May 2001 Course 3 **BEGINNING OF EXAMINATION** Prior to the medical breakthrough, s(x) followed de Moivre s law with ω =100 as the limiting age.

May 001 Course 3 **BEGINNING OF EXAMINATION** 1. For a given life age 30, it is estimated that an impact of a medical breakthrough will be an increase of 4 years in e o 30, the complete expectation of life. Prior to the medical breakthrough, s(x) followed de Moivre s law with ω =100 as the limiting age. Assuming de Moivre s law still applies after the medical breakthrough, calculate the new limiting age. (A) 104 (B) 105 (C) 106 (D) 107 (E) 108 COURSE 3: MAY 001-1 GO ON TO NEXT PAGE

. On January 1, 00, Pat, age 40, purchases a 5-payment, 10-year term insurance of 100,000: (i) (ii) Death benefits are payable at the moment of death. Contract premiums of 4000 are payable annually at the beginning of each year for 5 years. (iii) i = 0.05 (iv) L is the loss random variable at time of issue. Calculate the value of L if Pat dies on June 30, 004. (A) 77,100 (B) 80,700 (C) 8,700 (D) 85,900 (E) 88,000 COURSE 3: MAY 001 - GO ON TO NEXT PAGE

3. Glen is practicing his simulation skills. He generates 1000 values of the random variable X as follows: (i) He generates the observed value λ from the gamma distribution with α = and θ = 1 (hence with mean and variance ). (ii) He then generates x from the Poisson distribution with mean λ. (iii) (iv) He repeats the process 999 more times: first generating a value λ, then generating x from the Poisson distribution with mean λ. The repetitions are mutually independent. Calculate the expected number of times that his simulated value of X is 3. (A) 75 (B) 100 (C) 15 (D) 150 (E) 175 COURSE 3: MAY 001-3 GO ON TO NEXT PAGE

4. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins per minute. The denominations are randomly distributed: (i) 60% of the coins are worth 1; (ii) 0% of the coins are worth 5; (iii) 0% of the coins are worth 10. Calculate the variance of the value of the coins Tom finds during his one-hour walk to work. (A) 379 (B) 487 (C) 566 (D) 670 (E) 768 COURSE 3: MAY 001-4 GO ON TO NEXT PAGE

5. For a fully discrete 0-payment whole life insurance of 1000 on (x), you are given: (i) i = 0.06 (ii) q x+ 19 = 00154. (iii) The level annual benefit premium is 13.7. (iv) The benefit reserve at the end of year 19 is 34.03. Calculate 1000 P x+0, the level annual benefit premium for a fully discrete whole life insurance of 1000 on (x+0). (A) 7 (B) 9 (C) 31 (D) 33 (E) 35 COURSE 3: MAY 001-5 GO ON TO NEXT PAGE

6. For a multiple decrement model on (60): ( 1 (i) µ ) 60 ( t), t 0, follows the Illustrative Life Table. (ii) µ ( τ ) 1 ( t) = µ ( ) ( t), t 0 60 60 Calculate 10 q ( τ ) 60, the probability that decrement occurs during the 11 th year. (A) 0.03 (B) 0.04 (C) 0.05 (D) 0.06 (E) 0.07 COURSE 3: MAY 001-6 GO ON TO NEXT PAGE

7. A coach can give two types of training, light or heavy, to his sports team before a game. If the team wins the prior game, the next training is equally likely to be light or heavy. But, if the team loses the prior game, the next training is always heavy. The probability that the team will win the game is 0.4 after light training and 0.8 after heavy training. Calculate the long run proportion of time that the coach will give heavy training to the team. (A) 0.61 (B) 0.64 (C) 0.67 (D) 0.70 (E) 0.73 COURSE 3: MAY 001-7 GO ON TO NEXT PAGE

8. For a simulation of the movement of a stock s price: (i) The price follows geometric Brownian motion, with drift coefficient µ = 0. 01 and variance parameter σ = 00004.. (ii) The simulation projects the stock price in steps of time 1. (iii) Simulated price movements are determined using the inverse transform method. (iv) The price at t = 0 is 100. (v) The random numbers, from the uniform distribution on 01,, for the first steps are 0.1587 and 0.933, respectively. (vi) F is the price at t = 1; G is the price at t =. Calculate G F. (A) 1 (B) (C) 3 (D) 4 (E) 5 COURSE 3: MAY 001-8 GO ON TO NEXT PAGE

9. (x) and (y) are two lives with identical expected mortality. You are given: P x = P = 0.1 y P xy = 006., where P xy is the annual benefit premium for a fully discrete d = 006. b g. insurance of 1 on xy Calculate the premium P xy, the annual benefit premium for a fully discrete insurance of 1 on bxyg. (A) 0.14 (B) 0.16 (C) 0.18 (D) 0.0 (E) 0. COURSE 3: MAY 001-9 GO ON TO NEXT PAGE

10. For students entering a college, you are given the following from a multiple decrement model: (i) 1000 students enter the college at t = 0. (ii) Students leave the college for failure 1 b1 g b g (iii) µ t = µ 0 t 4 b g b g µ t = 004. 0 t < 4 b g or all other reasons b g. (iv) 48 students are expected to leave the college during their first year due to all causes. Calculate the expected number of students who will leave because of failure during their fourth year. (A) 8 (B) 10 (C) 4 (D) 34 (E) 41 COURSE 3: MAY 001-10 GO ON TO NEXT PAGE

