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Unformatted text preview: W’ filing? University of Waterloo Final Examination Term: Fall Year: 2002 Student Name: UW Student ID Number: Course Abbreviation and Number: Actsc 432/832 Course Title: Credibility and Rain Theory
Section(s): 001 Sections Combined Course(s): N/A Section Numbers of Combined Course(s): Instructor: Mary Hardy Date of Exam: December 16, 2002
Time Period: Start time: 2:00 pm End Time: 5:00 pm
Duration of Exam: 3 hours *
Number of Exam Pages: 10
(including this cover sheet)
Exam Type: Closed Book
Additional Materials Allowed: Calculator Markin Scheme: Question Score ( Question 1 
u
"
— Formulae:
Pareto Distribution: a A a
an — C;%%a: Fao=1—(A+m)
mm = ail a>1 Generalized Pareto Distribution: __ I‘(oz + T) 9“ {ET—1
f”) ” I‘(a) m) (a: + o)a+7 Lognormal distribution ﬁx) 1 _l (mgr—pf
mam EXP 2 a __ ekp+k202/2 = 82p+62 (602 __ Weibull distribution:
T (Er—1 e—(as/ey
f (x) = ‘77—”—
F(m) = 1 — €_(z/9)T
Poisson Distribution
k: —,\
Pr[N = k] = A I:
E[N] = A V[N] z A Ame): wmxé—n} Inverse Gamma Distribution: Parameters a, 6 > O 5a 66/2;
“35) = m
ELK} = a f 1
V X  62
{1 (asnaa—m Nonparametric empirical Bayes parameter estimation for the Buhlmann model: Estimators for the structural parameters are: l1 = '1' X1.
7"$21
1 T n 
13 = (X1 —X,)2
r(n—1)Z:1j; J
1 T — p
H Nonparametric empirical Bayes parameter estimation for the Buhlmann
Straub model: Estimators for the structural parameters are: ,. _ r=1 n i=1 Zz‘
{I = 3:: mil meme ‘ X02
i=_1(ni : 1)
a = 2:1 "112(er “ XV “ 17(7" — 1) m ~ 2le Critical points of the 98 distribution For each degree of freedom, and for each value of p, the table gives x where 1F(x)=p; F(x) is the cumuiative distribution
function of the x2 distribution. Degrees of 0005
0.95 0.50 0.20 0.10 0.05 0.01 «we.64 0.001 0.0001
freedom
1 0.004 0.455 1.642 2.706 3.841 6.635 7.879 10.827 15.134 2 0.103 1.386 3.219 4.605 5.991 9.210 10.597 13.815 18.425
3 0.352 2.366 4.642 6.251 7.815 11.345 12.838 16.266 21.104
4 0.711 3.357 5.989 7.779 9.488 13.277 14.860 18.466 23.506
5 1.145 4.351 7.289 9.236 11.070 15.086 16.750 20.515 25.751 6 1.635 5.348 8.558 10.645 12.592 16.812 18.548 22.457 27.853
7 2.167 6.346 9.803 12.017 14.067 18.475 20.278 24.321 29.881
8 2.733 « 7.344 11.030 13.362 15.507 20.090 21.955 26.124 31.827
9 3.325 8.343 12.242 14.684 16.919 21.666 23.589 27.877 33.725
10 3.940 9.342 13.442 15.987 18.307 23.209 25.188 29.588 35.557
11 4.575 10.341 14.631 17.275 19.675 24.725 26.757 31.264 37.365
12 5.226 11.340 15.812 18.549 21.026 26.217 28.300 32.909 39.131
13 5.892 12.340 16.985 19.812 22.362 27.688 29.819 34.527 40.873
14 6.571 13.339 18.151 21.064 23.685 29.141 31.319 36.124 42.575
15 7.261 14.339 19.31 1 22.307 24.996 30.578 32.801 37.698 44.260
16 7.962 15.338 20.465 23.542 26.296 32.000 34.267 39.252 45.926
17 8.672 16.338 21.615 24.769 27.587 33.409 35.718 40.791 47.559
18 9.390 17.338 22.760 25.989 28.869 34.805 37.156 42.312 49.185
19 10.117 18.338 23.900 27.204 30.144 36.191 38.582 43.819 50.787
20 10.851 19.337 25.038 28.412 31.410 37.566 39.997 45.314 52.383 Section I: Shorter Questions 1. The following list is ﬂ the 10 largest claims from a sample of 70 claim amounts paid
from a motor insurance portfolio. 6,644 7,059 7,061 7,323 8,249 8,357 11,597 14,281 18,249 20,946
Calculate the smoothed empirical 95th percentile.
[4 marks] 2. State which of the following two formulae is correct for the credibility factor Z for
the Buhlmann model of Greatest Accuracy Credibility Theory: TL
Z :
n+v/a
TL
Z =
n+a/v Justify your answer by discussing how Z should behave as a function of n, v and c3.
