SOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS

SOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS Copyright 5 by the Society of Actuaries Some of the questions in this study note are taken from past SOA examinations. M-9-5 PRINTED IN U.S.A.

. For two independent lives now age 3 and 34, you are given: x q x 3. 3. 3.3 33.4 34.5 35.6 36.7 37.8 Calculate the probability that the last death of these two lives will occur during the 3 rd year from now (i.e. q 3: 34 ). (A). (B).3 (C).4 (D).8 (E).4

. For a whole life insurance of on (x) with benefits payable at the moment of death: (i).4, < t δt =.5, < t.6, < t (ii) µ x () t =.7, < t Calculate the single benefit premium for this insurance. (A) 379 (B) 4 (C) 444 (D) 59 (E) 594 3. A health plan implements an incentive to physicians to control hospitalization under which the physicians will be paid a bonus B equal to c times the amount by which total hospital c. claims are under 4 ( ) The effect the incentive plan will have on underlying hospital claims is modeled by assuming that the new total hospital claims will follow a two-parameter Pareto distribution with α = and θ = 3. E( B ) = Calculate c. (A).44 (B).48 (C).5 (D).56 (E).6 3

4. Computer maintenance costs for a department are modeled as follows: (i) The distribution of the number of maintenance calls each machine will need in a year is Poisson with mean 3. (ii) The cost for a maintenance call has mean 8 and standard deviation. (iii) The number of maintenance calls and the costs of the maintenance calls are all mutually independent. The department must buy a maintenance contract to cover repairs if there is at least a % probability that aggregate maintenance costs in a given year will exceed % of the expected costs. Using the normal approximation for the distribution of the aggregate maintenance costs, calculate the minimum number of computers needed to avoid purchasing a maintenance contract. (A) 8 (B) 9 (C) (D) (E) 4

5. A whole life policy provides that upon accidental death as a passenger on an airplane a benefit of,, will be paid. If death occurs from other accidental causes, a death benefit of 5, will be paid. If death occurs from a cause other than an accident, a death benefit of 5, will be paid. You are given: (i) (ii) (iii) (iv) Death benefits are payable at the moment of death. () µ = /,, where () indicates accidental death as a passenger on an airplane. ( ) µ = / 5, where () indicates death from other accidental causes. ( 3) µ = /, where (3) indicates non-accidental death. (v) δ =.6 Calculate the single benefit premium for this insurance. (A) 45 (B) 46 (C) 47 (D) 48 (E) 49 5

6. For a special fully discrete whole life insurance of on (4): (i) The level benefit premium for each of the first years is π. (ii) The benefit premium payable thereafter at age x is vq x, x = 6, 6, 6, (iii) Mortality follows the Illustrative Life Table. (iv) i =.6 Calculate π. (A) 4.79 (B) 5. (C) 5.34 (D) 5.75 (E) 6.7 7. For an annuity payable semiannually, you are given: (i) Deaths are uniformly distributed over each year of age. (ii) q 69 =.3 (iii) i =.6 (iv) A 7 = 53 Calculate ( ) a. 69 (A) 8.35 (B) 8.47 (C) 8.59 (D) 8.7 (E) 8.85 6

8. For a sequence, uk ( ) is defined by the following recursion formula k (i) ( ) α k (ii) β ( k ) ( ) β( ) ( ) uk ( ) = α k + k u k for k =,, 3, q = p + i = p k (iii) u ( 7) =. k Which of the following is equal to u ( 4)? (A) A 3 (B) A 4 (C) A 4:3 (D) A 4:3 (E) A 4:3 7

9. Subway trains arrive at a station at a Poisson rate of per hour. 5% of the trains are express and 75% are local. The type of each train is independent of the types of preceding trains. An express gets you to the stop for work in 6 minutes and a local gets you there in 8 minutes. You always take the first train to arrive. Your co-worker always takes the first express. You both are waiting at the same station. Calculate the probability that the train you take will arrive at the stop for work before the train your co-worker takes. (A).8 (B).37 (C).5 (D).56 (E).75. For a fully discrete whole life insurance of on (4), the contract premium is the level annual benefit premium based on the mortality assumption at issue. At time, the actuary decides to increase the mortality rates for ages 5 and higher. You are given: (i) d =.5 (ii) Mortality assumptions: At issue 4 =., k =,,,...,49 k q Revised prospectively at time k q 5 =.4, k =,,,...,4 (iii) L is the prospective loss random variable at time using the contract premium. Calculate E[ LK(4) ] using the revised mortality assumption. 8

(A) Less than 5 (B) At least 5, but less than 5 (C) At least 5, but less than 75 (D) At least 75, but less than 3 (E) At least 3. For a group of individuals all age x, of which 3% are smokers and 7% are non-smokers, you are given: (i) δ =. (ii) (iii) (iv) (v) (vi) smoker A x =.444 non-smoker A x =.86 T is the future lifetime of (x). smoker Var a = 8.88 T non-smoker Var a = 8.53 T Calculate Var a T for an individual chosen at random from this group. (A) 8.5 (B) 8.6 (C) 8.8 (D) 9. (E) 9. 9

. T, the future lifetime of (), has a spliced distribution. (i) f () t follows the Illustrative Life Table. (ii) f () t follows DeMoivre s law with ω =. () kf t, t 5 (iii) ft () t =. f () t, 5 < t Calculate p 4. (A).8 (B).85 (C).88 (D).9 (E).96

3. A population has 3% who are smokers with a constant force of mortality. and 7% who are non-smokers with a constant force of mortality.. Calculate the 75 th percentile of the distribution of the future lifetime of an individual selected at random from this population. (A).7 (B). (C). (D).6 (E).8 4. Aggregate losses for a portfolio of policies are modeled as follows: (i) (ii) The number of losses before any coverage modifications follows a Poisson distribution with mean λ. The severity of each loss before any coverage modifications is uniformly distributed between and b. The insurer would like to model the impact of imposing an ordinary deductible, d( d b) on each loss and reimbursing only a percentage, c ( < c ), of each loss in excess of the deductible. It is assumed that the coverage modifications will not affect the loss distribution. The insurer models its claims with modified frequency and severity distributions. The modified claim amount is uniformly distributed on the interval,c( b d). Determine the mean of the modified frequency distribution. < <,

(A) (B) (C) (D) (E) λ λ c d λ b b d λ b b d λc b 5. The RIP Life Insurance Company specializes in selling a fully discrete whole life insurance of, to 65 year olds by telephone. For each policy: (i) The annual contract premium is 5. (ii) Mortality follows the Illustrative Life Table. (iii) i =.6 The number of telephone inquiries RIP receives follows a Poisson process with mean 5 per day. % of the inquiries result in the sale of a policy. The number of inquiries and the future lifetimes of all the insureds who purchase policies on a particular day are independent. Using the normal approximation, calculate the probability that S, the total prospective loss at issue for all the policies sold on a particular day, will be less than zero. (A).33 (B).5 (C).67 (D).84 (E).99

6. For a special fully discrete whole life insurance on (4): (i) The death benefit is for the first years; 5 for the next 5 years; thereafter. (ii) The annual benefit premium is P 4 for the first years; 5P 4 for the next 5 years; π thereafter. (iii) Mortality follows the Illustrative Life Table. (iv) i =.6 Calculate V, the benefit reserve at the end of year for this insurance. (A) 55 (B) 59 (C) 63 (D) 67 (E) 7 3

7. For a whole life insurance of on (4) with death benefit payable at the end of year of death, you are given: (i) i =.5 (ii) p 4 =.997 (iii) A4 A4 =.8 (iv) (v) A4 A4 =.433 Z is the present-value random variable for this insurance. Calculate Var(Z). (A).3 (B).4 (C).5 (D).6 (E).7 4

8. For a perpetuity-immediate with annual payments of : (i) The sequence of annual discount factors follows a Markov chain with the following three states: State number Annual discount factor, v.95.94.93 (ii) The transition matrix for the annual discount factors is:....9..... Y is the present value of the perpetuity payments when the initial state is. Calculate E(Y). (A) 5.67 (B) 5.7 (C) 5.75 (D) 6.8 (E) 6.86 5

9. A member of a high school math team is practicing for a contest. Her advisor has given her three practice problems: #, #, and #3. She randomly chooses one of the problems, and works on it until she solves it. Then she randomly chooses one of the remaining unsolved problems, and works on it until solved. Then she works on the last unsolved problem. She solves problems at a Poisson rate of problem per 5 minutes. Calculate the probability that she has solved problem #3 within minutes of starting the problems. (A).8 (B).34 (C).45 (D).5 (E).59. For a double decrement table, you are given: (i) (ii) (iii) () ( τ ) x x µ () t =. µ (), t t > ( τ ) µ x () t = k t, t > q ' =.4 () x Calculate (A).45 (B).53 (C).58 (D).64 (E).73 () q x. 6

. For (x): (i) K is the curtate future lifetime random variable. (ii) q =.( k+ ), k =,,,, 9 x+ k Calculate Var( K 3). (A). (B). (C).3 (D).4 (E).5. The graph of the density function for losses is: f(x )...8.6.4.. 8 Loss amount, x Calculate the loss elimination ratio for an ordinary deductible of. (A). (B).4 (C).8 (D).3 (E).36 7

3. Michel, age 45, is expected to experience higher than standard mortality only at age 64. For a special fully discrete whole life insurance of on Michel, you are given: (i) The benefit premiums are not level. (ii) The benefit premium for year, π 9, exceeds P 45 for a standard risk by.. (iii) Benefit reserves on his insurance are the same as benefit reserves for a fully discrete whole life insurance of on (45) with standard mortality and level benefit premiums. (iv) i =.3 (v) V 45 =.47 Calculate the excess of q 64 for Michel over the standard q 64. (A). (B).4 (C).6 (D).8 (E). 8

4. For a block of fully discrete whole life insurances of on independent lives age x, you are given: (i) i =.6 (ii) A x =.495 (iii) A =.9476 x (iv) π =.5, where π is the contract premium for each policy. (v) Losses are based on the contract premium. Using the normal approximation, calculate the minimum number of policies the insurer must issue so that the probability of a positive total loss on the policies issued is less than or equal to.5. (A) 5 (B) 7 (C) 9 (D) 3 (E) 33 9

5. Your company currently offers a whole life annuity product that pays the annuitant, at the beginning of each year. A member of your product development team suggests enhancing the product by adding a death benefit that will be paid at the end of the year of death. Using a discount rate, d, of 8%, calculate the death benefit that minimizes the variance of the present value random variable of the new product. (A) (B) 5, (C), (D) 5, (E), 6. A towing company provides all towing services to members of the City Automobile Club. You are given: Towing Distance Towing Cost Frequency -9.99 miles 8 5% -9.99 miles 4% 3+ miles 6 % (i) (ii) (iii) The automobile owner must pay % of the cost and the remainder is paid by the City Automobile Club. The number of towings has a Poisson distribution with mean of per year. The number of towings and the costs of individual towings are all mutually independent. Using the normal approximation for the distribution of aggregate towing costs, calculate the probability that the City Automobile Club pays more than 9, in any given year. (A) 3% (B) % (C) 5% (D) 9% (E) 97%

7. You are given: (i) Losses follow an exponential distribution with the same mean in all years. (ii) The loss elimination ratio this year is 7%. (iii) The ordinary deductible for the coming year is 4/3 of the current deductible. Compute the loss elimination ratio for the coming year. (A) 7% (B) 75% (C) 8% (D) 85% (E) 9% 8. For T, the future lifetime random variable for (): (i) ω > 7 (ii) 4 p =.6 (iii) E(T) = 6 (iv) E( T t) = t.5 t, < t < 6 Calculate the complete expectation of life at 4. (A) 3 (B) 35 (C) 4 (D) 45 (E) 5

9. Two actuaries use the same mortality table to price a fully discrete -year endowment insurance of on (x). (i) Kevin calculates non-level benefit premiums of 68 for the first year and 35 for the second year. (ii) Kira calculates level annual benefit premiums of π. (iii) d =.5 Calculateπ. (A) 48 (B) 489 (C) 497 (D) 58 (E) 57

3. For a fully discrete -payment whole life insurance of, on (x), you are given: (i) i =.5 (ii) q x + 9 =. (iii) q x + =. (iv) q x + =.4 (v) The level annual benefit premium is 78. (vi) The benefit reserve at the end of year 9 is 3,535. Calculate,A x+. (A) 34, (B) 34,3 (C) 35,5 (D) 36,5 (E) 36,7 3

3. You are given: (i) Mortality follows DeMoivre s law with ω = 5. (ii) (45) and (65) have independent future lifetimes. Calculate e. 45:65 (A) 33 (B) 34 (C) 35 (D) 36 (E) 37 4

3. Given: The survival function sx Calculate µ b4g. sx b g =, x< bg b g = x di { } b g, where sx = e /, x< 4. 5 sx, 4.5 x (A).45 (B).55 (C).8 (D). (E). 33. For a triple decrement table, you are given: (i) µ x t (ii) µ x t (iii) µ x t Calculate qbg x. (A).6 (B).3 (C).33 (D).36 (E).39 bg b g = 3., t > bg b g = 5., t > bg 3 b g = 7., t > 5

34. You are given: (i) the following select-and-ultimate mortality table with 3-year select period: (ii) i = 3. x q x q x + q x + q x+3 x + 3 6.9..3.5 63 6...4.6 64 6..3.5.7 65 63..4.6.8 66 64.3.5.7.9 67 Calculate A 6 on 6., the actuarial present value of a -year deferred -year term insurance (A).56 (B).6 (C).86 (D).9 (E).95 6

35. You are given: (i) (ii) µ x btg = t µ x btg =., 5 t., < 5 (iii) δ = 6. Calculate a x. (A).5 (B) 3. (C) 3.4 (D) 3.9 (E) 4.3 36. Actuaries have modeled auto windshield claim frequencies. They have concluded that the number of windshield claims filed per year per driver follows the Poisson distribution with parameter λ, where λ follows the gamma distribution with mean 3 and variance 3. Calculate the probability that a driver selected at random will file no more than windshield claim next year. (A).5 (B).9 (C). (D).4 (E).3 7

37. The number of auto vandalism claims reported per month at Sunny Daze Insurance Company (SDIC) has mean and variance 75. Individual losses have mean and standard deviation 7. The number of claims and the amounts of individual losses are independent. Using the normal approximation, calculate the probability that SDIC s aggregate auto vandalism losses reported for a month will be less than,. (A).4 (B).3 (C).36 (D).39 (E).49 38. For an allosaur with, calories stored at the start of a day: (i) (ii) (iii) (iv) The allosaur uses calories uniformly at a rate of 5, per day. If his stored calories reach, he dies. Each day, the allosaur eats scientist (, calories) with probability.45 and no scientist with probability.55. The allosaur eats only scientists. The allosaur can store calories without limit until needed. Calculate the probability that the allosaur ever has 5, or more calories stored. (A).54 (B).57 (C).6 (D).63 (E).66 8

39. Lucky Tom finds coins on his way to work at a Poisson rate of.5 coins/minute. The denominations are randomly distributed: (i) (ii) (iii) 6% of the coins are worth each % of the coins are worth 5 each % of the coins are worth each. Calculate the probability that in the first ten minutes of his walk he finds at least coins worth each, and in the first twenty minutes finds at least 3 coins worth each. (A).8 (B). (C).6 (D). (E).4 9

4. For a fully discrete whole life insurance of on (6), the annual benefit premium was calculated using the following: (i) i = 6. (ii) q 6 =. 376 (iii) A 6 = 369. 33 (iv) A 6 = 383. A particular insured is expected to experience a first-year mortality rate ten times the rate used to calculate the annual benefit premium. The expected mortality rates for all other years are the ones originally used. Calculate the expected loss at issue for this insured, based on the original benefit premium. (A) 7 (B) 86 (C) (D) 4 (E) 8 3