11. You are using the inverse transform method to simulate Z, the present value random variable for a special two-year term insurance on (70). You are given: (i) (70) is subject to only two causes of death, with k k q 70 ( 1) k q ( ) 70 0 0.10 0.10 1 0.10 0.50 (ii) Death benefits, payable at the end of the year of death, are: During year Benefit for Cause 1 Benefit for Cause 1 1000 1100 1100 100 (iii) i = 0. 06 (iv) For this trial your random number, from the uniform distribution on 0 1,, is 0.35. (v) High random numbers correspond to high values of Z. Calculate the simulated value of Z for this trial. (A) 943 (B) 979 (C) 1000 (D) 1038 (E) 1068 COURSE 3: MAY 001-11 GO ON TO NEXT PAGE

1. You are simulating one year of death and surrender benefits for 3 policies. Mortality follows the Illustrative Life Table. The surrender rate, occurring at the end of the year, is 15% for all ages. The simulation procedure is the inverse transform algorithm, with low random numbers corresponding to the decrement occurring. You perform the following steps for each policy: (1) Simulate if the policy is terminated by death. If not, go to Step ; if yes, continue with the next policy. () Simulate if the policy is terminated by surrender. The following values are successively generated from the uniform distribution on 0 1, : 0.3, 0.5, 0.1, 0.4, 0.8, 0., 0.3, 0.4, 0.6, 0.7,. You are given: Death Surrender Policy # Age Benefit Benefit 1 100 10 10 91 5 0 3 96 0 15 Calculate the total benefits generated by the simulation. (A) 30 (B) 35 (C) 40 (D) 45 (E) 50 COURSE 3: MAY 001-1 GO ON TO NEXT PAGE

13. Mr. Ucci has only 3 hairs left on his head and he won t be growing any more. (i) The future mortality of each hair follows k q = 01. k + 1, k = 0, 1,, 3 x b g and x is Mr. Ucci s age (ii) (iii) Hair loss follows the hyperbolic assumption at fractional ages. The future lifetimes of the 3 hairs are independent. Calculate the probability that Mr. Ucci is bald (has no hair left) at age x +.5. (A) 0.090 (B) 0.097 (C) 0.104 (D) 0.111 (E) 0.118 COURSE 3: MAY 001-13 GO ON TO NEXT PAGE

14. The following graph is related to current human mortality: 0 0 40 60 80 100 Age Which of the following functions of age does the graph most likely show? (A) µ bxg (B) l x x (C) (D) l x p x l x (E) l x COURSE 3: MAY 001-14 GO ON TO NEXT PAGE

15. An actuary for an automobile insurance company determines that the distribution of the annual number of claims for an insured chosen at random is modeled by the negative binomial distribution with mean 0. and variance 0.4. The number of claims for each individual insured has a Poisson distribution and the means of these Poisson distributions are gamma distributed over the population of insureds. Calculate the variance of this gamma distribution. (A) 0.0 (B) 0.5 (C) 0.30 (D) 0.35 (E) 0.40 COURSE 3: MAY 001-15 GO ON TO NEXT PAGE

16. A dam is proposed for a river which is currently used for salmon breeding. You have modeled: (i) For each hour the dam is opened the number of salmon that will pass through and reach the breeding grounds has a distribution with mean 100 and variance 900. (ii) The number of eggs released by each salmon has a distribution with mean of 5 and variance of 5. (iii) The number of salmon going through the dam each hour it is open and the numbers of eggs released by the salmon are independent. Using the normal approximation for the aggregate number of eggs released, determine the least number of whole hours the dam should be left open so the probability that 10,000 eggs will be released is greater than 95%. (A) 0 (B) 3 (C) 6 (D) 9 (E) 3 COURSE 3: MAY 001-16 GO ON TO NEXT PAGE

17. For a special 3-year term insurance on ( x ), you are given: (i) Z is the present-value random variable for the death benefits. (ii) qx+ k = 0. 0 ( k + 1 ) k = 0, 1, (iii) The following death benefits, payable at the end of the year of death: k b k+1 0 300,000 1 350,000 400,000 (iv) i = 0. 06 Calculate EbZg. (A) 36,800 (B) 39,100 (C) 41,400 (D) 43,700 (E) 46,000 COURSE 3: MAY 001-17 GO ON TO NEXT PAGE

18. For a special fully discrete 0-year endowment insurance on (55): (i) Death benefits in year k are given by bk = b 1 kg, k = 1,,, 0. (ii) The maturity benefit is 1. (iii) Annual benefit premiums are level. (iv) kv denotes the benefit reserve at the end of year k, k = 1,,, 0. (v) 10 V = 5.0 (vi) 19 V = 0.6 (vii) q 65 = 0.10 (viii) i = 0.08 Calculate 11 V. (A) 4.5 (B) 4.6 (C) 4.8 (D) 5.1 (E) 5.3 COURSE 3: MAY 001-18 GO ON TO NEXT PAGE

19. For a stop-loss insurance on a three person group: (i) (ii) Loss amounts are independent. The distribution of loss amount for each person is: Loss Amount Probability 0 0.4 1 0.3 0. 3 0.1 (iii) The stop-loss insurance has a deductible of 1 for the group. Calculate the net stop-loss premium. (A).00 (B).03 (C).06 (D).09 (E).1 COURSE 3: MAY 001-19 GO ON TO NEXT PAGE

0. An insurer s claims follow a compound Poisson claims process with two claims expected per period. Claim amounts can be only 1,, or 3 and these are equal in probability. Calculate the continuous premium rate that should be charged each period so that the adjustment coefficient will be 0.5. (A) 4.8 (B) 5.9 (C) 7.8 (D) 8.9 (E) 11.8 COURSE 3: MAY 001-0 GO ON TO NEXT PAGE

