy value at the rate of the annuity payment (plus any annuity–related expenses).
• Claims, plus claim–related expenses, decrease the amount of the policy value. The
expected extra amount payable in the time interval t to t + h is µ[x]+t h (St + Et − t V )
and so the rate of decrease at time t is µ[x]+t (St + Et − t V ).
Hence the total rate of increase of the policy value at time t is
δt t V + Pt − et − µ[x]+t (St + Et − t V ).
We can also relate formula (7.12) to equation (7.7) assuming that for some very small
value h,
1
d
( t+h V − t V ) ,
tV ≈
dt
h (7.13) leading to the relationship
(1 + δt h) t V + (Pt − et )h ≈ t+h V + hµ[x]+t (St + Et − t V ). 7.5. CONTINUOUS CASH FLOWS 311 Remembering that h is very small, the interpretation of the left hand side is that it is the
accumulation from time t to time t + h of the policy value at time t plus the accumulation
at time t + h of the premium income less premium–related expenses over the interval
(t, t + h). (Note that for very small h, sh ≈ h.) This total accumulation must provide
¯
the policy value at time t + h, and, if death occurs in the interval (t, t + h), it must also
provide the excess St + Et − t V over the policy value. The probability of death in the
interval (t, t + h) is approximately hµ[x]+t . 7.5.2 Numerical solution of Thiele’s diﬀerential equation In this section we show how we can evaluate policy values by solving Thiele’s diﬀerential
equation numerically. The key to this is to apply equation (7.13) as an identity rather
than an approximation, assuming that h is very small. This leads to
t+h V − t V = h(δt t V + Pt − et − µ[x]+t (St + Et − t V )). (7.14) The smaller the value of h, the better this approximation is likely to be. The values
of δt , Pt , et , µ[x]+t , St and Et are assumed to be known, so this equation allows us to
calculate t V provided we know the value of t+h V , or t+h V if we know the value of t V .
But we always know the value of t V as t approaches the end of the policy term since, in
the limit, it is the amount that should be held in respect of a policyholder who is still
alive. For an endowment policy with term n years and sum insured S , the policy value
builds up so that just before the maturity date it is exactly suﬃcient to pay the maturity
beneﬁt, that is
lim t V = S, t→n− for a term insurance with term n years and sum insured S , we have
lim t V = 0, t→n− 312 CHAPTER 7. POLICY VALUES and for a whole life insurance with sum insured S , we have
lim t V = S, t→ω − where ω is either the upper limit of the survival model, or a practical upper limit for
inﬁnite models.
Using the endowment policy with term n years and sum insured S as an example, formula
(7.14) with t = n − h gives us
S− n−h V = h δn−h n−h V + Pn−h − en−h − µ[x]+n−h (Sn−h + En−h − n−h V ), from which we can calculate n−h V . Another application of formula (7.14) with t = n − 2h
gives the value of n−2h V , and so on.
This method for the numerical solution of a diﬀerential equation is known as Euler’s
method. It is the continuous time version of the discrete time recursive method for
calculating reserves illustrated in Example 7.7.
Example 7.12 Consider a 20 year endowment insurance issued to a life aged 30. The
sum insured, $100 000, is payable immediately on death, or on survival to the end of the
term, whichever occurs sooner. Premiums are payable continuously at a constant rate of
$2 500 per year throughout the term of the policy. The policy value basis uses a constant
force of interest, δ , and makes no allowance for expenses.
(a) Evaluate 10 V . (b) Use Euler’s method with h = 0.05 years to calculate
Perform the calculations on the following basis:
Mortality: the standard select survival model
Interest: δ = 0.04 per year 10 V . 7.5. CONTINUOUS CASH FLOWS
Solution 7.12
10 V 313 (a) We have
¯
= 100 000A40:10 − 2 500¯40:10 ,
a and as
¯
A40:10 = 1 − δ a40:10 ,
¯
we can calculate
10 V 10 V as = 100 000 − (100 000δ + 2 500)¯40:10 .
a Using numerical integration or the three term Woolhouse formula, we get
a40:10 = 8.2167,
¯
and hence 10 V = 46 591. (b) For this example, δt = 0.04, et = 0 = Et , Pt = 2 500 and µ49.95 = 0.003204. Hence
100 000 − V19.95 = 0.05 × (0.04 × V19.95 + 2 500 − 0.003204 × (100 000 − V19.95 ))
and so
V19.95 = 99 676.
Calculating recursively V19.9 , V19.85 , . . ., we arrive at
10 V = 46 635. We note that the answer here is close to $46 591, the value calculated in part (a).
Using a value of h = 0.01 gives the closer answer of $46 600. 314 CHAPTER 7. POLICY VALUES We remarked earlier that a useful feature of setting up and numerically solving a diﬀerential equation for policy values is that the numerical solution gives policy values at a
variety of durations. We can see this in the above example. In part (a) we wrote down
an expression for 10 V and evaluated it using numerical integration. By contrast, in part
(b) with h = 0.05, as a by–product of our backwards recursive calculation of 10 V we also
obtained values of 10+h V , 10+2h V . . . ,20−h V .
Other major advantages of Thiele’s equation arise from its versatility and ﬂexibility. We
can easily accommodate variable premiums, beneﬁts and interest rates. We can also use
the equation to solve numerically for the premium given the beneﬁts, interest model and
boundary values for the policy values. 7.6 Policy alterations A life insurance policy is a contract between an individual, the policyholder, and the
insurance company. This contract places obligations on both parties; for example, the
policyholder agrees to pay regular premiums while he or she remains alive and the insurance company agrees to pay a sum insured, plus bonuses for a participating policy, on
the death of the policyholder. So far in this book we have assumed that the terms of the
contract are never broken or altered in any way. In practice, it is not uncommon, after
the policy has been in force for some time, for the policyholder to request a change in the
terms of the policy. Typical changes might be:
(1) The policyholder wishes to cancel the policy with immediate eﬀect. In this case,
it may be appropriate for the insurance company to pay a lump sum immediately
to the policyholder. This will be the case if the policy has a signiﬁcant investment
component – such as an endowment insurance, or a whole life insurance. Term
insurance contracts generally do not have an investment objective. A policy which
is cancelled at the request of the policyholder before the end of its originally agreed
term, is said to lapse or to be surrendered, and any lump sum payable by the 7.6. POLICY ALTERATIONS 315 insurance company for such a policy is called a surrender value or a cash value.
We tend to use the term lapse to indicate a voluntary cessation when no surrender
value is paid, and surrender when there is a return of assets of some amount to
the policyholder, but the words may be used interchangeably.
In the US and some other countries, insurers are required to oﬀer cash surrender values on certain contract types once they have been in force for one or two years. The
stipulation is known as the nonforfeiture law. The allowance for zero cash values
for early surrenders reﬂects the need of the insurers to recover the new business
strain associated with issuing the policy.
(2) The policyholder wishes to pay no more premiums but does not want to cancel the
policy, so that, in the case of an endowment insurance for example, a (reduced)
sum insured is still payable on death or on survival to the end of the original term,
whichever occurs sooner. Any policy for which no further premiums are payable
is said to be paid–up, and the reduced sum insured for a policy which becomes
paid–up before the end of its original premium paying term is called a paid–up
sum insured.
(3) A whole life policy may be converted to a paid–up term insurance policy for the
original sum insured.
(4) Many other types of alteration can be requested: reducing or increasing premiums;
changing the amount of the beneﬁts; converting a whole life insurance to an endowment insurance; converting a non–participating policy to a with proﬁt policy;
and so on. The common feature of these changes is that they are requested by the
policyholder and were not part of the original terms of the policy.
If the change was not part of the original terms of the policy, and if it has been requested by
the policyholder, it could be argued that the insurance company is under no obligation to
agree to it. However, when the insurer has issued a contract with a substantive investment
objective, rather than solely oﬀering protection against untimely death, then at least part 316 CHAPTER 7. POLICY VALUES of the funds should be considered to be the policyholder’s, under the stewardship of the
insurer. In the US the nonforfeiture law states that, for investment type policies, each
of (1), (2) and (3) would generally be available on pre–speciﬁed minimum terms. In
particular, ﬁxed or minimum cash surrender values, as a percentage of the sum insured,
are speciﬁed in advance in the contract terms for such policies.
For policies with pre–speciﬁed cash surrender values, let Ct denote the cash surrender
value at duration t. Where surrender values are not set in advance, the actuary would
determine an appropriate value for Ct at the time of alteration.
Starting points for the calculation of Ct could be the policy value at t, t V , if it is to be
calculated in advance, or the policy’s asset share, ASt , when the surrender value is not
pre–speciﬁed. Recall that ASt represents (approximately) the cash the insurer actually
has and t V represents the amount the insurer should have at time t in respect of the
original policy. Recall also that if the policy value basis is close to the actual experience,
then t V will be numerically close to ASt .
Setting Ct equal to either ASt or t V could be regarded as over–generous to the policyholder
for several reasons, including:
(1) It is the policyholder who has requested that the contract be changed. The insurer
will be concerned to ensure that surrendering policyholders do not beneﬁt at the
expense of the continuing policyholders – most insurers prefer the balance to go the
other way, so that policyholders who maintain their contracts through to maturity
achieve greater value than those who surrender early or change the contract. Another implication of the fact that the policyholder has called for alteration is that
the policyholder may be acting on knowledge that is not available to the insurer.
For example, a policyholder may alter a whole life policy to a term insurance (with
lower premiums or a higher sum insured) if he or she becomes aware that their
health is failing. This is called antiselection or selection against the insurer.
(2) The insurance company will incur some expenses in making the alterations to the
policy, and even in calculating and informing the policyholder of the revised values, 7.6. POLICY ALTERATIONS 317 which the policyholder may not agree to accept.
(3) The alteration may, at least in principle, cause the insurance company to realize
assets it would otherwise have held, especially if the alteration is a surrender. This
liquidity risk may lead to reduced investment returns for the company. Under US
nonforfeiture law, the insurer has six months to pay the cash surrender value, so
that it is not forced to sell assets at short notice.
For these reasons, Ct is usually less than 100% of either ASt or t V and may include an
explicit allowance for the expense of making the alteration.
For alterations other than cash surrenders, we can apply Ct as if it were a single premium,
or an extra preliminary premium, for the future beneﬁts. That is, we construct the
equation of value for the altered beneﬁts,
Ct + EPV at t of future premiums, altered contract
= EPV at t of future beneﬁts plus expenses, altered contract. (7.15) The numerical value of the revised beneﬁts and/or premiums calculated using equation
(7.15) depends on the basis used for the calculation, that is, the assumptions concerning
the survival model, interest rate, expenses and future bonuses (for a with proﬁts policy).
This basis may be the same as the premium basis, or the same as the policy value basis,
but in practice usually diﬀers from both of them.
The rationale behind equation (7.15) is the same as that which leads to the equivalence
principle for calculating premiums: together with the cash currently available (Ct ), the
future premiums are expected to provide the future beneﬁts and pay for the future expenses.
Example 7.13 Consider the policy discussed in Examples 7.4 and 7.9. You are given
that the insurer’s experience in the ﬁve years following the issue of this policy is as in 318 CHAPTER 7. POLICY VALUES Example 7.9. At the start of the sixth year, before paying the premium then due, the
policyholder requests that the policy be altered in one of the following three ways.
(a) The policy is surrendered immediately.
(b) No more premiums are paid and a reduced annuity is payable from age 60. In
this case, all premiums paid are refunded at the end of the year of death if the
policyholder dies before age 60.
(c) Premiums continue to be paid, but the beneﬁt is altered from an annuity to a lump
sum (pure endowment) payable on reaching age 60. Expenses and beneﬁts on death
before age 60 follow the original policy terms. There is an expense of $100 associated
with paying the sum insured at the new maturity date.
Calculate the surrender value (a), the reduced annuity (b) and the sum insured (c) assuming the insurer uses
(i) 90% of the asset share less a charge of $200, or
(ii) 85% of the policy value less a charge of $200
together with the assumptions in the policy value basis when calculating revised beneﬁts
and premiums.
Solution 7.13 We already know from Examples 7.4 and 7.9 that
5V = 65 470 and AS5 = 63 509. Hence, the amount C5 to be used in equation (7.15) is
(i) 0.9 × AS5 − 200 = 56 958,
(ii) 0.9 × 5 V − 200 = 58 723. 7.6. POLICY ALTERATIONS 319 (a) The surrender values are the cash values C5 , so we have
(i) $56 958,
(ii) $58 723.
(b) Let X denote the revised annuity amount. In this case, equation (7.15) gives
1
1
¨
C5 = 5 × 11 900A55:5 + 100A55:5 + (X + 25)v 5 5 p55 a60 . Using values calculated for the solution to Example 7.4, we can solve this equation
for the two diﬀerent values for C5 to give
(i) X = $4 859,
(ii) X = $5 012.
(c) Let S denote the new sum insured. Equation (7.15) now gives
C5 + 0.95 × 11 900¨55:5 = 11 900 (IA)1 5 + 5A1 5 + 100A1 5 + v 5 5 p55 (S + 100)
a
55:
55:
55:
which we solve using the two diﬀerent values for C5 to give
(i) S = $138 314,
(ii) S = $140 594. Example 7.14 Ten years ago a man now aged 40 purchased a withproﬁt whole life
insurance. The basic sum insured, payable at the end of the year of death, was $200 000.
Premiums of $1 500 were payable annually for life.
The policyholder now requests that the policy be changed to a withproﬁt endowment
insurance with a remaining term of 20 years, with the same premium payable annually,
but now for a maximum of 20 further years. 320 CHAPTER 7. POLICY VALUES The insurer uses the following basis for the calculation of policy values and policy alterations.
Mortality: the standard select survival model
Interest: 5% per year
Expenses: none
Bonuses: compound reversionary bonuses at rate 1.2% per year at the start of each
policy year, including the ﬁrst.
The insurer uses the full policy value less an expense of $1 000 when calculating revised
beneﬁts. You are given that the actual bonus rate declared in each of the past 10 years
has been 1.6%.
(a) Calculate the revised sum insured, to which future bonuses will be added, assuming
the premium now due has not been paid and the bonus now due has not been
declared.
(b) Calculate the revised sum insured, to which future bonuses will be added, assuming
the premium now due has been paid and the bonus now due has been declared to
be 1.6%
Solution 7.14 (a) Before the declaration of the bonus now due, the sum insured for
the original policy is
200 000 × 1.01610 = 234 405.
Hence, the policy value for the original policy,
10 V 10 V , is given by = 234 405A40 j − P a40
¨ where P = 1 500 and the subscript j indicates that the rate of interest to be used
is 3.75494% since
1.05/1.012 = 1.0375494. 7.6. POLICY ALTERATIONS 321 Let S denote the revised sum insured. Then, using equation (7.15)
10 V − 1 000 = S A40:20 j − P a40:20 .
¨ (7.16) A point to note here is that the life was select at the time the policy was purchased,
ten years ago. No further health checks are carried out at the time of a policy
alteration and so the policyholder is now assumed to be subject to the ultimate
part of the survival model.
You should check the following values
A40 j = 0.19569,
A40:20 j = 0.48233, a40 = 18.4578,
¨
a40:20 = 12.9935.
¨ Hence
S = $76 039.
(b) Let 10+ V denote the policy value just after the premium has been paid and the
bonus has been declared at time 10. The term A40 j used in the calculation of 10 V
assumed the bonus to be declared at time 10 would be 1.2%, so that the sum insured
in the 11th year would be 234 405 × 1.012, in the 12th year would be 234 405 × 1.0122 ,
and so on. Given that the bonus declared at time 10 is 1.6%, these sums insured
are now 234 405 × 1.016 (this value is known) and 234 405 × 1.016 × 1.012 (this is
an assumed value since it assumes the bonus declared at the start of the 12th year
will be 1.2%). Hence
10+ V = (1.016/1.012) × 234 405A40 j − P a40
= (1.016/1.012) × 234 405A40 j − P a40 + P.
¨ Let S denote the revised sum insured for the endowment policy in this case. Equation (7.15) now gives
10+ V − 1 000 = (S /1.012) A40:20 j − P a40:19 322 CHAPTER 7. POLICY VALUES
= (S /1.012) A40:20 j − P (¨40:20 − 1),
a
and hence
S = $77 331. Note that, in Example 7.14, the sum insured payable in the 11th year is S × 1.016 =
$149 295 in part (a) and $149 381 in part (b). The diﬀerence between these values is not
due to rounding – the timing of the request for the alteration has made a (small) diﬀerence
to the sum insured oﬀered by the insurer for the endowment insurance. This is caused
partly by the charge of $1 000 for making the alteration and partly by the fact that the
bonus rate in the 11th year is not as assumed in the policy value basis. In Example 7.14
we would have S = S × 1.012 if there were no charge for making the alteration and the
bonus rate declared in the 11th year were the same as the rate assumed in the reserve
basis (and the full policy value is still used in the calculation of the revised beneﬁt). 7.7 Retrospective policy value Our deﬁnition of a policy value is based on the future loss random variable. As noted
in Section 7.3.2, comment (ii), what we have called a policy value is called by some
authors a prospective policy value. Since prospective means looking to the future, this
name has some merit. Some authors also deﬁne what they call a retrospective policy value
at duration t, which is calculated by accumulating premiums received less beneﬁts paid up
to time t for a large group of identical policies, assuming the experience follows precisely
the assumptions in the policy value basis, and sharing the resulting fund equally among
the surviving policyholders. This is precisely the calculation detailed in the ﬁnal part
of Section 7.3.1 in respect of the policy studied in Example 7.1, so that the left hand
side of formula (7.2) is a formula for the retrospective policy value (at duration 10) for 7.8. NEGATIVE POLICY VALUES 323 this particular policy. These authors typically show that, under some conditions, the
retrospective and prospective policy values are equal. These conditions are conditions (a)
and (b) at the end of Section 7.3.1 – note that our condition (c) has already been used
to calculate the retrospective policy value. In this chapter we have not introduced the
retrospective policy value for the following reasons:
(1) When our conditions (a) and (b) in Section 7.3.1 do not hold, the retrospective
policy value is not equal to the prospective policy value.
(2) The retrospective policy value equals the asset share if the experience follows precisely the assumptions in the policy value basis. Otherwise, they are unlikely to be
equal. Since the asset share represents the amount the insurer actually has at time t
in respect of a policy still in force, it is a more useful quantity than the retrospective
policy value. 7.8 Negative policy values In all our examples in this chapter, the policy value was either zero or positive. It can
happen that a policy value is negative. In fact, negative policy values are not unusual in
the ﬁrst few months of a contract, after the initial expenses have been incurred, and before
suﬃcient premium is collected to defray these expenses. However, it would be unusual
for policy values to be negative after the early period of the contract. If we consider the
policy value equation
tV = EPV at t of Future Beneﬁts + Expenses − EPV at t of Future Premiums, then we can see that, since the future beneﬁts and premiums must both have nonnegative
EPVs, the only way for a negative policy value to arise is if the future beneﬁts are worth
less than the future premiums.
In practice, negative policy values would generally be set to zero when carrying out a
valuation of the insurance company. Allowing them to be entered as assets (negative 324 CHAPTER 7. POLICY VALUES liabilities) ignores the policyholder’s option to lapse the contract, in which case the excess
premium will not be received.
Negative policy values arise when a contract is poorly designed, so that the value of
beneﬁts in early years exceeds the value of premiums, followed by a period when the
order is reversed. If the policyholder lapses then the policyholder will have beneﬁtted
from the higher beneﬁts in the early years without waiting around to pay for the beneﬁt
in the later years. In fact, the policyholder may be able to achieve the same beneﬁt at a
cheaper price by lapsing and buying a new policy – called the lapse and reentry option. 7.9 Notes and further reading Thiele’s diﬀerential equation is named after the Danish actuary Thorvald N. Thiele (1838–
1910). For information about Thiele, see Hoem (1983).
Euler’s method for the numerical solution of a diﬀerential equation has the advantages
that it is relatively simple to implement and it relates to the recursive formulae for policy
values for policies with annual cash ﬂows. In practice, there are better methods for solving
such equations, for example the Runge–Kutta method. See Burden and Faires (2001).
Texts such as for example Neill (1977) and Bowers et al (1986) refer to retrospective
policy values. These references also contain standard actuarial notation for policy values. 7.10. EXERCISES 7.10 325 Exercises When a calculation is required in the following exercises, unless otherwise stated you
should assume that mortality follows the standard select survival model, as speciﬁed in
Example 3.13 in Section 3.9, and that the equivalence principle is used for the calculation
of premiums.
Exercise 7.1 You are given the following extract from a select life table with 4 year
select period. A select individual aged 41 purchased a 3 year term insurance with a sum
insured of $200 000, with premiums payable annually throughout the term.
[x ]
[40]
[41]
[42] l[x]
100 000
99 802
99 597 l[x]+1
l[x]+2
l[x]+3
lx+4
x+4
99 899 99 724 99 520 99 288
44
99 689 99 502 99 283 99 033
45
99 471 99 628 99 030 98 752
46 The basis for all calculations is an eﬀective rate of interest of 6% per year, and no expenses.
(a) Show that the premium for the term insurance is P = $323.59.
(b) Calculate the mean and standard deviation of the present value of future loss random
variable, L1 , for the term insurance.
(c) Calculate the sum insured for a 3 year endowment insurance for a select life age 41,
with the same premium as for the term insurance, P = $323.59.
(d) Calculate the mean and standard deviation of the present value of future loss random
variable, L1 , for the endowment insurance.
(e) Comment on the diﬀerences between the values for the term insurance and the
endowment insurance. 326 CHAPTER 7. POLICY VALUES Exercise 7.2 A whole life insurance with sum insured $100 000 is issued to a select life
aged 35. Premiums are paid annually in advance and the death beneﬁt is paid at the end
of the year of death.
The premium is calculated using the standard select survival basis, and assuming
Interest: 6% per year eﬀective
Initial Expenses: 40% of the gross premium plus $125
Renewal expenses: 5% of gross premiums plus $40, due at the start of each policy
year from the second onwards
(a) Calculate the gross premium.
(b) Calculate the net premium policy value at t = 1 using the premium basis.
(c) Calculate the gross premium policy value at t = 1 using the premium basis.
(d) Explain why the gross premium policy value is less than the net premium policy
value.
(e) Calculate the gross premium policy value at t = 1 assuming interest of 5.5% per
year. All other assumptions follow the premium basis.
(f) Calculate the asset share per policy at the end of the ﬁrst year of the contract if
experience exactly follows the premium basis.
(g) Calculate the asset share per policy at the end of the ﬁrst year of the contract if
the experienced mortality rate is given by q[35] = 0.0012, the interest rate earned on
assets was 10%, and expenses followed the premium basis, except that there was an
additional initial expense of $25 per policy. 7.10. EXERCISES 327 (h) Calculate the surplus at the end of the ﬁrst year per policy issued given that the
experience follows (g) and assuming the policy value used is as calculated in (c)
above.
(i) Analyze the surplus in (h) into components for interest, mortality and expenses. Exercise 7.3 A whole life insurance with reduced early sum insured is issued to a life
age 50. The sum insured payable at the end of the year of death in the ﬁrst 2 years is
equal to $1 000 plus the end year policy value in the year of death (that is, the policy
value that would have been required if the life had survived).
The beneﬁt payable at the end of the year of death in any subsequent year is $20 000. The
annual premium P is calculated using the equivalence principle. The insurer calculates
premiums and policy values using the standard select survival model, with interest at 6%
per year and no expenses.
(a) (i) Write down the equations for the recursive relationship between successive
policy values for the policy values in the ﬁrst two years of the contract, and
simplify as far as possible.
(ii) Write down an expression for the policy value at time 2, 2 V , in terms of the
premium P and standard actuarial functions.
(iii) Using (i) and (ii) above, or otherwise, calculate the annual premium and 2 V . (b) Calculate 2.25 V 1
, the policy value for the contract after 2 4 years. Exercise 7.4 A special deferred annuity issued to (30) provides the following beneﬁts: 328 CHAPTER 7. POLICY VALUES
A whole life annuity of $10 000 per year, deferred for 30 years, payable monthly in
advance.
The return of all premiums paid, without interest, at the moment of death, in the
event of death within the ﬁrst 30 years. Premiums are payable continuously for a maximum of 10 years.
(a) Write down expressions for
(i) the present value random variable for the beneﬁts, and
(ii) L0 , the loss at issue random variable for the contract.
(b) Write down an expression in terms of annuity and insurance functions for the net
annual premium rate, P , for this contract.
(c) Write down an expression for L5 , the net present value of future loss random variable
for a policy in force at duration 5.
(d) Write down an expression for 5 V , the net premium policy value at time 5 for the
contract, in terms of annuity and insurance functions, and the net annual premium
rate, P . Exercise 7.5 An insurer issues a 20 year term insurance policy to (35). The sum insured
of $100 000 is payable at the end of the year of death, and premiums are paid annually
throughout the term of the contract. The basis for calculating premiums and policy values
is:
Mortality:
Interest: Standard select survival model
5% per year eﬀective 7.10. EXERCISES
Expenses: 329
Initial:
Renewal: $200 plus 15% of the ﬁrst premium
4% of each premium after the ﬁrst (a) Show that the premium is $91.37 per year.
(b) Show that the policy value immediately after the ﬁrst premium payment is
0+ V = −$122.33. (c) Explain brieﬂy why the policy value in (b) is negative.
(d) Calculate the policy values at each year end for the contract, just before and just
after the premium and related expenses incurred at that time, and plot them on a
graph. At what duration does the policy value ﬁrst become strictly positive?
(e) Suppose now that the insurer issues a large number, N say, of identical contracts to
independent lives, all aged 35 and all with sum insured $100 000. Show that if the
experience exactly matches the premium/policy value basis, then the accumulated
value at (integer) time k of all premiums less claims and expenses paid out up to
time k , expressed per surviving policyholder, is exactly equal to the policy value at
time k . Exercise 7.6 Recalculate the analysis of surplus in Example 7.8 in the order: mortality,
interest, expenses. Check that the total proﬁt is as before and note the small diﬀerences
from each source. Exercise 7.7 Consider a 20 year endowment policy issued to (40), with premiums, P per
year payable continuously, and sum insured of $200 000 payable immediately on death.
Premiums and policy values are calculated assuming: 330 CHAPTER 7. POLICY VALUES Mortality:
Interest:
Expenses: Standard select survival model
5% per year eﬀective
None. (a) Show that the premium, P , is $6 020.40 per year.
(b) Show that the policy value at duration t = 4, 4 V , is $26 131.42.
(c) Assume that the insurer decides to change the valuation basis at t = 4 to Makeham’s
mortality with A = 0.0004, with B = 2.7 × 10−6 and c = 1.124 as before. Calculate
the revised policy value at t = 4 (using the premium calculated in part (a)).
(d) Explain why the policy value does not change very much.
(e) Now assume again that A = 0.00022 but that the interest assumption changes from
5% per year to 4% per year. Calculate the revised value of 4 V .
(f) Explain why the policy value has changed considerably.
(g) A colleague has proposed that policyholders wishing to alter their contracts to paidup status should be oﬀered a sum insured reduced in proportion to the number of
premiums paid. That is, the paid up sum insured after k years of premiums have
been paid, out of the original total of 20 years, should be S × k/20, where S is the
original sum insured. This is called the proportionate paid up sum insured.
Using a spreadsheet, calculate the EPV of the proportionate paid up sum insured
at each year end, and compare these graphically with the policy values at each year
end, assuming the original basis above is used for each. Explain brieﬂy whether you
would recommend the proportionate paid up sum insured for this contract. 7.10. EXERCISES 331 Exercise 7.8 Consider a whole life insurance policy issued to a select life aged x. Premiums of $P per year are payable continuously throughout the policy term, and the sum
insured of $S is paid immediately on death.
(a) Show that
V[Ln ]
t = P
S+
δ 2
2 ¯
¯
A[x]+t − A[x]+t 2 . (b) Assume the life is aged 55 at issue, and that premiums are $1 200 per year. Show
that the sum insured on the basis below is $77 566.44.
Mortality:
Interest:
Expenses: Standard select survival model
5% per year eﬀective
None. (c) Calculate the standard deviation of Ln , Ln and Ln . Comment brieﬂy on the results.
0
5
10 Exercise 7.9 For an n year endowment policy, level monthly premiums are payable
throughout the term of the contract, and the sum insured is payable at the end of the
month of death.
Derive the following formula for the net premium policy value at time t years, where t is
a premium date: (12)
a[x]+t:n−t
¨
1 −
.
tV = S
(12)
a[x]:n
¨ 332 CHAPTER 7. POLICY VALUES Exercise 7.10 A life aged 50 buys a participating whole life insurance policy with sum
insured $10 000. The sum insured is payable at the end of the year of death. The premium
is payable annually in advance. Proﬁts are distributed through cash dividends paid at
each year end to policies in force at that time.
The premium basis is:
Initial Expenses
Renewal Expenses
Interest
Mortality 22% of the annual gross premium plus $100
5% of the gross premium plus $10
4.5%
Standard select survival model (a) Show that the annual premium, calculated with no allowance for future bonuses, is
$144.63 per year.
(b) Calculate the policy value at each year end for this contract using the premium
basis.
(c) Assume the insurer earns interest of 5.5% each year. Calculate the dividend payable
each year assuming
(i) the policy is still in force at the end of the year
(ii) experience other than interest exactly follows the premium basis, and
(iii) that 90% of the proﬁt is distributed as dividends to policyholders.
(d) Calculate the expected present value of the proﬁt to the insurer per policy issued,
using the same assumptions as in (c).
(e) What would be a reasonable surrender beneﬁt for lives surrendering their contracts
at the end of the ﬁrst year? 7.10. EXERCISES 333 Exercise 7.11 A 10 year endowment insurance is issued to a life aged 40. The sum
insured is payable at the end of the year of death or on survival to the maturity date.
The sum insured is $20 000 on death, $10 000 on survival to age 50. Premiums are paid
annually in advance.
(a) The premium basis is:
Expenses:
Interest:
Mortality: 5% of each gross premium including the ﬁrst.
5%
Standard select survival model. Show that the gross premium is $807.71.
(b) Calculate the policy value on the premium basis just before the ﬁfth premium is
due.
(c) Just before the ﬁfth premium is due the policyholder requests that all future premiums, including the ﬁfth, be reduced to one half their original amount. The insurer
calculates the revised sum insured – the maturity beneﬁt still being half of the
death beneﬁt – using the policy value in part (b) with no extra charge for making
the change.
Calculate the revised death beneﬁt. Exercise 7.12 An insurer issues a whole life insurance policy to a life aged 40. The
death beneﬁt in the ﬁrst three years of the contract is $1 000. In subsequent years the
death beneﬁt is $50 000. The death beneﬁt is payable at the end of the year of death and
level premiums are payable annually throughout the term of the contract.
Basis for premiums and policy values: 334
Mortality
Interest:
Expenses: CHAPTER 7. POLICY VALUES
Standard select survival model
6% per year eﬀective
None (a) Calculate the premium for the contract.
(b) Write down the policy value formula for any integer duration t ≥ 3.
(c) Calculate the policy value at t = 3.
(d) Use the recurrence relation to determine the policy value after 2 years.
(e) The insurer issued 1 000 of these contracts to identical, independent lives age 40.
After 2 years there are 985 still in force. In the following year there were four further
deaths in the cohort, and the rate of interest earned on assets was 5.5%. Calculate
the proﬁt or loss from mortality and interest in the year. Exercise 7.13 A 20 year endowment insurance issued to a life aged 40 has level premiums
payable continuously throughout the term. The sum insured on survival is $60 000. The
sum insured payable immediately on death within the term is $20 000 if death occurs
within the ﬁrst 10 years and t V if death occurs after t years, 10 ≤ t < 20, where t V is the
policy value calculated on the premium basis.
Premium basis:
Mortality: Standard select survival model
Interest: δt = 0.06 − 0.001t per year
Expenses: None
(a) Write down the Thiele’s diﬀerential equation for t V , separately for 0 < t < 10 and
10 < t < 20, and give any relevant boundary conditions. 7.10. EXERCISES 335 (b) Determine the premium rate P by solving Thiele’s diﬀerential equation using Euler’s
method, with a time step h = 0.05.
(c) Plot the graph of t V for 0 < t < 20. Exercise 7.14 On 1 June 2008 an insurer issued a 20 year level term insurance to a life
then aged exactly 60. The single premium was paid on 1 June 2008. The beneﬁt is $1.
Let t V denote the policy value after t years.
(a) Suppose the death beneﬁt is paid at the year end. Write down and explain a
recurrence relation between t V and t+1 V for t = 0, 1, ..., 19.
(b) Suppose the beneﬁt is payable at the end of every h years, where h < 1. Write down
a recurrence relation between t V and t+h V for t = 0, h, 2h, ..., 20 − h.
(c) By considering the limit as h → 0, show that Thiele’s diﬀerential equation for the
policy value for a beneﬁt payable continuously is
d tV
= (µ60+t + δ )t V − µ60+t
dt
where δ is the force of interest, and state any boundary conditions.
(d) Show that
tV ¯1
= A[60]+t:20−t is the solution to the diﬀerential equation in (c). 336 CHAPTER 7. POLICY VALUES Exercise 7.15 An insurer issues identical deferred annuity policies to 100 independent
lives aged 60 at issue. The deferred period is 10 years, after which the annuity of $10 000
per year is paid annually in advance. Level premiums are payable annually throughout
the deferred period. The death beneﬁt during deferment is $50 000, payable at the end of
the year of death.
The basis for premiums and policy values is:
Mortality:
Interest:
Expenses: standard select survival model
6% per year
None (a) Calculate the premium for each contract.
(b) Write down the recursive relationship for the policy values, during and after the
deferred period.
(c) Calculate the death strain at risk in the third year of the contract, for each contract
still in force at the start of the third year.
(d) Calculate the death strain at risk in the 13th year of the contract, per contract in
force at the start of the year.
(e) Two years after the issue date, 97 policies remain in force. In the third year, 3
lives die. Calculate the total mortality proﬁt in the third year, assuming all other
experience follows the assumptions in the premium basis.
(f) Twelve years after the issue date 80 lives survive; in the thirteenth year there are 4
deaths. Calculate the total mortality proﬁt in the 13th year. Exercise 7.16 Consider Example 7.1. Calculate the policy values at intervals of h = 0.1
years from t = 0 to t = 2. 7.10. EXERCISES 337 Answers to selected exercises
7.1 (a) $323.59
(b) $116.68, $11 663.78 (c) $1 090.26
(d) $342.15, $15.73 7.2 (a) $469.81
(b) $381.39
(c) $132.91
(e) $168.38
(f) $132.91
(g) $25.10
(h) −$107.67
(i) $6.28, −$86.45, 7.3 (a) (iii) $185.08, −$27.50 $401.78 (b) $588.91
7.5 (a) $91.37
(b) −$122.33
(d) Selected values: 4 V = −$32.53,
4+ V = $55.18,
13 V = $238.95,
13+ V = $326.67
The policy value ﬁrst becomes positive at duration 4+.
7.6 −$26 504.04,
7.7 (a) $6 020.40 $51 011.26, −$5 588.00 338 CHAPTER 7. POLICY VALUES
(b) $26 131.42
(c) $26 348.41
(e) $36 575.95
(g) Selected values: t = 1 : $4 003.56,
t = 10 : $61 678.46,
$70 070.54 7.8 (c) $14 540.32, $16 240.72, $6 078.79 $17 619.98 7.10 (b) Selected values: 5 V = $509.93, 10 V = $1 241.77 (c) Selected values: Bonus at t = 5: $4.55
Bonus at t = 10: $10.96
(d) $263.37
(e) $0
7.11 (b) $3 429.68
(c) $14 565.95
7.12 (a) $256.07
(c) $863.45
(d) $558.58
(e) −$4 476.57
7.13 (b) $1 810.73
(c) Selected values: 5 V = $10 400.92,
7.15 (a) $7 909.25
(c) $23 671.76
(d) −$102 752.83
(e) −$61 294.26 10 V = $23 821.21, 15 V = $40 387.35 7.10. EXERCISES 339 (f) $303 485.21
7.16 Selected values: 0.5 V = $15 255.56,
2 V = $31 415.28 1V = $15 369.28, 1 .5 V = $30 962.03, 340 CHAPTER 7. POLICY VALUES Chapter 8
Multiple State Models
8.1 Summary In this chapter we reformulate the survival model introduced in Chapter 2 as an example
of a multiple state model. We then introduce several other multiple state models which
are useful as models for diﬀerent types of life insurance policies. A general deﬁnition of a
multiple state model, together with assumptions and notation, is given in Section 8.3. In
Section 8.4 we discuss the derivation of formulae for probabilities and in Section 8.5 the
numerical evaluation of these probabilities. This is extended in Section 8.6 to premium
calculation and in Section 8.7 to the numerical evaluation of reserves.
In the ﬁnal three sections we study in more detail some speciﬁc multiple state models
that are particularly useful – a multiple decrement model, the joint life and last survivor
model and a model where transitions can take place at speciﬁed ages.
341 342 CHAPTER 8. MULTIPLE STATE MODELS Alive
0 E Dead
1 Figure 8.1: The alive–dead model. 8.2 Examples of multiple state models Multiple state models are one of the most exciting developments in actuarial science in
recent years. They are a natural tool for many important areas of practical interest
to actuaries. They also simplify and provide a sound foundation for some traditional
actuarial techniques. In this section we illustrate some of the uses of multiple state
models using a number of examples which are common in actuarial practice. 8.2.1 The alive–dead model So far, we have modelled the uncertainty over the duration of an individual’s future lifetime by regarding the future lifetime as a random variable, Tx for an individual currently
aged x, with a given cumulative distribution function, Fx (t) (= Pr[Tx ≤ t]), and survival
function, Sx (t) = 1 − Fx (t). This is a probabilistic model in the sense that for an
individual aged x we have a single random variable, Tx , whose distribution, and hence all
associated probabilities, is assumed to be known.
We can represent this model diagrammatically as shown in Figure 8.1. Our individual
is, at any time, in one of two states, ‘Alive’ and ‘Dead’. For convenience we label these
states ‘0’ and ‘1’, respectively. Transition from state 0 to state 1 is allowed, as indicated
by the direction of the arrow, but transitions in the opposite direction cannot occur. This
is an example of a multiple state model with two states. 8.2. EXAMPLES OF MULTIPLE STATE MODELS 343 We can use this multiple state model to reformulate our survival model as follows. Suppose
we have a life aged x ≥ 0 at time t = 0. For each t ≥ 0 we deﬁne a random variable
Y (t) which takes one of the two values 0 and 1. The event ‘Y (t) = 0’ means that our
individual is alive at age x + t; ‘Y (t) = 1’ means that our individual died before age x + t.
The set of random variables {Y (t)}t≥0 is an example of a continuous time stochastic
process. A continuous time stochastic process is a collection of random variables indexed
by a continuous time variable. For all t, Y (t) is either 0 or 1, and Tx is connected to this
model as the time at which Y (t) jumps from 0 to 1, that is
Tx = max{t : Y (t) = 0}.
The alive–dead model represented by Figure 8.1 captures all the survival/mortality information for an individual that is necessary for calculating insurance premiums and reserves
for policies where payments – premiums, beneﬁts and expenses – depend only on whether
the individual is alive or dead at any given age, for example a term insurance or a whole
life annuity. More complicated forms of insurance require more complicated models. We
introduce more examples of such models in the remainder of Section 8.2 before giving a
formal deﬁnition of a multiple state model in Section 8.3. All these models consist of
a ﬁnite set of states with arrows indicating possible movements between some, but not
necessarily all, pairs of states. Each state represents the status of an individual or a set of
individuals. Loosely speaking, each model is appropriate for a given insurance policy in
the sense that the condition for a payment relating to the policy, for example a premium,
an annuity or a sum insured, is either that the individual is in a speciﬁed state at that
time or that the individual makes an instantaneous transfer between a speciﬁed pair of
states at that time. 8.2.2 Term insurance with increased beneﬁt on accidental death Suppose we are interested in a term insurance policy under which the death beneﬁt is
$100 000 if death is due to an accident during the policy term and $50 000 if it is due to 344 CHAPTER 8. MULTIPLE STATE MODELS Alive
0 d
d d d
d
Dead – Accident
1 Dead – Other Causes
2 Figure 8.2: The accidental death model. any other cause. The alive–dead model in Figure 8.1 is not suﬃcient for this policy since,
when the individual dies – that is, transfers from state 0 to state 1 – we do not know
whether death was due to an accident, and so we do not know the amount of the death
beneﬁt to be paid.
An appropriate model for this policy is shown in Figure 8.2. This model has three states,
labeled as shown, and we can deﬁne a continuous time stochastic process, {Y (t)}t≥0 ,
where each random variable Y (t) takes one of the three values 0, 1 and 2. Hence, for
example, the event ‘Y (t) = 1’ indicates that the individual, who is aged x at time t = 0,
has died from an accident before age x + t.
The model in Figure 8.2 is an extension of the model in Figure 8.1. In both cases an
individual starts by being alive, that is, starts in state 0, and, at some future time, dies.
The diﬀerence is that we now need to distinguish between deaths due to accident and
deaths due to other causes since the sum insured is diﬀerent in the two cases. Notice
that it is the beneﬁts provided by the insurance policy which determine the nature of the
appropriate model. 8.2. EXAMPLES OF MULTIPLE STATE MODELS Healthy E Disabled 0
d 345 1
d
d d
d
© Dead
2
Figure 8.3: The permanent disability model. 8.2.3 The permanent disability model Figure 8.3 shows a model appropriate for a policy which provides some or all of the
following beneﬁts:
• an annuity while permanently disabled,
• a lump sum on becoming permanently disabled, and,
• a lump sum on death,
with premiums payable while healthy. An important feature of this model is that disablement is permanent – there is no arrow from state 1 back to state 0. 8.2.4 The disability income insurance model Disability income insurance pays a beneﬁt during periods of sickness; the beneﬁt ceases on
recovery. Figure 8.4 shows an appropriate model for a policy which provides an annuity
while the person is sick, with premiums payable while the person is healthy. It could
also be used when there are lump sum payments on becoming sick or dying. The model 346 CHAPTER 8. MULTIPLE STATE MODELS
E Healthy
' 0
d Sick d
d d
d
1 © Dead
2
Figure 8.4: The disability income insurance model. represented by Figure 8.4 diﬀers from that in Figure 8.3 in only one respect: it is possible
to transfer from state 1 to state 0, that is, to recover from an illness.
This model illustrates an important general feature of multiple state models which was
not present for the models in Figures 8.1, 8.2 and 8.3. This feature is the possibility of
entering one or more states many times. In terms of our interpretation of the model, this
means that several periods of sickness could occur before death, with healthy (premium
paying) periods in between. 8.2.5 The joint life and last survivor model A joint life annuity is an annuity payable until the ﬁrst death among a group of lives.
A last survivor annuity is an annuity payable until the last death among a group of
lives. In principle, and occasionally in practice, the group could consist of three or more
lives. However, such policies are most commonly purchased by couples who are jointly
organising their ﬁnancial security and we will restrict our attention to the case of two
lives whom we will label, for convenience, ‘husband’ and ‘wife’.
A common beneﬁt design is an annuity payable at a higher rate while both partners are
alive and at a lower rate following the ﬁrst death. The annuity ceases on the second 8.3. ASSUMPTIONS AND NOTATION 347 death. This could be viewed as a last survivor annuity for the lower amount, plus a joint
life annuity for the diﬀerence.
A reversionary annuity is a life annuity that starts payment on the death of a speciﬁed
life, if his or her spouse is alive, and continues through the spouse’s lifetime. A pension
plan may oﬀer a reversionary annuity beneﬁt as part of the pension package, payable to
the pension plan member’s spouse for their remaining lifetime after the member’s death.
Couples may also be interested in joint life insurance, under which a death beneﬁt is
paid on the ﬁrst death of the husband and wife.
All of these beneﬁts may be valued using the model represented in Figure 8.5.
Let x and y denote the ages of the husband and wife, respectively, when the annuity or
insurance policy is purchased. For t ≥ 0, the event Y (t) = 0 indicates that both husband
and wife are alive at ages x + t and y + t, respectively; Y (t) = 1 indicates that the husband
is alive at age x + t and the wife died before age y + t; Y (t) = 2 indicates that the husband
died before age x + t and the wife is still alive at age y + t; Y (t) = 3 indicates that the
husband died before age x + t and the wife died before age y + t.
The multiple state models introduced above are all extremely useful in an insurance
context. We study in detail several of these models, and others, later in this chapter.
Before doing so, we need to introduce some assumptions and some notation. 8.3 Assumptions and notation In this section we consider a general multiple state model. We have a ﬁnite set of n + 1
states labelled 0, 1, . . . , n, with instantaneous transitions being possible between selected
pairs of states. These states represent diﬀerent conditions for an individual (as in Figures
8.2, 8.3 and 8.4) or groups of individuals (as in Figure 8.5). For each t ≥ 0, the random
variable Y (t) takes one of the values 0, 1, . . . , n, and we interpret the event Y (t) = i to
mean that the individual is in state i at age x + t, or, more generally as for the model 348 CHAPTER 8. MULTIPLE STATE MODELS Husband Alive
Wife Alive E Husband Alive
Wife Dead 0 1 c c Husband Dead
Wife Alive
2 E Husband Dead
Wife Dead
3 Figure 8.5: The joint life and last survivor model. in Figure 8.5, that the group of lives being modelled is in state i at time t. The set of
random variables {Y (t)}t≥0 is then a continuous time stochastic process.
The multiple state model will be an appropriate model for an insurance policy if the
payment of beneﬁts or premiums is dependent on being in a given state or moving between
a given pair of states at a given time, as illustrated in the examples in the previous section.
Note that in these examples there is a natural starting state for the policy, which we
always label state 0. This is the case for all examples based on multiple state models. For
example, a policy providing an annuity during periods of sickness in return for premiums
payable while healthy, as described in Section 8.2.4 and illustrated in Figure 8.4, would
be issued only to a person who was healthy at that time.
Assumption 1: We assume that for any states i and j and any times t and t + s, where
s ≥ 0, the conditional probability Pr[Y (t + s) = j  Y (t) = i] is well deﬁned in the sense
that its value does not depend on any information about the process before time t.
Intuitively, this means that the probabilities of future events for the process are completely 8.3. ASSUMPTIONS AND NOTATION 349 determined by knowing the current state of the process. In particular, these probabilities
do not depend on how the process arrived at the current state or how long it has been
in the current state. This property, that probabilities of future events depend on the
present but not on the past, is known as the Markov property. Using the language of
probability theory, we are assuming that {Y (t)}t≥0 is a Markov process.
Assumption 1 was not made explicitly for the models represented by Figures 8.1 and 8.2
since it was unnecessary given our interpretation of these models. In each of these two
cases, if we know that the process is in state 0 at time x (so that the individual is alive
at age x) then we know the past of the process (the individual was alive at all ages before
x). Assumption 1 is more interesting in relation to the models in Figures 8.3 and 8.4.
Suppose, for example, in the disability income insurance model (Figure 8.4) we know that
Y (t) = 1, so that we know that the individual is sick at time t. Then Assumption 1 says
that the probability of any future move after time t, either recovery or death, does not
depend on any further information, such as how long the life has been sick up to time t, or
how many diﬀerent periods of sickness the life has experienced up to time t. In practice,
we might believe that the probability of recovery in, say, the next week would depend
on how long the current sickness has already lasted. If the current sickness has already
lasted for, say, 6 months then it is likely to be a very serious illness and recovery within
the next week is possible but not likely; if the current sickness has lasted only one day so
far, then it may well be a trivial illness and recovery within a week could be very likely.
It is important to understand the limitations of any model and also to bear in mind that
no model is a perfect representation of reality.
Assumption 2: We assume that for any positive interval of time h,
Pr[2 or more transitions within a time period of length h] = o(h).
(Recall that any function of h, say g (h), is said to be o(h) if
g (h)
= 0.
h→0 h
lim Intuitively, a function is o(h) if, as h converges to 0, the function converges to zero faster 350 CHAPTER 8. MULTIPLE STATE MODELS than h.)
Assumption 2 tells us that for a small interval of time h, the probability of two or more
transitions in that interval is so small that it can be ignored. This assumption is unnecessary for the models in Figures 8.1 and 8.2 since in both cases only one transition can ever
take place. However, it is an assumption we need to make for technical reasons for the
models in Figures 8.3, 8.4 and 8.5. In these cases, given our interpretation of the models,
it is not an unreasonable assumption.
In Chapter 2 we introduced the standard actuarial notation for what we are now calling the
alive–dead model, as shown in Figure 8.1; speciﬁcally, t px , t qx and µx . For multiple state
models more complicated than that in Figure 8.1, we need a more ﬂexible notation. We
introduce the following notation for a general multiple state model to be used throughout
this chapter and in later chapters.
Notation: For states i and j in a multiple state model and for x, t ≥ 0, we deﬁne
ij
t px = Pr[Y (x + t) = j  Y (x) = i], (8.1) ii
t px = Pr [ Y (x + s) = i for all s ∈ [0, t]  Y (x) = i] , (8.2) so that t pij is the probability that a life aged x in state i is in state j at age x + t, where
x
j may be equal to i, while t pii is the probability that a life aged x in state i stays in state
x
i throughout the period from age x to age x + t.
For i = j we deﬁne
µij = lim+
x
h→0 ij
h px h i = j. (8.3) Assumption 3: For all states i and j and all ages x ≥ 0, we assume that t pij is a
x
diﬀerentiable function of t.
Assumption 3 is a technical assumption needed to ensure that the mathematics proceeds
smoothly. Consequences of this assumption are that the limit in the deﬁnition of µij
x
always exists and that the probability of a transition taking place in a time interval of 8.3. ASSUMPTIONS AND NOTATION 351 length t converges to 0 as t converges to 0. We also assume that µij is a bounded and
x
integrable function of x. These assumptions are not too restrictive in practice. However,
there are some circumstances where we need to put aside Assumption 3 and these are
discussed in the ﬁnal section of this chapter.
In terms of the alive–dead model represented by Figure 8.1, we can make the following
observations:
• t p00 is the same as t px in the notation of Chapter 2, and t p01 is the same as t qx .
x
x
• t p10 = 0 since backward transitions, ‘Dead’ to ‘Alive’, are not permitted in this
x
model.
• 0 pij equals 1 if i = j and zero otherwise.
x
• µ01 is the same as µx , the force of mortality at age x.
x
We use the following terminology for a general multiple state model.
Terminology: We refer to µij as the force of transition or transition intensity
x
between states i and j at age x.
Another way of expressing formula (8.3) is to write for h > 0
ij
h px = h µij + o(h).
x (8.4) From this formulation we can say that for small positive values of h
ij
h px ≈ h µij .
x (8.5) This is equivalent to formula (2.8) in Chapter 2 for the alive–dead model and will be very
useful to us.
Example 8.1 Explain why, for a general multiple state model, t pii is not equivalent to
x
pii . Write down an inequality linking these two probabilities and explain why
tx
ii
t px = t pii + o(t).
x (8.6) 352 CHAPTER 8. MULTIPLE STATE MODELS Solution 8.1 From formulae (8.1) and (8.2) we can see that t pii is the probability that
x
the process/individual does not leave state i between ages x and x + t, whereas t pii is the
x
probability that the process/individual is in state i at age x + t, in both cases given that
the process was in state i at age x. The important distinction is that t pii includes the
x
possibility that the process leaves state i between ages x and x + t, provided it is back
in state i at age x + t. For any individual state which either (a) can never be left or (b)
can never be reentered once it has been left, these two probabilities are equivalent. This
applies to all the states in the models illustrated in Figures 8.1, 8.2, 8.3, 8.4 and 8.5 except
states 0 and 1 in Figure 8.4.
The following inequality is always true since the left hand side is the probability of a set
of events which is included in the set of events whose probability is on the right hand side
ii
t px ≤ t pii .
x The diﬀerence between these two probabilities is the probability of those paths where the
process makes two or more transitions between ages x and x + t so that it is back in state
i at age x + t. From Assumption 2 we know that this probability is o(t). This gives us
formula (8.6). Example 8.2 Show that, for a general multiple state model and for h > 0,
n
ii
h px =1−h µij + o(h).
x (8.7) j =0,j =i Solution 8.2 First note that 1 − h pii is the probability that the process does leave state
x
i at some time between ages x and x + h, possibly returning to state i before age x + h. If
the process leaves state i between ages x and x + h then at age x + h it must be in some
state j (= i) or be in state i having made at least two transitions in the time interval of 8.4. FORMULAE FOR PROBABILITIES 353 length h. Using formula (8.4) and Assumption 2, the sum of these probabilities is
n µij + o(h),
x h
j =0,j =i which proves (8.7). 8.4 Formulae for probabilities In this section we regard the transition intensities as known and we show how to derive
formulae for all probabilities in terms of them. This is the same approach as we adopted in
Chapter 2, where we assumed the force of mortality, µx , was known and derived formula
(2.19) for t px in terms of the force of mortality.
The fact that all probabilities can be expressed in terms of the transition intensities is
important. It tells us that the transition intensities {µij ; x ≥ 0; i, j = 0, . . . , n, i = j } are
x
fundamental quantities which determine everything we need to know about a multiple
state model.
The ﬁrst result generalizes formula (2.19) from Chapter 2, and is valid for any multiple
state model. It gives a formula for t pii in terms of all the transition intensities out of state
x
i, µij .
x
For any state i in a multiple state model, t
ii
t px = exp − 0 n µij+s ds
x . (8.8) j =0,j =i We can derive this as follows. For any h > 0, consider the probability t+h pii . This is
x
the probability that the individual/process stays in state i throughout the time period 354 CHAPTER 8. MULTIPLE STATE MODELS [0, t + h], given that the process was in state i at age x. We can split this event into two
subevents:
• the process stays in state i from age x until (at least) age x + t, given that it was in
state i at age x, and
• the process stays in state i from age x + t until (at least) age x + t + h, given that
it was in state i at age x + t (note the diﬀerent conditioning).
The probabilities of these two subevents are t pii and h pii+t , respectively, and, using the
x
x
rules for conditional probabilities, we have
ii
t+h px = t pii h pii+t .
x
x Using the result in Example 8.2, this can be rewritten as
n
ii
ii
t+h px = t px 1−h µij+t + o(h)
x
j =0,j =i Rearranging this equation, we get
− t pii
x
= − t pii
x
h n ii
t+h px µij+t +
x
j =0,j =i and letting h → 0 we have
d ii
ii
t p = − t px
dt x n µij+t ,
x
j =0,j =i so that
d
log
dt n
ii
=−
t px µij+t .
x
j =0,j =i o(h)
,
h . 8.4. FORMULAE FOR PROBABILITIES 355 Integration over (0, t) gives
t log ii
t px ii
0 px − log =− n µij+r dr .
x 0 j =0,j =i So, by exponentiating both sides, we see that the solution to the diﬀerential equation is
t
ii
ii
t px = 0 px exp − n µij+s ds
x . 0 j =0,j =i Since 0 pii = 1, this proves (8.8).
x
We comment on this result after the next example.
Example 8.3 Consider the model for permanent disability illustrated in Figure 8.3. Explain why, for x ≥ 0 and t, h > 0,
01
t+h px = t p01 h p11 t + t p00 h µ01 t + o(h).
x
x+
x
x+ (8.9) Hence show that
d
dt t
01
t px exp
0 t µ12 s ds
x+ = t p00 µ01 t exp
x
x+ 0 µ12 s ds
x+ , (8.10) and hence that for u > 0
u
01
u px = 0 00
t px µ01 t u−t p11 t dt .
x+
x+ (8.11) Give a direct intuitive derivation of formula (8.11).
Solution 8.3 To derive (8.9), consider a life who is healthy at age x. The left hand side
of (8.9) is the probability that this life is alive and disabled at age x + t + h. We can write
down a formula for this probability by conditioning on which state the life was in at age
x + t. Either: 356 CHAPTER 8. MULTIPLE STATE MODELS • the life was disabled at age x + t (probability t p01 ) and remained disabled between
x
ages x + t and x + t + h (probability h p11 t ), or,
x+
• the life was healthy at age x + t (probability t p00 ) and then became disabled between
x
ages x + t and x + t + h (probability h µ01 t + o(h)).
x+
Combining the probabilities of these events gives (8.9). (Note that the probability of the
life being healthy at age x + t, becoming disabled before age x + t + h and then dying
before age x + t + h is o(h) since this involves two transitions in a time interval of length
h.)
Using Example 8.2, formula (8.9) can be rewritten as
01
t+h px = t p01 (1 − h µ12 t ) + t p00 h µ01 t + o(h).
x
x+
x
x+ (8.12) Rearranging, dividing by h and letting h → 0 gives
d 01
01 12
00 01
t p + t p x µ x+ t = t p x µ x+ t .
dt x
Multiplying all terms in this equation by exp
d
dt t
0 µ12 s ds , we have
x+ t
01
t px exp
0 t µ12 s ds
x+ = t p00 µ01 t exp
x
x+ 0 µ12 s ds
x+ . Integrating both sides of this equation from t = 0 to t = u, and noting that 0 p01 = 0, we
x
have
u
01
u px exp
0 u µ12 s
x+ ds =
0 Finally, dividing both sides by exp t
00 01
t p x µ x+ t
u
0 u
11
u−t px+t = exp − we have formula (8.11). t µ12 s ds
x+ , exp
0 µ12 s ds
x+ dt . µ12 s ds and noting that, using formula (8.8),
x+ 8.4. FORMULAE FOR PROBABILITIES 357 The intuitive derivation of (8.11) is as follows: for the life to move from state 0 to state
1 between ages x and x + u, the life must stay in state 0 until some age x + t, transfer
to state 1 between ages x + t and x + t + dt, where dt is small, and then stay in state 1
from age x + t + dt to age x + u. We can illustrate this event sequence using the time line
below. Time Age t 0 x Probability ¦ Event ¥
¦ x+t
00
t px in state 0
for t years t+dt u ¦
¥ x+u x+t+dt
µ01 t dt
x+
transition
to state 1 11
u − t p x+ t ¥ in state 1
for u − t years The inﬁnitesimal probability of this path is
00
t px µ01 t dt u−t p11 t
x+
x+ where we have written u−t p11 t instead of u−t−dt p11 t since the two are approximately equal
x+
x+
if dt is small. Since the age at transfer, x + t, can be anywhere between x and x + u,
the total probability, u p01 , is the ‘sum’(i.e. integral) of these probabilities from t = 0 to
x
t = u. We can make the following comments about formula (8.8) and Example 8.3.
(1) As we have already noted, formula (8.8) is an extension of formula (2.19) in Chapter
2 for t px . 358 CHAPTER 8. MULTIPLE STATE MODELS (2) Throughout Example 8.3 we could have replaced t pii by t pii for i = 0, 1, since, for
x
x
the disability insurance model, neither state 0 nor state 1 can be reentered once it
has been left. See the Solution to Example 8.1.
(3) Perhaps the most important point to note about formula (8.8) and Example 8.3
is how similar the derivations are in their basic approach. In particular, in both
cases we wrote down an expression for the probability of being in the required state
at age x + t + h by conditioning on the state occupied at age x + t. This led to
a formula for the derivative of the required probability which we were then able
to solve. An obvious question for us is, ‘Can this method be applied to a general
multiple state model to derive formulae for probabilities?’ The answer is, ‘Yes’.
This is demonstrated in Section 8.4.1. 8.4.1 Kolmogorov’s forward equations Let i and j be any two, not necessarily distinct, states in a multiple state model which
has a total of n + 1 states. For x, t, h ≥ 0, we derive the formula
n
ij
ij
t+h px = t px − h ij jk
t p x µ x+ t
k=0,k=j − t pik µkj t + o(h),
x x+ (8.13) and hence show the main result, that
n d ij
tp =
dt x
k=0,k=j ik kj
t p x µ x+ t − t pij µjk t .
x x+ (8.14) Formula (8.14) gives a set of equations for a Markov process known as Kolmogorov’s
forward equations.
To derive Kolmogorov’s forward equations, we proceed as we did in formula (8.8) and in
Example 8.3. We consider the probability of being in the required state, j , at age x + t + h, 8.5. NUMERICAL EVALUATION OF PROBABILITIES 359 and condition on the state of the process at age x + t: either it is already in state j , or it
is in some other state, say k , and a transition to j is required before age x + t + h. Thus,
we have
n
ij
t+h px = jj
ij
t p x h p x+ t kj
ik
t p x h p x+ t + . k=0,k=j Using formulae (8.6), (8.7) and (8.4), this can be rewritten as
n
ij
t+h px = ij
t px 1−h n µjk t
x+
k=0,k=j − o(h) ik
t px +h µkj t + o(h).
x+ k=0,k=j Rearranging the right hand side of this expression gives (8.13). Further rearranging,
dividing by h and letting h → 0 gives (8.14).
In the following section we give several examples of the application of the Kolmogorov forward equations as we use them to calculate probabilities for some of the models described
above. 8.5 Numerical evaluation of probabilities In this section we discuss methods for the numerical evaluation of probabilities for a
multiple state model given that all the transition intensities are known. In some cases,
the probabilities can be calculated directly from formulae in terms of integrals, as the
following example shows.
Example 8.4 Consider the permanent disability model illustrated in Figure 8.3.
(a) Suppose the transition intensities for this model are all constants, as follows
µ01 = 0.0279,
x
Calculate 00
10 p60 and µ02 = 0.0229,
x
01
10 p60 . µ12 = µ02 .
x
x 360 CHAPTER 8. MULTIPLE STATE MODELS (b) Now suppose the transition intensities for this model are as follows
µ01 = a1 + b1 exp{c1 x},
x
µ02 = a2 + b2 exp{c2 x},
x
µ12 = µ02 ,
x
x
where
a1 = 4 × 10−4 , b1 = 3.4674 × 10−6 , c1 = 0.138155,
a2 = 5 × 10−4 , b2 = 7.5858 × 10−5 , c2 = 0.087498.