4. For a fully discrete whole life insurance of on (4), you are given: (i) i = 6. (ii) Mortality follows the Illustrative Life Table. (iii) a 4 = 77. (iv) a 5 = 757. : : 4 (v) A : = 6. At the end of the tenth year, the insured elects an option to retain the coverage of for life, but pay premiums for the next ten years only. Calculate the revised annual benefit premium for the next years. (A) (B) 5 (C) 7 (D) 9 (E) 3

4. For a double-decrement table where cause is death and cause is withdrawal, you are given: (i) (ii) (iii) Deaths are uniformly distributed over each year of age in the single-decrement table. Withdrawals occur only at the end of each year of age. lbg τ x = (iv) qbg x = 4. bg x x (v) d = 45. d bg Calculate p x bg. (A).5 (B).53 (C).55 (D).57 (E).59 3

43. You intend to hire employees for a new management-training program. To predict the number who will complete the program, you build a multiple decrement table. You decide that the following associated single decrement assumptions are appropriate: (i) (ii) (iii) (iv) Of 4 hires, the number who fail to make adequate progress in each of the first three years is, 6, and 8, respectively. Of 3 hires, the number who resign from the company in each of the first three years is 6, 8, and, respectively. Of hires, the number who leave the program for other reasons in each of the first three years is,, and 4, respectively. You use the uniform distribution of decrements assumption in each year in the multiple decrement table. Calculate the expected number who fail to make adequate progress in the third year. (A) 4 (B) 8 (C) (D) 4 (E) 7 33

44. Bob is an overworked underwriter. Applications arrive at his desk at a Poisson rate of 6 per day. Each application has a /3 chance of being a bad risk and a /3 chance of being a good risk. Since Bob is overworked, each time he gets an application he flips a fair coin. If it comes up heads, he accepts the application without looking at it. If the coin comes up tails, he accepts the application if and only if it is a good risk. The expected profit on a good risk is 3 with variance,. The expected profit on a bad risk is with variance 9,. Calculate the variance of the profit on the applications he accepts today. (A) 4,, (B) 4,5, (C) 5,, (D) 5,5, (E) 6,, 45. Prescription drug losses, S, are modeled assuming the number of claims has a geometric distribution with mean 4, and the amount of each prescription is 4. Calculate EbS g. + (A) 6 (B) 8 (C) 9 (D) 4 (E) 46 34

46. For a temporary life annuity-immediate on independent lives (3) and (4): (i) Mortality follows the Illustrative Life Table. (ii) i = 6. Calculate a 3: 4:. (A) 6.64 (B) 7.7 (C) 7.88 (D) 8.74 (E) 9.86 47. For a special whole life insurance on (35), you are given: (i) (ii) (iii) The annual benefit premium is payable at the beginning of each year. The death benefit is equal to plus the return of all benefit premiums paid in the past without interest. The death benefit is paid at the end of the year of death. (iv) A 35 =. 4898 (v) biag 35 = 6676. (vi) i = 5. Calculate the annual benefit premium for this insurance. (A) 73.66 (B) 75.8 (C) 77.4 (D) 78.95 (E) 8.66 35

48. Subway trains arrive at a station at a Poisson rate of per hour. 5% of the trains are express and 75% are local. The types of each train are independent. An express gets you to work in 6 minutes and a local gets you there in 8 minutes. You always take the first train to arrive. Your co-worker always takes the first express. You both are waiting at the same station. Which of the following is true? (A) (B) (C) (D) (E) Your expected arrival time is 6 minutes earlier than your co-worker s. Your expected arrival time is 4.5 minutes earlier than your co-worker s. Your expected arrival times are the same. Your expected arrival time is 4.5 minutes later than your co-worker s. Your expected arrival time is 6 minutes later than your co-worker s. 49. For a special fully continuous whole life insurance of on the last-survivor of (x) and (y), you are given: (i) b g and Tbyg are independent. xbtg = ybtg = 7., t > Tx (ii) µ µ (iii) δ = 5. (iv) Premiums are payable until the first death. Calculate the level annual benefit premium for this insurance. (A).4 (B).7 (C).8 (D). (E).4 36

5. For a fully discrete whole life insurance of on (), you are given: (i) P = (ii) V = 49 (iii) V = 545 (iv) V = 65 (v) q 4 =. Calculate q 4. (A).4 (B).5 (C).6 (D).7 (E).8 5. For a fully discrete whole life insurance of on (6), you are given: (i) i = 6. (ii) Mortality follows the Illustrative Life Table, except that there are extra mortality risks at age 6 such that q 6 5 =.. Calculate the annual benefit premium for this insurance. (A) 3.5 (B) 3. (C) 3. (D) 33. (E) 33. 37

5. At the beginning of each round of a game of chance the player pays.5. The player then rolls one die with outcome N. The player then rolls N dice and wins an amount equal to the total of the numbers showing on the N dice. All dice have 6 sides and are fair. Using the normal approximation, calculate the probability that a player starting with 5, will have at least 5, after rounds. (A). (B).4 (C).6 (D).9 (E). 53. X is a discrete random variable with a probability function which is a member of the (a,b,) class of distributions. You are given: (i) P( X = )= P( X = )= 5. (ii) P( X = )= 875. Calculate PX= b 3g. (A). (B).5 (C).3 (D).35 (E).4 38

54. Nancy reviews the interest rates each year for a 3-year fixed mortgage issued on July. She models interest rate behavior by a Markov model assuming: (i) (ii) Interest rates always change between years. The change in any given year is dependent on the change in prior years as follows: from year t 3 to year t from year t to year t Probability that year t will increase from year t Increase Increase. Decrease Decrease. Increase Decrease.4 Decrease Increase.5 She notes that interest rates decreased from year to and from year to. Calculate the probability that interest rates will decrease from year 3 to 4. (A).76 (B).79 (C).8 (D).84 (E).87 39

55. For a -year deferred whole life annuity-due of per year on (45), you are given: (i) Mortality follows De Moivre s law with ω = 5. (ii) i = Calculate the probability that the sum of the annuity payments actually made will exceed the actuarial present value at issue of the annuity. (A).45 (B).45 (C).475 (D).5 (E).55 56. For a continuously increasing whole life insurance on bxg, you are given: (i) The force of mortality is constant. (ii) δ = 6. (iii) A x = 5. Calculate diai. x (A).889 (B) 3.5 (C) 4. (D) 4.667 (E) 5.5 4

57. XYZ Co. has just purchased two new tools with independent future lifetimes. Each tool has its own distinct De Moivre survival pattern. One tool has a -year maximum lifetime and the other a 7-year maximum lifetime. Calculate the expected time until both tools have failed. (A) 5. (B) 5. (C) 5.4 (D) 5.6 (E) 5.8 58. XYZ Paper Mill purchases a 5-year special insurance paying a benefit in the event its machine breaks down. If the cause is minor (), only a repair is needed. If the cause is major (), the machine must be replaced. Given: (i) (ii) The benefit for cause () is payable at the moment of breakdown. The benefit for cause () is 5, payable at the moment of breakdown. (iii) Once a benefit is paid, the insurance contract is terminated. (iv) µ bg b g (v) δ = 4. bg b g t =. and µ t =. 4, for t > Calculate the actuarial present value of this insurance. (A) 784 (B) 788 (C) 79 (D) 796 (E) 8 4

59. You are given: (i) R = e z (ii) S = e z µ bg x tdt c µ bg x t + k dt (iii) k is a constant such that S = 75. R h Determine an expression for k. b xg b xg c (A) n q /. 75q b xg b xg c (B) n. 75q / p b xg b xg c (C) n. 75p / p b xg b xg c (D) n p /. 75q b xg b xg c (E) n. 75q / q h h h h h 6. The number of claims in a period has a geometric distribution with mean 4. The amount of each claim X follows PX b = xg = 5., x = 34,,,. The number of claims and the claim amounts are independent. S is the aggregate claim amount in the period. b g. Calculate F s 3 (A).7 (B).9 (C).3 (D).33 (E).35 4

6. Insurance agent Hunt N. Quotum will receive no annual bonus if the ratio of incurred losses to earned premiums for his book of business is 6% or more for the year. If the ratio is less than 6%, Hunt s bonus will be a percentage of his earned premium equal to 5% of the difference between his ratio and 6%. Hunt s annual earned premium is 8,. Incurred losses are distributed according to the Pareto distribution, with θ = 5, and α =. Calculate the expected value of Hunt s bonus. (A) 3, (B) 7, (C) 4, (D) 9, (E) 35, 6. A large machine in the ABC Paper Mill is 5 years old when ABC purchases a 5-year term insurance paying a benefit in the event the machine breaks down. Given: (i) Annual benefit premiums of 6643 are payable at the beginning of the year. (ii) (iii) A benefit of 5, is payable at the moment of breakdown. Once a benefit is paid, the insurance contract is terminated. (iv) Machine breakdowns follow De Moivre s law with l x. (v) i = 6. Calculate the benefit reserve for this insurance at the end of the third year. (A) 9 (B) (C) 63 (D) 87 (E) 4 x = 43

63. For a whole life insurance of on x (i) The force of mortality is µ x t b g, you are given: b g. (ii) The benefits are payable at the moment of death. (iii) δ = 6. (iv) A x = 6. Calculate the revised actuarial present value of this insurance assuming µ x btg is increased by.3 for all t and δ is decreased by.3. (A).5 (B).6 (C).7 (D).8 (E).9 64. A maintenance contract on a hotel promises to replace burned out light bulbs at the end of each year for three years. The hotel has, light bulbs. The light bulbs are all new. If a replacement bulb burns out, it too will be replaced with a new bulb. You are given: (i) For new light bulbs, q =. q = 3. q = 5. (ii) Each light bulb costs. (iii) i = 5. Calculate the actuarial present value of this contract. (A) 67 (B) 7 (C) 73 (D) 76 (E) 8 44

65. You are given: R bg= S T µ x Calculate e 5: 5. (A) 4. (B) 4.4 (C) 4.8 (D) 5. (E) 5.6 4., < x < 4 5., x > 4 66. For a select-and-ultimate mortality table with a 3-year select period: (i) x q x q x + q x + q x+3 x + 3 6.9..3.5 63 6...4.6 64 6..3.5.7 65 63..4.6.8 66 64.3.5.7.9 67 (ii) White was a newly selected life on //. (iii) White s age on // is 6. (iv) P is the probability on // that White will be alive on //6. Calculate P. (A) P <.43 (B).43 P <.45 (C).45 P <.47 (D).47 P <.49 (E).49 P. 45

67. For a continuous whole life annuity of on ( x ) : (i) T( x) is the future lifetime random variable for ( x ). (ii) The force of interest and force of mortality are constant and equal. (iii) a x =. 5 Calculate the standard deviation of a T( x). (A).67 (B).5 (C).89 (D) 6.5 (E) 7. 68. For a special fully discrete whole life insurance on (x): (i) (ii) The death benefit is in the first year and 5 thereafter. Level benefit premiums are payable for life. (iii) q x = 5. (iv) v = 9. (v) a x = 5. (vi) V x =. (vii) V is the benefit reserve at the end of year for this insurance. Calculate V. (A) 795 (B) (C) 9 (D) 8 (E) 5 46

69. For a fully discrete -year term insurance of on (x): (i).95 is the lowest premium such that there is a % chance of loss in year. (ii) p x =.75 (iii) p x+ =.8 (iv) Z is the random variable for the present value at issue of future benefits. b g. Calculate Var Z (A).5 (B).7 (C).9 (D). (E).3 7. A group dental policy has a negative binomial claim count distribution with mean 3 and variance 8. Ground-up severity is given by the following table: Severity Probability 4.5 8.5.5.5 You expect severity to increase 5% with no change in frequency. You decide to impose a per claim deductible of. Calculate the expected total claim payment after these changes. (A) Less than 8, (B) At least 8,, but less than, (C) At least,, but less than, (D) At least,, but less than 4, (E) At least 4, 47

7. You own a fancy light bulb factory. Your workforce is a bit clumsy they keep dropping boxes of light bulbs. The boxes have varying numbers of light bulbs in them, and when dropped, the entire box is destroyed. You are given: Expected number of boxes dropped per month: 5 Variance of the number of boxes dropped per month: Expected value per box: Variance of the value per box: 4 You pay your employees a bonus if the value of light bulbs destroyed in a month is less than 8. Assuming independence and using the normal approximation, calculate the probability that you will pay your employees a bonus next month. (A).6 (B).9 (C).3 (D).7 (E).3 7. Each of independent lives purchase a single premium 5-year deferred whole life insurance of payable at the moment of death. You are given: (i) µ = 4. (ii) δ = 6. (iii) F is the aggregate amount the insurer receives from the lives. Using the normal approximation, calculate F such that the probability the insurer has sufficient funds to pay all claims is.95. 48

(A) 8 (B) 39 (C) 5 (D) 6 (E) 7 73. For a select-and-ultimate table with a -year select period: x p x p x + p x+ x+ 48.9865.984.973 5 49.9858.983.9698 5 5.9849.989.968 5 5.9838.983.9664 53 Keith and Clive are independent lives, both age 5. Keith was selected at age 45 and Clive was selected at age 5. Calculate the probability that exactly one will be alive at the end of three years. (A) Less than.5 (B) At least.5, but less than.5 (C) At least.5, but less than.35 (D) At least.35, but less than.45 (E) At least.45 49

74-75. Use the following information for questions 74 and 75. For a tyrannosaur with, calories stored: (i) (ii) (iii) (iv) The tyrannosaur uses calories uniformly at a rate of, per day. If his stored calories reach, he dies. The tyrannosaur eats scientists (, calories each) at a Poisson rate of per day. The tyrannosaur eats only scientists. The tyrannosaur can store calories without limit until needed. 74. Calculate the probability that the tyrannosaur dies within the next.5 days. (A).3 (B).4 (C).5 (D).6 (E).7 75. Calculate the expected calories eaten in the next.5 days. (A) 7,8 (B) 8,8 (C) 9,8 (D),8 (E),8 5

76. A fund is established by collecting an amount P from each of independent lives age 7. The fund will pay the following benefits:, payable at the end of the year of death, for those who die before age 7, or P, payable at age 7, to those who survive. You are given: Mortality follows the Illustrative Life Table. (i) i =.8 Calculate P, using the equivalence principle. (A).33 (B).38 (C) 3. (D) 3.7 (E) 3.55 77. You are given: (i) P x =. 9 (ii) nv x = 563. (iii) P xn : =. 864 Calculate P xn :. (A).8 (B).4 (C).4 (D).65 (E).85 5

78. You are given: (i) Mortality follows De Moivre s law with ω =. (ii) i = 5. (iii) The following annuity-certain values: a a 4 5 = 7. 58 = 8. 7 a = 9. 4 6 Calculate VcA 4 h. (A).75 (B).77 (C).79 (D).8 (E).83 79. For a group of individuals all age x, you are given: (i) 3% are smokers and 7% are non-smokers. (ii) The constant force of mortality for smokers is.6. (iii) The constant force of mortality for non-smokers is.3. (iv) δ = 8. F H Calculate Var a T x (A) 3. bg I K for an individual chosen at random from this group. (B) 3.3 (C) 3.8 (D) 4. (E) 4.6 5

8. For a certain company, losses follow a Poisson frequency distribution with mean per year, and the amount of a loss is,, or 3, each with probability /3. Loss amounts are independent of the number of losses, and of each other. An insurance policy covers all losses in a year, subject to an annual aggregate deductible of. Calculate the expected claim payments for this insurance policy. (A). (B).36 (C).45 (D).8 (E).96 8. A Poisson claims process has two types of claims, Type I and Type II. (i) The expected number of claims is 3. (ii) The probability that a claim is Type I is /3. (iii) Type I claim amounts are exactly each. (iv) The variance of aggregate claims is,,. Calculate the variance of aggregate claims with Type I claims excluded. (A),7, (B),8, (C),9, (D),, (E),, 53