1-. Use the following information for questions 1 and. The Simple Insurance Company starts at time t = 0 with a surplus of S = 3. At the beginning of every year, it collects a premium of P =. Every year, it pays a random claim amount: Claim Amount Claim amounts are mutually independent. Probability of Claim Amount 0 0.15 1 0.5 0.50 4 0.10 If, at the end of the year, Simple s surplus is more than 3, it pays a dividend equal to the amount of surplus in excess of 3. If Simple is unable to pay its claims, or if its surplus drops to 0, it goes out of business. Simple has no administrative expenses and its interest income is 0. 1. Determine the probability that Simple will ultimately go out of business. (A) 0.00 (B) 0.01 (C) 0.44 (D) 0.56 (E) 1.00 COURSE 3: MAY 001-1 GO ON TO NEXT PAGE

1-. (Repeated for convenience) Use the following information for questions 1 and. The Simple Insurance Company starts at time t = 0 with a surplus of S = 3. At the beginning of every year, it collects a premium of P =. Every year, it pays a random claim amount: Claim Amount Claim amounts are mutually independent. Probability of Claim Amount 0 0.15 1 0.5 0.50 4 0.10 If, at the end of the year, Simple s surplus is more than 3, it pays a dividend equal to the amount of surplus in excess of 3. If Simple is unable to pay its claims, or if its surplus drops to 0, it goes out of business. Simple has no administrative expenses and its interest income is 0.. Calculate the expected dividend at the end of the third year. (A) 0.115 (B) 0.350 (C) 0.414 (D) 0.458 (E) 0.550 COURSE 3: MAY 001 - GO ON TO NEXT PAGE

3. A continuous two-life annuity pays: 100 while both (30) and (40) are alive; 70 while (30) is alive but (40) is dead; and 50 while (40) is alive but (30) is dead. The actuarial present value of this annuity is 1180. Continuous single life annuities paying 100 per year are available for (30) and (40) with actuarial present values of 100 and 1000, respectively. Calculate the actuarial present value of a two-life continuous annuity that pays 100 while at least one of them is alive. (A) 1400 (B) 1500 (C) 1600 (D) 1700 (E) 1800 COURSE 3: MAY 001-3 GO ON TO NEXT PAGE

4. For a disability insurance claim: (i) (ii) (iii) The claimant will receive payments at the rate of 0,000 per year, payable continuously as long as she remains disabled. The length of the payment period in years is a random variable with the gamma distribution with parameters α = and θ = 1. Payments begin immediately. (iv) δ = 005. Calculate the actuarial present value of the disability payments at the time of disability. (A) 36,400 (B) 37,00 (C) 38,100 (D) 39,00 (E) 40,000 COURSE 3: MAY 001-4 GO ON TO NEXT PAGE

5. For a discrete probability distribution, you are given the recursion relation b g b g p k = * p k 1, k = 1,,. k Determine pb4g. (A) 0.07 (B) 0.08 (C) 0.09 (D) 0.10 (E) 0.11 COURSE 3: MAY 001-5 GO ON TO NEXT PAGE

6. A company insures a fleet of vehicles. Aggregate losses have a compound Poisson distribution. The expected number of losses is 0. Loss amounts, regardless of vehicle type, have exponential distribution with θ = 00. In order to reduce the cost of the insurance, two modifications are to be made: (i) (ii) a certain type of vehicle will not be insured. It is estimated that this will reduce loss frequency by 0%. a deductible of 100 per loss will be imposed. Calculate the expected aggregate amount paid by the insurer after the modifications. (A) 1600 (B) 1940 (C) 50 (D) 300 (E) 3880 COURSE 3: MAY 001-6 GO ON TO NEXT PAGE

7. An actuary is modeling the mortality of a group of 1000 people, each age 95, for the next three years. The actuary starts by calculating the expected number of survivors at each integral age by l95+ k = 1000k p95, k = 1,, 3 The actuary subsequently calculates the expected number of survivors at the middle of each year using the assumption that deaths are uniformly distributed over each year of age. This is the result of the actuary s model: Age Survivors 95 1000 95.5 800 96 600 96.5 480 97 -- 97.5 88 98 -- The actuary decides to change his assumption for mortality at fractional ages to the constant force assumption. He retains his original assumption for each k p 95. Calculate the revised expected number of survivors at age 97.5. (A) 70 (B) 73 (C) 76 (D) 79 (E) 8 COURSE 3: MAY 001-7 GO ON TO NEXT PAGE

8. For a population of individuals, you are given: (i) Each individual has a constant force of mortality. (ii) The forces of mortality are uniformly distributed over the interval (0,). Calculate the probability that an individual drawn at random from this population dies within one year. (A) 0.37 (B) 0.43 (C) 0.50 (D) 0.57 (E) 0.63 COURSE 3: MAY 001-8 GO ON TO NEXT PAGE

9-30. Use the following information for questions 9 and 30. You are the producer of a television quiz show that gives cash prizes. The number of prizes, N, and prize amounts, X, have the following distributions: b g x PrbX = xg n Pr N = n 1 0.8 0 0. 0. 100 0.7 1000 0.1 9. Your budget for prizes equals the expected prizes plus the standard deviation of prizes. Calculate your budget. (A) 306 (B) 316 (C) 416 (D) 510 (E) 518 COURSE 3: MAY 001-9 GO ON TO NEXT PAGE