Calculate 00
10 p60 and 01
10 p60 . Solution 8.4 For this model, neither state 0 nor state 1 can be reentered once it has
been left, so that
ii
t px ≡ t pii
x for i = 0, 1 and any x, t ≥ 0. See the solution to Example 8.1.
(a) Using formula (8.8) we have
t
00
t p60 ≡ t p00 = exp −
60 (0.0279 + 0.0229) ds
0 giving
00
10 p60 = exp{−0.508} = 0.60170. Similarly
11
10−t p60+t = exp{−0.0229(10 − t)}, = exp{−0.0508t}, (8.15) 8.5. NUMERICAL EVALUATION OF PROBABILITIES
and we can calculate 01
10 p60 361 using formula (8.11) as 10
01
10 p60 00
t p60 =
0 µ01 t 10−t p11 t dt
60+
60+ 10 =
0 exp{−0.0508t} × 0.0279 × exp{−0.0229(10 − t)} dt = 0.19363.
(b) In this case
t
00
t p60 = exp − (µ01 r + µ02 r ) dr
60+
60+ 0 = exp − (a1 + a2 )t + b2
b1 60 c1 c1 t
(e − 1) + e60 c2 (ec2 t − 1)
e
c1
c2 and
t
11
t p60 = exp − 0 µ12 r dr
60+ = exp − a2 t + b2 60 c2 c2 t
e
(e − 1)
c2 . Hence
00
10 p60 = 0.58395. Substituting the expressions for t p00 and 10−t p11 t and the formula for µ01 t into
60
60+
60+
formula (8.11) gives an integrand that cannot be integrated analytically. However,
we can integrate it numerically, obtaining
01
10 p60 = 0.20577. 362 CHAPTER 8. MULTIPLE STATE MODELS Probabilities of the form t pii can be evaluated analytically provided the sum of the relevant
x
intensities can be integrated analytically. In other cases numerical integration can be used.
However, the approach used in Example 8.4 part (b) to calculate a more complicated
probability, 10 p01 – deriving an integral formula for the probability which can then be
60
integrated numerically – is not tractable except in the simplest cases. Broadly speaking,
this approach works if the model has relatively few states and if none of these states
can be reentered once it has been left. These conditions are met by the permanent
disability model, illustrated in Figure 8.3 and used in Example 8.4, but are not met, for
example, by the disability income insurance model illustrated in Figure 8.4 since states
0 and 1 can both be reentered. This means, for example, that t p01 , is the sum of the
x
probabilities of exactly one transition (0 to 1), plus 3 transitions (0 to 1, then 1 to 0, then
0 to 1 again), plus 5 transitions and so on. A probability involving k transitions involves
multiple integration with k nested integrals.
Euler’s method, introduced in Chapter 6, can be used to evaluate probabilities for all
models in which we are interested. The key to using this method is formula (8.13) and
we illustrate it by applying it in the following example. Example 8.5 Consider the disability income insurance model illustrated in Figure 8.4.
Suppose the transition intensities for this model are as follows
µ01 = a1 + b1 exp{c1 x},
x
µ10 = 0.1 µ01 ,
x
x
µ02 = a2 + b2 exp{c2 x},
x
µ12 = µ02 ,
x
x 8.5. NUMERICAL EVALUATION OF PROBABILITIES 363 where a1 , b1 , c1 , a2 , b2 and c2 are as in Example 8.4, part (b) (though this is a diﬀerent
model).
Calculate 10 p00 and 10 p01 using formula (8.13) with a step size of h = 1/12 years (we use
60
60
a monthly time step, because this generates values of t p00 and t p01 for t = 0, 1, 2, . . . , 120
60
60
months, which we use in Example 8.6).
Solution 8.5 For this particular model, formula (8.13) gives us the two formulae
00
t+h p60 = t p00 − h t p00 µ01 t + µ02 t + h t p01 µ10 t + o(h)
60
60
60+
60+
60 60+ 01
t+h p60 = t p01 − h t p01 µ12 t + µ10 t + h t p00 µ01 t + o(h).
60 60+
60
60
60+
60+ and As in Chapter 6, we choose a small step size h, ignore the o(h) terms and regard the
resulting approximations as exact formulae. This procedure changes the above formulae
into
00
t+h p60 = t p00 − h t p00 µ01 t + µ02 t + h t p01 µ10 t
60
60
60+
60+
60 60+ 01
t+h p60 = t p01 − h t p01 µ12 t + µ10 t + h t p00 µ01 t .
60
60
60+
60+
60 60+ and By choosing successively t = 0, h, 2h, . . . , 10 − h, we can use these formulae, together with
the initial values 0 p00 = 1 and 0 p01 = 0, to calculate h p00 , h p01 , 2h p00 , 2h p01 and so on
60
60
60
60
60
60
00
until we have a value for 10 p60 , as required. These calculations are very well suited to a
spreadsheet. For a step size of h = 1/12 years, selected values are shown in Table 8.1.
Note that the calculations have been carried out using more signiﬁcant ﬁgures than are
shown in this table. Note that the implementation of Euler’s method in this example diﬀers in two respects
from the implementation in Example 7.10: 364 CHAPTER 8. MULTIPLE STATE MODELS
t
0
1
12
2
12
3
12 .
.
.
1
.
.
. 9 11
12
10 µ01 t
60+
0.01420
0.01436
0.01453
0.01469
.
.
. µ02 t
60+
0.01495
0.01506
0.01517
0.01527
.
.
. µ10 t
60+
0.00142
0.00144
0.00145
0.00147
.
.
. µ12 t
60+
0.01495
0.01506
0.01517
0.01527
.
.
. 00
t p60 01
t p60 1.00000
0.99757
0.99512
0.99266
.
.
. 0.00000
0.00118
0.00238
0.00358
.
.
. 0.01625
.
.
. 0.01628
.
.
. 0.00162
.
.
. 0.01628
.
.
. 0.96977
.
.
. 0.01479
.
.
. 0.05473
0.05535 0.03492
0.03517 0.00547
0.00554 0.03492
0.03517 0.59189
0.58756 0.20061
0.20263 Table 8.1: Calculation of 00
10 p60 and 01
10 p60 using a step size h = 1/12 years. (1) We work forward recursively from initial values for the probabilities rather than
backwards from the ﬁnal value of the policy value. This is determined by the
boundary conditions for the diﬀerential equations.
(2) We have two equations to solve simultaneously rather than a single equation. This
is a typical feature of applying Euler’s method to the calculation of probabilities
for multiple state models. In general, the number of equations increases with the
number of states in the model. 8.6 Premiums So far in this chapter we have shown that multiple state models are a natural way of
modelling cash ﬂows for insurance policies and we have also shown how to evaluate probabilities for such models given only the transition intensities between pairs of states. The
next stage in our study of multiple state models is to calculate premiums and policy values
for a policy represented by such a model and to show how we can evaluate them. 8.6. PREMIUMS 365 To do this we can generalize our deﬁnitions of insurance and annuity functions to a multiple state framework. We implicitly use the indicator function approach, which leads
directly to intuitive formulae for the expected present values, but does not give higher
moments. There is no standard notation for annuity and insurance functions in the multiple state model framework. The notation used in this chapter generalizes the notation
introduced in Chapters 4 and 5.
Suppose we have a life aged x currently in state i of a multiple state model. We wish
to value an annuity of 1 per year payable continuously while the life is in some state j
(which may be equal to i).
The EPV of the annuity, at force of interest δ per year, is
∞ aij = E
¯x 0
∞ =
0 ∞ =
0 e−δ t I (Y (t) = j Y (0) = i)dt e−δ t E [I (Y (t) = j Y (0) = i)] dt
e−δ t t pij dt,
x where I is the indicator function.
Similarly, if the annuity is payable at the start of each year, from the current time,
conditional on the life being in state j , given that the life is currently in state i, the
expected present value is
aij =
¨x ∞ v k k pij .
x k=0 Annuity beneﬁts payable more frequently can be valued similarly.
For insurance beneﬁts, the payment is usually conditional on making a transition. A death
beneﬁt is payable on transition into the dead state; a critical illness insurance policy might 366 CHAPTER 8. MULTIPLE STATE MODELS pay a sum insured on death or earlier diagnosis of one of a speciﬁed group of illnesses.
Suppose a unit beneﬁt is payable immediately on each future transfer into state k , given
that the life is currently in state i (which may be equal to k ). Then the expected present
value of the beneﬁt is
¯
Aik =
x ∞
0 e−δt t pij µjk t dt.
x x+ (8.16) j =k To derive this, we consider payment in the interval (t, t + dt);
• the amount of the payment is 1,
• the discount factor (for suﬃciently small dt) is e−δt , and
• the probability that the beneﬁt is paid is the probability that the life transfers into
state k in (t, t + dt), given that the life is in state i at time 0. In order to transfer
into state k in (t, t + dt), the life must be in some state j that is not k immediately
before (the probability of two transitions in inﬁnitesimal time being negligible),
with probability t pij , then transfer from j to k during the interval (t, t + dt), with
x
probability (loosely) µjk t dt.
x+
Summing (that is, integrating) over all the possible time intervals gives equation (8.16).
Other beneﬁts and annuity designs are feasible; for example, a lump sum beneﬁt might
be paid on the ﬁrst transition from healthy to sick, or premiums may be paid only during
the ﬁrst sojourn in state 0. Most practical cases can be managed from ﬁrst principles
using the indicator function approach.
In general, premiums are calculated using the equivalence principle and we assume that
lives are in state 0 at the policy inception date.
Example 8.6 An insurer issues a 10 year disability income insurance policy to a healthy
life aged 60. Calculate the premiums for the following two policy designs using the model 8.6. PREMIUMS 367 and parameters from Example 8.5. Assume an interest rate of 5% per year eﬀective, and
that there are no expenses.
(a) Premiums are payable continuously while in the healthy state. A beneﬁt of $20 000
per year is payable continuously while in the disabled state. A death beneﬁt of
$50 000 is payable immediately on death.
(b) Premiums are payable monthly in advance conditional on the life being in the healthy
state at the premium date. The sickness beneﬁt of $20 000 per year is payable
monthly in arrear, if the life is in the sick state at the payment date. A death
beneﬁt of $50 000 is payable immediately on death.
Solution 8.6 (a) We equate the EPV of the premiums with the EPV of the beneﬁts. The computation of the EPV of the beneﬁts requires numerical integration. All
values below have been calculated using the repeated Simpson’s rule, with h = 1/12
(where h is as in Section B.1.2 in Appendix B), using Table 8.1.
Let P denote the annual rate of premium. Then the EPV of the premium income
is
10 P a00 10 = P
¯60: 0 e−δ t t p00 dt
60 and numerical integration gives a00 10 = 6.5714.
¯60:
Next, the EPV of the sickness beneﬁt is
10 20 000 a01 10 = 20 000
¯60: 0 e−δ t t p01 dt,
60 and numerical integration gives a01 10 = 0.66359.
¯60:
Last, the EPV of the death beneﬁt is
¯
50 000 A02 10 = 50 000
60: 10
0 e−δ t 00
t p60 µ02 t + t p01 µ12 t dt.
60+
60 60+ 368 CHAPTER 8. MULTIPLE STATE MODELS
¯
Using numerical integration, we ﬁnd A01 10 = 0.16231.
60:
Hence, the annual premium rate is
P= ¯
20 000 a01 10 + 50 000 A01 10
¯60:
60:
a00 10
¯60: = $3 254.65. (b) We now need to ﬁnd the EPV of annuities payable monthly, and we can calculate
these from Table 8.1. First, to ﬁnd the EPV of premium income we calculate
(12) 00 a60:10
¨ = 1
1+
12 1 1
12 p00 v 12 +
60 2 2
12 p00 v 12 +
60 3 3
12 p00 v 12 + ... +
60 00
9 11 p60
12 11 v 9 12 = 6.5980,
and to ﬁnd the EPV of the sickness beneﬁt we require (12) 01 a60:10 = 1
12 1 1
12 p01 v 12 +
60 2 2
12 p01 v 12 +
60 3 3
12 p01 v 12 + ... +
60 01
10 p60 v 10 = 0.66877.
Note that the premiums are payable in advance, so that the ﬁnal premium payment
date is at time 9 11 . However, the disability beneﬁt is payable in arrear so that a
12
payment will be made at time 10 if the policyholder is disabled at that time.
The death beneﬁt is unchanged from part (a), so the premium is $3 257.20 per year,
or $271.43 per month. We explore insurance and annuity functions, as well as premium calculation, in more
detail in Sections 8.8, 8.9 and 8.10 for the models that are most common in actuarial
applications. 8.7. POLICY VALUES AND THIELE’S DIFFERENTIAL EQUATION 8.7 369 Policy values and Thiele’s diﬀerential equation The deﬁnition of the time t policy value for a policy modelled using a multiple state model
is exactly as in Chapter 7 – it is the expected value at that time of the future loss random
variable – with one additional requirement. For a policy described by a multiple state
model, the future loss random variable, and hence the policy value, at duration t years
depends on which state of the model the policyholder is in at that time. We can express
this formally as follows: a policy value is the expected value at that time of the future
loss random variable conditional on the policy being in a given state at that time. We
use the following notation for policy values.
Notation: t V (i) denotes the policy value at duration t for a policy which is in state i at
that time.
This additional feature was not necessary in Chapter 7 since all policies discussed in that,
and earlier, chapters were based on the ‘alive–dead’ model illustrated in Figure 8.1, and
for that model the policyholder was either dead at time t, in which case no policy value
was required, or was in state 0.
As in Chapter 7, a policy value depends numerically on the basis used in its calculation,
that is
(a) the transition intensities between pairs of states as functions of the individual’s age,
(b) the force of interest,
(c) the assumed expenses, and
(d) the assumed bonus rates in the case of participating policies.
The key to calculating policy values is Thiele’s diﬀerential equation, which can be solved
numerically using Euler’s, or some more sophisticated, method. To establish some ideas we
start by considering a particular example represented by the disability income insurance
model, Figure 8.4. We then consider the general case. 370 CHAPTER 8. MULTIPLE STATE MODELS 8.7.1 The disability income model Consider a policy with a term of n years issued to a life aged x. Premiums are payable
continuously throughout the term at rate P per year while the life is healthy, an annuity
beneﬁt is payable continuously at rate B per year while the life is sick and a lump sum,
S , is payable immediately on death within the term. Recovery from sick to healthy is
possible and the disability income insurance model, Figure 8.4, is appropriate.
We are interested in calculating policy values for this policy and also in calculating the
premium using the equivalence principle. For simplicity we ignore expenses in this section,
but these could be included as extra ‘beneﬁts’ or negative ‘premiums’ provided only that
they are payable continuously at a constant rate while the life is in a given state and/or
are payable as lump sums immediately on transition between pairs of states. Also for
simplicity, we assume that the premium, the beneﬁts and the force of interest, δ per year,
are constants rather than functions of time.
Example 8.7
tV (0) (a) Show that for 0 ≤ t < n
¯
= B a01 t:n−t + S A02 t:n−t − P a00 t:n−t
¯ x+
¯ x+
x+ (8.17) and derive a similar expression for t V (1) .
(b) Show that, for 0 ≤ t < n
d
(0)
= δ t V (0) + P − µ01 t
tV
x+
dt tV (1) − t V (0) − µ02 t S − t V (0)
x+ (8.18) d
(1)
= δ t V (1) − B − µ10 t
tV
x+
dt tV (0) − t V (1) − µ12 t S − t V (1) .
x+ (8.19) and (c) Suppose that
x = 40, n = 20, δ = 0.04, B = $100 000, S = $500 000 8.7. POLICY VALUES AND THIELE’S DIFFERENTIAL EQUATION 371 and
µ01 = a1 + b1 exp{c1 x},
x
µ10 = 0.1 µ01 ,
x
x
µ02 = a2 + b2 exp{c2 x},
x
µ12 = µ02 ,
x
x
where a1 , b1 , c1 , a2 , b2 and c2 are as in Example 8.4.
(i) Calculate 10 V (0) , 10 V (1) and 0 V (0) for n = 20 using Euler’s method with a step
size of 1/12 years given that
1. P = $5 500, and
2. P = $6 000.
(ii) Calculate P using the equivalence principle.
Solution 8.7 (a) The policy value t V (0) equals EPV of future beneﬁts − EPV of future premiums
conditional on being in state 0 at time t
= EPV of future disability income beneﬁt + EPV of future death beneﬁt
− EPV of future premiums
conditional on being in state 0 at time t
This leads directly to formula (8.17). 372 CHAPTER 8. MULTIPLE STATE MODELS
The policy value for a life in state 1 is similar, but conditioning on being in state 1
at time t, so that
tV (1) ¯
= B a11 t:n−t + S A12 t:n−t − P a10 t:n−t
¯ x+
¯ x+
x+ (8.20) where the annuity and insurance functions are deﬁned as in Section 8.6.
(b) We could derive formula (8.18) by diﬀerentiating formula (8.17) but it is more
instructive and quicker to derive it directly. To do this it is helpful to think of t V (0)
as the amount of cash the insurer is holding at time t, given that the policyholder
is in state 0 and that, in terms of expected values, this amount is exactly suﬃcient
to provide for future losses.
Let h be such that t < t + h < n and let h be small. Consider what happens between
times t and t + h. Premiums received and interest earned will increase the insurer’s
cash to
tV (0) δ h e + P sh .
¯ Recall that eδh = 1 + δ h + o(h) and sh = (eδh − 1)/δ = h + o(h), so that
¯
tV (0) δ h e + P sh = t V (0) (1 + δ h) + P h + o(h).
¯ This amount must be suﬃcient to provide the amount the insurer expects to need
at time t + h. This amount is a policy value of t+h V (0) and possible extra amounts
of
(i) S −
and
(ii) t+h V (0) if the policyholder dies: the probability of which is h µ02 t + o(h),
x+ (1)
− t+h V (0)
t+h V
h µ01 t + o(h).
x+ if the policyholder falls sick: the probability of which is Hence
tV (0) (1 + δ h) + P h 8.7. POLICY VALUES AND THIELE’S DIFFERENTIAL EQUATION
= t+h V (0) + h µ02 t S −
x+ t+h V (0) + µ01 t
x+ t+h V (1) − t+h V (0) 373
+ o(h). Rearranging, dividing by h and letting h → 0 gives formula (8.18).
Formula (8.19) is derived similarly.
(c) (i) Euler’s method for the numerical evaluation of t V (0) and t V (1) is based on
replacing the diﬀerentials on the left hand sides of formulae (8.18) and (8.19)
by discrete time approximations based on a step size h, which are correct up
to o(h). We could write, for example,
d
(0)
= ( t+h V (0) − t V (0) )/h + o(h)/h.
tV
dt
Putting this into formula (8.18) would give a formula for t+h V (0) in terms of
(0)
and t V (1) . This is not ideal since the starting values for using Euler’s
tV
method are n V (0) = 0 = n V (1) and so we will be working backwards, calculating successively policy values at durations n − h, n − 2h, . . . , h, 0. For this
reason, it is more convenient to have formulae for t−h V (0) and t−h V (1) in terms
of t V (0) and t V (1) . We can achieve this by writing
d
(0)
(0)
(0)
= (Vt − Vt−h )/h + o(h)/h
tV
dt
and
d
(1)
(1)
= (Vt −
tV
dt t−h V (1) )/h + o(h)/h. Putting these expressions into formulae (8.18) and (8.19), multiplying through
by h, rearranging and ignoring terms which are o(h), gives the following two
(approximate) equations
(0) Vt−h = tV (0) (1 − δ h) − P h + hµ01 t ( t V (1) − t V (0) ) + hµ02 t (S − t V (0) )
x+
x+
(8.21) 374 CHAPTER 8. MULTIPLE STATE MODELS
and
(1) Vt−h = tV (1) (1 − δ h) + Bh + hµ10 t ( t V (0) − t V (1) ) + hµ12 t (S − t V (1) ).
x+
x+
(8.22) These equations, together with the starting values at time n and given values
of the step size, h, and premium rate, P , can be used to calculate successively
(0) (1) (0) (1) (0) (1) (0) Vn−h , Vn−h , Vn−2h , Vn−2h , . . . , V10 , V10 , . . . , V0 .
1. For n = 20, h = 1/12 and P = $5 500, we get
(0) V10 = $18 084, (1) V10 = $829 731, (0) V0 = $3 815. 2. For n = 20, h = 1/12 and P = $6 000, we get
(0) V10 = $14 226, (1) V10 = $829 721, (0) V0 = −$2 617. (ii) Let P ∗ be the premium calculated using the equivalence principle. Then for
(0)
this premium we have by deﬁnition V0 = 0. Using the results in part (i) and
(0)
assuming V0 is (approximately) a linear function of P , we have
0 − 3 815
P ∗ − 5 500
≈
6 000 − 5 500
−2 617 − 3 815
∗
so that P ≈ $5 797. Using Solver or Goal Seek in Excel, setting 0 V (0) to be equal to zero, by varying
P , the equivalence principle premium is $5 796.59.
Using the techniques of Example 8.6 gives
a00 20 = 12.8535,
¯40: a01 20 = 0.31593,
¯40: ¯
A02 20 = 0.08521,
40: and hence an equivalence principle premium of $5 772.56. The diﬀerence arises
because we are using two diﬀerent approximation methods. 8.7. POLICY VALUES AND THIELE’S DIFFERENTIAL EQUATION 375 The above example illustrates why, for a multiple state model, the policy value at duration
t depends on the state the individual is in at that time. If, in this example, the individual
is in state 0 at time 10, then it is quite likely that no beneﬁts will ever be paid and so
only a modest policy value is required. On the other hand, if the individual is in state 1,
it is very likely that beneﬁts at the rate of $100 000 per year will be paid for the next 10
years and no future premiums will be received. In this case, a substantial policy value is
(0)
(1)
required. The diﬀerence between the values of V10 and V10 in part (c), and the fact that
the latter are not much aﬀected by the value of the premium, demonstrate this point. 8.7.2 Thiele’s diﬀerential equation – the general case Consider an insurance policy issued at age x and with term n years described by a multiple
state model with n + 1 states, labelled 0, 1, 2, . . . , n. Let
µij denote the transition intensity between states i and j at age y ,
y
δt denote the force of interest per year at time t,
(i) Bt (ij ) St denote the rate of payment of beneﬁt while the policyholder is in state i, and
denote the lump sum beneﬁt payable instantaneously at time t on transition from
state i to state j . i
We assume that δt , Bt and Stij are continuous functions of t. Note that premiums are
included within this model as negative beneﬁts and expenses can be included as additions
to the beneﬁts. For this very general model, Thiele’s diﬀerential equation is as follows.
For i = 0, 1, . . . , n and 0 ≤ t ≤ n,
n d
(i)
(i)
= δt t V (i) − Bt −
tV
dt
j =0, (ij ) µij+t St
x
j =i + t V (j ) − t V (i) . (8.23) 376 CHAPTER 8. MULTIPLE STATE MODELS Formula (8.23) can be interpreted in exactly the same way as formula (7.12). At time t
the policy value for a policy in state i, t V (i) , is changing as a result of
interest being earned at rate δt t V (i) , and
(i) beneﬁts being paid at rate Bt .
The policy value will also change if the policyholder jumps from state i to any other state
j at this time. The intensity of such a jump is µij+t and the eﬀect on the policy value will
x
be
(ij ) a decrease of St as the insurer has to pay any lump sum beneﬁt contingent on
jumping from state i to state j ,
a decrease of t V (j ) as the insurer has to set up the appropriate policy value in the
new state, and
an increase of t V (i) as this amount is no longer needed.
Formula (8.23) can be derived more formally by writing down an integral equation for
(i)
and diﬀerentiating it. See Exercise 8.3.
tV
We can use formula (8.23) to calculate policy values exactly as we did in Example 8.7.
We choose a small step size h and replace the left hand side by
( t V (i) − t−h V (i) + o(h))/h Multiplying through by h, rearranging and ignoring terms which are o(h), we have a
(i)
formula for Vt−h , i = 0, . . . , n, in terms of the policy values at duration t. We can then
(i)
use Euler’s method, starting with Vn = 0, to calculate the policy values at durations
n − h, n − 2h, . . . , h, 0. 8.8. MULTIPLE DECREMENT MODELS 377
µ01
x E
t
t
t
02
t
µx
t
s
t
t
t
0n
t µx
t
t
t
t
Alive
0 Exit
1
Exit
2
.
.
.
.
.
.
Exit
n Figure 8.6: A general multiple decrement model 8.8 Multiple decrement models Multiple decrement models are special types of multiple state models which occur frequently in actuarial applications. A multiple decrement model is characterized by having
a single starting state and several exit states with a possible transition from the starting
state to any of the exit states, but no further transitions. Figure 8.6 illustrates a general
multiple decrement model. The accidental death model, illustrated in Figure 8.2, is an
example of such a model with two exit states.
Calculating probabilities for a multiple decrement model is relatively easy since only
one transition can ever take place. For such a model we have for i = 1, 2, . . . , n and
j = 0, 1, . . . , n (j = i),
t
00
t px ≡ 00
t px = exp − 0 n µ0i+s ds ,
x
i=1 378 CHAPTER 8. MULTIPLE STATE MODELS
t
0i
=
t px 0 ii
0 px µ0i+s ds,
x = 1, ij
0 px 00
s px = 0. Assuming we know the transition intensities as functions of x, we can calculate t p00 and
x
0i
t px using numerical or, in some cases, analytic integration.
The following example illustrates a feature which commonly occurs when a multiple decrement model is used. We discuss the general point after completing the example.
Example 8.8 A 10 year term insurance policy is issued to a life aged 50. The sum
insured, payable immediately on death, is $200 000 and premiums are payable continuously
at a constant rate throughout the term. No beneﬁt is payable if the policyholder lapses,
that is, cancels the policy during the term.
Calculate the annual premium rate using the following two sets of assumptions.
(a) The force of interest is 2.5% per year.
The force of mortality is given by µx = 0.002 + 0.0005(x − 50).
No allowance is made for lapses.
No allowance is made for expenses.
(b) The force of interest is 2.5% per year.
The force of mortality is given by µx = 0.002 + 0.0005(x − 50).
The transition intensity for lapses is a constant equal to 0.05.
No allowance is made for expenses.
Solution 8.8 (a) Since lapses are being ignored, an appropriate model for this policy
is the ‘alive–dead’ model shown in Figure 8.1. 8.8. MULTIPLE DECREMENT MODELS 379 The annual premium rate, P , calculated using the equivalence principle, is given by
P = 200 000 ¯
A01 10
50:
a00 10
¯50: where
¯
A01 10 =
50: 10
0
10 a00 10 =
¯50: 0 e−δt t p00 µ01 t dt,
50 50+ e−δt t p00 dt
50 and
00
t p50 = exp{−0.002t − 0.00025t2 }. Using numerical integration to calculate the integrals, we ﬁnd
P = 200 000 × 0.03807/8.6961 = $875.49.
(b) To allow for lapses, the model should be as in Figure 8.7. Note that this has the
same structure as the accidental death model illustrated in Figure 8.2 – a single
starting state and two exit states – but with diﬀerent labels for the states. Using
this model, the formula for the premium, P , is still P = 200 000 ¯
A01 10
50:
a00 10
¯50: but now
00
t p50 = exp{−0.052t − 0.00025t2 }, which gives
P = 200 000 × 0.02890/6.9269 = $834.54. 380 CHAPTER 8. MULTIPLE STATE MODELS Active
0 d
d d d
d
Dead
1 Lapsed
2 Figure 8.7: The insurance–with–lapses model.
We make the following observations about Example 8.8.
(1) The premium allowing for lapses is a little lower than the premium which does not
allow for lapses. This was to be expected. The insurer will make a proﬁt from any
lapses in this example because, without allowing for lapses, the policy value at any
duration is positive and a lapse (with no beneﬁt payable) releases this amount as
proﬁt to the insurer. If the insurer allows for lapses, these proﬁts can be used to
reduce the premium.
(2) In practice, the insurer may prefer not to allow for lapses when pricing policies
if, as in this example, this leads to a higher premium. The decision to lapse is
totally at the discretion of the policyholder and depends on many factors, both
personal and economic, beyond the control of the insurer. Where lapses are used to
reduce the premium, the business is called lapse supported. Because lapses are
unpredictable, lapse supported pricing is considered somewhat risky and has proved
to be a controversial technique.
(3) Note that two diﬀerent models were used in the example to calculate a premium 8.8. MULTIPLE DECREMENT MODELS 381 for the policy. The choice of model depends on the terms of the policy and on the
assumptions made by the insurer.
(4) The two models used in this example are clearly diﬀerent, but they are connected.
The diﬀerence is that the model in Figure 8.7 has more exit states; the connections
between the models are that the single exit state in Figure 8.1, ‘Dead’, is one of the
exit states in Figure 8.7 and the transition intensity into this state, µ01 , is the same
x
in the two models.
(5) The probability that the policyholder, starting at age 50, ‘dies’, that is enters state
1, before age 50 + t is diﬀerent for the two models. For the model in Figure 8.1 this
is
t exp{−0.002r − 0.00025r2 } (0.002 + 0.0005r)dr, 0 whereas for the model in Figure 8.7 it is
t exp{−0.052r − 0.00025r2 } (0.002 + 0.0005r)dr. 0 The explanation for this is that for the model in Figure 8.7, we interpret ‘dies’ as
dying before lapsing. The probability of this is aﬀected by the intensity of lapsing.
If we increase this intensity, the probability of dying (before lapsing) decreases, as
more lives lapse before they die.
Points (4) and (5) illustrate common features in the application of multiple decrement
models. When working with a multiple decrement model we are often interested in a
simpler model with only one of the exit states and with the same transition intensity into
this state. For exit state j , the reduced model is called the related single decrement
model for decrement j . Using the notation in Figure 8.6, the related single decrement
model for decrement j is shown in Figure 8.8.
Starting in state 0 at age x, the probability of being in state j = 0 at age x + t is
t
0j
t px =
0 s exp − 0 n µ0i+u du
x
i=1 µ0j s ds
x+ 382 CHAPTER 8. MULTIPLE STATE MODELS
Active
0 µ0 j E
x Exit
j Figure 8.8: Independent single decrement model, exit j . and
t
00
t px = exp − n µ0i+u du
x 0 i=1 for the multiple decrement model in Figure 8.6, and
t
0j
t px =
0 s exp − µ0j u du
x+ µ0j s ds
x+ 0 for j = 0 and
s
00
t px = exp − µ0j u du
x+ 0 for the related single decrement model in Figure 8.8. The ﬁrst two of these probabilities
are called the dependent probabilities of exiting by decrement j before age x + t, or
of surviving in force to age x + t because the values depend on the values of the other
transition intensities; the p probabilities are called the independent probabilities of
exiting or surviving for decrement j because the values are independent of any other
transition intensities. The purpose of identifying the independent probabilities is usually
associated with changing assumptions. 8.9. JOINT LIFE AND LAST SURVIVOR BENEFITS Husband Alive
Wife Alive µ01 t:y+t
x+ E 383 Husband Alive
Wife Dead
1 0
µ02 t:y+t
x+ µ13 t
x+ c Husband Dead
Wife Alive c µ23 t
y+ E 2 Husband Dead
Wife Dead
3 Figure 8.9: The joint life and last survivor model. 8.9
8.9.1 Joint life and last survivor beneﬁts
The model and assumptions In this section we consider the valuation of beneﬁts and premiums for an insurance policy
where these payments depend on the survival or death of two lives. Such policies are
very common. Policies relating to three or more lives also exist, but are far less common.
For conciseness, we refer to the two lives as ‘husband’ and ‘wife’. It is often the case in
practice that the two lives are social partners, but they need not be; for example, they
may be business partners. We consider future payments from a time when both husband
and wife are alive and are aged x and y , respectively.
We need to evaluate probabilities of survival/death for our two lives, and these probabilities must come from a model. The model we use is illustrated in Figure 8.9 and has the
same structure as the model in Figure 8.5. 384 CHAPTER 8. MULTIPLE STATE MODELS Our model incorporates the following assumptions and notational changes.
(1) The intensity of mortality for each life depends on whether the other partner is
still alive. If the partner is alive, the intensity depends on the exact age of the
partner, as well as the age of the life being considered, and our notation is adjusted
appropriately. For example, µ01 t:y+t is the intensity of mortality for the wife when
x+
she is aged y + t given that her husband is still alive and that he is aged x + t.
However, if one partner, say the husband, has died, the intensity of mortality for
the wife depends on her current age, and the fact that her husband has died, but
not on how long he has been dead. Since the age at death of the husband is assumed
not to aﬀect the transition intensity from state 2 to state 3, this intensity is denoted
µ23 t , where y + t is the current age of the wife.
y+
(2) Our notation for probabilities for this model diﬀers slightly from our usual notation
for multiple state models and is consistent with the notation we adopt for transition
intensities. Hence, for i = 0, 1, 2, 3, t p0i denotes the probability that at time t the
xy
‘process’ is in state i given that the husband and wife are now both alive and are
aged x and y , respectively, whereas, for example, t p13 u denotes the probability that
x+
the husband, who is now aged x + u and whose wife has already died, dies before
reaching age x + u + t. For this latter probability, the exact age at which the wife
died is assumed to be irrelevant and so is not part of the notation. 8.9.2 Joint life and last survivor probabilities Notation
The standard actuarial notation for joint life beneﬁts diﬀers from the general multiple
state model notation that we have been using previously in this chapter, because joint
life policies have been around from before the time when multiple state models were
developed. We therefore need to introduce this new notation, which is consistent with
the notation of Chapters 2, 4 and 5. 8.9. JOINT LIFE AND LAST SURVIVOR BENEFITS 385 In the list below we give the new notation, followed by its deﬁnition as a probability,
followed, in some cases, by the equivalent multiple state model notation.
t pxy = Pr[(x) and (y ) are both alive in t years]= t p00 .
xy t qxy = Pr[(x) and (y ) are not both alive in t years]= t p01 + t p02 + t p03 .
xy
xy
xy 1
t qxy = Pr[(x) dies before (y ) and within t years]. 2
t qxy = Pr[(x) dies after (y ) and within t years]. t pxy = Pr[at least one of (x) and (y ) is alive in t years]= t p00 + t p01 + t p02 .
xy
xy
xy t qxy = Pr[ (x) and (y ) are both dead in t years]= t p03 .
xy We refer to the right subscript, x y or x y as a status. The q type probabilities are
associated with the failure of the status – the joint life status x y fails on the ﬁrst death
of (x) and (y ), and the last survivor status x y fails on the last death of (x) and (y ).
The joint life status in particular is important in life insurance. The standard actuarial
notation µx+t:y+t denotes the total force of transition out of state 0 at time t, that is
µx+t:y+t = µ01 t:y+t + µ02 t:y+t .
x+
x+ (8.24) 1
The ‘1’ over x in, for example, t qxy indicates that we are interested in the probability of
1
(x) dying ﬁrst. We have already used this notation, in Ax:n , where the beneﬁt is only
paid if x dies ﬁrst, before the term, n years, expires. Note that in cases where it makes the notation clearer, we put a colon between the ages
in the right subscript. For example, we write t p30:40 rather than t p30 40 .
The probabilities listed above do not all correspond to t pij type probabilities. We examine
two in more detail in the following example.
1
Example 8.9 (a) Explain why t qxy is not the same as t p02 , and write down an integral
xy
1
equation for t qxy . 386 CHAPTER 8. MULTIPLE STATE MODELS 2
(b) Write down an integral equation for t qxy .
1
Solution 8.9 (a) The probability t qxy is the probability that (x) dies within t years,
and that (y ) is alive at the time of (x)’s death. The probability t p02 is the probability that (x) dies within t years, and that (y ) is
xy
alive at time t years. So the ﬁrst probability allows for the possibility that (y ) dies
after (x), within t years, and the second does not.
The probability that (x) dies within t years, and that (y ) is alive at the time of the
death of (x) can be constructed by summing (integrating) over all the inﬁnitesimal
intervals in which (x) could die, conditioning on the survival of both (x) and (y ) up
to that time, so that
t
1
t qxy 00
r pxy =
0 µ02 r:y+r dr .
x+ 2
(b) The probability t qxy is the probability that the husband dies within t years and that
the wife is already dead when the husband dies, conditional on the husband and
wife both being alive now, time 0, and aged x and y , respectively. In terms of the
model in Figure 8.9, the process must move into state 1 and then into state 3 within
t years, given that it starts in state 0 at time 0. Summing all the probabilities of
such a move over inﬁnitesimal intervals, we have
t
2
t qxy = 8.9.3 0 01
r pxy µ13 r dr .
x+ Joint life and last survivor annuity and insurance functions We consider the EPVs, using a constant force of interest δ per year, of the following
payments. In each case the deﬁnition of the cash ﬂow is preceded by the actuarial notation
for its EPV. 8.9. JOINT LIFE AND LAST SURVIVOR BENEFITS 387 axy Joint life annuity: a continuous payment at unit rate per year while both husband
¯
and wife are still alive. This is the same as a00 in the multiple state model notation.
¯xy
If there is a maximum period, n years, for the annuity, then we refer to a ‘temporary
joint life annuity’ and the notation for the EPV is axy:n . If the annuity is payable
¯
mthly in advance, with payments of 1/m every 1/m years, the EPV is denoted
(m)
axy .
¨
¯
Axy Joint life insurance: a unit payment immediately on the death of the ﬁrst to die
of the husband and wife.
axy Last survivor annuity: a continuous payment at unit rate per year while at least
¯
one of the two lives is still alive. In multiple state model notation we have
axy = a00 + a01 + a02 .
¯
¯xy ¯xy ¯xy
¯
Axy Last survivor insurance: a unit payment immediately on the death of the second
to die of the husband and wife. In multiple state model notation we have
¯
¯xy
Axy = A03 .
axy Reversionary annuity: a continuous payment at unit rate per year while the wife
¯
is alive provided the husband has already died. In multiple state model notation we
have
axy = a02 .
¯
¯xy
¯xy
A1 Contingent insurance: a unit payment immediately on the death of the husband
provided he dies before his wife. If there is a time limit on this payment, say n
years, then it is called a ‘temporary contingent insurance’ and the notation for the
¯1
EPV is Axy:n .
We also need the following EPVs, which have the same meanings as in Chapter 4. 388 CHAPTER 8. MULTIPLE STATE MODELS ay Single life annuity: a continuous payment at unit rate per year while the wife is
¯
still alive.
¯
Ax Whole life insurance: a unit payment immediately on the death of the husband.
Although we have deﬁned these functions in terms of continuous beneﬁts, the annuity and
insurance functions can easily be adapted for payments made at discrete points in time.
(12)
For example, the EPV of a monthly term joint life annuitydue would be denoted axy .
¨
For annuities we can write down the following formulae, given (x) and (y ) are alive at
time t = 0:
∞ ay =
¯
0 ∞ ax =
¯
0 ∞ axy =
¯
0 ∞ axy =
¯
0 ax  y =
¯ ∞
0 e−δt ( t p00 + t p02 ) dt ,
xy
xy
e−δt ( t p00 + t p01 ) dt ,
xy
xy
e−δt t p00 dt ,
xy
e−δt ( t p00 + t p01 + t p02 ) dt ,
xy
xy
xy
e−δt t p02 dt .
xy By manipulating the probabilities in the integrands in these formulae we can derive the
following important formulae
axy = ax + ay − axy
¯
¯
¯
¯ (8.25) axy = ay − axy .
¯
¯
¯ (8.26) and 8.9. JOINT LIFE AND LAST SURVIVOR BENEFITS 389 Formulae (8.25) and (8.26) can be explained in words as follows.
• The payment stream for the last survivor annuity is equivalent to continuous payments at unit rate per year to both husband and wife while each of them is alive
minus a continuous payment at unit rate per year while both are alive. This gives a
net payment at unit rate per year while at least one of them is alive, which is what
we want.
• If we pay one unit per year continuously while the wife is alive but take this amount
away while the husband is also alive, we have a continuous payment at unit rate per
year while the wife is alive but the husband is dead. This is what we want for the
reversionary annuity. For the EPVs of the lump sum payments we have the following formulae:
∞ ¯
Ay =
0 ∞ ¯
Ax =
0 ∞ ¯
Axy =
0 ∞ ¯
Axy =
0 ¯xy
A1 =
¯
A1 :n
xy ∞
0
n =
0 e−δt ( t p00 µ01 t:y+t + t p02 µ23 t ) dt,
xy x+
xy y +
e−δt ( t p00 µ02 t:y+t + t p01 µ13 t ) dt,
xy x+
xy x+
e−δt t p00 (µ01 t:y+t + µ02 t:y+t ) dt,
xy
x+
x+
e−δt (t p01 µ13 t + t p02 µ23 t ) dt,
xy x+
xy y +
e−δt t p00 µ02 t:y+t dt,
xy x+ e−δt t p00 µ02 t:y+t dt.
xy x+ 390 CHAPTER 8. MULTIPLE STATE MODELS From these formulae we can derive the important formula
¯
¯
¯
¯
Axy = Ax + Ay − Axy (8.27) which can explained in the same way as formulae (8.25) and (8.26) by considering cash
ﬂows.
¯
Note that the relationship between ax and Ax in equation (5.14) follows to the joint life
¯
case, so that axy =
¯ ¯
1 − Axy
.
δ (8.28) The proof of this is left to Exercise 8.4.
The formulae for EPVs have been written in terms of probabilities derived from our model.
Since none of the states in the model can be reentered once it has been left, we have
ii
t pxy ≡ t pii for i = 0, 1, 2, 3
xy so that using formula (8.8)
t
00
t pxy = exp − (µ01 s:y+s + µ02 s:y+s +) ds ,
x+
x+ 0 (8.29) t
11
t px = exp − µ13 s ds ,
x+ 0
t 22
t py = exp − 0 µ23 s ds ,
y+ and, for example,
t
01
t pxy =
0 00
s pxy µ01 s:y+s t−s p11 s ds.
x+
x+ (8.30) 8.9. JOINT LIFE AND LAST SURVIVOR BENEFITS 391 Assuming as usual that we know the transition intensities, probabilities for the model can
be evaluated either by starting with Kolmogorov’s forward equations, (8.14), and then
using Euler’s, or some more sophisticated, method, or, alternatively, by starting with
formulae corresponding to (8.29) and (8.30) and integrating, probably numerically.
Example 8.10 (a) Derive the following expression for the probability that the husband
has died before reaching age x + t
t
0 t
00 02
s pxy µx+s:y +s ds + 0 s
u=0 00
u pxy µ01 u:y+u s−u p11 u du µ13 s ds.
x+
x+
x+ (b) Now suppose that µ01 t:y+t = µ23 t and µ02 t:y+t = µ13 t for all t ≥ 0. Show that
x+
x+
x+
y+
(i) the probability that both husband and wife are alive at time t is
t exp − t µ13 s ds exp −
x+ 0 0 µ23 s ds ,
y+ (8.31) (ii) the probability that the husband is alive and the wife is dead at time t is
t exp − t µ13 s ds
x+ 0 1 − exp − 0 µ23 s ds
y+ , (8.32) (iii) the probability that the husband is alive at time t is
t exp − 0 µ13 s ds ,
x+ (8.33) (iv) the probability that both the husband and the wife are dead at time t is
t t 1 − exp −
Solution 8.10
to 0 µ13 s ds
x+ 1 − exp − 0 µ23 s ds
y+ . (8.34) (a) For the husband to die before time t we require the process either 392 CHAPTER 8. MULTIPLE STATE MODELS
– enter state 2 from state 0 at some time s (0 < s ≤ t), or
– enter state 1 (the wife dies while the husband is alive) at some time u
(0 < u ≤ t) and then enter state 3 at some time s (u < s ≤ t).
The total probability of these events, integrating over the time of death of (x), is
t
0 t
00
s pxy µ02 s:y+s
x+ 01
s pxy ds +
0 µ13 s ds
x+ where
s
00
s pxy = exp − 0 (µ01 u:y+u + µ02 u:y+u )du ,
x+
x+ s
01
s pxy = 0 00
u pxy µ01 u:y+u s−u p11 u du,
x+
x+ and
11
s−u px+u = exp − s−u
0 µ13 u+r dr .
x+ This gives the formula in part (a).
(b) (i) The required probability is t p00 , which can be written as
xy
t
00
t pxy = exp − (µ01 s:y+s + µ02 s:y+s ) ds
x+
x+ 0
t = exp − t µ01 s:y+s ds exp −
x+ 0
t = exp − 0 0 µ02 s:y+s ds
x+ t µ13 s ds exp −
x+ 0 µ23 s ds .
y+ 8.9. JOINT LIFE AND LAST SURVIVOR BENEFITS 393 (ii) The probability that the husband is alive and the wife is dead at time t is t p01 .
xy
Integrating over the age (y + s) at which the wife dies and using the formulae
for s p00 and t−s p11 t , gives
xy
x+
t
01
t pxy = 00
s pxy 0 µ01 s:y+s
x+ t
0 (µ01 u:y+u + µ02 u:y+u ) du
x+
x+ 0 × exp −
t =
0 ds s exp − = 11
t−s px+s t−s
0 µ13 s+u du
x+ µ01 s:y+s
x+ ds s exp − 0 (µ23 u + µ13 u ) du
y+
x+ µ23 s
y+ t × exp − s µ13 u du
x+ t = exp − t µ13 u du
x+ 0 ds 0 s exp − 0 µ23 u du
y+ t = exp − 0 t µ13 u du
x+ 1 − exp − 0 as required.
(iii) The probability that the husband is alive at time t is
00
t pxy + t p01 .
xy Using the results from parts (i) and (ii), we have
t
00
t pxy + t p01 = exp −
xy 0 (µ13 s + µ23 s ) ds
x+
y+ µ23 u du
y+ µ23 s ds
y+ 394 CHAPTER 8. MULTIPLE STATE MODELS
t + exp − 0 t µ13 s ds
x+ 1 − exp − 0 µ23 s ds
y+ t = exp − 0 µ13 s ds .
x+ (iv) The required probability, t p03 , can be written as
xy
03
t pxy = 1 − t p00 − t p01 − t p02 .
xy
xy
xy The result follows from formulae (8.31), (8.32) (and the corresponding formula
for t p02 ) and (8.34).
xy 8.9.4 An important special case: independent survival models In our model for the two lives, the mortality of each life depends on the survival or death
of the other life through the assumption that the intensity of mortality of, for example,
the husband depends on whether or not the wife is still alive. A special case of this model,
which is important because it is often used in practice, makes the following simplifying
assumptions, which were used in part (b) of Example 8.10:
µ01 t:y+t = µ23 t = µyf t
+
x+
y+
and
µ02 t:y+t = µ13 t = µxm t ,
x+
x+
+
where µyf t and µxm t are the individual forces of mortality for the wife (female) and
+
+
husband (male), respectively, from the twostate, alivedead models for their individual
mortality.
These equivalencies tell us that, with these assumptions, the mortality of each life does
not depend on whether the partner is still alive. These assumptions remove any link 8.9. JOINT LIFE AND LAST SURVIVOR BENEFITS 395 between the survival/mortality of the two lives so that, in terms of survival, they are
probabilistically independent. This independence is illustrated in formulae (8.31), (8.32),
(8.33) and (8.34) where probabilities of joint events are the product of the probabilities
of events for each life separately and probabilities for the separate lives are derived from
the two individual ‘alivedead’ models for the husband and wife.
The connection between the individual models and the joint model is illustrated in Figure
8.10, where we show that each transition depends only on the single life force of mortality. Husband Alive
Wife Alive µyf t
+ E 0 Husband Alive
Wife Dead
1 µxm t
+ µxm t
+ c Husband Dead
Wife Alive c µyf t
+ 2 E Husband Dead
Wife Dead
3 Figure 8.10: The independent joint life and last survivor model. In particular, in standard actuarial notation, assuming independence of the two lives
means that
t pxy = t px t py (8.35) which is the same result as equation (8.31),
t pxy = 1 − (1 − t px )(1 − t py ) (8.36) 396 CHAPTER 8. MULTIPLE STATE MODELS which is derived from equation (8.34), and
f
µm:y = µm + µf .
x
y
x (8.37) Example 8.11 A husband, currently aged 55, and his wife, currently aged 50, have just
purchased an annuity policy. Level premiums are payable monthly for at most 10 years
but only if both are alive. If either dies within 10 years, a sum insured of $200 000 is
payable at the end of the year of death. If both lives survive 10 years, an annuity of
$50 000 per year is payable monthly in advance while both are alive, reducing to $30 000
per year, still payable monthly, while only one is alive. The annuity ceases on the death
of the last survivor.
Calculate the monthly premium on the following basis:
Mortality: Both lives are subject to the standard select survival model and may be
considered independent with respect to mortality. They are select at the time the
policy is purchased.
Interest: 5% per year eﬀective.
Expenses: Nil.
Solution 8.11 Since the two lives are independent with respect to mortality, we can use
the results in Example 8.10(b) to write the probability that they both survive t years as
t p[55] t p[50] and the probability that, for example, the husband dies within t years but the wife is still
alive as
(1 − t p[55] ) t p[50]
where each single life probability is calculated using the select survival model in Example
3.13. 8.9. JOINT LIFE AND LAST SURVIVOR BENEFITS 397 Let P denote the annual amount of the premium. Then the EPV of the premiums is
P
12 119
t v 12 t
12 p[55] t
12 p[50] = $7.7786 P. t=0 The EPV of the death beneﬁt is
9 200 000
t=0 v t+1 t p[55] t p[50] (1 − p[55]+t p[50]+t ) = $7 660. To ﬁnd the EPV of the annuities we note that if both lives are alive at time 10 years, the
EPV of the payment at time t/12 years from time 10 is
t 50 000 v 12 t
12 p65 t t
12 p60 + 30 000 v 12 t t
= v 12 30 000( 12 p65 + t
12 t t
12 t
p65 1 − 12 p60 + 30 000 v 12 t
p60 ) − 10 000 12 p65 t
12 t
12 t
p60 1 − 12 p65 p60 . Thus, the EPV of the annuities is
1 10
v 10 p[55] 10 p[50]
12 12(ω −65) t t
v 12 30 000( 12 p65 + t=0 t
12 t
p60 ) − 10 000 12 p65 t
12 p60 = $411 396. Hence the monthly premium, $P/12, is given by
P/12 = (7 660 + 411 396)/(12 × 7.7786) = $4 489.41. Note that in the above solution we can write the formula for the monthly premium as
follows
(12) P/12 = (12) (12) 7 660 + v 10 10 p[55] 10 p[50] (30 000(¨65 + a60 ) − 10 000 a65:60 )
a
¨
¨
(12) 12 a[55]:[50]:10
¨ . (8.38) 398 CHAPTER 8. MULTIPLE STATE MODELS As we know the force of mortality at all ages for each life, we can calculate the EPVs
of the annuities exactly. However, we have noted in earlier chapters that it is sometimes
the case in practice that the only information available to us to calculate the EPV of
an annuity payable more frequently than annually is a life table. In Section 5.12 we
illustrated methods of approximating the EPV of an annuity payable m times per year,
and these methods can also be applied to joint life annuities. To illustrate, consider the
annuity EPVs in equation (8.38). These can be approximated from the corresponding
annual values using UDD as
(12) a65
¨ ≈ α(12) a65 − β (12)
¨
= 1.000197 × 13.5498 − 0.466508
= 13.0860, (12) a60
¨ ≈ α(12) a60 − β (12)
¨
= 1.000197 × 14.9041 − 0.466508
= 14.4405, (12) a65:60 ≈ α(12) a65:60 − β (12)
¨
¨
= 1.000197 × 12.3738 − 0.466508
= 11.9097, 8.10. TRANSITIONS AT SPECIFIED AGES 399 and
(12) a[55]:[50]:10
¨ ≈ α(12) a[55]:[50]:10 − β (12)(1 −
¨ 10 p[55] 10 p[50] v 10 ) = 1.000197 × 7.9716 − 0.466508 × 0.41790
= 7.7782.
The approximate value of the monthly premium is then
P/12 ≈ 32 715 + 0.54923 × (30 000(13.0860 + 14.4405) − 10 000 × 11.9097)
= $4 489.33.
12 × 7.7782 An important point to appreciate about applying UDD as we just have is that under
UDD, we have, for example,
a(m) = α(m)¨x − β (m)
¨x
a
but for a joint life status, under the assumption of UDD for each life we do not get a
(m)
simple exact relationship between, for example axy and axy . It is, however, true that
¨
¨
a(m) ≈ α(m)¨xy − β (m).
¨xy
a (8.39) Our calculations above illustrate the general point that this approximation is usually very
accurate. See Exercise 8.17.
In Exercise 8.18 we illustrate how Woolhouse’s formula can be applied to ﬁnd the EPV
of a joint life annuity payable m times per year. 8.10 Transitions at speciﬁed ages A feature of all the multiple state models considered so far in this chapter is that transitions take place in continuous time so that the probability of a transition taking place in a 400 CHAPTER 8. MULTIPLE STATE MODELS E Retired
1 c
E Employee
0
E Dead
3
T Withdrawn
2 Figure 8.11: A withdrawal/retirement model. time interval of length h converges to 0 as h converges to 0. This follows from Assumption
3 in Section 8.3. In practice, there are situations, particularly in the context of pension
plans, where this assumption is not realistic.
The following example illustrates such a situation and the solution shows how this feature
can be incorporated in our calculation of probabilities and EPVs.
Example 8.12 The employees of a large corporation can leave the corporation in three
ways: they can withdraw from the corporation, they can retire or they can die while they
are still employees. Figure 8.11 illustrates this set–up.
Our model is speciﬁed as follows.
• The force of mortality depends on the individual’s age but not on whether the
individual is an employee, has withdrawn or is retired, so that for all ages x
µ03 ≡ µ13 ≡ µ23 = µx , say.
x
x
x
• Withdrawal can take place at any age up to age 60 and the intensity of withdrawal 8.10. TRANSITIONS AT SPECIFIED AGES 401 is a constant denoted µ02 . Hence
µ02 for x < 60,
0
for x ≥ 60. µ02 =
x • Retirement can take place only on an employee’s 60th, 61st, 62nd, 63rd, 64th or 65th
birthday. It is assumed that 40% of employees reaching exact age 60, 61, 62, 63 and
64 retire at that age and 100% of employees who reach age 65 retire immediately.
The corporation oﬀers the following beneﬁts to the employees:
• For those employees who die while still employed, a lump sum of $200 000 is payable
immediately on death.
• For those employees who retire, a lump sum of $150 000 is payable immediately on
death after retirement.
Show that the EPVs, calculated using a constant force of interest δ per year, of these
¯
beneﬁts to an employee currently aged 40 can be written as follows, where A65 and 25 E40
are standard single life functions based on the force of mortality µx .
Death in service beneﬁt
5 ¯1
200 000 A40:20 + 20 E40 −20µ02 e k=1 ¯1
0.6k k−1 A60:1 . Death after retirement beneﬁt
4
−20µ02 150 000 20 E40 e 0.4
k=0 ¯
¯
0.6k k A60 + 0.65 5 A60 . ¯
Solution 8.12 First, note that Ax , the EPV of a unit payment immediately on the death
of a life now aged x, does not depend on whether this individual is still an employee, has 402 CHAPTER 8. MULTIPLE STATE MODELS withdrawn or has retired. This is because the intensity of mortality is the same from
states 0, 1 and 2 in our model.
The novel feature in this example is the nonzero probability of transitions at speciﬁed
ages, in this case retirement on birthdays from ages 60 to 65. For these transitions the
transition intensity does not exist because the limit in formula (8.3) is inﬁnite at these
speciﬁed ages. We need to be able to calculate probabilities for such models and we
can do this by breaking the probability up into the part before the speciﬁed age, the part
relating to transition at the speciﬁed age, and the part after. For example, the probability
of surviving in employment to just before age 60 from age 40 is, say, 20−p00 , where
40
20 p00 = exp −
40 20− (µ02 + µ03 t ) dt
40+ 0
20 = exp −
= 20 p40 µ40+t dt e−20 µ 02 0
02 e−20µ . At exact age 60, 40% of the survivors retire, so the probability of surviving to just after
age 60, 20+p00 say, can be written
40
00
20+p40 = 0.6 20−p00 .
40 Between ages 60 and 61 the only cause of decrement is mortality. So, the probability of
surviving in employment from age 40 to just before age 61 is
00
21−p40 = 00
20+p40 p60 . Then at age 61, another 40% exit, so the probability of being in employment just after
age 61 is
00
21+p40 = 0.6 21−p00 = 0.62
40 21 p40 02 e−20µ , 8.10. TRANSITIONS AT SPECIFIED AGES 403 and so on.
Consider the beneﬁt on death after retirement. Retirement can take place only at exact
ages 60, 61, 62, . . . , 65. If the employee retires at age x, the EPV of the beneﬁt from that
age is
¯
150 000 Ax .
The probability that an employee currently aged 40 will retire at 60 is
00
20− p40 × 0.4 = 0.4 20 p40 e−20µ02 . Hence the EPV at age 40 of the retirement beneﬁt from age 60 is
¯
150 000 A60 e−20δ 0.4 20 p40 e−20µ02
02 ¯
= 150 000 e−20µ 0.4 20 p40 e−20δ A60 .
The probability of retiring at age 61 is the product of the probabilities of the following
events:
surviving in employment to age 60− ,
not retiring at age 60,
surviving from age 60+ to age 61− , and
retiring at age 61.
This probability is equal to
00
20− p40 × 0.6 × 1 p60 × 0.4 = −20µ02
21 p60 e (0.6 × 0.4). Continuing in this way, the probability that the 40 year old employee retires at age 65 is
00
20− p40 = × 0.6 × 1 p60 × 0.6 × 1 p61 × 0.6 × 1 p62 × 0.6 × 1 p63 × 0.6 × 1 p64 25 p40 02 e−20µ 0.65 . 404 CHAPTER 8. MULTIPLE STATE MODELS Hence the EPV of the beneﬁt payable on death after retirement is
4 20+k 02 150 000 exp{−20µ } k=0
25 5 + 0.6 exp − 0 0.6k × 0.4 exp − 0 ¯
µ03 t dt e−(20+k)δ A60+k
40+ ¯
µ03 t dt e−25δ A65
40+
4 −20µ02 = 150 000 20 E40 e 0.4
k=0 ¯
¯
0.6k k A60 + 0.65 5 A60 . The EPV of the lump sum payable on death as an employee can be expressed as the sum
of the EPV of any beneﬁt payable before age 60, the EPV of any beneﬁt payable between
60 and 61, and so on up to the value of any beneﬁt payable between 64 and 65. As with the
death after retirement beneﬁt, we need to split the probabilities after age 60 into up to, at
and after the year end exits. Recalling that, in this example, the probability of surviving
in employment between exact age retirements is the ordinary survival probability 1 px , the
EPV is
20 200 000 −δ t e
0 00
t p40 µ03 t
40+ −20δ dt + e + e−21δ 20− p00 × 0.6 × 1 p60 × 0.6
40 1
00
20− p40
1
0 × 0.6 0 e−δt t p00+ µ03 t dt
60+
60 e−δt t p00+ µ03 t dt
61+
61 +...
+ e−24δ 20− p00 × 0.6 × 1 p60 × 0.6 × 1 p61 × 0.6 × 1 p62 × 0.6 × 1 p63 × 0.6
40
which can be written more neatly as
5 ¯1
200 000 A40:20 + 20 E40 −20µ02 e k=1 ¯1
0.6k k−1 A60:1 . 1
0 e−δt t p00+ µ03 t dt
64+
64 8.11. NOTES AND FURTHER READING 405 We note that in this example considerable simpliﬁcation was possible because the force
of withdrawal was constant, and because the transition intensities to state 3 were the
same. For other examples these assumptions may not hold. The important element of
this example is the technique of splitting up survival probabilities when a transition can
occur at a speciﬁed age. 8.11 Notes and further reading Multiple state models are known to probabilists as Markov processes with discrete states
in continuous time. The processes of interest to actuaries are time–inhomogeneous since
the transition intensities are functions of time/age. Good references for such processes
are Cox and Miller (1965) and Ross (1995). Rolski et al (1999) provide a brief treatment
of such models within an insurance context.
Andrei Andreyevich Markov (1865–1922) was a Russian mathematician best known for
his work in probability theory. Andrei Nikolaevich Kolmogorov (1903–1987) was also a
Russian mathematician. He made many fundamental contributions to probability theory
and is generally credited with putting probability theory on a sound mathematical basis.
The application of multiple state models to problems in actuarial science goes back at
least to Sverdrup (1965). Hoem (1988) provides a very comprehensive treatment of the
mathematics of such models. Multiple state models are not only a natural framework for
modelling conventional life and health insurance policies, they are also a valuable research
tool in actuarial science. See, for example, Macdonald et al (2003a & 2003b).
Norberg (1995) shows how to calculate the k − th moment, k = 1, 2, 3, . . ., for the present
value of future cash ﬂows from a very general multiple state model. He also reports that
the transition intensities used in part (b) of Example 8.4, and subsequent examples, are
those used at that time by Danish insurance companies.
In Section 8.4 we remarked that the transition intensities are fundamental quantities 406 CHAPTER 8. MULTIPLE STATE MODELS which determine everything we need to know about a multiple state model. They are also
in many insurance–related contexts the natural quantities to estimate from data. See, for
example, Sverdrup (1965) or Waters (1984).
We can extend multiple state models in various ways. One way is to allow the transition
intensities out of a state to depend not only on the individual’s current age but also
on how long they have been in the current state. This breaks the Markov assumption
and the new process is known as a semiMarkov process. This could be appropriate for
the disability income insurance process (Figure 8.4) where the intensities of recovery and
death from the sick state could be assumed to depend on how long the individual had
been sick, as well as on current age. Precisely this model has been applied to UK insured
lives data. See CMIR (1991).
