PSTAT 172A: ACTUARIAL STATISTICS FINAL EXAM

Transcription

1 PSTAT 172A: ACTUARIAL STATISTICS FINAL EXAM March 17, 2009 This exam is closed to books and notes, but you may use a calculator. You have 3 hours. Your exam contains 7 questions and 11 pages. Please make sure you print your name and sign the honor code below. Name: Solutions I acknowledge that I have neither given nor received aid on this examination nor have I concealed any violation of the Honor Code. (Signed) 1

2 1. Mortality is based on the following extract from a life table and the interest rate is i = 6%. x l x d x p x q x 30 10, , (a) (5 pts.) A payment of one unit is made at the end of the year in which (31) dies or at the end of three years, whichever comes first. Find the actuarial present value (APV) of the payment. A 31:3 = vq 31 + v 2 p 31 q 32 + v 3 p 31 p 32 = (0.98) (1.06) 2 (1.06) 2 = (b) (5 pts.) A payment of one unit is made at the beginning of each year for 3 years until (31) dies. Find the APV of the payments. ä 31:3 = 1 + vp 31 + v 2 p 31 p 32 = (0.98)2 (1.06) 2 =

3 1. Mortality is based on the following extract from a life table and the interest rate is i = 6%. x l x d x p x q x 30 10, , (c) (5 pts.) A payment of one unit is made at the beginning of each year for 3 years until (31) dies. Find the probability that the sum of the payments will exceed the APV of the payments. Let Y be the sum of the payments. 1, K(31) = 0 Y = 1 + v = , K(31) = v + v 2 = , K(31) 2 Pr(Y > ) = Pr(K(31) 2) = 2 p 31 = (0.98) 2 = (d) (5 pts.) A payment of one unit is made at the end of each year for 3 years if (31) is still alive at the beginning of that year. Calculate the variance of the present value of the payments. Let Y be the present value of the payments. v, K(31) = 0 Y = v + v 2, K(31) = 1 v + v 2 + v 3, K(31) 2 E[Y ] = v ä 31:3 = E[Y 2 ] = v 2 q 31 + (v + v 2 ) 2 p 31 q 32 + (v + v 2 + v 3 ) 2 p 31 p 32 = Var[Y ] = E[Y 2 ] E[Y ] 2 =

4 1. Mortality is based on the following extract from a life table and the interest rate is i = 6%. x l x d x p x q x 30 10, , (e) (5 pts.) Calculate P :3 P31: 3 1 P 1 31:3 + P31: 1 3 = A 1 A 31:3 31: ä 31:3 ä 31:3 = A 31:3 ä 31:3 (a)(b) ===== = (f) (5 pts.) For a fully discrete 3-year endowment insurance of $1 on (31), find the smallest annual benefit premium such that there is a 0% chance of loss. Let L be the loss-at-issue random variable with the annual benefit premium π. v π, K(31) = 0 L = v 2 π(1 + v), K(31) = 1 v 3 π(1 + v + v 2 ), K(31) 2 Pr(L > 0) = 0 or Pr(L 0) = 1 π = v =

5 2. Rui, age 25, purchases a 10-year continuous payment, continuous whole life insurance policy with a benefit of $1. He is subject to a constant force of mortality equal to 0.04 and a constant force of interest equal to The annual benefit premium for a fully continuous 10-payment whole life insurance policy is determined based on the equivalence principle. (a) (5 pts.) Calculate the annual benefit premium rate for this policy. Ā 25 = ā 25:10 = P ( Ā 25 ) = e 0.06t 0.04e 0.04t dt = = 0.4 e 0.06t e 0.04t dt = 10(1 e 1 ) = Ā25 = 0.4 ā 25: = (b) (7 pts.) Find the variance of the loss associated with this policy. Let L be the loss-at-issue random variable. { V T 10 P ( Ā 25 ) ā L = T = e 0.06T , T < 10 V T 10 P ( Ā 25 ) ā 10 = e 0.06T , T 10 Var(L) ====== E(L)=0 E(L 2 ) = ( e 0.06T ) e 0.04t dt 10 = (e 0.06T ) e 0.04t dt 5