8. Don, age 5, is an actuarial science professor. His career is subject to two decrements: (i) (ii) Decrement is mortality. The associated single decrement table follows De Moivre s law with ω =. Decrement is leaving academic employment, with bg btg µ 5 = 5., t Calculate the probability that Don remains an actuarial science professor for at least five but less than ten years. (A). (B).5 (C).8 (D).3 (E).34 83. For a double decrement model: (i) In the single decrement table associated with cause (), q bg 4 =. and decrements are uniformly distributed over the year. (ii) In the single decrement table associated with cause (), q bg 4 = 5. and all decrements occur at time.7. Calculate qbg 4. (A).4 (B).5 (C).6 (D).7 (E).8 54

84. For a special -payment whole life insurance on (8): (i) Premiums of π are paid at the beginning of years and 3. (ii) (iii) (iv) The death benefit is paid at the end of the year of death. There is a partial refund of premium feature: If (8) dies in either year or year 3, the death benefit is + π. Otherwise, the death benefit is. Mortality follows the Illustrative Life Table. (v) i =.6 Calculate π, using the equivalence principle. (A) 369 (B) 38 (C) 397 (D) 49 (E) 45 85. For a special fully continuous whole life insurance on (65): 4. t (i) The death benefit at time t is b = e, t. t (ii) Level benefit premiums are payable for life. b g = (iii) µ 65 t., t (iv) δ = 4. Calculate V, the benefit reserve at the end of year. (A) (B) 9 (C) 37 (D) 6 (E) 83 55

86. You are given: (i) A x = 8. (ii) A x+ = 4. (iii) A x: = 5. (iv) i = 5. Calculate a x:. (A). (B). (C).7 (D). (E).3 87. On his walk to work, Lucky Tom finds coins on the ground at a Poisson rate. The Poisson rate, expressed in coins per minute, is constant during any one day, but varies from day to day according to a gamma distribution with mean and variance 4. Calculate the probability that Lucky Tom finds exactly one coin during the sixth minute of today s walk. (A). (B).4 (C).6 (D).8 (E).3 56

88. The unlimited severity distribution for claim amounts under an auto liability insurance policy is given by the cumulative distribution: b g =. x. x Fx 8. e. e, x The insurance policy pays amounts up to a limit of per claim. Calculate the expected payment under this policy for one claim. (A) 57 (B) 8 (C) 66 (D) 5 (E) 4 89. A machine is in one of four states (F, G, H, I) and migrates annually among them according to a Markov process with transition matrix: F G H I F..8.. G.5..5. H.75...5 I.... At time, the machine is in State F. A salvage company will pay 5 at the end of 3 years if the machine is in State F. Assuming v = 9., calculate the actuarial present value at time of this payment. (A) 5 (B) 55 (C) 6 (D) 65 (E) 7 57

9. The claims department of an insurance company receives envelopes with claims for insurance coverage at a Poisson rate of λ = 5 envelopes per week. For any period of time, the number of envelopes and the numbers of claims in the envelopes are independent. The numbers of claims in the envelopes have the following distribution: Number of Claims Probability..5 3.4 4.5 Using the normal approximation, calculate the 9 th percentile of the number of claims received in 3 weeks. (A) 69 (B) 7 (C) 73 (D) 75 (E) 77 9. You are given: x (i) The survival function for males is sx bg=, < x < 75. 75 (ii) (iii) Female mortality follows De Moivre s law. At age 6, the female force of mortality is 6% of the male force of mortality. For two independent lives, a male age 65 and a female age 6, calculate the expected time until the second death. (A) 4.33 (B) 5.63 (C) 7.3 (D).88 (E) 3.7 58

9. For a fully continuous whole life insurance of : (i) µ = 4. (ii) δ = 8. (iii) L is the loss-at-issue random variable based on the benefit premium. Calculate Var (L). (A) (B) (C) (D) (E) 5 4 3 93. The random variable for a loss, X, has the following characteristics: x Fx b g Eb X xg.. 9.6 53. 33 Calculate the mean excess loss for a deductible of. (A) 5 (B) 3 (C) 35 (D) 4 (E) 45 59

94. WidgetsRUs owns two factories. It buys insurance to protect itself against major repair costs. Profit equals revenues, less the sum of insurance premiums, retained major repair costs, and all other expenses. WidgetsRUs will pay a dividend equal to the profit, if it is positive. You are given: (i) Combined revenue for the two factories is 3. (ii) (iii) Major repair costs at the factories are independent. The distribution of major repair costs for each factory is k Prob (k).4.3. 3. (iv) (v) At each factory, the insurance policy pays the major repair costs in excess of that factory s ordinary deductible of. The insurance premium is % of the expected claims. All other expenses are 5% of revenues. Calculate the expected dividend. (A).43 (B).47 (C).5 (D).55 (E).59 6

95. For watches produced by a certain manufacturer: (i) Lifetimes follow a single-parameter Pareto distribution with α > and θ = 4. (ii) The expected lifetime of a watch is 8 years. Calculate the probability that the lifetime of a watch is at least 6 years. (A).44 (B).5 (C).56 (D).6 (E).67 96. For a special 3-year deferred whole life annuity-due on (x): (i) i = 4. (ii) The first annual payment is. (iii) (iv) (v) Payments in the following years increase by 4% per year. There is no death benefit during the three year deferral period. Level benefit premiums are payable at the beginning of each of the first three years. (vi) e x = 5. is the curtate expectation of life for (x). (vii) k 3 k p x.99.98.97 Calculate the annual benefit premium. (A) 65 (B) 85 (C) 35 (D) 35 (E) 345 6

97. For a special fully discrete -payment whole life insurance on (3) with level annual benefit premium π : (i) The death benefit is equal to plus the refund, without interest, of the benefit premiums paid. (ii) A 3 =. (iii) A 3 =. 88 (iv) biag 3 = 78 :. (v) a 3 = 7. 747 : Calculate π. (A) 4.9 (B) 5. (C) 5. (D) 5. (E) 5.3 6

98. For a given life age 3, it is estimated that an impact of a medical breakthrough will be an increase of 4 years in e 3, the complete expectation of life. Prior to the medical breakthrough, s(x) followed de Moivre s law with ω = as the limiting age. Assuming de Moivre s law still applies after the medical breakthrough, calculate the new limiting age. (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 99. On January,, Pat, age 4, purchases a 5-payment, -year term insurance of,: (i) (ii) Death benefits are payable at the moment of death. Contract premiums of 4 are payable annually at the beginning of each year for 5 years. (iii) i =.5 (iv) L is the loss random variable at time of issue. Calculate the value of L if Pat dies on June 3, 4. (A) 77, (B) 8,7 (C) 8,7 (D) 85,9 (E) 88, 63

. Glen is practicing his simulation skills. He generates values of the random variable X as follows: (i) He generates the observed value λ from the gamma distribution with α = and θ = (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) (C) 5 (D) 5 (E) 75. Lucky Tom finds coins on his way to work at a Poisson rate of.5 coins per minute. The denominations are randomly distributed: (i) 6% of the coins are worth ; (ii) % of the coins are worth 5; (iii) % of the coins are worth. 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) 67 (E) 768 64

. For a fully discrete -payment whole life insurance of on (x), you are given: (i) i =.6 (ii) q x+ 9 =. 54 (iii) The level annual benefit premium is 3.7. (iv) The benefit reserve at the end of year 9 is 34.3. Calculate P x+, the level annual benefit premium for a fully discrete whole life insurance of on (x+). (A) 7 (B) 9 (C) 3 (D) 33 (E) 35 3. For a multiple decrement model on (6): (i) µ () 6 (), t t, follows the Illustrative Life Table. ( τ ) ( ) 6 6 (ii) µ () t = µ (), t t Calculate q ( τ ) 6, the probability that decrement occurs during the th year. (A).3 (B).4 (C).5 (D).6 (E).7 65

4. (x) and (y) are two lives with identical expected mortality. You are given: Px = Py =. P xy = 6., where P xy is the annual benefit premium for a fully discrete insurance of on xy b g. d = 6. Calculate the premium P xy, the annual benefit premium for a fully discrete insurance of on bxyg. (A).4 (B).6 (C).8 (D). (E). 5. For students entering a college, you are given the following from a multiple decrement model: (i) students enter the college at t =. (ii) Students leave the college for failure bg b g (iii) µ t = µ t 4 bg b g µ t = 4. t < 4 b g or all other reasons (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) (C) 4 (D) 34 (E) 4 b g. 66

6. The following graph is related to current human mortality: 4 6 8 Which of the following functions of age does the graph most likely show? (A) µ bxg (B) lxµbxg Age (C) (D) lp x x l x (E) l x 7. 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. and variance.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). (B).5 (C).3 (D).35 (E).4 67

68

8. 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 and variance 9. (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, eggs will be released is greater than 95%. (A) (B) 3 (C) 6 (D) 9 (E) 3 9. 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=. ( k + ) k =,, (iii) The following death benefits, payable at the end of the year of death: (iv) i = 6. k b k+ 3, 35, 4, 69

Calculate EbZg. (A) 36,8 (B) 39, (C) 4,4 (D) 43,7 (E) 46,. For a special fully discrete -year endowment insurance on (55): (i) Death benefits in year k are given by bk = b kg, k =,,,. (ii) The maturity benefit is. (iii) Annual benefit premiums are level. (iv) kv denotes the benefit reserve at the end of year k, k =,,,. (v) V =5. (vi) 9 V =.6 (vii) q 65 =. (viii) i =.8 Calculate V. (A) 4.5 (B) 4.6 (C) 4.8 (D) 5. (E) 5.3 7

. For a stop-loss insurance on a three person group: (i) Loss amounts are independent. (ii) (iii) The distribution of loss amount for each person is: Loss Amount Probability.4.3. 3. The stop-loss insurance has a deductible of for the group. Calculate the net stop-loss premium. (A). (B).3 (C).6 (D).9 (E).. A continuous two-life annuity pays: while both (3) and (4) are alive; 7 while (3) is alive but (4) is dead; and 5 while (4) is alive but (3) is dead. The actuarial present value of this annuity is 8. Continuous single life annuities paying per year are available for (3) and (4) with actuarial present values of and, respectively. Calculate the actuarial present value of a two-life continuous annuity that pays while at least one of them is alive. (A) 4 (B) 5 (C) 6 (D) 7 (E) 8 7

3. For a disability insurance claim: (i) The claimant will receive payments at the rate of, per year, payable continuously as long as she remains disabled. (ii) (iii) The length of the payment period in years is a random variable with the gamma distribution with parameters α = and θ =. Payments begin immediately. (iv) δ = 5. Calculate the actuarial present value of the disability payments at the time of disability. (A) 36,4 (B) 37, (C) 38, (D) 39, (E) 4, 4. For a discrete probability distribution, you are given the recursion relation Determine pb4g. bg b g pk = * pk, k =,,. k (A).7 (B).8 (C).9 (D). (E). 7

5. A company insures a fleet of vehicles. Aggregate losses have a compound Poisson distribution. The expected number of losses is. Loss amounts, regardless of vehicle type, have exponential distribution with θ =. In order to reduce the cost of the insurance, two modifications are to be made: (i) a certain type of vehicle will not be insured. It is estimated that this will reduce loss frequency by %. (ii) a deductible of per loss will be imposed. Calculate the expected aggregate amount paid by the insurer after the modifications. (A) 6 (B) 94 (C) 5 (D) 3 (E) 388 6. 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 (,). Calculate the probability that an individual drawn at random from this population dies within one year. (A).37 (B).43 (C).5 (D).57 (E).63 73

7. 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 Prb X xg n Pr N = n =.8...7. Your budget for prizes equals the expected prizes plus the standard deviation of prizes. Calculate your budget. (A) 36 (B) 36 (C) 46 (D) 5 (E) 58 8. For a special fully discrete 3-year term insurance on x (i) b g : Level benefit premiums are paid at the beginning of each year. (ii) k b k+ q x+ k,.3 5,.6,.9 (iii) i =.6 Calculate the initial benefit reserve for year. (A) 6,5 (B) 7,5 (C) 8, (D) 9,4 (E),3 74

75

9. For a special fully continuous whole life insurance on (x): (i) The level premium is determined using the equivalence principle. t (ii) Death benefits are given by bt = + ig where i is the interest rate. b (iii) (iv) L is the loss random variable at t = 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 A x xh h c hc h (B) ν T Ax + Ax (C) c c ν T A + A x xh h c hc h (D) ν T Ax Ax (E) dv c T + A + A x xi h 76

. For a 4-year college, you are given the following probabilities for dropout from all causes: q q q q 3 = 5. =. = 5. =. Dropouts are uniformly distributed over each year. Compute the temporary.5-year complete expected college lifetime of a student entering the second year, e 5 :.. (A).5 (B).3 (C).35 (D).4 (E).45 77

. Lee, age 63, considers the purchase of a single premium whole life insurance of, 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 =.5 The company calculates contract premiums as % 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).3 (B).35 (C).4 (D).45 (E).5. You have calculated the actuarial present value of a last-survivor whole life insurance of 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 = 6.. (iii) δ = 5. 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.. Calculate the increase in the actuarial present value over what you originally calculated. 78

(A). (B).39 (C).93 (D).9 (E).63 3. The number of accidents follows a Poisson distribution with mean. Each accident generates,, or 3 claimants with probabilities, 3, 6, respectively. Calculate the variance in the total number of claimants. (A) (B) 5 (C) 3 (D) 35 (E) 4 79

4. For a claims process, you are given: m b g r (i) The number of claims Nt, t is a nonhomogeneous Poisson process with intensity function: λ( t) = R S T, t <, t < 3, t (ii) 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 [,8]. (iv) The random variable P is the number of claims with claim amount less than 5 by time t = 3. (v) The random variable Q is the number of claims with claim amount greater than 5 by time t = 3. (vi) R is the conditional expected value of P, given Q = 4. Calculate R. (A). (B).5 (C) 3. (D) 3.5 (E) 4. 8

5. 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., the death benefit is. With probability.8, the death benefit is. (ii) (iii) (iv) At the start of each year, including the first, while (5) is alive, there is a random drawing. With probability.8, the level premium π is paid. With probability., no premium is paid. The random drawings are independent. Mortality follows the Illustrative Life Table. (v) i = 6. (vi) π is determined using the equivalence principle. Calculate the benefit reserve at the end of year. (A).5 (B).5 (C) 3.75 (D) 4. (E) 5.5 6. A government creates a fund to pay this year s lottery winners. You are given: (i) There are winners each age 4. (ii) (iii) (iv) Each winner receives payments of per year for life, payable annually, beginning immediately. Mortality follows the Illustrative Life Table. The lifetimes are independent. (v) i =.6 (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%. 8

Calculate the initial amount of the fund. (A) 4,8 (B) 4,9 (C) 5,5 (D) 5,5 (E) 5,5 7. For a special fully discrete 35-payment whole life insurance on (3): (i) (ii) (iii) The death benefit is for the first years and is 5 thereafter. The initial benefit premium paid during the each of the first years is one fifth of the benefit premium paid during each of the 5 subsequent years. Mortality follows the Illustrative Life Table. (iv) i = 6. (v) A 3 =. 337 : (vi) a 3 35 = 4. 835 : Calculate the initial annual benefit premium. (A). (B).5 (C). (D).5 (F).3 8