As noted at the end of Chapter 7, there are more sophisticated ways of solving systems of
diﬀerential equations than Euler’s method. Waters and Wilkie (1988) present a method
speciﬁcally designed for use with multiple state models. For a discussion on how to use
mathematical software to tackle the problems discussed in this chapter see Dickson (2006). 8.12. EXERCISES 8.12 407 Exercises Exercise 8.1 Consider the accidental death model illustrated in Figure 8.2. Let
µ01 = A + Bcx
x and µ02 = 10−5 for all x
x and assume A = 5 × 10−4 , B = 7.6 × 10−5 and c = 1.09.
(a) Calculate
(i) 00
10 p30 , (ii) 01
10 p30 , (iii) 02
10 p30 . and (b) An insurance company uses the model to calculate premiums for a special 10 year
term life insurance policy. The basic sum insured is $100 000, but the death beneﬁt
doubles to $200 000 if death occurs as a result of an accident. The death beneﬁt is
payable immediately on death. Premiums are payable continuously throughout the
term. Using an eﬀective rate of interest of 5% per year and ignoring expenses, for a
policy issued to a life aged 30
(i) calculate the annual premium for this policy, and
(ii) calculate the policy value at time 5. 408 CHAPTER 8. MULTIPLE STATE MODELS Exercise 8.2 Consider the following model for an insurance policy combining disability
income insurance beneﬁts and critical illness beneﬁts. Healthy
0 E Sick
1 '
d d d d ©
d
c ' Dead
2 c Critically ill
3 The transition intensities are as follows:
µ01 = a1 + b1 exp{c1 x},
x
µ12 = µ02 ,
x
x
µ10 = 0.1µ01 ,
x
x µ02 = a2 + b2 exp{c2 x},
x µ32 = 1.2µ02 ,
x
x
µ03 = 0.05µ01 ,
x
x µ13 = µ03
x
x where
a1 = 4 × 10−4 , b1 = 3.5 × 10−6 , c1 = 0.14,
a2 = 5 × 10−4 , b2 = 7.6 × 10−5 , c2 = 0.09.
(a) Using Euler’s method with a step size of
12
0, 12 , 12 , . . . , 35. 1
,
12 calculate values of t p00 for t =
30 (b) An insurance company issues a policy with term 35 years to a life aged 30 which 8.12. EXERCISES 409 provides a death beneﬁt, a disability income beneﬁt, and a critical illness beneﬁt as
follows:
• a lump sum payment of $100 000 is payable immediately on the life becoming
critically ill,
• a lump sum payment of $100 000 is payable immediately on death, provided
that the life has not already been paid a critical illness beneﬁt,
• a disability income annuity of $75 000 per year payable whilst the life is disabled.
Premiums are payable monthly in advance provided that the policyholder is healthy.
(i) Calculate the monthly premium for this policy on the following basis:
Transition intensities: as in (a)
Interest: 5% per year eﬀective
Expenses: Nil
Use numerical integration with the repeated Simpson’s rule with h = 1
.
12 (ii) Suppose that premium is payable continuously rather than monthly. Use
Thiele’s diﬀerential equation to solve for the total premium per year, using
1
Euler’s method with a step size of h = 12 .
(iii) Using your answer to part (ii), ﬁnd the policy value at time 10 for a healthy
life. Exercise 8.3 In Section 8.7.2 Thiele’s diﬀerential equation for a general multiple state
model was stated as
n d
(i)
(i)
= δt t V (i) − Bt −
tV
dt
j =0, (ij ) µij+t St
x
j =i + t V (j ) − t V (i) . 410 CHAPTER 8. MULTIPLE STATE MODELS tV (i) t
δ ds}.
0s Explain why n (a) Let v (t) = exp{− v (t + s)
(ij )
St+s + t+s V (j )
v (t) ∞ =
j =0, j =i
∞ +
0 0 ii
s px+t µij+t+s ds
x v (t + s) (i) ii
Bt+s s px+t ds.
v (t) (b) Using the techniques introduced in Section 7.5.1, diﬀerentiate the above expression
to obtain Thiele’s diﬀerential equation. Exercise 8.4 (a) Write down the Kolmogorov forward diﬀerential equation for t p00 in
x
the joint life model illustrated in Figure 8.9.
(b) Using (a), or otherwise, prove that
axy =
¯ ¯
1 − Axy
.
δ Exercise 8.5 Figure 8.12 illustrates the common shock model. This is the joint life and
last survivor model, adjusted to allow for the possibility that the husband and wife die at
the same time (for example as the result of a car accident).
An insurance company issues a joint life insurance policy to a married couple. The
husband is aged 28 and his wife is aged 27. The policy provides a beneﬁt of $500 000
immediately on the death of the husband provided that he dies ﬁrst. The policy terms
stipulate that if the couple die at the same time, the elder life is deemed to have died
ﬁrst. Premiums are payable annually in advance while both lives are alive for at most 30
years. 8.12. EXERCISES 411 Husband Alive
Wife Alive
0 E
Husband Alive
Wife Dead
1
c
s c E Husband Dead
Wife Dead Husband Dead
Wife Alive 3 2 Figure 8.12: The common shock model.
Calculate the annual net premium using an eﬀective rate of interest of 5% per year and
transition intensities of
µ01 = A + Bcy ,
xy µ02 = A + Dcx ,
xy µ03 = 5 × 10−5 ,
xy where A = 0.0001, B = 0.0003, c = 1.075 and D = 0.00035. Exercise 8.6 In a double decrement model (i.e. the model depicted in Figure 8.6 with
n = 2), let µ01 = µ and µ02 = θ for 0 ≤ x ≤ 1.
x
x
(a) Find expressions for 1 p00 , 1 p01 and 1 p02 .
x
x
x
(b) Let θ = nµ. Deduce that
1
(1 − 1 p00 )
x
n+1
and explain this result by general reasoning.
01
1 px = 412 CHAPTER 8. MULTIPLE STATE MODELS Exercise 8.7 Consider the insurance–with–lapses model illustrated in Figure 8.7. Suppose that this model is adjusted to include death after withdrawal, i.e. the transition
intensity µ21 is introduced into the model.
x
(a) Show that if withdrawal does not aﬀect the transition intensity to state 1 (i.e. that
µ21 = µ01 ), then the probability that an individual aged x is dead by age x + t is
x
x
the same as that under the ‘alive–dead’ model with the transition intensity µ01 .
x
(b) Why is this intuitively obvious? Exercise 8.8 An insurer prices critical illness insurance policies on the basis of a double
decrement model, in which there are two modes of decrement – death (state 1) and
becoming critically ill (state 2). For all x ≥ 0, µ01 = A + Bcx where A = 0.0001,
x
B = 0.00035 and c = 1.075, and µ02 = 0.05µ01 . On the basis of interest at 4% per
x
x
year eﬀective, calculate the monthly premium, payable for at most 20 years, for a life
aged exactly 30 at the issue date of a policy which provides $50 000 immediately on
death, provided that the critical illness beneﬁt has not already been paid, and $75 000
immediately on becoming critically ill, should either event occur within 20 years of the
policy’s issue date. Ignore expenses. Exercise 8.9 In a certain country, members of its regular armed forces can leave active
service (state 0) by transfer (state 1), by resignation (state 2) or by death (state 3). The
transition intensities are
µ01 = 0.001x,
x
µ02 = 0.01,
x 8.12. EXERCISES 413 µ03 = A + Bcx ,
x
where A = 0.001, B = 0.0004 and c = 1.07. New recruits join only at exact age 25.
(a) Calculate the probability that a new recruit
(i) is transferred before age 27,
(ii) dies aged 27 last bithday, and
(iii) is in active service at age 28.
(b) New recruits who are transferred within three years of joining receive a lump sum
payment of $10 000 immediately on transfer. This sum is provided by a levy on all
recruits in active service on the ﬁrst and second anniversary of joining. On the basis
of interest at 6% per year eﬀective, calculate the levy payable by a new recruit.
(c) Those who are transferred enter an elite force. Members of this elite force are
subject to a force of mortality at age x equal to 1.5µ03 , but are subject to no other
x
decrements. Calculate the probability that a new recruit into the regular armed
forces dies before age 28 as a member of the elite force. Exercise 8.10 Two lives aged 30 and 40 are independent with respect to mortality, and
each is subject to Makeham’s law of mortality with A = 0.0001, B = 0.0003 and c = 1.075.
Calculate
(a) 10 p30:40 , (b) 1
10 q30:40 , (c) 2
10 q30:40 , and 414
(d) CHAPTER 8. MULTIPLE STATE MODELS
10 p30:40 . Exercise 8.11 Two independent lives, (x) and (y ), experience mortality according to
Gompertz’ law, that is, µx = Bcx .
(a) Show that t pxy = t pw for w = log(cx + cy )/ log c.
(b) Show that
Ax
1 y cx
= w Aw
c Exercise 8.12 Smith and Jones are both aged exactly 30. Smith is subject to Gompertz’
law of mortality with B = 0.0003 and c = 1.07, and Jones is subject to a force of mortality
at all ages x of Bcx + 0.039221. Calculate the probability that Jones dies before reaching
age 50 and before Smith dies. Assume that Smith and Jones are independent with respect
to mortality. Exercise 8.13 Two lives aged 25 and 30 are independent with respect to mortality, and
each is subject to Makeham’s law of mortality with A = 0.0001, B = 0.0003 and c = 1.075.
Using an eﬀective rate of interest of 5% per year, calculate
(a) a25:30 ,
¨
(b) a25:30 ,
¨
(c) a2530 ,
¨ 8.12. EXERCISES 415 ¯
(d) A25:30 ,
¯1
(e) A25:30:10 , and
¯2
(f) A25:30 . Exercise 8.14 Bob and Mike are independent lives, both aged 25. They eﬀect an insurance policy which provides $100 000, payable at the end of the year of Bob’s death,
provided Bob dies after Mike. Annual premiums are payable in advance throughout Bob’s
lifetime. Calculate
(a) the net annual premium, and
(b) the net premium policy value after 10 years (before the premium then due is payable)
if
(i) only Bob is then alive, and
(ii) both lives are then alive.
Basis:
Mortality: Gompertz’ law, with B = 0.0003 and c = 1.075 for both lives
Interest: 5% per year eﬀective
Expenses: None Exercise 8.15 Ryan is entitled to an annuity of $100 000 per year at retirement, paid
monthly in advance, and the normal retirement age is 65. Ryan’s wife, Lindsay, is two
years younger than Ryan. 416 CHAPTER 8. MULTIPLE STATE MODELS (a) Calculate the EPV of the annuity at Ryan’s retirement date.
(b) Calculate the revised annual amount of the annuity (payable in the the ﬁrst year)
if Ryan chooses to take a beneﬁt which provides Lindsay with a monthly annuity
following Ryan’s death equal to 60% of the amount payable whilst both Ryan and
Lindsay are alive.
(c) Calculate the revised annual amount of the annuity (payable in the the ﬁrst year) if
Ryan chooses the beneﬁt in part (b), with a ‘popup’ – that is, the annuity reverts
to the full $100 000 on the death of Lindsay if Ryan is still alive. (Note that under a
‘popup’, the beneﬁt reverts to the amount to which Ryan was originally entitled.)
Basis:
Male mortality before and after widowerhood:
Makeham’s law, A = 0.0001, B = 0.0004 and c = 1.075
Female mortality before widowhood:
Makeham’s law, A = 0.0001, B = 0.00025 and c = 1.07
Female mortality after widowhood:
Makeham’s law, A = 0.0001, B = 0.0003 and c = 1.072
Interest:
5% per year eﬀective Exercise 8.16 A man and his wife are aged 28 and 24 respectively. They are about to
eﬀect an insurance policy that pays $100 000 immediately on the ﬁrst death. Calculate
the premium for this policy, payable monthly in advance as long as both are alive and
limited to 25 years, on the following basis:
Male mortality: Makeham’s law, with A = 0.0001, B = 0.0004 and c = 1.075
Female mortality: Makeham’s law, with A = 0.0001, B = 0.0003 and c = 1.07 8.12. EXERCISES 417 Interest: 5% per year eﬀective
Initial expenses: $250
Renewal expenses: 3% of each premium
Assume that this couple are independent with respect to mortality. Exercise 8.17 Let Axy denote the EPV of a beneﬁt of 1 payable at the end of the year
(m)
in which the ﬁrst death of (x) and (y ) occurs, and let Axy denote the EPV of a beneﬁt of
1
1 payable at the end of the m th of a year in which the ﬁrst death of (x) and (y ) occurs.
(a) As an EPV, what does
m v t/m
t=1 (t−1)/m pxy − t/m pxy represent?
(m) (b) Write down an expression for Axy in summation form by considering the insurance
beneﬁt as comprising a series of deferred one year term insurances with the beneﬁt
1
payable at the end of the m th of a year in which the ﬁrst death of (x) and (y )
occurs.
(c) Assume that two lives (x) and (y ) are independent with respect to mortality. Show
that under the UDD assumption,
(t−1)/m pxy − t/m pxy = 1
m − 2t + 1
(1 − pxy ) +
q x qy
m
m and that
m v t/m
t=1 (t−1)/m pxy − t/m pxy = (1 − pxy ) iv
i(m) m v t/m + qx qy
t=1 m − 2t + 1
.
m2 418 CHAPTER 8. MULTIPLE STATE MODELS (d) Deduce that under the assumptions in part (c),
A(m) ≈
xy Exercise 8.18 i
i(m) Axy . (a) Show that dt
v t px t py = −δ v t t px t py − v t t px t py (µx+t:y+t ) .
dt
(b) Use Woolhouse’s formula to show that,
a(m) ≈ axy −
¨xy
¨ m − 1 m2 − 1
−
(δ + µx:y )
2m
12m2 Exercise 8.19 Consider a husband (x) and wife (y ). Let Tx and Ty denote their respective future lifetimes.
Let Txy denote the time to failure of the joint life status, xy , and let Txy denote the time
to failure of the last survivor status, xy .
(a) Write down expressions for Txy and Txy in terms of the state process {Y (t)}t≥0 , as
deﬁned in the joint life and last survivor model in Figure 8.9.
(b) Show that
Txy + Txy = Tx + Ty .
(c) The force of mortality associated with the joint life status is µxy , deﬁned in formula
(8.24). Show that
¯
Axy = E v Txy . 8.12. EXERCISES 419 (d) For independent lives (x) and (y ), show that
¯
¯
Cov v Txy , v Txy = Ax − Axy ¯
¯
Ay − Axy . Exercise 8.20 A husband and wife, aged 65 and 60 respectively, purchase an insurance
policy, under which the beneﬁts payable on ﬁrst death are a lump sum of $10 000, payable
immediately on death, plus an annuity of $5 000 per year payable continuously throughout
the lifetime of the surviving spouse. A beneﬁt of $1 000 is paid immediately on the second
death. Premiums are payable continuously until the ﬁrst death.
¯
¯
¯
You are given that A60 = 0.353789, A65 = 0.473229 and that A60:65 = 0.512589 at 4% per
year eﬀective rate of interest. The lives are assumed to be independent.
(a) Calculate the EPV of the lump sum death beneﬁts, at 4% per year interest.
(b) Calculate the EPV of the reversionary annuity beneﬁt, at 4% per year interest.
(c) Calculate the annual rate of premium, at 4% per year interest.
(d) Ten years after the contract is issued the insurer is calculating the policy value.
(i) Write down an expression for the policy value at that time assuming that both
lives are still surviving.
(ii) Write down an expression for the policy value assuming that the husband has
died but the wife is still alive.
(iii) Write down Thiele’s diﬀerential equation for the policy value assuming (1) both
lives are still alive, and (2) only the wife is alive. 420 CHAPTER 8. MULTIPLE STATE MODELS Exercise 8.21 Consider Example 8.12, and suppose that
µx = A + Bcx and µ02 = 0.02, where A = 0.0001, B = 0.0004 and c = 1.07.
A corporation contributes $10 000 to a pension fund when an employee joins the corporation and on each anniversary of that person joining the corporation, provided the person
is still an employee. On the basis of interest at 5% per year eﬀective, calculate the EPV
of contributions to the pension fund in respect of a new employee aged 30. Exercise 8.22 A university oﬀers a four year degree course. Semesters are half a year
in length and the probability that a student progresses from one semester of study to the
next is 0.85 in the ﬁrst year of study, 0.9 in the second year, 0.95 in the third, and 0.98
in the fourth. All students entering the ﬁnal semester obtain a degree. Students who fail
in any semester may not continue in the degree.
Students pay tuition fees at the start of each semester. For the ﬁrst semester the tuition
fee is $10 000. Allowing for an increase in fees of 2% each semester, and assuming interest
at 5% per year eﬀective, calculate the EPV of fee income to the university for a new
student aged 19. Assume that the student is subject to a constant force of mortality
between integer ages x and x + 1 of 5x × 10−5 for x = 19, 20, 21 and 22, and that there
are no means of leaving the course other than by death or failure. Exercise 8.23 An insurance company sells 10 year term insurance policies with sum
insured $100 000 payable immediately on death to lives aged 50. Calculate the monthly
premium for this policy on the following basis.
Mortality: Makeham’s law, with A = 0.0001, B = 0.0004 and c = 1.075 8.12. EXERCISES 421 Lapses: 2% of policyholders lapse their policy on each of the ﬁrst two policy
anniversaries
Interest: 5% per year eﬀective
Initial expenses: $200
Renewal expenses: 2.5% of each premium (including the ﬁrst)
Value the death beneﬁt using the UDD assumption. 422 CHAPTER 8. MULTIPLE STATE MODELS Answers to selected exercises
8.1 (a) (i) 0.979122
(ii) 0.020779
(iii) 0.000099
(b) (i) $206.28
(ii) $167.15
8.2 (a) 00
35 p30 = 0.581884 (b) (i) $206.56
(ii) $2 498.07
(iii) $16 925.88
8.5 $4 948.24
8.8 $28.01
8.9 (a) (i) 0.050002
(ii) 0.003234
(iii) 0.887168
(b) $397.24
(c) 0.000586
8.10 (a) 0.886962
(b) 0.037257
(c) 0.001505
(d) 0.997005
8.12 0.567376 8.12. EXERCISES
8.13 (a) 15.8901
(b) 18.9670
(c) 1.2013
(d) 0.2493
(e) 0.0208
(f) 0.0440
8.14 (a) $243.16
(b) (i) $18 269.42
(ii) $2 817.95
8.15 (a) $802 639
(b) $76 837
(c) $73 933
8.16 $161.78
8.20 (a) $5 440.32
(b) $26 262.16
(c) $2 470.55
8.21 $125 489.33
8.22 $53 285.18
8.23 $225.95 423 424 CHAPTER 8. MULTIPLE STATE MODELS Chapter 9
Pension Mathematics
9.1 Summary In this chapter we introduce some of the notation and concepts of pension plan valuation
and funding. We discuss the diﬀerence between deﬁned beneﬁt (DB) and deﬁned contribution (DC) pension plans. We introduce the salary scale function, and show how to
calculate an appropriate contribution rate in a DC plan to meet a target level of pension
income.
We then deﬁne the service table, which is a summary of the multiple state model appropriate for a pension plan. Using the service table and the salary scale, we can value the
beneﬁts and contributions of a pension plan, using the same principles as we have used
for valuing beneﬁts under an insurance policy. 9.2 Introduction The pension plans we discuss in this chapter are typically employer sponsored plans,
designed to provide employees with retirement income.
425 426 CHAPTER 9. PENSION MATHEMATICS Employers sponsor plans for a number of reasons, including
• Competition for new employees;
• To facilitate turnover of older employees by ensuring that they can aﬀord to retire;
• To provide incentive for employees to stay with the employer;
• Pressure from trade unions;
• To provide a tax eﬃcient method of remunerating employees;
• Responsibility to employees who have contributed to the success of the company.
The plan design will depend on which of these motivations is most important to the
sponsor. If competition for new employees is the most important factor, for example,
then the employer’s plan will closely resemble other employer sponsored plans within the
same industry. Ensuring turnover of older employees, or rewarding longer service might
lead to a diﬀerent beneﬁt design.
The two major categories of employer sponsored pension plans are deﬁned contribution
(DC) and deﬁned beneﬁt (DB).
The deﬁned contribution pension plan speciﬁes how much the employer will contribute, as
a percentage of salary, into a plan. The employee may also contribute, and the employer’s
contribution may be related to the employee’s contribution (for example, the employer
may agree to match the employee’s contribution up to some maximum). The contributions
are accumulated in a notional account, which is available to the employee when he or she
leaves the company. The contributions may be set to meet a target beneﬁt level, but
the actual retirement income may be well below or above the target, depending on the
investment experience.
The deﬁned beneﬁt plan speciﬁes a level of beneﬁt, usually in relation to salary near
retirement (ﬁnal salary plans), or to salary throughout employment (career average salary 9.3. THE SALARY SCALE FUNCTION 427 plans). The contributions, from the employer and, possibly, employee are accumulated to
meet the beneﬁt. If the investment or demographic experience is adverse, the contributions
can be increased; if experience is favourable, the contributions may be reduced. The
pension plan actuary monitors the plan funding on a regular basis to assess whether the
contributions need to be changed.
The beneﬁt under a DB plan, and the target under a DC plan, are set by consideration
of an appropriate replacement ratio. The pension plan replacement ratio is deﬁned as
R= pension income in the year after retirement
salary in the year before retirement where we assume the plan member survives the year following retirement. The target
for the plan replacement ratio depends on other post retirement income, such as government beneﬁts. A total replacement ratio, including government beneﬁts and personal
savings, of around 70% is often assumed to allow retirees to maintain their preretirement
lifestyle. Employer sponsored plans often target 50% to 70% as the replacement ratio for
an employee with a full career in the company. 9.3 The salary scale function The contributions and the beneﬁts for most employer sponsored pension plans are related
to salaries, so we need to model the progression of salaries through an individual’s employment. We use a deterministic model based on a salary scale, {sy }y≥x0 , where x0 is some
suitable initial age. The value of sx0 can be set arbitrarily, and then for any x, y (≥ x0 )
we deﬁne
sy
salary received in year of age y to y + 1
=
sx
salary received in year of age x to x + 1
where we assume the individual remains in employment throughout the periods x to x + 1
and y to y + 1. The salary scale may be deﬁned as a continuous function of age, or may
be summarized in a table of integer age values. Future changes in salary cannot usually 428 CHAPTER 9. PENSION MATHEMATICS be predicted with the certainty a deterministic salary scale implies. However, this model
is almost universally used in practice and a more realistic model would complicate the
presentation in this chapter.
Salaries usually increase as a result of promotional increases and inﬂation adjustments.
We assume in general that the salary scale allows for both forces, but it is straightforward
to manage these separately.
Example 9.1 The ﬁnal average salary for the pension beneﬁt provided by a pension plan
is deﬁned as the average salary in the three years before retirement. Members’ salaries
are increased each year 6 months before the valuation date.
(a) A member aged exactly 35 at the valuation date received $75 000 in salary in the
year to the valuation date. Calculate his predicted ﬁnal average salary assuming
retirement at age 65.
(b) A member aged exactly 55 at the valuation date was paid salary at a rate of $100 000
at that time. Calculate her predicted ﬁnal average salary assuming retirement at
age 65.
Assume
(i) a salary scale where sy = 1.04y , and
(ii) the integer age salary scale in Table 9.1. Solution 9.1 (a) The member is aged 35 at the valuation date, so that the salary in
the previous year is the salary from age 34 to age 35. The predicted ﬁnal average
salary in the three years to age 65 is then
s62 + s63 + s64
75 000
3 s34
which gives $234 019 under assumption (i) and $201 067 under assumption (ii). 9.3. THE SALARY SCALE FUNCTION
x
30
31
32
33
34
35
36
37
38
39 sx
1.000
1.082
1.169
1.260
1.359
1.461
1.566
1.674
1.783
1.894 x
40
41
42
43
44
45
46
47
48
49 sx
2.005
2.115
2.225
2.333
2.438
2.539
2.637
2.730
2.816
2.897 429
x
50
51
52
53
54
55
56
57
58
59 sx
2.970
3.035
3.091
3.139
3.186
3.234
3.282
3.332
3.382
3.432 x
60
61
62
63
64 sx
3.484
3.536
3.589
3.643
3.698 Table 9.1: Salary scale for Example 9.1. (b) The current annual salary rate of $100 000 is the salary which will be earned in the
year of age 54.5 to 55.5, so the ﬁnal average salary is
100 000 s62 + s63 + s64
3 s54.5 Under assumption (i) this is $139 639. Under assumption (ii) we need to estimate
s54.5 , which we would normally do using linear interpolation, so that
s54.5 = (s54 + s55 )/2 = 3.210,
giving a ﬁnal average salary of $113 499. Example 9.2 The current annual salary rate of an employee aged exactly 40 is $50 000.
Salaries are revised continuously. Using the salary scale {sy }, where sy = 1.03y , estimate
(a) the employee’s salary between ages 50 and 51, and 430 CHAPTER 9. PENSION MATHEMATICS (b) the employee’s annual rate of salary at age 51.
In both cases, you should assume the employee remains in employment until at least age
51.
Solution 9.2 The salary scale, as deﬁned above, relates to earnings over years of age.
The information we are given in this example is that the employee’s current annual rate of
salary is $50 000 and that salaries are increased continuously. This is a common situation
in practice. We make the reasonable assumption that the current annual rate of salary is
approximately the earnings between ages 39.5 and 40.5 (assuming the employee remains
in employment until at least age 40.5).
(a) Given our assumption, the estimated earnings between ages 50 and 51 are given by
50 000 × s50
= 50 000 × 1.0310.5 = $68 196.
s39.5 (b) We assume that the annual rate of salary at age 51 is approximately the earnings
between ages 50.5 and 51.5. This is consistent with the assumption above. Hence,
the estimated salary rate at age 51 is given by
50 000 × 9.4 s50.5
= 50 000 × 1.0311 = $69 212.
s39.5 Setting the DC contribution To set the contribution rate for a DC plan to aim to meet a target replacement ratio for
a ‘model’ employee, we need
• the target replacement ratio and retirement age, 9.4. SETTING THE DC CONTRIBUTION 431 • assumptions on the rate of return on investments, interest rates at retirement, a
salary scale and a model for postretirement mortality, and
• the form the beneﬁts should take.
With this information we can set a contribution rate that will be adequate if experience
follows all the assumptions. We might also want to explore sensitivity to the assumptions,
to assess a possible range of outcomes for the plan member’s retirement income. The
following example illustrates these points.
Example 9.3 An employer establishes a DC pension plan. On withdrawal from the plan
before retirement age, 65, for any reason, the proceeds of the invested contributions are
paid to the employee or the employee’s survivors.
The contribution rate is set using the following assumptions.
• The employee will use the proceeds at retirement to purchase a pension for his
lifetime, plus a reversionary annuity for his wife at 60% of the employee’s pension.
• At age 65, the employee is married, and the age of his wife is 61.
• The target replacement ratio is 65%.
• The salary scale is given by sy = 1.04y and salaries are assumed to increase continuously.
• Contributions are payable monthly in arrear at a ﬁxed percentage of the salary rate
at that time.
• Contributions are assumed to earn investment returns of 10% per year.
• Annuities purchased at retirement are priced assuming an interest rate of 5.5% per
year. 432 CHAPTER 9. PENSION MATHEMATICS • Male mortality: Makeham’s law, with A = 0.0004, B = 4 × 10−6 , c = 1.13.
• Female mortality: Makeham’s law, with A = 0.0002, B = 10−6 , c = 1.135.
• Members and their spouses are independent with respect to mortality.
Consider a male new entrant aged 25.
(a) Calculate the contribution rate required to meet the target replacement ratio for
this member.
(b) Assume now that the contribution rate will be 5.5% of salary, and that over the
member’s career, his salary will actually increase by 5% per year, investment returns
will be only 8% per year and the interest rate for calculating annuity values at
retirement will be 4.5% per year. Calculate the actual replacement ratio for the
member.
Solution 9.3 (a) First, we calculate the accumulated DC fund at retirement. Mortality
is not relevant here, as in the event of the member’s death, the fund is paid out
anyway; the DC fund is more like a bank account than an insurance policy.
We then equate the accumulated fund with the expected present value at retirement
of the pension beneﬁts.
Suppose the initial salary rate is $S . As everything is described in proportion
to salary, the amount assumed does not matter. Then the annual salary rate at
age x > 25 is S (1.04)x−25 , which means that the contribution at time t, where
t = 1/12, 2/12, ..., 40, is
c
S (1.04t )
12
where c is the contribution rate per year. Hence, the accumulated amount of con 9.4. SETTING THE DC CONTRIBUTION 433 tributions at retirement is
cS
12 480 40 40 k
k
1.1 − 1.04 = 719.6316cS.
1.04 12 1.140− 12 = cS 1
1.1 12
12 1.04
−1
k=1 The salary received in the year prior to retirement, under the assumptions, is
s64
S = 1.0439.5 S = 4.7078S.
s24.5
Since the target replacement ratio is 65%, the target pension beneﬁt per year is
0.65 × 4.7078S = 3.0601S.
The EPV at retirement of a beneﬁt of 3.0601S per year to the member, plus a
reversionary beneﬁt of 0.6 × 3.0601S per year to his wife, is
(12) (12) 65 6561 3.0601S ( a m + 0.6 a m f )
¨
¨
where the m and f scripts indicate male and female mortality, respectively.
Using the given survival models and an interest rate of 5.5% per year, we have
(12) am
¨ 65 (12) am
¨ f = 10.5222,
(12) = af
¨ (12) −am
¨ 6561 61 af
¨ (12) = , = 13.9194, (12) f 65:61 61 am
¨ f 65:61 ∞
k=0 1k
v 12
12 = 10.0066, k
12 p m 65+k k
12 p f 61+k (9.1) 434 CHAPTER 9. PENSION MATHEMATICS
giving
(12) am
¨ f = 3.9128. 6561 Note that we can write the joint life survival probability in formula (9.1) as the
product of the single life survival probabilities using the independence assumption,
as in Section 8.9.4.
Hence, the value of the beneﬁt at retirement is
3.0601S (10.5222 + 0.6 × 3.9128) = 39.3826S.
Equating the accumulation of contributions to age 65 with the EPV of the beneﬁts
at age 65 gives
c = 5.4726% per year.
(b) We now repeat the calculation, using the actual experience rather than estimates.
We use an annual contribution rate of 5.5%, and solve for the amount of beneﬁt
funded by the accumulated contributions, as a proportion of the ﬁnal year’s salary.
The accumulated contributions at age 65 are now 28.6360S , and the annuity values
at 4.5% per year interest are
(12) a m = 11.3576,
¨
65 (12) af
¨ 61 = 15.4730, (12) am
¨ f = 10.7579. 65:61 Thus, the EPV of a beneﬁt of X per year to the member and of 0.6 X reversionary
beneﬁt to his spouse is 14.1867X . Equating the accumulation of contributions to
age 65 with the EPV of beneﬁts at age 65 gives X = 2.0185S .
The ﬁnal year salary, with 5% per year increases, is 6.8703S . Hence, the replacement
ratio is
2.0185S
R=
= 29.38%.
6.8703S 9.5. THE SERVICE TABLE 435 We note that apparently quite small diﬀerences between the assumptions used to set the
contribution and the experience can make a signiﬁcant diﬀerence to the level of beneﬁt,
in terms of the preretirement income. This is true for both DC and DB beneﬁts. In
the DC case, the risk is taken by the member, who takes a lower beneﬁt, relative to
salary, than the target. In the DB case, the risk is usually taken by the employer, whose
contributions are usually adjusted when the diﬀerence becomes apparent. If the diﬀerences
are in the opposite direction, then the member beneﬁts in the DC case, and the employer
contributions may be reduced in the DB case. 9.5 The service table The demographic elements of the basis for pension plan calculations include assumptions
about survival models for members and their spouses, and about the exit patterns from
employment.
There are several reasons why a member might exit the plan. At early ages, the employee
might withdraw to take another job with a diﬀerent employer. At later ages, employees
may be oﬀered a range of ages at which they may retire with the pension that they have
accumulated. A small proportion of employees will die while in employment, and another
group may leave early through disability retirement.
In a DC plan, the beneﬁt on exit is the same, regardless of the reason for the exit, so
there is no need to model the member employment patterns.
In a DB plan diﬀerent beneﬁts may be payable on the diﬀerent forms of exit. In the UK
it is common on the death in service of a member for the pension plan to oﬀer both a
lump sum and a pension beneﬁt for the member’s surviving spouse. In North America,
any lump sum beneﬁt is more commonly funded through separate group life insurance,
and so the liability does not fall on the plan. There may be a contingent spouse’s beneﬁt.
The extent to which the DB plan actuary needs to model the diﬀerent exits depends on 436 CHAPTER 9. PENSION MATHEMATICS how diﬀerent the values of beneﬁts are from the values of beneﬁts for people who do not
leave until the normal retirement age.
For example, if an employer oﬀers a generous beneﬁt on disability (or ill health) retirement,
that is worth substantially more than the beneﬁt that the employee would have been
entitled to if they had remained in good health, then it is necessary to model that exit
and to value that beneﬁt explicitly. Otherwise, the liability will be understated. On the
other hand, if there is no beneﬁt on death in service (for example, because of a separate
group life arrangement), then to ignore mortality before retirement would overstate the
liabilities within the pension plan.
If all the exit beneﬁts have roughly the same value as the normal age retirement beneﬁt,
the actuary may assume that all employees survive to retirement. It is not a realistic
assumption, but it simpliﬁes the calculation and is appropriate if it does not signiﬁcantly
over or under estimate the liabilities.
It is relatively common to ignore withdrawals in the basis, even if a large proportion of
employees do withdraw, especially at younger ages. It is reasonable to ignore withdrawals
if the eﬀect on the valuation of beneﬁts is small, compared with allowing explicitly for
withdrawals. By ignoring withdrawals, we are implicitly assuming (loosely) that the
lives who withdraw instead take age retirement beneﬁts; this is reasonable if the age
retirement beneﬁts have similar value to the withdrawal beneﬁts, which is often the case.
For example, in a ﬁnal salary plan, if withdrawal beneﬁts are increased in line with
inﬂation, the value of withdrawal and age beneﬁts will be similar. Even if the diﬀerence is
relatively large, withdrawals may be ignored to provide an implicit margin in the valuation
if withdrawal beneﬁts are generally less expensive than retirement beneﬁts, which is often
the case. An additional consideration is that withdrawals are notoriously unpredictable,
as they are strongly aﬀected by economic and social factors, so that historical trends may
not provide a good indicator of future exit patterns.
When the actuary does model the exits from a plan, an appropriate multiple decrement
model could be similar to the one shown in Figure 9.1. All the model assumptions 9.5. THE SERVICE TABLE 437 Member
0
~ ©
A
z Disability
Retirement
2 Withdrawn
1 Age
Retirement
3 Died in Service
4 Figure 9.1: A multiple decrement model for a pension plan. of Chapter 8 apply to this model, except that some age retirements will be exact age
retirements, as discussed in Section 8.10.
Example 9.4 A pension plan member is entitled to a lump sum beneﬁt on death in
service of four times the salary paid in the year up to death.
Assume the appropriate multiple decrement model is as in Figure 9.1, with 0.1 for x < 35, 0.05 for 35 ≤ x < 45,
01
w
µx ≡ µx = 0.02 for 45 ≤ x < 60, 0
for x ≥ 60,
µ02 ≡ µi = 0.001,
x
x
µ03 ≡ µr =
x
x 0
for x < 60,
0.1 for 60 < x < 65. In addition, 30% of the members surviving in employment to age 60 retire at that time,
and 100% of the lives surviving in employment to age 65 retire at that time. For transitions
to state 4,
µ04 ≡ µd = A + Bcx ; with A = 0.00022, B = 2.7 × 10−6 , c = 1.124.
x
x 438 CHAPTER 9. PENSION MATHEMATICS (This is the standard ultimate survival model from Section 4.3.)
(a) For a member aged 35, calculate the probability of retiring at age 65.
(b) For each mode of exit, calculate the probability that a member currently aged 35
exits employment by that mode.
Solution 9.4 (a) Since all surviving members retire at age 65, the probability can be
written 30 p00 . To calculate this, we need to consider separately the periods before
35
and after the jump in the withdrawal transition intensity, and before and after the
exact age retirements at age 60.
For 0 < t < 10,
t
00
t p35 = exp − 0 µw s + µi s + µd s ds
35+
35+
35+ = exp − (A + 0.05 + 0.001)t + B 35 t
c (c − 1)
log c , giving
00
10 p35 = 0.597342. For 10 ≤ t < 25,
00
t p35 = = t−10 00
10 p35 exp − 00
10 p35 exp − (A + 0.02 + 0.001)(t − 10) + 0 µw s + µi s + µd s ds
45+
45+
45+ giving
25− p35 = 0.597342 × 0.712105 = 0.425370. B 45 t−10
c (c
− 1)
log c 9.5. THE SERVICE TABLE 439 At t = 25, 30% of the survivors retire, so at t = 25+ we have
00
25+ p35 = 0.7 25− p00 = 0.297759.
35 For 25 < t < 30,
00
t p35 = 00
25+ p35 t−25 exp − 0 µr s + µi s + µd s ds
60+
60+
60+ = 0.297759 exp − (A + 0.1 + 0.001)(t − 25) + B 60 t−25
c (c
− 1)
log c giving
00
30− p35 = 0.297759 × 0.590675 = 0.175879. The probability of retirement at exact age 65 is then 0.1759.
(b) We know that all members leave employment by or at age 65.
All withdrawals occur by age 60. To compute the probability of withdrawal, we
split the period into before and after the change in the withdrawal force at age 45.
The probability of withdrawal by age 45 is
10
01
10 p35 =
0 10
00
t p35 µw t
35+ dt = 0.05
0 00
t p35 dt which we can calculate using numerical integration to give
01
10 p35 = 0.05 × 7.8168 = 0.3908. The probability of withdrawal between ages 45 and 60 is
10
00
01
10 p35 15 p45 = 0.597342
0 15
00
t p45 µw t dt = 0.597342 × 0.02
45+ 0 00
t p45 dt 440 CHAPTER 9. PENSION MATHEMATICS
which, again using numerical integration, gives
00
01
10 p35 15 p45 = 0.597342 × 0.02 × 12.7560 = 0.1524. So, the total probability of withdrawal is 0.5432.
We calculate the probability of disability retirement similarly. The probability of
disability retirement by age 45 is
10
02
10 p35 = 10
00 i
t p35 µ35+t dt = 0.001 0 0 00
t p35 dt = 0.001 × 7.8168 = 0.0078, and the probability of disability retirement by age 60 is
10
00
02
10 p35 15 p45 = 0.597342
0 15
00
t p45 µi t
45+ dt = 0.597342 × 0.001 0 00
t p45 dt = 0.597342 × 0.001 × 12.7560 = 0.0076.
The probability of disability retirement in the ﬁnal ﬁve years is
5
00
02
25+ p35 5 p60 = 0.297759
0 00
t p60 µi t dt = 0.297759 × 0.001 × 3.8911 = 0.0012.
60+ So, the total probability of disability retirement is 0.0078+0.0076+0.0012 = 0.0166.
The probability of age retirement is the sum of the probabilities of exact age retirements and retirements between ages 60 and 65.
The probability of exact age 60 retirement is
0.3 25− p35 = 0.1276, and the probability of exact age 65 retirement is
30− p35 = 0.1759. 9.5. THE SERVICE TABLE 441 The probability of retirement between exact ages 60 and 65 is
5
25+ p00 5 p03 = 0.297759
35
60 00
t p60 0 µr t dt = 0.297759 × 0.1 × 3.8911 = 0.1159.
60+ So, the total age retirement probability is 0.4194.
We could infer the death in service probability, by the law of total probability, but
we instead calculate it directly as a check on the other results. We use numerical
integration for all these calculations.
The probability of death in the ﬁrst 10 years is
10
04
10 p35 = 0 00
t p35 µd t dt = 0.0040,
35+ and the probability of death in the next 15 years is
15
00
04
10 p35 15 p45 00
t p45 = 0.59734
0 µd t dt = 0.0120.
45+ The probability of death in the ﬁnal 5 years is
5
25+ p00 5 p04 = 0.297759
35
60 0 00
t p60 µr t dt = 0.297759 × 0.016323 = 0.0049.
60+ So the total death in service probability is 0.0208.
We can check our calculations by summing the probabilities of exiting by each mode.
This gives a total of 1 (= 0.5432 + 0.0166 + 0.4194 + 0.0208), as it should. Often the multiple decrement model is summarized in tabular form at integer ages, in
the same way that a life table summarizes a survival model. Such a summary is called a
pension plan service table. We start at some minimum integer entry age, x0 , by deﬁning 442 CHAPTER 9. PENSION MATHEMATICS an arbitrary radix, for example, lx0 = 1 000 000. Using the model of Figure 9.1, we then
deﬁne for integer ages x0 + k (k = 0, 1, . . .)
wx0 +k = lx0 k p00 p01 +k
x0 x0
ix0 +k = lx0 k p00 p02 +k
x0 x0
rx0 +k = lx0 k p00 p03 +k
x0 x0
dx0 +k = lx0 k p00 p04 +k
x0 x0
lx0 +k = lx0 k p00
x0
Since the probability that a member aged x0 withdraws between ages x0 + k and x0 + k +1
is k p00 p01 +k , we can interpret wx0 +k as the number of members expected to withdraw
x0 x0
between ages x0 + k and x0 + k + 1 out of lx0 members aged exactly x0 ; ix0 +k , rx0 +k and
dx0 +k can be interpreted similarly. We can interpret lx0 +k as the expected number of lives
who are still plan members at age x0 + k out of lx0 members aged exactly x0 . We can
extend these interpretations to say that for any integer ages x and y (> x), wy is the
number of members expected to withdraw between ages y and y + 1 out of lx members
aged exactly x and ly is the expected number of members at age y out of lx members
aged exactly x. These interpretations are precisely in line with those for a life table – see
Section 3.2.
Note that, using the law of total probability, we have the following identity for any integer
age x
lx = lx−1 − wx−1 − ix−1 − rx−1 − dx−1 . (9.2) A service table summarizing the model in Example 9.4 is shown in Table 9.2 from age 20
with the radix l20 = 1 000 000. This service table has been constructed by calculating, for
each integer age x (> 20), wx , ix , rx and dx as described above. The value of lx shown 9.5. THE SERVICE TABLE 443 in the table is then calculated recursively from age 20. The table is internally consistent
in the sense that identity (9.2) holds for each row of the table. However, this does not
appear to be the case in Table 9.2 for the simple reason that all values have been rounded
to the nearer integer. The exact age exits at ages 60 and 65 are shown in the rows labeled
60− and 65− .
We remark that in all subsequent calculations based on Table 9.2, we use the exact values
rather than the rounded ones.
Having constructed a service table, the calculation of the probability of any event between
integer ages can be performed relatively simply. To see this, consider the calculations
required for Example 9.4. For part (a), the probability that a member aged 35 survives
in service to age 65, calculated using Table 9.2, is
l65 /l35 = 38 488/218 834 = 0.1759.
For part (b), the probability that a member aged 35
withdraws is
(10 665 + 10 131 + . . . + 1 930 + 1 884)/218 834 = 0.5432,
retires in ill health is
(213 + 203 + . . . + 45 + 41)/218 834 = 0.0166,
retires for age reasons is
(27 926 + 6 188 + 5 573 + 5 018 + 4 515 + 4 061 + 38 488)/218 834 = 0.4194,
dies in service is
(83 + 84 + . . . + 214 + 215)/218 834 = 0.0208. 444 CHAPTER 9. PENSION MATHEMATICS x
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43 lx
1 000 000
903 707
816 684
738 038
666 962
602 728
544 677
492 213
444 800
401 951
363 226
328 228
296 599
268 014
242 181
218 834
207 872
197 455
187 555
178 147
169 206
160 708
152 631
144 954 wx
ix rx
95 104 951 0
85 946 859 0
77 670 777 0
70 190 702 0
63 430 634 0
57 321 573 0
51 800 518 0
46 811 468 0
42 301 423 0
38 226 382 0
34 543 345 0
31 215 312 0
28 207 282 0
25 488 255 0
23 031 230 0
10 665 213 0
10 131 203 0
9 623 192 0
9 141 183 0
8 682 174 0
8 246 165 0
7 832 157 0
7 438 149 0
7 064 141 0 dx
x
237 44
218 45
200 46
184 47
170 48
157 49
145 50
134 51
125 52
117 53
109 54
102 55
96
56
91
57
86
58
83
59
84 60−
84 60+
85
61
86
62
87
63
89
64
90 65−
93 lx
137 656
130 719
127 904
125 140
122 428
119 763
117 145
114 572
112 042
109 553
107 102
104 688
102 308
99 960
97 642
95 351
93 085
65 160
58 700
52 860
47 579
42 805
38 488 wx
6 708
2 586
2 530
2 476
2 422
2 369
2 317
2 266
2 216
2 166
2 118
2 070
2 023
1 976
1 930
1 884
0
0
0
0
0
0
0 Table 9.2: Pension plan service table. ix
rx
134
0
129
0
127
0
124
0
121
0
118
0
116
0
113
0
111
0
108
0
106
0
103
0
101
0
99
0
96
0
94
0
0 27 926
62 6 188
56 5 573
50 5 018
45 4 515
41 4 061
0 38 488 dx
95
100
106
113
121
130
140
151
163
176
190
206
224
243
264
288
0
210
212
213
214
215
0 9.5. THE SERVICE TABLE 445 Example 9.5 Employees in a pension plan pay contributions of 6% of their previous
month’s salary at each month end. Calculate the EPV at entry of contributions for a new
entrant aged 35, with a starting salary rate of $100 000, using
(a) exact calculation using the multiple decrement model speciﬁed in Example 9.4, and
(b) the values in Table 9.2, adjusting EPV of an annuity payable annually in the same
way as under the UDD assumption in Chapter 5.
Other assumptions:
Salary Scale: Salaries increase at 4% per year continuously
Interest: 6% per year
Solution 9.5 (a) The EPV is 0.06 × 100 000
12
0.06 × 100 000
=
12 299 360
k k
12 k p00 (1.04) 12 v 12 +
35 25− k=1 k
12 00
k p35
12 k
12 vj + 360
25− p00
35 25
vj + k k
12 k p00 (1.04) 12 v 12
35 k=301 299 k=1 k p00 (1.04)25 v 25 +
35 p00 vj12
35 k=301 = 6 000 × 13.3529
= $80 117
where j = 0.02/1.04 = 0.01923 and where we have separated out the term relating
to age 60 to emphasize the point that contributions would be paid by all employees
reaching ages 60 and 65, even those who retire at those ages.
(b) Recall from Chapter 5 that the UDD approximation to the EPV of a term annuity
(12)
payable monthly in arrear ax:n , in terms of the corresponding value for annual 446 CHAPTER 9. PENSION MATHEMATICS
payments in advance, ax:n , is
¨
(12) ax:n ≈ α(12) ax:n − β (12) +
¨ 1
12 (1 − v n n px ). This approximation will work for the monthly multiple decrement annuity, which we
00(12)
will denote a x :n , provided that the decrements, in total, are approximately UDD.
This is not the case for our service table, because between ages 60− and 61, the
vast majority of decrements occur at exact age 60. We can take account of this by
splitting the annuity into two parts, up to age 60− and from age 60+ , and applying
a UDDstyle adjustment to each part as follows:
a 00(12)
35:30 =a 00(12)
35:25 + l60+ 25 00(12)
va
l35 j 60+ :5 ≈ α(12) a00 25 − β (12) +
¨ 35:
+ l60+ 25
v
l35 j 1
12 1− l60− 25
v
l35 j α(12) a00 + :5 − β (12) +
¨ 60 1
12 1− l65− 5
v
l60+ j As a00 25 = 13.0693 and a00+ :5 = 3.9631 we ﬁnd that
¨35:
¨60
00(12) 6 000 a35:30 ≈ $80 131. Using the service table and the UDDbased approximation has resulted in a relative error
of the order of 0.03% in this example. This demonstrates again that the service table
summarizes the underlying multiple decrement model suﬃciently accurately for practical
purposes.
In applying this adjustment we are eﬀectively saying that the arguments we applied to
deaths in Chapter 5 can be applied to total decrements. However, just as in Section 8.9.4, 9.6. VALUATION OF BENEFITS 447 if we were to assume a uniform distribution of decrements in each of the related single
decrement models, we would ﬁnd that there is not a uniform distribution of the overall
decrements. Nevertheless, the assumption of a uniform distribution of total decrements
provides a useful, and relatively accurate, means of calculating the EPV of an annuity
payable m times a year from a service table.
It is very common in pension plan valuation to use approximations, primarily because of
the long term nature of the liabilities and the huge uncertainty in the parameters of the
models used. To calculate values with great accuracy when there is so much uncertainty
involved would be spurious. While this argument is valid, one needs to ensure that the
approximation methods do not introduce potentially signiﬁcant biases in the ﬁnal results,
for example, by systematically underestimating the value of liabilities. 9.6 Valuation of beneﬁts 9.6.1 Final salary plans In a DB ﬁnal salary pension plan, the basic annual pension beneﬁt is equal to
n SFin α
where
n is the total number of years of service,
SFin is the average salary in a speciﬁed period before retirement; for example, in the
three years preceding exit, and
α is the accrual rate, typically between 0.01 and 0.02. For an employee who has
been a member of the plan all her working life, say n = 40 years, this typically gives
a replacement ratio in the range 40 – 80%. 448 CHAPTER 9. PENSION MATHEMATICS We interpret this beneﬁt formula to mean that the employee earns a pension of 100α% of
ﬁnal average salary for each year of employment.
Consider a member who is currently aged y , who joined the pension plan at age x (≤ y )
and for whom the normal retirement age is 60. Our estimate of her annual pension at
retirement is
ˆ
(60 − x) SFin α
ˆ
where SF in is the current estimate of SFin . This estimate is calculated using her current
salary and an appropriate salary scale. We can split this annual amount into two parts
ˆ
ˆ
ˆ
(60 − x) SFin α = (y − x) SFin α + (60 − y ) SFin α.
The ﬁrst part is related to her past service, and is called the accrued beneﬁt. The
second part is related to future service. Note that both parts use an estimate of the ﬁnal
ˆ
average salary at retirement, SFin .
The employer who sponsors the pension plan retains the right to stop oﬀering pension
beneﬁts in the future. If this were to happen, the ﬁnal beneﬁt would be based on the
member’s past service at the windup of the pension plan; in this sense, the accrued
beneﬁts (also known as the past service beneﬁts) are already secured. The future service
beneﬁts are more of a statement of intent, but do not have the contractual nature of the
accrued beneﬁts.
In valuing the plan liabilities then, modern valuation approaches often consider only the
accrued beneﬁts, even when the plan is valued as a going concern.
Example 9.6 The pension plan in Example 9.4 oﬀers an age retirement pension of 1.5%
of ﬁnal average salary for each year of service, where ﬁnal average salary is deﬁned as the
earnings in the three years before retirement.
Use Table 9.4 to estimate the EPV of the accrued age retirement pension for a member
aged 55 with 20 years of service, whose salary in the year prior to the valuation date was
$50 000. 9.6. VALUATION OF BENEFITS 449 The pension beneﬁt is paid monthly in advance for life, with no spouse’s beneﬁt.
Other assumptions:
Salary Scale: From Table 9.1
PostRetirement Mortality: Standard ultimate survival model from Section 4.3
Interest: 5% per year Solution 9.6 Age retirement can take place at exact age 60, at exact age 65, or at
any age in between. We assume that midyear age retirements (the retirements that do
not occur at exact age 60 or 65) that are assumed to occur between ages x and x + 1
(x = 60, 61, . . . , 64) take place at age x + 0.5 exact. This is a common assumption in
pensions calculation and is a similar approach to the claims acceleration approach for
continuous beneﬁts in Section 4.5. The assumption considerably simpliﬁes calculations
for complex beneﬁts, as it converts a continuous model for exits into a discrete model, more
suitable for eﬃcient spreadsheet calculation, and the inaccuracy introduced is generally
small.
Suppose retirement takes place at age y . Then the projected ﬁnal average salary is
50 000 zy
s54 where
zy = (sy−3 + sy−2 + sy−1 )/3
and where we use the values in Table 9.1 and linear interpolation to calculate, for example,
s58.5 . The function zy is the averaging function for the salary scale to give the ﬁnal average
salary, so that if we multiply the salary in the year of age x to x + 1 by zy /sx we get the
ﬁnal average salary on exit at exact age y .
If the member retires at exact age 60, the accrued beneﬁt, based on 20 years’ past service
and an accrual rate of 1.5%, is a pension payable monthly in advance from age 60 of 450 CHAPTER 9. PENSION MATHEMATICS annual amount
50 000 z60
× 20 × 0.015 = $15 922.79
s54 To value this we need to use life annuity values from the age at exit. The annuity values
used below have been calculated accurately, but interpolating values from Table 6.1 with
a UDD adjustment gives similar results.
The EPV of the accrued age retirement pension is then
50 000 × 0.015 × 20 r60− z60 5 (12)
r60+ z60.5 5.5 (12)
r61 z61.5 6.5 (12)
v a60 +
¨
v a60.5 +
¨
v a61.5
¨
l55 s54
l55 s54
l55 s54
+ ... + r64 z64.5 9.5 (12) r65− z65 10 (12)
v a64.5 +
¨
v a65
¨
l55 s54
l55 s54 = $137 508.
This type of repetitive calculation is ideally suited to spreadsheet software. Withdrawal Pension
When an employee leaves employment before being eligible to take an immediate pension,
the usual beneﬁt (subject to some minimum period of employment) in a DB plan is a
deferred pension. The beneﬁt would be based on the same formula as the retirement
pension, that is, Accrual Rate × Service × Final Average Salary, but would not be paid
until the member attains the normal retirement age. Note that Final Average Salary here
is based on earnings in the years immediately preceding withdrawal.
The deferred period could be very long, perhaps 35 years for an employee who changes
jobs at age 30. If the deferred beneﬁt is not increased during the deferred period, then 9.6. VALUATION OF BENEFITS 451 inﬂation, even at relatively low levels, will have a signiﬁcant eﬀect on the purchasing
power of the pension. In some plans the withdrawal beneﬁt is adjusted through the
deferred period to make some, possibly partial, allowance for inﬂation. Such adjustments
are called cost of living adjustments, or COLAs. In the UK, some inﬂation adjustment is
mandatory. Some plans outside the UK do not guarantee any COLA but apply increases
on a discretionary basis.
Example 9.7 A ﬁnal salary pension plan oﬀers an accrual rate of 2%, and the normal
retirement age is 65. Final average salary is the average salary in the three years before
retirement or withdrawal. Pensions are paid monthly in advance for life from age 65, with
no spouse’s beneﬁt, and are guaranteed for ﬁve years.
(a) Estimate the EPV of the accrued withdrawal pension for a life now aged 35 with 10
years of service whose salary in the past year was $100 000
(i) with no COLA, and
(ii) with a COLA in deferment of 3% per year.
(b) On death during deferment, a lump sum beneﬁt of ﬁve times the accrued annual
pension, including a COLA of 3% per year, is paid immediately. Estimate the EPV
of this beneﬁt.
Basis:
Service Table: From Table 9.4
Salary Scale: From Table 9.1
Post–Withdrawal Mortality: Standard ultimate survival model from Section 4.3
Interest: 5% per year Solution 9.7 According to our service table assumptions, the member can withdraw at
any age up to 60. There are no ‘exact age’ withdrawals, unlike age retirements, so if 452 CHAPTER 9. PENSION MATHEMATICS the member withdraws between ages x and x + 1 (x = 35, 36, . . . , 59) we assume that
withdrawal takes place at age x + 0.5.
Since the deferred pension is based on Final Average Salary, which is deﬁned as the
average annual salary in the three years before withdrawal, we deﬁne
zy = (sy−3 + sy−2 + sy−1 )/3
as we did in Example 9.6.
(12) (a) The guaranteed annuity EPV factor at age 65 is a , which can be evaluated as
¨
65:5
follows
a
¨ (12)
65:5 (12) = a5
¨ (12) + 5 p65 v 5 a70
¨ = 4.4459 + 0.75455 × 11.5451
= 13.1573
where the annuity and the survival probability are calculated using the standard
survival model, as set out in Section 4.3.
(i) If the member withdraws between integer ages 35+ t and 35+ t +1, the accrued
withdrawal pension, with no COLA, payable from age 65, is estimated to be
100 000 × z35+t+0.5
× 10 × 0.02
s34 The EPV of this at age 65 is
100 000 × z35+t+0.5
(12)
× 10 × 0.02 × a
¨
65:5
s34 and the EPV at age 35 + t + 0.5 is
100 000 × z35+t+0.5
(12)
× 10 × 0.02 a
¨
( 29.5−t p35+t+0.5 ) v 29.5−t
65:5
s34 9.6. VALUATION OF BENEFITS 453 where the t px factor is for survival only, not for the multiple decrement, as we
are applying it to a life who has just withdrawn.
The probability that the member withdraws between integer ages 35 + t and
35 + t + 1 is w35+t /l35 . Applying this probability and the discount factor v t+0.5
we obtain the EPV at age 35 of the accrued withdrawal beneﬁt as
100 000 × 10 × 0.02
l35 s34 24 w35+t z35+t+0.5 a
¨ (12) 65:5 t=0 ( 29.5−t p35+t+0.5 ) v 30 which is $48 246.
(ii) To allow for a COLA at 3% per year during deferement, the above formula for
the EPV of the accrued withdrawal beneﬁt must be adjusted by including a
term 1.0329.5−t , so that it becomes
100 000 × 10 × 0.02
l35 s34 24 w35+t z35+t+0.5 1.0329.5−t a
¨ (12) 65:5 t=0 ( 29.5−t p35+t+0.5 ) v 30 which is $88 853.
(b) Suppose the member withdraws between integer ages 35 + t and 35 + t + 1; the
probability that this happens is w35+t /l35 . The estimated initial annual accrued
pension is
0.02 × 10 × 100 000 z35+t+0.5
s34 and the sum insured on death before age 65 is ﬁve times this annual amount increased by the COLA. Hence the EPV of the beneﬁt on death after withdrawal
is
24 EPV =
t=0 5 × 0.02 × 10 × 100 000 v t+0.5 z35+t+0.5 w35+t+0.5 ¯ 1
A35+t+0.5: 65−(35+t+0.5) j
s34
l35 = 5 × 0.02 × 10 × 100 000 × 0.01813 454 CHAPTER 9. PENSION MATHEMATICS
= $1 813 where the subscript j indicates that the rate of interest used to calculate the term
insurances is j = 0.02/1.03.
Throughout this section, we have assumed that the accrued beneﬁt allows fully for future
salary increases. However, as for the future service beneﬁt, future salary increases are not
guaranteed and there is a case for omitting them from the accrued liabilities. When a
salary increase is actually declared, then it would be brought into the liability valuation.
The approach which uses salaries projected to the exit date is called the projected unit
method. Valuing the accrued beneﬁts with no allowance for future salary increases is the
traditional unit or current unit approach. Each has its adherents.
zx+t
To adapt the methodology above to the current unit approach, the
factors would be
sx
zx
replaced by
in the valuation formulae, or by the actual average pensionable earnings
sx
at valuation. That is, if the pension calculation uses the average of three year earnings
to retirement, the current unit valuation could use the average of the three year earnings
to the valuation date. This would need to be adjusted for a member with less than three
years’ service. For simplicity, the valuation in a current unit approach may use the current
salary at valuation. 9.6.2 Career average earnings plans Under a career average earnings (CAE) deﬁned beneﬁt pension plan, the beneﬁt formula
is based on the average salary during the period of pension plan membership, rather than
the ﬁnal average salary. Suppose a plan member retires at age xr with n years of service
and total pensionable earnings during their service of (TPE)xr . Then their career average
earnings are (TPE)xr /n. So a CAE plan with an accrual rate of α would provide a pension 9.6. VALUATION OF BENEFITS 455 beneﬁt on retirement at age xr, for a member with n years of service, of
αn (TPE)xr
= α (TPE)xr .
n Under a career average earnings plan, the accrued or past service beneﬁt that we value
at age x is α (TPE)x , where (TPE)x denotes the total pensionable earnings up to age x.
The methods available for valuing such beneﬁts are the same as for a ﬁnal salary beneﬁt.
A popular variation of the career average earnings plan is the career average revalued
earnings plan, in which an inﬂation adjustment of the salary is made before averaging.
The accrual principle is the same. The accrued beneﬁt is based on the total past earnings
after the revaluation calculation.
Example 9.8 A pension plan oﬀers a retirement beneﬁt of 4% of career average earnings
for each year of service. The pension beneﬁt is paid monthly in advance for life, guaranteed
for ﬁve years, with no spousal beneﬁt. On withdrawal, a deferred pension is payable from
age 65. The multiple decrement model in Example 9.4 is appropriate for this pension
plan, including the assumption that members can retire at exact age 60 and at exact age
65.
Consider a member now aged 35 who has 10 years of service, with total past earnings of
$525 000.
(a) Write down an integral formula for an accurate calculation of the EPV of his accrued
age and withdrawal beneﬁts.
(b) Use Table 9.2 to estimate the EPV of his accrued age and withdrawal beneﬁts.
Other assumptions:
Post–Retirement/Withdrawal Mortality: Standard ultimate survival model
from
Section 4.3 456 CHAPTER 9. PENSION MATHEMATICS
Interest: 5% per year Solution 9.8 (a) The EPV of the accrued age retirement beneﬁt is
25− 0.04 (TPE)35 0 00
t p35 µ03 t v t a
¨
35+ (12) 35+t:5 30− +
25+ 00
t p35 dt + 0.3 25 p00 v 25 a
¨
35 µ03 t v t a
¨
35+ (12) 60:5 (12)
35+t:5 dt + 00 30 (12)
¨
30 p35 v a
65:5 where the second and fourth terms allow for the exact age retirements.
The EPV of the accrued withdrawal beneﬁt is
0.04 (TPE)35 v 30 a
¨ 30 (12)
65:5 00
t p35 0 µ01 t
35+ 30−t p35+t dt where the survival probability 30−t p35+t is calculated using a mortality assumption
appropriate for members who have withdrawn.