[4 marks] 3. The following data are claims observed from different contracts in a household prop—
erty portfolio. The losses involved are assumed to be independent and identically
distributed, but the claims are not identically distributed because of differing de
ductibles and policy limits. Claims are assumed to be independent. The table gives
the claim size and the associated deductibles and policy limits. (Note: the policy limit is the maximum claim paid; for a deductible D and a policy
limit L and a loss X, the claim paid is L if X > L + D.) Deductible O
200
100 0
100 Write down an expression for the likelihood function in terms of the density and
distribution functions for the original loss random variable, fX and FX (2:). [5 marks] 4. You have ﬁtted the models in the table below to a data set ofmll BOiindependent
observations, using maximum likelihood. The table gives the maximum value for the loglikelihood for the data. Model Maximum Loglikelihood
Pareto 832.1
Lognormal 830.7
Generalized Pareto —829.8 (a) Which model is preferred using the Akaike Information Criterion? (b) Which model is preferred using the Schwartz Bayes Criterion (with 0.510g(n)
penalty for each parameter)? (c) Which model is preferred using the Likelihood Ratio Test? (Explain carefully
your assumptions). [6 marks] 5. You have a set of loss data. You wish to calculate the maximum likelihood estimators .
for the Weibull distribution using these data. Explain carefully how you would use
Solver in Excel to ﬁnd the maximum likelihood parameter estimates. [5 marks] Section II: Longer Questions 6. (a) Your boss did his actuarial exams before the Bayesian approach was included;
he thinks it can’t be right, as it is clear that a distribution parameter is ﬁxed,
and is not a random variable. Is he right? Justify your answer with a brief
discussion of the role of the prior distribution. {3 marks] (b) Suppose that, given 1/, the random variable X has a normal distribution with
parameters [1. and 2/. That is, the density for X [1/ is f(:vlv)= game; (as—Mr} The parameter pi is assumed known, and 1/ is to be estimated using a Bayesian
approach, treating 1/ as a random variable. The prior distribution is the Inverse
Gamma distribution, with parameters or and 6, as given in the formula sheet. (i) Given a = 73 and 6 = 300,000 calculate the prior mean and standard
deviation for 1/. [1 mark]
(ii) Show that the inverse gamma distribution is conjugate for 1/, given a sample (£171,1E2, ..., :13").
[6 marks] (iii) You are given a sample of 60 observations of X. These have mean a’: =
1010.0 and Z (x, — “)2 = 154, 500 i=1 Show that the Bménnﬁstimator for 1/ (with respect to quadratic loss) is m Calculate the standard deviation of the posterior distribution of 1/.
36"”?! [4 marks] (iv) You are given that the maximum likelihood estimate of 1/ is Discuss brieﬂy the difference between the maximum likelihood estimate of 1/ and the Bayesian estimate of 1/. Which would you use? [3 marks]
[Total 17 marks] 7. Let {Xj}, j = 1,2, ..., n be a random sample, where all the variables Xj depend on
some parameter 0. Let be known risk volumes associated with the values X j. (a) Write down the assumptions of the Buhlmann—Straub model of Greatest Ac— curacy Credibility Theory in terms of X j, 9, and m,, and deﬁne the structural
parameters ,u, v and a. (b) Derive expressions in terms of the structural parameters and the risk volumes
for: (i) The covariance of Xi and Xj, 2' % j.
(ii) Vlel (iii) V[X] where
— m1X1+m2X2+m+man X 2
m1 +m2+...+mn
(iv) E[X1 X] where X is as in part (iii) above.
(v) E[Xn+1 X] where X is as in part (iii) above. {12 marks] 8. A set of random variables {X,j}, where i = 1,2, ...,r and j = 1,2, ...,n represents
the aggregate claims in n successive years from 7‘ risks are assumed to satisfy the assumptions for nonparametric parameterwestimation for the Buhlmann model of
greatest accuracy credibility theory, that is: o Xijw, are independent and identically distributed for j = 1, 2, ..., n. o Xij and Xkl are independent and identically distributed for 2' 75 k and for all j
and l. o 6, are independent and identically distributed for z' = 1, 2, ..., r. (a) Show that the estimating functions given below are unbiased estimators of the
structural parameters a, v and a for the Buhlmann model: r n
Z'ZXU i=1 j=1 73  lis2
_ TL, 1—1
A T — ——2 73
a “'7;
_ rel
[9marks] (b) The table below gives sample means and variances for the aggregate claims
experience of each of 4 risks in a collective of workers compensation insurance.
For each risk the sample statistics are based on 7 years of data. Use these to
calculate the credibility premium for risk number 4. l A_1
M~—TTZ 101.8 120.9 [6 marks] (c) The insurer has one more risk in the collective, risk 5. For this risk, only 4
years of data are available, with mean 755 = 90.2 and variance 3% = 1010.
Show how this additional data can be incorporated into the estimation of the
structural parameters, and use the revised structural parameters to re—calculate the credibility premium for risk number 4. [7 marks] [Total 22 marks] 9. (a) Claim frequency for risks in a motor insurance portfolio are assumed to follow
a Poisson distribution with parameter A. Given a sample of n policies, with mean claim frequency is, show that the
maximum likelihood estimate of A is k.
[4 marks] (b) You are given the following sample of claim frequency data, with nk observa
tions of It claims for a total of n = 10, 000 policies. No of claims No of observations k nk 0 8775 1 1111 2 100 3 10 4 4
Total 10,000 The mean of the sample is 19 = 0.1357. (i) Conduct a goodness of ﬁt test using the MLE estimate of A, and state
clearly your conclusion. (ii) Calculate the maximum log—likelihood for the sample with the Poisson
distribution. [7 marks] (0) Assume now that the claim frequency follows a zero modiﬁed Poisson distri bution, with a new Poisson parameter X, and modiﬁed probability at zero
p8” (i) Write down the likelihood function for the zero modiﬁed Poisson distribu— tion parameters.
[2 marks] (ii) Derive and simplify as far as possible an equation for the maximum like
lihood estimate of X. Hence, show that X = 0.20828 is the maximum
likelihood estimate of X. [6 marks] (iii) Estimate the asymptotic variance for X. [6 marks]
[Total 25 marks] 10 ...
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