9-30. (Repeated for convenience) Use the following information for questions 9 and 30. You are the producer of a television quiz show that gives cash prizes. The number of prizes, N, and prize amounts, X, have the following distributions: b g x PrbX = xg n Pr N = n 1 0.8 0 0. 0. 100 0.7 1000 0.1 30. You buy stop-loss insurance for prizes with a deductible of 00. The cost of insurance includes a 175% relative security load. Calculate the cost of the insurance. (A) 04 (B) 7 (C) 45 (D) 73 (E) 357 COURSE 3: MAY 001-30 GO ON TO NEXT PAGE

31. For a special fully discrete 3-year term insurance on bxg : (i) Level benefit premiums are paid at the beginning of each year. (ii) (iii) i = 0.06 k b k+1 q x+ k 0 00,000 0.03 1 150,000 0.06 100,000 0.09 Calculate the initial benefit reserve for year. (A) 6,500 (B) 7,500 (C) 8,100 (D) 9,400 (E) 10,300 COURSE 3: MAY 001-31 GO ON TO NEXT PAGE

3. For a special fully continuous whole life insurance on (x): (i) The level premium is determined using the equivalence principle. (ii) t Death benefits are given by bt = b1 + ig where i is the interest rate. (iii) (iv) L is the loss random variable at t = 0 for the insurance. T is the future lifetime random variable of (x). Which of the following expressions is equal to L? (A) c c ν T A 1 A x xh h c hc h (B) ν T Ax 1+ Ax (C) c c ν T A 1+ A x xh h c hc h (D) ν T Ax 1 Ax (E) dv c T + A 1+ A x xi h COURSE 3: MAY 001-3 GO ON TO NEXT PAGE

33. For a 4-year college, you are given the following probabilities for dropout from all causes: q q q q 0 1 3 = 015. = 010. = 005. = 001. Dropouts are uniformly distributed over each year. Compute the temporary 1.5-year complete expected college lifetime of a student entering the second year, e o 115 :.. (A) 1.5 (B) 1.30 (C) 1.35 (D) 1.40 (E) 1.45 COURSE 3: MAY 001-33 GO ON TO NEXT PAGE

34. Lee, age 63, considers the purchase of a single premium whole life insurance of 10,000 with death benefit payable at the end of the year of death. The company calculates benefit premiums using: (i) mortality based on the Illustrative Life Table, (ii) i = 0.05 The company calculates contract premiums as 11% of benefit premiums. The single contract premium at age 63 is 533. Lee decides to delay the purchase for two years and invests the 533. Calculate the minimum annual rate of return that the investment must earn to accumulate to an amount equal to the single contract premium at age 65. (A) 0.030 (B) 0.035 (C) 0.040 (D) 0.045 (E) 0.050 COURSE 3: MAY 001-34 GO ON TO NEXT PAGE

35. You have calculated the actuarial present value of a last-survivor whole life insurance of 1 on (x) and (y). You assumed: (i) (ii) The death benefit is payable at the moment of death. The future lifetimes of (x) and (y) are independent, and each life has a constant force of mortality with µ = 0. 06. (iii) δ = 005. Your supervisor points out that these are not independent future lifetimes. Each mortality assumption is correct, but each includes a common shock component with constant force 0.0. Calculate the increase in the actuarial present value over what you originally calculated. (A) 0.00 (B) 0.039 (C) 0.093 (D) 0.109 (E) 0.163 COURSE 3: MAY 001-35 GO ON TO NEXT PAGE

36. The number of accidents follows a Poisson distribution with mean 1. Each accident generates 1,, or 3 claimants with probabilities 1, 1 3, 1 6, respectively. Calculate the variance in the total number of claimants. (A) 0 (B) 5 (C) 30 (D) 35 (E) 40 COURSE 3: MAY 001-36 GO ON TO NEXT PAGE

37. For a claims process, you are given: m b g r (i) The number of claims N t, t 0 is a nonhomogeneous Poisson process with intensity function: (ii) λ( t) = R S T 1, 0 t < 1, 1 t < 3, t Claims amounts Y i are independently and identically distributed random variables that are also independent of N( t). (iii) Each Y i is uniformly distributed on [00,800]. (iv) The random variable P is the number of claims with claim amount less than 500 by time t = 3. (v) The random variable Q is the number of claims with claim amount greater than 500 by time t = 3. (vi) R is the conditional expected value of P, given Q = 4. Calculate R. (A).0 (B).5 (C) 3.0 (D) 3.5 (E) 4.0 COURSE 3: MAY 001-37 GO ON TO NEXT PAGE

38. Lottery Life issues a special fully discrete whole life insurance on (5): (i) At the end of the year of death there is a random drawing. With probability 0., the death benefit is 1000. With probability 0.8, the death benefit is 0. (ii) (iii) (iv) At the start of each year, including the first, while (5) is alive, there is a random drawing. With probability 0.8, the level premium π is paid. With probability 0., no premium is paid. The random drawings are independent. Mortality follows the Illustrative Life Table. (v) i = 0. 06 (vi) π is determined using the equivalence principle. Calculate the benefit reserve at the end of year 10. (A) 10.5 (B) 0.50 (C) 30.75 (D) 41.00 (E) 51.5 COURSE 3: MAY 001-38 GO ON TO NEXT PAGE

39. A government creates a fund to pay this year s lottery winners. You are given: (i) There are 100 winners each age 40. (ii) (iii) (iv) Each winner receives payments of 10 per year for life, payable annually, beginning immediately. Mortality follows the Illustrative Life Table. The lifetimes are independent. (v) i = 0.06 (vi) The amount of the fund is determined, using the normal approximation, such that the probability that the fund is sufficient to make all payments is 95%. Calculate the initial amount of the fund. (A) 14,800 (B) 14,900 (C) 15,050 (D) 15,150 (E) 15,50 COURSE 3: MAY 001-39 GO ON TO NEXT PAGE