(b) The EPV of the accrued age retirement beneﬁt is estimated as
0.04 (TPE)35
(12)
(12)
(12)
(12)
r60− v 25 a
¨
+ r60+ v 25.5 a
¨
+ r61 v 26.5 a
¨
+ . . . + r64 v 29.5 a
¨
60:5
60.5:5
61.5:5
65.5:5
l35
+ r65 v 30 a
¨ (12)
65:5 = $31 666.
Note the exact age retirement terms for ages 60 and 65.
The EPV of the accrued withdrawal beneﬁt is
0.04 (TPE)35 v 30 a
¨
l35
= $33 173. (12)
65:5 w35 (29.5 p35.5 ) + w36 (28.5 p36.5 ) + . . . + w59 (5.5 p59.5 ) 9.7. FUNDING PLANS 9.7 457 Funding plans In a typical DB pension plan the employee pays a ﬁxed contribution, and the balance of
the cost of the employee beneﬁts is funded by the employer. The employer’s contribution
is set at the regular actuarial valuations, and is expressed as a percentage of salary.
With an insurance policy, the policyholder pays for a contract typically through a level,
regular premium or a single premium. The nature of the pension plan is that there is
no need for the funding to be constant, as contributions can be adjusted from time to
time. The level of contribution from the employer is not usually a part of the contract,
the way that the premium is speciﬁed in the insurance contract. Nevertheless, because
the employer will have an interest in smoothing its contributions, there is some incentive
for the funding to be reasonably smooth and predictable.
We assume that the beneﬁt valuation approach from the previous section is used to
establish a reserve level at the start of the year. The reserve refers to the assets set aside
to meet the accrued liabilities as they fall due in the future. So, the reserve at time t, say,
is the sum of the EPVs of all the accrued beneﬁts at that time, taking into consideration
all the appropriate beneﬁts. We denote this reserve t V . It is also called the Actuarial
Liability.
We then set the funding level for the year to be the amount required to be paid such that,
together with the fund value at the start of the year, the assets are exactly suﬃcient to
pay the expected cost of any beneﬁts due during the year, and to pay the expected cost
of establishing the new actuarial liability at the year end.
We assume that (i) all employer contributions are paid at the start of the year, (ii) there
are no employee contributions, and (iii) any beneﬁts payable during the year are paid
exactly halfway through the year. These are simplifying assumptions that make the
development of the principles and formulae clearer, but they can be relaxed quite easily.
With these assumptions, the normal contribution due at the start of the year t to t + 1
for a member aged x at time t, denoted Ct , is found from 458 CHAPTER 9. PENSION MATHEMATICS tV + Ct = EPV of beneﬁts for midyear exits + v 1 p00 t+1 V.
x (9.3) That is
Ct = v 1 p00 t+1 V + EPV of beneﬁts for midyear exits − t V.
x
By EPV of beneﬁts for midyear exits we mean the EPV at the start of the year of
beneﬁts that would be payable given that the life exits during the year, multiplied by the
probability of exit during the year.
The funding equation (9.3) is interpreted as follows: the start of year actuarial liability
plus normal contributions must be suﬃcient, on average, to pay for the beneﬁts if the
member exits during the following year, or to fund the value of the actuarial liability
at the year end if the member remains in employment. The ideas, which are similar to
those developed when we discussed policy values in Section 7.3.3, are demonstrated in the
following example.
Example 9.9 A member aged 50 has 20 years past service. His salary in the year to
valuation was $50 000. Calculate the value of his accrued pension beneﬁt and the normal
contribution due at the start of the year assuming (a) projected unit credit (PUC) funding,
and (b) traditional unit credit (TUC) funding, assuming valuation uses ‘ﬁnal pensionable
earnings’ at the valuation date
You are given the pension plan information and valuation assumptions below.
• Accrual Rate: 1.5%
• Final salary plan
• Pension based on earnings in the year before age retirement
• Normal retirement age 65 9.7. FUNDING PLANS 459 • The pension beneﬁt is a life annuity payable monthly in advance
• There is no beneﬁt due on death in service
Assumptions:
No exits other than by death before normal retirement age.
Interest rate: 5% per year eﬀective.
Salaries increase at 4% per year (Projected Unit Credit).
Mortality before and after retirement follows the Standard survival model from
Section 4.3.
Solution 9.9 (a) Using the projected unit credit approach, the funding and valuation
are based on projected ﬁnal average earnings, so
SFin = 50 000s64 /s49 = 50 000(1.04)15 = 90 047.
The actuarial liability is the value at the start of the year of the accrued beneﬁts,
which is
0V (12) = 0.015 × 20 × SFin × 15 p50 × v 15 × a65 = 163 161
¨
(12) (Note that a65 = 13.0870.)
¨
The value at the start of the following year of the accrued beneﬁts, assuming the
member is still alive, is
1V (12) = 0.015 × 21 × SFin × 14 p51 × v 14 × a65
¨ and we take the value at time 0 of this liability,
(12) v p50 1 V = 0.015 × 21 × SFin × 15 p50 × v 15 × a65 =
¨ 21
0 V.
20 460 CHAPTER 9. PENSION MATHEMATICS
In this example there are no beneﬁts payable on midyear exit, so the funding
equation is
0V +C = 21
0V
20 which gives
C= 0V 20 = 8 158 or 16.3% of salary in the previous year.
This contribution formula can be explained intuitively: the normal contribution in
the year of age x to x + 1 must be suﬃcient to fund one extra year of accrual, on
average.
(b) Using the traditional unit credit approach, the valuation at time t is based on the
ﬁnal average earnings at time t. At the start of the year, the salary for valuation
is $50 000; at the year end the projected salary is $50 000 × 1.04 = $52 000. Let Sx
denote the salary earned (or projected) in the year of age x to x + 1. Then
(12) 0V = 0.015 × 20 × S49 × 15 p50 × v 15 × a65
¨ 1V = 0.015 × 21 × S50 × 14 p51 × v 14 × a65 .
¨ and
(12) So
(12) v p50 1 V = 0.015 × 21 × S50 × 15 p50 × v 15 × a65 = 0 V
¨
Hence
C = 0V 21 S50
−1
20 S49 = 8 335 or 16.7% of the previous year’s salary. 21 S50
.
20 S49 9.7. FUNDING PLANS 461 We can decompose the normal contribution here as
C = 0V S50
S50 1
− 1 + 0V
.
S49
S49 20 The ﬁrst term represents the contribution required to adjust the previous valuation
for the increase in salary over the year, and the second term represents the contribution required for the extra year’s accrual. The ﬁrst term is required here because the
TUC valuation does not allow for future salary increases, so they must be funded
year by year, through the contributions, as the salaries increase each year. Note that in this example the normal contributions are similar, though the valuation
liability under the TUC approach is rather less than that under the PUC approach.
In fact, under both funding approaches the contribution rate tends to increase as the
member acquires more service, and gets closer to retirement. The TUC contribution starts
rather smaller than the PUC contribution, and rises more steeply, ending at considerably
more than the PUC contribution. Ultimately, if all the assumptions in the basis are
realized, both methods generate exactly the same amount at the normal retirement age
(12)
for surviving members, speciﬁcally 0.015 × 35 × SFin × a65 , which is exactly enough to
¨
fund the retirement beneﬁt at that time.
In the example above, we showed how the PUC and TUC funding plans allow for the
normal contribution to fund the extra year of accrual (and the salary increase, in the
TUC case). The situation is slightly more complicated when there are beneﬁts payable
on exit during the year, as discussed in the next example. However, if the employee
leaves before the year end, then the normal contribution only has to fund the additional
accrual up to exit. We typically assume that midyear exits occur, on average, halfway
through the valuation year, in which case the members leaving accrue an extra half year
of beneﬁts. We explore this in the following example.
Example 9.10 A pension plan oﬀers a pension beneﬁt of $1 000 for each year of service, 462 CHAPTER 9. PENSION MATHEMATICS with fractional years counting prorata. A member aged 61 has 35 years past service.
Value the accrued age retirement beneﬁt and determine the normal contribution rate
payable in respect of age retirement beneﬁts using the following plan information and
valuation assumptions.
• Age retirements are permitted at any age between 60 and 65.
• The pension is paid monthly in advance for life.
• Contributions are paid annually at the start of each year.
• The unit credit funding method is used. We do not need to specify whether we use
projected or traditional unit credit as this is not a ﬁnal salary plan.
Assumptions:
Exits follow the service table given in Table 9.2.
Interest rate: 6% per year eﬀective.
All lives taking age retirement exit exactly halfway through the year of age (except
at age 65).
Mortality after retirement follows the standard survival model from Section 4.3.
Solution 9.10 Apart from the diﬀerent pension beneﬁt, this example diﬀers from the
previous one because we need to allow for midyear exits. We noted above that the
contribution under a unit credit approach pays for the extra one year of accrued beneﬁt
for the lives who stay, and pays for an extra half year’s accrued beneﬁt (on average) for
the lives who leave. We have 0V = 1000 × 35 × r61 0.5 (12) r62 1.5 (12) r63 2.5 (12) r64 3.5 (12) r65− 4 (12)
v a61.5 +
¨
v a62.5 +
¨
v a63.5 +
¨
v a64.5 +
¨
v a65
¨
l61
l61
l61
l61
l61 9.8. NOTES AND FURTHER READING 463 = 345 307
and
v p00 1 V
61 = 1000 × 36 × r62 1.5 (12) r63 2.5 (12) r64 3.5 (12) r65− 4 (12)
v a62.5 +
¨
v a63.5 +
¨
v a64.5 +
¨
v a65
¨
l61
l61
l61
l61 = 312 863.
Note the exact age retirement terms for age 65.
The EPV of the beneﬁts for lives exiting by age retirement in the middle of the valuation
year is
1000 × 35.5 × r61 0.5 (12)
v a61.5 = 41 723.
¨
l61 Hence, the normal contribution required at the start of the year is C where
0V + C = EPV beneﬁts to midyear exits + p00 v 1 V
61 giving
C = 41 723 + 312 863 − 345 307 = 9 278. 9.8 Notes and further reading In this chapter we have introduced some of the language and concepts of pension plan
funding and valuation. The presentation has been relatively simpliﬁed to bring out some of
the major concepts, in particular, accruals funding principles. In North America, what we
have called the normal contribution is called the normal cost. The diﬀerence between
the normal contribution and the actual contribution paid represents a paying down of 464 CHAPTER 9. PENSION MATHEMATICS surplus or deﬁcit. Such practical considerations are beyond the scope of this book – we
are considering pensions here in the speciﬁc context of the application of life contingent
mathematics. For more information on pension plan design and related issues, texts such
as McGill et al (2005) and Blake (2006) are useful. 9.9. EXERCISES 9.9 465 Exercises Where an Exercise uses the service table speciﬁed in Example 9.5, the calculations are
based on the exact values underlying Table 9.2. Using the integer–rounded values presented in Table 9.2 may result in very slight diﬀerences from the numerical answers printed
at the end of this chapter.
The standard survival model is the model speciﬁed in Section 4.3.
Exercise 9.1 In order to value the beneﬁts in a ﬁnal salary pension scheme as at 1
January 2008, a salary scale, sx , has been deﬁned so that sx+t /sx is the ratio of a member’s
total earnings between ages x + t and x + t + 1 to the member’s total earnings between
ages x and x + 1. Salary increases take place on 1 July every year.
One member, whose date of birth is 1 April 1961, has an annual salary rate of $75 000 on
the valuation date. Using the salary scale in Table 9.1, estimate the member’s expected
earnings during 2008. Exercise 9.2 Assume the salary scale given in Table 9.1 and a valuation date of 1 January.
(a) A plan member aged 35 at valuation received $75 000 in salary in the year to the
valuation date. Given that ﬁnal average salary is deﬁned as the average salary in the
four years before retirement, calculate the member’s expected ﬁnal average salary
assuming retirement at age 60.
(b) A plan member aged 55 at valuation was paid salary at a rate of $100 000 per year
at the valuation date. Salaries are increased on average halfway through each year.
Calculate the expected average salary earned in the two years before retirement at
age 65. 466 CHAPTER 9. PENSION MATHEMATICS Exercise 9.3 A pension plan member is aged 55. One of the plan beneﬁts is a death in
service beneﬁt payable on death before age 60.
(a) Calculate the probability that the employee dies in service before age 60.
(b) Assuming that the death in service beneﬁt is $200 000, and assuming that the death
beneﬁt is paid immediately on death, calculate the EPV at age 55 of the death in
service beneﬁt.
(c) Now assume that the death in service beneﬁt is twice the annual salary rate at death.
At age 55 the member’s salary rate is $85 000 per year. Assuming that deaths occur
evenly throughout the year, estimate the EPV of the death in service beneﬁt.
Basis:
Service Table from Example 9.4.
Interest rate 6% per year eﬀective.
Salary scale follows Table 9.1; all salary increases occur halfway through the
year of age, on average. Exercise 9.4 A new member aged 35 exact, expecting to earn $40 000 in the next 12
months, has just joined a pension plan. The plan provides a pension on age retirement
of 1/60th of ﬁnal pensionable salary for each year of service, with fractions counting
proportionately, payable monthly in advance for life. There are no spousal beneﬁts.
Final pensionable salary is deﬁned as the average salary over the three years prior to
retirement. Members contribute a percentage of salary, the rate depending on age. Those
under age 50 contribute 4% and those aged 50 and over contribute 5%.
The employer contributes a constant multiple of members’ contributions to meet exactly
the expected cost of pension beneﬁts. Calculate the multiple needed to meet this new 9.9. EXERCISES 467 member’s age retirement beneﬁts. Assume all contributions are paid exactly half way
through the year of age in which they are paid.
Basis: Service Table:
Mortality after retirement:
Interest: from Example 9.4
Standard survival model
4% per year eﬀective Exercise 9.5 (a) A new employee age 25 joins a DC pension plan. Her starting salary
is $40 000 per year. Her salary is assumed to increase continuously at a rate of 7%
per year for the ﬁrst 20 years of her career and 4% per year for the following 15
years.
At retirement she is to receive a pension payable monthly in advance, guaranteed
for 10 years. She plans to retire at age 60, and she wishes to achieve a replacement
ratio of 70% through the pension plan. Using the assumptions below, calculate the
level annual contribution rate c (% of salary) that would be required to achieve this
replacement ratio.
Assumptions
 Interest rate 7% per year eﬀective before retirement, 5% per year eﬀective after
retirement.
 Mortality follows the standard survival model.
(b) Now assume that this contribution rate is paid, but her salary increases at a rate of
5% throughout her career, and interest is earned at 6% on her contributions, rather
than 7%. In addition, at retirement, interest rates have fallen to 4.5% per year.
Calculate the replacement ratio achieved using the same mortality assumptions. 468 CHAPTER 9. PENSION MATHEMATICS Exercise 9.6 A pension plan member aged 61 has 35 years of past service at the funding
valuation date. His salary in the year to the valuation date was $50 000.
The death in service beneﬁt is 10% of salary at death for each year of service. Calculate
the value of the accrued death in service beneﬁt and the normal contribution rate for the
death in service beneﬁt.
Basis:
Service Table from Example 9.4.
Interest rate 6% per year eﬀective.
Salary scale follows Table 9.1; all salary increases take place on the valuation date.
Projected unit credit funding method. Exercise 9.7 A new company employee is 25 years old. Her company oﬀers a choice of a
deﬁned beneﬁt or a deﬁned contribution pension plan. All contributions are paid by the
employer, none by the employee.
Her starting salary is $50 000 per year. Salaries are assumed to increase at a rate of 5%
per year, increasing at each year end.
Under the deﬁned beneﬁt plan her ﬁnal pension is based on the salary received in the
year to retirement, using an accrual rate of 1.6% for each year of service. The normal
retirement age is 65. The pension is payable monthly in advance for life.
Under the deﬁned contribution plan, contributions are deposited into the member’s account at a rate of 12% of salary per year. The total accumulated contribution is applied
at the normal retirement age to purchase a monthly life annuitydue.
(a) Assuming the employee chooses the deﬁned beneﬁt plan and that that she stays in
employment through to age 65, calculate her projected annual rate of pension. 9.9. EXERCISES 469 (b) Calculate the contribution, as a percentage of her starting salary, for the retirement
pension beneﬁt for this life, for the year of age 25 to 26, using the projected unit
credit method. Assume no exits except mortality, and that the survival probability
is 40 p25 = 0.80. The valuation interest rate is 6% per year eﬀective. The annuity
(12)
factor a65 is expected to be 11.00.
¨
(c) Now assume that the employee joins the deﬁned contribution plan. Contributions
(12)
are expected to earn a rate of return of 8% per year. The annuity factor a65 is
¨
expected to be 11.00. Assuming the employee stays in employment through to age
65, calculate (i) the projected fund at retirement and (ii) her projected annual rate
of pension, payable from age 65.
(d) Explain brieﬂy why the employee might choose the deﬁned beneﬁt plan even though
the projected pension is smaller.
(e) Explain brieﬂy why the employer might prefer the deﬁned contribution plan even
though the contribution rate is higher. Exercise 9.8 In a pension plan, a member who retires before age 65 has their pension
reduced by an actuarial reduction factor. The factor is expressed as a rate per month,
k , say, and is then applied to reduce the member’s pension to (1 − t × k ) B , where B is the
normal accrued beneﬁt that the member would be entitled to if they had already reached
age 65, and t is the time in months from the actual retirement age to age 65.
The plan sponsor wishes to calculate k such that the EPV at early retirement of the
reduced pension beneﬁt is the same as the EPV of the accrued beneﬁt payable at age 65,
assuming no exits from mortality or any other decrement before age 65, and ignoring pay
increases up to age 65. The pension is assumed to be paid monthly in advance for the
member’s lifetime. 470 CHAPTER 9. PENSION MATHEMATICS Calculate k for a retiree who entered the plan at age 25 and wishes to retire at age (i) 55
and (ii) 60, using the following further assumptions:
Mortality after retirement:
Interest rate Standard survival model
6% per year eﬀective. Exercise 9.9 A pension plan has only one member, who is aged 35 at the valuation
date, with 5 years past service. The plan beneﬁt is $350 per year pension for each year of
service, payable monthly in advance. There is no actuarial reduction for early retirement.
Calculate the actuarial liability and the normal contribution rate for the age retirement
beneﬁt for the member. Use the service table from Example 9.4. Postretirement mortality
follows the standard survival model. Assume 6% per year interest and use the unit credit
funding method. Exercise 9.10 An employer oﬀers a career average pension scheme, with accrual rate
2.5%. A plan member is aged 35 with 5 years past service, and total past salary $175 000.
His salary in the year to valuation is $40 000.
Using the service table from Example 9.4, calculate the actuarial liability and the normal
contribution rate for the age retirement beneﬁt for the member. There is no actuarial
reduction for early retirement. Postretirement mortality follows the standard survival
model. Assume 6% per year interest and use the unit credit funding method. Exercise 9.11 9.9. EXERCISES 471 • Allison is a member of a pension plan. At the valuation date, 31 December 2008,
she is exactly 45.
• Her salary in the year before valuation is $100 000.
• The ﬁnal average salary is deﬁned as the average salary in the two years before exit.
• Salaries are revised annually on 1st July each year in line with the salary scale in
Table 9.1.
The pension plan provides a beneﬁt of 1.5% of ﬁnal average salary for each year of service.
The beneﬁts are valued using the standard survival model, using an interest rate of 5%
per year eﬀective. Allison has 15 years service at the valuation date. She is contemplating
three possible retirement dates.
• She could retire at 60.5, with an actuarial reduction applied to her pension of 0.5%
per month up to age 62.
(That means her beneﬁt would be (1 − (18)(0.005))B where B would be the usual
ﬁnal salary beneﬁt, calculated as B = n SF in α.)
• She could retire at age 62 with no actuarial reduction.
• She could retire at age 65 with no actuarial reduction.
(a) Calculate the replacement rate provided by the pension for each of the retirement
dates.
(b) Calculate the expected present value of Allison’s retirement pension for each of
the possible retirement dates. Assume mortality is the only decrement and use an
interest rate of 5% per year. The basic pension beneﬁt is a single life annuity, paid
monthly in advance.
(c) Now assume Allison leaves the company and withdraws from active membership
of the pension plan immediately after the valuation. Her total salary in the two 472 CHAPTER 9. PENSION MATHEMATICS
years before exit is $186 000. She is entitled to a deferred pension of 1.5% of her
ﬁnal average earnings in the two years before withdrawal for each year of service,
payable at age 62. There is no COLA for the beneﬁt. Calculate the EPV of the
withdrawal beneﬁt using the valuation assumptions. Exercise 9.12 Using the unit credit method, calculate the actuarial liability and the
normal contribution for the following pension plan.
Beneﬁt
Normal Retirement Age
Mortality
Interest
Pension
Preretirement exits $300 per year pension for each year of service
60
Standard survival model
6% per year eﬀective
payable weekly, guaranteed for ﬁve years
mortality only Active membership data at valuation
Age
25
35
45
55 Service
Number of employees
for each employee
0
3
10
3
15
1
25
1 Inactive membership data at valuation
Age
35
75 Service
7
25 Number of employees
1 (deferred pensioner)
1 (pension in payment) 9.9. EXERCISES 473 Exercise 9.13 A deﬁned beneﬁt pension plan oﬀers an annual pension of 2% of the ﬁnal
year’s salary for each year of service payable monthly in advance. Using the following
information:
Interest rate
Salary growth rate 4% per year
Salary Scale follows Table 9.1
all increases occur on 31 December each year
65
None
standard survival model Retirement age
Preretirement exits
Retirement mortality Membership:
Name Age at entry Giles
Faith 30
30 (a) Age at
1 January 2009
35
60 Salary at
1 January 2008
38 000
47 000 Salary at
1 January 2009
40 000
50 000 (i) Calculate the actuarial liability at 1 January 2009 using the projected unit
credit method.
(ii) Calculate the normal contribution rate in 2009 separately for Giles and Faith,
as a proportion of their 2009 salary, using the projected unit credit funding
method. (b) (i) Calculate the actuarial liability at 1 January 2009 using the traditional unit
credit method.
(ii) Calculate the normal contribution rate in 2009 separately for Giles and Faith,
as a proportion of their 2009 salary, using the traditional unit credit funding
method. (c) Comment on your answers. 474 CHAPTER 9. PENSION MATHEMATICS Answers to selected exercises
9.1 $77 706.
9.2 (a) $172 331.
(b) $114 346. 9.3 (a) 0.01171
(b) $2 011.21.
(c) $1 678.64.
9.4 3.11
9.5 (a) 20.32%
(b) 55.21%.
9.6 Accrued Death Beneﬁt Value: $5 030.78
Normal Contribution $123.11
9.7 (a) $214 552.
(b) 9.18%.
(c) (i) $3 053 123; (ii)$277 466.
9.8 (a) 0.448%
(b) 0.550%
9.9 Actuarial Liability: $1 815.27
Normal Contribution: $363.05
9.10 Actuarial Liability: $4 538.17
Normal Contribution: $1 115.15. 9.9. EXERCISES
9.11 (a) 41.3% 475
47.6% (b) $383 700 $346 160 52.1%
$308 025. (c) $104 816.
9.12 Total Actuarial Liability: $197 856
Total Normal Contribution: $8 625.
9.13 (a) (i) $422 201.
(ii) Giles: 22.5% Faith: 25.1%. (b) (i) $350 945.
(ii) Giles: 11.1% Faith: 66.3%. 476 CHAPTER 9. PENSION MATHEMATICS Chapter 10
Interest Rate Risk
10.1 Summary In this chapter we consider the eﬀect on annuity and insurance valuation of interest
rates varying with the duration of investment, as summarized by a yield curve, and of
uncertainty over future interest rates, which we will model using stochastic interest
rates.
We introduce the concepts of diversiﬁable and nondiversiﬁable risk and give conditions under which mortality risk can be considered to be diversiﬁable.
In the ﬁnal section we demonstrate the use of Monte Carlo methods to explore distributions of uncertain cash ﬂows and loss random variables through simulation of both future
lifetimes and future interest rates. 10.2 The yield curve In practice, at any given time interest rates vary with the duration of the investment;
that is, a sum invested for a period of, say, 5 years, would typically earn a diﬀerent rate
477 478 CHAPTER 10. INTEREST RATE RISK of interest than a sum invested for a period of 15 year or a sum invested for a period of
6 months.
Let v (t) denote the current market price of a t year zerocoupon bond; that is, the current
market price of an investment which pays a unit amount with certainty t years from now.
Note that, at least in principle, there is no uncertainty over the value of v (t) although
this value can change at any time as a result of trading in the market. The t year spot
rate of interest, denoted yt , is the yield per year on this zerocoupon bond, so that
v (t)(1 + yt )t = 1 ⇐⇒ v (t) = (1 + yt )−t . (10.1) The term structure of interest rates describes the relationship between the term of the
investment and the interest rate on the investment, and it is expressed graphically by the
yield curve, which is a plot of {yt }t>0 against t. Figures 10.1 to 10.4 show diﬀerent
yield curves, derived using government issued bonds from the UK, the US and Canada,
at various dates from relatively recent history. The UK issues longer bonds than most
other countries, so the UK yield curve is longer.
These ﬁgures illustrate some of the shapes a yield curve can have. Figure 10.1 shows
a relatively ﬂat curve, so that interest rates vary little with the term of the investment.
Figure 10.2 shows a falling curve. Both of these shapes are relatively uncommon; the most
common shape is that shown in Figures 10.3 and 10.4, a rising yield curve, ﬂattening out
after 1015 years, with spot rates increasing at a decreasing rate.
Previously in this book we have assumed a ﬂat term structure. This assumption has
allowed us to use v t or e−δ t as discount functions for any term t, with v and δ as constants.
When we relax this assumption, and allow interest rates to vary by term, the v t discount
function is no longer appropriate. Figure 10.3 shows that the rate of interest on a 1 year
US government bond in Jan 2002 was 1.6% per year and on a 20 year bond was 5.6%. The
diﬀerence of 4% may have a signiﬁcant eﬀect on the valuation of an annuity or insurance
beneﬁt. The present value of a 20 year annuitydue of $1 per year payable in advance,
valued at 1.6%, is $17.27; valued at 5.6% it is $12.51. The value of the annuity should be
the amount required to be invested now to produce payments of 1 at the start of each of 10.2. THE YIELD CURVE 479 5.0 Spot Rate of Interest (%) 4.0 3.0 2.0 1.0 0.0
0 5 10 15 20 25 30 Term (Years) Figure 10.1: Canadian government bond yield curve (spot rates), May 2007. 6.0 Spot Rate of Interest (%) 5.0 4.0 3.0 2.0 1.0 0.0
0 5 10 15 20 25 30 35 40 45 Term (Years) Figure 10.2: UK government bond yield curve (spot rates), November 2006. 480 CHAPTER 10. INTEREST RATE RISK 6.0 Spot Rates of Interest (%) 5.0 4.0 3.0 2.0 1.0 0.0
0 5 10 15 20 25 30 Term (years) Figure 10.3: US government bond yield curve (spot rates), January 2002. 5.0 Spot Rates of Interest (%) 4.0 3.0 2.0 1.0 0.0
0 5 10 15 20 25 30 Term (years) Figure 10.4: US government bond yield curve (spot rates), November 2008. 10.2. THE YIELD CURVE 481 the next 20 years – this is how we have been implicitly valuing annuities when we discount
at the rate of interest on assets. When we have a term structure this means we should
discount each future payment using the spot interest rate appropriate to the term until
that payment is due. This is a replication argument: the present value of any cash ﬂow
is the cost of purchasing a portfolio which exactly replicates the cash ﬂow.
Since an investment now of amount v (t) in a t year zerocoupon bond will accumulate to
1 in t years, v (t) can be interpreted as a discount function which generalizes v t .
9
The price of the 20 year annuitydue with this discount function is 1=0 v (t) which means
t
that the price of the annuitydue is the cost of purchasing 20 zerocoupon bonds, each
with $1 face value, with maturity dates corresponding to the annuity payment dates. The
spot rates underlying the yield curve in Figure 10.3 give a value of $13.63 for the 20 year
annuitydue, closer to, but signiﬁcantly higher than the cost using the long term rate of
5.6%. At any given time the market will determine the price of zerocoupon bonds and this will
determine the yield curve. These prices also determine forward rates of interest at
that time. Let f (t, t + k ) denote the forward rate, contracted at time zero, eﬀective from
time t to t + k , expressed as an eﬀective annual rate. This represents the interest rate
contracted at time 0 earned on an investment made at t, maturing at t + k . To determine
forward rates in terms of spot rates of interest, consider two diﬀerent ways of investing
$1 for t + k years. Investing for the whole period, the t + k year spot rate, yt+k , gives the
accumulation of this investment as (1 + yt+k )t+k . On the other hand, if the unit sum is
invested ﬁrst for t years at the t year spot rate, then reinvested for k years at the k year
forward rate starting at time t, the accumulation will be (1 + yt )t (1 + ft,t+k )k . Since there
is no uncertainty involved in either of these schemes – note that yt+k , yt and f (t, t + k ) are
all known now – the accumulation at t + k under these two schemes must be the same.
That is
(1 + yt+k )t+k
v (t)
(1 + f (t, t + k ))k =
=
.
t
(1 + yt )
v (t + k )
This is (implicitly) a no arbitrage argument, which, essentially, says in this situation that 482 CHAPTER 10. INTEREST RATE RISK we should not be able to make money from nothing in risk free bonds by disinvesting and
then reinvesting. The no arbitrage assumption is discussed further in Chapter 13. 10.3 Valuation of insurances and life annuities The present value random variable for a life annuitydue with annual payments, issued to
a life aged x, given a yield curve {yt }, is
Kx Y= v (k ) (10.2) k=0 where v (k ) = (1 + yk )−k . The expected present value of the annuity, denoted a(x)y , can
¨
be found using the paymentbypayment (or indicator function) approach, so that
a(x)y =
¨ ∞ k px v (k ). (10.3) k=0 Similarly, the present value random variable for a whole life insurance for (x), payable
immediately on death, is
Z = v (Tx ) (10.4) and the expected present value is
∞ ¯
A(x)y = v (t) t px µx+t dt. (10.5) 0 Note that we have to depart from International Actuarial Notation here as it is deﬁned
in terms of interest rates that do not vary by term, though we will retain the spirit of the
notation.
By allowing for a nonconstant yield curve we lose many of the relationships that we have
developed for ﬂat interest rates, such as the equation linking ax and Ax .
¨ 10.3. VALUATION OF INSURANCES AND LIFE ANNUITIES 483 Example 10.1 You are given the following spot rates of interest per year.
y1
y2
y3
y4
y5
y6
y7
y8
y9
y10
0.032 0.035 0.038 0.041 0.043 0.045 0.046 0.047 0.048 0.048
(a) Calculate the discount function v (t) for t = 1, 2, ..., 10.
(b) A survival model follows Makeham’s law with A = 0.0001, B = 0.00035 and c =
1.075. Calculate the net level annual premium for a 10 year term insurance policy,
with sum insured $100 000 payable at the end of the year of death, issued to a life
aged 80
(i) using the spot rates of interest in the table above, and,
(ii) using a level interest rate of 4.8% per year eﬀective.
Solution 10.1 (a) Use equations (10.1) for the discount function values and (2.26) for
the Makeham survival probabilities. The following table summarizes some of the
calculations.
t
0
1
2
3
4
5
6
7
8
9
10 v (t)
1.0000
0.9690
0.9335
0.8941
0.8515
0.8102
0.7679
0.7299
0.6925
0.6558
0.6257 p80+t
0.88845
0.88061
0.87226
0.86337
0.85391
0.84387
0.83320
0.82188
0.80988
0.79718
0.78374 t p80 1.00000
0.88845
0.78237
0.68243
0.58919
0.50312
0.42456
0.35374
0.29073
0.23546
0.18770 484
(b) CHAPTER 10. INTEREST RATE RISK
(i) The expected present value for the 10 year life annuitydue is
9 a(80 : 10 ) =
¨ v (k )k p80 = 5.0507.
k=0 The expected present value for the term insurance beneﬁt is
1 9
k p80 (1 100 000 A(80 : 10 ) =
k=0 − p80+k )v (k + 1) = 66 739. So the annual premium is $13 213.72.
(ii) Assuming a 4.8% per year ﬂat yield curve gives a premium of $13 181.48. In general, life insurance contracts are relatively long term. The inﬂuence of the yield
curve on long term contracts may not be very great since the yield curve tends to ﬂatten
out after around 15 years. It is common actuarial practice to use the long term rate in
traditional actuarial calculations, and in many cases, as in the example above, the answer
will be close. However, using the long term rate overstates the interest when the yield
curve is rising, which is the most common shape. Overstating the interest results in a
premium that is lower than the true premium. An insurer that systematically charges
premiums lower than the true price, even if each is only a little smaller, may face solvency
problems in time. With a rising yield curve, if a level interest rate is assumed, it should
be a little less than the long term rate. 10.3.1 Replicating the cash
participating product ﬂows of a traditional non In this section we continue Example 10.1. Recall that the forward rate is contracted at
the inception of the contract. We show that, if we take the premium and the cash ﬂow 10.3. VALUATION OF INSURANCES AND LIFE ANNUITIES 485 brought forward each year and invest them at the forward rate, then there is exactly
enough to fund the sums insured, provided that the mortality and survival experience
follows the assumptions. This demonstrates replication – if the premiums and cash ﬂows,
are completely predictable, and are invested in the forward rates each year, the resulting
cash ﬂows exactly match the claims outgo.
We will assume each year end cash ﬂow for the policy is invested at the 1year forward
rate applying at the year end. We assume further that the assumed rates of mortality
are exactly experienced – that is, we make no allowance here for mortality variation or
uncertainty. The cash ﬂows (in $000s) for a portfolio of N = 100 000 contracts are given
in Table 10.1. The entries in the table are calculated as follows, where P is the premium
of $13,213.72, S is the sum insured of $100 000 and N is the number of contracts.
The premium income at time k , denoted Pk , is k p80 N P because at time k we have
k p80 N survivors with our deterministic model for mortality.
The total claim amount paid at time k , denoted Ck , is k−1 p80 q80+k−1 N S. The net cashfow carried forward at time k + 1, denoted CFk+1 , is
(Pk + CFk )(1 + f (k, k + 1)) − Ck+1 .
So, in the ﬁrst year the insurer receives a premium P from each policyholder at the start
of the year. Over the course of the year interest is earned at rate f (0, 1). At the year
end N q80 claims each of amount S are paid and the excess of premiums and interest over
claims is carried forward to be combined with premiums the following year.
What this shows is that if the cash ﬂows are certain, and if the policy is not so long
that it extends beyond the scope of risk free investments, then there is no need for the
policy to involve interest rate uncertainty. At the inception of the contract, we can lock
in forward rates that will exactly replicate the required cash ﬂows. Another interesting
point to note is that the net cash ﬂow carried forward at each year end represents the
total policy value for all the contracts on the premium basis. To get the policy value 486 CHAPTER 10. INTEREST RATE RISK Year
k →k+1
0→1
1→2
2→3
3→4
4→5
5→6
6→7
7→8
8→9
9 → 10 Expected Premium
Income
Pk
1 321 372
1 173 970
1 033 806
901 744
778 537
664 803
561 004
467 426
384 166
311 127 Forward Rate
f (k, k + 1)
0.0320
0.0380
0.0440
0.0501
0.0510
0.0551
0.0520
0.0540
0.0560
0.0480 Expected Claims Net Cash Flow
Outgo
carried forward
Ck+1
CFk+1
1 115 528
248 129
1 060 741
415 409
999 431
513 587
932 420
553 752
860 726
539 560
785 539
485 133
708 189
392 368
630 102
276 143
552 746
144 563
477 564
0 Table 10.1: Cashﬂow table for the term insurance policy from Example 10.1; 100,000
contracts, in $000s. per surviving policyholder at k , divide the net cash ﬂow at k by the assumed number of
survivors, 100 000 k px . So, the policy value at t at time 1 for each contract in the term
insurance example would be 248 129 000/(100 0001 p80 ) which is $2 792.83.
This example raises two immediate questions.
First, we know that mortality is uncertain, so that the mortality related cash ﬂows are not
certain. To what extent does this invalidate the replication argument? The answer is that,
if the portfolio of life insurance policies is suﬃciently large, and, crucially, if mortality can
be treated as diversiﬁable, then it is reasonable to treat the life contingent cash ﬂows as
if they were certain. In Section 10.4.1 we discuss in detail what we mean by diversiﬁability,
and under what conditions it might be a reasonable assumption for mortality.
The second question is, what risks are incurred by an insurer if it chooses not to replicate, 10.4. DIVERSIFIABLE AND NONDIVERSIFIABLE RISK 487 or is unable to replicate for lack of appropriate risk free investments? If the insurer does
not replicate the cash ﬂows, then interest rate risk is introduced, and must be modelled
and managed. Interest rate risk is inherently nondiversiﬁable, as we shall discuss in
Section 10.4.2. 10.4 Diversiﬁable and nondiversiﬁable risk Consider a portfolio consisting of N life insurance policies. We can model as a random
variable, Xi , i = 1, . . . , N , many quantities of interest for the ith policy in this portfolio.
For example, Xi could take the value 1 if the policyholder is still alive, say, 10 years after
the policy was issued and the value zero otherwise. In this case, N Xi represents the
i=1
number of survivors after 10 years. Alternatively, Xi could represent the present value
of the loss on the ith policy so that N Xi represents the present value of the loss on
i=1
the whole portfolio. Suppose for convenience that the Xi s are identically distributed with
common mean µ and standard deviation σ . Let ρ denote the correlation coeﬃcient for
any pair Xi and Xj (i = j ). Then
N E N Xi = N µ and V i=1 i=1 Xi = N σ 2 + N (N − 1)ρσ 2 . Suppose now that the Xi s are independent, so that ρ is zero. Then
N Xi = N σ 2 V
i=1 and the Central Limit Theorem (which is described in Appendix A) tells us that, provided
N is reasonably large,
N i=1 Xi ∼ N (N µ, N σ 2 ) ⇒ N
i=1 Xi − N µ
√
∼ N (0, 1).
Nσ 488 CHAPTER 10. INTEREST RATE RISK In this case, the probability that N Xi /N deviates from its expected value decreases
i=1
1
to zero as N increases . More precisely, for any k > 0
N N Pr
i=1 Xi /N − µ ≥ k = Pr = Pr i=1 Xi − N µ ≥ kN √
Xi − N µ
kN
√
≥
.
σ
Nσ N
i=1 If we now let N → ∞, so that we can assume from the Central Limit Theorem that
√
( N Xi − N µ)/( N σ ) is normally distributed, then the probability can be written as
i=1
√
kN
lim Pr Z  ≥
N →∞
σ = lim 2Φ
N →∞ √
kN
−
σ = 0, where Z ∼ N (0, 1).
So, as N increases, the variation of the mean of the Xi s from their expected value will
tend to zero, if V[ N Xi ] is linear in N . In this case we can reduce the risk measured by
i=1
Xi , relative to its mean value, by increasing the size of the portfolio. This result relies on
the fact that we have assumed that the Xi are independent; it is not generally true if ρ is
not equal to zero, as in that case V[ N Xi ] is of order N 2 , which means that increasing
i=1
the number of policies increases the risk relative to the mean value.
So, we say that the risk within our portfolio, as measured by the random variable Xi , is
said to be diversiﬁable if the following condition holds lim N →∞ V[ N
i=1 N Xi ] = 0. A risk is nondiversiﬁable if this condition does not hold. In simple terms, a risk is
diversiﬁable if we can eliminate it (relative to its expectation) by increasing the number of
1 We are using a special case of the Law of Large Numbers here. 10.4. DIVERSIFIABLE AND NONDIVERSIFIABLE RISK 489 policies in the portfolio. An important aspect of ﬁnancial risk management is to identify
those risks which can be regarded as diversiﬁable and those which can not. Diversiﬁable
risks are generally easier to deal with than those which are not. 10.4.1 Diversiﬁable mortality risk In Section 10.2 we employed the no arbitrage principle to argue that the value of a
deterministic payment stream should be the same as the price of the zerocoupon bonds
that replicate that payment stream. In Section 10.3.1 we explore the replication idea
further. To do this we need to assume that the mortality risk associated with a portfolio
is diversiﬁable and we discuss conditions for this to be a reasonable assumption.
Consider a group of N lives all now aged x who have just purchased identical insurance or
annuity policies. We will make the following two assumptions throughout the remainder
of this chapter, except where otherwise stated.
(i) The N lives are independent with respect to their future mortality.
(ii) The survival model for each of the N lives is known.
We also assume, for convenience, that each of the N lives has the same survival model.
The cash ﬂow at any future time t for this group of policyholders will depend on how many
are still alive at time t and on the times of death for those not still alive. These quantities
are uncertain. However, with the two assumptions above the mortality risk is diversiﬁable.
This means that, provided N is large, the variability of, say, the number of survivors at
any time relative to the expected number is small so that we can regard mortality, and
hence the cash ﬂows for the portfolio, as deterministic. This is demonstrated in the
following example.
Example 10.2 For 0 ≤ t ≤ t + s, let Nt,s denote the number of deaths between ages 490 CHAPTER 10. INTEREST RATE RISK x + t and x + t + s. Show that lim N →∞ V[Nt,s ]
= 0.
N Solution 10.2 The random variable Nt,s has a binomial distribution with parameters N
and t px (1 − s px+t ). Hence
V[Nt,s ] = N t px (1 − s px+t ) (1 − t px (1 − s px+t ))
⇒ V[Nt,s ]
=
N ⇒ lim N →∞ (1 − s px+t )(1 −t px (1 − s px+t )
N V[Nt,s ]
= 0.
N In practice most insurers sell so many contracts over all their life insurance or annuity
portfolios that mortality risk can be treated in many situations as fully diversiﬁed away.
There are exceptions; for example, for very old age mortality, where the number of policyholders tends to be small, or where an insurance has a very high sum at risk, in which
case the outcome of that particular contract may have a signiﬁcant eﬀect on the portfolio
as a whole, or where the survival model for the policyholders cannot be predicted with
conﬁdence.
If mortality risk can be treated as fully diversiﬁed then we can assume that the mortality
experience is deterministic – that is, we may assume that the number of claims each
year is equal to the expected number. In the following section we use this deterministic
assumption for mortality to look at the replication of the term insurance cash ﬂows in
Example 10.1 above. 10.4. DIVERSIFIABLE AND NONDIVERSIFIABLE RISK 10.4.2 491 Nondiversiﬁable risk In practice, many insurers do not replicate with forward rates or zerocoupon bonds either
because they choose not to or because there are practical diﬃculties in trying to do so.
By locking into forward rates at the start of a contract, the insurer can remove (much
of) the investment risk, as shown in Table 10.1. However, while this removes the risk of
losses, it also removes the possibility of proﬁts. Also there may be practical constraints.
For example, in some countries it may not be possible to ﬁnd risk free investments for
terms longer than around 20 years, which is often not long enough. A whole life insurance
contract issued to a life aged 40 may not expire for 50 years. The rate of interest that
would be appropriate for an investment to be made over 20 years ahead could be very
diﬃcult to predict.
If an insurer does not lock into the forward rates at inception, there is a risk that interest
rates will move, resulting in premiums that are either too low or too high. The risk that
interest rates are lower than those expected in the premium calculation is an example
of nondiversiﬁable risk. Suppose an insurer has a large portfolio of whole life insurance
policies issued to lives aged 40, with level premiums payable throughout the term of the
contract, and that mortality risk can be considered diversiﬁed away. The insurer decides
to invest all premiums in 10 year bonds, reinvesting when the bonds mature. The price
of 10 year bonds at each of the future premium dates is unknown now. If the insurer
determines the premium assuming a ﬁxed interest rate of 6% per year, and the actual
interest rate earned is 5% per year, then the portfolio will make a substantial loss, and in
fact each individual contract is expected to make a loss. Writing more contracts will only
increase the loss, because each policy experiences the same interest rates. The key point
here is that the policies are not independent of each other with respect to the interest
rate risk.
Previous chapters in this book have focused on the mortality risk in insurance, which,
under the conditions discussed in Section 10.4.1 can be considered to be diversiﬁable.
However, nondiversiﬁable risk is, arguably, even more important. Most life insurance 492 CHAPTER 10. INTEREST RATE RISK company failures occur because of problems with nondiversiﬁable risk related to assets.
Note also that not all mortality risk is diversiﬁable. In Example 10.4 below, we look at
a situation where the mortality risk is not fully diversiﬁable. First, in Example 10.3 we
look at an example of nondiversiﬁable interest rate risk.
Example 10.3 An insurer issues a whole life insurance policy to (40), with level premiums payable continuously throughout the term of the policy, and with sum insured $50 000
payable immediately on death. The insurer assumes that an appropriate survival model
is given by Makeham’s law with parameters A = 0.0001, B = 0.00035 and c = 1.075.
(a) Suppose the insurer prices the policy assuming an interest rate of 5% per year
eﬀective. Show that the annual premium rate is P = $1 010.36.
(b) Now suppose that the eﬀective annual interest rate is modelled as a random variable,
denoted i, with the following distribution. 4% with probability 0.25 i=
5% with probability 0.5 6% with probability 0.25
Calculate the expected value and the standard deviation of the present value of the
future loss on the contract. Assume that the future lifetime is independent of the
interest rate. Solution 10.3 (a) At 5% we have
a40 = 14.49329
¯ and giving a premium of
P = 50 000 ¯
A40
= $1 010.36.
a40
¯ ¯
A40 = 0.29287 10.4. DIVERSIFIABLE AND NONDIVERSIFIABLE RISK 493 (b) Let S = 50 000, P = 1 010.36, T = T40 . The present value of the future loss on the
policy, L0 , is given by
L0 = S viT − P aT i .
¯
To calculate the moments of L0 we condition on the value of i and then use iterated
expectation (see Appendix A for a review of conditional expectation). As
L0 i = S viT − P aT i ,
¯ (10.6) ¯
E[L0 i] = (S A40 − P a40 )i
¯ 1 587.43
=
0 −1 071.49 with probability 0.25 (i = 4%),
with probability 0.50 (i = 5%),
with probability 0.25 (i = 6%). (10.7) (10.8) So
E[L0 ] = E [E[L0 i]] = 0.25 (1 587.43) + 0.5 (0) + 0.25 (−1 071.49) = $128.99. (10.9) For the standard deviation, we use
V[L0 ] = E[V[L0 i]] + V[E[L0 i]]. (10.10) We can interpret the ﬁrst term as the risk due to uncertainty over the future lifetime and
the second term as the risk due to the uncertain interest rate.
Now
L0 i = S viT − P aT i =
¯ S+ P
δi viT − P
δi 494 CHAPTER 10. INTEREST RATE RISK so
V[L0 i] = Hence P
S+
δi 14 6752 =
14 0142 13 3162 2
2 ¯
¯40
A40 − A2 i with probability 0.25 (i = 4%)
with probability 0.5 (i = 5%)
with probability 0.25 (i = 6%) E[V[L0 i]] = $196 364 762.
Also, from equation (10.7),
V [E[L0 i]] = (1 587.432 ) 0.25+(02 ) 0.5+(−1 071.492 ) 0.25 −128.992 = 900 371 = $948.882 . So
V[L0 ] = 196 364 762 + 900 371 = 197 265 133 = $14 0452 . (10.11) Comments
This example illustrates some important points.
• The ﬁxed assumption, 5% in this example, is what is often called the ‘best estimate’
assumption – it is the expected value, as well as the most likely value, of the future
interest rate. It is tempting to calculate the premium using the best estimate assumption, but this example illustrates that doing so may lead to systematic losses.
In this example, using a 5% per year interest assumption to price the policy leads
to an expected loss of $128.99 on every policy issued. 10.4. DIVERSIFIABLE AND NONDIVERSIFIABLE RISK 495 • Breaking the variance down into two terms separates the diversiﬁable risk from the
nondiversiﬁable risk. Consider a portfolio of, say, N contracts. Let L0,j denote the
present value of the loss at inception on the j th policy and let
N L= L0,j
j =1 so that L denotes the total future loss random variable.
Following formula (10.10), and noting that, given our assumptions at the start of this
section, the random variables {L0,j }N are independent and identically distributed,
j =1
we can write
V[L] = E[V[Li]] + V[E[Li]]
N =EV
j =1 N L0,j i +V E
j =1 L0,j i = E[N V[L0 i]] + V[N E[L0 i]]
= 196 364 762 N + 900 371 N 2 .
Now consider separately each component of the variance of L. The ﬁrst term represents diversiﬁable risk since it is a multiple of N and the second term represents
nondiversiﬁable risk since it is a multiple of N 2 . We can see that, for an individual
policy (N =1), the future lifetime uncertainty is very much more inﬂuential than the
interest rate uncertainty, as the ﬁrst term is much greater than the second term.
But, for a large portfolio, the contribution of the interest uncertainty to the total
standard deviation is far more important than the future lifetime uncertainty.
The conclusion above, that for large portfolios, interest rate uncertainty is more important
than mortality uncertainty relies on the assumption that the future survival model is 496 CHAPTER 10. INTEREST RATE RISK known and that the separate lives are independent with respect to mortality. The following
example shows that if these conditions do not hold, mortality risk can be nondiversiﬁable.
Example 10.4 A portfolio consists of N identical one year term insurance policies issued
simultaneously. Each policy was issued to a life aged 70, has a sum insured of $50 000
payable at the end of the year of death and was purchased with a single premium of
$1 300. The insurer uses an eﬀective interest rate of 5% for all calculations but is unsure
about the mortality of this group of policyholders over the term of the policies. The
probability of dying within the year, regarded as a random variable q70 , is assumed to
have the following distribution 0.022 with probability 0.25, q70 =
0.025 with probability 0.5, 0.028 with probability 0.25.
The value of q70 is the same for all policies in the portfolio and, given this value, the
policies are independent with respect to mortality.
(a) Let D(N ) denote the number of deaths during the one year term. Show that
lim N →∞ V[D(N )]
= 0.
N (b) Let L(N ) denote the present value of the loss from the whole portfolio. Show that
lim N →∞ Solution 10.4 V[L(N )]
= 0.
N
(a) We have V[D(N )] = V[E[D(N )q70 ]] + E[V[D(N )q70 ]].
Now
V[E[D(N )q70 ]] = 0.25((0.022 − 0.025)N )2 + 0 + 0.25((0.028 − 0.025)N )2 10.4. DIVERSIFIABLE AND NONDIVERSIFIABLE RISK
= 4.5 × 10−6 N 2
and
E[V[D(N )q70 ]] = 0.25 × 0.022(1 − 0.022)N
+0.5 × 0.025(1 − 0.025)N
+0.25 × 0.028(1 − 0.028)N
= 0.0243705N.
Hence
V[D(N )] = 4.5 × 10−6 N 2 + 0.0243705N
and so
lim N →∞ V[D(N )]
= 0.002121.
N (b) The arguments are as in part (a). We have
V[L(N )] = E[V[L(N )q70 ] + V[E[L(N )q70 ].
As
L(N ) = 50 000vD(N ) − 1 300N,
we have
V[L(N )q70 ] = (50 000v )2 V[D(N )q70 ]
= (50 000v )2 N q70 (1 − q70 ) 497 498 CHAPTER 10. INTEREST RATE RISK
and
E[L(N )q70 ] = 50 000vN q70 − 1 300N.
Thus
E[V[L(N )q70 ]] = (50 000v )2 N (E[q70 ] − E[q70 2 ])
= (50 000v )2 N (0.025 − 0.0006295)
and
V[E[L(N )q70 ]] = (50 000v )2 N 2 V[q70 ]
= (50 000v )2 N 2 × 4.5 × 10−6 .
Hence
lim N →∞ 10.5 V[L(N )]
= 50 000v
N V[q70 ] = 101.02. Monte Carlo simulation Suppose we wish to explore a more complex example of interest rate variation than in
Example 10.3. If the problem is too complicated, for example if we want to consider both
lifetime variation and the interest rate uncertainty, then the numerical methods used in
previous chapters may be too unwieldy. An alternative is Monte Carlo, or stochastic, simulation. Using Monte Carlo techniques allows us to explore the distributions of present 10.5. MONTE CARLO SIMULATION 499 values for highly complicated problems, by generating a random sample2 from the distribution. If the sample is large enough, we can get good estimates of the moments of the
distribution, and, even more interesting, the full picture of a loss distribution. Appendix
C gives a brief review of Monte Carlo simulation.
In this section we demonstrate the use of Monte Carlo methods to simulate future lifetimes
and future rates of interest, using a series of examples based on the following deferred
annuity policy issued to a life aged 50.
• Policy terms:
An annuity of $10 000 per year is payable continuously from age 65 contingent
on the survival of the policyholder.
Level premiums of amount P = $4 447 per year are payable continuously
throughout the period of deferment.
If the policyholder dies during the deferred period, a death beneﬁt equal to the
total premiums paid (without interest) is due immediately on death.
• Basis for all calculations:
The survival model follows Gompertz’ law with parameters B = 0.0004 and
c = 1.07.
The force of interest during deferment is δ = 5% per year.
The force of interest applying at age 65 is denoted r.
In the next three examples we will assume that r is ﬁxed and known. In the ﬁnal example
we will assume that r has a ﬁxed but unknown value.
Example 10.5 Assume the force of interest from age 65 is 6% per year, so that r = 0.06.
Technically, a deterministic (nonrandom) sample that is statistically indistinguishable from a random
sample.
2 500 CHAPTER 10. INTEREST RATE RISK (a) Calculate the expected present value of the loss on the contract.
(b) Calculate the probability that the expected present value of the loss on the policy
will be positive.
Solution 10.5 (a) The expected present value of the loss on this contract is ¯¯ 1
10 000 15 E50 a∗ − P (I A)50: 15 − P a50: 15
¯65
¯
where ∗ denotes calculation using a force of interest 6% per year and all other
functions are calculated using a force of interest 5% per year. This gives the expected
present value of the loss as
10 000 × 0.34773 × 8.51058 + 4 447 × 1.32405 − 4 447 × 9.49338 = −$6 735.38.
(b) The present value of the loss, L, can be written in terms of the expected future
lifetime, T50 , as follows
L= P T50 v T50 − P aT50
¯
∗
10 000 aT −15 v 15 − P a15
¯
¯
50 if T50 ≤ 15,
if T50 > 15. By looking at the relationship between L and T50 we can see that the policy generates
a proﬁt if the life dies in the deferred period, or in the early years of the annuity
payment period, and that
Pr[L > 0] = Pr 10 000 e−15δ aT50 −15 6% − P a15 5% > 0
¯
¯
= Pr T50 > 15 − 1
P
log 1 − 4 e15(0.05) a15 5% (0.06)
¯
0.06
10 = Pr[T50 > 30.109] = 30.109 p50 = 0.3131. 10.5. MONTE CARLO SIMULATION 501 Example 10.6 Use the three U (0, 1) random variates below to simulate values for T50
and hence values for the present value of future loss, L0 , for the deferred annuity contract.
Assume that the force of interest from age 65 is 6% per year:
u1 = 0.16025, u2 = 0.51720, u3 = 0.99855. Solution 10.6 Let FT be the distribution function of T50 . Each simulated uj generates
a simulated future lifetime tj through the inverse transform algorithm, where
uj = FT (tj ).
See Appendix C. Hence
u = FT (t)
= 1 − e−(B/ log(c))c 50 (ct −1) −
⇒ t = FT 1 (u) = 1
log(c) log 1 − log(c)(log(1 − u))
B c50 . (10.12) So
−
t1 = FT 1 (0.16025) = 10.266,
−
t2 = FT 1 (0.5172) = 24.314,
−
t3 = FT 1 (0.9985) = 53.969. These simulated lifetimes can be checked by noting in each case that tj q50 = uj .
We can convert the sample lifetimes to the corresponding sample of the present value of
future loss random variable, L0 , as follows. If (50) dies after exactly 10.266 years, then 502 CHAPTER 10. INTEREST RATE RISK death occurs during the deferred period, the death beneﬁt is 10.266P , the present value
of the premiums paid is P a10.266 , and so the present value of the future loss is
¯
L0 = 10.266 P e−10.266 δ − P a10.266 δ = −$8 383.80.
¯
Similarly, the other two simulated future lifetimes give the following losses
L0 = 10 000e−15δ a9.314 r=6% − P a15 δ = −$13 223.09,
¯
¯
L0 = 10 000e−15δ a38.969 r=6% − P a15 δ = $24 202.36.
¯
¯
The ﬁrst two simulations generate a proﬁt, and the third generates a loss.
Example 10.7 Repeat Example 10.6, generating 5 000 values of the present value of
future loss random variable. Use the simulation output to:
(a) Estimate the expected value and the standard deviation of the present value of the
future loss from a single policy.
(b) Calculate a 95% conﬁdence interval for the expected value of the present value of
the loss.
(c) Estimate the probability that the contract generates a loss.
(d) Calculate a 95% conﬁdence interval for the probability that the contract generates
a loss.
Solution 10.7 Use an appropriate random number generator to produce a sequence of
5 000 U (0, 1) random numbers, {uj }. Use equation (10.12) to generate corresponding
values of the future lifetime, {tj }, and the present value of the future loss for a life with
future lifetime ti , say {L0,j }, as in Example 10.6.
The result is a sample of 5 000 independent values of the future loss random variable. Let
¯ and sl represent the mean and standard deviation of the sample.
l 10.5. MONTE CARLO SIMULATION 503 (a) The precise answers will depend on the random number generator (and seed value)
used. Our calculations gave
¯ = −$6 592.74;
l sl = $15 733.98. (b) Let µ and σ denote the (true) mean and standard deviation of the present value of
the future loss on a single policy. Using the Central Limit Theorem, we can write
1
5 000 5 000 j =1 L0,j ∼ N (µ, σ 2 /5 000). Hence
σ
Pr µ − 1.96 √
5 000 ≤ 1
5, 000 5 000 L0,j
j =1 ≤ µ + 1.96 √ σ
= 0.95.
5 000 Since ¯ and sl are estimates of µ and σ , respectively, a 95% conﬁdence interval for
l
the mean loss is
¯ − 1.96 √ sl , ¯ + 1.96 √ sl
l
l
5 000
5 000 . Using the values of ¯ and sl from part (a) gives (−7 028.86, − 6 156.61) as a 95%
l
conﬁdence interval for µ.
(c) Let L− denote the number of simulations which produce a loss, that is, the number
for which L0,i is positive. Let p denote the (true) probability that the present value
of the loss on a single policy is positive. Then
L− ∼ B (5 000, p)
and our estimate of p, denoted p, is given by
ˆ
p=
ˆ l−
5 000 504 CHAPTER 10. INTEREST RATE RISK
where l− is the simulated realization of L− , that is, the number of losses which are
positive out of the full set of 5 000 simulated losses. Using a normal approximation,
we have
L−
p(1 − p)
∼ N p,
5 000
5 000
and so an approximate 95% conﬁdence interval for p is
p − 1.96
ˆ p(1 − p)
ˆ
ˆ
, p + 1.96
ˆ
5 000 p(1 − p)
ˆ
ˆ
5 000 where we have replaced p by its estimate p. Our calculations gave a total of 1 563
ˆ
simulations with a positive value for the expected present value of the future loss.
Hence
p = 0.3126
ˆ
and an approximate 95% conﬁdence interval for this probability is
(0.2998, 0.3254).
Diﬀerent sets of random numbers would result in diﬀerent values for each of these
quantities. In fact it was not necessary to use simulation to calculate µ or p in this example As
we have seen in Example 10.5, the values of µ and p can be calculated as −$6 735.38
and 0.3131, respectively. The 95% conﬁdence intervals calculated in Example 10.7 parts
(b) and (d) comfortably span these true values. We used simulation in this example to
illustrate the method and to show how accurate we can be with 5 000 simulations.
An advantage of Monte Carlo simulation is that we can easily adapt the simulation to
model the eﬀect of a random force of interest from age 65, which would be less tractable,
analytically. The next example demonstrates this in the case where the force of interest
from age 65 is ﬁxed but unknown. 10.5. MONTE CARLO SIMULATION 505 Example 10.8 Repeat Example 10.7, but now assuming that r is a random variable with
a N (0.06, 0.0152 ) distribution. Assume the random variables T50 and r are independent.
Solution 10.8 For each of the 5 000 simulations generate both a value for T50 , as in the
previous example, and also a value of r from the N (0.06, 0.0152 ). Let tj and rj denoted
the simulated values of T50 and r, respectively, for the j th simulation. The simulated
value of the present value of the loss for this simulation, L0,j , is
L0,j = P ti v tj − P atj
¯
10 000 a∗ −15
¯t
j v 15 − P a15
¯ if tj ≤ 15,
if tj > 15. where ∗ now denotes calculation at the simulated force of interest rj . The remaining steps
in the solution are as in Example 10.7.
Our simulation gave the following results.
¯ = −$6 220.5; sl = $16 903.1; L− = 1 502.
l
Hence, an approximate 95% conﬁdence interval for the mean loss is
(−6 689, − 5 752).
An estimated probability that a policy generates a loss is
p = 0.3004,
ˆ
with an approximate 95% conﬁdence interval for this probability of
(0.2877, 0.3131).
Note that allowing for the future interest variability has reduced the expected proﬁt and
increased the standard deviation. The probability of loss is not signiﬁcantly diﬀerent from
the ﬁxed interest case. 506 10.6 CHAPTER 10. INTEREST RATE RISK Notes and further reading The simple interest rate models we have used in this chapter are useful for illustrating
the possible impact of interest rate uncertainty, but developing more realistic interest rate
models is a major topic in its own right, beyond the scope of this text. Some models are
presented in McDonald (2006) and a comprehensive presentation of the topic is available
in Cairns (2004).