8. For independent lives (x) and (y): (i) q x = 5. (ii) q y =. (iii) Deaths are uniformly distributed over each year of age. Calculate 75. q xy. (A).88 (B).97 (C).6 (D).6 (E).5 9. In a clinic, physicians volunteer their time on a daily basis to provide care to those who are not eligible to obtain care otherwise. The number of physicians who volunteer in any day is uniformly distributed on the integers through 5. The number of patients that can be served by a given physician has a Poisson distribution with mean 3. Determine the probability that or more patients can be served in a day at the clinic, using the normal approximation with continuity correction. b g b g b g b g b g (A) Φ. 68 (B) Φ. 7 (C) Φ. 93 (D) Φ 33. (E) Φ 36. 83

3. A person age 4 wins, in the actuarial lottery. Rather than receiving the money at once, the winner is offered the actuarially equivalent option of receiving an annual payment of K (at the beginning of each year) guaranteed for years and continuing thereafter for life. You are given: (i) i = 4. (ii) A 4 = 3. (iii) A 5 = 35. (iv) A 4 : = 9. Calculate K. (A) 538 (B) 54 (C) 545 (D) 548 (E) 55 3. Mortality for Audra, age 5, follows De Moivre s law with ω =. If she takes up hot air ballooning for the coming year, her assumed mortality will be adjusted so that for the coming year only, she will have a constant force of mortality of.. Calculate the decrease in the -year temporary complete life expectancy for Audra if she takes up hot air ballooning. (A). (B).35 (C).6 (D).8 (E). 84

3. For a 5-year fully continuous term insurance on (x): (i) δ =. (ii) (iii) All the graphs below are to the same scale. All the graphs show µ x t b g on the vertical axis and t on the horizontal axis. Which of the following mortality assumptions would produce the highest benefit reserve at the end of year? (A) (B).8.8.6.6.4.4.. 3 4 5 3 4 5 (C) (D).8.8.6.6.4.4... 3 4 5 3 4 5 (E).8.6.4. 3 4 5 85

33. For a multiple decrement table, you are given: (i) Decrement withdrawal. () q 6 (ii) =. ( ) q 6 (iii) =. 5 ( 3) q 6 (iv) =. b g is death, decrement bg is disability, and decrement b3g is (v) (vi) Withdrawals occur only at the end of the year. Mortality and disability are uniformly distributed over each year of age in the associated single decrement tables. ( 3 Calculate q ) 6. (A).88 (B).9 (C).94 (D).97 (E). 86

34. The number of claims, N, made on an insurance portfolio follows the following distribution: n Pr(N=n).7. 3. If a claim occurs, the benefit is or with probability.8 and., respectively. The number of claims and the benefit for each claim are independent. Calculate the probability that aggregate benefits will exceed expected benefits by more than standard deviations. (A). (B).5 (C).7 (D).9 (E). 87

35. For a special whole life insurance of, on (x), you are given: (i) δ = 6. (ii) (iii) (iv) The death benefit is payable at the moment of death. If death occurs by accident during the first 3 years, the death benefit is doubled. µ τ bg b g = x t. 8, t (v) µ bg x btg =., t, where µ bg x is the force of decrement due to death by accident. Calculate the single benefit premium for this insurance. (A),765 (B),95 (C),6 (D) 3,44 (E) 3,35 36. You are given the following extract from a select-and-ultimate mortality table with a - year select period: x l x l x + l x+ x + 6 8,65 79,954 78,839 6 6 79,37 78,4 77,5 63 6 77,575 76,77 75,578 64 Assume that deaths are uniformly distributed between integral ages. Calculate 9. q 6 +. 6. (A). (B).3 (C).4 (D).5 (E).6 88

37. A claim count distribution can be expressed as a mixed Poisson distribution. The mean of the Poisson distribution is uniformly distributed over the interval [,5]. Calculate the probability that there are or more claims. (A).6 (B).66 (C).7 (D).76 (E).8 89

38. For a double decrement table with lbg τ 4 = : Calculate lbg τ 4. x qbg x qbg x 4 4.4 --. -- q x bg q x bg.5. y y (A) 8 (B) 8 (C) 84 (D) 86 (E) 88 39. For a fully discrete whole life insurance of, on (3): (i) (ii) (iii) π denotes the annual premium and L π variable for this insurance. Mortality follows the Illustrative Life Table. i=.6 b g denotes the loss-at-issue random Calculate the lowest premium, π, such that the probability is less than.5 that the loss L π b g is positive. (A) 34.6 (B) 36.6 (C) 36.8 (D) 39. (E) 39. 9

4. Y is the present-value random variable for a special 3-year temporary life annuity-due on (x). You are given: (i) t px t = 9., t (ii) (iii) K is the curtate-future-lifetime random variable for (x). Y = R S T., K = 87., K = 7., K = 3,,... Calculate Var(Y). (A).9 (B).3 (C).37 (D).46 (E).55 4. A claim severity distribution is exponential with mean. An insurance company will pay the amount of each claim in excess of a deductible of. Calculate the variance of the amount paid by the insurance company for one claim, including the possibility that the amount paid is. (A) 8, (B) 86, (C) 9, (D) 99, (E),, 9

4. For a fully continuous whole life insurance of on (x): (i) π is the benefit premium. (ii) L is the loss-at-issue random variable with the premium equal to π. (iii) L* is the loss-at-issue random variable with the premium equal to.5 π. (iv) a x = 5. (v) δ = 8. (vi) Var L b g = 565. Calculate the sum of the expected value and the standard deviation of L*. (A).59 (B).7 (C).86 (D).89 (E). 43. Workers compensation claims are reported according to a Poisson process with mean per month. The number of claims reported and the claim amounts are independently distributed. % of the claims exceed 3,. Calculate the number of complete months of data that must be gathered to have at least a 9% chance of observing at least 3 claims each exceeding 3,. (A) (B) (C) 3 (D) 4 (E) 5 9

44. For students entering a three-year law school, you are given: (i) The following double decrement table: Academic Year 3 For a student at the beginning of that academic year, probability of Withdrawal for Survival All Other Through Reasons Academic Year Academic Failure.4 -- --..3 -- -- --.6 (ii) Ten times as many students survive year as fail during year 3. (iii) The number of students who fail during year is 4% of the number of students who survive year. Calculate the probability that a student entering the school will withdraw for reasons other than academic failure before graduation. (A) Less than.35 (B) At least.35, but less than.4 (C) At least.4, but less than.45 (D) At least.45, but less than.5 (E) At least.5 93

45. Given: (i) (ii) µ Superscripts M and N identify two forces of mortality and the curtate expectations of life calculated from them. N 5 bg t = R S T (iii) e M 5 =. µ µ M 5 M 5 b g b g t +.* t t btg t > Calculate e5. N (A) 9. (B) 9.3 (C) 9.4 (D) 9.5 (E) 9.6 46. A fund is established to pay annuities to independent lives age x. Each annuitant will receive, per year continuously until death. You are given: (i) δ = 6. (ii) A x = 4. (iii) A x = 5. Calculate the amount (in millions) needed in the fund so that the probability, using the normal approximation, is.9 that the fund will be sufficient to provide the payments. (A) 9.74 (B) 9.96 (C).3 (D).64 (E). 94

47. Total hospital claims for a health plan were previously modeled by a two-parameter Pareto distribution with α = and θ = 5. The health plan begins to provide financial incentives to physicians by paying a bonus of 5% of the amount by which total hospital claims are less than 5. No bonus is paid if total claims exceed 5. Total hospital claims for the health plan are now modeled by a new Pareto distribution with α = and θ = K. The expected claims plus the expected bonus under the revised model equals expected claims under the previous model. Calculate K. (A) 5 (B) 3 (C) 35 (D) 4 (E) 45 48. A decreasing term life insurance on (8) pays (-k) at the end of the year of death if (8) dies in year k+, for k=,,,,9. You are given: (i) i=.6 (ii) For a certain mortality table with q 8 =., the single benefit premium for this insurance is 3. (iii) For this same mortality table, except that q 8 =., the single benefit premium for this insurance is P. Calculate P. (A). (B).4 (C).7 (D). (E).3 95

49. Job offers for a college graduate arrive according to a Poisson process with mean per month. A job offer is acceptable if the wages are at least 8,. Wages offered are mutually independent and follow a lognormal distribution with µ =. and σ =.. Calculate the probability that it will take a college graduate more than 3 months to receive an acceptable job offer. (A).7 (B).39 (C).45 (D).58 (E).6 5. For independent lives (5) and (6): bg= µ x, x x < Calculate e. 5: 6 (A) 3 (B) 3 (C) 3 (D) 33 (E) 34 96

5. For an industry-wide study of patients admitted to hospitals for treatment of cardiovascular illness in 998, you are given: (i) (ii) Duration In Days Number of Patients Remaining Hospitalized 4,386, 5,46,554 486,739 5 6,8 53,488 5 7,384 3 5,349 35,337 4 Discharges from the hospital are uniformly distributed between the durations shown in the table. Calculate the mean residual time remaining hospitalized, in days, for a patient who has been hospitalized for days. (A) 4.4 (B) 4.9 (C) 5.3 (D) 5.8 (E) 6.3 97

5. For an individual over 65: (i) The number of pharmacy claims is a Poisson random variable with mean 5. (ii) The amount of each pharmacy claim is uniformly distributed between 5 and 95. (iii) The amounts of the claims and the number of claims are mutually independent. Determine the probability that aggregate claims for this individual will exceed using the normal approximation. (A) Φ 33. b b b b b (B) Φ 66. g g g g g (C) Φ. 33 (D) Φ. 66 (E) Φ 3. 33 53. For a fully discrete three-year endowment insurance of, on (5), you are given: (i) i = 3. (ii) q 5 = 8. 3 (iii) q 5 = 9. (iv) V = 39, 5: 3 (v) V = 6539, 5: 3 (vi) L is the prospective loss random variable at issue, based on the benefit premium. Calculate the variance of L. (A) 77, (B) 33, (C) 357, (D) 43, (E) 454, 98

54. For a special 3-year deferred annual whole life annuity-due of on (35): (i) If death occurs during the deferral period, the single benefit premium is refunded without interest at the end of the year of death. (ii) a 65 = 99. (iii) A 35 3 =. : (iv) A 35 3 =. : 7 Calculate the single benefit premium for this special deferred annuity. (A).3 (B).4 (C).5 (D).6 (E).7 55. Given: x b g = + (i) µ x F e, x (ii) 4. p = 5. Calculate F. (A) -. (B) -.9 (C). (D).9 (E). 99

56-57 Use the following information for questions 56 and 57. An insurer has excess-of-loss reinsurance on auto insurance. You are given: (i) Total expected losses in the year are,,. (ii) In the year individual losses have a Pareto distribution with bg= Fx F I x HG x + K J, >. (iii) Reinsurance will pay the excess of each loss over 3. (iv) (v) (vi) Each year, the reinsurer is paid a ceded premium, C year, equal to % of the expected losses covered by the reinsurance. Individual losses increase 5% each year due to inflation. The frequency distribution does not change. 56. Calculate C. (A),, (B) 3,3, (C) 4,4, (D) 5,5, (E) 6,6, 57. Calculate C / C. (A).4 (B).5 (C).6 (D).7 (E).8

SOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE SOLUTIONS Copyright 5 by the Society of Actuaries Some of the questions in this study note are taken from past SOA examinations. M-9-5 PRINTED IN U.S.A.

Question # Key: E q = p p 3:34 3:34 3 3:34 p p p 3 34 3:34 p 3:34 p p p p 3 3 3 34 3 3:34 3 3:34 ( )( ) ( )( ) ( )( ) =.9.8 =.7 =.5.4 =. =.7. =.44 =.7 +..44 =.776 ( )( ) ( )( ) ( )( ) =.7.7 =.54 =..3 =.6 =.54.6 =.34 =.54 +.6.34 =.53376 q =.776.53376 3:34 =.44 Alternatively, q = q + q q = p q + p q p p 3: 34 3 34 3: 34 3 3 34 36 3: 34 3: 36 b g = (.9)(.8)(.3) + (.5)(.4)(.7) (.9)(.8)(.5)(.4) [-(.7)(.3)] =.6 +.4.44(.79) =.44 Alternatively, q = q q q q 3:34 3 3 3 34 3 34 ( 3 p3 )( 3 p34 ) ( p3 )( p34 ) (.54)(.6) (.7)(.) = = =.44 (see first solution for p 3, p 34, 3 p 3, 3 p 34 )

Question # Key: E Ax = A + : A x x.4t.6t.4.6.5t.7t = e e (.6) dt+ e e e e (.7) dt.t.t =.6 e dt+ e (.7) e dt =.6 + (.7) e e e...t.t.6.7. = e + e e.. =.3797 +.46 = 593.87 ( ) Because this is a timed exam, many candidates will know common results for constant force and constant interest without integration. ( E ) For example A x: = µ + δ x E ( µ + δ) = e A x x µ µ = µ + δ With those relationships, the solution becomes Ax = A E : x A + x x+.6 (.6+.4) (.6+.4).7 = ( e ) + e.6.4.7.5 + + = (.6)( e ) +.5833e = 593.86

Question #3 Key: A ( ) c 4 x x< 4 B = x 4 = E B = c 4 ce X 4 ( ) i ( ) 3 = ci4 ci3 3 + 4 4 = c 4 3i 7 c = =.44 8.6 Question #4 Key: C Let N = # of computers in department Let X = cost of a maintenance call Let S = aggregate cost ( X) ( X) ( ) λ ( ) ( ) λ ( ) Var = Standard Deviation = = 4, E( X ) = Var( X) + E( X) = 4, + 8 = 46, 4 E S = N E X = N 3 8 = 4N Var S = N E X = N 3 46, 4 = 39, N We want. Pr >. ( S E( S) ) ( ) ( ) S E S.E S. 4N Pr >.8 =Φ.9 39, N 39, N 373. N.8 373. N = 99.3 48 ( )

Question #5 Key: B t ( τ ) ( ) ( ) ( 3) = + + =.45 µ µ µ µ x x x x ( τ ) p = e x.45t δ t ( τ ) ( ) APV Benefits = e,, p µ dt + + t ( ) ( ) e δ τ 5, p µ dt ( ) ( 3) e δτ τ, p µ dt ( ) t x x t x x,,.645t 5,.645t 5,.645t = e dt e dt e dt,, + 5, +, = 7.5 6.6377 = 457.54 t x x Question #6 Key: B 4: 4: APV Benefits = A + E vq APV Premiums = π a + E vq k k= 4 4+ k 4 4+ k + 4: k 4 4+ k = π + 4: k 4 4+ k k= π = A / a 4: 4: A E vq a E vq k k= Benefit premiums Equivalence principle ( )( ) ( )( ) 6.3.744 369.3 = 4.866.744.454 = 5. While this solution above recognized that π = P and was structured to take advantage of that, it wasn t necessary, nor would it save much time. Instead, you could do: 4: APV Benefits = A = 6.3 4

APV Premiums = π a + E E vq 4: 4: 4 k 6 6+ k k= = π a + E A 4 6 ( )( ) ( )( ) = π 4.866.744.454 +.744 369.3 =.76π +.9.76π +.9 = 6.3 6.3.9 π = = 5..76 Question #7 Key: C ( ) ( ) ln.6 A7 = δ A7 =.53 =.547 i.6 A7.547 a 7 = = = 8.5736 d.6/.6.97 a 69 = + vp69a 7 = + ( 8.5736) = 8.8457.6 ( ) a 69 = α( ) a 69 β( ) = (.)( 8.8457).5739 = 8.59 ( m) ( m ) Note that the approximation a x a x works well (is closest to the exact answer, m only off by less than.). Since m =, this estimate becomes 8.8457 = 8.5957 4 Question #8 Key: C The following steps would do in this multiple-choice context:. From the answer choices, this is a recursion for an insurance or pure endowment.. Only C and E would satisfy u(7) =.. + i u k = u k 3. It is not E. The recursion for a pure endowment is simpler: ( ) ( ) 4. Thus, it must be C. More rigorously, transform the recursion to its backward equivalent, u k in terms of u k : ( ) ( ) pk

q + i p p k ( ) = + u( k ) u k k k k ( ) = k + ( + ) ( ) ( ) = + ( ) p u k q i u k u k vq vp u k k k This is the form of (a), (b) and (c) on page 9 of Bowers with x = k. Thus, the recursion could be: or or A = vq + vp A + x x x x xy : = x + x x x : y x A vq vp A + A = vq + vp A + xy : x x x x : y x Condition (iii) forces it to be answer choice C ( ) u k = A fails at x = 69 since it is not true that x ( ) A vq vp ( )() 69 = 69 + 69 u k = A x: y x fails at x = 69 since it is not true that 69: A = vq + vp ( ) : ( )() 69 69 u k = A is OK at x= 69 since 69: xy x A = vq + vp ( )() 69 69 Note: While writing recursion in backward form gave us something exactly like page 9 of Bowers, in its original forward form it is comparable to problem 8.7 on page 5. Reasoning from that formula, with π h = and bh+ =, should also lead to the correct answer.