6 3. For a fully discrete whole life insurance of $1000 on (60), the annual benefit premium is determined based on the equivalence principle. Since Alisha, age 60, is recovering from surgery, she is expected to have extra mortality risks at age 60. Her force of mortality during the year of age 60 to 61 is obtained by doubling the force of mortality of the Illustrative Life Table. After age 60, her mortality rates, q x, at any other ages are the same as the ones given in the life table. The interest rate is i = 6%. (a) (7 pts.) Calculate the annual benefit premium for a fully discrete whole life insurance policy on Alisha. Since µ A 60(t) = 2 µ 60 (t), ( p A 60 = (p 60 ) 2 = ) 2 = A A 60 = v q60 A + v p A 60 A 61 ( ) ( ) ( ) = = P A 60 = 1000AA 60 ä A 60 = 1000d AA 60 1 A A 60 = (b) (5 pts.) Find the variance of the loss associated with a fully discrete whole life insurance for Alisha. 2 A A 60 = v 2 q60 A + v 2 p A 60 2A 61 ( ) ( ) ( ) = = ( Var(L) = P ) 60 A 2 ( 2 A A 60 (A A d 60) 2) =

7 4. The value of an appliance is given by V (t) = e 7 0.2t, where t denotes the number of years since purchase. If the appliance fails within seven years of purchase, a warranty pays the owner the value of the appliance at the time of failure. After seven years, the warranty pays nothing. The time until failure of the appliance has an exponential distribution with mean 10. The force of interest rate is δ = (a) (5 pts.) Calculate the expected payment from the warranty. 7 0 e 7 0.2t e 0.06t 0.1e 0.1t dt = e t dt = 0.1e (1 e ) = e 7 = (b) (5 pts.) Calculate the probability that the payment actually made from the warranty will exceed the expected payment. Pr ( e 7 0.2T e 0.06T > e 7) = Pr ( e 0.26T > ) = Pr(T < ) = 1 e =

8 5. You are retiring at age 60 with a pension of $15, 000 annually payable at the beginning of each year. You will also receive a Social Security benefit of $11, 000 annually, starting in 2 years. Your human resources department has offered to pay you an increased pension for 2 years, until your Social Security payments start, and then a reduced pension. Under this new plan, you will receive $23, 000 per year for 2 years, and then a lifetime pension of $12, 000 per year. This benefit of $12, 000 after 2 years, combined with Social Security s $11, 000, will give you $23, 000 annually. Mortality follows the Illustrative Life Table and the interest rate is i = 6%. (a) (5 pts.) Would you accept the new pension plan from the department of human resources? Original Plan: 15000ä 60 = = New Plan: 23000ä E 60 ä 62 ( ) ( ) = = (b) (7 pts.) Find the probability that the original pension plan is better than the new one. Pr (15000ä K+1 > 12000ä K (1 + v), K(60) 2) = Pr (3000ä K+1 > 11000(1 + v), K(60) 2) ( 1 v K+1 = Pr > 11 ) (1 + v), K(60) 2 d 3 = Pr (K > ) = 8 p 60 = =

9 6. (12 pts.) Tai, age 35, wishes to provide cash for her son Neo, currently age 10, to go to college. Tai purchases a 10-year deferred 4-year temporary life annuity-due contingent on Neo s survival. This life annuity will pay Neo $25, 000 per year for 4 years starting at his age of 20 while he is alive. Tai will make 5 equal annual premium payments beginning today. The 5 premium payments take the form of a 5-year temporary life annuity-due contingent on Tai s survival. She will pay annual premium payments for 5 years as long as she is alive. Mortality follows the Illustrative Life Table and the interest rate is i = 6%. What is the amount of each annual premium payment based on the equivalence principle? Let π be the annual benefit premium. π ä 35:5 = E 10 ä 20:4 π = E 10 (ä 20 4 E 20 ä 24 ) ä 35 5 E 35 ä 40 = =

10 7. (12 pts.) A special fully discrete 5-payment 10-year deferred whole life insurance policy on (50) provides that if death occurs at any time within 10 years, then all premium payments are refunded with interest at the rate of 6% at the end of the year of death. If death occurs after 10 years, the benefit of $10, 000 is paid at the end of the year of death. The premiums for this policy are to be paid annually at the beginning of each year for 5 years. Mortality follows the Illustrative Life Table and the interest rate is i = 6%. Find the level annual benefit premium for the policy based on the equivalence principle. Let L be the loss-at-issue random variable with the annual premium π. { 0, K(50) < 10 L = 10000v K+1 π ä 5, K(50) 10 E(L) = E 50 A 60 = πä 5 10 p 50 π = 10000v10 A 60 ä 50 =