We have shown in this chapter that uncertainty in the mortality experience is a source of
nondiversiﬁable risk. This is important because improving mortality has been a feature
in many countries and the rate of improvement has been diﬃcult to predict. See, for
example, Willets et al (2004). In these circumstances, the assumptions about the survival
model in Section 10.4.1 may not be reasonable and so a signiﬁcant aspect of mortality risk
is nondiversiﬁable. Note that in Examples 10.6 to 10.8 we simulated the future lifetime
random variable T50 assuming the survival model and its parameters were known. Monte
Carlo methods could be used to model uncertainty about the survival model; for example,
by assuming that the two parameters in the Gompertz formula were unknown but could
be modelled as random variables with speciﬁed distributions.
Monte Carlo simulation is a key tool in modern risk management. A general introduction
is presented in e.g. Ross (2006), and Glasserman (2004) oﬀers a text more focussed on
ﬁnancial modelling. Algorithms for writing your own generators are given in the Numerical
Recipes reference texts, such as Press et al (2007). 10.7. EXERCISES 10.7 507 Exercises Exercise 10.1 You are given the following zero coupon bond prices:
Term, t
(years)
1
2
3
4
5 P (t) as
% of face value
94.35
89.20
84.45
79.95
75.79 (a) Calculate the annual eﬀective spot rates for t = 1, 2, 3, 4, 5.
(b) Calculate the 1 year forward rates, at t = 0, 1, 2, 3, 4.
(c) Calculate the EPV of a 5 year term life annuitydue of $1 000 per year, assuming
that the probability of survival each year is 0.99. Exercise 10.2 Consider an endowment insurance with sum insured $100 000 issued to a
life aged 45 with term 15 years under which the death beneﬁt is payable at the end of the
year of death. Premiums, which are payable annually in advance, are calculated using the
standard ultimate survival model, assuming a yield curve of eﬀective annual spot rates
given by
yt = 0.035 + √ t
.
200 (a) Show that the net premium for the contract is $4 207.77. 508 CHAPTER 10. INTEREST RATE RISK (b) Calculate the net premium determined using a ﬂat yield curve with eﬀective rate of
interest i = y15 and comment on the result.
(c) Calculate the net policy value for a policy still in force three years after issue, using
the rates implied by the original yield curve, using the premium basis.
(d) Construct a table showing the expected cash ﬂows for the policy, assuming a premium of $4 207.77. Use this table to verify the net policy value calculation in (c). Exercise 10.3 An insurer issues a portfolio of identical 5 year term insurance policies to
independent lives aged 75. One half of all the policies have a sum insured of $10 000, and
the other half have a sum insured of $100 000. The sum insured is payable immediately
on death.
The insurer wishes to measure the uncertainty in the total present value of claims in the
portfolio. The insurer uses the standard ultimate survival model, and assumes an interest
rate of 6% per year eﬀective.
(a) Calculate the standard deviation of the present value of the beneﬁt for an individual
policy, chosen at random.
(b) Calculate the standard deviation of the total present value of claims for the portfolio
assuming that 100 contracts are issued.
(c) By comparing the portfolio of 100 policies with a portfolio of 100 000 policies, demonstrate that the mortality risk is diversiﬁable. Exercise 10.4 (a) The coeﬃcient of variation for a random variable X is deﬁned as
the ratio of the standard deviation of X to the mean of X . Show that for a random 10.7. EXERCISES 509 N
variable X =
j =1 Xj , if the risk is diversiﬁable, then the limiting value of the
coeﬃcient of variation, as N → ∞, is zero. (b) An insurer issues a portfolio of identical 15 year term insurance policies to independent lives age 65. The sum insured for each policy is $100 000, payable at the end
of the year of death.
The mortality for the portfolio is assumed to follow Makeham’s law with A =
0.00022, B = 2.7 × 10−6 . The insurer is uncertain whether the parameter c for
Makeham’s mortality law is 1.124, as in the standard ultimate survival model, or
1.114. The insurer models this uncertainty assuming that there is a 75% probability
that c = 1.124 and a 25% probability that c = 1.114. Assume the same mortality
applies to each life in the portfolio. The eﬀective rate of interest is assumed to be
6% per year.
(i) Calculate the coeﬃcient of variation of the present value of the beneﬁt for an
individual policy.
(ii) Calculate the coeﬃcient of variation of the total present value of beneﬁts for
the portfolio assuming that 10 000 policies are issued.
(iii) Demonstrate that the mortality risk is not fully diversiﬁable, and ﬁnd the
limiting value of the coeﬃcient of variation. Exercise 10.5 An insurer issues a 25 year endowment insurance policy to (40), with level
premiums payable continuously throughout the term of the policy, and with sum insured
$100 000 payable immediately on death or at the end of the term. The insurer calculates
the premium assuming an interest rate of 7% per year eﬀective, and using the standard
ultimate survival model. 510 CHAPTER 10. INTEREST RATE RISK (a) Calculate the annual net premium payable.
(b) Suppose that the eﬀective annual interest rate is a random variable, i, with the
following distribution: 5% with probability 0.5, i=
7% with probability 0.25, 11% with probability 0.25. Write down the EPV of the net loss at issue on the policy using the mean interest
rate. (c) Calculate the EPV of the net loss at issue on the policy using the modal interest
rate.
(d) Calculate the EPV and the standard deviation of the present value of the future
loss on the policy. Use the premium from (a) and assume that the future lifetime is
independent of the interest rate.
(e) Comment on the results. Exercise 10.6 An insurer issues 15 year term insurance policies to lives aged 50. The
sum insured of $200 000 is payable immediately on death. Level premiums of $550 per
year are payable continuously throughout the term of the policy. The insurer assumes the
lives are subject to Gompertz’ law of mortality with B = 3 × 10−6 and c = 1.125, and
that interest rates are constant at 5% per year.
(a) Generate 1 000 simulations of the future loss.
(b) Using your simulations from (a), estimate the mean and variance of the future loss
random variable. 10.7. EXERCISES 511 (c) Calculate a 90% conﬁdence interval for the mean future loss.
(d) Calculate the true value of the mean future loss. Does it lie in your conﬁdence
interval in (c)?
(e) Repeat the 1 000 simulations 20 times. How often does the conﬁdence interval
calculated from your simulations not contain the true mean future loss?
(f) If you calculated a 90% conﬁdence interval for the mean future loss a large number
of times from 1,000 simulations, how often (as a percentage) would you expect the
conﬁdence interval not to contain the true mean?
(g) Now assume interest rates are unknown. The insurer models the interest rate on all
policies, I , as a lognormal random variable, such that
1 + I ∼ LN (0.0485, 0.02412 ).
Reestimate the 90% conﬁdence interval for the mean of the future loss random
variable, using Monte Carlo simulation. Comment on the eﬀect of interest rate
uncertainty. Exercise 10.7 An actuary is concerned about the possible eﬀect of pandemic risk on
the term insurance portfolio of her insurer. She assesses that in any year there is a 1%
probability that mortality at all ages will jump by 25%, for that year only.
(a) State, with explanation, whether pandemic risk is diversiﬁable or nondiversiﬁable.
(b) Describe how the actuary might quantify the possible impact of pandemic risk on
her portfolio. 512 CHAPTER 10. INTEREST RATE RISK Answers to selected exercises
10.1 (a) (0.05988, 0.05881, 0.05795, 0.05754, 0.05701)
(b) (0.05988, 0.05774, 0.05625, 0.05629, 0.05489)
(c) $4 395.73 10.2 (b) $4 319.50
(c) $13 548
(d) We show the ﬁrst three rows of the cash ﬂow table. Year
k →k+1
0
1
2 Expected premium
income
Pk
4207.77
4204.52
4200.99 10.3 (a) $19 784
(b) $193 054 10.4 (b) (i) 2.2337
(ii) 0.2204
(iii) 0.2192 10.5 (a) $1 608.13 Forward rate
f (k, k + 1)
1.0400
1.0441
1.0468 Expected claims
Net cash ﬂow
outgo
carried forward
Ck+1
CFk+1
77.11
4 298.97
83.88
8 795.01
91.47
13 513.33 10.7. EXERCISES 513 (b) $0
(c) $7 325.40
(d) $2 129.80, $5 629.76 10.6 (d) −$184.07
(f) 10% of sets of simulated values should generate a 90% conﬁdence interval that
does not contain the true mean.
(g) Term insurance is not very sensitive to interest rate uncertainty, as the standard
deviation of outcomes with interest rate uncertainty is very similar to that
without interest rate uncertainty. 514 CHAPTER 10. INTEREST RATE RISK Chapter 11
Emerging Costs for Traditional Life
Insurance
11.1 Summary In this chapter we introduce emerging costs, or cash ﬂow analysis for traditional life
insurance contracts. This is often called proﬁt testing when applied to life insurance.
Traditional actuarial analysis focusses on determining the EPV of a cash ﬂow series, usually under a constant interest rate assumption. This emphasis on the EPV was important
in an era of manual computation, but with powerful computers available we can do better. Using cash ow projections to model risk oﬀers much more ﬂexibility than the EPV
approach and provides actuaries with a better understanding of the liabilities under their
management and the relationship between the liabilities and the corresponding assets.
We introduce proﬁt testing in two stages. First we consider only those cash ﬂows generated
by the policy, then we introduce reserves to complete the cash ﬂow analysis.
We deﬁne several measures of the proﬁtability of a contract: internal rate of return,
expected present value of future proﬁt (net present value), proﬁt margin and discounted
515 516 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE payback period. We show how cash ﬂow analysis can be used to set premiums to meet a
given measure of proﬁt.
We restrict our attention in this chapter to deterministic proﬁt tests, and introduce
stochastic proﬁt tests in Chapter 12. 11.2 Proﬁt testing for traditional life insurance 11.2.1 The net cash ﬂows for a policy We introduce proﬁt testing by studying in detail a 10 year term insurance issued to a life
aged 60. The details of the policy are as follows. The sum insured, denoted S , is $100 000,
payable at the end of the year of death. Level annual premiums, denoted P , of amount
$1 500 are payable throughout the term.
We want to analyze the cash ﬂows from this policy at discrete intervals throughout its
term. It would be very common to choose one month as the interval since in practice
premiums are often paid monthly. However, to illustrate more clearly the mechanics of
proﬁt testing, we use a time interval of one year for this example, taking time 0 to be the
moment when the policy is issued.
The purpose of a proﬁt test is to identify the proﬁt which the insurer can claim from
the contract at the end of each time period, in this case at the end of each year. To do
this, the insurer needs to make assumptions about the expenses which will be incurred,
the survival model for the policyholder, the rate of interest to be earned on cash ﬂows
within each time period before the proﬁt is released and possibly other items such as
an assessment of the probability that the policyholder surrenders the policy. For ease of
presentation, we ignore the possibility of lapsing in this example. The set of assumptions
used in the proﬁt test is called the proﬁt test basis.
For this example, we use the following proﬁt test basis. 11.2. PROFIT TESTING FOR TRADITIONAL LIFE INSURANCE
Interest:
Initial Expenses:
Renewal Expenses:
Survival model: 517 5.5% per year eﬀective on all cash ﬂows.
$400 plus 20% of the ﬁrst premium.
3.5% of premiums.
q60+t = 0.01 + 0.001 t for t = 0, 1, . . . , 9. The initial expenses represent the acquisition costs for the policy. These are paid by
the insurer when, or even just before, the policy is issued, that is, at time t = 0. For each
year that the policy is still in force, cash ﬂows contributing to the surplus emerging at
the end of that year are the premium less any renewal expense, interest earned on this
amount and the expected cost of a claim at the end of the year. The calculations of the
emerging surplus, called the net cash ﬂows for the policy, are summarized in Table 11.1.
For time t = 0 the only entry is the total initial expenses for the policy, $(400 + 0.2 P ).
These expenses are assumed to occur and be paid at time 0, so no interest accrues on
them.
For the ﬁrst policy year there is a premium payable at time 0, but no expenses since these
are included in the row for t = 0. Interest is earned at 5.5% and the expected death
claims, payable at time 1, are q60 S = 0.01 × 100 000 = 1 000. Hence the emerging surplus,
or net cash ﬂow, at time 1 is
1 500 + 82.5 − 1 000 = 582.5.
For subsequent policy years, the net cash ﬂows are calculated assuming the policy is still
in force at the start of the year. For example, the net cash ﬂow at time 7 is calculated as
1 500 − 0.035 × 1 500+0.055 × (1 500 − 0.035 × 1 500) − 100 000 × (0.01+6 × 0.001) = −72.89.
In Table 11.1, E0 denotes the initial expenses incurred at time 0 and for t = 1, 2, . . . , 10,
Et denotes the renewal expenses incurred at the start of the year from t − 1 to t. 518 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE Time
t
0
1
2
3
4
5
6
7
8
9
10 Premium
at t − 1
1 500
1 500
1 500
1 500
1 500
1 500
1 500
1 500
1 500
1 500 Expenses Interest
Expected
Et
death claims
700.00
0.00
82.50
1 000
52.50
79.61
1 100
52.50
79.61
1 200
52.50
79.61
1 300
52.50
79.61
1 400
52.50
79.61
1 500
52.50
79.61
1 600
52.50
79.61
1 700
52.50
79.61
1 800
52.50
79.61
1 900 Surplus
emerging at t
−700.00
582.50
427.11
327.11
227.11
127.11
27.11
−72.89
−172.89
−272.89
−372.89 Table 11.1: Net cash ﬂows for the 10 year term insurance in Section 11.2. 11.2. PROFIT TESTING FOR TRADITIONAL LIFE INSURANCE 11.2.2 519 Reserves Table 11.1 reveals a typical feature of net cash ﬂows: several of the net cash ﬂows in
later years are negative. This occurs because the level premium is more than suﬃcient
to pay the renewal expenses and expected death claims in the early years, but, with an
increasing probability of death, is not suﬃcient in the later years. The expected cash ﬂow
values in the ﬁnal column of Table 11.1 have been calculated in the same way and show
the same general features as the values illustrated in Figures 6.1 and 6.2.
In Chapter 7 we explained why the insurer needed to set aside assets to cover negative
expected future cash ﬂows. The policy values that we calculated in that chapter represented the amount that would, in expectation be suﬃcient with the future premiums to
meet future beneﬁts. In modelling cash ﬂows, we use reserves rather than policy values.
The reserve is the actual amount of money held by the insurer to meet future liabilities.
The reserve may be equal to the policy value, or may be some diﬀerent amount. It should
not be less than the policy value, but may be greater than the policy value to allow for
uncertainty or adverse experience. Usually, though, for traditional insurance, the policy
value calculation will be used to set reserves, perhaps using a conservative basis. Note
that the negative cash ﬂow at time 0 in Table 11.1 does not require a reserve since it will
have been paid as soon as the policy was issued.
The amount of the reserves is determined by a process separate from the proﬁt test and is
based on a set of assumptions, the reserve basis, which may be diﬀerent from the proﬁt
test basis. In practice the reserve basis is likely to be more conservative than the proﬁt
test basis.
Suppose that the insurer sets reserves at the start of each year for this policy equal to the
net premium policy values on the following (reserve) basis.
Interest:
Survival model: 4% per year eﬀective on all cash ﬂows.
q60+t = 0.011 + 0.001 t for t = 0, 1, . . . , 9. 520 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE Then the reserve required at the start of the (t + 1)th year, i.e. at time t, is
1
100 000 A60+t:10−t − P a60+t:10−t
¨ where the net premium, P , is calculated as
P = 100 000 1
A60:10 a60:10
¨ = $1 447.63, and all functions are calculated using the reserve basis. The values for the reserves are
shown in Table 11.2.
t
0
1
2
3
4 tV 0.00
410.05
740.88
988.90
1 150.10 t
5
6
7
8
9 tV 1 219.94
1 193.37
1 064.74
827.76
475.45 Table 11.2: Reserves for the 10 year term insurance in Section 11.2. The reserves shown in Table 11.2 are amounts that the insurer needs to assign from its
assets to support the policy. We need to include in our proﬁt test the cost of assigning
these amounts. To see how to do this, consider, for example, the reserve required at
time 1, 1 V = 410.05. This amount is required for every policy still in force at time 1. The
cost to the insurer of setting up this reserve is assigned to the previous time period and
this cost is
1V p60 = 410.05 × (1 − 0.01) = 405.95. The cost includes the factor p60 since all costs relating to the previous time period are per
policy in force at the start of that time period, that is, at time 0. The expected proportion
of policyholders surviving to the start of the following time period, i.e. to age 61, is p60 . 11.2. PROFIT TESTING FOR TRADITIONAL LIFE INSURANCE 521 Note that p60 is evaluated on the proﬁt test basis. In general, the cost at the end of the
year from t − 1 to t of setting up a reserve of amount t V at time t for each policy still in
force at time t is t V p60+t−1 .
The proﬁt test calculations, including reserves, are set out in Table 11.3. Here It denotes
the interest earned in the year from t − 1 to t, E0 denotes the initial expenses incurred
at time 0 and for t = 1, 2, . . . , 10, Et denotes the renewal expenses incurred at the start
of the year from t − 1 to t. Note that the cost of setting up a reserve, t V p60+t−1 is a
cost to the insurer at the end of the year from t − 1 to t, whereas the reserve t V is a
positive asset at the start of the following year. Hence, for example, the calculation of
the proﬁt emerging at the end of the seventh year per policy in force at the start of the
year, denoted Pr7 , is
Pr7 = P + 6 V − E7 + i(P + 6 V − E7 ) − Sq66 − 7 V p66
= 1 500 + 1 193.37 − 0.035 × 1 500 + 0.055(0.965 × 1 500 + 1 193.37)
− 100 000 × 0.016 − 1 064.74 × 0.984
= $138.41.
For t = 1, 2, . . . , 10, the calculation of Prt in Table 11.3 is given by
Prt = ( t−1 V + P − Et )(1 + i) − Sq60+t−1 − t V p60+t−1 .
Many actuaries prefer to write this in the equivalent form
Prt = (P − Et )(1 + i) + ∆ t V − Sq60+t−1 ,
where ∆ t V is called the change in reserve in year t and is deﬁned as
∆ t V = (1 + i) t−1 V − t V p60+t−1 . 522 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE t
0
1
2
3
4
5
6
7
8
9
10 t−1 V P 0.00
410.05
740.88
988.90
1 150.10
1 219.94
1 193.37
1 064.74
827.76
475.45 1 500
1 500
1 500
1 500
1 500
1 500
1 500
1 500
1 500
1 500 Et
700.0
0.0
52.50
52.50
52.50
52.50
52.50
52.50
52.50
52.50
52.50 It
82.50
102.17
120.36
134.00
142.87
146.71
145.25
138.17
125.14
105.76 S q60+t−1
1 000
1 100
1 200
1 300
1 400
1 500
1 600
1 700
1 800
1 900 tV p60+t−1 405.59
732.73
977.04
1 135.15
1 202.86
1 175.47
1 047.70
813.69
466.89
0.00 Prt
−700.00
176.55
126.99
131.70
135.26
137.61
138.68
138.41
136.72
133.52
128.71 Table 11.3: Emerging surplus, per policy in force at start of year, for the 10 year term
insurance in Section 11.2.
This alternative approach reﬂects the diﬀerence between the reserves and the other cash
ﬂows. The incoming and outgoing reserves each year are not real income and outgo in
the same way as premiums, claims and expenses, but accounting transfers.
The vector Pr = (Pr0 , . . . , Pr10 ) is called the proﬁt vector for the contract. The elements
of Pr are the expected proﬁt at the end of each year given that the policy is in force at
the start of the year. Multiplying Prt by t p60 gives a vector each of whose elements is the
expected proﬁt at the end of each year given only that the contract was in force at age
60. With this in mind, we deﬁne
Π0 = Pr0 ; Πt = t−1 p60 Prt for t = 1, 2, . . . , 10. (11.1) The vector Π, where
Π = (Π0 , Π1 , . . . , Π10 ) = (Pr0 , Pr1 , 1 p60 Pr2 , 2 p60 Pr3 , . . . , 9 p60 Pr10 ) (11.2) 11.3. PROFIT MEASURES 523 is called the proﬁt signature for the contract. The proﬁt signature is the key to assessing
the proﬁtability of the contract. For this example, the proﬁt signature is
(−700, 176.55, 125.72, 128.96, 130.84, 131.39, 130.56, 128.35, 124.76, 119.75, 113.37) . 11.3 Proﬁt measures Once we have projected the cash ﬂows, we need to assess whether the emerging proﬁt is
adequate. There are a number of ways to measure proﬁt, all based on the proﬁt signature.
The internal rate of return (IRR) is the interest rate j such that the present value of
the expected cash ﬂows is zero. Given a proﬁt signature (Π0 , Π1 , . . . , Πn ) for an n year
contract, the internal rate of return is j where
n
t
Πt vj = 0. (11.3) t=0 For the example in Section 11.2, the internal rate of return is j = 14.24%.
The insurer may set a minimum hurdle rate or risk discount rate for the internal rate
of return, so that the contract is deemed adequately proﬁtable if the IRR exceeds the
hurdle rate.
One problem with the internal rate of return is that there may be no real solution to
equation (11.3), or there may be several. However, we can still use the risk discount rate
to calculate the expected present value of future proﬁt (EPVFP), also called the
net present value (NPV) of the contract. Let r be the risk discount rate. Then the
NPV is the present value, at rate r, of the projected proﬁt signature cash ﬂows, so that
n
t
Πt vr . NPV =
t=0 For the example in Section 11.2, suppose the insurer uses a risk discount rate of 10% per
year. Then the NPV of the contract is $124.48. 524 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE The proﬁt margin is the NPV expressed as a proportion of the EPV of the premiums,
evaluated at the risk discount rate. For a contract with level premiums of P per year
payable mthly throughout an n year contract issued to a life aged x, the proﬁt margin is Proﬁt Margin = NPV
(m) P ax : n
¨ (11.4) using the risk discount rate for all calculations.
For the example in Section 11.2, the proﬁt margin using a risk discount rate of 10% is
NPV
124.48
=
= 1.29%.
P a60:10
¨
9 684
Another proﬁt measure is the NPV as a proportion of the acquisition costs. For the
example in Section 11.2, the acquisition costs are $700, so the NPV is 17.8% of the total
acquisition costs.
Our ﬁnal proﬁt measure is the discounted payback period (DPP), also known as the
breakeven period. This is calculated using the risk discount rate, r, and is the smallest
value of m such that
m t=0 t
Πt vr ≥ 0. The DPP represents the time until the insurer starts to make a proﬁt on the contract.
For the example in Section 11.2, the DPP is 8 years.
None of these measures of proﬁt explicitly takes into consideration the risk associated
with the contract; most of the inputs we have used in the emerging surplus calculation
are in practice uncertain, for example we do not usually know what interest rates and
mortality rates will be. If the experience is adverse, the proﬁt will be smaller, or there
could be signiﬁcant losses.
These measures of proﬁtability can be used to calculate a premium. For example, suppose
the insurer requires a proﬁt margin of 10% for the term insurance studied in Section 11.2. 11.4. A FURTHER EXAMPLE OF A PROFIT TEST 525 The premium would have to increase $1 663.45, which gives the revised proﬁt signature
Π=
(−732.69, 348.99, 290.46, 291.88, 291.82, 290.27, 287.21, 282.65, 276.59, 269.01, 259.94) .
This gives an internal rate of return of 40.4% per year and the following values for measures
of proﬁtability using a risk discount rate of 10%: NPV = $1 073.97, proﬁt margin = 10%,
NPV as a percentage of acquisition costs = 146.6%, DPP = 3 years.
It is interesting to see how the reserve basis aﬀects the proﬁtability of the contract.
Suppose that in our example the insurer uses an interest rate of 3% rather than 4% to
calculate reserves. This will have the eﬀect of increasing the size of the reserves required,
so that, for example 3 V = 1 001.94 and 7 V = 1 065.13 rather than 988.90 and 1 064.74,
respectively. The NPV, using an annual premium of $1 500, decreases from $124.48 to
$122.88. On the other hand, weakening the reserve basis by using an interest rate of
5% gives a higher NPV of $126.11. By increasing the size of the reserves, the insurer is
being required to assign more of its assets to the policy. These assets are assumed to
earn interest at the rate assumed in the proﬁt test basis, 5.5% per year in our example.
This is lower than the risk discount rate, 10% in our example, at which cash ﬂows are
discounted. The intuition is that the reserve is assumed to be invested conservatively,
so higher reserves means tying up more assets in conservative investments, reducing the
proﬁtability. 11.4 A further example of a proﬁt test The term insurance example used throughout Section 11.2 was useful in terms of introducing proﬁt testing concepts. The policy itself was relatively uncomplicated – term
insurance, level annual premiums, sum insured payable at the end of the year of death –
and we assessed its proﬁtability assuming no allowance for withdrawals and by calculating
cash ﬂows at annual intervals. The following example is based on a more complicated 526 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE policy structure, involving disability beneﬁt, monthly premiums – and is more realistic
as it allows for withdrawals and calculates cash ﬂows at monthly intervals. However, the
basic principles are unchanged.
Example 11.1 A special 10 year endowment insurance is issued to a healthy life aged
55. The beneﬁts under the policy are
• $50 000 if at the end of a month the life is disabled, having been healthy at the start
of the month,
• $100 000 if at the end of a month the life is dead, having been healthy at the start
of the month,
• $50 000 if at the end of a month the life is dead, having been disabled at the start
of the month,
• $50 000 if the life survives as healthy to the end of the term.
On withdrawal at any time, a surrender value equal to 80% of the net premium policy
value is paid, and level monthly premiums are payable throughout the term while the life
is healthy.
The survival model used for proﬁt testing is shown in Figure 11.1. The transition intensities µ01 , µ02 , µ03 and µ12 are constant for all ages x with values per year as follows:
x
x
x
x
µ01 = 0.01,
x µ02 = 0.015,
x µ03 = 0.01,
x µ12 = 0.03.
x Other elements of the proﬁt testing basis are as follows.
• Interest: 7% per year.
• Expenses: 5% of each gross premium, including the ﬁrst, together with an additional
initial expense of $1 000. 11.4. A FURTHER EXAMPLE OF A PROFIT TEST 527 • The beneﬁt on withdrawal is payable at the end of the month of withdrawal and
is equal to 80% of the sum of the reserve held at the start of the month and the
premium paid at the start of the month.
• Reserves are set equal to the net premium policy values.
• The gross premium and net premium policy values are calculated using the same
survival model as for proﬁt testing except that withdrawals are ignored, so that
µ03 = 0 for all x.
x
• The net premium policy values are calculated using an interest rate of 5% per year.
The monthly gross premium is calculated using the equivalence principle on the following
basis:
Interest: 5.25% per year.
Expenses: 5% of each premium, including the ﬁrst, together with an additional
initial expense of $1 000. (a) Calculate the monthly premium on the net premium policy value basis.
(b) Calculate the reserves at the start of each month for both healthy lives and for
disabled lives.
(c) Calculate the monthly gross premium.
(d) Project the emerging surplus using the proﬁt testing basis.
(e) Calculate the internal rate of return.
(f) Calculate the NPV, the proﬁt margin (using the EPV of gross premiums), the NPV
as a percentage of the acquisition costs, and the discounted payback period for the
contract, in all cases using a risk discount rate of 15% per year. 528 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE Healthy 0
r rr
r c Withdrawn 3 E rr
r
r
r
j Disabled 1 c Dead 2 Figure 11.1: Multiple state model for Example 11.1. Before solving this example, we remark that in practice it would be very unlikely that a
policyholder would withdraw late into the term of a policy such as this one. However, for
ease of presentation we have assumed in our survival model that withdrawal is possible
at any time within the policy term. This assumption simpliﬁes the formulae we require
to conduct the proﬁt test. Solution 11.1 We have a survival model, as shown in Figure 11.1, with two parameterizations, one for proﬁt testing and one for the calculation of the gross premium and
the reserves. The diﬀerence between the parameterizations is that the former allows for
withdrawals, whereas the latter does not.
Let t pij ∗ denote the probability that a life in state i at age x will be in state j at age x + t
x
given the parameterizations allowing for withdrawals and let t pij denote the corresponding
x
probability when withdrawals are not included. The following probabilities are useful in
our calculations. 11.4. A FURTHER EXAMPLE OF A PROFIT TEST
For 0 ≤ t ≤ 10,
00∗
t p55 t = exp − (µ01 s + µ02 s + µ03 s ) ds
55+
55+
55+ 0 = exp{−0.035t}, t
00
= exp −
t px 0 (µ01 s + µ02 s ) ds
x+
x+ 11∗
t px = exp{−0.03t} = t p11 ,
x 01∗
t p55 = = exp{−0.025t}, t
0 00∗
s p55 µ01 s t−s p11∗ s ds
55+
55+ t exp{−0.035s} 0.01 exp{−0.03(t − s)} ds =
0 = 2(exp{−0.03t} − exp{−0.035t}),
t
01
t p55 = 00
s p55 0 µ01 s t−s p11 s ds
55+
55+ t =
0 exp{−0.025s} 0.01 exp{−0.03(t − s)} ds = 2(exp{−0.025t} − exp{−0.03t}),
12∗
t px = 1 − exp{−0.03t} = t p12 .
x Further, for 55 ≤ x ≤ 64 11 ,
12
1
12 p12∗ = 1 − exp{−0.03/12} =
x 1
12 p12 ,
x 529 530 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE 1
12 1
12 1
12 1
12 5
(1 − exp{−0.035/12}) − 2(exp{−0.03/12} − exp{−0.035/12}),
7 p02∗ =
x p02 = 1 − 3 exp{−0.025/12} + 2 exp{−0.03/12},
x
2
(1 − exp{−0.035/12}),
7 p03∗ =
x p01∗ = 2(exp{−0.03/12} − exp{−0.035/12}),
x 1
12 p01 = 2(exp{−0.025/12} − exp{−0.03/12}).
x (a) Let P denote the monthly premium calculated on the net premium policy value
basis, so that withdrawals are ignored. Then
119 119
t P t
12 p00 v 12 = 50 000 10 p00 v 10 + 50 000
55
55 t=0 t
12 p00
55 1
12 p01 t +
55+ t=0
119 +100 000 t
12 t=0 p00
55 1
12 p02 t v
55+
12 t+1
12 . Using the formulae we have developed, we can calculate that
119
t t
12 p00 v 12 = 85.13
55 t
12 p01 v 12 = 3.65
55 t=0 and
119
t t=0 giving P = $452.00. 12 t
12 p01
55 1
12 p12 t
55+ 12 v t+1
12 11.4. A FURTHER EXAMPLE OF A PROFIT TEST 531 (b) Let t V (0) and t V (1) denote the net premium policy values at policy duration t years
given that the policyholder is healthy and disabled, respectively. If t is an exact
number of months, then the policy value is calculated before payment of a premium
and after payment of any beneﬁts due at that time.
Then
10 V (0) = 10 V (1) = 0 1
11
and we can calculate the policy values recursively for t = 9 12 , 9 10 , . . . , 12 , 0 from
12
these starting values using the formulae
1 (t V (0) + P ) 1.05 12 = 1
12 p00 t
55+ 1
t+ 12 V (0) + 1 p01 t (50 000 + t+ 1 V (1) ) + 100 000
55+
12 12 1
12 p02 t
55+ and
tV (1) 1 1.05 12 = 1
12 p11 t
55+ 1
t+ 12 V (1) + 50 000 1
12 p12 t .
55+ Policy values for a selection of durations are shown in Table 11.4.
t years
0 tV (0) tV (1) 0.00
−
279.32 10 301.49
560.40 10 244.19
...
...
15 237.52 7 234.67
15 613.44 7 157.17
15 991.75 7 079.16 1
12
2
12 ...
3 11
12
4
1
4 12 t years
7 11
12
8
1
8 12
...
9 10
12
9 11
12
10 tV (0) tV (1) 36 252.19 2 876.14
36 761.39 2 769.93
37 273.82 2 663.02
...
...
48 818.44
247.86
49 407.35
124.34
0.00
0.00 Table 11.4: Net premium policy values for Example 11.1. (c) Let P denote the monthly gross premium. Then, using the equivalence principle,
119 119
00
tp
12 55 0.95 P
t=0 v t
12 = 50 000 10 p00
55 v 10 + 50 000 t
12 t=0 p00
55 1
12 p01 t +
55+
12 t
12 p01
55 1
12 p12 t
55+ 12 v t+1
12 532 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE
119 +100 000 t
12 t=0 p00
55 1
12 p02 t v
55+
12 t+1
12 + 1 000 where the rate of interest is now 5.25% per year. Hence, we now have
119
t t
12 p00 v 12 = 84.26
55 t
12 p01 v 12 = 3.59,
55 t=0 and
119
t t=0 giving P = $484.27.
(d) The emerging surplus at the end of each month is calculating in two parts: ﬁrst we
assume the life is healthy at the start of the month and then we assume the life is
disabled at the start of the month. Parts of the calculation are shown in Tables 11.5
and 11.6.
The key to the columns in Table 11.5 is as follows.
1
(1) denotes the time interval from t − 12 to t, measured in years, except that t = 0
denotes time 0. (2) denotes the reserve held at the start of the time interval for a life who is healthy
1
at that time, t− 1 V (0) . No reserve is required for t = 0 or t = 12 since 0 V (0) = 0.
12 (3) denotes the gross premium, $484.27, payable at the start of the month.
(4) denotes the expenses payable in the time interval. The entry for t = 0 includes
all the initial expenses, so that
1 024.21 = 1 000 + 0.05 × P = 1 000 + 0.05 × 484.27.
1
The entry for t = 12 is zero since all the initial expenses have been assigned to
the row for t = 0. For other rows the expense is 0.05P , which is assumed to
be incurred at the start of the month. 11.4. A FURTHER EXAMPLE OF A PROFIT TEST
t
years
(1)
0
1
12
2
12 ...
9 10
12
9 11
12
10 1
t− 12 V (0) P Et (2) (3) 0.00
279.32
...
48 233.24
48 818.44
49 407.35 484.27
484.27
...
484.27
484.27
484.27 533 It (4)
1 024.21
0.00
24.21
...
24.21
24.21
24.21 Dis. Dis. Death With’l Healthy (5) ben.
(6) res.
(7) ben.
(8) ben.
(9) res.
(10) 2.74
4.18
...
275.32
278.63
281.96 41.55
41.55
...
41.55
41.55
41.55 8.56
8.51
...
0.21
0.10
0.00 124.92
124.92
...
124.92
124.92
124.92 0.32
0.51
...
32.43
32.82
33.21 (11)
−1 024.21
278.50
33.15
558.77
9.29
...
...
48 676.26
93.24
49 263.46
94.27
49 854.38
95.30 Table 11.5: Emerging surplus for Example 11.1 assuming the life is healthy at the start
of the month.
(5) denotes the interest earned during the month at the assumed rate of 7% per
year eﬀective, so that
1 It = (1.07 12 − 1)(t− 1 V (0) + P − Et ).
12 (6) is the expected disability beneﬁt payable at the end of the month,
50 000 1 p01∗ t , in respect of a life who was healthy at the start of the month
55+
12
but disabled at the end of the month.
(7) is the expected cost of setting up the required reserve at the end of the month
for a life who was healthy at the start of the month but disabled at the end of
the month. This expected cost is 1 p01∗ t t V (1) .
55+
12 (8) is the expected death beneﬁt payable at the end of the month for a life who
was healthy at the start of the month. This expected cost is 100 000 1 p02∗ t .
55+
12 (9) is the expected cost of the withdrawal beneﬁt. This is
1
12 p03∗ t
55+ 1
t− 12 V (0) + P × 0.8. (0) Prt 534 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE
(10) denotes the expected cost of setting up the reserve required at the start of the
following month for a life who remains healthy throughout the month. This
expected cost is 1 p00∗ t t V (0) .
55+
12 (11) denotes the expected surplus emerging at the end of the month in respect of a
policyholder who was healthy at the start of the month, so that
(0) Prt = (2) + (3) – (4) + (5) – (6) – (7) – (8) – (9) – (10).
t
years
(1)
1
12
2
12
3
12 ...
9 10
12
9 11
12
10 1
t− 12 V (1) (2)
0.00
10 031.49
10 244.19
...
370.58
247.86
124.34 It (1) Death Disabled Prt ben.
(3)
(4)
0.00
0.00
58.25 124.84
57.92 124.84
...
...
2.10 124.84
1.40 124.84
0.70 124.84 res.
(5)
0.00
10 218.61
10 161.08
...
247.24
124.03
0.00 (6)
0.00
16.28
16.19
...
0.59
0.39
0.20 Table 11.6: Emerging surplus for Example 11.1 assuming the life is disabled at the start
of the month.
The key to the columns in Table 11.6 is as follows.
(1) denotes the time interval from t − 1
12 to t, measured in years. (2) denotes the reserve held at the start of the time interval for a life who is
disabled at that time, t− 1 V (1) . No cash ﬂows are included for the ﬁrst month,
12
1
corresponding to t = 12 , since the life is healthy when the policy is issued.
(3) denotes the interest earned during the month, at the rate of 7% per year
eﬀective, on the reserve held at the beginning of the month. 11.4. A FURTHER EXAMPLE OF A PROFIT TEST 535 (4) is the expected death beneﬁt payable at the end of the month for a life who
was disabled at the start of the month. This expected cost is 50 000 1 p12∗ t .
55+
12 (5) denotes the expected cost of setting up the reserve required at the start of the
following month for a life who remains disabled throughout the month. This
expected cost is 1 p11∗ t t V (1) .
55+
12 (6) denotes the expected surplus emerging at the end of the month in respect of a
policyholder who was disabled at the start of the month, so that
(1) Prt = (2) + (3) – (4) – (5).
(e) The proﬁt signature vector, Π = (Π0 , Π 1 , . . . , Π9 11 , Π10 ) , is calculated as
12 12 (0) Π0 = Pr0
and for t = 12
, , . . . , 10,
12 12
(0) Πt = Prt 00∗
1
t− 12 p55 (0) 00∗
1
t− 12 p55 , Values of Prt ,
Table 11.7. (1) + Prt
(1) Prt , 01∗
1
t− 12 p55 . 01∗
1
t− 12 p55 and Πt for selected values of t are shown in The internal rate of return is the rate of interest, r, per year such that
120
k Π k (1 + r)− 12 = 0.
12 k=0 This gives an internal rate of return of 32.7% per year.
(f) The net present value, NPV, is given by
120
k Π k (1 + 0.15)− 12 = $992.29. NPV = 12 k=0 536 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE
(0) t years
0
1
12
2
12 ...
3 11
12
4
1
4 12
...
7 11
12
8
1
8 12
...
9 10
12
9 10
12
10 Prt −1 024.21
33.15
9.29
...
34.82
35.48
36.13
...
71.38
72.27
73.16
...
93.24
94.27
95.30 (1) 00∗
1
t− 12 p55 Prt −
0.00
16.28
...
11.55
11.43
11.31
...
4.71
4.54
4.38
...
0.59
0.39
0.20 1.00
1.00
0.9971
...
0.8744
0.8719
0.8694
...
0.7602
0.7580
0.7558
...
0.7109
0.7088
0.7067 01∗
1
t− 12 p55 Πt 0.00 −1 024.21
0.00
33.15
0.0008
9.27
...
...
0.0338
30.84
0.0345
31.33
0.0351
31.81
...
...
0.0607
54.55
0.0612
55.06
0.0617
55.56
...
...
0.0710
66.33
0.0714
66.85
0.0719
67.37 Table 11.7: Calculation of the proﬁt signature for Example 11.1.
The proﬁt margin, i.e. the NPV as a percentage of the EPV of gross premiums, is
119
k NPV/ P k
12 p00∗ (1 + 0.15)− 12
55 = 3.84%, k=0 and the NPV as a percentage of the acquisition costs is
NPV/ (0.05 P + 1 000) = 97.0%.
The discounted payback period is m/12 years, where m is the smallest integer such
that
m
k Π k (1 + r)− 12 ≥ 0.
12 k=0 11.5. NOTES AND FURTHER READING 537 This gives a discounted payback period of 5 years and 5 months. 11.5 Notes and further reading In Example 11.1 we used three diﬀerent bases in our calculations: the reserve basis, the
premium basis and the proﬁt test basis. This is a common feature of proﬁt tests in
practice. The reserve basis is usually a little more conservative than the premium basis
and the proﬁt test basis is the most realistic, incorporating the current best estimates of
each factor: withdrawal rates, interest rates and so on. In Example 11.1 the reserve basis
is more conservative than the premium basis since the reserves are greater than the gross
premium policy values at all durations for both healthy and disabled lives.
For each of the policies considered in this chapter, beneﬁts are payable at the end of a
time period. However, in practice, beneﬁts are usually payable on, or shortly after, the
occurrence of a speciﬁed event. For example, for the term insurance policy considered
in Section 11.2, the death beneﬁt is payable at the end of the year of death. If, instead,
the death beneﬁt had been payable immediately on death, then we could allow for this
in our proﬁt test by assuming all deaths occurred in the middle of the year. Taking this
approach, the expected death claims in Table 11.1 would all be adjusted by multiplying
by a factor of 1.0551/2 .
Throughout this chapter we have used deterministic assumptions for all the factors. By
doing this we gain no insight into the eﬀect of uncertainty on the results. In Chapter 12
we describe how we might use stochastic scenarios for emerging cost analysis for equity
linked contracts. Stochastic scenarios can also be used for traditional insurance. 538 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE 11.6 Exercises Exercise 11.1 A 5 year policy with annual cash ﬂows issued to a life (x) produces the
proﬁt vector
Pr = (−360.98, 149.66, 14.75, 273.19, 388.04, 403.00),
where Pr0 is the proﬁt at time 0 and Prt (t = 1, 2, . . . , 5) is the proﬁt at time t per policy
in force at time t − 1.
The survival model used in the proﬁt test is given by qx+t = 0.0085 + 0.0005t.
(a) Calculate the proﬁt signature for this policy.
(b) Calculate the NPV for this policy using a risk discount rate of 10% per year.
(c) Calculate the NPV for this policy using a risk discount rate of 15% per year.
(d) Comment brieﬂy on the diﬀerence between your answers to parts (b) and (c).
(e) Calculate the IRR for this policy. Exercise 11.2 A 10 year term insurance issued to a life aged 55 has sum insured $200 000
payable immediately on death and monthly premiums of $100 payable throughout the
term. Initial expenses are $500 plus 50% of the ﬁrst monthly premium; renewal expenses
are 5% of each monthly premium after the ﬁrst. The insurer earns interest at 6% per year
on all cash ﬂows and assumes the policyholder is subject to Makeham’s law of mortality
with parameters A = 0.00022, B = 2.7 × 10−6 and c = 1.124.
Calculate the proﬁt vector at monthly intervals for this policy, assuming deaths occur at
the midpoint of each month. 11.6. EXERCISES 539 Exercise 11.3 An insurer issues a 4 year term insurance contract to a life aged 60. The
sum insured, $100 000, is payable at the end of the year of death. The gross premium for
the contract is $1 100 per year. The reserve at each year end is 30% of the gross premium.
The company uses the following assumptions to assess the proﬁtability of the contract:
Mortality
Interest:
Initial expense:
Renewal expenses:
Claim expenses:
Lapses: q60 = 0.008, q61 = 0.009, q62 = 0.010, q63 = 0.012
8% eﬀective per year
30% of the ﬁrst gross premium
2% of each gross premium after the ﬁrst
$60
None (a) Calculate the proﬁt vector for the contract.
(b) Calculate the proﬁt signature for the contract.
(c) Calculate the net present value of the contract using a risk discount rate of 12% per
year.
(d) Calculate the proﬁt margin for the contract using a risk discount rate of 12% per
year.
(e) Calculate the discounted payback period using a risk discount rate of 12% per year.
(f) Determine whether the internal rate of return for the contract exceeds 50% per year.
(g) If the insurer has a ‘hurdle rate’ of 15% per year, is this contract satisfactory? Exercise 11.4 A life oﬃce issues a 20 year endowment insurance policy to a life aged
exactly 55. The sum insured is $100 000, payable at the end of the year of death or on 540 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE survival to age 75. Premiums are payable annually in advance for at most 10 years. The
oﬃce assumes that initial expenses will be $300, and renewal expenses, which are incurred
at the beginning of the second and subsequent years in which a premium is payable, will
be 2.5% of the gross premium. The funds invested for the policy are expected to earn
interest at 7.5% per year. The oﬃce holds net premium reserves, using an interest rate
of 6% per year. The survival model used to calculate the premium and the net premium
reserves follows Makeham’s Law with parameters A = 0.00022, B = 2.7 × 10−6 and
c = 1.124.
The oﬃce sets premiums so that the proﬁt margin on the contract is 15%, using a risk
discount rate of 12% per year.
Calculate the gross annual premium. Exercise 11.5 Repeat Exercise 2 assuming that the sum insured is paid immediately on
death, premiums are payable monthly for at most 10 years and expenses are $300 initially
and then 2.5% of each monthly premium after the ﬁrst. Exercise 11.6 A life insurance company issues a special 10 year term insurance policy to
two lives aged exactly 50 at the issue date, in return for the payment of a single premium.
The following beneﬁts are payable under the contract.
• In the event of either of the lives dying within 10 years, a sum insured of $100 000
is payable at the year end.
• In the event of the second death within 10 years, a further sum insured of $200 000
is payable at the year end. (If both lives die within 10 years and in the same year,
a total of $300 000 is paid at the end of the year of death.) 11.6. EXERCISES 541 The basis for the calculation of the premium and the reserves is as follows.
Survival model: Assume the two lives are independent with respect to survival and the
model for each follows Makeham’s Law with parameters A = 0.00022, B = 2.7 × 10−6
and c = 1.124.
Interest:
4% per year
Expenses: 3% of the single premium at the start of each year that the contract is in
force.
(a) Calculate the single premium using the equivalence principle.
(b) Calculate the reserves on the premium basis assuming that
(i) only one life is alive, and
(ii) both lives are still alive.
(c) Using the premium and reserves calculated, determine the proﬁt signature for the
contract assuming:
Survival model: As for the premium basis
Interest: 8% per year
Expenses: 1.5% of the premium at issue, increasing at 4% per year Exercise 11.7 A life insurance company issues a reversionary annuity policy to a husband and wife, both of whom are aged exactly 60. The annuity commences at the end of
the year of death of the wife and is payable subsequently while the husband is alive, for
a maximum period of 20 years after the commencement date of the policy. The annuity
is payable annually at $10 000 per year. The premium for the policy is payable annually
while the wife and husband are both alive and for a maximum of ﬁve years.
The basis for calculating the premium and reserves is as follows.
Survival model: Assume the two lives are independent with respect to survival and the 542 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE model for each follows Makeham’s Law with parameters A = 0.00022, B = 2.7 × 10−6
and c = 1.124.
Interest: 4% per year
Expenses: Initial expense of $300 and an expense of 2% of each annuity payment whenever
an annuity payment is made.
(a) Calculate the annual premium.
(b) Calculate the NPV for the policy assuming:
a risk discount rate of 15% per year,
expenses and the survival model are as in the premium basis, and
interest is earned at 6% per year on cash ﬂows. Exercise 11.8 A life aged 60 purchases a deferred life annuity, with a ﬁve year deferred
period. At age 65 the annuity vests, with payments of $20 000 per year at each year
end, so that the ﬁrst payment is on the 66th birthday. All payments are contingent on
survival. The policy is purchased with a single premium.
If the policyholder dies before the ﬁrst annuity payment, the insurer returns her gross
premium, with interest of 5% per year, at the end of the year of her death.
(a) Calculate the single premium using the following premium basis:
Survival model: µx = 0.9(0.00022 + 2.7 × 10−6 × 1.124x ) for all x
Interest: 6% per year before vesting; 5% per year thereafter
Expenses: $275 at issue plus $20 with each annuity payment
(b) Gross premium reserves are calculated using the premium basis. Calculate the year
end reserves (after the annuity payment) for each year of the contract. 11.6. EXERCISES 543 (c) The insurer conducts a proﬁt test of the contract assuming the following basis:
Survival model: µx = 0.00022 + 2.7 × 10−6 × 1.124x for all x
Interest: 8% per year before vesting; 6% per year thereafter
Expenses: $275 at issue plus $20 with each annuity payment
(i) Calculate the proﬁt signature for the contract.
(ii) Calculate the proﬁt margin for the contract using a risk discount rate of 10%
per year. 544 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE Answers to selected exercises
1. (a) (−360.98, 149.66, 14.62, 268.43, 377.66, 388.29)
(b) $487.88
(c) $365.69
(e) 42.7%
2. Selected values are Pr30 = 54.53 and Pr84 = 28.75, measuring time in months
3. (a) (−300.00, 60.16, 295.23, 195.50, 322.08)
(b) (−300.00, 60.16, 292.87, 192.19, 313.46)
(c) $323.19
(d) 8.7%
(e) 3 years
(f) 48%
(g) Yes
4. $4 553.75
5. $394.27 (per month)
6. (a) $4 180.35
(b) Selected values are (i) 4 V = $3 126.04, and (ii) 4 V = $3 146.06
(c) Selected values are Π0 = −$62.71, Π5 = $177.35 and Π10 = $63.42
7. (a) $1 832.79
(b) $779.26
8. (a) $192 805.84 11.6. EXERCISES
(b) Selected values are 4 V = $243 148.51 and 545
10 V = $226 245.94 (c) (i) Selected values are Π4 = $4 538.90 and Π10 = $2 429.55
(ii) 14.8% 546 CHAPTER 11. EMERGING COSTS FOR TRADITIONAL LIFE INSURANCE Chapter 12
Emerging Costs for EquityLinked
Insurance
12.1 Summary In this chapter we introduce equitylinked insurance contracts. We explore deterministic emerging costs techniques with examples, and demonstrate that deterministic proﬁt
testing cannot adequately model these contracts.
We introduce stochastic cash ﬂow analysis, which gives a fuller picture of the characteristics of the equitylinked cash ﬂows, particularly when guarantees are present, and we
demonstrate how stochastic cash ﬂow analysis can be used to determine better contract
design.
Finally we discuss the use of quantile and conditional tail expectation reserves for equitylinked insurance.
547 548 12.2 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE Equitylinked insurance In Chapter 1 we described some modern insurance contracts where the main purpose
of the contract is investment. These contracts include some life contingent guarantees,
predominantly as a way of distinguishing them from pure investment products.
These contracts are called unitlinked insurance in the UK and parts of Europe, variable annuities in the USA (though there is often no actual annuity component) and
segregated funds in Canada. All fall under the generic title of equitylinked insurance. The basic premise of these contracts is that a policyholder pays a single or regular
premium which, after deducting expenses, is invested on the policyholder’s behalf. The accumulating premiums form the policyholder’s fund. Regular management charges
are deducted from the fund by the insurer and paid into the insurer’s fund to cover
expenses and insurance charges.
On survival to the end of the contract term the beneﬁt may be just the policyholder’s fund
and no more, or there may be a guaranteed minimum maturity beneﬁt (GMMB).
On death during the term of the policy, the policyholder’s estate would receive the policyholder’s fund, possibly with an extra amount – for example, a death beneﬁt of 110% of the
policyholder’s fund means an additional payment of 10% of the policyholder’s fund at the
time of death. There may also be a guaranteed minimum death beneﬁt (GMDB).
Some conventions and jargon have developed around these contracts, particularly in the
UK where the policyholder is deemed to buy units in an underlying asset fund (hence ‘unitlinked’). One example is the bidoﬀer spread. If a contract is sold with a bidoﬀer spread
of, say, 5%, only 95% of the premium paid is actually invested in the policyholder’s fund;
the remainder goes to the insurer’s fund. There may also be an allocation percentage;
if 101% of the premium is allocated to units at the oﬀer price, and there is a 5% bidoﬀer
spread, then 101% of 95% of the premium (that is 95.95%) goes to the policyholder’s
fund and the rest goes to the insurer’s fund. The bidoﬀer spread mirrors the practice in
unitised investment funds that are major competitors for policyholders’ investments. 12.3. DETERMINISTIC PROFIT TESTING 12.3 549 Deterministic proﬁt testing for equitylinked insurance Equitylinked insurance policies are usually analyzed using emerging surplus techniques.
The cash ﬂows can be separated into those that are in the policyholder’s fund and those
that are income or outgo for the insurer. It is the insurer’s cash ﬂows that are important
in pricing and reserving, but since the insurer’s income and outgo depend on how much is
in the policyholder’s fund, we must ﬁrst project the cash ﬂows for the policyholder’s fund
and then use these to project the cash ﬂows for the insurer’s fund. The projected cash
ﬂows for the insurer’s fund can then be used to calculate the proﬁtability of the contract
using the proﬁt vector, proﬁt signature, and perhaps the NPV, IRR, proﬁt margin and
discounted payback period, in the same way as in Chapter 11.
The following two examples illustrate these calculations.
Example 12.1 A 10 year equitylinked contract is issued to a life aged 55 with the
following terms.
The policyholder pays an annual premium of $5 000. The insurer deducts a 5%
expense allowance from the ﬁrst premium and a 1% allowance from subsequent
premiums. These amounts, known as unallocated premiums, are paid into the
insurer’s fund; the remaining amounts, the allocated premiums, are paid into the
policyholder’s fund.
At the end of each year a management charge of 0.75% of the policyholder’s fund
is transferred from the policyholder’s fund to the insurer’s fund.
If the policyholder dies during the contract term, a beneﬁt of 110% of the value
of the policyholder’s year end fund (after management charge deductions) is paid
at the end of the year of death. This is paid partly from the policyholder’s fund
(100%) and partly from the insurer’s fund (10%). 550 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
If the policyholder surrenders the contract, he receives the value of the policyholder’s
fund at the year end, after management charge deductions. This does not result in
a cost to the insurer’s fund.
If the policyholder holds the contract to the maturity date, he receives the greater
of the value of the policyholder’s fund and the total of the premiums paid. This is
a GMMB which results in a cost to the insurer’s fund if the total of the allocated
premiums exceeds the value of the policyholder’s fund. (a) Assume the policyholder’s fund earns interest at 9% per year. Project the year end
fund values for a contract that remains in force for 10 years.
(b) Calculate the proﬁt vector for the contract using the following basis.
Survival model: The probability of dying in any year is 0.005.
Lapses: 10% of lives surrender in the ﬁrst year of the contract, 5% in the second year
and none in subsequent years. All surrenders occur at the end of a year immediately
after the management charge deduction.
Initial expenses: 10% of the ﬁrst premium plus $150.
Renewal expenses: 0.5% of the second and subsequent premiums.
Interest: The insurer’s funds earn interest at 6% per year.
Reserves: The insurer holds no reserves for the contract.
(c) Calculate the proﬁt signature for the contract.
(d) Calculate the NPV using a risk discount rate of 15% per year eﬀective.
Solution 12.1 (a) The projection of the policyholder’s fund is shown in Table 12.1.
The key to the columns of Table 12.1 is as follows.
(1) The entries for t are the years of the contract, from time t − 1 to time t.
(2) This shows the allocated premium invested in the policyholder’s fund at time
t − 1. 12.3. DETERMINISTIC PROFIT TESTING 551 (3) This shows the fund brought forward from the previous year end.
(4) This shows the interest income on the combined premium and fund brought
forward at the rate assumed for the policyholder’s fund, 9% per year.
(5) This is the premium plus fund brought forward plus interest, and shows the
amount in the policyholder’s fund at the year end, just before the annual
management charge is deducted.
(6) This shows the management charge, at 0.75% of the previous column.
(7) This shows the remaining fund, which is carried forward to the next year.
t Allocated
premium
(1)
(2)
1
4750
2
4950
3
4950
4
4950
5
4950
6
4750
7
4950
8
4950
9
4950
10
4950 Fund
b/f
(3)
0.00
5 138.67
10 914.17
17 162.26
23 921.60
31 234.01
39 144.77
47 702.83
56 961.14
66 977.02 Interest
(4)
427.50
907.98
1 427.78
1 990.10
2 598.44
3 256.56
3 968.53
4 738.75
5 572.00
6 473.43 Fund
at t−
(5)
5 177.50
10 996.65
17 291.95
24 102.36
31 470.04
39 440.58
48 063.30
57 391.58
67 483.15
78 400.45 Management
charge
(6)
38.83
82.47
129.69
180.77
236.03
295.80
360.47
430.44
506.12
588.00 Fund
c/f
(7)
5 138.67
10 914.17
17 162.26
23 921.60
31 234.01
39 144.77
47 702.83
56 961.14
66 977.02
77 812.45 Table 12.1: Projection of policyholder’s fund for Example 12.1. (b) The emerging surplus is shown in Table 12.2.
The key to the information in Table 12.2 is as follows. 552 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
t Unallocated
premium
(1)
(2)
0
0.00
1
250.00
2
50.00
3
50.00
4
50.00
5
50.00
6
50.00
7
50.00
8
50.00
9
50.00
10
50.00 Expenses
(3)
650.00
0.00
25.00
25.00
25.00
25.00
25.00
25.00
25.00
25.00
25.00 Interest
(4)
0.00
15.00
1.50
1.50
1.50
1.50
1.50
1.50
1.50
1.50
1.50 Management
Expected
Prt
charge
Death Beneﬁt
(5)
(6)
(7)
0.00
0.00
−650.00
38.83
2.57
301.26
82.47
5.46
103.52
129.69
8.58
147.61
180.77
11.96
195.31
236.03
15.62
246.91
295.80
19.57
302.73
360.47
23.85
363.12
430.44
28.48
428.46
506.12
33.49
499.14
588.00
38.91
575.60 Table 12.2: Emerging surplus for Example 12.1. (1) The entries for t are the years of the contract, from time t − 1 to time t, except
for t = 0 which represents the issue date.
(2) This shows the unallocated premium.
(3) This shows the insurer’s expenses at the start of the year. All expenses incurred
at time 0 are shown in the entry for t = 0 and are not included in the entries for
t = 1. This is consistent with our calculation of emerging costs for traditional
policies in Chapter 11.
(4) This shows the interest earned in the year on the unallocated premium received
less expenses paid at the start of the year at the rate assumed for the insurer’s
fund, 6% per year.
(5) This shows the management charge, which is taken directly from Table 12.1. 12.3. DETERMINISTIC PROFIT TESTING
t
0
1
2
3
4
5 Probability
Πt
in force
1.00000
−650.00
1.00000
301.26
0.89550
92.70
0.84647
124.95
0.84224
164.50
0.83803
206.92 553
t Probability
in force
6
0.83384
7
0.82967
8
0.82552
9
0.82139
10
0.81729 Πt
252.43
301.27
353.70
409.99
470.43 Table 12.3: Calculation of the proﬁt signature for Example 12.1. (6) This shows the expected death beneﬁt, which is 10% of the year end fund
value from Table 12.1, multiplied by the mortality probability. We need only
10% of the fund as the rest is paid from the policyholder’s fund – in the
insurer’s cash ﬂows we consider only income and outgo that are not covered
by the policyholder’s fund.
(7) Prt is the proﬁt emerging at time t and is calculated as
Prt = unallocated premium − expenses + interest
+ management charge − expected death beneﬁt.
Note that there is no projected cost in Table 12.2 for the GMMB as the ﬁnal
projected fund value, $77 812.45, is greater than the guarantee, 10 × 5 0 = $50 000.
(c) For the proﬁt signature we multiply the tth element of the proﬁt vector, Prt , by
the probability that the contract is still in force at the start of the year for t =
1, 2, . . . , 10. (For t = 0, the required probability is 1.) The values are shown in
Table 12.3. 554 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE (d) The NPV is calculated by discounting the proﬁt signature at the risk discount rate
of interest, r = 15%, so that
10 Πt (1 + r)−t = $531.98. NPV =
t=0 Example 12.2 The terms of a 5 year equitylinked insurance policy issued to a life aged
60 are as follows.
The policyholder pays a single premium of $10 000, of which 3% is taken by the insurer for expenses and the remainder, the allocated premium, is invested in suitable
assets.
At the start of the second and subsequent months, a management charge of 0.06%
of the policyholder’s fund is transferred to the insurer’s fund.
If the policyholder dies during the term, the policy pays out 101% of all the money
in her fund. In addition, the insurer guarantees a minimum beneﬁt. The guaranteed
minimum death beneﬁt in the tth year is 10 000 (1.05t−1 ), where t = 1, 2, . . . , 5.
If the policyholder surrenders the contract during the ﬁrst year, she receives 90%
of the money in the policyholder’s fund. In the second year a surrendered contract
pays 95% of the policyholder’s fund. If the policyholder surrenders the contract
after the second policy anniversary, she receives 100% of the policyholder’s fund.
If the policyholder holds the contract to the maturity date, she receives the money
in the policyholder’s fund with a guarantee that the payout will not be less than
$10 000.
The insurer assesses the proﬁtability of the contract by projecting cash ﬂows on a monthly
basis using the following assumptions. 12.3. DETERMINISTIC PROFIT TESTING 555 Survival model: The force of mortality is constant for all ages and equal to 0.006 per year.
Death beneﬁt: This is paid at the end of the month in which death occurs.
Lapses: Policies are surrendered only at the end of a month. The probability of surrendering at the end of any particular month is 0.004 in the ﬁrst year, 0.002 in the second
year and 0.001 in each subsequent year.
Interest: The policyholder’s fund earns interest at 8% per year eﬀective. The insurer’s
fund earns interest at 5% per year eﬀective.
Initial expenses: 1% of the single premium plus $150.
Renewal expenses: 0.008% of the single premium plus 0.01% of the policyholder’s funds
at the end of the previous month. Renewal expenses are payable at the start of each
month after the ﬁrst.
(a) Calculate the probabilities that a policy in force at the start of a month is still in
force at the start of the next month.
(b) Construct a table showing the projected policyholder’s fund assuming the policy
remains in force throughout the term.
(c) Construct a table showing the projected insurer’s fund.
(d) Calculate the NPV for the contract using a risk discount rate of 12% per year.
Solution 12.2 (a) The probability of not dying in any month is exp{−0.006/12} = 0.9995.