40. For a special fully discrete 35-payment whole life insurance on (30): (i) (ii) (iii) The death benefit is 1 for the first 0 years and is 5 thereafter. The initial benefit premium paid during the each of the first 0 years is one fifth of the benefit premium paid during each of the 15 subsequent years. Mortality follows the Illustrative Life Table. (iv) i = 0. 06 (v) A 30 0 = 03307. : (vi) a&& 30 35 = 14835. : Calculate the initial annual benefit premium. (A) 0.010 (B) 0.015 (C) 0.00 (D) 0.05 (E) 0.030 **END OF EXAMINATION** COURSE 3: MAY 001-40 STOP

COURSE 3 MAY 001 MULTIPLE-CHOICE ANSWER KEY 1 E 6 B A 7 D 3 C 8 D 4 E 9 E 5 D 30 D 6 B 31 D 7 C 3 A 8 D 33 D 9 C 34 A 10 A 35 B 11 B 36 E 1 A 37 C 13 E 38 A 14 B 39 E 15 A 40 B 16 B 17 A 18 E 19 C 0 C 1 E D 3 A 4 B 5 C COURSE 3: MAY 001-41 STOP

SOLUTIONS FOR MAY 001 COURSE 3 EXAM Test Question: 1 Key: E For de Moivre s law, z o ω 30 t e 30 = F1 0 ω 30 L N M HG t = t ω 30 b O gq P 0 I K J dt ω 30 ω 30 = 100 30 Prior to medical breakthrough ω = 100 e o 30 = = 35 After medical breakthrough e 30 = e30+ 4 = 39 o ω so = = 30 e 30 39 ω = 108 o o

Test Question: Key: A 0. 5 L = 100, 000v 4000a&& = 77,079 3 @5%

Test Question: 3 Key: C Ε Ν = Ε Ε Ν Λ = Ε Λ = Λ Var Ν = Ε Var Ν Λ + Var E N Λ Λ =Ε Λ + Var Λ = + = 4 Λ Λ Λ Distribution is negative binomial (Loss Models, 3.3.) Per supplied tables mean = rβ = b g Var = rβ 1+ β = 4 b 1+ β = β = 1 rβ = r = g From tables 3 3 rbr + 1gb r + gβ bgb 3gb4g1 4 p3 = = = = 015. r + 3 5 3! b1 + βg 3! 3 1000 p 3 = 15 Λ

Test Question: 4 Key: E E N b gb g b. gb g b. gb g b. gb g. b gb g b gb g b gb g = Var N = 60 05. = 30 E X = 0 6 1 + 0 5 + 0 10 = 36 E X = 06. 1 + 0. 5 + 0. 100 = 56. Var X = 5. 6 36. = 1. 64 For any compound distribution, per Loss Models Var S = E N Var X + Var N ce X h = (30) (1.64) + (30) d36. i = 768 For specifically Compound Poisson, per Probability Models Var S = λt E X = (60) (0.5) (5.6) = 768 Alternatively, consider this as 3 Compound Poisson processes (coins worth 1; worth 5; worth b g = 0, thus for each VarbSg 10), where for each Var X Processes are independent, so total Var is b g = Var N E X b gb gb g b gb gb g b gb gb gb g Var = 60 05. 0. 6 1 + 60 05. 0. 5 + 60 05. 0. 10 = 768.

Test Question: 5 Key: D 0 d ib g b g 1000 106 1000 0 19V x+ 0Px. qx+ 19 1000 0V x = 1000Ax+ 0 = px+ 19 b34. 03+ 137. gb1. 06g 0. 0154b 1000g = = 36918. 098746. 1 0 36918 ax 0 =. && + = 111445. 006. / 106. so b Ax + 0 36918. 1000Px + 0 = 1000 = = 331. a&& 111445. x+ 0 g

Test Question: 6 Key: B k k bτg k p e e k F = e z0 b g b g b1g x b g τ µ x t dt µ 0 0 x = z = z = p 10 60 p 11 60 HG b p µ g b g 1 x t dt b g t dt I KJ where p is from Illustrative Life Table, since 1 µ follows I.L.T. k x k x 6, 616, 155 = = 08080. 8188,, 074 6, 396, 609 = = 0. 7811 8188,, 074 b g b g b g τ τ τ 10 q60 = 10p60 11p60 = b 10 p60g b11 p60g from I.L.T. = 08080. 07811. = 0. 046 b g

Test Question: 7 Key: C State 1: State : light Training heavy Training P 11 = 04. 05. + 0. 6 0 = 0. P1 = 04. 05. + 0. 6 1 = 08. P = 08. 05. + 0. 0 = 0. 4 P 1 P = L N M = 08. 05. + 0. 1= 0. 6 0. 08. 0. 4 06. O Q P π = 0. π + 04. π 1 1 π = 08. π1+ 0. 6π π1 + π = 1 8 1 π = 0. b1 π g + 04.. π = 0. + 0. π 1. π = 08. π = = 1 3 Note: the notation in Probability Models would label the states 0 and 1, and would label the top row and left column of the matrix P with subscript 0. The underlying calculations are the same. The matrix P would look different, but the result would be the same, if you chose to make heavy the lower-numbered state.