Question #9 Key: A You arrive first if both (A) the first train to arrive is a local and (B) no express arrives in the minutes after the local arrives. PA= ( ).75 Expresses arrive at Poisson rate of (.5)( ) = 5 per hour, hence per minutes. e f ( ) = =.368! A and B are independent, so P A and B =.75.368 =.76 ( ) ( )( ) Question # Key: E d =.5 v =.95 At issue ( ) ( ) ( ) ( ) 49 k+ 4 k 4 5 5 k= A = v q =. v +... + v =.v v / d =.3576 and a = A / d =.3576 /.5 =.9848 4 4 A 35.76 4 so P4 = = = 7.3 a 4.9848 Revised Revised ( ( ) ) 5 4 5 ( )( ) E L K 4 = A P a = 549.8 7.3 9.64 = 35.6 where Revised ( ) ( ) ( ) ( ) 4 Revised k+ Revised 5 k 5 k= 5 5 Revised a 5 = A5 d = = A = v q =.4 v +... + v =.4v v / d =.5498 and /.5498 /.5 9.64

Question # Key: E Let NS denote non-smokers and S denote smokers. The shortest solution is based on the conditional variance formula Var ( X ) = E ( Var ( X Y )) + Var( E ( X Y )) Let Y = if smoker; Y = if non-smoker S S Ax E( a Y = T ) = ax = δ.444 = = 5.56..86 Similarly E( a Y = ) = = 7.4 T. E E a Y = E E a Prob Y= + E E a Prob Y= ( ( )) ( ) ( ) ( ) ( ( )) ( ) T T T ( 7.4)(.7) ( 5.56)(.3) = + = 6.67 E E a Y = + = 44.96 ( ( )) ( 7.4 )(.7 ) ( 5.56 )(.3 T ) ( ( T )) ( ) Var E a Y = 44.96 6.67 =.47 ( Var ) ( 8.53 )(.7 ) ( 8.88 )(.3 T ) E a Y = + = 8.6 Var a = 8.6 +.47 = 9.7 ( ) T Alternatively, here is a solution based on ( ) ( ) Var( Y) = E Y E Y, a formula for the variance of any random variable. This can be transformed into E( Y ) Var ( Y) E( Y) E( ( a ) NS ) = Var ( a NS ) + E( a NS ) T T T ( ) ( ) = + which we will use in its conditional form Var a E T = a E a T T E a E a S Prob S E a NS Prob NS T = + T T [ ] [ ]

S x =.3a +.7a NS x S NS ( Ax ) ( Ax ).3.7 = +...3(.444) +.7(.86) = = +. =.67 + 5. = 6.67 (.3)( 5.56) (.7)( 7.4) ( ) = S Prob[ S] + NS Prob[ NS] E a E a E a T T T ( ( a ) ( E a T T ) ( ( a T ) E( a T ) ) =.3 Var S + S +.7 Var NS + NS ( ) ( ) =.3 8.88 + 5.56 +.7 8.53+ 7.4.99 + 4.638 = 53.557 ( ) Var a T = 53.557 6.67 = 9. v Alternatively, here is a solution based on a = T δ T v Var ( a ) = Var T δ δ T v = Var since Var ( X + constant) = Var ( X) δ T Var ( v ) = since Var constant = constant Var δ Ax ( Ax) = which is Bowers formula 5..9 δ ( X ) ( X ) A a A T x This could be transformed into x δ Var ( ) NS S x and x A A. T = +, which we will use to get

= T Ax E v A x Var T ( ) E v ( ) T = E v NS Prob NS + S Prob S ( a ) ( A x ) ( ) NS = δ Var NS + Prob NS T ( a ) ( A x ) ( ) S + δ Var S + Prob S T ( )( ) =. 8.53.86 +.7 ( )( ) +. 8.88.444 +.3 = + (.6683)(.7) (.853)(.3) =.38 T = E v T ( ) E v ( ) T = E vns Prob NS S + Prob S = + (.86)(.7) (.444)(.3) =.3334 ( a ) T = A x δ ( A ) x.38.3334 = = 9.. Question # Key: A To be a density function, the integral of f must be (i.e., everyone dies eventually). The solution is written for the general case, with upper limit. Given the distribution of f t, we could have used upper limit here. () Preliminary calculations from the Illustrative Life Table: l l l l 5 4 =.895 =.933

5 () () () = f t dt = k f t dt +. f t dt T 5 5 (). () ( ) = k f t dt+ f t dt 5 ( 5). ( ) ( 5) ( 5 p ).(.5) (.895).6 = kf + F F = k + = k +.6 k = = 3.83.895 For x 5, F ( x) = 3.83 f ( t) dt = 3.83F ( x) F F T T T x l 4 = 3.83 = 3.83.933 =.6 l 4 ( ) ( ) l 5 = 3.83 = 3.83.895 =.4 l FT ( 5).4 p4 = = =.83 F 4.6 5 ( ) ( ) T ( ) Question #3 Key: D Let NS denote non-smokers, S denote smokers. ( T < t) = ( T < t ) ( ) + ( T < t ) ( ) Prob Prob NS Prob NS P rob S P rob S.t.t ( e ) ( e ) =.7 +.3.t.t =.7e.3e.. ().3 t t = +.7 S t e e Want ˆt such that.75 = S( tˆ) or.5 = S( tˆ) ( ) tˆ.tˆ.tˆ.tˆ.5 =.3e +.7e =.3 e +.7e ˆ Substitute: let x= e.t.3x +.7x.5 =

( )( ) ( ).7 ±.49 +.3.5 4 This is quadratic, so x =.3 e x =.347.tˆ ˆ =.347 so t =.56 Question #4 Key: D The modified severity, X*, represents the conditional payment amount given that a payment occurs. Given that a payment is required (X > d), the payment must be uniformly distributed between and c ( b d). The modified frequency, N*, represents the number of losses that result in a payment. b d The deductible eliminates payments for losses below d, so only Fx ( d) = of b losses will require payments. Therefore, the Poisson parameter for the modified b d frequency distribution is λ. (Reimbursing c% after the deductible affects only the b payment amount and not the frequency of payments). Question #5 Key: C Let N = number of sales on that day S = aggregate prospective loss at issue on those sales K = curtate future lifetime ( ) [ ] [ ] K + [ ] N Poisson.*5 E N = Var N = L=,v 5a E L =,A 5a K + 65 65 5 K + 5 5 L=, + v Var[ L] =, + A65 ( A65) d d d = + + + E S E N E L S L L... L N [ ] = [ ] [ ] [ ] = [ ] [ ] + ( [ ]) Var[ ] Var S Var L E N E L N

Pr( S < ) = Pr Z < [ ] [ ] E S Var S Substituting d =.6/(+.6), A 65 =.363, A65 =.4398 and a 65 = 9.8969 yields E [ L] [ L] [ ] [ S] = 55.45 Var = 5,, E S = 554.5 Var = 54,5, Std Dev (S) =,46 S + 554.5 554.5 Pr( S < ) = Pr <,46,46 = Pr Z <.443 =.67 ( ) With the answer choices, it was sufficient to recognize that:.6554 = φ.4 < φ.443 < φ.5 =.695 ( ) ( ) ( ) By interpolation, φ(.443) (.43) φ(.5) + (.57) φ(.4) = (.43)(.695) + (.57)(.6554) =.679

Question #6 Key: A A 6.3 4 P4 = = =.89 a 4 4.866 = a =.454 = V + 5 P ( + i) 5q 6 V 4 47.78 a 4 4.866 V = ( ) 4 6 P 6 ( 47.78 + (5)(.89)).6 5(.376) = = 55.376 [Note: For this insurance, V = V 4 because retrospectively, this is identical to whole life] Though it would have taken much longer, you can do this as a prospective reserve. The prospective solution is included for educational purposes, not to suggest it would be suitable under exam time constraints. P 4 =.89 as above A 4 E A P 5P E a E E a where 4 + 4 = 4 + 4 4 + π 6:5 6:5 4 5 6 65 A = A 6:5 6 5 E6 A65 =.6674 a = a 4: 4 E4 a 6 =.76 a = a E a = 4.347 6:5 6 5 6 65 ( ) + ( )( )( ) = = (.89)(.76) + ( 5)(.89)(.744)( 4.347) + π (.744)(.68756)( 9.8969).63 4.744.6674 6.3 + 73.8 8.8 64.79 π =.86544 =.3 Having struggled to solve for π, you could calculate V prospectively then (as above) calculate V recursively. V = 4A + A 6:5 6 5P4 a π 6:5 5 E6 a 65 ( 4)(.6674) 369.3 ( 5)(.89)( 4.347) (.3)(.68756)( 9.8969) = + = 47.86 (minor rounding difference from V ) 4

Or we can continue to V prospectively V = 5A + 6:4 4 E6 A65 5P4 a π 6:4 4 E6 a 65 65 4 where E v ( ) 4 6 l 7,533,964 = =.799.73898 l = 8,75, 43 6 6:4 A = A6 4 E6 A65 =.3879.73898.4398 =.5779 a = a 6:4 6 4 E6 a 65 =.94.73898 9.8969 = 3.595 ( )( ) ( )( )( ) ( 5)(.89)( 3.595).3(.73898)( 9.8969) V = 5.5779 +.73898.4398 = 55 Finally. A moral victory. Under exam conditions since prospective benefit reserves must equal retrospective benefit reserves, calculate whichever is simpler. Question #7 Key: C ( ) = ( ) Var Z A4 A4 A4 A4 =.8 = A4 ( vq4 + vp4a4 ) = A4 (.8/.5 + (.997 /.5) A4) A =.65 4 ( ) ( ) A4 A4 =.433 = A4 v q4 + v p4 A4 = A4 + A4 A4 =.793 ( Z ) = Var.793.65 =.544 (.8/.5.997/.5 )

Question #8 Key: D This solution looks imposing because there is no standard notation. Try to focus on the big picture ideas rather than starting with the details of the formulas. Big picture ideas:. We can express the present values of the perpetuity recursively.. Because the interest rates follow a Markov process, the present value (at time t ) of the future payments at time t depends only on the state you are in at time t, not how you got there. 3. Because the interest rates follow a Markov process, the present value of the future payments at times t and t are equal if you are in the same state at times t and t. Method : Attack without considering the special characteristics of this transition matrix. Let k s = state you are in at time k ( thus s =, or ) k Let Y k = present value, at time k, of the future payments. Y k is a random variable because its value depends on the pattern of discount factors, which are random. The expected value of Y k is not constant; it depends on what state we are in at time k. Recursively we can write ( ) Yk = v + Y k +, where it would be better to have notation that indicates the v s are not constant, but are realizations of a random variable, where the random variable itself has different distributions depending on what state we re in. However, that would make the notation so complex as to mask the simplicity of the relationship. Every time we are in state we have E Yk sk = =.95 + E Yk+ sk = ( ) ( {( E Yk+ sk+ ) ) ( k+ sk ]) + ( E Yk+ sk+ = ) Prob( sk+ = sk = ] ) = ( E Yk+ sk+ = ) Prob sk+ = sk = ] =.95 + = Pr ob s = = ( E Yk+ sk+ ) =.95 + = ( )}

That last step follows because from the transition matrix if we are in state, we always move to state one period later. Similarly, every time we are in state we have E Yk sk = =.93 ( + E Yk+ sk = ) =.93 ( + E Yk+ sk+ = ) That last step follows because from the transition matrix if we are in state, we always move to state one period later. Finally, every time we are in state we have E Yk sk = =.94 ( + E Yk+ sk = ) =.94 + E Yk+ sk+ = Pr sk+ = sk = + E Yk+ sk+ = Pr sk+ = sk = } ( { ) ( { E Yk+ sk+ E Yk+ sk+ }) =.94 + =.9 + =.. Those last two steps follow from the fact that from state we always go to either state (with probability.9) or state (with probability.). Now let s write those last three paragraphs using this shorter notation: xn = E Yk sk = n. We can do this because (big picture idea #3), the conditional expected value is only a function of the state we are in, not when we are in it or how we got there. ( x ) ( ) ( x ) x =.95 + x =.94 +.9x +.x x =.93 + That s three equations in three unknowns. Solve (by substituting the first and third into the second) to get x = 6.8. That s the answer to the question, the expected present value of the future payments given in state. The solution above is almost exactly what we would have to do with any 3 3 transition matrix. As we worked through, we put only the non-zero entries into our formulas. But if for example the top row of the transition matrix had been (.4.5. ), then the first of our three equations would have become x =.95( +.4x +.5x+.x), similar in structure to our actual equation for x. We would still have ended up with three linear equations in three unknowns, just more tedious ones to solve. Method : Recognize the patterns of changes for this particular transition matrix.