Hence, allowing for lapses, the probability that a policy in force at the start of a
month is still in force at the start of the following month is
(1 − 0.004) exp{−0.006/12} = 0.9955 in the ﬁrst year, (1 − 0.002) exp{−0.006/12} = 0.9975 in the second year, 556 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
(1 − 0.001) exp{−0.006/12} = 0.9985 in subsequent years. (b) Table 12.4 shows the projected policyholder’s fund at selected durations assuming
the policy remains in force throughout the 5 years. Note that in this example the
management charge is deducted at the start of the month rather than the end. The
minimum death beneﬁt is also given in the table – in the ﬁrst year this is the full
premium and it increases by 5% at the start of each year.
The entries for time t in Table 12.4 show the cash ﬂows for the policyholder’s fund
1
in the month from time t − 12 to time t.
(c) The projected cash ﬂows for the insurer’s fund are shown in Table 12.5. Consider,
3
for example, the entries for t = 12 . These are the cash ﬂows for the time period
starting 2 months after the issue of the policy.
Since the policy is purchased with a single premium payable at the start of the ﬁrst
month, there is no premium paid by the policyholder, and hence no unallocated
premium paid into the insurer’s fund, in this period.
The amount of the management charge, $5.89, is taken directly from Table 12.4 and
is paid into the insurer’s fund at the start of the period.
The expenses, $1.78, are calculated as 0.00008 × 10 000 + 0.0001 × 9 819.33 and are
paid from the insurer’s fund at the start of the month.
The interest is calculated as (1.051/12 − 1)(5.89 − 1.78) = $0.02.
The basic death beneﬁt, payable at the end of the month, is 101% of the policyholder’s fund at the end of the month. The insurer’s fund has to pay the extra 1%,
so the expected cost, $0.05, is (1 − exp{−0.006/12}) × 0.01 × 9 876.57. However,
there is a guaranteed minimum death beneﬁt of $10 000, which, for this month is
greater than 101% of the policyholder’s fund at the end of the month; the expected 12.3. DETERMINISTIC PROFIT TESTING
t 2
1
2 12
.
.
. 0
0
.
.
. 3
1
3 12
.
.
. 0
0
.
.
. 4
1
4 12
.
.
. 0
0
.
.
. Fund
Management Interest
Fund
Minimum
b/f
charge
c/f
DB
0.00
0.00
62.41 9 762.41 10 000.00
9 762.41
5.86
62.77 9 819.33 10 000.00
9 819.33
5.89
63.14 9 876.57 10 000.00
9 876.57
5.93
63.51 9 934.16 10 000.00
9 934.16
5.96
63.88 9 992.07 10 000.00
9 992.07
6.00
64.25 10 050.33 10 000.00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
10 346.74
6.21
66.53 10 407.07 10 000.00
10 407.07
6.24
66.92 10 467.74 10 500.00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
11 094.29
6.66
71.34 11 158.97 10 500.00
11 158.97
6.70
71.75 11 224.03 11 025.00
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
11 895.85
7.14
76.49 11 965.20 11 025.00
11 965.20
7.18
76.94 12 034.96 11 576.25
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
12 755.32
7.65
82.02 12 829.68 11 576.25
12 829.68
7.70
82.50 12 904.48 12 155.06
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. 5 0 13 676.89 1
12
2
12
3
12
4
12
5
12
6
12 .
.
.
1
1
1 12
.
.
. Allocated
premium
9 700
0
0
0
0
0
.
.
. 557 0
0
.
.
. 8.21 87.94 13 756.62 12 155.06 Table 12.4: Deterministic projection of the policyholder’s fund for Example 12.2. extra cost, $0.03, is paid by the insurer’s fund at the end of the month and is calculated as (1 − exp{−0.006/12}) × max(0, 10 000 − 1.01 × 9 876.57). The expected
cost of this GMDB is zero after three months since, using the assumptions in the 558 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
proﬁt testing basis, 101% of the policyholder’s fund is greater than the minimum
death beneﬁt thereafter.
Lapses in the ﬁrst two years are a source of income for the insurer’s fund since, on
surrendering her policy, the policyholder receives less than the full amount of the
policyholder’s fund. The expected income from lapses at the end of 3 months, $3.95,
is calculated as exp{−0.006/12} × 0.004 × 0.1 × 9 876.57.
The expected proﬁt at the end of the month per policy in force at the start of the
month, Prt , is calculated as
Prt = Unallocated premium + Management charge − Expenses
+ Interest − 1% DB − Minimum DB + Lapses. (d) Table 12.6 shows for selected durations the expected proﬁt at the end of each month
per policy in force at the start of the tth month (Prt ), the probability that the policy
is in force at the start of the month (given only that it was in force at time 0) and
the proﬁt signature, Πt , which is the product of these two elements.
The net present value for this policy is calculated by summing the elements of the
proﬁt signature discounted to time 0 at the risk discount rate, r. Hence
60
k NPV = Π k (1 + r) 12 = $302.42.
12 k=0 In both the examples in this section, the beneﬁt involved a guarantee. In the ﬁrst example
the guarantee had no eﬀect at all on the calculations, and in the second the eﬀect was
negligible. This does not mean that the guarantees are costfree. In practice, even though
the policyholder’s fund may earn on average a return of 9% or more, the return could 12.3. DETERMINISTIC PROFIT TESTING
t 559 Unallocated
premium
0
300
0
0
0
0
0
.
.
. Management
charge
0.00
0.00
5.86
5.89
5.93
5.96
6.00
.
.
. Expenses Interest 1% DB 250.00
0.80
1.78
1.78
1.79
1.79
1.80
.
.
. 0.00
1.22
0.02
0.02
0.02
0.02
0.02
.
.
. 0
0
.
.
. 6.21
6.24
.
.
. 1.83
1.84
.
.
. 2
1
2 12
.
.
. 0
0
.
.
. 6.66
6.70
.
.
. 3
1
3 12
.
.
. 0
0
.
.
. 4
1
4 12
.
.
.
5 0
1
12
2
12
3
12
4
12
5
12
6
12 .
.
.
1
1
1 12
.
.
. Lapses Prt 0.00
0.05
0.05
0.05
0.05
0.05
0.05
.
.
. Minimum
DB
0.00
0.07
0.04
0.01
0.00
0.00
0.00
.
.
. 0.00
3.90
3.93
3.95
3.97
3.99
4.02
.
.
. −250.00
304.20
7.93
8.01
8.08
8.13
8.18
.
.
. 0.02
0.02
.
.
. 0.05
0.05
.
.
. 0.00
0.00
.
.
. 4.16
1.05
.
.
. 8.50
5.42
.
.
. 1.91
1.92
.
.
. 0.02
0.02
.
.
. 0.06
0.06
.
.
. 0.00
0.00
.
.
. 1.12
0.00
.
.
. 5.83
4.74
.
.
. 7.14
7.18
.
.
. 1.99
2.00
.
.
. 0.02
0.02
.
.
. 0.06
0.06
.
.
. 0.00
0.00
.
.
. 0.00
0.00
.
.
. 5.11
5.14
.
.
. 0
0
.
.
. 7.65
7.70
.
.
. 2.08
2.08
.
.
. 0.02
0.02
.
.
. 0.06
0.06
.
.
. 0.00
0.00
.
.
. 0.00
0.00
.
.
. 5.54
5.57
.
.
. 0 8.21 2.17 0.02 0.00 0.00 0.00 5.99 Table 12.5: Deterministic projection of the insurer’s fund for Example 12.2. be very volatile. A few years of poor returns could generate a signiﬁcant cost for the
guarantee. We can explore the sensitivity of the emerging proﬁt to adverse scenarios by 560 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
t Prt 0 −250.00
304.20
7.93
8.01
8.08
8.13
8.18
.
.
. 1
12
2
12
3
12
4
12
5
12
6
12 .
.
.
1
1
1 12
.
.
. 8.50
5.42
.
.
. Probability
Πt
in force
1.0000
250.00
1.0000
304.20
0.9955
7.90
0.9910
7.94
0.9866
7.97
0.9821
7.98
0.9777
8.00
.
.
.
.
.
.
0.9516
8.09
0.9492
5.14
.
.
.
.
.
. 2
1
2 12
.
.
. 5.83
4.74
.
.
. 0.9235
0.9221
.
.
. 5.38
4.37
.
.
. 3
1
3 12
.
.
. 5.11
5.14
.
.
. 0.9070
0.9056
.
.
. 4.63
4.66
.
.
. 4
1
4 12
.
.
.
5 5.54
5.57
.
.
.
5.99 0.8908
0.8895
.
.
.
0.8749 4.93
4.96
.
.
.
5.24 Table 12.6: Calculation of the proﬁt signature for Example 12.2. using stress testing.
In Example 12.1 there is a GMMB – the ﬁnal payout is guaranteed to be at least the 12.4. STOCHASTIC PROFIT TESTING 561 total amount invested, $50 000. Assume as an adverse scenario that the return on the
policyholder’s fund is only 5% rather than 9%. The result is that the GMMB still has
no eﬀect, and the NPV changes from $531.98 to $417.45. We must reduce the return
assumption to 1% or lower for the guarantee to have any cost. However, under the
deterministic model there is no way to turn this analysis into a price for the guarantee.
Furthermore, the deterministic approach does not reﬂect the potentially huge uncertainty
involved in the income and outgo for equitylinked insurance. The insurer’s cash ﬂows
depend on the policyholder’s fund, and the policyholder’s fund depends on market conditions.
The deterministic proﬁt tests described in this section can be quite misleading. The investment risks in equitylinked insurance cannot be treated deterministically. It is crucial
that the uncertainty is properly taken into consideration for adequate pricing, reserving
and risk management. In the next section we develop the methodology introduced in this
section to allow appropriately for uncertainty. 12.4 Stochastic proﬁt testing For traditional insurance policies we often assume that the demographic uncertainty dominates the investment uncertainty – which may be a reasonable assumption if the underlying assets are invested in low risk ﬁxed interest securities of appropriate duration. As
discussed in Chapter 10, the demographic uncertainty can be related to the size of the
portfolio and can often be assumed to be diversiﬁed away. The uncertainty involved in
equitylinked insurance is very diﬀerent. The mortality element is assumed diversiﬁable
and is not the major factor. The uncertainty in the investment performance is a far more
important element, and it is not diversiﬁable. Selling 1 000 equitylinked contracts with
GMMBs to identical lives is almost the same as issuing one big contract; when one policyholder’s fund dips in value, then all dip, increasing the chance that the GMMB will
cost the insurer money for every contract. 562 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE Using a deterministic proﬁt test does not reﬂect the reality of the situation adequately in
most cases. The EPV of future proﬁt – expected in terms of demographic uncertainty only
– does not contain any information about the uncertainty from investment returns. The
proﬁt measure for an equitylinked contract is modelled more appropriately as a random
variable rather than a single number. This is achieved by stochastic proﬁt testing.
The good news is that we have done much of the work for stochastic proﬁt testing in the
deterministic proﬁt testing of the previous section. The diﬀerence is that in the earlier
section we assumed deterministic interest and demographic scenarios. In this section
we replace the deterministic investment scenarios with stochastic scenarios. The most
common practical way to do this is with Monte Carlo simulation, which we introduced in
Section 10.5, and used already for this purpose with interest rates in Chapter 10.
Using Monte Carlo simulation, we generate a large number of outcomes for the investment
return on the policyholder’s fund. The simulated returns are used in place of the constant
investment return assumption in the deterministic case. The proﬁt test proceeds exactly as
described in the deterministic approach, except that we repeat the test for each simulated
investment return outcome, so we generate a random sample of outcomes for the contract,
which we can use to determine the probability distribution for each proﬁt measure for a
contract.
Typically, the policyholder’s fund may be invested in a mixed fund of equities or equities
and bonds. The policyholder may have a choice of funds available, involving greater or
lesser amounts of uncertainty.
A very common assumption for returns on equity portfolios is the independent lognormal
assumption. This assumption, which is very important in ﬁnancial modelling, can be expressed as follows. Let R1 , R2 , . . . be a sequence of random variables, where Rt represents
the accumulation at time t of a unit amount invested in an equity fund at time t − 1, so
that Rt − 1 is the rate of interest earned in the year. These random variables are assumed
to be mutually independent, and each Rt is assumed to have a lognormal distribution (see
2
Appendix A). Note that if Rt has a lognormal distribution with parameters µt and σt , 12.4. STOCHASTIC PROFIT TESTING 563 then
2
log Rt ∼ N (µt , σt ). Hence, values for Rt can be simulated by simulating values for log Rt and exponentiating.
We now demonstrate stochastic proﬁt testing for equitylinked insurance by considering
further the 10 year policy discussed in Example 12.1. In the discussion of Example 12.1
in Section 12.3 we assumed a rate of return of 9% per year on the policyholder’s fund.
This resulted in a zero cost for the GMMB. We now assume that the accumulation factor
for the policyholder’s fund over the tth policy year is Rt , where the sequence {Rt }10
t=1
satisﬁes the independent lognormal assumption. To simplify our presentation we further
assume that these random variables are identically distributed, with Rt ∼ LN (µ, σ 2 ),
where µ = 0.074928 and σ 2 = 0.152 . Note that the expected accumulation factor each
year is
E [Rt ] = eµ+σ 2 /2 = 1.09, which is the same as under the deterministic assumption in Section 12.3.
Table 12.7 shows the results of a single simulation of the investment returns on the
policyholder’s fund for the policy in Example 12.1.
The values in column (2), labelled z1 , ..., z10 , are simulated values from a N (0, 1) distribution. These values are converted to simulated values from the speciﬁed lognormal
distribution using rt = exp{0.074928 + 0.15zt }, giving the annual accumulation factors
shown in column (3). The values {rt }10 are a single simulation of the random variables
t=1
10
{Rt }t=1 . These simulated annual accumulation factors should be compared with the value
1.09 used in the calculation of Table 12.1. The values in columns (4) and (5) are calculated
in the same way as those in columns (6) and (7) in Table 12.1, using the annual interest
rate rt − 1 in place of 0.09. Note that in some years, for example the second policy year,
the accumulation factor for the policyholder’s fund is less than one. The values in column
(6) are calculated in the same way as those in column (7) in Table 12.2 except that there 564 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
t Simulated Simulated
zt
rt
(1)
(2)
(3)
0
1
0.95518
1.24384
2
−2.45007
0.74633
3
−1.23376
0.89571
4
0.55824
1.17194
5
−0.62022
0.98206
6
0.01353
1.08000
7
−1.22754
0.89655
8
0.07758
1.09042
9
−0.61893
0.98225
10 −0.25283
1.03770 Management
charge
(4)
44.31
60.53
87.07
144.78
177.57
230.44
238.33
298.41
327.38
375.70 Fund c/f
(5)
5 863.94
8 010.27
11 521.61
19 159.03
23 498.89
30 494.26
31 539.16
39 490.18
43 323.89
49 717.95 Prt Πt (6)
(7)
−650.00 −650.00
306.38
306.38
83.03
74.35
107.80
90.80
161.70
135.51
192.32
160.37
241.69
200.52
249.06
205.61
305.17
250.66
332.22
271.52
96.71
78.64 Table 12.7: A single simulation of the proﬁt test. is an extra deduction in the calculation of Pr10 of amount
p54 max(50 000 − F10 , 0)
where F10 denotes the ﬁnal fund value. This deduction was not needed in our calculations
in Section 12.3 since, with the deterministic interest assumption, the ﬁnal fund value,
$77 812.45, was greater than the GMMB. For this simulation, F10 is less than the GMMB
so there is a deduction of amount
0.995 × (50 000 − 49 717.95) = $280.64.
The values for Πt are calculated by multiplying the corresponding value of Prt by the
probability of the policy being force, as shown in Table 12.3. The values for Prt and Πt
shown in Table 12.7 should be compared with the corresponding values in Tables 12.2 12.4. STOCHASTIC PROFIT TESTING 565 and 12.3, respectively. Using a risk discount rate of 15% per year, the NPV using this
single simulation of the investment returns on the policyholder’s fund is $232.09.
To measure the eﬀect of the uncertainty in rates of return, we generate a large number, N ,
of sets of rates of return and for each set carry out a proﬁt test as above. Let NPVi denote
the net present value calculated from the ith simulation, for i = 1, 2, . . . , N . Then the net
present value for the policy, NPV, is being modelled as a random variable and {NPVi }N
i=1
is a set of N independent values sampled from the distribution of NPV. From this sample
we can estimate the mean, standard deviation and percentiles of this distribution. We
can also count the number of simulations for which NPVi is negative, denoted N − , and
the number of simulations, denoted N ∗ , for which the ﬁnal fund value is greater than
$50 000, so that there is no liability for the GMMB.
Let m and s be the estimates of the mean and standard deviation of NPV. Since N is
large, we can appeal to the Central Limit Theorem to say that a 95% conﬁdence interval
(CI) for E[NPV] is given by
s
s
m − 1.96 √ , m + 1.96 √
N
N . It is important whenever reporting summary results from a stochastic simulation to give
some measure of the variability of the results, such as a standard deviation or a conﬁdence
interval.
Calculations by the authors using N = 1 000 gave the results shown in Table 12.8. To
calculate the median and the percentiles we arrange the simulated values of NPV in
000
ascending or descending order. Let {NPV(i) }1=1 denote the simulated values for NPV
i
arranged in ascending order. Then the median is estimated as (NPV(500) + NPV(501) )/2,
so that 50% of the observations lie above the estimated median, and 50% lie above.
This would be true for any value lying between NPV(500) and NPV(501) , and taking the
mid point is a conventional approach. Similarly the 5th percentile value is estimated as
(NPV(50) + NPV(51) )/2 and the 95th percentile is estimated as (NPV(950) + NPV(951) )/2.
The results in Table 12.8 put a very diﬀerent light on the proﬁtability of the contract. 566 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
E[NPV]
SD[NPV]
95% CI for E[NPV]
5th percentile
Median of NPV
95th percentile
N−
N∗ 380.91
600.61
(343.28, 417.74)
−859.82
498.07
831.51
87
897 Table 12.8: Results from 1 000 simulations of the net present value. Under the deterministic analysis, the proﬁt test showed no liability for the guaranteed
minimum maturity beneﬁt, and the contract appeared to be proﬁtable overall – the net
present value was $531.98. Under the stochastic analysis, the GMMB plays a very important role. The value of N ∗ shows that in most cases the GMMB liability is zero and
so it does not aﬀect the median. However, it does have a signiﬁcant eﬀect on the mean,
which is considerably lower than the median. From the 5th percentile ﬁgure, we see that
very large losses are possible; from the 95th percentile we see that there is somewhat less
upside potential with this policy. Note also that an estimate of the probability that the
net present value is negative, calculated using a risk discount rate of 15% per year, is
N − /N = 0.087, indicating a probability of around 9% that this apparently proﬁtable contract actually
makes a loss.
This proﬁt test reveals what we are really doing with the deterministic test, which is,
approximately at least, projecting the median result. Notice how close the median value
of NPV is to the deterministic value. 12.5. STOCHASTIC PRICING 12.5 567 Stochastic pricing Recall from Chapter 6 that the equivalence principle premium is deﬁned such that the expected value of the present value of the future loss at the issue of the policy is zero. In fact,
the expectation is usually taken over the future lifetime uncertainty (given ﬁxed values for
the mortality rates), not the uncertainty in investment returns or nondiversiﬁable mortality risk. This is an example of an expected value premium principle, where premiums
are set considering only the expected value of future loss, not any other characteristics of
the loss distribution.
The example studied in Section 12.4 above demonstrates that incorporating a guarantee
may add signiﬁcant risk to a contract and that this only becomes clear when modelled
stochastically. The risk cannot be quantiﬁed deterministically. Using the mean of the
stochastic output is generally not adequate as it fails to protect the insurer against signiﬁcant nondiversiﬁable risk of loss.
For this reason it is not advisable to use the equivalence premium principle when there is
signiﬁcant nondiversiﬁable risk. Instead we can use stochastic simulation with diﬀerent
premium principles.
The quantile premium principle was introduced (as the ‘portfolio percentile premium
principle’) in Section 6.8. This principle is based on the requirement that the policy
should generate a proﬁt with a given probability. We can extend this principle to the
pricing of equitylinked policies. For example, we might be willing to write a contract if,
using a given risk discount rate, the lower 5th percentile point of the net present value is
positive and the expected net present value is at least 65% of the acquisition costs.
The example studied throughout Section 12.4 meets neither of these requirements; the
lower 5th percentile point is −$859.82 and the expected net present value, $380.91, is
58.6% of the acquisition costs, $650.
We cannot determine a premium analytically for this contract which would meet these
requirements. However, we can investigate the eﬀects of changing the structure of the 568 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE policy. For the example studied in Section 12.4, Table 12.9 shows results in the same
format as in Table 12.8 for four changes to the policy structure. These changes are as
follows.
(1) Increasing the premium from $5 000 to $5 500, and hence increasing the GMMB to
$55 000 and the acquisition costs to $700.
(2) Increasing the annual management charge from 0.75% to 1.25%.
(3) Increasing the expense deductions from the premiums from 5% to 6% in the ﬁrst
year and from 1% to 2% in subsequent years.
(4) Decreasing the GMMB from 100% to 90% of premiums paid.
In each of the four cases, the remaining features of the policy are as described in Example
12.1.
Change
(1)
E[N P V ]
433.56
SD[N P V ]
660.67
95% CI for E[N P V ] (392.61, 474.51)
5%ile
−930.81
Median of N P V
562.87
95%ile
929.66
N−
86
∗
N
897 (2)
939.60
725.97
(894.60, 984.60)
−617.22
1 065.66
1 625.44
78
882 (3)
594.68
619.75
(556.27, 633.09)
−724.40
721.74
1 051.78
80
894 (4)
460.33
384.96
(436.47, 484.19)
145.29
500.00
831.51
46
939 Table 12.9: Results from changing the structure of the policy. Increasing the premium, change (1), makes little diﬀerence in terms of our chosen proﬁt
criterion. The lower 5th percentile point is still negative – the increase in the GMMB 12.6. STOCHASTIC RESERVING 569 means that even larger losses can occur – and the expected net proﬁt is still less than
65% of the increased acquisition costs. The premium for an equitylinked contract is not
like a premium for a traditional contract, since most of it is unavailable to the insurer.
The role of the premium in a traditional policy – to compensate the insurer for the risk
coverage oﬀered – is taken in equitylinked insurance by the management charge on the
policyholder’s funds and any loading taken from the premium before it is invested.
Increasing the management charge, change (2), or the expense loadings, change (3), does
increase the expected net present value to the required level but the probability of a loss
is still greater than 5%.
The one change that meets both parts of our proﬁt criterion is change (4), reducing the
level of the maturity guarantee. This is a demonstration of the important principle that
risk management begins with the design of the beneﬁts.
An alternative, and in many ways more attractive, method of setting a premium for such
a contract is to use modern ﬁnancial mathematics to both price the contract and reduce
the risk of making a loss. We return to this topic in Chapter 14. 12.6 Stochastic reserving 12.6.1 Reserving for policies with nondiversiﬁable risk In Chapter 7 we deﬁned a policy value as the EPV of the future loss from the policy
(using a deterministic interest rate assumption). This, like the use of the equivalence
principle to calculate a premium, is an example of the application of the expected value
principle. When the risk is almost entirely diversiﬁable, the expected value principle works
adequately. When the risk is nondiversiﬁable, which is usually the case for equitylinked
insurance, the expected value principle is inadequate both for pricing, as discussed in
Section 12.5, and for calculating appropriate reserves.
Consider the further discussion of Example 12.1 in Section 12.4. On the basis of the 570 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE assumptions in that section, there is a 5% chance that the insurer will make a loss in
excess of $859.82, in present value terms calculated using the risk discount rate of 15%
per year, on each policy issued. If the insurer has issued a large number of these policies,
such losses could have a disastrous eﬀect on its solvency, unless the insurer has anticipated
the risk by reserving for it, by hedging it in the ﬁnancial markets (which we explain in
Chapter 14) or by reinsuring it (which means passing the risk on by taking out insurance
with another insurer).
Calculating reserves for policies with signiﬁcant nondiversiﬁable risk requires a methodology that takes account of more than just the expected value of the loss distribution.
Such methodologies are called risk measures. A risk measure is a functional that is
applied to a random loss to give a reserve value that reﬂects the riskiness of the loss.
There are two common risk measures used to calculate reserves for nondiversiﬁable risks:
the quantile reserve and the conditional tail expectation reserve. 12.6.2 Quantile reserving A quantile reserve (also known as ValueatRisk, or VaR) is deﬁned in terms of a parameter
α, where 0 ≤ α ≤ 1. Suppose we have a random loss L. The quantile reserve with
parameter α represents the amount which, with probability α, will not be exceeded by
the loss.
If L has a continuous distribution function, FL , the αquantile reserve is Qα , where
Pr [L ≤ Qα ] = α, (12.1) so that
−
Qα = FL 1 (α). If FL is not continuous, so that L has a discrete or a mixed distribution, Qα needs to be
deﬁned more carefully. In the example below (which continues in the next section) we
assume that FL is continuous. 12.6. STOCHASTIC RESERVING 571 To see how to apply this in practice, consider again Example 12.1 as discussed in Section
12.4. Suppose that immediately after issuing the policy, and paying the acquisition costs
of $650, the insurer wishes to set up a 95% quantile reserve, denoted 0 V . In other words,
after paying the acquisition costs the insurer wishes to set aside an amount of money, 0 V ,
so that, with probability 0.95, it will be able to pay its liabilities.
We need some notation. Let j denote the rate of interest per year assumed to be earned
on reserves. In practice, j will be a conservative rate of interest, probably much lower
than the risk discount rate. Let t p00 denote the probability that a policy is still in force
55
at duration t. This is consistent with our notation from Chapter 8 since our underlying
model for the policy contains three states – in force (which we denote by 0), lapsed and
dead.
The reserve, 0 V , is calculated by simulating N sets of future accumulation factors for the
policyholder’s fund, exactly as in Section 12.4, and for each of these we calculate Prt,i , the
proﬁt emerging at time t, t = 1, 2, . . . , 10 for simulation i, per policy in force at duration
t − 1. For simulation i we calculate the EPV of the future loss, say Li , as
10 Li = − t=1 00
t−1 p55 Prt,i
.
(1 + j )t (12.2) Note that in the deﬁnition of Li we are considering future proﬁts at times t = 1, 2, . . . , 10,
and we have not included Pr0,i in the deﬁnition.
Then 0 V is set equal to the upper 95th percentile point of the empirical distribution of
L obtained from our simulations, provided that the upper 95th percentile is positive, so
that the reserve is positive. If the upper 95th percentile point is negative, 0 V is set equal
to zero.
Calculations by the authors, with N = 1 000 and j = 0.06, gave a value for 0 V of $1 259.56.
Hence, if, after paying the acquisition costs, the insurer sets aside a reserve of $1 259.56
for each policy issued, it will be able to meet its future liabilities with probability 0.95
provided all the assumptions underlying this calculation are realized. These assumptions
relate to 572 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
expenses,
lapse rates,
the survival model, and, in particular, the diversiﬁcation of the mortality risk,
the interest rate earned on the insurer’s fund,
the interest rate earned on the reserve,
the interest rate model for the policyholder’s fund,
the accuracy of our estimate of the upper 95th percentile point of the loss distribution. The reasoning underlying this calculation assumes that no adjustment to this reserve will
be made during the course of the policy. In practice, the insurer will review its reserves at
regular intervals, possibly annually, during the term of the policy and adjust the reserve if
necessary. For example, if after one year the rate of return on the policyholder’s fund has
been low and future expenses are now expected to be higher than originally estimated,
the insurer may need to increase the reserve. On the other hand, if the experience in the
ﬁrst year has been favourable, the insurer may be able to reduce the reserve. The new
reserve would be calculated by simulating the present value of the future loss from time
t = 1, using the information available at that time, and setting the reserve equal to the
greater of zero and the upper 95th percentile of the simulated loss distribution.
In our example, the initial reserve, 0 V = $1 259.45, is around 25% of the annual premium,
$5 000. This amount is expected to earn interest at a rate, 6%, considerably less than
the insurer’s risk discount rate, 15%. Setting aside substantial reserves, which may not
be needed when the policy matures, will have a serious eﬀect on the proﬁtability of the
policy. 12.6. STOCHASTIC RESERVING 12.6.3 573 CTE reserving The quantile reserve assesses the ‘worst case’ loss, where worst case is deﬁned as the
event with a 1 − α probability. One problem with the quantile approach is that it does
not take into consideration what the loss will be if that 1 − α worst case event actually
occurs. In other words, the loss distribution above the quantile does not aﬀect the reserve
calculation. The Conditional Tail Expectation (or CTE) was developed to address some
of the problems associated with the quantile risk measure. It was proposed more or
less simultaneously by several research groups, so it has a number of names, including
Tail Value at Risk (or TailVaR), Tail Conditional Expectation (or TCE) and Expected
Shortfall.
As for the quantile reserve, the CTE is deﬁned using some conﬁdence level α, where
0 ≤ α ≤ 1, which is typically 90%, 95% or 99% for reserving.
In words, the CTEα is the expected loss given that the loss falls in the worst 1 − α part
of the loss distribution, L. The worst 1 − α part of the loss distribution is the part above
the αquantile, Qα . If Qα falls in a continuous part of the loss distribution, that is, not
in a probability mass, then we can deﬁne the CTE at conﬁdence level α as
CTEα = E [LL > Qα ] . (12.3) If L has a discrete or a mixed distribution, then more care needs to be taken with the
deﬁnition. If Qα falls in a probability mass, that is, if there is some > 0 such that
Qα+ = Qα , then if we consider only losses strictly greater than Qα , we are using less
than the worst 1 − α of the distribution; if we consider losses greater than or equal to Qα ,
we may be using more than the worst 1 − α of the distribution. We therefore adapt the
formula of equation (12.3) as follows. Deﬁne β = max{β : Qα = Qβ }. Then
CTEα = (β − α)Qα + (1 − β ) E [LL > Qα ]
.
1−α (12.4) It is worth noting that, given that the CTEα is the mean loss given that the loss lies
above the VaR at level α, (at least when the VaR does not lie in a probability mass) then 574 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE CTEα is always greater than or equal to Qα , and usually strictly greater. Hence, for a
given value of α, the CTEα reserve is generally considerably more conservative than the
Qα quantile reserve.
Suppose the insurer wishes to set a CTE0.95 reserve, just after paying the acquisition costs,
for the policy studied in Example 12.1 and throughout Sections 12.4, 12.5 and 12.6.2. We
proceed by simulating a large number of times the present value of the future loss using
formula (12.2), with the rate of interest j per year we expect to earn on reserves, exactly
as in Section 12.6.2. From our calculations in Section 12.6.2 with N = 1 000 and j = 0.06,
the 50 worst losses, that is, the 50 highest values of Li , ranged in value from $1 260.76
to $7 512.41, and the average of these 50 values is $3 603.11. Hence we set the CTE0.95
reserve at the start of the ﬁrst year equal to $3 603.11.
The same remarks that were made about quantile reserves apply equally to CTE reserves.
(1) The CTE reserve in our example has been estimated using simulations based only
on information available at the start of the policy.
(2) In practice, the CTE reserve would be updated regularly, perhaps yearly, as more
information becomes available, particularly about the rate of return earned on the
policyholder’s fund. If the returns are good in the early years of the contract, then
it is possible that the probability that the guarantee will cost anything reduces, and
part of the reserves can be released back to the insurer before the end of the term.
(3) Holding a large CTE reserve, which earns interest at a rate lower than the insurer’s
risk discount rate, and which may not be needed when the policy matures, will have
an adverse eﬀect on the proﬁtability of the policy. 12.6.4 Comments on reserving The examples in this chapter illustrate an important general point. Financial guarantees
are risky and can be expensive. Several major life insurance companies have found their 12.7. NOTES AND FURTHER READING 575 solvency at risk through issuing guarantees that were not adequately understood at the
policy design stage, and were not adequately reserved for thereafter. The method of
covering that risk by holding a large quantile or CTE reserve reduces the risk, but at
great cost in terms of tying up amounts of capital that are huge in terms of the contract
overall. This is a passive approach to managing the risk and is usually not the best way
to manage solvency or proﬁtability.
Using modern ﬁnancial theory we can take an active approach to ﬁnancial guarantees that
for most equitylinked insurance policies oﬀers less risk, and, since the active approach
requires less capital, it generally improves proﬁtability when the required risk discount
rate is large enough to make carrying capital very expensive.
The active approach to risk mitigation and management comes from option pricing theory.
We utilize the fact that the guarantees in equitylinked insurance are ﬁnancial options
embedded in insurance contracts. There is an extensive literature available on the active
risk management of ﬁnancial options. In Chapter 13 we review the science of option
risk management, at an introductory level, and in Chapter 14 we apply the science to
equitylinked insurance. 12.7 Notes and further reading A practical feature of equitylinked contracts in the UK which complicates the analysis a
little is capital and accumulation units. The premiums paid at the start of the contract,
which are notionally invested in capital units, are subject to a signiﬁcantly higher annual
management charge than later premiums, which are invested in accumulation units. This
contract design has been developed to defray at an early stage the insurer’s acquisition
costs.
Stochastic proﬁt testing can also be used for traditional insurance. We would generally
simulate values for the interest earned on assets, and we might also simulate expenses and
withdrawal rates. Exercise 12.2 demonstrates this. 576 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE For shorter term insurance, the sensitivity of the proﬁt to the investment assumptions may
not be very great. The major risk for such insurance is misestimation of the underlying
mortality rates. This is also nondiversiﬁable risk, as underestimating the mortality rates
aﬀects the whole portfolio. It is therefore useful with term insurance to treat the force of
mortality as a stochastic input.
The CTE has become a very important risk measure in actuarial practice. It is intuitive,
easy to understand and to apply with simulation output. As a mean, it is more robust
with respect to sampling error than a quantile. The CTE is used for stochastic reserving
and solvency testing for Canadian and US equitylinked life insurance.
Hardy (2003) discusses risk measures, quantile reserves and CTE reserves in the context
of equitylinked life insurance. In particular, she gives full deﬁnitions of quantile and
CTE reserves, covering the cases when the loss distribution is not continuous and also
shows how to simulate the emerging costs and calculate proﬁt measures when stochastic
reserving is used. 12.8. EXERCISES 12.8 577 Exercises Exercise 12.1 An insurer sells a one year variable annuity contract. The policyholder
deposits $100, and the insurer deducts 3% for expenses and proﬁt. The expenses incurred
at the start of the year are 2.5% of the premium.
The remainder of the premium is invested in an investment fund. At the end of one year
the policyholder receives the fund proceeds; if the proceeds are less than the initial $100
investment the insurer pays the diﬀerence.
Assume that a unit investment in the fund accumulates to R after 1 year, where R ∼
LN (0.09, 0.182 ).
Let F1 denote the fund value at the year end. Let L0 denote the present value of future
outgo minus the margin oﬀset income random variable, assuming a force of interest of 5%
per year, i.e.
L0 = max(100 − 97 R, 0) e−0.05 − (3 − 2.5).
(a) Calculate Pr[F1 < 100].
(b) Calculate E[F1 ].
(c) Show that the 5th percentile of the distribution of R is 0.81377.
(d) Hence, or otherwise, calculate Q0.99 (L0 ).
(e) Let f be the probability density function of a lognormal random variable with
parameters µ and σ 2 . Use the result (which is derived in Appendix A)
A x f (x)dx = eµ+σ
0 2 /2 Φ log A − µ − σ 2
σ , where Φ is the standard normal distribution function, to calculate
(i) E[L0 ], and 578 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE
(ii) CTE0.95 (L0 ). (f) Now simulate the year end fund, using 100 projections. Compare the results of your
simulations with the accurate values calculated in (a)–(e). Exercise 12.2 A life insurer issues a special 5 year endowment insurance policy to a life
aged 50. The death beneﬁt is $10 000 and is payable at the end of the year of death,
if death occurs during the 5 year term. The maturity beneﬁt on survival to age 55 is
$20 000. Level annual premiums are payable in advance.
Reserves are required at integer durations for each policy in force, are independent of the
premium, and are as follows:
0V = 0, 1 V = 3 000, 2 V = 6 500, 3 V = 10 500, 4 V = 15 000, 5 V = 0. The company determines the premium by projecting the emerging cash ﬂows according
to the projection basis given below. The proﬁt objective is that the EPV of future proﬁt
must be 1/3 of the gross annual premium, using a risk discount rate of 10% per year.
Projection Basis
Initial Expenses:
Renewal Expenses:
Mortality:
Interest on all funds: 10% of the gross premium plus $100
6% of the second and subsequent gross premiums
Standard ultimate survival model
8% per year (a) Calculate the annual premium.
(b) Generate 500 diﬀerent scenarios for the cash ﬂow projection, assuming a premium
of $3 740, and assuming interest earned follows a lognormal distribution, such that 12.8. EXERCISES 579 if It denotes the return in the tth year,
(1 + It ) ∼ LN (0.07, 0.132 ).
(i) Estimate the probability that the policy will make a loss in the ﬁnal year, and
calculate a 95% conﬁdence interval for this probability.
(ii) Calculate the exact probability that the policy will make a loss in the ﬁnal year,
assuming mortality exactly follows the projection basis, so that the interest
rate uncertainty is the only source of uncertainty. Compare this with the 95%
conﬁdence interval for the probability determined from your simulations.
(iii) Estimate the probability that the policy will achieve the proﬁt objective, and
calculate a 95% conﬁdence interval for this probability. Exercise 12.3 An insurer issues an annual premium unitlinked contract with a 5 year
term. The policyholder is aged 60 and pays an annual premium of $100. A management
charge of 3% per year of the policyholder’s fund is deducted annually in advance.
The death beneﬁt is the greater of $500 and the amount of the fund, payable at the end
of the year of death. The maturity beneﬁt is the greater of $500 and the amount of the
fund, paid on survival to the end of the 5 year term.
d
d
d
d
Mortality rates assumed are: q60 = 0.0020, q61 = 0.0028, q62 = 0.0032, q63 = 0.0037 and
d
q64 = 0.0044. There are no lapses. (a) Assuming that interest of 8% per year is earned on the policyholder’s fund, project
the policyholder’s fund values for the term of the contract and hence calculate the
insurer’s management charge income.
(b) Assume that the insurer’s fund earns interest of 6% per year. Expenses of 2% of the
policyholder’s funds are incurred by the insurer at the start of each year. Calculate
the proﬁt signature for the contract assuming that no reserves are held. 580 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE (c) Explain why reserves may be established for the contract even though no negative
cash ﬂows appear after the ﬁrst year in the proﬁt test.
(d) Explain how you would estimate the 99% quantile reserve and the 99% CTE reserve
for this contract.
(e) The contract is entering the ﬁnal year. Immediately before the ﬁnal premium payment the policyholder’s fund is $485.
Assume that the accumulation factor for the policyholder’s fund each year is lognormally distributed with parameters µ = 0.09 and σ 2 = 0.182 . Let L4 represent
the present value of future loss random variable at time 4, using an eﬀective rate of
interest of 6% per year.
(i) Calculate the probability of a payment under the maturity guarantee.
(ii) Calculate Q99% (L4 ) assuming that insurer’s funds earn 6% per year as before. Exercise 12.4 An insurer used 1 000 simulations to estimate the present value of future
loss distribution for a segregated fund contract. The table shows the largest 100 simulated
values of L0 .
6.255 6.321 6.399 6.460 6.473
6.556
6.865 6.918 6.949 7.042 7.106
7.152
7.585 7.614 7.717 7.723
7.847 7.983
8.416 8.508
8.583 8.739 8.895
8.920
9.477 9.555 9.651 9.675 9.872
9.972
10.284 10.814 10.998 11.170 11.287 11.314
11.840 11.867 11.966 12.586 12.662 12.792
14.322 14.327 14.404 14.415 14.625 14.733 6.578
7.337
8.051
8.981
10.010
11.392
13.397
14.925 6.597
7.379
8.279
9.183
10.199
11.546
13.822
15.076 6.761
7.413
8.370
9.335
10.216
11.558
13.844
15.091 6.840
7.430
8.382
9.455
10.268
11.647
14.303
15.343 12.8. EXERCISES
15.490 15.544 15.617 15.856 16.369
17.357 17.774 18.998 19.200 21.944 581
16.458
21.957 17.125 17.164 17.222 17.248
22.309 24.226 24.709 26.140 (a) Estimate Pr[L0 > 10].
(b) Calculate an approximate 99% conﬁdence interval for Pr[L0 > 10].
(c) Estimate Q0.99 (L0 ) from these simulations.
(d) Estimate CT E0.99 (L0 ) from these simulations. Exercise 12.5 A life insurance company issues a 5 year unitlinked endowment policy to
a life aged 50 under which level premiums of $750 are payable yearly in advance throughout
the term of the policy or until earlier death.
In the ﬁrst policy year, 25% of the premium is allocated to the policyholder’s fund, followed
by 102.5% in the second and subsequent years. The units are subject to a bidoﬀer spread
of 5% and an annual management charge of 1% of the bid value of units is deducted at
the end of each policy year. Management charges are deducted from the unit fund before
death, surrender and maturity beneﬁts are paid.
If the policyholder dies during the term of the policy, the death beneﬁt of $3 000 or the
bid value of the units, whichever is higher, is payable at the end of the policy year of
death. The policyholder may surrender the policy only at the end of each policy year. On
surrender, the bid value of the units is payable at the end of the policy year of exit. On
maturity, 110% of the bid value of the units is payable. The company uses the following
assumptions in carrying out proﬁt tests of this contract: Rate of growth on assets 582 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE in the policyholder’s fund:
Rate of interest on
insurer’s fund cash ﬂows:
Mortality:
Initial expenses
Renewal expenses:
Initial commission:
Renewal commission:
Risk discount rate:
Surrenders: 6.5% per year
5.5% per year
Standard ultimate survival model
$150
$65 per year on the second
and subsequent premium dates
10% of ﬁrst premium
2.5% of the second
and subsequent years premiums
8.5% per year
10% of policies in force
at the end of each of the ﬁrst 3 years. (a) Calculate the proﬁt margin for the policy on the assumption that the company does
not hold reserves.
(b) (i) Explain brieﬂy why it would be suitable to establish reserves for this policy.
(ii) Calculate the eﬀect on the proﬁt margin of a reserve requirement of $400 at
the start of the second, third and fourth years, and $375 at the start of the
ﬁfth year. There is no initial reserve requirement. (c) An actuary has suggested the proﬁt test should be stochastic, and has generated a
set of random accumulation factors for the policyholder’s funds. The ﬁrst stochastic
scenario of annual accumulation factors for each of the ﬁve years is generated under the assumption that the accumulation factors are lognormally distributed with
parameters µ = 0.07 and σ 2 = 0.22 . Using the random standard normal deviates
given below, conduct the proﬁt test using your simulated accumulation factors, and 12.8. EXERCISES 583 hence calculate the proﬁt margin, allowing for the reserves as in (b):
−0.71873, −1.09365, 0.08851, 0.67706, 1.10300. 584 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE Answers to selected exercises
12.1 (a) 0.37040
(b) $107.87
(d) $19.54
(e) (i) $3.46
(ii) $24.83 12.2 (a) $3 739.59
(b) Based on one set of 500 projections
(i) 0.342, (0.300,0.384)
(ii) 0.519, (0.484,0.572)
(iii) 0.488, (0.444, 0.532) 12.3 (a) (3.00, 6.14, 9.44, 12.88, 16.50)
(b) (0.27, 1.37, 2.77, 4.33, 5.76)
(e) (i) 0.114
(ii) $80.50 12.4 (a) 0.054
(b) (0.036, 0.072)
(c) $17.30
(d) $21.46 12.8. EXERCISES
12.5 (a) 1.56%
(b) (ii) Reduces to 0.51%
(c) −1.43% 585 586 CHAPTER 12. EMERGING COSTS FOR EQUITYLINKED INSURANCE Chapter 13
Option Pricing
13.1 Summary In this chapter we review the basic ﬁnancial mathematics behind option pricing. First,
we discuss the no arbitrage assumption, which is the foundation for all modern ﬁnancial mathematics. We present the binomial model of option pricing, and illustrate the
principles of the risk neutral and real world measures, and of pricing by replication.
We discuss the BlackScholesMerton option pricing formula, and, in particular, demonstrate how it may be used both for pricing and risk management. 13.2 Introduction In Section 12.4 we discussed the problem of nondiversiﬁable risk in connection with
equitylinked insurance policies. A methodology for managing this risk, stochastic pricing
and reserving, was set out in Sections 12.5 and 12.6. However, as we explained there, this
methodology is not entirely satisfactory since it often requires the insurer to set aside large
amounts of capital as reserves to provide some protection against adverse experience. At
587 588 CHAPTER 13. OPTION PRICING the end of the contract, the capital may not be needed, but having to maintain large
reserves is expensive for the insurer. If experience is adverse, there is no assurance that
reserves will be suﬃcient.
Since the nondiversiﬁable risks in equitylinked contracts and some pension plans typically arise from ﬁnancial guarantees on maturity or death, and since these guarantees
are very similar to the guarantees in exchange traded ﬁnancial options, we can use the
BlackScholesMerton theory of option pricing to price and actively manage these risks.
When an ﬁnancial guarantee is a part of the beneﬁts under an insurance policy, we call
it an embedded option.
There are several reasons why it is very helpful for an insurance company to understand
option pricing and ﬁnancial engineering techniques. The insurer may buy options from a
third party such as a bank or a reinsurer to oﬀset the embedded options in their liabilities;
a good knowledge of derivative pricing will be useful in the negotiations. Also, by understanding ﬁnancial engineering methods an insurer can make better risk management
decisions. In particular, when an option is embedded in an insurance policy, the insurer
must make an informed decision whether to hedge the products inhouse or subcontract
the task to a third party.
There are many diﬀerent types of ﬁnancial guarantees in insurance contracts. This chapter
contains suﬃcient introductory material on ﬁnancial engineering to enable us to study in
Chapter 14 the valuation and hedging of options embedded within insurance policies that
can be viewed as relatively straightforward European put or call options. 13.3 The ‘no arbitrage’ assumption The ‘no arbitrage’ assumption is the foundation of modern valuation methods in ﬁnancial
mathematics. The assumption is more colloquially known as the ‘no free lunch’ assumption, and states quite simply that you cannot get something for nothing. 13.3. THE ‘NO ARBITRAGE’ ASSUMPTION 589 An arbitrage opportunity exists if an investor can construct a portfolio that costs zero
at inception and generates positive proﬁts with a nonzero probability in the future, with
no possibility of incurring a loss at any future time.
If we assume that there are no arbitrage opportunities in a market, then it follows that
any two securities or combinations of securities that give exactly the same
payments must have the same price. For example, consider two assets priced at $A
and $B which produce the same future cash ﬂows. If A = B , then an investor could buy
the asset with the lower price and sell the more expensive one. The cash ﬂows purchased
at the lower price would exactly match the cash ﬂows sold, so the investor would make a
risk free proﬁt of the diﬀerence between A and B .
The no arbitrage assumption is very simple and very powerful. It enables us to ﬁnd the
price of complex ﬁnancial instruments by ‘replicating’ the payoﬀs. Replication is a crucial
part of the framework. This means that if we can construct a portfolio of assets with
exactly the same payments as the investment in which we are interested, then the price
of the investment must be the same as the price of the ‘replicating portfolio’.
For example, suppose an insurer incurs a liability, under which it must deliver the price
of one share in Superior Life Insurance Company in one year’s time, and the insurer
wishes to value this liability. The traditional way to value this might be by constructing
a probability distribution for the future value – suppose the current value is $400 and the
insurer assumes the share price in one year’s time will follow a lognormal distribution,
with parameters µ = 6.07 and σ = 0.162 . Then the mean value of the share price in one
2
year’s time is eµ+σ /2 = $438.25.
The next step is to discount to current values, at, say 6% per year (perhaps using the
long term bond yield), to give a present value of $413.45.
So we have a value for the liability, with an implicit risk management plan of putting the
$413.45 in a bond, which in one year will pay $438.25, which may or may not be suﬃcient
to buy the share to deliver to the creditor. It will almost surely be either too much or
not enough. 590 CHAPTER 13. OPTION PRICING A better approach is to replicate the payoﬀ, and value the cost of replication. In this simple
case, that means holding a replicating portfolio of one share in Superior Life Insurance
Company. The cost of this now is $400. In one year, the portfolio is exactly suﬃcient to
pay the creditor, whatever the outcome. So, since it costs $400 to replicate the payoﬀ,
that is how much the liability is worth. It cannot be worth $413.45 – that would allow
the company to sell the liability for $413.45, and replicate it for $400, giving a risk free
proﬁt (or arbitrage) of $13.45.
Replication does not require a model; we have eliminated the uncertainty in the payoﬀ,
and we implicitly have a risk management strategy – buy the share and hold it until the
liability falls due.
Although this is an extreme example, the same argument will be applied in this chapter
and the next even when ﬁnding the replicating portfolio is a more complicated process.
In practice, in most securities markets, arbitrage opportunities arise from time to time
and are very quickly eliminated as investors spot them and trade on them. Since they
exist only for very short periods, assuming that they do not exist at all is suﬃciently close
to reality for most purposes. 13.4 Options Options are very important ﬁnancial contracts, with billions of dollars of trades in options
daily around the world. In this section we introduce the language of options and explain
how some option contracts operate. European options are perhaps the most straightforward type of options, and the most basic forms of these are a European call option
and a European put option.
The holder of a European call option on a stock has the right (but not the obligation) to
buy an agreed quantity of that stock at a ﬁxed price, known as the strike price, at a
ﬁxed date, known as the expiry or maturity date of the contract. 13.4. OPTIONS 591 Let St denote the price of the stock at time t. The holder of a European call option on this
stock with strike price K and maturity date T would exercise the option only if ST > K ,
in which case the option is worth ST − K to the option holder at the maturity date. The
option would not be exercised at the maturity date in the case when ST < K , since the
stock could then be bought for a lower price in the market at that time. Thus, the payoﬀ
at time T under the option is
(ST − K )+ = max(ST − K, 0).
The holder of a European put option on a stock has the right (but not the obligation) to
sell an agreed quantity of that stock at a ﬁxed strike price, at the maturity date of the
contract. The holder of a European put option would exercise the option only if ST < K ,
since the holder of the option could sell the stock at time T for K then buy the stock
at the lower price of ST in the market and hence make a proﬁt of K − ST . In this case
the option is worth K − ST to the option holder at the maturity date. The option would
not be exercised at the maturity date in the case when ST > K , since the option holder
would then be selling stock at a lower price than could be obtained by selling it in the
market. Thus, the payoﬀ at time T under a European put option is
(K − ST )+ = max(K − ST , 0).
In making all of the above statements, we are assuming that people act rationally when
they exercise options. We can think of options as providing guarantees on prices. For
example, a call option guarantees that the holder of the option pays no more than the
strike price to buy the underlying stock at the maturity date.
American options are deﬁned similarly, except that the option holder has the right
to exercise the option at any time before the maturity date. The names ‘European’
and ‘American’ are historical conventions, and do not signify where these options are
sold – both European and American options are sold worldwide. In this book we are
concerned only with European options which are signiﬁcantly more straightforward to
price than American options. Many of the options embedded in life insurance contracts
are Europeanstyle. 592 CHAPTER 13. OPTION PRICING If at any time t prior to the maturity date the stock price St is such that the option would
mature with a nonzero value if the stock price did not change, we say that the option
is ‘inthemoney’; so, a call option is inthemoney when St > K , and a put option is
inthemoney when K > St . When K = St , or even when K is close to St , we say the
option is ‘atthemoney’. Otherwise it is ‘outofthemoney’. 13.5 The binomial option pricing model 13.5.1 Assumptions Throughout Section 13.5 we use the no arbitrage principle together with a simple discrete
time model of a stock price process called the binomial model to price options.
Although the binomial model is simple, and not very realistic, it is useful because the
techniques we describe below carry through to more complicated models for a stock price
process.
We make the following assumptions.
• There is a frictionless ﬁnancial market in which there exists a risk free asset (such
as a zerocoupon bond) and a risky asset, which we assume here to be a stock. The
market is free of arbitrage.
• The ﬁnancial market is modelled in discrete time. Trades only occur at speciﬁed
time points. Changes in asset prices and the exercise date for an option can occur
only at these same dates.
• In each unit of time the stock price either moves up by a predetermined amount or
moves down by a predetermined amount. This means there are just two possible
states one period later if we start at a given time and price. Over this time unit the
market is said to be complete since the number of assets is equal to the number of
states. 13.5. THE BINOMIAL OPTION PRICING MODEL 593 • Investors can buy and sell assets without cost. These trades do not impact the
prices.
• Investors can short sell assets, so that they can hold a negative amount of an asset.
This is achieved by selling an asset they do not own, so the investor ‘owes’ the asset
to the lender. We say that an investor is long in an asset if the investor has a positive
holding of the asset, and is short in the asset if the investor has a negative holding.
We start by considering the pricing of an option over a single time period. We then extend
this to two time periods. 13.5.2 Pricing over a single time period To illustrate ideas numerically, consider a stock whose current price is $100 and whose
price at time t = 1 will be either $105 or $90. We assume that the continuously compounded risk free rate of interest is r = 0.03 per unit of time. Note that we must have
90 < 100er < 105
since otherwise arbitrage is possible. To see this, suppose 100er > 105. In this case an
investor could receive $100 by short selling one unit of stock at time t = 0 and invest this
for one unit of time at the risk free rate of interest. At time t = 1 the investor would then
have $100er , part of which would be used to buy one unit of stock in the market to wipe
out the negative holding, leaving a proﬁt of either $(100er − 105) or $(100er − 90), both of
which are positive. Similarly, if 100er < 90 (which means a negative risk free rate) selling
the risk free asset short and buying the stock will generate an arbitrage.
Now, consider a put option on this stock which matures at time t = 1 with a strike price
of K = $100. The holder of this option will exercise the option at time t = 1 only if the
stock price goes down, since by exercising the option the option holder will get $100 for
a stock worth $90. As we are assuming that there are no trading costs in buying and 594 CHAPTER 13. OPTION PRICING selling stocks, the option holder could use the sale price of $100 to buy stock at $90 at
time t = 1 and make a proﬁt of $10.
The seller of the put option will have no liability at time t = 1 if the stock price rises,
since the option holder will not sell a stock for $100 when it is worth $105 in the market.
However, if the stock price falls, the seller of the put option has a liability of $10.
We use the concept of replication to price this put option. This means that we look for a
portfolio of assets at time t = 0 that will exactly match the payoﬀ under the put option
at time t = 1. Since our market comprises only the risk free asset and the stock, any
portfolio at time t = 0 must consist of some amount, say $a, in the risk free asset and
some amount, $100b, in the stock (so that b units of stock are purchased). Then at time
t = 1, the portfolio is worth
aer + 105b
if the stock price goes up, and is worth
aer + 90b
if the stock price goes down. If this portfolio replicates the payoﬀ under the put option,
then the portfolio must be worth 0 at time t = 1 if the stock price goes up, and $10 at
time t = 1 if the stock price goes down. To achieve this we require that
aer + 105b = 0,
aer + 90b = 10.
Solving these equations we obtain b = −2/3 and a = 67.9312. We have shown that a
portfolio consisting of $67.9312 of the risk free asset and a short holding of −2/3 units
of stock exactly matches the payoﬀ under the put option at time t = 1, regardless of the
stock price at time t = 1. This portfolio is called the replicating, or hedge, portfolio.
The no arbitrage principle tells us that if the put option and the replicating portfolio have
the same value at time t = 1, they must have the same value at time t = 0, and this then 13.5. THE BINOMIAL OPTION PRICING MODEL 595 must be the price of the option, which is
a + 100b = $1.26.
We can generalize the above arguments to the case when the stock price at time t = 0
is S0 , the stock price at time t = 1 is uS0 if the stock price goes up and dS0 if the stock
price goes down, and the strike price for the put option is K . We note here that under
the no arbitrage assumption, we must have dS0 < S0 er < uS0 . Similarly, we must also
have dS0 < K < uS0 for a contract to be feasible.
The hedge portfolio consists of $a in the risk free asset and $bS0 in stock. Since the payoﬀ
at t = 1 from this portfolio replicates the option payoﬀ, we must have
aer + buS0 = 0,
aer + bdS0 = K − dS0
giving
a= ue−r (K − dS0 )
u−d and b= dS0 − K
.
S0 (u − d) The option price at time 0 is a + bS0 , the value of the hedge portfolio, which we can write
as
e−r q (K − dS0 ) (13.1) where
q= u − er
.
u−d (13.2) Note that, from our earlier assumptions,
0 < q < 1.
An interesting feature of expression (13.1) for the price of the put option is that, if we
were to treat q as the probability of a downward movement in the stock price and 1 − q as 596 CHAPTER 13. OPTION PRICING the probability of an upward movement, then formula (13.1) could be thought of as the
discounted value of the expected payoﬀ under the option. If the stock price moves down,
the payoﬀ is K − dS0 , with discounted value e−r (K − dS0 ). If q were the probability of
a downward movement in the stock price, then qe−r (K − dS0 ) would be the EPV of the
option payoﬀ. Recall that these parameters, q and 1 − q are not the true ‘up’ an ‘down’
probabilities. In fact, nowhere in our determination of the price of the put option have we
needed to know the probabilities of the stock price moving up or down. The parameter
q comes from the binomial framework, but it is not the ‘real’ probability of a downward
movement; it is just convenient to treat it as such, as it allows us to use the conventions
and notation of probability. It is important to remember though that we have not used a
probabilistic argument here, we have used instead a replication argument.
It turns out that the price of an option in the binomial framework can always be expressed
as the discounted value of the option’s ‘expected’ payoﬀ using the artiﬁcial probabilities of
upward and downward price movements, 1 − q and q respectively. The following example
demonstrates this for a general payoﬀ.
Example 13.1 Consider an option over one time period which has a payoﬀ Cu if the
stock price at the end of the period is uS0 , and has a payoﬀ Cd if the stock price at the
end of the period is dS0 . Show that the option price is
e−r (Cu (1 − q ) + Cd q )
where q is given by formula (13.2).
Solution 13.1 We construct the replicating portfolio which consists of $a in the risk free
asset and $bS0 in stock so that
aer + buS0 = Cu ,
aer + bdS0 = Cd , 13.5. THE BINOMIAL OPTION PRICING MODEL 597 giving
b= Cu − Cd
(u − d)S0 and
a = e−r Cu − u
= e−r Cu − Cd
u−d u
d
Cd −
Cu .
u−d
u−d Hence the option price is
a + bS0 = e−r = Cu
−r =e u
d
Cd −
Cu
u−d
u−d
1 − de−r
u−d
Cu + Cd er − d
u−d + Cu − Cd
u−d ue−r − 1
u−d + Cd u − er
u−d = e−r (Cu (1 − q ) + Cd q ) . In the above example, if we treat q as the probability that the stock price at time t = 1
is dS0 , then the expected payoﬀ under the option at time t = 1 is
Cu (1 − q ) + Cd q,
and so the option price is the discounted expected payoﬀ. Note that q has not been
deﬁned as the probability that the stock price is equal to dS0 at time t = 1, and, in 598 CHAPTER 13. OPTION PRICING general, will not be equal to this probability. We emphasize that the probability q is an
artiﬁcial construct, but a very useful one.
Under the binomial framework that we use here, there is some real probability that
the stock price moves down or up. We have not needed to identify it here. The true
distribution is referred to by diﬀerent names, the physical measure, the real world
measure, the subjective measure or nature’s measure. In the language of probability
theory, it is called the P measure. The artiﬁcial distribution that arises in our pricing of
options is called the risk neutral measure, and in the language of probability theory
is called the Qmeasure. The term ‘measure’ can be thought of as interchangeable with
‘probability distribution’. In what follows, we use EQ to denote expectation with respect
to the Qmeasure. The Qmeasure is called the risk neutral measure since, under the
Qmeasure, the expected return on every asset in the market (risky or not) is equal to the
risk free rate of interest, as if investors in this hypothetical world were neutral as to the
risk in the assets. We know that in the real world investors require extra expected return
for extra risk. We demonstrate risk neutrality in the following example.
Example 13.2 Show that if S1 denotes the stock price at time t = 1, then under our
model EQ [e−r S1 ] = S0 .
Solution 13.2 Under the Qmeasure,
S1 = uS0 with probability 1 − q,
dS0 with probability q. Then
EQ [e−r S1 ] = e−r ((1 − q )uS0 + qdS0 )
= e−r
= S0 . er − d
u−d uS0 + u − er
u−d dS0 13.5. THE BINOMIAL OPTION PRICING MODEL 599 The result in Example 13.2 shows that under the risk neutral measure, the stock price at
time t = 0 is the EPV under the Qmeasure of the stock price at time t = 1. We also see
that the expected accumulation factor of the stock price over a unit time interval is er , the
same as the risk free accumulation factor. Under P measure we expect the accumulation
factor to exceed er on average, as a reward for the extra risk. 13.5.3 Pricing over two time periods In the previous section we considered a single period of time and priced the option by
ﬁnding the replicating portfolio at time t = 0. We now extend this idea to pricing an
option over two time periods. This involves the idea of dynamic hedging, which we
introduce by extending the numerical example of the previous section.
Let us now assume that in each of our two time periods, the stock price can either increase
by 5% of its value at the start of the time period, or decrease by 10% of its value. We
assume that the stock price movement in the second time period is independent of the
movement in the ﬁrst time period.
As before, we consider a put option with strike price $100, but this time the exercise date
is at the end of the second time period. As illustrated in Figure 13.1, the stock price at
time t = 2 is $110.25 if the stock price moves up in each time period, $94.50 if the stock
price moves up once and down once, and $81.00 if the stock price moves down in each
time period. This means that the put option will be exercised if at time t = 2 the stock
price is $94.50 or $81.00.