Test Question: 8 Key: D σ = 0. 0004 = 0. 0 Yb1g is normal b0. 01, 0. 0004g 0.1587 corresponds to -1 standard deviation Y (1) = 0.01-(1)(0.0) = -0.01 Y( ) Y( 1) is normal (0.01, 0.0004) 0.933 corresponds to +1.5 standard deviation Y( ) = Y( 1) + 0. 01+ ( 15. )( 00. ) = 0. 01+ 001. + 0. 03 b g b g = 0. 03 F = 100e = 100e = 99. 00 G = 100e = 100e = 103. 05 G F = 405. Y 1 0. 01 Y 0. 03

Test Question: 9 Key: C P s = 1 d a, where s can stand for any of the statuses under consideration. && s a&& a&& a&& s x xy 1 = P + d s 1 = a&& y = = 65. 01. + 0. 06 1 = = 8. 333 0. 06+ 006. a&& + a&& = a&& + a&& xy xy x y a&& xy = 65. + 65. 8333. = 4167. 1 P xy = 0. 06 = 018. 4167.

Test Question: 10 Key: A bτ g z 1 bµ + 004. gt 0 0 d = 1000 e µ + 0. 04 dt b µ e µ + 004. = 1000 1 e = 48 0. 04 g e + = 0. 95 µ + 004. = ln 0. 95 = 0049. µ = 0. 009 b1g z 4 0. 049t 3 3 b b g b gj d = 1000 e 0. 009 dt e b b gb g b gb g 0 049 3 0 049 4 1000 0009... = e e = 7. 6 0049. g g j

Test Question: 11 Key: B b g τ p70 = 1 01. 01. 01. 05. = 0. b g τ p70 F( 0) = = 00. 1 b g b g b g 0 70 1 d1100 i = b979g = b943g+ b g 1 70 = 0. 40 b1100 g = b1038g = ( 979) + b g 0 70 = 0. 50 d i b g b g b g 1 70. good; must have bmaximum possibleg F( 1000v) = F 943 = F 0 + q = 0. 30 F v F F q F v F F q F 100v = F 1068 = F 1038 + q = 100 F = 1 F(943) < random number < F(979), so choose 979

Test Question: 1 Key: A Let Z i be random variable indicating death; W i be random variable indicating lapse for policy. Let U denote the random number used. policy # 1: q 100 = 0. 4081 from Illustrative Life Table U = 03. < 0. 4081 Z 1 = 1 W 1 = 0 policy # : q 91 = 0. 0493 from Illustrative Life Table U = 05. > 0. 0493 Z = 0 next checking lapse U = 01. < 015. (surrender rate) W = 1 policy # 3 q 96 = 030445. U = 04. > 0. 30445 Z 3 = 0 next checking lapse U = 08. > 015. W3 = 0 total Death and Surrender Benefits = 10+0+0 = 30

Test Question: 13 Key: E 3 p p x x = 1 01. 0. = 07. = 0. 7 03. = 04. Use l x = 1 (arbitrary, doesn t affect solution) so lx+ = 07. lx+ 3 = 04. 1 By hyperbolic =. 5 1 +. 5 1 l l l x+. 5 x+ x+ 3. 5. 5 = + = 19643.. 7. 4 lx+. 5 = 05091. = 5. p. 5q x = 1 05091. = 04909. Prob (all 3 failed) = 04909. = 0118. b x 3 g

Test Question: 14 Key: B This is a graph of l µbxg. x b g would be increasing in the interval 80, 100 µ x b g. The graphs of l p x x, l x and l x would be decreasing everywhere. The graph shown is comparable to Figure 3.3. on page 65 of Actuarial Mathematics

Test Question: 15 Key: A Using the conditional mean and variance formulas: E N = E Λ N Λ c h Λd c hj Λd c h Var N = Var E N Λ + E Var N Λ Since N, given lambda, is just a Poisson distribution, this simplifies to: E N = E Λ Λ b g ΛbΛg Var N = Var + E Λ bλg We are given that E N = 0. and Var N = 04., subtraction gives Var Λ j b g = 0.

Test Question: 16 Key: B N = number of salmon X = eggs from one salmon S = total eggs. E(N) = 100t Var(N) = 900t b g b g b g b g b g b g b g b g F > 10 000 500, b, g = > HG 3, 000, E S = E N E X = 500t Var S = E N Var X + E X Var N = 100t 5+ 5 900t = 3, 000t P S P S t t 10 000 500t 3 000t 10, 000 500t = 1645. 3000 t = 50 t 40 t = d i t t 40 = 0 1 1 30 t = ± + = 4. 73 4 t =. 4 round up to 3 t I KJ =. 95

Test Question: 17 Key: A + A P V (x s benefits) = v k 1 b k + 1 k p x q k= 0 = 1000 300v 0. 0 + 350v 098. 004. + 400v 098. 0. 96 006. = 36, 89 x+ k 3 b g b gb g b gb gb g

Test Question: 18 Key: E π denotes benefit premium 19 V = APV future benefits - APV future premiums 1 0. 6 = π π = 036. 108. b 10V + πgb108. g bq65gb10g 11V = p = 65 b5. 0+ 036. gb108. g b010. gb10g =5.8 1 010.

Test Question: 19 Key: C X = losses on one life E X = 03. 1 + 0. + 01. 3 b gb g b gb g b gb g = 1 S = total losses E S = 3E X = 3 b g+ c sb g E S 1 = E S 1 1 F 0 b gc b gd = E S 1 1 f 0 = 3 1 1 04. = 3 0. 936 =. 064 3 h sb gh i

Test Question: 0 Key: C M r = E e x b g rx r r 3r e + e + e = 3 0. 5 15. e + e + e M xb05. g = =. 95 3 1 3 p1 = E X = + + = 3 xb g = b g b g λ M r 1 cr Since λ = and r = 0.5, M x 05. 1 = 05. c. 95 1 = 05. c 39. = 05. c c = 7. 8 = premium rate per period

Test Question: 1 Key: E Simple s surplus at the end of each year follows a Markov process with four states: State 0: out of business State 1: ending surplus 1 State : ending surplus State 3: ending surplus 3 (after dividend, if any) State 0 is absorbing (recurrent). All the other states are transient states. Thus eventually Simple must reach state 0.

