exam mlc may solutions

exam mlc may solutions - SPRING 2005 EXAM M SOLUTIONS...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
SPRING 2005 EXAM M SOLUTIONS Question #1 Key: B Let K be the curtate future lifetime of ( x + k ) 1 :3 1 1000 1000 K k x K Lv P a + + =− × ±± When (as given in the problem), ( x ) dies in the second year from issue, the curtate future lifetime of () 1 x + is 0, so 1 :3 1 1000 1000 x LvP a 1000 279.21 1.1 629.88 630 =≈ The premium came from x x x A P a = 1 x x Ad a 1 1 279.21 x x xx da Pd aa == =
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Question #2 Key: E Note that above 40, decrement 1 is DeMoivre with omega = 100; decrement 2 is DeMoivre with omega = 80. That means () ( ) 12 40 40 20 1/ 40 0.025; 20 1/ 20 0.05 µµ == 40 20 0.025 0.05 0.075 τ µ =+ = Or from basic definition of , 2 40 60 40 2400 100 60 40 2400 t tt t t p −− + =×= 40 / 100 2 / 2400 t dp d t t =− + at 20 t = gives 60/ 2400 0.025 −= 20 40 40 40 20 40 2/3 * 1/2 1/3 20 / / 0.025/ 1/3 0.075 t p t p ττ ⎡⎤ = = ⎢⎥ ⎣⎦
Background image of page 2
Question #3 Key: B 5 35 45 5 35 5 45 5 35 45 5 35 40 5 45 50 5 35 45 40 50 5 35 40 5 45 50 5 35 5 45 40 50 5 35 40 5 45 50 5 35 5 45 40 50 1 1 09 03 08 005 09 08 1 097 095 0 01048 qq q q pq p q p p p p p pp : : :: : .. . . . . . =+− × × =+ = b g bg bgbg b g bg b g Alternatively, () ( ) ( ) 63 5 53 5 4 0 6 45 5 45 50 0.90 1 0.03 0.873 0.80 1 0.05 0.76 p p = = = = 56 5 35:45 35:45 35:45 5 35 5 45 5 35:45 6 35 6 45 6 35:45 qpp ppp =− −+ ( ) 53 5 54 5 5 64 5 63 5 p pppp + ×−+ × ( ) 0.90 0.80 0.90 0.80 0.873 0.76 0.873 0.76 ×− + −× 0.98 0.96952 0.01048 =
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Question #4 Key: D Let G be the expense-loaded premium. Actuarial present value (APV) of benefits = 35 100,000 A APV of premiums = 35 Ga ±± APV of expenses = () ( ) 35 0.1 25 2.50 100 ++ ⎡⎤ ⎣⎦ Equivalence principle: 35 35 35 35 35 100,000 0.1 25 250 100,000 0.1 275 Ga A G a A GG a =+ + + + ( ) 35 0.9 100,000 275 100 8.36 275 0.9 1234 GP G + = = Question #5 Key: D Poisoned wine glasses are drunk at a Poisson rate of 2 × 0.01 = 0.02 per day. Number of glasses in 30 days is Poisson with 0.02 30 0.60 λ = ×= 0.60 00 . 5 5 fe ==
Background image of page 4
Question #6 Key: E View the compound Poisson process as two compound Poisson processes, one for smokers and one for non-smokers. These processes are independent, so the total variance is the sum of their variances. For smokers, () ( ) 0.2 1000 200 λ == Var(losses) () () 2 Var XE X ⎡⎤ =+ ⎢⎥ ⎣⎦ 2 200 5000 100 3,000,000 = For non-smokers, ( ) 0.8 1000 800 Var(losses) 2 Var X 2 800 8000 100 14,400,000 = Total variance = 3,000,000 + 14,400,000 = 17,400,000
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Question #7 Key: E [ ] x EZ bA = since constant force / x A µ µδ =+ E ( Z ) () 0.02 /3 0.06 b b b == = + [ ] ( ) T2 2 2 2 2 2 22 Var Var Var 2 21 4 10 9 45 x x Z bv b v b A A b bb µµ δ Τ ⎡⎤ = ⎣⎦ ⎛⎞ ⎜⎟ =− ++ ⎝⎠ = ⎢⎥ () () Var Z EZ = 2 4 45 3 41 3.75 45 3 b b = =⇒=
Background image of page 6
Question #8 Key: A () 12 3 30 30 30 30:3 23 1000 500 250 1 1.53 1 1.61 1 1.70 1000 500 0.99847 250 0.99847 0.99839 1.06 1000 1.06 1000 1.06 1000 Av q v q v q =+ + ⎛⎞ + ⎜⎟ ⎝⎠ 1.4434 0.71535 0.35572 2.51447 + = ( ) 1 2 2 11 1 22 2 3 0 30:1 1 1 1 0.00153 1 0.97129 1 1.06 2 2 2 0.97129 0.999235 0.985273 aq = ±± Annualized premium 2.51447 0.985273 = 2.552 = Each semiannual premium 2.552 2 = 1.28 =
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Question #9 Key: B Ex dx d ⎡− > ⎤ ⎣⎦ is the expected payment per payment with an ordinary deductible of d It can be evaluated (for Pareto) as () ( ) () 1 1 11 1 d Ex Ex d Fd d α θθ θ αα ⎡⎤ ⎛⎞ −− ⎢⎥ ⎜⎟ + ⎝⎠ −∧ = −⎡ + 1 1 d d αθ −+ = + 1 d + = d =+ in this problem, since 2 = 5 3 5 3 100 100 50 50 100 50 Ex x x > ⎤= += + 300 3 250 5 25 + == 150 150 150 150 25 175 x > + =
Background image of page 8
Question #10 Key: D Let S = score () 75 ES EES E Var S E Var S Var E S θθ θ == = ⎡⎤ =+ ⎣⎦ 2 8 EV a r 2 64 6 100 = S
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 30

exam mlc may solutions - SPRING 2005 EXAM M SOLUTIONS...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online