In order to price the option, we use the same replication argument as in the previous
section, but now we must work backwards from time t = 2. Suppose ﬁrst that at time
t = 1 the stock price is $105. We can establish a portfolio at time t = 1 that replicates
the payoﬀ under the option at time t = 2. Suppose this portfolio contains $au of the risk
free asset and bu units of stock, so that the replicating portfolio is worth $(au + 105bu ). 600 CHAPTER 13. OPTION PRICING Time 0 Time 1 Time 2
¨
¨¨
¨
¨¨ ¨
¨¨ S0 = 100 ¨
rr ¨
¨¨ rr
r ¨
¨¨ ¨¨
P 0 rrr
r ¨
¨ rr
r
r ¨
¨
¨¨ rr
r r
r 105 ¨
¨¨
r
rr
r rr
r ¨
¨¨ 90 Pu ¨
¨¨
r
rr
r ¨
¨¨
rr
r rr
¨
¨¨ rr
r ¨
¨¨
rr
r ¨
¨¨ Pd 110.25 ¨
¨¨
rr
r
rr
r rr ¨
¨¨ rr
r
¨
¨¨ rr
r 94.50 Stock price 81.00 0 5.50 19 Figure 13.1: Two period binomial model. Option payoﬀ 13.5. THE BINOMIAL OPTION PRICING MODEL 601 Then at time t = 2, the value of the portfolio should be 0 if the stock price moves up in
the second time period since the option will not be exercised, and the value should be
$5.50 if the stock price moves down in the second time period since the option will be
exercised in this case. The equations that determine au and bu are
au er + 110.25bu = 0,
au er + 94.5bu = 5.50,
giving bu = −0.3492 and au = 37.3622. This shows that the replicating portfolio at time
t = 1, if the stock price at that time is 105, has value Pu = $0.70.
Similarly, if at time t = 1 the stock price is $90, we can ﬁnd the replicating portfolio
whose value at time t = 1 is $(ad + 90bd ), where the equations that determine ad and bd
are
ad er + 94.5bd = 5.50,
ad er + 81bd = 19,
since if the stock price rises to $94.50, the payoﬀ under the put option is $5.50, and if the
stock price falls to $81, the payoﬀ under the option is $19. Solving these two equations
we ﬁnd that bd = −1 and ad = 97.0446. Thus, the replicating portfolio at time t = 1, if
the stock price at that time is $90, has value Pd = $7.04.
We now move back to time t = 0. At this time point we want to ﬁnd a portfolio that
replicates the possible amounts required at time t = 1, namely $0.70 if the stock price goes
up to $105 in the ﬁrst time period, and $7.04 if it goes down to $90. This portfolio consists
of $a in the risk free asset and b units of stock, so that the equations that determine a
and b are
aer + 105b = 0.70,
aer + 90b = 7.04, 602 CHAPTER 13. OPTION PRICING giving b = −0.4233 and a = 43.8049. The replicating portfolio has value P0 at time t = 0,
where
P0 = a + 100b = $1.48
and, by the no arbitrage principle, this is the price of the option.
There are two important points to note about the above analysis. The ﬁrst is a point
we noted about option pricing over a single period – we do not need to know the true
probabilities of the stock price moving up or down in any time period in order to ﬁnd the
option price. The second point is that the replicating portfolio is self ﬁnancing. The
initial portfolio of $43.80 in the risk free asset and a short holding of −0.4233 units of
stock is exactly suﬃcient to provide the replicating portfolio at time t = 1 regardless of
the stock price movement in the ﬁrst time period. The replicating portfolio at time t = 1
then matches exactly the option payoﬀ at time t = 2. Thus, once the initial portfolio has
been established, no further injection of funds is required to match the option payoﬀ at
time t = 2.
What we have done in this process is an example of dynamic hedging. At time t = 1 we
established what portfolios were required to replicate the possible payoﬀs at time t = 2,
then at time t = 0 we worked out what portfolio was required to provide the portfolio
values required at time t = 1. This process works for any number of steps, but if there
is a large number of time periods it is a time consuming process to work backwards
through time to construct all the hedging strategies. However, if all we want to work out
is the option price, the result we saw for a single time period, that the option price is
the discounted value of the expected payoﬀ at the expiry date under the Qmeasure, also
holds when we are dealing with multiple time periods.
In our analysis we have u = 1.05, d = 0.9 and r = 0.03. From formula (13.2), the
probability of a downward movement in the stock price under the Qmeasure is
q= 1.05 − e0.03
= 0.1303,
1.05 − 0.9 13.5. THE BINOMIAL OPTION PRICING MODEL 603 and so the expected payoﬀ at the expiry date is
19(q )2 + 5.5 × 2(1 − q )q = $1.5962.
This gives the option price as
1.5962e−0.06 = $1.48. 13.5.4 Summary of the binomial model option pricing technique • We use the principle of replication; we construct a portfolio that replicates the
option’s payoﬀ at maturity. The value of the option is the cost of purchasing the
replicating (or hedge) portfolio.
• We use dynamic hedging – replication requires us to rebalance the portfolio at each
time step according to the movement in the stock price in the previous time step.
• We do not use any argument involving the true probabilities of upward or downward
movements in the stock price. However, there are important links between the real
world (P measure) model and the risk neutral (Qmeasure) model. We started by
assuming a twopoint distribution for the stock price after a single time period in
the real world. From this we showed that in the risk neutral world the stock price
after a single time period also has a twopoint distribution with the same possible
values, uS0 and dS0 , but the probabilities of moving up or down are not linked to
those of the real world model.
• Our valuation can be written in the form of an EPV, using artiﬁcial probabilities that
are determined by the possible changes in the stock price. This artiﬁcial distribution
is called the risk neutral measure because the mean accumulation of a unit of stock
under this distribution is exactly the accumulated value of a unit investment in the
risk free asset. Thus, an investor would be indiﬀerent between investment in the
risk free asset and investment in the stock, under the risk neutral measure. 604 CHAPTER 13. OPTION PRICING The binomial model option pricing framework is clearly not very realistic, but we can
make it more ﬂexible by increasing the number of steps in a unit of time, as discussed
below. If we do this, the binomial model converges to the BlackScholes Merton model,
which is described in the following section. 13.6 The BlackScholesMerton model 13.6.1 The model Under the BlackScholesMerton model, we make the following assumptions.
• The market consists of zero coupon bonds (the risk free asset) and stocks (the risky
asset).
• The stock does not pay any dividends, or, equivalently, any dividends are immediately reinvested in the stock. This assumption simpliﬁes the presentation but can
easily be relaxed if necessary.
• Portfolios can be rebalanced (that is, stocks and bonds can be bought and sold) in
continuous time. In the two period binomial example we showed how the replicating
portfolio was rebalanced (costlessly) after the ﬁrst time unit. In the continuous
time model the stock price moves are continuous, so the rebalancing is (at least in
principle) continuous.
• There are no transactions costs associated with trading the stocks and bonds.
• The continuously compounded risk free rate of interest, r per unit time, is constant
and the yield curve is ﬂat.
• Stocks and bonds can be bought or sold in any quantities, positive or negative;
we are not restricted to integer units of stock, for example. Selling or buying can 13.6. THE BLACKSCHOLESMERTON MODEL 605 be transacted at any time without restrictions on the amounts available, and the
amount bought or sold does not aﬀect the price.
• In the real world, the stock price, denoted St at time t, follows a continuous time
lognormal process with some parameters µ and σ . This process, also called geometric
Brownian motion, is the continuous time version of the lognormal model for one year
accumulation factors introduced in Chapter 12.
Clearly these are not realistic assumptions. Continuous rebalancing is not feasible, and
although major ﬁnancial institutions like insurance companies can buy and sell assets
cheaply, transactions costs will arise. We also know that yield curves are rarely ﬂat.
Despite all this, the BlackScholesMerton model works remarkably well, both for determining the price of options and for determining risk management strategies. The BlackScholesMerton theory is extremely powerful and has revolutionized risk management for
nondiversiﬁable ﬁnancial risks.
A lognormal stochastic process with parameters µ and σ has the following characteristics.
• Over any ﬁxed time interval, say (t, t + τ ) where τ > 0, the stock price accumulation
factor, St+τ /St , has a lognormal distribution with parameters µτ and σ 2 τ , so that
St+τ
∼P LN (µτ , σ 2 τ ),
St (13.3) which implies that
log St+τ
∼P N (µτ , σ 2 τ ).
St We have added the subscript P as a reminder that these statements refer to the
real world, or P measure, model. Our choice of parameters µ and σ here uses
the standard statistical parameterizations. Some authors, particularly in ﬁnancial
mathematics, use the same σ , but use a diﬀerent location parameter µ , say, such
that µ = µ + σ 2 /2. It is important to check what µ represents when it is used as a
parameter of a lognormal distribution. 606 CHAPTER 13. OPTION PRICING
We call log(St+τ /St ) the logreturn on the stock the time period (t, t + τ ). The
parameter µ is the mean logreturn over a unit of time, and σ is the standard
deviation of the logreturn over a unit of time. We call σ the volatility, and it is
common for the unit of time to be one year so that these parameters are expressed as
annual rates. Some information on the lognormal distribution is given in Appendix
A. • Stock price accumulation factors over nonoverlapping time intervals are independent of each other. (This is the same as in the binomial model, where the stock
price movement in any time interval is independent of the movement in any other
time interval.) Thus, if Su /St and Sw /Sv represent the accumulation factors over
the time intervals (t, u) and (v, w) where t < u ≤ v < w, then these accumulation
factors are independent of each other.
The lognormal process assumed in the BlackScholesMerton model can be derived as the
continuous time limit, as the number of steps increases, of the binomial model of the
previous sections. The proof requires mathematics beyond the scope of this book, but we
give some references in Section 13.7 for interested readers. 13.6.2 The BlackScholesMerton option pricing formula Under the BlackScholesMerton model assumptions we have the following important
results.
There is a unique risk neutral distribution, or Qmeasure, for the stock price process,
under which the stock price process, {St }t≥0 , is a lognormal process with parameters
r − σ 2 /2 and σ 2
For any European option on the stock, with payoﬀ function h(ST ) at maturity date
T , the value of the option at time t ≤ T is v (t), can be found as the expected present 13.6. THE BLACKSCHOLESMERTON MODEL 607 value of the payoﬀ under the risk neutral distribution (Qmeasure)
v (t) = EQ e−r(T −t) h(ST ) ,
t (13.4) where EQ denotes expectation using the risk neutral (or Q) measure, using all the
t
information available up to time t. This means, in particular, that valuation at time
t assumes knowledge of the stock price at time t.
Important points to note about this result are:
• Over any ﬁxed time interval, say (t, t + τ ) where τ > 0, the stock price accumulation factor, St+τ /St , has a lognormal distribution in the risk neutral world with
parameters (r − σ 2 /2)τ and σ 2 τ , so that
St+τ
∼Q LN ((r − σ 2 /2)τ , σ 2 τ ),
St (13.5) which implies that
log St+τ
∼Q N ((r − σ 2 /2)τ , σ 2 τ ).
St We have added the subscript Q as a reminder that these statements refer to the risk
neutral, or Qmeasure model.
• The expected Qmeasure present value (rate r per year) of the future stock price,
St+τ , is the stock price now, St . This follows from the previous point since
EQ [St+τ /St ] = exp((r − σ 2 /2)τ + τ σ 2 /2) = er τ .
t
This is the result within the BlackScholesMerton framework which corresponds to
the result in Example 13.2 for the binomial model.
• The Qmeasure is related to the corresponding P measure in two ways. 608 CHAPTER 13. OPTION PRICING
– Under the Qmeasure, the stock price follows a lognormal process, as it does
in the real world.
– The volatility parameter, σ , is the same for both measures.
The ﬁrst of these connections should not surprise us since the real world model, the
lognormal process, can be regarded as the limit of a binomial process, for which, as
we have seen in Section 13.5, the corresponding risk neutral model is also binomial;
the limit as the number of steps increases in the (risk neutral) binomial model is
then also a lognormal process. The second connection does not have any simple
explanation. Note that the parameter µ, the mean logreturn per unit time for the
P measure, does not appear in the speciﬁcation of the Qmeasure. This should not
surprise us: the real world probabilities of upward and downward movements in the
binomial model did not appear in the corresponding Qmeasure probability, q . • Formula (13.4) is the continuoustime extension of the same result for the single
period binomial model (Example 13.1) and the twoperiod binomial model (Section 13.5.3). In both the binomial and BlackScholesMerton models, we take the
expectation under the Qmeasure of the payoﬀ discounted at the risk free force of
interest.
• A mathematical derivation of the Qmeasure and of formula (13.4) is beyond the
scope of this book. Interested readers should consult the references in Section 13.7.
Now consider the particular case of a European call option with strike price K . The
option price at time t is c(t), where
c(t) = EQ e−r(T −t) (ST − K )+ .
t
To evaluate this price, ﬁrst we write it as
c(t) = e−r(T −t) St EQ (ST /St − K/St )+ .
t (13.6) 13.6. THE BLACKSCHOLESMERTON MODEL 609 Now note that, under the Qmeasure,
ST /St ∼ LN ((r − σ 2 /2)(T − t), σ 2 (T − t)).
So, letting f and F denote the lognormal probability density function and distribution
function, respectively, of ST /St , under the Qmeasure, we have
c(t) = e−r(T −t) St ∞
K/St = e−r(T −t) St (x − K/St )f (x) dx ∞
K/St x f (x) dx − K
(1 − F (K/St )) .
St (13.7) In Appendix A we derive the formula
a
0 xf (x)dx = exp{µ + σ 2 /2} Φ log a − µ − σ 2
σ for a lognormal random variable with parameters µ and σ 2 . Since the mean of this random
variable is
∞
0 xf (x)dx = exp{µ + σ 2 /2}, we have
∞
a xf (x)dx = exp{µ + σ 2 /2} 1 − Φ
= exp{µ + σ 2 /2} Φ log a − µ − σ 2
σ − log a + µ + σ 2
σ , where Φ denotes the standard normal distribution function. Applying this to formula
(13.7) for c(t) gives
c(t) = e−r(T −t) St er(T −t) Φ − log(K/St ) + (r − σ 2 /2)(T − t) + σ 2 (T − t)
√
σ T −t 610 CHAPTER 13. OPTION PRICING − e−r(T −t) K 1 − Φ
= St Φ log(K/St ) − (r − σ 2 /2)(T − t)
√
σ T −t log(St /K ) + (r + σ 2 /2)(T − t)
√
σ T −t
− e−r(T −t) K Φ log(St /K ) + (r − σ 2 /2)(T − t)
√
σ T −t , which we usually write as
c(t) = St Φ (d1 (t)) − Ke−r(T −t) Φ (d2 (t)) , (13.8) where d1 (t) = log(St /K ) + (r + σ 2 /2)(T − t)
√
σ T −t √
and d2 (t) = d1 (t) − σ T − t. (13.9) Since the stock price St appears (explicitly) only in the ﬁrst term of formula (13.8) and
r appears only in the second term, this formula suggests that the replicating portfolio at
time t for the call option comprises
Φ (d1 (t)) units of the stock, with total value at time t
St Φ (d1 (t)) ,
plus
a short holding of Φ (d2 (t)) units of zero coupon bonds with face value K , maturing
at time T , with a value at time t of
−Ke−r(T −t) Φ (d2 (t)) . 13.6. THE BLACKSCHOLESMERTON MODEL 611 Indeed, this is the selfﬁnancing replicating portfolio required at time t. We note though
that the derivation is not quite as simple as it looks, as Φ (d1 (t)) and Φ (d2 (t)) both depend
on the current stock price and time.
If the strike price is very small relative to the stock price we see that Φ (d1 (t)) tends to
one and Φ (d2 (t)) tends to zero. The replicating portfolio tends to a long position in the
stock and zero in the bond.
For a European put option, with strike price K , the option price at time t is p(t), where
p(t) = EQ e−r(T −t) (K − ST )+ ,
t
which, after working through the integration, becomes
p(t) = Ke−r(T −t) Φ (−d2 (t)) − St Φ (−d1 (t)) , (13.10) where d1 (t) and d2 (t) are deﬁned as before.
The replicating portfolio for the put option comprises
Φ (−d2 (t)) units of zero coupon bonds with face value K , maturing at time T , with
value at time t
Ke−r(T −t) Φ (−d2 (t)) ,
plus
a short holding of Φ (−d1 (t)) units of the stock, with total value at time t
−St Φ (−d1 (t)) .
For the European call and put options, we can show that
St d
d
c(t) = St Φ(d1 (t)) and St
p(t) = −St Φ(−d1 (t)).
dSt
dSt 612 CHAPTER 13. OPTION PRICING You are asked to prove the ﬁrst of these formulae as Exercise 13.1. These two formulae
show that, for these options, the replicating portfolio has a portion St dv (t)/dSt invested
in the stock, and hence a portion v (t) − St dv (t)/dSt invested in the bond, where v (t) is
the value of the option at time t.
This result holds generally for any option valued under the BlackScholesMerton framework and the quantity dv (t)/dSt is known as the delta of the option at time t.
Example 13.3 Let p(t) and c(t) be the prices at time t for a European put and call,
respectively, both with strike price K and remaining term to maturity T − t.
(a) Use formulae (13.8) and (13.10) to show that, using the BlackScholesMerton framework,
c(t) + K e−r(T −t) = p(t) + St . (13.11) (b) Use a noarbitrage argument to show that formula (13.11) holds whatever the model
for stock price movements between times t and T .
Solution 13.3 (a) From formulae (13.8) and (13.10), and using the fact that Φ(z ) =
1 − Φ(−z ) for any z , we have
c(t) = St Φ(d1 (t)) − Ke−r(T −t) Φ(d2 (t))
= St (1 − Φ(−d1 (t))) − Ke−r(T −t) (1 − Φ(−d2 (t)))
= St − Ke−r(T −t) + p(t)
which proves (13.11).
(b) To prove this result without specifying a model for stock price movements, consider
two portfolios held at time t. The ﬁrst comprises the call option plus a zero coupon 13.6. THE BLACKSCHOLESMERTON MODEL 613 bond with face value K maturing at time T ; the second comprises the put option
plus one unit of the stock. These two portfolios have current values
c(t) + K e−r(T −t) and p(t) + St respectively. At time T the ﬁrst portfolio will be worth K if ST ≤ K , since the
call option will then be worthless and the bond will pay K , and it will be worth
ST if ST > K , since then the call option would be exercised and the proceeds from
the bond would be used to purchase one unit of stock. Now consider the second
portfolio at time T . This will be worth K if ST ≤ K , since the put option would
be exercised and the stock would be sold at the exercise price, K , and it will be
worth ST if ST > K , since the put option will then be worthless and the stock
will be worth ST . Since the two portfolios have the same value at time T under all
circumstances, they must have the same value at all other times, in particular at
time t. This gives equation (13.11).
This important result is known as putcall parity. Example 13.4 An insurer oﬀers a two year contract with a guarantee under which the
policyholder invests a premium of $1 000. The insurer keeps 3% of the premium to cover all
expenses, then invests the remainder in a mutual fund. (A mutual fund is an investment
that comprises a diverse portfolio of stocks and bonds. In the UK similar products are
called unit trusts or investment trusts.) The mutual fund investment value is assumed to
follow a lognormal process, with parameters µ = 0.085 and σ 2 = 0.22 per year. The mutual
fund does not pay out dividends; any dividends received from the underlying portfolio are
reinvested. The risk free rate of interest is 5% per year compounded continuously. The
insurer guarantees that the payout at the maturity date will not be less than the original
$1 000 investment.
(a) Show that the 3% expense loading is not suﬃcient to fund the guarantee. 614 CHAPTER 13. OPTION PRICING (b) Calculate the real world probability that the guarantee applies at the maturity date.
(c) Calculate the expense loading that would be exactly suﬃcient to fund the guarantee.
Solution 13.4 (a) The policyholder has, through the insurer, invested $970 in the
mutual fund. This will accumulate over the two years of the contract to some
random amount, S2 , say. If S2 < $1 000 then the insurer’s guarantee bites, and the
insurer must make up the diﬀerence. In other words, the policyholder has the right
at the maturity date to receive a price of $1 000 from the insurer for the mutual
fund stocks. This is a two year put option, with payoﬀ at time T = 2 of
(1 000 − S2 )+ .
If the mutual fund stocks are worth more than $1 000, then the policyholder just
takes the proceeds and the insurer has no further liability.
In terms of option pricing, we have a strike price K = $1 000, a mutual fund stock
price at time t = 0 of S0 = $970, and a risk free rate of interest of 5%. So the price
of the put option at inception is
p(0) = Ke−rT Φ (−d2 (0)) − S0 Φ (−d1 (0))
where
log(S0 /K ) + (r + σ 2 /2)T
√
= 0.3873 ⇒ Φ(−d1 (0)) = 0.3493,
σT
√
d2 (0) = d1 (0) − σ T = 0.1044 ⇒ Φ(−d2 (0)) = 0.4584,
d1 (0) = giving
p(0) = 414.786 − 338.794 = $75.99.
So the 3% expense charge, $30, is insuﬃcient to fund the guarantee cost. The cost
of the guarantee is actually 7.599% of the initial investment. However, if we actually 13.6. THE BLACKSCHOLESMERTON MODEL 615 set 7.599% as the expense loading, the price of the guarantee would be even greater,
as we would invest less money in the mutual fund at inception whilst keeping the
same strike price.
(b) The real world distribution of S2 /S0 is LN (2µ, 2σ 2 ). This means that
log(S2 /S0 ) ∼ N (2µ, 2σ 2 ),
which implies that
log S2 ∼ N (log S0 + 2µ, 2σ 2 ).
Then
S2 ∼ LN (log S0 + 2µ, 2σ 2 ),
which implies that
Pr[S2 < 1 000] = Φ log 1 000 − log S0 − 2µ
√
2σ = 0.311. That is, the probability of a payoﬀ under the guarantee is 0.311.
(c) Increasing the expense loading increases the cost of the guarantee, and there is no
analytic method to ﬁnd the expense loading, $E , which pays for the guarantee with
an initial investment of $(1 000 − E ). Figure 13.2 shows a plot of the expense loading
against the cost of the guarantee (shown as a solid line). Where this line crosses
the line x = y (shown as a dotted line) we have a solution. From this plot we
see that the solution is around 10.72% (i.e. the expense loading is around $107.2).
Alternatively, Excel Solver gives the solution that an expense loading of 10.723%
exactly funds the resulting guarantee. 616 CHAPTER 13. OPTION PRICING 111 109 Guarantee cost 107 105 103 101 99
100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 Expense loading Figure 13.2: Expense loading plotted against option cost for Example 13.4.
Finding the price is only the ﬁrst step in the process. The beauty of the BlackScholesMerton approach is that it gives not only the price but also directs us in what we can do
with the price to manage the guarantee risk. In part (a) of Example 13.4, the guarantee
payoﬀ can be replicated by investing $414.79 in two year zero coupon bonds and short
selling $338.79 of the mutual fund stock, with a net cost of $75.99. If we continuously
rebalance such that at any time t the bond position has value 1 000e−r(2−t) Φ(−d2 (t))
and the short stock position has value − St Φ(−d1 (t)), then this will exactly pay oﬀ the
guarantee liability at the maturity date.
In practice, continuous rebalancing is impossible. Rebalancing at discrete intervals is
possible but introduces some additional cash ﬂows, and in the next example we explore
this issue.
Example 13.5 Let us continue Example 13.4 above, where an insurer has issued a guarantee which matures in two years. The initial investment (net of expenses) is $970 and 13.6. THE BLACKSCHOLESMERTON MODEL 617 the maturity guarantee is $1 000.
In Table 13.1 you are given the monthly values for the underlying mutual fund stock price
for the two year period, assuming a starting price of $970.
Assume, as in Example 13.4, that the continuously compounded risk free rate, r, is 5%
per year. Determine the cash ﬂows arising assuming that the insurer
(a) invests the entire option cost in the risk free asset,
(b) invests the entire option cost in the mutual fund asset,
(c) allocates the initial option cost to bonds and the mutual fund at the outset, according to the BlackScholesMerton model, that is $414.79 to zero coupon bonds and
−$338.79 to the mutual fund shares, and then
(i) never subsequently rebalances the portfolio,
(ii) rebalances only once, at the end of the ﬁrst year, and
(iii) rebalances at the end of each month. Solution 13.5 We note that the guarantee ends inthemoney, with a liability under the
put option of $(1 000 − 766.66) = $233.34 at the maturity date.
(a) If the option cost is invested in the risk free asset, it accumulates to 75.99e2r =
$83.98. This leaves a shortfall at maturity of $(233.34 − 83.98) = $149.36.
(b) If the option cost is invested in the mutual fund asset, it will accumulate to 75.99 ×
(766.66/970) = $60.06 leaving a shortfall at maturity of $173.28.
(c) (i) If the insurer invests in the initial hedge portfolio, but never rebalances,
◦ the bond part of the hedge accumulates at the risk free rate for the whole
two year period to an end value of $458.41; 618 CHAPTER 13. OPTION PRICING Time, t
St
(months)
$
0
970.00
1
950.07
2
959.99
3
940.93
4
921.06
5
967.25
6
1 045.15
7
1 007.59
8
945.97
9
913.77
10
932.99
11
951.11
12
906.11
13
824.86
14
831.08
15
797.99
16
785.86
17
724.36
18
707.43
19
713.87
20
715.14
21
690.74
22
675.80
23
699.71
24
766.66
Table 13.1: Table of mutual fund stock prices for Example 13.5. 13.6. THE BLACKSCHOLESMERTON MODEL 619 ◦ the stock part of the hedge accumulates in proportion to the mutual fund
share price, with ﬁnal value −338.79 × (766.66/970) = −$267.77; and
◦ the hedge portfolio value at maturity is then worth 458.41 − 267.77 =
$190.64, which means that the insurer is liable for an additional cash ﬂow
at maturity of $42.70, as the hedge portfolio value is less than the option
guarantee cost. In this case the total cost of the guarantee is the initial
hedge cost of $75.99 plus a ﬁnal balancing payment of $42.70.
(ii) If the insurer rebalances only once, at the end of the ﬁrst year, the value of the
initial hedge portfolio at that time is
Bonds: 414.79er = $436.05.
Mutual Fund: −338.79 (906.11/970) = −$316.48.
So the value of the hedge portfolio immediately before rebalancing is $119.57.
The rebalanced hedge is found from formula (13.10) with t = 1 year as
p(1) = Ke−r(2−1) Φ (−d2 (1)) − S1 Φ (−d1 (1))
= 603.26 − 504.56
= 98.70.
This means there is a cash ﬂow of $(119.57 − 98.70) = $20.87 back to the
insurer, as the value of the initial hedge more than pays for the rebalanced
hedge.
We now track the new hedge through to the maturity date.
Bonds: 603.26er = $634.19.
Mutual Fund: −504.56 × (766.66/906.11) = −$426.91.
Total hedge portfolio value: $207.28.
We need $233.34 to pay the guarantee liability, so the insurer is liable for an
additional cash ﬂow of $26.06. 620 CHAPTER 13. OPTION PRICING
So, in tabular form we have the following cash ﬂows, where a positive value is
a cash ﬂow out and negative value is a cash ﬂow back to the insurer.
Time
Value of hedge
Cost of
Final
Net cash ﬂow
(years) brought forward new hedge guarantee cost
$
0
0
75.99
–
75.99
1
119.57
98.70
–
−20.87
2
207.28
–
233.34
26.06
(iii) Here, we repeat the exercise in (b) but we now accumulate and rebalance each
month. The results are given in Table 13.2. The second, third and fourth
columns show the bond part, the mutual fund part and the total cost of the
hedge required at the start of each month. In the ﬁnal month, the total reﬂects
the cost of the guarantee payoﬀ. The ﬁfth column shows the value of the hedge
brought forward, and the diﬀerence between the new hedge cost and the hedge
brought forward is the cash ﬂow required at that time.
We see how the rebalancing frequency aﬀects the cash ﬂows; with a monthly
rebalancing frequency, all the cash ﬂows required are relatively small, after the
initial hedge cost. The fact that these cash ﬂows are nonzero indicates that
the original hedge is not self ﬁnancing with monthly rebalancing. However, the
amounts are small, demonstrating that if the insurer follows this rebalancing
strategy, there is little additional cost involved after the initial hedge cost, even
though the ﬁnal guarantee payout is substantial. The total of the additional
cash ﬂows after the initial hedge cost is −$12.26 in this case. It can be shown
that the expected value of the additional cash ﬂows using the P measure is
zero. This example demonstrates that in this case, where the option matures inthemoney,
the dynamic hedge is remarkably eﬃcient at converging to the payoﬀ with only small
adjustments required each month. If we were to rebalance more frequently still, the 13.7. NOTES AND FURTHER READING 621 rebalancing cash ﬂows would converge to zero. In practice, many hedge portfolios are
rebalanced daily or even several times a day.
Of course, this guarantee might well end up outofthemoney, in which case the hedge
portfolio would be worth nothing at the maturity date, and the insurer would lose the
cost of establishing the hedge portfolio in the ﬁrst place. The hedge is a form of insurance,
and, as with all insurance, there is a cost even when there is no claim. 13.7 Notes and further reading This chapter oﬀers a very brief introduction to an important and exciting area. For a
much more comprehensive introduction, see for example Hull (2005) or McDonald (2006).
For a description of the history of options and option pricing, see Boyle and Boyle (2001).
The proof that the binomial model converges to the lognormal model as the time unit,
h, tends to zero is somewhat technical. The original proof is given in Cox et al (1979);
another method is in Hsia (1983).
We assumed from Section 13.6.1 onwards that the stock did not pay any dividends. Adapting the model and results for dividends is explained in Hull (2005) and McDonald (2006). 622 Time
(months)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24 CHAPTER 13. OPTION PRICING New hedge portfolio
Bonds Mutual Fund Total
414.79
−338.79
75.99
446.09
−363.17
82.92
437.17
−358.37
78.80
469.69
−383.83
85.86
505.72
−411.67
94.05
441.15
−366.59
74.56
332.07
−283.53
48.54
388.22
−329.43
58.79
492.86
−411.51
81.35
557.18
−461.25
95.94
531.28
−445.30
85.97
505.60
−428.81
76.78
603.26
−504.56
98.70
769.54
−617.58
151.96
776.58
−628.41
148.17
847.22
−671.33
175.88
882.11
−693.88
188.22
948.97
−700.74
248.24
965.94
−697.59
268.35
973.54
−707.77
265.76
981.09
−712.67
268.42
987.44
−690.59
296.84
991.70
−675.80
315.90
995.84
−699.71
296.13
233.34 Old hedge
Net cash ﬂow
brought forward
$
0.00
75.99
84.68
−1.76
80.99
−2.19
87.74
−1.88
95.93
−1.88
75.52
−0.96
46.88
1.66
60.11
−1.33
80.56
0.79
97.41
−1.48
88.56
−2.59
79.54
−2.76
99.18
−0.48
146.46
5.50
150.52
−2.35
176.43
−0.55
189.62
−1.40
246.21
2.03
268.58
−0.23
266.03
−0.27
268.57
−0.15
296.83
0.01
315.90
0.00
296.13
0.00
233.34
0.00 Table 13.2: Cash ﬂow calculations for Example 13.5. 13.8. EXERCISES 13.8 623 Exercises Exercise 13.1 Let c(t) denote the price of a call option on a nondividend paying stock,
using the BlackScholes equation (13.6). Show that
dc(t)
= Φ(d1 (t)).
dSt
Hint: remember that d1 (t) is a function of St . Exercise 13.2 (a) Show that, under the binomial model of Section 13.5, E Q [Sn ] = S0 ern .
(b) Show that, under the BlackScholesMerton model,
E Q [Sn ] = S0 ern . Exercise 13.3 A binomial model for a nondividend paying security with price St at
time t is as follows:
S0 = 100,
St+1 = 1.1St if the stock price rises,
0.9St if the stock price falls. Zero coupon bonds are available for all integer durations, with a risk free rate of interest
of 6% compounded continuously.
A derivative security pays $20 at a speciﬁed maturity date if the stock price has increased
from the start value, and pays $0 if the stock price is at or below the start value at
maturity. 624 CHAPTER 13. OPTION PRICING (a) Find the price and the replicating portfolio for option assuming it is issued at t = 0
and matures at t = 1.
(b) Now assume the option is issued at t = 0 and matures at t = 2. Find the price and
the replicating portfolio at t = 0 and at t = 1. Exercise 13.4 Consider a twoperiod binomial model for a non dividend paying security
with price St at time t, where S0 = 1.0,
St+1 = 1.2St if the stock price rises,
0.95St if the stock price falls. At time t = 2 option A pays $3 if the stock price has risen twice, $2 if it has risen once
and fallen once and $1 if it has fallen twice.
At time t = 2 option B pays $1 if the stock price has risen twice, $2 if it has risen once
and fallen once and $3 if it has fallen twice.
The risk free force of interest is 4.879% per period. You are given that the true probability
that the price rises each period is 0.5.
(a) Calculate the EPV (under the P measure) of option A and show that it is the same
as the EPV of option B.
(b) Calculate the price of option A and show that it is diﬀerent to the price of option
B.
(c) Comment on why the prices diﬀer even though the expected payout is the same.
Exercise 13.5 A stock is currently priced at $400. The price of a 6 month European
call option with a strike price of $420 is $41. The risk free rate of interest is 7% per year,
compounded continuously. 13.8. EXERCISES 625 Assume the BlackScholes pricing formula applies.
(a) Calculate the current price of a 6 month European put option with the same exercise
price. State the assumptions you make in the calculation.
(b) Estimate the implied volatility of the stock.
(c) Calculate the delta of the option.
(d) Find the hedging portfolio of stock and risk free zero coupon bonds that a writer of
10 000 units of the call option should hold. Exercise 13.6 A binomial model for a nondividend paying security with price St at
time t is as follows: the price at time t + 1 is either 1.25 St or 0.8 St . The risk free rate of
interest is 10% per time unit eﬀective.
(a) Calculate the risk neutral probability measure for this model.
The value of S0 is 100. A derivative security with price Dt at time t pays the following
returns at time 2: 1 if S2 = 156.25, D2 =
2 if S2 = 100, 0 if S2 = 64.
(b) Determine D1 when S1 = 125 and when S1 = 80 and hence calculate the value of
D0 . (c) Derive the corresponding hedging strategy, i.e. the combination of the underlying
security and the risk free asset required to hedge an investment in the derivative
security. 626 CHAPTER 13. OPTION PRICING (d) Comment on your answer to (c) in the light of your answer to part (b). Exercise 13.7 A nondividend paying stock has a current price of $8.00. In any unit of
time (t, t + 1) the price of the stock either increases by 25% or decreases by 20%. $1 held
in cash between times t and t + 1 receives interest to become $1.04 at time t + 1. The
stock price after t time units is denoted by St .
(a) Calculate the riskneutral probability measure for the model.
(b) Calculate the price (at time t = 0) of a derivative contract written on the stock with
expiry date t = 2 which pays $10.00 if and only if S2 is not $8.00 (and otherwise
pays 0). 13.8. EXERCISES 627 Answers to selected exercises
13.3 (a) $15.24
(b) $11.61 13.4 (a) $1.81
(b) $2.00
(c) Option A: $1.633, Option B: $1.996 13.5 (a) $46.55
(b) 38.6%
(c) 53.42%
(d) Long 5 342.5 shares of stock and short 17 270 bonds, where each bond is worth
$100 at time zero 13.6 (a) 2
3 (increase), 1
3 (decrease) (b) D0 = 1.1019 13.7 (a) 0.5333 (increase),
(b) $4.6433 0.4667 (decrease) 628 CHAPTER 13. OPTION PRICING Chapter 14
Embedded Options
14.1 Summary In this chapter we describe ﬁnancial options embedded in insurance contracts, focusing
in particular on the most straightforward options which appear as guaranteed minimum
death and maturity options in equitylinked life insurance policies eﬀected by a single
premium. We investigate pricing, valuation and risk management for these guarantees,
performing our analysis under the BlackScholesMerton framework described in Chapter
13. 14.2 Introduction The guaranteed minimum payments under an equitylinked contract usually represent
a relatively minor aspect of the total payout under the policy, because the guarantees
are designed to apply only in the most extreme situation of very poor returns on the
policyholders’ funds. Nevertheless, these guarantees are not negligible – failure to manage
the risk from apparently innocuous guarantees has led to signiﬁcant ﬁnancial problems
629 630 CHAPTER 14. EMBEDDED OPTIONS for some insurers.
In Chapter 12 we described proﬁt testing of equitylinked contracts with guarantees, where
the only risk management involved was a passive strategy of holding capital reserves in
case the experience is adverse – or, even worse, holding no capital in the expectation that
the guarantee will never apply. However, in the case when the equitylinked contract
incorporates ﬁnancial guarantees that are essentially the same as the ﬁnancial options
discussed in Chapter 13, we can use the more sophisticated techniques of Chapter 13
to price and manage the risks associated with the guarantees. These techniques are
preferable to those of Chapter 12 because they mitigate the risk that the insurer will have
insuﬃcient funds to pay for the guarantees when necessary.
To show how the guarantees can be viewed as options, recall Example 12.2 in Chapter 12,
where we described an equitylinked insurance contract, paid for with a single premium P ,
with a guaranteed minimum maturity beneﬁt (GMMB) and a guaranteed minimum death
beneﬁt (GMDB). Consider, for now, the GMMB only. After some expense deductions a
single premium is invested in an equity fund, or perhaps a mixed equity/bond fund. The
fund value is variable, moving up and down with the underlying assets. At maturity, the
insurer promises to pay the greater of the actual fund value and the original premium
amount.
Let Ft denote the value of the policyholder’s fund at time t. Suppose that, as in Example 12.2, the beneﬁt for policies still in force at the maturity date, say at time n,
(the term is n = 5 years in Example 12.2, but more typically it would be ten years or
longer) is max(P, Fn ). As the policyholder’s fund contributes the amount Fn , the insurer’s
additional liability is h(n), where
h(n) = max(P − Fn , 0).
The total beneﬁt paid for such a contract in force at the maturity date is
Fn + h(n).
Recognizing that the fund value process {Ft }t≥0 may be considered analogous to a stock 14.2. INTRODUCTION 631 price process, and that P is a ﬁxed, known amount, the guarantee payoﬀ h(n) is identical
to the payoﬀ under an n year European put option with strike price $P , as described in
Section 13.4. So, while in Chapter 12 we modelled this contract with cash ﬂow projection,
we have a more appropriate technique for pricing and valuation from Chapter 13, using
the BlackScholesMerton framework.
Similarly, the guaranteed minimum death beneﬁt in an equitylinked insurance contract
oﬀers a payoﬀ that can be viewed as an option – often a put option similar to that under
a GMMB.
There are a few diﬀerences between the options embedded in equitylinked contracts and
standard options traded in markets. Two important diﬀerences are as follows.
1. The options embedded in equitylinked contracts have random terms to maturity. If
the policyholder surrenders the contract, or dies, before the expiry date, the GMMB
will never be paid. The GMDB expires on the death of the policyholder, if that
occurs during the term of the contract.
2. The options embedded in equitylinked contracts depend on the value of the policyholder’s fund at death or maturity. The underlying risky asset process represents
the value of a traded stock or stock index. The fund value at time t, Ft , is related
to the risky asset price, St , since we assume the policyholder’s fund is invested in
a fund with returns following traded stocks, but with the important diﬀerence that
regular management fees are being deducted from the policyholder’s fund.
These diﬀerences mean that we must adapt the BlackScholesMerton theory of Chapter
13 in order to apply it to equitylinked insurance.
Throughout this chapter we consider equitylinked contracts paid for by a single premium,
P , which, after the deduction of any initial charges, is invested in the policyholder’s fund.
This fund, before allowing for the deduction of any management charges, earns returns
following the underlying stock price process, {St }t≥0 . We make all the assumptions in
Section 13.6.1 relating to the BlackScholesMerton framework. In particular, we assume 632 CHAPTER 14. EMBEDDED OPTIONS the stock price process is a lognormal process with volatility σ per year, and also that
there is a risk free rate of interest, r per year, continuously compounded. 14.3 Guaranteed minimum maturity beneﬁts 14.3.1 Pricing From Chapter 13 we know that the price of an option is the EPV of the payoﬀ under the
risk neutral probability distribution, discounting at the risk free rate. Suppose a GMMB
under a single premium contract guarantees that the payout at maturity, n years after
the issue date of the contract, will be at least equal to the single premium, P .
Then the option payoﬀ, as mentioned above, is h(n) = max(P − Fn , 0), because the
remainder of the beneﬁt, Fn , will be paid from the policyholder’s fund.
This payoﬀ is conditional on the policy remaining in force until the maturity date. In
order to price the guarantee we assume that the survival of a policyholder for n years,
taking account of mortality and lapses, is independent of the fund value process and is a
diversiﬁable risk. For simplicity here we ignore surrenders and assume all policyholders
are aged x at the commencement of their policies, and are all subject to the same survival
model. Under these assumptions, the probability that a policy will still be in force at the
end of the term is n px .
Consider the situation at the issue of the contract. If the policyholder does not survive n
years, the GMMB does not apply at time n, and so the insurer does not need to fund the
guarantee in this case. If the policyholder does survive n years, the GMMB does apply
at time n, and we know that the amount required at the issue of the contract to fund this
guarantee is
EQ e−rn (P − Fn )+ .
0
Thus, the expected amount (with respect to mortality and lapses) required by the insurer 14.3. GUARANTEED MINIMUM MATURITY BENEFITS 633 at the time of issue per contract issued is π (0), where
π (0) = n px EQ e−rn (P − Fn )+ .
0
Note that we are adopting a mixture of two diﬀerent methodologies here. The nondiversiﬁable risk from the stock price process, which channels through to Fn , is priced
using the methodology of Chapter 13, whereas the mortality risk, which we have assumed
to be diversiﬁable, is priced using the expected value principle.
Suppose that the total initial expenses are a proportion e of the single premium, and the
management charge is a proportion m of the policyholder’s fund, deducted at the start of
each year after the ﬁrst. Then
Fn = P (1 − e)(1 − m)n−1 Sn
.
S0 Since we are interested in the relative increase in St , we can assume S0 = 1 without any
loss of generality. (We interpret the stock price process {St }t≥0 as an index for the fund
assets; as an index, we can arbitrarily set S0 to any convenient value.) Then
Fn = P (1 − e)(1 − m)n−1 Sn .
The value of the guarantee can be written
π (0) = n px EQ e−rn (P − P (1 − e)(1 − m)n−1 Sn )+
0 = P n px EQ e−rn 1 − (1 − e)(1 − m)n−1 Sn
0
= P n px ξ EQ e−rn ξ −1 − Sn
0 + + where the expense factor ξ = (1 − e)(1 − m)n−1 is a constant. We can now apply formula
(13.10) for the price of a put option, setting the strike price for the option, K = ξ −1 . Then
the price at the issue date of a GMMB, guaranteeing a return of at least the premium P , 634 CHAPTER 14. EMBEDDED OPTIONS is
π (0) = P n px ξ ξ −1 e−rn Φ(−d2 (0)) − S0 Φ(−d1 (0)) (14.1) = P n px e−rn Φ(−d2 (0)) − ξ S0 Φ(−d1 (0))
where
d1 (0) = log S0 ξ + (r + σ 2 /2)n
√
σn and
√
d2 (0) = d1 (0) − σ n.
The return of premium guarantee is a common design for a GMMB, but many other
designs are sold. Any guarantee can be viewed as a ﬁnancial option. Suppose h(n)
denotes a general payoﬀ function for a GMMB when it matures at time n years. In
equation (14.1) the payoﬀ function is h(n) = (P − Fn )+ . In other cases when the only
random quantity in the payoﬀ function is the fund value at maturity, we can use exactly
the same approach as in equation (14.1), so that the value of the GMMB is always
π (0) = n px EQ e−rn h(n) .
0
Example 14.1 Consider a 10 year equitylinked contract issued to a life aged 60, with a
single premium of P = $10 000. After a deduction of 3% for initial expenses, the premium
is invested in an equity fund. An annual management charge of 0.5% is deducted from
the fund at the start of every year except the ﬁrst.
The contract carries a guarantee that the maturity beneﬁt will not be less than the single
premium, P .
The risk free rate of interest is 5% per year, continuously compounded, and stock price
volatility is 25% per year. 14.3. GUARANTEED MINIMUM MATURITY BENEFITS 635 (a) Calculate the cost at issue of the GMMB as a percentage of the single premium,
assuming there are no lapses and that the survival model is Makeham’s law with
A = 0.0001, B = 0.00035 and c = 1.075.
(b) Now suppose that, allowing for mortality and lapses, the insurer expects only 55%
of policyholders to maintain their policies to maturity. Calculate the revised cost
at issue of the GMMB as a percentage of the single premium, commenting on any
additional assumptions required.
Solution 14.1 (a) With n = 10 we have
ξ = (1 − 0.03)(1 − 0.005)9 = 0.927213,
d1 (0) = log ξ + (r + σ 2 /2)n
√
= 0.932148,
σn √
d2 (0) = d1 (0) − σ n = 0.141579,
EQ e−10r h(10)
0 = 0.106275 P and
10 p60 = 0.673958, so that
π (0) = 0.0716P.
That is, the option cost, assuming no lapses, is 7.16% of the single premium.
(b) If we assume that precisely 55% of policies issued reach maturity, the option value
per policy issued is reduced to 0.55 EQ [e−10r h(10)] = 0.55 × 0.106275 = 5.85% of
0
the single premium. 636 CHAPTER 14. EMBEDDED OPTIONS The assumption that 55% of policies reach maturity is reasonable if we assume that
survival, allowing for mortality and lapses, is a diversiﬁable risk which is independent of the stock price process. In practice, lapse rates may depend on the fund’s
performance so that this assumption may not be reasonable. 14.3.2 Reserving We have already deﬁned the reserve for an insurance contract as the capital set aside
during the term of a policy to meet future obligations under the policy. In Chapter 12 we
demonstrated a method of reserving for ﬁnancial guarantees using a stochastic projection
of the net present value of future outgo minus income, where we set the reserve to provide
adequate resources in the event that investment experience for the portfolio was adverse.
Using the BlackScholesMerton approach, the value of the guarantee is interpreted as
the value of the portfolio of assets that hedges, or replicates, the payoﬀ under the guarantee. The insurer may use the cost of the guarantee to purchase appropriate options
from another ﬁnancial institution. If the mortality and lapse experiences follow the basis
assumptions, the payoﬀs from the options will be precisely the amounts required for the
guarantee payments. There is usually no need to hold further reserves since any reserve
would cover only the future net expenses of maintaining the contract, which are, usually,
fully defrayed by the future management charge income.
Increasingly, insurers are hedging their own guarantees. This should be less expensive than
buying options from a third party, but requires the insurer to have the necessary expertise
in ﬁnancial risk management. When the insurer retains the risk, the contribution to the
policy reserve for the guarantee will be the cost of maintaining the replicating portfolio.
We saw in Chapter 13 that the cost of the replicating portfolio at some time t, before an
option matures, is the price of the option at time t.
Suppose we consider the GMMB from Section 14.3.1, where the guarantee liability for 14.3. GUARANTEED MINIMUM MATURITY BENEFITS 637 the insurer at maturity, time n, is (P − Fn )+ , and where the issue price was π (0) from
equation (14.1). The contribution to the reserve at time t, where 0 ≤ t ≤ n, for the
GMMB, assuming the contract is still in force at time t, is
π (t) = P n−t px+t e−r(n−t) Φ(−d2 (t)) − ξ St Φ(−d1 (t)) where
d1 (t) = log(ξ St ) + (r + σ 2 /2)(n − t)
√
σ n−t √
and d2 (t) = d1 (t) − σ n − t. Note here that the expense factor ξ = (1 − e)(1 − m)n−1 does not depend on t, but the
reserve at time t does depend on the stock price at time t, St .
For a more general GMMB, with payoﬀ h(n) on survival to time n, the contribution to
the reserve is
π (t) = n−t px+t EQ e−r(n−t) h(n) ,
t where EQ denotes the expectation at time t with respect to the Qmeasure. In particular,
t
Q
Et assumes knowledge of the stock price process at t, St .
In principle, the hedge for the maturity guarantee will (under the basis assumptions)
exactly pay oﬀ the guarantee liability, so there should be no need to apply stochastic
reserving methods. In practice though, it is not possible to hedge the guarantee perfectly,
as the assumptions of the BlackScholesMerton formula do not apply exactly. The insurer
may hold an additional reserve over and above the hedge cost to allow for unhedgeable
risk and for the risk that lapses, mortality and volatility do not exactly follow the basis
assumptions. Determining an appropriate reserve for the unhedgeable risk is beyond the
scope of this book, but could be based on the stochastic methodology described in Chapter
12.
Example 14.2 Assume that the policy in Example 14.1 is still in force six years after it
was issued to a life aged 60. Assuming there are no lapses, calculate the contribution to 638 CHAPTER 14. EMBEDDED OPTIONS the reserve from the GMMB at this time given that, since the policy was purchased, the
value of the stock has
(a) increased by 45%, and
(b) increased by 5%.
Solution 14.2 (a) Recall that in the option valuation we have assumed that the return
on the fund, before management charge deductions, is modelled by the index {St }t≥0 ,
where S0 = 1. We are given that S6 = 1.45. Then
π (6) = P 4 p66 e−4r Φ(−d2 (6)) − ξ S6 Φ(−d1 (6))
where
d1 (t) = log(ξ St ) + (r + σ 2 /2)(10 − t)
√
σ 10 − t √
and d2 (t) = d1 (t) − σ 10 − t, and ξ = 0.927213 as in Example 14.1. So
d1 (6) = 1.241983,
d2 (6) = 0.741983,
4 p66 = 0.824935 and hence
π (6) = 0.035892P = $358.92.
(b) For S6 = 1.05 we have π (6) = $905.39.
A lower current fund value means that the guarantee is more likely to mature inthemoney
and so a larger reserve is required. 14.4. GUARANTEED MINIMUM DEATH BENEFIT 14.4 Guaranteed minimum death beneﬁt 14.4.1 639 Pricing Not all equitylinked insurance policies carry GMMBs, but most carry GMDBs of some
kind to distinguish them from regular investment products. The most common guarantees
on death are a ﬁxed or an increasing minimum death beneﬁt. In Canada, for example,
contracts typically oﬀer a minimum death beneﬁt of the total amount of premiums paid.
In the USA, the guaranteed minimum payout on death might be the accumulation at
some ﬁxed rate of interest of all premiums paid. In the UK, the beneﬁt might be the
greater of the total amount of premiums paid and, say, 101% of the policyholder’s fund.
We approach GMDBs in the same way as we approached GMMBs. Consider an n year
policy issued to a life aged x under which the payoﬀ under the GMDB is h(t) if the life
dies at age x + t, where t < n. If the insurer knew at the issue of the policy that the life
would die at age x + t, the insurer could cover the guarantee by setting aside
v (0, t) = EQ e−rt h(t)
0
at the issue date, where Q is again the risk neutral measure for the stock price process
that underlies the policyholder’s fund.
We know from Chapter 2 that the probability density associated with death at age x + t
for a life now aged x is t px µx+t , and so the amount that should be set aside to cover the
GMDB, denoted π (0), is found by averaging over the possible ages at death, x + t, so that
n π (0) = v (0, t)t px µx+t dt. (14.2) 0 If the death beneﬁt is payable at the end of the month of death rather than immediately,
the value of the guarantee becomes
12n π (0) = v (0, j/12)
j =1 j −1
12  1 qx .
12 (14.3) 640 CHAPTER 14. EMBEDDED OPTIONS Notice that (14.2) and (14.3) are similar to formulae we have met in earlier chapters. For
example, the EPV of a term insurance beneﬁt of $S payable immediately on the death
within n years of a life currently aged x is
n Sv t t px µx+t dt. (14.4) 0 There are similarities and diﬀerences between (14.2) and (14.4). In each expression we
are ﬁnding the expected amount required at time 0 to provide a death beneﬁt (and in
each case we require 0 at time n with probability n px ). In expression (14.4) the amount
required if death occurs at time t is the present value of the payment at time t, namely
Sv t , whereas in expression (14.2) v (0, t) is the amount required at time 0 in order to
replicate the (possible) payment at time t.
Example 14.3 An insurer issues a 5 year equitylinked insurance policy to a life aged 60.
A single premium of P = $10 000 is invested in an equity fund. Management charges of
0.25% are deducted at the start of each month. At the end of the month of death before
age 65, the death beneﬁt is the accumulated amount of the investment with a GMDB
equal to the accumulated amount of the single premium, with interest at 5% per year
compounded continuously.
Calculate the value of the guarantee on the following basis.
Survival model: Makeham’s law with A = 0.0001, B = 0.00035 and c = 1.075
Risk free rate of interest: 5% per year, continuously compounded
Volatility: 25% per year
Solution 14.3 As in previous examples, let {St }t≥0 be an index of prices for the equity
fund, with S0 = 1, and let m = 0.0025 denote the monthly management charge. Then
1
12
the payo if death occurs in the month from time k − 12 to k , for k = 12 , 12 , . . . , 60 , is
12
h(k ) = max(P e0.05k − Fk , 0) 14.4. GUARANTEED MINIMUM DEATH BENEFIT 641 where
Fk = P Sk (1 − m)12k ,
so that
h(k ) = P (1 − m)12k max e0.05k
− Sk , 0 .
(1 − m)12k 12
For any value of k (k = 12 , 12 , . . . , 60 ), the payoﬀ is a multiple of the payoﬀ under a put
12
option with strike price e0.05k /(1 − m)12k . Before applying formula (13.10) to value this
option, it is convenient to extend the notation for d1 (t) and d2 (t) in formula (13.9) to
include the maturity date, so we now write these as d1 (t, T ) and d2 (t, T ) where T is the
maturity date. We can now apply formula (13.10) with strike price e0.05k /(1 − m)12k , which we discount at the risk free rate of r = 0.05, to obtain the ﬁrst term in formula (13.10) as
Φ(−d2 (0, k ))/(1 − m)12k . Thus, if v (0, k ) denotes the value at time 0 of the guarantee at
time k , then
v (0, k ) = P (1 − m)12k Φ(−d2 (0, k ))
− S0 Φ(−d1 (0, k ))
(1 − m)12k = P Φ(−d2 (0, k )) − (1 − m)12k Φ(−d1 (0, k ))
where, from (13.9),
d1 (0, k ) = log((1 − m)k /e0.05k ) + (r + σ 2 /2) k
√
σk and d2 (0, k ) = d1 (0, k ) − σ √ k, with σ = 0.25.
Table 14.1 shows selected values from a spreadsheet containing deferred mortality probabilities and option prices for each possible month of death. Using these values in formula
(14.3), the value of this GMDB is 2.7838% of the single premium, or $278.38. 642 CHAPTER 14. EMBEDDED OPTIONS  1 qx k (years) d1 (0, k ) d2 (0, k ) v (0, k ) 1/12
2/12
3/12
4/12
5/12
6/12
7/12
.
.
. 0.001400
0.001980
0.002425
0.002800
0.003130
0.003429
0.003704
.
.
. −0.070769
−0.100082
−0.122575
−0.141538
−0.158244
−0.173347
−0.187237
.
.
. 300.16
431.43
534.79
623.65
703.20
776.12
843.99
.
.
. 0.002248
0.002257
0.002265
0.002273
0.002282
0.002290
0.002299
.
.
. 56/12
57/12
58/12
59/12
60/12 0.010477
0.010570
0.010662
0.010754
0.010844 −0.529585
−0.534293
−0.538959
−0.543585
−0.548173 2 708.30
2 735.70
2 762.88
2 789.86
2 816.63 0.002702
0.002709
0.002717
0.002725
0.002732 k−1
12 12 Table 14.1: Spreadsheet excerpt for the GMDB in Example 14.3. 14.4. GUARANTEED MINIMUM DEATH BENEFIT 14.4.2 643 Reserving We now apply the approach of the previous section to reserving for a GMDB on the
assumption that the insurer is internally hedging. Consider a policy issued to a life aged
x with a term of n years and with a GMDB which is payable immediately on death if
death occurs at time s where 0 < s < n. Suppose that the payoﬀ function under the
guarantee at time s is h(s). Let v (t, s) denote the price at time t for an option with payoﬀ
h(s) at time s, where 0 ≤ t ≤ s, assuming the policyholder dies at age x + s. Then
v (t, s) = EQ e−r(s−t) h(s) .
t
Hence, the value of the GMDB for a policy in force at time t (< n) is π (t), where
n π (t) = v (t, s) s−t px+t µx+s ds t n−t = v (t, w + t) w px+t µx+t+w dw, 0 when the beneﬁt is paid immediately on death, and
π (t) = 12(n−t)
j =1 v (t, t + j/12) j −1
12  1
12 q x+ t , when the beneﬁt is paid at the end of the month of death.
Example 14.4 Assume that the policy in Example 14.3 is still in force 3 years and 6
months after the issue date. Calculate the contribution of the GMDB to the reserve if
the stock price index of the underlying fund assets
(a) has grown by 50% since inception, so that S3.5 = 1.5, and
(b) is the same as the initial value, so that S3.5 = 1.0. 644 CHAPTER 14. EMBEDDED OPTIONS Solution 14.4 Following the solution to Example 14.3, the strike price for an option
expiring at time s is e0.05s /(1 − m)12s . Since we are valuing the option at time t < s, the
time to expiry is now s − t. Thus, applying formula (13.10) we have v (t, s) = P (1 − m)12s e0.05s e−0.05(s−t)
Φ(−d2 (t, s)) − St Φ(−d1 (t, s))
(1 − m)12s = P e0.05t Φ(−d2 (t, s)) − St (1 − m)12s Φ(−d1 (t, s))
where
d1 (t, s) =
and log(St (1 − m)12s /e0.05s ) + (r + σ 2 /2)(s − t)
√
σ s−t d2 (t, s) = d1 (t, s) − σ √ s − t. 7
8
7
9
For the valuation at t = 3.5, we calculate v (3.5, s) for s = 3 12 , 3 12 , 3 12 , 3 12 , . . . , 5 and
multiply each value by the mortality probability, s−t− 1  1 q63.5 . The resulting valuation is
12 12 (a) $30.55 when S3.5 = 1.5, and
(b) $172.05 when S3.5 = 1. Example 14.5 An insurer oﬀers a 10 year equitylinked policy with a single premium.
An initial expense deduction of 4% of the premium is made, and the remainder of the
premium is invested in an equity fund. Management charges are deducted daily from the
policyholder’s account at a rate of 0.6% per year. On death before the policy matures
a death beneﬁt of 110% of the fund value is paid. There is no guaranteed minimum
maturity beneﬁt. 14.4. GUARANTEED MINIMUM DEATH BENEFIT 645 (a) Calculate the price at issue of the excess amount of the death beneﬁt over the fund
value at the date of death for a life aged 55 at the purchase date, as a percentage
of the single premium.
(b) Calculate the value of the excess amount of the death beneﬁt over the fund value at
the date of death six years after the issue date, as a percentage of the policyholder’s
fund at that date. You are given that the policy is still in force at the valuation
date.
Basis:
Survival model: Makeham’s law, with A = 0.0001, B = 0.00035 and c = 1.075
Risk free rate of interest: 5% per year, continuously compounded
Volatility: 25% per year
Solution 14.5 (a) First, we note that the daily management charge can be treated as
a continuous deduction from the fund, so that, for a unit premium,
Ft = 0.96e−0.006t St .
Second, we note that the excess amount of the death beneﬁt over the fund value at
the date of death can be viewed as a GMDB equal to 10% of the fund value at the
date of death. For a unit premium, the payoﬀ function h(s) if death occurs at time
s, is
h(s) = 0.1 Fs = 0.096 e−0.006s Ss .
The value at issue of the death beneﬁt payable if the policyholder dies at time s is
v (0, s) = EQ e−rs h(s) = EQ e−rs 0.096 e−0.006s Ss .
0
0
In the previous chapter we saw that under the risk neutral measure the EPV of a
stock price at a future point in time is the stock price now. Thus
Q
E0 e−rs Ss = S0 . 646 CHAPTER 14. EMBEDDED OPTIONS Since S0 = 1, we have
v (0, s) = S0 × 0.096 e−0.006s = 0.096 e−0.006s .
The GMDB value at issue is then
10 π (0) = v (0, s) s p55 µ55+s ds
0
10 = 0.096 e−0.006s s p55 µ55+s ds 0 ¯
= 0.096 A1 10 δ=0.6%
55: (14.5) = 0.02236.
So the value of the GMDB at the inception of the policy is 2.24% of the single
premium.
(b) The value at time t < s of the option that would be needed to fund the GMDB if
the policyholder were to die at time s, given that the policy is in force at t, is, for
a unit premium,
v (t, s) = EQ e−r(s−t) h(s) = 0.1 × 0.96St e−0.006s .
t
The total contribution to the reserve for the GMDB for a policy still in force at time
t, with original premium P , is then
10−t π (t) = P v (t, w + t) w p55+t µ55+t+w dw 0
10−t = 0.096P St
0 e−0.006(w+t) w p55+t µ55+t+w dw 14.5. PRICING METHODS FOR EMBEDDED OPTIONS
10−t = 0.096 P St e−0.006t 647 e−0.006 w w p55+t µ55+t+w dw 0 ¯1
= 0.096 P St e−0.006t A55+t:10−t δ=0.6% .
So, at time t = 6, given that the policy is still in force, the contribution to the
reserve from the GMDB, per unit premium, is
¯
π (6) = 0.096 P S6 e−0.006×6 A1 4 δ=0.6%
61:
= 0.096 P S6 e−0.036 × 0.12403.
The fund value at time t = 6 is
F6 = 0.96 P S6 × e−0.036 ,
and so the reserve, as a proportion of the fund value, is
¯
0.096 P S6 e−0.036 A1 4 δ=0.6%
61:
0.96 P S6 e−0.036 ¯
= 0.1A1 4 δ=0.6% = 0.0124.