Test Question: Key: D (See solution to problem 1 for definition of states). t = 0 0 0 0 1 L N M 10. 0. 0 000. 000. O P Q 01. 05. 05. 015. 00. 01. 00. 09. 1 01. 0. 0 050. 0. 40 P = at t = 0. 0 01. 000. 090. t = 1 00. 01. 0. 0 09. L N M 10. 00. 0. 00 000. O P Q 01. 05. 0. 5 015. 0. 01 014. 0. 05 085. 01. 00. 050. 0. 40 P = at t = 00. 01. 0. 00 090. Expected dividend at the end of the third year = 3 k= 0 (probability in state k at t = ) (expected dividend if in state k) 0.01*0 + 0.14*0 + 0.05(0*0.85 + 1*0.15) + 0.85*(0*0.6 +1*0.5 + *0.15) = 0.4575

Test Question: 3 Key: A 1180 = 70a + 50a 0a 1180 = 70 1 + 50 10 0a a 30: 40 = 8 a = a + a a = 1 + 10 8 = 14 100a = 1400 30 40 30: 40 30: 40 30 40 30: 40 30: 40 b gb g b gb g 30: 40

Test Question: 4 Key: B a = a f t dt zo b g = t z 1 0. 05t e o 005 z. Γb g b L NM b g 1 L F = H G I K J N 0 05M 1 1. 105. 1 = te t te 005. o 1 t te dt 105. t F HG O Q P = i dt 1 t t 1 = t + 1 e + + 0. 05 105.. 185941. I K J 105. t 105 e O QP 0 0, 000 185941. = 37, 188

Test Question: 5 Key: C b g b g p k = k p k 1 = L 0+ O k p b k 1 NM QP g Thus an (a, b, 0) distribution with a = 0, b =. Thus Poisson with λ =. b g = p 4 4 e 4! = 0. 09

Test Question: 6 Key: B By the memoryless property, the distribution of amounts paid in excess of 100 is still exponential with mean 00. With the deductible, the probability that the amount paid is 0 is F Thus the average amount paid per loss is (0.393) (0) + (0.607) (00) = 11.4 The expected number of losses is (0) (0.8) = 16. The expected amount paid is (16) (11.4) = 194. 100/ 00 b g = = 100 1 e 0. 393.

Test Question: 7 Key: D l96 + l97 From UDD l96. 5 = 600 + l97 480 = l97 = 360 Likewise, from l = 360 and l. = 88,we get l 98 = 16 97 97 5 For constant force, e l µ p 05. 97 l98 16 = = = 0. 6 l 360 97 1. 5µ = e = b0. 6g = 07746. b g b gb g = 0. 7746 l = 0. 7746 360 = 78. 86 97. 5 97

Test Question: 8 Key: D Let M = the force of mortality of an individual drawn at random; and T = future lifetime of the individual. n Pr T 1 = E Pr T 1 M z = Pr T 1 M = µ f Mbµ gdµ 0 z z 1 µ = µ e t dt 1 d µ 0 0 z = 1 µ 1 1 = + = + 1 1 e du e 1 e 0 s d i d i d i = 056767.

Test Question: 9 Key: E E N E N E X VarbNg Var X E S b gb g b gb g b g b gb g = 08. 1 + 0. = 1. = 08. 1+ 0. 4 = 16. = 16. 1. = 016. = 70 + 100 = 170 b g = E X E X = b7000 + 100, 000g 170 = 78100, = E N E X = 1. b170g = 04 b g E N b g E X b g b g b g b g = 98, 344 = 3136. Var S = Var X + Var N = 1. 78, 100 + 170 016. = 98, 344 Std dev S So B = 04 + 314 = 518

Test Question: 30 Key: D f f f s s s b1000g = b08. gb01. g + b0. gbgb 0. gb01. g = 0088. b1100g = b0. gbgb07. gb 01. g = 008. b000g = b0. gb01. g = 0. 00 b 00g b0. 088gb800g b0. 08gb900g b0. 00gb1800g E S = + + + = 99. With 175% relative security loading, cost = (.75) (99.) = 7.8 Alternatively, f F f 0 = F 0 = 08. 0. + 0. 0. = 0168. s s sb g b gb g b gb gb gb g s b g b g b gb g b gb g 100 = 08. 07. + 0. 0. 07. = 0. 616 b g 100 = 0168. + 0616. = 0. 784 E S = 04 [from problem 9] b g+ b g+ b gc sb gh = E S b100gc1 Fsb0gh b100gc1 Fsb100g E S 00 = E S 100 100 1 F 100 = 99. cost = (.75) (99.) = 7.8 b gb g b gb g = 04 100 1 0. 168 100 1 0. 784 h

Test Question: 31 Key: D Let π = benefit premium Actuarial present value of benefits = = 0. 03 00, 000 v + 0. 97 0. 06 150, 000 v + 0. 97 0. 94 0. 09 100, 000 v b gb g b gb gb g b gb gb gb g = 5660. 38 + 7769. 67 + 6890. 08 = 0, 3013. Actuarial present value of benefit premiums = a&& π x: 3 = 1+ 0. 97v + 097. 0. 94 v =. 766π b gb g 0, 3013. π = = 745. 55. 766 b745. 55gb 106. g b00, 000gb0. 03g 1V = 1 003. = 1958. 46 Initial reserve, year = 1 V + π = 1958.56 + 745.55 = 9411.01 π 3