This particular transition matrix has a recurring pattern that leads to a much quicker solution. We are starting in state and are guaranteed to be back in state two steps later, with the same prospective value then as we have now. Thus, EY = EY first move is to Pr first move is to + EY first move is to Pr first move is to [ ] [ ] [ ] ( ( EY [ ]) ( ( EY [ ]) =.94 +.95 +.9 +.94 +.93 +. (Note that the equation above is exactly what you get when you substitute x and x into the formula for x in Method.) [ ] (.837.874) [ ] EY [ ] =.6497 +.837EY +.84 +.874.6497 +.84 EY = = 6.8 Question #9 Key: E The number of problems solved in minutes is Poisson with mean. If she solves exactly one, there is /3 probability that it is #3. If she solves exactly two, there is a /3 probability that she solved #3. If she solves #3 or more, she got #3. f() =.353 f() =.77 f() =.77 P = (.77) + (.77) + (.353.77.77) =.594 3 3

Question # Key: D ( τ ) ( = ) () t + x x x µ µ µ ( ) () t ( τ ) ( =.µ () ) x t + µ x () t ( ) () t =.8 µ µ x ( τ ) () t x k. ' () ' ().kt dt q = p = e = e 3 =.4 x x ( ) ( ) k 3 ln.4 /. =.4 k =.63 ( ) ( ) ( ) ( ) ( q = p τ µ dt =.8 p τ µ τ) () t dt x t x x t x x ( x ) ( τ) ( τ) =.8 q =.8 p x ( τ ) µ () p = e x = e 8k 3 x t dt kt dt = e ( 8) (.63) = e 3 =.9538 ( ) q x =.8(.9538) =.644 Question # Key: A k 3 k f(k) f ( k) ( k 3) f ( k) ( k 3). (.9)(.) =.8.8.8 (.7)(.3) =.6.43.864 3+ 3 -.-.8-.6 =.54.5 4.536.4 5.58

( ) ( K ) E K 3 =.4 E ( ) 3 = 5.58 Var( K 3) = 5.58.4 =.7 e if the life is age x. x:3 Problem 3.7 in Bowers, pages 86 and 87, gives an alternative formula for the variance, Note that E[ K 3] is the temporary curtate life expectancy, basing the calculation on k p x rather than k q x. Question # Key: E f( x) =., x 8 =..5( x 8) =.3.5 x, 8 < x 8 ( ) Ex ( ) =.xdx+.3x.5x dx 8 x 8 x 3 x 8 8..3.5 = + 3 = 3 +.33 = 5.66667.x ( ) ( ) EX ( ) = EX ( ) xf xdx ( f xdx) + ( x ) = 5.6667. = 5.6667 (.8) = 3.6667 3.6667 Loss Elimination Ratio = =.3553 5.6667

Question #3 Key: D Let q 64 for Michel equal the standard q 64 plus c. We need to solve for c. Recursion formula for a standard insurance: ( )(.3) ( ) V = V + P q V 45 9 45 45 64 45 Recursion formula for Michel s insurance ( )( ) ( ) V = V + P +..3 q + c ( V ) 45 9 45 45 64 45 The values of 9 V 45 and V 45 are the same in the two equations because we are told Michel s benefit reserves are the same as for a standard insurance. Subtract the second equation from the first to get: ( ) ( ) ( V ) =.3 (.) + c( V45 ) (.3). c = 45.3 =.47 =.8 Question #4 Key: B K is the curtate future lifetime for one insured. L is the loss random variable for one insurance. L is the aggregate loss random variables for the individual insurances. AGG σ AGG is the standard deviation of AGG M is the number of policies. L. K+ π K+ L= v π a = v π K + + d d ( Ax ) E[ L] = ( Ax πa x) = Ax π d.7595 =.495.5 =.868.5664

( ) π.5 Var[ L] = + ( Ax Ax) = +.9476 (.495) =.6834 d.5664 [ AGG ] = [ ] = [ ] [ ] E L ME L.868M Var L = M Var L = M(.6834) σ =.6833 M AGG AGG [ ] ( ) L E L E L Pr[ LAGG > ] = > σ AGG σ.868m Pr N(,) > M (.6833).868 M.645 =.6833 M = 6.97 AGG AGG AGG minimum number needed = 7 AGG Question #5 Key: D Annuity benefit: Death benefit: New benefit: Z K + v =, for K =,,,... d K Z = Bv + for K =,,,... K + v K + Z = Z+ Z =, + Bv d,, K = + B v d d + K + ( ), Var( Z) = B Var v d, Var ( Z) = if B= = 5,..8 In the first formula for Var ( Z ), we used the formula, valid for any constants a and b and random variable X, Var ( a+ bx) = b Var( X)

Question #6 Key: B First restate the table to be CAC s cost, after the % payment by the auto owner: Towing Cost, x p(x) 7 5% 9 4% 44 % Then E( X ) =.5*7 +.4*9 +.*44 = 86.4 ( ) E X =.5*7 +.4*9 +.*44 = 795.6 Var ( X ) = 795.6 86.4 = 44.64 Because Poisson, E( N) = ( N) = E( S) = E( X) E( N) = 86.4* = 86,4 Var Var( S) = E( N)Var( X) + E( X) Var( N) = * 44.64 + 86.4 * = 7,95,6 S E( S) 9, 86, 4 Pr( S > 9,) + Pr > = Pr( Z >.8) = Φ(.8) =. Var( S) 7,95,6 Since the frequency is Poisson, you could also have used Var S = λ E X = 795.6 = 7,95,6 ( ) ( ) ( )( ) That way, you would not need to have calculated Var ( X ). Question #7 Key: C ( d) E( X) d / θ ( e ) E X θ LER = = = e θ d / θ d / θ Last year.7 = e d = θ log.3 Next year: d = θ log( LER ) new Hence θ ( ) new 4 log LER new = dnew = θ log.3 3 log LER =.653 ( new ).653 ( ) e LER = =. new LER =.8 new

Question #8 Key: E E( X) = e( d) S( d) + E( X d) [Klugman Study Note, formula 3.] 4 4 ( ) 6 = e p + E T 4 e4 6 = ( )(.6) + 4 (.5)(4 ) 4 e e4 =.6 + 3 ( 6 3) = = 5.6 The first equation, in the notation of Bowers, is e = e 4 4 p + e :4. The corresponding formula, with i >, is a very commonly used one: a = a + E a + x xn : n x x n Question #9 Key: B d =.5 v=.95 Step Determine p from Kevin s work: x vpx vqx v px( px+ qx+ ) ( ) px ( )( px) ( ) px( ) p ( p ) p 68 + 35 = + + 68 + 35.95 =.95 +.95 68 + 33.5 x = 95 x + 9.5 x p = 34/38 =.9 x Step Calculate P x:, as Kira did: 68 + 35.95.9 = P.95.9 x: + [ 99.5 + 68] P = = 489.8 x:.855 ( )( ) ( )( ) The first line of Kira s solution is that the actuarial present value of Kevin s benefit premiums is equal to the actuarial present value of Kira s, since each must equal the actuarial present value of benefits. The actuarial present value of benefits would also have been easy to calculate as.95. +.95.9 = 97.5 ( )( )( ) ( )( )( )

Question #3 Key: E Because no premiums are paid after year for (x), V x = A x+ ( V + )( + i) b q Rearranging 8.3. from Bowers, we get h+ V = p ( ) ( ) h π h h+ x+ h 3,535 +,78.5,. V = = 35,635.64.989 ( 35,635.64 + ) (.5),. V = = 36,657.3 = A x +.988 x+ h Question #3 Key: B For De Moivre s law where s( x) ω x t ex = and t px = ω x 5 45 e45 = = 3 5 65 e65 = = x = ω : 4 4 6 t 4 t e 45:65 = t p 45:65 dt = dt 6 4 F HG + = + 6 4 6 4 6 4 t t t 3 3 I K J 4 = 556. e45: 65 = e45+ e65 e45: 65 = 3 + 556. = 34 In the integral for e 45:65, the upper limit is 4 since 65 (and thus the joint status also) can survive a maximum of 4 years.

Question #3 Answer: E b g b g b g µ 4 = s' 4 / s 4 d 4 = e / 4 e / 4 e / = 4 e / 4 e = e 4 = 553. i Question # 33 Answer: A q bg i x = q bg bg τ x L NM τ µ = q x µ ln px ln p bg i bg τ bg i bg τ x O P Q q P = bg τ x L NM bg bg bg bg τ 3 x x x x µ = µ + µ + µ = bg τ x bg µ τ q = e = e =.7769 bg q x 5. bg i µ ln e µ bg ln e - τ 5. b. 7769gµ bg b5. gb. 7769g µ bg τ 5. = = =. 59 O QP

Question # 34 Answer: D 6 3 A = v p q + 6 6 B B B + pay at end live then die of year 3 years in year 3 4 + v p q + 3 6 6 3 pay at end live then die of year 4 3 years in year 4 = 9.. 3. + 9.. 3. 5. 3 4 3. 3. b g b gb gb g b g b gb gb gb g = 9. Question # 35 Answer: B ax = a + E a x x x+ a 5 x: 5 :5 5 5 bg 7. 5 e = = 4. 9 7. bg 7. 5 Ex = e =. 75, where 7. = µ + δ for t < 5 a x+ 5 = =. 5, where 8. = µ + δ for t 5 8. b gb g a x = 4. 9 +. 75. 5 = 33.

Question # 36 Answer: E The distribution of claims (a gamma mixture of Poissons) is negative binomial. Ebg N = E decn Λhi= E bg Λ = 3 Varbg N = E dvarcn Λhi+ Var decn Λ hi = E bλg + Var b Λg = 6 rβ = 3 rβb+ βg = 6 b + βg = 6/ 3 = ; β = rβ = 3 r = 3 r b g. p = + β = 5 rβ p = = 875. r+ + β g b g = p + Pr ob at most p =. 35 Question # 37 Answer: A ES b g = EN b g EX b g =, =, Varbg S = Ebg N Varbg X + Ebg X Varbg N = 7 + 75 = 99, 689, 75 b g = 3, 6 bs < g = cz < b g = Pr bz <. 7g =. 4 Std Dev S Pr, Pr,, / 3, 6h where Z has standard normal distribution

Question # 38 Answer: C This is just the Gambler s Ruin problem, in units of 5, calories. Each day, up one with p = 45. ; down with q = 55. Will Allosaur ever be up before being down? e b g j 55. / 45. P = =. 598 e b55. / 45. g 3j Or, by general principles instead of applying a memorized formula: Let P = probability of ever reaching 3 (5, calories) if at (5, calories). Let P = probability of ever reaching 3 (5, calories) if at (, calories). From either, we go up with p = 45., down with q = 55. P(reaching 3) = P(up) P(reaching 3 after up) + P(down) P(reaching 3 after down) P = 45. + 55. P P =. 45 P + 55. =. 45 P P =. 45+ 55. P =. 45+ 55.. 45 P =. 45+. 475P P =. 45 /. 475 =. 598 b g Here is another approach, feasible since the number of states is small. Let states,,,3 correspond to ; 5,;,; ever reached 5, calories. For purposes of this problem, state 3 is absorbing, since once the allosaur reaches 5, we don t care what happens thereafter. The transition matrix is L N M 55. 45..55.45 Starting with the allosaur in state ; [ ] at inception [.55.45] after [.35.475.45] after [.35.36.564] after 3 [.3774.6.564] after 4 [.3774.337.5889] after 5 [.3959.5.5889] after 6 By this step, if not before, Prob(state 3) must be converging to.6. It s already closer to.6 than.57, and its maximum is.5889 +.5 O Q P

Question # 39 Answer: D b gb gb g = Per minutes, find coins worth exactly at Poisson rate 5.. b g Per minutes, f =. 3679 F =. 3679 f g =. 3679 Fbg =. 7358 fbg = 839. Fbg =. 997 f 3 =. 63 F 3 =. 98 b g b g b g Let Period = first minutes; period = next. Method, succeed with 3 or more in period ; or exactly, then one or more in period P = c Fbg h + fbg c Fbg h = b. 997g+ b839. gb. 3679g = 965. Method : fail in period if < ; Pr ob = F =. 7358 fail in period if exactly in period, then ; Pr ob = fbg fbg = b839. gb. 3679g =. 677 Succeed if fail neither period; Pr ob =. 7358. 677 = 965. b g (Method is attacking the problem as a stochastic process model; method attacks it as a ruin model.) Question # 4 Answer: D Use Mod to designate values unique to this insured. b g b g b g b g 6 6 6 b g a = A / d =. 36933 /. 6 / 6. = 48. 6 6 P = A / a =. 36933 / 48. = 335. d i b gb g Mod Mod Mod A6 = v q6 + p6 A6 = 376. 864 383 444 6. +.. =.

Mod Mod d 6 i b g a = A / d =. 444 /. 6 / 6. = 9. 8684 Mod Mod Mod 6 6 6 E L = A P a d =. 444. 335 9. 8684 = 4. 7 i b g Question # 4 Answer: D The prospective reserve at age 6 per of insurance is A 6, since there will be no future premiums. Equating that to the retrospective reserve per of coverage, we have: A P s 4: Mod 6 = 4 + P5 s k 5: 4 E A 6 5 A a a A 4 4: Mod 5: : = + P5 a E E E E 4 4 5 5 4 4 63. 77. 757. 6.. 3693 = + P Mod 5 4. 866 53667. 58. 58.. 744 b gb g. 3693 =. 358 + 4. 896. 887 P Mod 5 = 9. 4 P Mod 5 Alternatively, you could equate the retrospective and prospective reserves at age 5. Your equation would be: A a A Mod 4 4: : A5 P5 a = 5: a E E 4 4 4 4 where A 4 : = A4 E4 A5 = 63.. 53667. 495 =. 766 b gb g

d ib g 63. 77.. 766. 495 P Mod 5 7. 57 = 4. 866 53667. 53667. bgb4437 P Mod. g 5 = = 9. 7 757. Alternatively, you could set the actuarial present value of benefits at age 4 to the actuarial present value of benefit premiums. The change at age 5 did not change the benefits, only the pattern of paying for them. Mod 4 4 4: 5 4 5: A = P a + P E a 63. 63. 4. 866 F = H G I K J + bgb. 7748g b7. 7g dp5 ib53667. gb7. 57g Mod P Mod 5 = = 9. 7 4. 66 Question # 4 Answer: A bg bg bg τ x x x d = q l = 4 d x bg b g bg =. 45 4 = 8 qx = l bg dx bg x dbg. τ x bg =. 488 =. 5 p x 4 = = 488 8 Note: The UDD assumption was not critical except to have all deaths during the year so that - 8 lives are subject to decrement.

Question #43 Answer: D Use age subscripts for years completed in program. E.g., p applies to a person newly hired ( age ). Let decrement = fail, = resign, 3 = other. Then q bg =, q bg =, q bg = 4 5 3 q bg = q = 5, bg 3, q bg= 8 q bg=, q bg=, q bg= This gives p = / 4 / 5 / = 54. p = / 5 / 3 / 9 = 474. p = / 3 / 8 / 4 =. 438 bg b g So,. 54 8, and = 8. 474 = 5. 3 3 3 9 4 bg τ b gb gb g bg τ b gb gb g bg τ b gb gb g bg τ bg τ τ = = b g = τ τ bg bg bg bg q = log p / log p q bg ch b g = log 3 / log. 438. 438 q = b. 45 /. 86gb56. g =. 76 bg bg bg = τ d l q = 5.. 76 = 4 b gb g Question #44 Answer: C Let: N = number X = profit S = aggregate profit subscripts G = good, B = bad, AB = accepted bad c hb g c hc 3hb6g (If you have trouble accepting this, think instead of a heads-tails λ G = 3 6 = 4 λ AB = = rule, that the application is accepted if the applicant s government-issued identification number, e.g. U.S. Social Security Number, is odd. It is not the same as saying he automatically alternates accepting and rejecting.) b Gg b Gg b Gg b Gg b Gg Var S = E N Var X + Var N E X

= b4gb, g + b4gd3 i = 4,, b ABg b ABg b ABg b ABg b ABg Var S = E N Var X + Var N E X b gb g b gb g = 9, + =,, S G and S AB are independent, so bg b g b g Var S = Var SG + Var SAB = 4,, +,, = 5,, If you don t treat it as three streams ( goods, accepted bads, rejected bads ), you can compute the mean and variance of the profit per bad received. λ B = c 3hb6g = If all bads were accepted, we would have E X = Var X + E X d Bi b Bg b Bg b g = 9, + =, Since the probability a bad will be accepted is only 5%, E XB = Prob accepted E XB accepted + Prob not accepted E XB not accepted = 5. + 5. = 5 b g b g c h b g c h b gb g b gb g d i= b5. gb, g + b5. gbg = 5, E X B Likewise, Now Var S = E N Var X + Var N E X b Bg b Bg b Bg b Bg b Bg b gb g b gd i = 47, 5 + 5 =,, S G and S B are independent, so Varbg S = VarbSGg+ VarbSBg= 4,, +,, = 5,,

Question # 45 Answer: C Let N = number of prescriptions then S bg E N E S 8 + = 4= F j j= c bgh n f n 3 = N 4 Nb g FNbng FNbng..6.8.4 b g = 4 E bn g + = 4 c Fbg jh j= L = O 4 NM c Fb jgh c Fb jgh j= j= QP = 4b4 44. g = 4. 56 =. 4 b g = 4 EbN 3g = 4 + c Fbjgh j= 3 L = O 4 NM c Fb jgh c Fb jgh j= j= QP = 4b4 95. g = 4. 48 = 89. E S +..36.488.594 Since no values of S between 8 and are possible, b g E S = + Alternatively, E S 8 + 8 E S.8.64.5.496 b g b g+ b g b g +. b g+ b g Nb g Nb g Nbg Nb g j= E S = 4 j f j + f + 6 f + f = 96 (The correction terms are needed because b4 j g would be negative for j =,, ; we need to add back the amount those terms would be negative) Nbg Nbg b gb g b gb g b gb g j= j= = 4 j f j f j +. + 6. 6 + 8. b g.. = 4 E N + + 9 6 + 56 = 6 67. 84 = 9. 6

Question #46 Answer: B 3: 4 3 4 d 3 id 4 ib g b E3gb E4gb ig E = p p v = p v p v + i = + = b. 54733gb. 53667gb7985. g = 564. The above is only one of many possible ways to evaluate p 3 p 4 v, all of which should give.564 a = a E a 3: 4: 3: 4 3: 4 3+ : 4+ b g b gb g = a 3: 4. 564 a 4: 5 = b3. 68g b. 564gb4784. g = 7. 687 Question #47 Answer: A Equivalence Principle, where π is annual benefit premium, gives d b g A + IA π = a x π 35 35 i π = d A35. 4898 = a IA ( 9943. 6676. ) b g 35 35 i 48. 98 = 5838. = 7366. We obtained a 35 from a 35 A35. 4898 = = = 9943. d. 4769

Question #48 Answer: C Time until arrival = waiting time plus travel time. Waiting time is exponentially distributed with mean λ. The time you may already have been waiting is irrelevant: exponential is memoryless. You: E (wait) = hour = 3 minutes E (travel) = b. 5gb6g + b. 75gb8g = 5 minutes E (total) = 8 minutes Co-worker: E (wait) = 5 hour = minutes E (travel) = 6 minutes E (total) = 8 minutes Question #49 Answer: C µ xy = µ x + µ y =4. µ Ax = Ay = + = 7. = 5833. µ δ 7. + 5. µ xy Axy = axy + = 4. 4. = = 7368 = µ δ 4+ 5 9 µ + δ =. and... 4. + 5. xy Axy A + A A P = = a a xy x y xy xy b g 5833.. 7368 = =. 87 563. xy = 563.