61: That is, the GMDB reserve would be 1.24% of the policyholder’s fund value. 14.5 Pricing methods for embedded options In discussing pricing above, we have expressed the price of a GMMB and a GMDB as
a percentage of the initial premium. This is appropriate if the option is funded by a
deduction from the premium at the inception of the policy. That is, the price of the
option would come from the initial deduction of eP in the notation of Section 14.3.1
above. This sum could then be invested in the hedge portfolio for the option. 648 CHAPTER 14. EMBEDDED OPTIONS A relatively large expense deduction at inception, called a frontendload, is common
for UK policies, but less common in North America. A more common expense loading in
North America is a management charge, applied as a regular percentage deduction from
the policyholder’s fund.
If the guarantee is to be funded through a regular management charge, rather than a
deduction from the single premium as in Sections 14.3.1 and 14.4.1, we need a way to
express the cost in terms of this charge.
Consider a single premium equitylinked policy with a term of n years issued to a life aged
x. We assume, for simplicity, that there are no lapses and no initial expenses, so that
e = 0 in the notation of Section 14.3.1. Also, we assume that mortality is a diversiﬁable
risk which is independent of the stock price process.
Let π (0) denote the cost at inception of the guarantees embedded in the policy, as derived
in Sections 14.3.1 and 14.4.1. Suppose these guarantees consist of a payment of amount
h(t) if the life dies at time t (< n) and a payment h(n) if the life survives to the end of
the term. The value of each of these guarantees is
EQ [h(t) e−rt ]
0
given that the life does die at time t, and
EQ [h(n) e−rn ]
0
given that the life does survive to time n. Allowing for the probabilities of death and
survivorship, we have
n π (0) =
0 EQ [h(t) e−rt ] t px µx+t dt + n px EQ [h(n) e−rn ].
0
0 We interpret π (0) as the cost at time 0 of setting up the replicating portfolios to pay the
guarantees.
Let c denote the component of the management charge that is required to fund the
guarantees from a total (ﬁxed) management charge of m (> c) per year. We call c the
risk premium for the guarantees. 14.5. PRICING METHODS FOR EMBEDDED OPTIONS 649 Assume that the management charge is deducted daily, which we treat as a continuous
deduction. With these assumptions, the fund value at time t for a policy still in force at
that time, Ft , can be written
Ft = P St e−mt .
Hence, the risk premium received in the time interval t to t + dt for a policy still in force
is (loosely) c P St e−mt dt. Ignoring survivorship for the moment, the value at time 0 of
this payment can be calculated as the cost of setting up a replicating portfolio which will
pay this amount at time t. This cost is c P e−mt dt since an investment of this amount
at time 0 in the stock will accumulate to c P St e−mt dt at time t (recall that S0 = 1).
Allowing for survivorship, the value at time 0 of the risk premium received in the time
interval t to t + dt is c P e−mt dt t px and so the value at time 0 of the total risk premiums
to be received is
n c P e−mt t px dt = c P ax:n δ=m .
¯ 0 The risk premium c is chosen so that the value to the insurer of the risk premiums to be
received is equal to the cost at time 0 of setting up the replicating portfolios to pay the
guarantees, so that
c = π (0)/(P ax:n δ=m ).
¯
Calculating c from this formula is a slightly circular process. The risk premium c is a
component of the total management charge m, but we need to know m to calculate the
right hand side of this equation for c. In practice, we may need to iterate through the
calculations a few times to determine the value of c. In some cases there may be no
solution. For example, increasing the total management charge m may increase the cost
of the guarantees, therefore requiring a higher value for the risk premium c, which may
in turn require a higher value for m.
If the management charge is deducted less frequently, say annually in advance, we can use
the same principles as above to derive the value of the risk premiums. The cost at time 650 CHAPTER 14. EMBEDDED OPTIONS 0 of setting up the replicating portfolios which will provide exactly for the guarantees is
still π (0). Ignoring survivorship, the amount of the risk premium to be received at time t
(t = 0, 1, . . . , n − 1) is c Ft = c P (1 − m)t St and the value of this at time 0 is c P (1 − m)t .
Allowing for survivorship, this value is c P (1 − m)t t px and so the value at time 0 of all
the risk premiums to be received is
n−1 t px t=0 c P (1 − m)t = c P ax:n i∗
¨ where
i∗ = m/(1 − m) so that 1/(1 + i∗ ) = 1 − m.
Example 14.6 In Example 14.3 the monthly management charge, m, was 0.25% of the
fund value and the GMDB option price was determined to be 2.7838% of the single
premium.
You are given that 0.20% per month is allocated to commission and administrative expenses. Determine whether the remaining 0.05% per month is suﬃcient to cover the risk
premium for the option.
Use the same basis as in Example 14.3.
Solution 14.6 The risk neutral value of the risk premium of c per month is
EQ cF0 + cF1/12 e−r/12 1/12 p60 + . . . + c F59/12 e−59r/12 59/12 p60
0
= c P S0 1 + (1 − m) 1/12 p60 + (1 − m)2 2/12 p60 + . . . + (1 − m)59 59/12 p60
(12) = 12 c P S0 a60:5
¨
where the annuity interest rate is i such that
1/12 vi = (1 − m) ⇒ i = (1 − m)−12 − 1 = 3.0493% per year. 14.6. RISK MANAGEMENT 651 The annuity value is 4.32662, so the value of the risk premium of 0.05% per month is
$259.60.
The value of the guarantee at the inception date, from Example 14.3, is $278.38 so the
risk premium of 0.05% per month is not suﬃcient to pay for the guarantee. The insurer
needs to revise the pricing structure for this product. 14.6 Risk management The option prices derived in this chapter are the cost of either buying the appropriate
options in the market, or internally hedging the options. If the insurer does not plan
to purchase or hedge the options, then the price or reserve amount calculated may be
inadequate. It would be inappropriate to charge an option premium using the BlackScholesMerton framework, and then invest the premium in bonds or stocks with no
consideration of the dynamic hedging implicit in the calculation of the cost. Thus, the
decision to use BlackScholesMerton pricing carries with it the consequential decision
either to buy the options or to hedge using the BlackScholesMerton framework.
Under the assumptions of the BlackScholesMerton model, and provided the mortality
and lapse experience is as assumed, the hedge portfolio will mature to the precise cost of
the guarantee. In reality the match will not be exact but will usually be very close. So
hedging is a form of risk mitigation. Choosing not to hedge may be a very risky strategy
– with associated probabilities of severe losses. (Generally, if the risk is not hedged,
the reserves required using the stochastic techniques of Chapter 12 will be considerably
greater than the hedge costs.)
One of the reasons why the hedge portfolio will not exactly meet the cost of the guarantee
is that under the BlackScholesMerton assumptions, the hedge portfolio should be continuously rebalanced. In reality, the rebalancing will be less frequent. A large portfolio 652 CHAPTER 14. EMBEDDED OPTIONS might be rebalanced daily, a smaller one at weekly or even monthly intervals.
If the hedge portfolio is rebalanced at discrete points in time (e.g. monthly), there will
be small costs (positive or negative) incurred as the previous hedge portfolio is adjusted
to create the new hedge portfolio. See Example 13.5.
The hedge portfolio value required at time t for an n year GMMB is, from Section 14.3.2,
π (t) = n−t px+t EtQ [e−r(n−t) h(n)] = n−t px+t v (t, n) where, as above, v (t, n) is the value at time t of the option maturing at time n, unconditional on the policyholder’s survival.
The hedge portfolio is invested partly in zero coupon bonds, maturing at time n, and
partly (in fact, a negative amount, i.e. a short sale) in stocks. The value of the stock part
of the hedge portfolio is
d
v (t, n) St
dSt n−t px+t and the value of the zero coupon bond part of the hedge portfolio is
π (t) − n−t px+t d
v (t, n)St .
dSt For a GMDB, the approach is identical, but the option value is a weighted average of
options of all possible maturity dates, so the hedge portfolio is a mixture of zero coupon
bonds of all possible maturity dates, and (short positions in) stocks. For example, when
the beneﬁt is payable immediately on death, the value at time t of the option is π (t),
where
n−t π (t) = v (t, w + t) w px+t µx+t+w dw. 0 The stock part of the hedge portfolio has value
n−t
0 St d
v (t, w + t)
dSt w p x+ t µx+t+w dw. 14.7. EMERGING COSTS 653 The value of the bond part of the hedge portfolio is the diﬀerence between π (t) and the
value of the stock part, so that the amount invested in a w year zero coupon bond at time
t is (loosely)
v (t, t + w) − St d
v (t, t + w)
dSt w p x+ t µx+t+w dw. The hedge strategy described in this section, which is called a deltahedge, uses only zero
coupon bonds and stocks to replicate the guarantee payoﬀ. More complex strategies are
also possible, bringing options and futures into the hedge, but these are beyond the scope
of this book.
The BlackScholesMerton valuation can be interpreted as a marketconsistent valuation,
by which we mean that the option sold in the ﬁnancial markets as a stand alone product
(rather than embedded in life insurance) would have the same value. Many jurisdictions
are moving towards market consistent valuation for accounting purposes, even where the
insurers do not use hedging. 14.7 Emerging costs Whether the insurer is hedging internally or buying the options to hedge, the proﬁt testing
of an equitylinked policy proceeds as described in Chapter 12. The insurer might proﬁt
test deterministically, using best estimate scenarios, and then stress test using diﬀerent
scenarios, or might test stochastically, using Monte Carlo simulation to generate the
scenarios for the increase in the stock prices in the policyholder’s fund. In this section,
we ﬁrst explore deterministic proﬁt testing, and then discuss how to make the proﬁt test
stochastic.
The cash ﬂow projection depends on the projected fund values. Suppose we are projecting
the emerging cash ﬂows for a single premium equitylinked policy with a term of n years
and with a GMDB and/or a GMMB, for a given stock price scenario. We assume all cash
ﬂows occur at intervals of 1/m years. 654 CHAPTER 14. EMBEDDED OPTIONS Assuming the insurer hedges the options internally, the income to and outgo from the
insurer’s fund for this contract arise as follows:
Income: + Initial frontendload expense deduction.
+ Regular management charge income.
+ Investment return on income over the 1/m year period. Outgo: − Expenses.
− Initial hedge cost, at t = 0.
− After the ﬁrst month, the hedge portfolio needs to be rebalanced; the cost is
the diﬀerence between the hedge value brought forward and the hedge required
to be carried forward.
− If the policyholder dies, there may be a GMDB liability.
− If the policyholder survives to maturity, there may be a GMMB liability. The part of this that diﬀers from Chapter 12 is the cost of rebalancing the hedge portfolio.
In Example 13.5, for a standard put option, we looked at calculating rebalancing errors
for a hedge portfolio adjusted monthly. The hedge portfolio adjustment in this chapter
follows the same principles, but with the complication that the option is contingent on
survival. As in Example 13.5, we assume that the hedge portfolio value is invested in a
delta hedge. If rebalancing is continuous (in practice, one or more times daily), then the
hedge adjustment will be (in practice, close to) zero, and the emerging guarantee cost
will be zero given that the experience in terms of stock price movements and survival is
in accordance with the models used. Under the model assumptions, the hedge is selfﬁnancing and exactly meets the guarantee costs. Also, if the hedge cost is used to buy
options in the market, there will be no hedge adjustment cost and no guarantee cost once
the options are purchased.
If the rebalancing takes place every 1/m years, then we need to model the rebalancing
costs. We break the hedge portfolio down into the stock part, assumed to be invested in 14.7. EMERGING COSTS 655 the underlying index {St }t≥0 , and the bond part, invested in a portfolio of zero coupon
bonds. Suppose the values of these two parts are Ψt St and Υt , respectively, so that
π (t) = Υt + Ψt St .
Then 1/m years later, the bond part of the hedge portfolio has appreciated by a factor
er/m and the stock part by a factor St+1/m /St . This means that, before rebalancing, the
1
value of the hedge portfolio is, say, π bf (t + m ), where
π bf (t + = Υt er/m + Ψt St+1/m . 1
)
m 1
The rebalanced hedge portfolio required at time t +1/m has value π (t + m ), but is required
1
only if the policyholder survives. If the policyholder dies, the guarantee payoﬀ is h(t + m ).
So the total cost at time t + 1/m of rebalancing the hedge, given that the policy was in
force at time t, is π (t + 1
)
m 1
m px+t − π bf (t + 1
)
m and the cost of the GMDB is
h(t + 1
)
m 1
m q x+ t . Note that these formulae need to be adjusted for the costs at the ﬁnal maturity date, n:
π (n) is zero since there is no longer any need to set up a hedge portfolio, and the cost of
the GMMB is h(n) 1 px+n− 1 .
m m If lapses are explicitly allowed for, then the mortality probability would be replaced by
an inforce survival probability.
In the following example, all of the concepts introduced in this chapter are illustrated as
we work through the process of pricing and proﬁttesting an equitylinked contract with
both a GMDB and a GMMB. 656 CHAPTER 14. EMBEDDED OPTIONS Example 14.7 An insurer issues a 5 year equitylinked policy to a life aged 60. The
single premium is P = $1 000 000. The beneﬁt on maturity or death is a return of the
policyholder’s fund, subject to a minimum of the initial premium. The death beneﬁt is
paid at the end of the month of death and is based on the fund value at that time.
Management charges of 0.3% per month are deducted from the fund at the start of each
month.
(a) Calculate the monthly risk premium (as part of the overall management charge)
required to fund the guarantees, assuming
(i) volatility is 25% per year, and
(ii) volatility is 20% per year.
Basis:
Survival model: Makeham’s law with A = 0.0001, B = 0.00035 and c = 1.075
Lapses: None
Risk free rate of interest: 5% per year, continuously compounded (b) The insurer is considering purchasing the options for the guarantees in the market; in
this case the price for the options would be based on the 25% volatility assumption.
Assuming that the monthly risk premium based on the 25% volatility assumption
is used to purchase the options for the GMDB and GMMB liabilities, proﬁt test the
contract for the two stock price scenarios below, using a risk discount rate of 10%
per year eﬀective, and using monthly time intervals. Use the basis from part (a),
assuming, additionally, that expenses incurred at the start of each month are 0.01%
of the fund, after deducting the management charge, plus $20. The two stock price
scenarios are
(i) stock prices in the policyholder’s fund increase each month by 0.65%, and
(ii) stock prices in the policyholder’s fund decrease each month by 0.05%. 14.7. EMERGING COSTS 657 (c) The alternative strategy for the insurer is to hedge internally. Calculate all the cash
1
2
ﬂows to and from the insurer’s fund at times 0, 12 and 12 per policy issued for the
following stock price scenarios
(i) stock prices in the policyholder’s fund increase each month by 0.65%,
(ii) stock prices in the policyholder’s fund decrease each month by 0.05%, and
(iii) S 1 = 1.0065, S 2 = 0.9995.
12 12 Assume that
the hedge cost is based on the 20% volatility assumption,
the hedge portfolio is rebalanced monthly,
expenses incurred at the start of each month are 0.025% of the fund, after
deducting the management charge, and
the insurer holds no additional reserves apart from the hedge portfolio for the
options.
Solution 14.7 (a) The payoﬀ function, h(t), for t = 1
, 2,
12 12 . . . , 59 ,
12 60
,
12 is h(t) = (P − Ft )+
where
Ft = P St (1 − m)12t
and m = 0.003. Let v (t, s) denote the value at t of the option given that it matures
at s (> t). Then
v (t, s) = EQ e−r(s−t) h(s)
t
= EQ e−r(s−t) P − P Ss (1 − m)12s
t + 658 CHAPTER 14. EMBEDDED OPTIONS = P e−r(s−t) Φ(−d2 (t, s)) − St (1 − m)12s Φ(−d1 (t, s))
where
d1 (t, s) = log(St (1 − m)12s ) + (r + σ 2 /2)(s − t)
√
σ s−t and
d2 (t, s) = d1 (t, s) − σ √ s − t. The option price at issue is
1
π (0) = v (0, 12 ) 1
12 2
qx + v (0, 12 ) ... 1
12 3
 1 qx + v (0, 12 )
12 2
12  1 qx +
12 + v (0, 60 ) 59  1 qx + v (0, 60 ) 60 px .
12
12
12 12 12 This gives the option price as
(i) 0.145954 P for σ = 0.25 per year, and
(ii) 0.112710 P for σ = 0.20 per year.
Next, we convert the premium to a regular charge on the fund, c, using
(12) π (0) = 12 c P a60:5
¨
where the interest rate for the annuity is i = (1 − m)−12 − 1 = 3.6712%, which gives
(12)
a60:5 = 4.26658. The charge on the fund is then
¨
(i) c = 0.00285 for σ = 0.25, and
(ii) c = 0.00220 for σ = 0.20. 14.7. EMERGING COSTS 659 (b) Following the convention of Chapter 12, we use the stock price scenarios to project
the policyholder’s fund value assuming that the policy stays in force throughout the
5 year term of the contract. From this projection we can project the management
charge income to the insurer’s fund at the start of each month. Outgo at the start
of the month comprises the risk premium for the option (which is paid to the option
provider), and the expenses.
The steps in this calculation are as follows. At time t = k/12, where k = 0, 1, . . . , 59,
assuming the policy is still in force,
– The policyholder’s fund, just before the deduction of the management charge,
is Ft , where
Ft = P (1 + g )k (1 − 0.003)k
and g is the rate of growth of the stock price.
– The amount transferred to the insurer’s fund in respect of the management
charge is
0.003 Ft .
– The insurer’s expenses, excluding the risk premium, are
0.0001 (1 − 0.003) Ft + 20.
– The risk premium is
0.00285 (1 − 0.003) Ft .
– The proﬁt to the insurer is
Prt = (0.003 − (1 − 0.003)(0.0001 + 0.00285)) Ft − 20. 660 CHAPTER 14.
Time, t Management
(months)
charge
0
3 000.00
1
3 010.44
2
3 020.92
.
.
.
.
.
.
58
3 669.78
59
3 682.55 Expenses Prt EMBEDDED OPTIONS 119.70
120.05
120.40
.
.
. Risk
premium
2 842.63
2 852.52
2 862.45
.
.
. t/12 p60 Πt 37.67
37.87
38.08
.
.
. 1
37.67
0.99775 37.79
0.99550 37.90
.
.
. 141.96
142.38 3 477.27
3 489.37 50.55
50.79 0.85582
0.85309 43.26
43.33 Table 14.2: Proﬁt test for Example 14.7 part (b), ﬁrst stock price scenario. – The proﬁt to the insurer, allowing for survivorship to time t, is
Πt = t p60 ((0.003 − (1 − 0.003)(0.0001 + 0.00285)) Ft − 20).
– The net present value of the proﬁt using a risk discount rate of 10% per year
is
59
k Π k 1.1− 12 . NPV = 12 k=0 Because the insurer is buying the options, there is no outgo for the insurer in respect
of the guarantees on death or maturity – the purchased options are assumed to cover
any liability. As there is no residual liability for the insurer for the contract, there
is no need to hold reserves. There are no end of month cash ﬂows, so we calculate
the proﬁt vector using cash ﬂows at the start of the month. Hence, Prt is the proﬁt
to the insurer at time t, assuming the policy is in force at that time, and Πt is the
proﬁt at time t assuming only that the policy was in force at time 0.
Some of the calculations for the scenario where the stock price grows at 0.65% per
month are presented in Table 14.2. 14.7. EMERGING COSTS 661 The net present value for this contract, using the 10% risk discount rate and the
ﬁrst stock price scenario, is $1 940.11.
The second stock price scenario, with stock prices falling by 0.05% each month,
gives a NPV of $1 463.93.
1
(c) The items of cash ﬂow for the insurer’s fund at times 0, 12 and
are shown in Table 14.3. The individual items are as follows: 2
,
12 per policy issued, Income: the management charge (1).
Outgo:
the insurer’s expenses (2),
the amount, if any, needed to increase death or maturity beneﬁts to the
guaranteed amount (3),
the amount needed to set up, or rebalance, the hedge portfolio (4), and
the net cash ﬂow (5), calculated as
(5) = (1) − (2) − (3) − (4).
The individual cash ﬂows at time t, per policy issued, are calculated as follows.
(1) Management charge
t p60 106 St × 0.99712t × 0.003. (2) Expenses
t p60 106 St × 0.99712t+1 × 0.00025. (3) Death beneﬁt (for t > 0)
( t− 1 p60 − t p60 ) 106 (1 − St × 0.99712t )+ .
12 662 CHAPTER 14. Time,
t Scenario Expenses
(2)
249.25
249.25
249.25 GMDB and
GMMB
(3)
0
0
0 EMBEDDED OPTIONS 0 (i)
(ii)
(iii) Management
charge
(1)
3 000
3 000
3 000 Cost of
Net cash
hedge
ﬂow
(4)
(5)
112 709.54 −109 958.79
112 709.54 −109 958.79
112 709.54 −109 958.79 1
12 (i)
(ii)
(iii) 3 003.67
2 982.78
3 003.67 249.56
247.82
249.56 0
7.87
0 −1 380.84
−1 380.40
−1 380.84 4 134.96
4 107.50
4 134.96 2
12 (i)
(ii)
(iii) 3 007.31
2 965.63
2 967.11 249.86
246.39
246.52 0
15.76
14.64 −1 388.47
−1 394.21
−1 352.25 4 145.92
4 097.68
4 058.20 Table 14.3: Cash ﬂows for Example 14.7 part (c). 14.7. EMERGING COSTS 663 (4) The cost of setting up the hedge portfolio at time 0 is the same for each stock
1
price scenario and is equal to 106 π (0). At time t = 12 the value of the hedge
portfolio is
106 (Υ0 e0.05/12 + Ψ0 S 1 ).
12 The cost of setting up the new hedge portfolio for each policy still in force is
1
106 π ( 12 ). Hence, the net cost of rebalancing the hedge portfolio at this time
per policy originally issued is
106 ( 1
12 p60 π (1/12) − (Υ0 e0.05/12 + Ψ0 S 1 )).
12 Similarly, the net cost of rebalancing the hedge portfolio at time
originally issued is
106 ( 2
12 p60 π (2/12) − 1
12 2
12 per policy p60 (Υ 1 e0.05/12 + Ψ 1 S 2 )).
12 12 12 The values of π (t), Υt and Ψt are shown in Table 14.4. We note several important points about this example.
1. Stock price scenarios (i) and (ii) used in parts (b) and (c) are not realistic, and
lead to unrealistic ﬁgures for the NPV. This is particularly true for the internal
hedging case, part (c). The NPV values for scenarios (i) and (ii), assuming internal
hedging and a risk discount rate of 10% per year, can be shown to be $99 944 and
$73 584, respectively. If the lognormal model for stock prices is appropriate, then
the expected present value (under the P measure) of the hedge rebalancing costs
will be close to zero. Under both scenarios (i) and (ii) in Example 14.7 the present
value is signiﬁcant and negative, meaning that the hedge portfolio value brought
forward each month is more than suﬃcient to pay for the guarantee and new hedge 664 CHAPTER 14. Time
t
0 π (t)
Υt
Ψt St EMBEDDED OPTIONS Investment scenario
(i)
(ii)
(iii)
112 710
112 710
112 710
0.417174
0.417174
0.417174
−0.304465 −0.304465 −0.304465 1
12 π (t)
Υt
Ψt St 111 342
113 478
111 342
0.415700
0.421369
0.415700
−0.304358 −0.307891 −0.304358 2
12 π (t)
Υt
Ψt St 109 956
114 253
114 097
0.414172
0.425626
0.425216
−0.304216 −0.311373 −0.311119 Table 14.4: Hedge portfolios for Example 14.7 part(c). 14.7. EMERGING COSTS 665 portfolio at the month end. This is because more realistic scenarios involve far more
substantial swings in stock price values, and it is these that generate positive hedge
portfolio rebalancing costs.
2. The comment above is more clearly illustrated when the proﬁt test is used with
stochastic stock price scenarios. In the table below we show some summary statistics
for 500 simulations of the NPV for part (c), again calculated using a risk discount
rate of 10% per year. The stock price scenarios were generated using a lognormal
model, with parameters µ = 8% per year, and volatility σ = 0.20 per year.
Mean NPV
$31 684 Standard Deviation
$37 332 5% quantile
−$23 447 50% quantile
$28 205 95% quantile
$99 861 We note that the NPV value for scenario (i) falls outside the 90% conﬁdence interval
for the net present value generated by stochastic simulation. This is because this
scenario is highly unrepresentative of the true stock price process. Overreliance
on deterministic scenarios can lead to poor risk management.
3. If we run a stochastic proﬁt test under part (b), where the option is purchased in the
market, the variability of simulated NPVs is very small. The net management charge
income is small, and the variability arising from the guarantee cost has been passed
on to the option provider. The mean NPV over 500 simulations is approximately
$2 137, and the standard deviation of the NPV is approximately $766, assuming the
same parameters for the stock price process as for (c) above.
4. If we neither hedge nor reserve for this option, and instead use the methods from
Chapter 12, the two deterministic scenarios give little indication of the variability of
the net present value. Using the ﬁrst scenario (increasing prices) generates a NPV
of $137 053 and using the second gives $2 381. Using stochastic simulation generates
a mean NPV of around $100 000 with a 5% quantile of approximately −$123 000. 666 14.8 CHAPTER 14. EMBEDDED OPTIONS Notes and further reading There is a wealth of literature on pricing and hedging embedded options. Hardy (2003)
gives some examples and information on practical ways to manage the risks. The options illustrated here are relatively straightforward. Much more convoluted options are
sold, particularly in association with variable annuity policies. For example, a guaranteed minimum withdrawal beneﬁt allows the policyholder the right to withdraw some
proportion of the fund for a ﬁxed time, even if the fund is exhausted. Also, the guarantee
may specify that after an introductory period, the policyholder could withdraw 5% of the
initial premium per year for 20 years. Other complicating features include resets where
the policyholder has the right to set the guarantee at the current fund value at certain
times during the contract. New variants are being created regularly, reﬂecting the strong
interest in these products in the market.
In Section 14.2 we noted three diﬀerences between options embedded in insurance policies
and standard options commonly traded in ﬁnancial markets. The ﬁrst was the life contingent nature of the beneﬁt and the second was the fact that the option is based on the fund
value rather than the underlying stocks. Both of these issues have been addressed in this
chapter. The third issue is the fact that embedded options are generally much longer term
than traded options. One of the implications is that the standard models for short term
options may not be appropriate over longer terms. The most important area of concern
here is the lognormal model for stock prices. There is considerable empirical evidence
that the lognormal model is not a good ﬁt for stock prices in the long run. This issue is
not discussed further here, but is important for more advanced treatment of equitylinked
insurance risk management. Sources for further information include Hardy (2003) and
Møller (1998).
The ﬁrst applications of modern ﬁnancial mathematics to equitylinked insurance can be
found in Brennan and Schwartz (1976) and Boyle and Schwartz (1977).
In some countries annual premium equitylinked contracts are common. We have not 14.8. NOTES AND FURTHER READING 667 discussed these in this chapter, as the valuation and risk management is more complicated
and requires more advanced ﬁnancial mathematics. Bacinello (2003) discusses an Italian
style annual premium policy.
Ledlie et al (2008) give an introduction to some of the issues around equitylinked insurance, including a discussion of a guaranteed minimum income beneﬁt, another more
complex embedded option. 668 14.9 CHAPTER 14. EMBEDDED OPTIONS Exercises Exercise 14.1 An insurer is designing a 10 year single premium variable annuity policy
with a guaranteed maturity beneﬁt of 85% of the single premium.
(a) Calculate the value of the GMMB at the issue date for a single premium of $100.
(b) Calculate the value of the GMMB as a regular annual deduction from the fund.
(c) Calculate the value of the GMMB 2 years after issue, assuming that the policy is still
in force, and that the underlying stock prices have decreased by 5% since inception.
Basis and policy information:
Age at issue:
Front end expense loading:
Annual management charge:
Mortality:
Lapses:
Risk free rate:
Volatility: 60
2%
2% at each year end
Standard ultimate survival model
5% at each year end
4% per year, continuously compounded
20% per year Exercise 14.2 An insurer issues a 10 year equitylinked insurance policy to a life aged
60. A single premium of $10 000 is invested in an equity fund. Management charges at a
rate of 3% per year are deducted daily. At the end of the month of death before age 70,
the death beneﬁt is 105% of the policyholder’s fund subject to a minimum of the initial
premium.
(a) Calculate the price of the death beneﬁt at issue. 14.9. EXERCISES 669 (b) Express the cost of the death beneﬁt as a continuous charge on the fund.
Basis:
Mortality:
Risk free rate:
Volatility:
Lapses: Standard ultimate survival model
4% per year, continuously compounded
25% per year
None Exercise 14.3 An insurer issues a range of 10 year variable annuity guarantees. Assume an investor deposits a single premium of $100 000. The policy carries a guaranteed
minimum maturity beneﬁt of 100% of the premium.
(a) Calculate the probability that the guaranteed minimum maturity beneﬁt will mature
inthemoney (i.e. the probability that the fund at the maturity date is worth less
than 100% of the single premium) under the P measure.
(b) Calculate the probability that the guaranteed minimum maturity beneﬁt will mature
inthemoney under the Qmeasure.
(c) Calculate the expected present value of the option payoﬀ under the P measure,
discounting at the risk free rate.
(d) Calculate the price of the option.
(e) A colleague has suggested the value of the option should be the EPV of the guarantee
under the P measure, analogous to the value of term insurance liabilities. Explain
why this value would not be suitable.
(f) For options that are complicated to value analytically we can use Monte Carlo
simulation to ﬁnd the value. We simulate the payoﬀ under the risk neutral measure, 670 CHAPTER 14. EMBEDDED OPTIONS discount at the risk free rate and take the mean value to estimate the Qmeasure
expectation. Use Monte Carlo simulation to estimate the value of this option with
1 000 scenarios, and comment on the accuracy of your estimate.
Basis:
Mortality:
Stock price appreciation:
Risk free rate of interest:
Management charges: None
Lognormallty distributed, with µ = 0.08 per year,
σ = 0.25 per year
4% per year, continuously compounded
3% of the fund per year, in advance Exercise 14.4 An insurer issues a single premium variable annuity contract with a 10
year term. There is a guaranteed minimum maturity beneﬁt equal to the initial premium
of $100.
After ﬁve years the policyholder’s fund value has increased to 110% of the initial premium.
The insurer oﬀers the policyholder a reset option, under which the policyholder may reset
the guarantee to the current fund level, in which case the remaining term of the policy
will be increased to 10 years.
(a) Determine which of the original guarantee and the reset guarantee has greater value
at the reset date.
(b) Determine the threshold value for F5 (i.e. the fund at time 5) at which the option
to reset becomes more valuable than the original option.
Basis:
Mortality:
Volatility: None
σ = 0.18 per year 14.9. EXERCISES
Risk free rate:
Management charges:
Frontendload charge: 671
5% per year, continuously compounded
1% of the fund per year, in advance
3% Exercise 14.5 An insurer issues a 5 year single premium equitylinked insurance policy
to (60) with guaranteed minimum maturity beneﬁt of 100% of the initial premium. The
premium is $100 000. Management fees of 0.25% of the fund are deducted at the start of
each month.
(a) Verify that the guarantee cost expressed as a monthly deduction is 0.17% of the
fund.
(b) The actuary is proﬁt testing this contract using a stochastic proﬁt test. The actuary
ﬁrst works out the hedge rebalancing cost each month then inserts that into the
proﬁt test.
The stock price ﬁgures in Table 14.5 represent one randomly generated scenario. The
table shows the stock price index values for each month in the 60 month scenario.
(i) Table 14.6 shows the ﬁrst two rows of the hedge rebalancing cost table. Use
the stock price scenario in Table 14.5 to complete this table. Calculate the net
present value of the hedge rebalance costs at an eﬀective rate of interest of 5%
per year.
(ii) Table 14.7 shows the ﬁrst two rows of the proﬁt test for this scenario. The
insurer uses the full cost of the option at the start of the contract to pay for
the hedge portfolio.
Complete the proﬁt test and determine the proﬁt margin (NPV as a percentage
of the single premium) for this scenario. 672 CHAPTER 14. EMBEDDED OPTIONS (iii) State with reasons whether you would expect this contract to be proﬁtable, on
average, over a large number of simulations.
Basis for hedging and proﬁt
Mortality:
Lapses:
Risk free rate:
Volatility:
Incurred expenses – Initial:
Incurred expenses – Renewal:
Risk discount rate: test calculations:
Standard ultimate survival model
None
5% per year, continuously compounded
20% per year
1% of the premium
0.065% of the fund before management charge deduction,
monthly in advance from the second month
10% per year 14.9. EXERCISES t
0
1
2
3
4
5
6
7
8
9
10
11
12 St
1.00000
0.95449
0.96745
0.97371
1.01158
1.01181
0.93137
0.98733
0.89062
0.91293
0.90374
0.88248
0.92712 673 t St t St t St t St 13
14
15
16
17
18
19
20
21
22
23
24 0.92420
0.95545
1.02563
1.13167
1.25234
1.10877
1.10038
0.99481
1.04213
1.07980
1.14174
1.12324 25
26
27
28
29
30
31
32
33
34
35
36 1.09292
1.17395
1.27355
1.32486
1.31999
1.24565
1.20481
1.18405
1.23876
1.15140
1.09478
1.03564 37
38
39
40
41
42
43
44
45
46
47
48 1.09203
1.10988
1.05115
1.05659
1.18018
1.20185
1.34264
1.37309
1.39327
1.40633
1.41652
1.43076 49
50
51
52
53
54
55
56
57
58
59
60 1.34578
1.42368
1.50309
1.63410
1.45134
1.46399
1.40476
1.44512
1.39672
1.30130
1.25762
1.19427 Table 14.5: Single scenario of stock prices for stochastic proﬁt test for Exercise 14.5. Time
St
Option cost Stock part
(months)
at t
of hedge at t
0
1.00000
0.09927
−0.17978
1
0.95449
0.11317
−0.19904
2
0.96745
0.10960
−0.19570
.
.
.
59
60 1.25762
1.19427 0.00000
0.00000 −0.00001
– Bond part
of hedge at t
0.27905
0.31221
0.30530 Hedge b/f 0.00001
– 0.00006
0.00000 0
0.10862
0.11177 Table 14.6: Hedge rebalance table for Exercise 14.5, in $100 000s. Hedging
Rebalance cost
–
0.00455
−0.00217
−0.00006
0.00000 674 CHAPTER 14. Time, t
Ft Management
(months)
Costs
0
100 000.00
250.00
1
95 210.22
238.03
2
96 261.70
240.65
.
.
. Expenses
1 000.00
61.89
62.57 EMBEDDED OPTIONS Hedge costs Prt 9 927.14 −10 677.14
454.94
−278.81
−217.31
395.40 Table 14.7: Proﬁt test table for Exercise 14.5, in $s. 14.9. EXERCISES 675 Answers to selected exercises
14.1 (a) $6.47
(b) 0.72%
(c) $8.63 14.2 (a) $742.63
(b) 0.88% 14.3 (a) 0.26545
(b) 0.60819
(c) $6 033
(d) $18 429 14.4 (a) The original option value is $4.95 and the reset option value is $6.14.
(b) At F5 = 1.05 both options have value $5.86 14.5 (b) (i) The NPV of rebalancing costs is −$4 315
(ii) 2.226%. (iii) We note that the initial hedge cost plus initial expenses converts to a
monthly outgo of 0.19% of the fund; adding the incurred expenses, this
comes to 0.255%, compared with income of 0.25% of the fund. Overall we
would not expect this contract to be proﬁtable on these terms. 676 CHAPTER 14. EMBEDDED OPTIONS Appendix A
Probability theory
A.1 Probability distributions In this appendix we give a very brief description of the probability distributions used in
this book. Derivations of the results quoted in this appendix can be found in standard
introductory textbooks on probability theory. A.1.1 Binomial distribution If a random variable X has a binomial distribution with parameters n and p, where n is
a positive integer and 0 < p < 1, then its probability function is
Pr[X = x] = nx
p (1 − p)n−x
x for x = 0, 1, 2, ..., n. This distribution has mean np and variance np(1 − p), and we write
X ∼ B (n, p).
The moment generating function is
MX (t) = (pet + 1 − p)n . (A.1)
677 678 APPENDIX A. PROBABILITY THEORY A.1.2 Uniform distribution If a random variable X has a uniform distribution on the interval (a, b), then it has
distribution function
Pr[X ≤ x] = x−a
b−a for a ≤ x ≤ b, and has probability density function
f (x) = 1
b−a for a < x < b. This distribution has mean (a + b)/2 and variance (b − a)2 /12, and we
write X ∼ U (a, b). A.1.3 Normal distribution If a random variable X has a normal distribution with parameters µ and σ 2 then its
probability density function is
1
f (x) = √ exp
σ 2π −(x − µ)2
2σ 2 for −∞ < x < ∞, where −∞ < µ < ∞ and σ > 0. This distribution has mean µ and
variance σ 2 , and we write X ∼ N (µ, σ 2 ).
The random variable Z deﬁned by the transformation Z = (X − µ)/σ has mean 0 and
variance 1 and is said to have a standard normal distribution. A common notation
is Pr[Z ≤ z ] = Φ(z ), and as the probability density function is symmetric about 0,
Φ(z ) = 1 − Φ(−z ).
The traditional approach to computing probabilities for a normal random variable is to
use the relationship
Pr[X ≤ x] = Pr[Z ≤ (x − µ)/σ ] A.1. PROBABILITY DISTRIBUTIONS 679 and to ﬁnd the right hand side from tables of the standard normal distribution, or from
an approximation such as
Φ(x) ≈ 1 − 1
2 1 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + a6 x 6 −16 for x ≥ 0 where
a1 = 0.0498673470, a4 = 0.0000380036, a2 = 0.0211410061, a5 = 0.0000488906, a3 = 0.0032776263, a6 = 0.0000053830. The absolute value of the error in this approximation is less than 1.5 × 10−7 .
There are plenty of software packages that compute values of the normal distribution
function. For example, in Excel we can ﬁnd Pr[X ≤ x] from the NORMDIST command
as
= NORMDIST( x, µ, σ, TRUE )
where the value TRUE for the ﬁnal parameter indicates that we want to obtain the distribution function. (Changing this parameter to FALSE gives the value of the probability
density function at x.)
Similarly we can ﬁnd percentiles of a normal distribution using either approximations
or software. Suppose we want to ﬁnd the value xp such that Pr[Z > xp ] = p where
Z ∼ N (0, 1) and 0 < p ≤ 0.5. We can ﬁnd this approximately as
xp = t −
where t = a0 + a1 t + a2 t2
1 + d1 t + d2 t2 + d3 t3 log(1/p2 ) and a0 = 2.515517, b1 = 1.432788, 680 APPENDIX A. PROBABILITY THEORY
a1 = 0.802853, b1 = 0.189269, a2 = 0.010328, b2 = 0.001308. The absolute value of the error in this approximation is less than 4.5 × 10−4 . Using the
symmetry of the normal distribution we can deal with the case p > 0.5, but in practical
actuarial applications this case rarely arises.
In Excel, we use the NORMINV command to ﬁnd percentiles. Speciﬁcally, we can ﬁnd
x such that Pr[X ≤ x] = p using
= NORMINV (p, µ, σ ). A.1.4 Lognormal distribution If a random variable X has a lognormal distribution with parameters µ and σ 2 then its
probability density function is
f (x) = 1
√ xσ 2π −(log x − µ)2
2σ 2 exp for x > 0, where −∞ < µ < ∞ and σ > 0. This distribution has mean exp{µ + σ 2 /2}
and variance exp{2µ + σ 2 }(exp{σ 2 } − 1), and we write X ∼ LN (µ, σ 2 ).
We can calculate probabilities as follows. We know that
x Pr[X ≤ x] = 0 1
√ y σ 2π exp −(log y − µ)2
2σ 2 dy. Now substitute z = log y , so that the range of the integral changes to (−∞, log x), with
dz = dy/y . Then
log x Pr[X ≤ x] = −∞ 1
√ exp
σ 2π −(z − µ)2
2σ 2 dz A.1. PROBABILITY DISTRIBUTIONS
(log x−µ)/σ = −∞ = Pr Z ≤ 1
√ exp{−t2 /2}dt
2π log x − µ
σ log x − µ
σ =Φ 681 , where Z has a standard normal distribution. Thus, we can compute probabilities for a
lognormally distributed random variable from the standard normal distribution.
The above argument also shows that if X ∼ LN (µ, σ 2 ), then log X ∼ N (µ, σ 2 ).
In Chapter 13 we used the result that if X ∼ LN (µ, σ 2 ) then
a log a − µ − σ 2
σ xf (x)dx = exp{µ + σ 2 /2}Φ 0 . To show this, we ﬁrst note that
a a xf (x)dx =
0 0 1
√ exp
σ 2π −(log x − µ)2
2σ 2 dx, and the substitution z = log x gives
a log a xf (x)dx =
−∞ 0 1
√ exp
σ 2π −(z − µ)2
2σ 2 exp{z }dz. Combining the exponential terms, the exponent becomes
z− (z − µ)2
−1 2
=
z − 2µz + µ2 − 2σ 2 z
2
2
2σ
2σ
= −1 2
z − 2(µ + σ 2 )z + µ2
2
2σ = −1 2
z − 2(µ + σ 2 )z + (µ + σ 2 )2 + µ2 − (µ + σ 2 )2
2
2σ (A.2) 682 APPENDIX A. PROBABILITY THEORY = −1
(z − (µ + σ 2 ))2 − 2µσ 2 − σ 4
2
2σ
2 − (z − (µ + σ 2 ))
σ2
=
+µ+ .
2σ 2
2
This technique is known as ‘completing the square’ and is very useful in problems involving
normal or lognormal random variables. We can now write
a log a xf (x)dx =
−∞ 0 − (z − (µ + σ 2 ))
2σ 2 1
√ exp
σ 2π σ2
= exp µ +
2 log a
−∞ 1
√ exp
σ 2π 2 exp µ + σ2
2 − (z − (µ + σ 2 ))
2σ 2 dz 2 dz. Now the integrand is the probability density function of normal random variable with
mean µ + σ 2 and variance σ 2 , and so
log a
−∞ 1
√ exp
σ 2π − (z − (µ + σ 2 ))
2σ 2 2 dz = Φ log a − µ − σ 2
σ , giving formula (A.2). We note that lim Φ a→∞ log a − µ − σ 2
σ = 1, and from this result and formula (A.2) we see that the mean of the lognormal distribution
with parameters µ and σ 2 is exp µ + σ2
2 . A.2. THE CENTRAL LIMIT THEOREM A.2 683 The central limit theorem The central limit theorem is a very important result in probability theory. Suppose that
X1 , X2 , X3 , . . . is a sequence of independent and identically distributed random variables,
each having mean µ and variance σ 2 . Now deﬁne the sum Sn = n=1 Xi so that E[Sn ] =
i
nµ and V[Sn ] = nσ 2 . The central limit theorem states that
lim Pr n→∞ Sn − nµ
√
≤ x = Φ(x)
σn where Φ is the standard normal distribution function.
The central limit theorem can be used to justify approximating the distribution of a
(ﬁnite) sum of independent and identically distributed random variables by a normal
distribution. For example, suppose that each Xi has a Bernoulli distribution (i.e. a B (1, p)
distribution). Then using moment generating functions we see that the distribution of Sn
is B (n, p) since
E[exp{tSn }] = E[exp{t(X1 + X2 + ... + Xn )}]
n =
i=1 E[exp{tXi }] n =
i=1 (pet + 1 − p) = (pet + 1 − p)n .
(Here we have used in order: independence, identical distribution and formula (A.1) with
n = 1.) The uniqueness of moment generating functions tells us that Sn ∼ B (n, p). Thus
we can think of a binomial random variable as the sum of Bernoulli random variables,
and, provided the number of variables being summed is large, we can approximate the
distribution of this sum by a normal distribution. 684 APPENDIX A. PROBABILITY THEORY A.3 Functions of a random variable In many places in this book we have considered functions of a random variable. For
example, in Chapter 4 we considered v Tx where Tx is a random variable representing future
lifetime. We have also evaluated the expected value and higher moments of functions of
a random variable. Here, we brieﬂy review the theory that we have applied, considering
separately random variables that follow discrete, continuous and mixed distributions. We
quote results only, giving references for these results in Section A.5. A.3.1 Discrete random variables We ﬁrst consider a discrete random variable, X, with probability function Pr[X = x] for
x = 0, 1, 2, . . . . Let g be a function and let Y = g (X ), so that the possible values for Y
are g (0), g (1), g (2), . . . . Then for x = 0, 1, 2, . . . , Y takes the value g (x) if X takes the
value x. Thus,
Pr[Y = g (x)] = Pr[X = x],
and so
∞ E[Y ] = g (x) Pr[Y = g (x)] = x=0 ∞ g (x) Pr[X = x]. (A.3) x=0 Thus, we can compute E[Y ] in terms of the probability function of X . Higher moments
are similarly computed. For r = 1, 2, 3, . . . we have
E[Y r ] = ∞ g (x)r Pr[X = x]. x=0 For example, suppose that X has probability function
Pr[X = x] = pq x−1 A.3. FUNCTIONS OF A RANDOM VARIABLE 685 for x = 1, 2, 3, . . . , and deﬁne Y = v X where 0 < v < 1. Then g (x) = v x and
r E[Y ] = ∞ v xr pq x−1 = x=1 A.3.2 pv r
.
1 − qv r Continuous random variables We next consider the situation of a continuous random variable, X , distributed on (0, ∞)
with probability density function f (x) for x > 0. Consider a function g, let g −1 denote
the inverse of this function, and deﬁne Y = g (X ). Then we can compute the expected
value of Y as
∞ E[Y ] = E[g (X )] = g (x)f (x)dx. (A.4) 0 As in the case of discrete random variables, the expected value of Y can be found without
explicitly stating the distribution of Y , and higher moments can be found similarly. Note
the analogy with equation (A.3) – probability function has been replaced by probability
density function, and summation by integration.
It can be shown that Y has a probability density function, which we denote h, given by
h(y ) = f g −1 (y ) d −1
g (y )
dy (A.5) provided that g is a monotone function. However, formula (A.4) allows us to compute
the expected value of Y without ﬁnding its probability density function.
For example, suppose that X has an exponential distribution with parameter λ. Now
deﬁne Y = e−δX , where δ > 0. Then by formula (A.4) with g (y ) = e−δy ,
∞ E[Y ] =
0 e−δy λe−λy dy = λ
.
λ+δ The alternative (and more complicated) approach to ﬁnding E[Y ] is to ﬁrst identify the
distribution of Y , then ﬁnd its mean. To follow this approach, we ﬁrst note that if 686 APPENDIX A. PROBABILITY THEORY g (y ) = e−δy , then g −1 (y ) = (−1/δ ) log y and so
d −1
−1
g (y ) =
.
dy
δy
By formula (A.5), Y has probability density function h(y ), which is deﬁned for 0 < y < 1
(since X > 0 implies that 0 < e−δX < 1 as δ > 0), with
h(y ) = λ exp{(λ/δ ) log y } 1
δy λ (λ/δ)−1
y
.
δ =
Thus 1 E[Y ] =
0 λ
yh(y )dy =
δ 1 y
0 λ/δ λ y (λ/δ)+1
dy =
δ (λ/δ ) + 1 1 =
0 λ
.
λ+δ We could also have evaluated this integral by noting that Y has a beta distribution with
parameters λ/δ and 1. In any event, the key point is that a function of a random variable
is itself a random variable with its own distribution, but because of formula (A.4) it is
not necessary to ﬁnd this distribution to evaluate its moments. A.3.3 Mixed random variables Most of the mixed random variables we have encountered in this book have a probability
density function over an interval and a mass of probability at one point only. For example,
under an endowment insurance with term n years, there is probability density associated
with payment of the sum insured at time t for 0 < t < n, and a mass of probability
associated with payment at time n. In that situation we deﬁned the random variable (see
Section 4.4.7)
Z= v Tx
vn if Tx < n,
if Tx ≥ n. A.4. CONDITIONAL EXPECTATION AND CONDITIONAL VARIANCE 687 More generally, suppose that X is a random variable with probability density function
f over some interval (or possibly intervals) which we denote by I , and has masses of
probability, Pr[X = xi ], at points x1 , x2 , x3 , . . . . Then if we deﬁne Y = g (X ), we have
E[Y ] = g (x)f (x)dx +
I g (xi ) Pr[X = xi ].
i For example, suppose that Pr[X ≤ x] = 1 − e−λx for 0 < x < n, and Pr[X = n] = e−λn .
Then X has probability density function f (x) = λe−λx for 0 < x < n, and has a mass of
probability of amount e−λn at n. If we deﬁne Y = e−δX , then
n E[Y ] = e−δx λe−λx + e−δn e−λn 0 =
= A.4 λ
1 − e−(λ+δ)n + e−(λ+δ)n
λ+δ
1
λ + δ e−(λ+δ)n .
λ+δ Conditional expectation and conditional variance Consider two random variables X and Y whose ﬁrst two moments exist. We can ﬁnd the
mean and variance of Y in terms of the conditional mean and variance of Y given X . In
particular,
E[Y ] = E [E[Y X ]] (A.6) V[Y ] = E [V[Y X ]] + V[E[Y X ]]. (A.7) and 688 APPENDIX A. PROBABILITY THEORY These formulae hold generally, but to prove them we restrict ourselves here to the situation
when both X and Y are discrete random variables. Consider ﬁrst expression (A.6). We
note that for a function g of X and Y , we have
E[g (X, Y )] = g (x, y ) Pr[X = x, Y = y ]
x (A.8) y (this is just the bivariate version of formula (A.3)). By the rules of conditional probability,
Pr[X = x, Y = y ] = Pr[Y = y X = x] Pr[X = x].
Then setting g (X, Y ) = Y and using (A.8) and (A.9) we obtain
E[Y ] =
x = y y Pr[Y = y X = x] Pr[X = x] Pr[X = x]
x =
x y y Pr[Y = y X = x] Pr[X = x]E[Y X = x] = E [E[Y X ]] .
To obtain formula (A.7) we have
V[Y ] = E[Y 2 ] − E[Y ]2
= E[E[Y 2 X ]] − E[Y ]2
= E V[Y X ] + E[Y X ]2 − E[Y ]2
= E [V[Y X ]] + E E[Y X ]2 − E [E[Y X ]]2
= E [V[Y X ]] + V [E[Y X ]] . (A.9) A.5. NOTES AND FURTHER READING A.5 689 Notes and further reading Further details on the probability theory contained in this appendix can be found in texts
such as Grimmett and Welsh (1986) and Hogg and Tanis (2005). The approximations for
the standard normal distribution can be found in Abramovitz and Stegun (1965). 690 APPENDIX A. PROBABILITY THEORY Appendix B
Numerical techniques
B.1 Numerical integration In this section we illustrate two methods of numerical integration. The ﬁrst, the trapezium rule has the advantage of simplicity, but its main disadvantage is the amount of
computation involved for the method to be very accurate. The second, repeated Simpson’s rule, is not quite as straightforward, but is usually more accurate. We now outline
each method, and give numerical illustrations of both. Further details can be found in
the references in Section B.3.
Our aim in the next two sections is to evaluate numerically b I= f (x)dx
a for some function f .
691 692 APPENDIX B. NUMERICAL TECHNIQUES B.1.1 The trapezium rule Under the trapezium rule, the interval (a, b) is split into n intervals, each of length h =
(b − a)/n. Thus, we can write I as
a+h I= a+2h f (x)dx +
a = a+nh f (x)dx + . . . +
a+h n−1
j =0 f (x)dx
a+(n−1)h a+(j +1)h f (x)dx.
a+jh We obtain the value of I under the trapezium rule by assuming that f is a linear function
in each interval so that under this assumption
a+(j +1)h f (x)dx =
a+jh h
2 (f (a + jh) + f (a + (j + 1)h)) , and hence
I=h
=h 1
f (a)
2 + f (a + h) + f (a + 2h) + . . . + f (a + (n − 1)h) + 1 f (b)
2 1
f (a)
2 + n−1 f (a + jh) + 1 f (b) .
2 j =1 To illustrate the application of the trapezium rule, consider
20 I∗ = e−0.05x dx. 0 We have chosen this integral as we can evaluate it exactly as
1
1 − e−0.05×20 = 12.64241,
0.05
and hence we can compare evaluation by numerical integration with the true value. We
have a = 0 and b = 20, and for our numerical illustration we have set n = 20, 40, 80, 160
I∗ = B.1. NUMERICAL INTEGRATION 693 n
I∗
20 12.64504
40 12.64307
80 12.64258
160 12.64245
320 12.64242
Table B.1: Values of I ∗ under the trapezium rule.
and 320, so that the values of h are 1, 0.5, 0.25, 0.125 and 0.0625. Table B.1 shows the
results. We see that in this example we need a small value of h to obtain an answer that
is correct to 4 decimal places, but we note that the percentage error is small in all cases. B.1.2 Repeated Simpson’s rule This rule is based on Simpson’s rule which gives the following approximation:
a+2h f (x)dx ≈ a h
3 (f (a) + 4f (a + h) + f (a + 2h)) . This approximation arises by approximating the function f by a quadratic function that
goes through the three points (a, f (a)), (a + h, f (a + h)) and (a + 2h, f (a + 2h)). Repeated
application of this result leads to the repeated Simpson’s rule, namely
n b
a f (x)dx ≈ h
3 f (a) + 4 where h = (b − a)/2n.
Let us again consider
20 I∗ =
0 e−0.05x dx. j =1 f (a + (2j − 1)h) + 2 n−1
j =1 f (a + 2jh) + f (b) 694 APPENDIX B. NUMERICAL TECHNIQUES
n
10
20
40 I∗
12.6424116
12.6424112
12.6424112 Table B.2: Values of I ∗ under repeated Simpson’s rule.
To seven decimal places, I ∗ = 12.6424112 and Table B.2 shows numerical values for I ∗
when n = 10, 20 and 40.
We see from Table B.2 that the calculations are considerably more accurate than under
the trapezium rule. The reason for this is that the error in applying the trapezium rule is
(b − a)3
f (c)
12n2
for some c, where a < c < b, whilst under repeated Simpson’s rule the error is
(b − a)5 (4)
f (c)
2, 880n4
for some c, where a < c < b. B.1.3 Integrals over an inﬁnite interval Many situations arise under which we have to ﬁnd the numerical value of an integral over
the interval (0, ∞). For example, we saw in Chapter 2 that the complete expectation of
life is given by
∞ ◦ ex = t px dt. 0 To evaluate such integrals numerically, it usually suﬃces to take a pragmatic approach.
For example, looking at the integrand in the above expression, we might say that the
probability of a life aged x surviving a further 120 − x years is very small, and so we might B.1. NUMERICAL INTEGRATION
m
60
70
80
90
100 695
Im
34.67970
34.75059
34.75155
34.75155
34.75155 Table B.3: Values of Im .
replace the upper limit of integration by 120 − x, and perform numerical integration over
the ﬁnite interval (0, 120 − x). We could then assess our answer by considering a wider
interval, say (0, 130 − x).
To illustrate this idea, consider the following integral from Section 2.6.2 where we com◦
puted ex for a range of values for x in Table 2.2. Table B.3 shows values of
m Im = t p40 dt 0 for a range of values for m. These values have been calculated using repeated Simpson’s
rule. We set n = 120 for m = 60, then changed the value of n for each subsequent value
of m in such a way that the value of h was unchanged. For example, with m = 70, setting
n = 140 results in h = 0.25, which is the same value of h obtained when m = 60 and
n = 120. This maintains consistency between successive calculations of Im values. For
example,
70 I70 = I60 + t p40 dt, 60 and setting n = 140 to compute I70 then gives the value we computed for I60 with n = 120.
◦
From this table our conclusion is that, to 5 decimal places, e40 = 34.75155. 696 APPENDIX B. NUMERICAL TECHNIQUES B.2 Woolhouse’s formula Woolhouse’s formula was used in Chapter 4. Here we give an indication of how this
formula arises. We use the EulerMaclaurin formula which is concerned with numerical
integration. This formula gives a series expansion for the integral of a function, assuming
that the function is diﬀerentiable a certain number of times. For a function f , the EulerMaclaurin formula can be written as
N b f (x)dx = h
a i=0 f (a + ih) − 1
(f (a) + f (b))
2 2 + h (f (a) − f (b)) −
12 h4
720 (f (a) − f (b)) + . . . , (B.1) where h = (b − a)/N , N is an integer, and the terms we have omitted involve higher
derivatives of f . We shall apply this formula twice, in each case ignoring second and
higher order derivatives of f .
First, setting a = 0 and b = N = n (so that h = 1), the left hand side of (B.1) is
n i=0 f (i) − 1 (f (0) + f (n)) +
2 1
12 (f (0) − f (n)) . (B.2) Second, setting a = 0, b = n and N = mn for some integer m > 1 (so that h = 1/m), the
left hand side of (B.1) is
1
m mn i=0 f (i/m) − 1 (f (0) + f (n))
2 + 1
12m2 (f (0) − f (n)) . (B.3) As each of (B.2) and (B.3) approximates the same quantity, we can obtain an approxi1
mation to m mn f (i/m) by equating them, so that
i=0
1
m mn i=0 n f (i/m) ≈ i=0 f (i) − m−1
m2 − 1
(f (0) + f (n)) +
(f (0) − f (n)) .
2m
12m2 B.3. NOTES AND FURTHER READING 697
(B.4) The left hand side of formula (B.4) gives the ﬁrst three terms of Woolhouse’s formula,
and in actuarial applications it usually suﬃces to apply only these terms. B.3 Notes and further reading A list of numerical integration methods is given in Abramovitz and Stegun (1965). Details of the derivation of the trapezium rule and repeated Simpson’s rule can be found
in standard texts on numerical methods such as Burden et al (1978) and Ralston and
Rabinowitz (1978). 698 APPENDIX B. NUMERICAL TECHNIQUES Appendix C
Simulation
C.1 The inverse transform method The inverse transform method allows us to simulate observations of a random variable,
X , when we have a uniform U (0, 1) random number generator available.
The method states that if F (x) = Pr[X ≤ x] and u is a random drawing from the U (0, 1)
distribution, then
x = F −1 (u)
is our simulated value of X .
The result follows for the following reason: if U ∼ U (0, 1), then F −1 (U ) has the same
distribution as X . To show this, we assume for simplicity that the distribution function
F is continuous – this is not essential for the method, it just gives a simpler proof. First,
we note that as the distribution function F is continuous, it is a monotonic increasing
function. Next, we know from the properties of the uniform distribution on (0, 1) that for
0 ≤ y ≤ 1,
Pr[U ≤ y ] = y.
699 700 APPENDIX C. SIMULATION ˜
Now let X = F −1 (U ). Then
˜
Pr[X ≤ x] = Pr[F −1 (U ) ≤ x]
= Pr[U ≤ F (x)]
since F is a monotonic increasing function. As Pr[U ≤ F (x)] = F (x), we have
˜
Pr[X ≤ x] = F (x) = Pr[X ≤ x]
˜
which shows that X and X have the same distribution function.
Example C.1 Simulate three values from an exponential distribution with mean 100
using the three random drawings
u1 = 0.1254, u2 = 0.4529, u3 = 0.7548, from the U (0, 1) distribution.
Solution C.1 Let F denote the distribution function of an exponentially distributed
random variable with mean 100, so that
F (x) = 1 − exp{−x/100}.
Then setting u = F −1 (x) gives
x = −100 log(1 − u),
and hence our three simulated values from this exponential distribution are
−100 log 0.8746 = 13.399,
−100 log 0.5471 = 60.312, C.2. SIMULATION FROM A NORMAL DISTRIBUTION 701 −100 log 0.2452 = 140.57. C.2 Simulation from a normal distribution In Chapter 10 we noted that Excel could be used to generate random numbers from a
normal distribution. In many situations, for example if we wish to create a large number
of simulations of an insurance portfolio over a long time period, it is much more eﬀective in
terms of computing time to use a programming language rather than a spreadsheet. Most
programming languages do not have an inbuilt function to generate random numbers
from a normal distribution, but do have a random number generator, that is they have
an inbuilt function to generate (pseudo)random numbers from the U (0, 1) distribution.
Without going into details, we now state the two most common approaches to simulating
values from a standard normal distribution. The detail behind these ‘recipes’ can be
found in the references in Section C.3. C.2.1 The BoxMuller method The BoxMuller method is to ﬁrst simulate two values, u1 and u2 , from a U (0, 1) distribution, then to compute the pair
x= −2 log u1 cos(2π u2 ) y= −2 log u1 sin(2π u2 ) which are random drawings from the standard normal distribution.
For example, if u1 = 0.643 and u2 = 0.279, we ﬁnd that x = −0.1703 and y = 0.9242. 702 APPENDIX C. SIMULATION C.2.2 The polar method From a computational point of view, the weakness of the BoxMuller method is that we
have to compute trigonometric functions to apply it. This issue can be avoided by using
the polar method which says that if u1 and u2 are as above, then set
v1 = 2u1 − 1,
v2 = 2u2 − 1,
2
2
s = v1 + v2 . If s < 1, we compute
x = v1 −2 log s
,
s y = v2 −2 log s
s which are random drawings from the standard normal distribution. However, should
the computed value of s exceed 1, we discard the random drawings from the U (0, 1)
distribution and repeat the procedure until the computed value of s is less than 1.
For example, if u1 = 0.643 and u2 = 0.279, we ﬁnd that v1 = 0.2860, v2 = −0.4420 and
hence s = 0.2772. As the value of s is less than 1, we proceed to compute x = 0.8703 and
y = −1.3450. C.3. NOTES AND FURTHER READING C.3 703 Notes and further reading Details of all the above methods can be found in standard texts on simulation, e.g.
Ross (2006), or on probability theory, e.g. Borovkov (2003). 704 APPENDIX C. SIMULATION Bibliography
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This note was uploaded on 09/30/2011 for the course ACTSC 232 taught by Professor Matthewtill during the Winter '08 term at Waterloo.
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