Test Question: 3 Key: A Let π denote the premium. L b v a i v a a T T T T T T T = = + = π π π 1 1 b g E L a x ax = = = 1 0 1 π π = = = = = L a a a a v a v a a v A A T T x x T x T x x T x x 1 1 1 1 1 π δ δ δ δ d i b g

Test Question: 33 Key: D (0, 1) (1, 0.9) (1.5, 0.8775) (, 0.885) tp 1 1 t p 1 1 1 = ( 1 01. ) = 0. 9 b gb p = 09. 1 005. = 0855. since uniform, 15. p 1 = 09. + 0855. / = 08775. g b g o e 115 :. = Area between t = 0 and t = 15. F1+ 0. 9I 0. 9 + 08775. = 1 05. HG K J b g F I + HG K Jb g = 0. 95+ 0. 444 = 1394. Alternatively, z z o 15. e 115 :. = t p1dt 0 z 1 0. 5 = tp1dt+ p x p dx 0 1 1 0 z z 1 0. 5 0 0 1 05. = t 01. t x 005 x + 09.. 0 0 b g b g = 1 01. t dt+ 09. 1 0. 05x dx = 095. + 0. 444 = 1394.

Test Question: 34 Key: A 10, 000A63b11. g = 533 A = 0. 467 A x+ 1 A 63 64 b Ax 1+ i qx = p b x g gb g 0467. 105. 001788. = 1 0. 01788 = 04813. b gb g 04813. 105. 0. 0195 A65 = 1 00195. = 04955. Single contract premium at 65 = (1.1) (10,000) (0.4955) = 5550 b g. 5550 5550 1+ i = i = 533 533 1 = 0 0984

Test Question: 35 Key: B Original Calculation (assuming independence): µ µ µ A A A x y xy x y xy = 006. = 006. = 0. 06 + 006. = 01. = µ x + = 006. = 054545. µ δ 0. 06 + 0. 05 x µ y = + = 0. 06 = 054545. µ δ 006. + 0. 05 y µ xy = + = 01. = 0. 70588 µ δ 01. + 005. xy A = A + A A = 054545. + 054545. 070588. = 03850. xy x y xy Revised Calculation (common shock model): µ = 006., µ = 0. 04 x T* x x µ = 006., µ = 004. y b g b g T* y y µ = µ + µ + µ +. +. +. =. A A A xy x y xy = b g b g 004 0 04 0 0 010 T* x T* y Z x y µ x + = 006. = 0. 54545 µ δ 0. 06 + 005. x µ y 0. 06 = = = 054545. 006. + 005. µ y+ δ µ xy = + = 010. = 0. 66667 µ δ 010. + 005. xy A = A + A A = 054545. + 054545. 066667. = 0443. xy x y xy Difference = 0. 443 03850. = 00391.

Test Question: 36 Key: E Treat as three independent Poisson variables, corresponding to 1, or 3 claimants. rate rate rate Var Var Var 1 6 3 1 3 1 = = 1 = 4 = = 6 L NM = 16 = 4 = 18 total Var = 6 + 16 + 18 = 40, since independent. Alternatively, d i = + + = E X 1 3 3 6 10 3 O QP For compound Poisson, Var S = E N E X = b gf H G I K J = 1 10 3 40

Test Question: 37 Key: C z3 λbtgdt = 6 so Nb3g is Poisson with λ = 6. 0 b g g P is Poisson with mean 3 (with mean 3 since Prob y i < 500 = 05. P and Q are independent, so the mean of P is 3, no matter what the value of Q is.

Test Question: 38 Key: A At age x: Actuarial Present value (APV) of future benefits = 4 APV of future premiums = F I 5 &&a x K J π HG F HG 1 5 A x I K J 1000 1000 A 5 1000 A 4 a&& 4 = π a&& by equivalence principle 5 1 8165. = π π = = 158. 4 16. 4 5 5 5 5 10V = APV (Future benefits) APV (Future benefit premiums) 1000 4 = A35 π a&& 35 5 5 1 = 5 18 7 4 b. g b158. gb15396. g 5 = 105.

Test Question: 39 Key: E Let E Y Var Y E S Var Y = present value random variable for payments on one life S = Y = present value random variable for all payments = 10a&& = 148166. 40 = 10 d A d = 100 0. 04863 01613. 1. 06/ 0. 06 = 70555. 40 A40 i d ib g = 100E Y = 14, 816. 6 S = 100Var Y = 70, 555 Standard deviation S = 70, 555 = 656. By normal approximation, need E [S] + 1.645 Standard deviations = 14,816.6 + (1.645) (65.6) = 15,54

Test Question: 40 Key: B 5A 4 A1 Initial Benefit Prem = 5a&& 4a&& Where e e 30 30: 0 30: 35 30: 0 b. g b. g 5 01048 4 00933 = 5 14. 835 4 11959. = b g b g j 0514. 01173. 039508. = = 0015. 74175. 47. 836 6339. j 1 1 A30 : 0 = A30 : A 0 30: = 03307. 09374. = 00933. 0 and 1 A30: 0 1 0. 3307 a&& = = = 11959. 30: 0 d F 0. 06 H G I 106. K J Comment: the numerator could equally well have been calculated as A E A 30 0 30 50 + 4 = 0.1048 + (4) (0.9374) (0.4905) = 0.39510