Question #5 Answer: E b gb g b g b gb i g b g 545 + b + ig b. gb. g. = V + P + i q V = V 4. 49 +. +. 545. = 545. =. 5. b gb g b g b gb g q 4b g V + P + i q V = V 4 545. +... 65 =. 65. 665. 65 q 4 =. 395 =. 8 Question #5 Answer: E P = A / a 6 6 6 b 6 6 6g b 6 6g b 6 6 6g b 6 6g = vq + p A / + p va = q + p A / 6. + p a c b gb gh c b gb gh = 5+. 985 38. 79 / 6. +. 985. 94 = 33.

Question #5 Answer: E Method : In each round, N = result of first roll, to see how many dice you will roll X = result of for one of the N dice you roll S = sum of X for the N dice b g = EbNg =35. b g = VarbNg =. 967 b g = b g* b g =. 5 b g = b g b g + b g b g b35. gb. 967g b. 967gb35. g E X Var X ES EN EX Var S E N Var X Var N E X = + = 45. 938 Let S = the sum of the winnings after rounds ES b g= *. 5 =, 5 Stddev S = sqrt * 45. 938 = 4. 33 b g b g After rounds, you have your initial 5,, less payments of,5, plus winnings of S. Since actual possible outcomes are discrete, the solution tests for continuous outcomes greater than 5-.5. In this problem, that continuity correction has negligible impact. b Pr 5 5 + S > 4999. 5 = c b g b g = Pr S 5 / 4. 33 > 4999. 5 5 5 / 4. 33 = b g = Φ 7. =. Method g Realize that you are going to determine N times and roll the sum of those N s dice, adding the numbers showing. Let N = sum of those N s h

b g b g b gb g E E N 35 35 Varb = Ng =. = = Var( N ) = 96. 7 ES = EN EX = 35 35. =. 5 b g b g b g b gb g b g= b g b g+ b g b g = b gb. g+ b. gb. g =. Var S E N Var X Var N E X 35 967 96 7 35 45 938 StddevbS g= 4. 33 Now that you have the mean and standard deviation of S (same values as method ), use the normal approximation as shown with method. Question #53 Answer: B p k F bi = a+ k K J p k HG b g F bi = + HG K J F bi a HG K J = 5. = a+ b 5. a+ b= 875. 5.. 5 875. b = 5. a = 5. F I HG K J = p 3 5 5. =. + 3 875. 5.

Question #54 Answer: B Transform these scenarios into a four-state Markov chain, where the final disposition of rates in any scenario is that they decrease, rather than if rates increase, as what is given. P from year t 3 from year t Probability that year t will State to year t to year t decrease from year t - Decrease Decrease.8 Increase Decrease.6 Decrease Increase.75 3 Increase Increase.9 Transition matrix is + P = 8. * 8. +. * 75. = 79. L N M 8.... 6.. 4... 75.. 5.. 9... For this problem, you don t need the full transition matrix. There are two cases to consider. Case : decrease in 3, then decrease in 4; Case : increase in 3, then decrease in 4. For Case : decrease in 3 (following decreases) is.8; decrease in 4 (following decreases is.8. Prob(both) =.8.8 =.64 For Case : increase in 3 (following decreases) is.; decrease in 4 (following a decrease, then increase) is.75. Prob(both) =..75 =.5 Combined probability of Case and Case is.64 +.5 =.79 O Q P

Question #55 Answer: B lx = ω x = 5 x P45 = l45 / l45 = 6 t / 6 t + t Let K be the curtate future lifetime of (45). Then the sum of the payments is if K 9 and is K 9 if K. a 45 = b 6 = K= F 6 KI HG 6 K J 4 + 39+... + 4 4 = = 366. 6 6 g b gb g b g Hence, ProbcK 9 > 3. 66 h = ProbcK > 3. 66 h b g b g = Prob K 33 since K is an integer = Prob T 33 l78 = 33p45 = = l =. 45 45 7 6

Question #56 Answer: C µ A x = = 5. µ = 4. µ + δ µ A x = µ + δ = diai = 4. x z s x z sex Axds d =. s e z = F e b4. ghg A ds ib4. g ds. s. I 4. = = 4 KJ. Alternatively, using a more fundamental formula but requiring more difficult integration. c h z z bg δ t x t x x 4. t 6. t = b g z. t = 4. te dt IA = t p µ t e dt te 4. e dt (integration by parts, not shown) F t = HG I. t 4. K J e.. 4. = = 4.

Question #57 Answer: E Subscripts A and B here just distinguish between the tools and do not represent ages. ο We have to find e AB e ο A F I HG K J = = = z t t = dt t 5 5 F I 7 HG K J = = 49 49 = 35. 4 F ti t dt οhg K JF I HG K J z 7 7F = + HG t e ο t = z7 B dt t 7 4 e ο AB z t t t = 7 7 7 3 t t t = t + 4 = 7 49 49 343 + = 4 e ο AB = e ο A+ e ο B e ο AB = 5+ 35.. 683 = 587. 7. 683 I KJ dt

Question #58 Answer: A bg btg =. +. 4 = 4. µ τ x t p bg = 4. τ x e t Actuarial present value (APV) = APV for cause + APV for cause. z5 z 4. t. 4 t 5. 4 t. 4 t z5. 44t.,. b g b g e e. dt + 5, e e. 4 dt c b g b gh e dt = + 5 4 e bgj = = 44. 44 5 e. 784 Question #59 Answer: A R= p = q x x S = p e since e = e x b g z k µ x t + k dt µ x t dt k dt So S = 75. R p e = 75. q x e e k k = e z z d bg i bg k x bg µ x tdt qx = 75. px px qx = = 75. q 75. q L NM qx k = ln 75. q x x O QP x e z z kdt

Question #6 Answer: E k k + p k β = mean = 4; = β / + β f n PN= n b g b g x fbg bxg fbg bxg fbg 3 bxg..6.5.8.5.65 3.4 3.5.5.56 bg k bxg = probability that, given exactly k claims occur, that the aggregate amount is x. bg b g b g x k k f x = f x ; the claim amount distribution for a single claim bg bg b g e bgj b g f x = f j x f x j j= x sb g b g bg k b g k = k f x = P N = k f x ; upper limit of sum is really, but here with smallest possible claim size, f bg b x g = for k > x g =. g =. *. =. =. *. +. *. =. b g = + + = b g =. +. +. +. =. f s f s 6 5 4 f s 6 5 8 65 48 f s 3 6. *. 5 8. * 5. 4. *. 56. 576 F s 3 4 48 576 346

Question #6 Answer: E Let L = incurred losses; P = earned premium = 8, F LI Bonus = 5. 6. HG K J P if positive P = 5. 6. P Lg if positive = 5. b48, Lg if positive = 5. c48, bl 48, gh E (Bonus) =.5 (48, EcL 48, gh From Appendix A..3. =.5{48, [5, ( (5, / (48,+5,)))]} = 35,65 Question #6 Answer: D 8: z δ t 7 A = e dt d i b g δ = e =. 6 since δ = ln 6. =. 587 7δ F 7 a = + v 933. 8: H G I 7 K J = 3V = 5, A 6643 a 8: 8: = 87 Question #63 Answer: D Let A x and a x be calculated with µ x t * * b g and δ = 6. b g Let Ax and ax be the corresponding values with µ x t increased by.3 and δ decreased by.3

a a x * x Ax = 4. = = 6. 667 δ 6. = a x L N M * * t * µ x s + 3. ds 3. t x Proof: a = e e dt = = z z z = a x z z z d t µ x sds 3. t 3. t e e e dt t µ x Ax = 3. ax = 3. ax =. 3 6. 667 = 8. Question #64 Answer: A bg bg bg sds 6. t e e dt b gb g i bulb ages Year 3 # replaced - 9 +7 9 63 8 3 8+7+35 37 The diagonals represent bulbs that don t burn out. E.g., of the initial,, (,) (-.) = 9 reach year. (9) (-.3) = 63 of those reach year. Replacement bulbs are new, so they start at age. At the end of year, that s (,) (.) = At the end of, it s (9) (.3) + () (.) = 7 + At the end of 3, it s (8) (.) + (9) (.3) + (63) (.5) = 37 8 37 Actuarial present value = + + 3 5. 5. 5. = 6688

Question #65 Key: E Model Solution: z z z 5 HG KJ z. 6. 6L. 5 d i d i e5: 5 = t p5dt+ p t p dt 5 5 4 5 5. 4t F. 4ds. 5t = e dt+ ez I e dt e e e 4 = + 5.. = 797. + 4. 387 = 5. 6 NM O QP Question #66 Key: C Model Solution: p = 5 6 + e q 6 + je q 6 + jb q63gb q64gb q65g b gb gb gb gb g =. 89. 87. 85. 84. 83 =. 4589 Question # 67 Key: E Model Solution:. 5 = a x = 8. 4. + µ + δ µ δ = µ = δ = A A x x µ = + = 5. µ δ µ = = µ + δ 3

Var a e T j = x Ax A δ = 3 4 = 5. 83. 6 S.D. = 5. 83 = 7. 7 Question # 68 Key: D Model Solution: v = 9. d =. A = da =. 5 = 5. x x b gb g 5Ax 5vqx Benefit premium π = a x 5. 5 5. 9. 5 = 5 a Vx = a x a. 5 A x+ x+ = ax+ = b gb g b gb g 4 = 455 = da x =. 4 = 6. V = 5A π a = 5. 6 455 4 = 8 x+ + x+ x+ b gb g b gb g b gb g Question #68 Key: D Model Solution: v is the lowest premium to ensure a zero % chance of loss in year (The present value of the payment upon death is v, so you must collect at least v to avoid a loss should death occur). Thus v =.95.

bg x x x+ b g E Z = vq + v p q =.95.5 +.95.75. =.379 4 4 d i x x x+ b g b g E Z = vq + vpq =.95.5 +.95.75. =.3478 b g d i c b gh b g Var Z = E Z E Z =.3478.379 =. Question # 7 Key: D Model Solution: Severity after increase Severity after increase and deductible 6 8 8 3 Expected payment per loss =. 5 +. 5 +. 5 8 +. 5 = 75 Expected payments = Expected number of losses Expected payment per loss = 75 3 =,5 Question # 7 Key: A Model Solution: E (S) = E (N) E (X) = 5 =, VarbSg = EbNgVarb Xg + Eb Xg VarbNg = 5 4 + = bgb gd ib g 4,,

b g F HG b 8,, Pr S < 8, = Pr Z < 4,, = Pr Z <.998 6% g I KJ Question #7 Key: A Model Solution: Let Z be the present value random variable for one life. Let S be the present value random variable for the lives. δ t µ t EZ bg = z e e µ dt 5 µ = e bδ + µ g5 δ + µ =. 46 5 EZ d i F µ I eb δ + µ = g HG δ + µ KJ F.4 = H G I.8 K J de i= 33..6 b g = d i c b gh VarZ EZ EZ bg = 33.. 46 = 5348. bg ES = EZ = 4. 6 Varbg S = Varbg Z = 534. 8 F 4. 6 = 645. F = 8 534. 8 Question #73 Key: D Model Solution: Prob{only survives} = -Prob{both survive}-prob{neither survives}

b ge j b gb gb gb gb gb g b gb g = p p p p 3 5 3 5 3 5 3 5 =.973.9698.968.9849.989.968.9.9363 =.446 =.9.9363 Question # 74 Key: C Model Solution: The tyrannosaur dies at the end of the first day if it eats no scientists that day. It dies at the end of the second day if it eats exactly one the first day and none the second day. If it does not die by the end of the second day, it will have at least, calories then, and will survive beyond.5. b g b g b g b gb g Prob (ruin) = f + f f =. 368 +. 368. 368 =. 53 e since fbg = =. 368! e fbg = =. 368! Question #75 Key: B Model Solution: Let X = expected scientists eaten. For each period, EX = EXdead Prob already dead + EXalive Prob alive = Probbdeadg + EX alive Probbaliveg Day, E X = b g fbg b g Prob (dead at end of day ) =.53 [per problem ] e Prob dead at end of day = = =. 368! Day, E X =. 368 +. 368 =. 63 b g b g

b g Day.5, E X 5. =. 53 +. 5. 53 =. 49 where EX 5. alive = 5. since only day in period. E X = E X + E X + E X5. = +. 63 +. 49 = 88. E, X = 8, 8 Question # 76 Key: C Model Solution: This solution applies the equivalence principle to each life. Applying the equivalence principle to the life group just multiplies both sides of the first equation by, producing the same result for P. b g b g 7 7 7 7 7 bgb. 338g bgb. 338gb. 366g Pb. 338gb. 366g APV Prems = P = APV Benefits = q v + p q v + Pp p v P = + 8. 8. =. 37 +. 36 +. 7988P. 678 P = = 3.. (APV above means Actuarial Present Value). + 8. Question #77 Key: E Model Solution: One approach is to recognize an interpretation of formula 7.4. or exercise 7.7a: Level benefit premiums can be split into two pieces: one piece to provide term insurance for n years; one to fund the reserve for those who survive. If you think along those lines, you can derive formula 7.4.: P = P + P V x xn : xn : n x

And plug in to get. 9 = P +. 864. 563 P xn : xn : =. 85 b gb g Another approach is to think in terms of retrospective reserves. Here is one such solution: V = P P s n x x xn : xn : = P P x = P P = e e e e x P x e P P xn : xn : xn : xn : j j j j j a n P xn : E x a xn : a xn : xn :. 563 =. 9 P /. 864 P xn : e =. 9. 864. 563 =. 85 xn : j b gb g Question #78 Key: A Model Solution: b g δ= ln 5. =. 4879 z z µ ω x A = p t e dt x t x x bg ω x = x e ω = x a ω x ω δt δt dt for DeMoivre

c From here, many formulas for V A 4 hcould be used. One approach is: Since so P A A A 5 4 c 4h a 5 8. 7 A5 = = =. 374 so a5 =. 83 5 5 δ a 6 9. 4 A4 = = =. 333 so a4 = 387. 6 6 δ. 333 = =. 33 387. V A = A P A a =. 374. 33. 83 =. 75. 4 5 4 5 F HG F HG I = KJ I = KJ c h c h b gb g Question #79 Key: D Model Solution: A = E v = E v NS Prob NS + E v S Prob S x T x T x T x F b g F bg bg bg 3. I = HG + K J 7. + 3. 8. =. 395 F Similarly, 3. A x I = 7 HG 3+ 6 K J. +.. F H Var a bg T x I K = x Ax A δ 93.. 395 = 8. 6. I HG + K J 3. 6. 8. F 6. I HG 6+ 6 K J =.. = 4.. b g. 3 93.. Question #8 Key: B Model Solution: Let S denote aggregate losses before deductible.

ES = = 4, since mean severity is. f f Sbg = = S e! e! 353., since must have number to get aggregate losses =. 3 bg= F H G I K J F H G I K J =. 9, since must have loss whose size is to get aggregate losses =. b g Sb g S Sb g Sbg ES = f + f + f f b g = 353. +. 9 + 353.. 9 = 639. E S = E S E S + = 4 639. =. 368 c b h g Question #8 Key: D Model Solution: Poisson processes are separable. The aggregate claims process is therefore equivalent to two F independent processes, one for Type I claims with expected frequency 3 HG I 3K J b g= and one for Type II claims. Let S I = aggregate Type I claims. N I = number of Type I claims. X I = severity of a Type I claim (here = ). b g b g Since X =, aconstant, E X = ;Var X =. I I I b Ig b Ig b Ig b Ig b Ig Var S = E N Var X + Var N E X bgbg bgbg = + =,

bg b g b g Var S = Var SI + Var SII since independent,, =, + VarbSIIg Var S =,, b IIg Question #8 Key: A Model Solution: bg bg bg τ 5p5 = 5p5 5p5 F HG b gb I K J 55 5. 5 = e 5 =. 9. 7788 =. 79 g b gb g Similarly 6 5 pbg τ F I 5 e b. = gb g HG 5 K J =. 8. 665 =. 485 b gb bg τ bg τ bg τ 55q5 = 5p5 p5 =. 79. 485 =. 57 g Question #83 Key: C Model Solution: Only decrement operates before t =.7 bg 7q4 b7gqbg 4 b7gbg 7 since UDD. =. =.. =. Probability of reaching t =.7 is -.7 =.93 Decrement operates only at t =.7, eliminating.5 of those who reached.7 bg b gb g q 4 =. 93 5. = 65.

Question #84 Key: C Model Solution: e j 3 8 8 π π π 8 vq8 v p q + p v = A8 + + F F H G b g b g b7468. g 665. 75 πb. 756g b6754. g = + = 665. 75 π. 839 I. 83. 839. 956 + 665 75 π HG 6 K J =. + + 6 3.. 6. π π π = 397. 4 Where p = 3, 84, 54 8 3 94 365 =.,, 839 b gb g Or p 8 =. 83. 8764 =. 839 Question #85 Key: E Model Solution: At issue, actuarial present value (APV) of benefits t = bv p µ t dt = z z b g t t 65 65 4. t 4. t de ide it p t dt 65 µ 65bg z t 65 65b g 65 = p µ t dt = q = I APV of premiums = π = π a HG + K J = π 65 6. 667 4.. Benefit premium π = / 6. 667 = 6 z z u + u u 67 65 4. b+ ug 4. u u z 8. u 67 65 V = b v p µ + u du πa b g = e e p µ + u du 6 6. 667 F 67 65 = e p + u du 67 I K J b g b gb g = 83. 9 q = 83. 9 = 83. 9 67 µ b g

Question #86 Key: B Model Solution: () a = a + E Ax: () a x : = d (3) A = A + A x: x: x x: x: x: : (4) Ax = A + E A x x x+ x: 8. = A + 5. 4. A x: = 8. b gb g Now plug into (3): A x: = 8. + 5. = 43. Now plug into (): 43. a x : = 5. / 5. b g = 97. Now plug into (): a x: = 97. +. 5 =. Question #87 Key: A Model Solution: EN = E ENΛ = E Λ = Λ Λ Var N = E Var N Λ + Var E N Λ Λ = E Λ + Var Λ = + 4= 6 Λ Distribution is negative binomial. Λ Λ

Per supplied tables: p rβ = = =. r+!+β 3 b g b gb g bgbg rβ= rβ + β = 6 and b g + β = 3 β = rβ = r = Alternatively, if you don t recognize that N will have a negative binomial distribution, derive gamma density from moments (hoping α is an integer). Mean = θα = Var = E Λ E Λ = θ α + α θ α θα 4 θ = = = θα α = θα = θ p = p λ f λ dλ = e j = θα= 4 z z λ λ e λ λ e c h bg b g b / / g! λγ bg z α = λe dλ Integrate by parts; not shown F HG λ 4 = λe e 3 9 = =. 9 3 3 λ dλ I KJ

Question #88 Key: C Model Solution: Limited expected value = z z. x. x. x. x c Fbg x hdx = d. 8e +. e idx = d 4e e i = 4 + 6. 4 = 66.4 Question #89 Key: E Model Solution: M T = Initial state matrix = = One year transition matrix = L N M..8.5.5.75.5. O Q P cb b M T = g h M T T = g M T T T =. 8. 44. 6. 4.. 468. 35. 8. Probability of being in state F after three years =.468. 3 d ib g Actuarial present value =. 468v 5 = 7 Notes:. Only the first entry of the last matrix need be calculated (verifying that the four sum to is useful quality control. ). Compare this with solution 3. It would be valid to calculate T 3 here, but advancing M one year at a time seems easier.

Question #9 Key: B Model Solution: Let Y i be the number of claims in the ith envelope. Let Xbg 3 be the aggregate number of claims received in 3 weeks. EY EY b g b g b g b g b g b g b g b g bg b g i i =. + 5. + 3 4. + 4 5. = 5. =. + 4 5. + 9 4. + 6 5. = 7. EX3 = 5 3. 5 = 65 Var X 3 = 5 3 7. = 468 ProbmXbg 3 Zr=. 9 = Φb8. g R X 3 65 S b g U Prob 8. T V 468 W X 3 7.7 bg Note: The formula for Var Xbg 3 took advantage of the frequency s being Poisson. The more general formula for the variance of a compound distribution, Var S = E N Var X + Var N E X, would give the same result. bg bg bg bgbg Question #9 Key: E Model Solution: µ t t µ M p p F b g b g M 65 F 6 6 = = = ω 6 75 6 5 6 3 = = = ω = 85 ω 6 5 5 5 t = t = 5 Let x denote the male and y denote the female.

ex ey exy = 5 = = = z z F HG c c F HG F HG mean for uniform distribution over, =. 5 mean for uniform distribution over,5 I K JF HG t t 5 7 t t + 5 5 3 I K J I KJ I dt dt 7 t 7 t t + = + 75KJ 75 4 = 7 + = 3 3 3 b b gh gh + e = e e e xy x+ y xy = 5 + 5 3 3 75 6 = = 37. 3 6 Question #9 Key: B Model Solution: A A x x µ = µ + δ = µ = = µ + δ 3 5 PcA x h= µ = 4. Var L F P Ax bg c hi = + H G K J e δ F I F = + HG K J F HG H G I 4. K J 8. 5 3 F = H G 3I K J F H G 4 I K J = 5 45 Ax Ax I KJ j

Question #93 Key: B Model Solution: bg g b g E X E X Mean excess loss = F 33 9 = = 3 8. bg b g b g E X = E X since F =. Question #94 Key: E Model Solution: Expected insurance benefits per factory = E X + =. +. = 4.. Insurance premium = (.) ( factories) (.4 per factory) =.88. Let R = retained major repair costs, then f f f R R R b g b g b g = 4. = 6. = 4. 6. = 48. = 6. = 36. Dividend = 3 88. 5. 3, = 67. R, if positive E Dividend R b gbg if positive b g b gb g b gb g b gb g = 6. 67. +. 48 67. +. 36 =. 5888 [The (.36)() term in the last line represents that with probability.36, b67. Rg is negative so the dividend is.] b g

Question #95 Key: A Model Solution: αθ EX = = 4α = 8 α α 4α = 8α 8 α = F 4 Fbg= 6 H G θi 6 K J F = H G I α 6K J b g b g =. 555 s 6 = F 6 =. 444 Question #96 Key: B Model Solution: ex = px+ px+ 3px +... = 5. Annuity = v p + v p 4. +... = 3 4 3 x 4 k = 3 = v b 4. 3 k k = 3 g p x k 3 k v k p x x 3 3 F = = H G I vbex. 99. 98g K J. = b g 4. 9 8 87 Let π = benefit premium. e π + 99. v+ 98. v = 87 j. 858π = 87 π = 84

Question #97 Key B Model Solution: 3 : 3 b g3 : b ge 3j πa = A + P IA + π A A3 π = a IA A b g 3 : 3 : 3 b g b g. = 7. 747. 78. 88 = 6. 789 = 5. 4 Test Question: 98 Key: E For de Moivre s law, z ω 3 t e 3 F = ω 3 L M HG t = t N bω 3 ω 3 = O gq P I K J dt ω 3 Prior to medical breakthrough ω = = e 3 After medical breakthrough e 3 = e3+ 4= 39 ω so = = 3 e 3 39 ω = 8 3 = 35

Test Question: 99 Key: A 5. L=, v 4a = 77,79 3 @5% Test Question: Key: C ΕΝ = Ε ΕΝΛ = Ε Λ = Λ Var Ν = ΕΛVar ΝΛ + VarΛ E N Λ = Ε Λ + Var Λ = + = 4 Λ Λ Λ Distribution is negative binomial (Loss Models, 3.3.) Per supplied tables mean = rβ = Var = rβb+ βg = 4 b+ βg = β = rβ = r = From tables 3 3 rr b + gbr+ gβ bgbgbg 3 4 4 p3 = = = = 5. r+ 3 5 3! b + βg 3! 3 p 3 = 5

Test Question: Key: E b gb g b gb g b gb g b gb g b gb g b gb g b gb g E N = Var N = 6. 5 = 3 E X = 6. +. 5 +. = 36. E X =. 6 +. 5 +. = 56. Var X = 5. 6 36. =. 64 For any compound distribution, per Loss Models Var S = E N Var X + Var N ce X h = (3) (.64) + (3) d36. i = 768 For specifically Compound Poisson, per Probability Models Var S = λt E X = (6) (.5) (5.6) = 768 Alternatively, consider this as 3 Compound Poisson processes (coins worth ; worth 5; worth ), where for each Varb Xg =, thus for each VarbSg = VarbNgE X. Processes are independent, so total Var is Var = b6gb5. gb. 6g + b6gb5. gb. g5 + b6gb5. gb. gbg = 768 Test Question: Key: D d ib g b g 6 9Vx+ Px. qx+ 9 Vx = Ax+ = px+ 9 b34. 3+ 3. 7gb6. g. 54bg = = 3698.. 98746 3698 ax =. + = 445. 6. / 6. so b Ax + 3698. Px + = = = 33. a 445. x+ g

Test Question: 3 Key: B k bg p e e F = e z bg bg bg bg k τ k τ µ x tdt µ x tdt x = z = z p = 6 p 6 HG b p k µ g bg x bg tdt I KJ where p is from Illustrative Life Table, since µ follows I.L.T. k x k x 6, 66, 55 = = 88. 8, 88, 74 6, 396, 69 = =. 78 8, 88, 74 bg bg bg τ τ τ q6 = p6 p6 = b p6g b p6g from I.L.T. = 88.. 78 =. 46 bg Test Question: 4 Key: C P s = d a, where s can stand for any of the statuses under consideration. s a a a s x xy = P + d s = a y = = 65.. + 6. = = 8. 333 6. + 6. a + a = a + a xy xy x y a xy = 6. 5+ 6. 5 8. 333 = 467. P xy = 6. = 8. 467.

Test Question: 5 Key: A bg z τ bµ + 4. gt d = e µ +. 4 dt b e µ + 4. = e = 48 g µ 4. e + =. 95 µ +. 4 = ln. 95 =. 49 µ =. 9 bg z 4 49. t 3 3 b b g b gj d = e. 9 dt e b b gb g b gb g 49 3 49 4. 9.. = e e = 76.. 49 g g j Test Question: 6 Key: B This is a graph of lxµbxg. µ bxg would be increasing in the interval b8, g. The graphs of l x p 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: 7 Key: A Using the conditional mean and variance formulas: EN = E Λ NΛ c h d c hj d c hj Var N = VarΛ E N Λ + EΛVar N Λ Since N, given lambda, is just a Poisson distribution, this simplifies to: EN = E Λ Var N = VarΛbΛg + EΛbΛg We are given that EN =. and Var N = 4., subtraction gives Var bλg =. Test Question: 8 Key: B

N = number of salmon X = eggs from one salmon S = total eggs. E(N) = t Var(N) = 9t b g b g b g b g b g b g b g b g F > 5, b, g = > HG 3,, ES = ENEX =5 t Var S = E N Var X + E X Var N = t 5+ 5 9t = 3, t PS P S t t 5t 3 t, 5t = 645. 3 t = 5 t 4 t = t d i t t 4= 3 t = ± + = 473. 4 t =. 4 round up to 3 I KJ =. 95 Test Question: 9 Key: A k + A P V (x s benefits) = v bk+ kpxq k = = 3v. + 35v. 98. 4 + 4v. 98. 96. 6 = 36, 89 x+ k b g b gb g 3b gb gb g

Test Question: Key: E π denotes benefit premium 9V = APV future benefits - APV future premiums 6. = π π =. 36 8. bv + πgb8. g bq65gbg V = p = 65 b5. +. 36gb8. g b. gbg =5.8. Test Question: Key: C X = losses on one life E X = 3. +. +. 3 = b gb g b gb g b gb g S = total losses ES = 3EX = 3 b g+ c sb g E S = E S F b gc bgd = ES f = 3 4. = 3. 936 =. 64 3 h sb gh i

Test Question: Key: A 8 = 7a + 5a a 8 = 7 + 5 a a 3: 4 = 8 a = a + a a = + 8 = 4 a = 4 3 4 3: 4 3: 4 3 4 3: 4 3: 4 b gb g b gb g 3: 4 Test Question: 3 Key: B a = a f t dt z bg = t z 5. t e 5 z. Γb g b L NM b g L F = H G I K J 5NM. 5. = te t te 5. t te dt 5. t F HG O QP = i dt t t = t+ e + + 5. 5.. 8594. I K J 5. t e 5 O QP, 8594. = 37, 88 Test Question: 4 Key: C b g = k pk b g L O NM QP b pk = + k pk g Thus an (a, b, ) distribution with a =, b =. Thus Poisson with λ =. 4 e pb4g = 4! = 9.

Test Question: 5 Key: B By the memoryless property, the distribution of amounts paid in excess of is still exponential with mean. With the deductible, the probability that the amount paid is is / F e. 393. b g = = Thus the average amount paid per loss is (.393) () + (.67) () =.4 The expected number of losses is () (.8) = 6. The expected amount paid is (6) (.4) = 94. Test Question: 6 Key: D Let M = the force of mortality of an individual drawn at random; and T = future lifetime of the individual. n Pr T = E Pr T M z = Pr T M = µ fmbµ g dµ z z µ = µ e t dt d µ z = µ = + = + e du e e s d i d i d i = 56767.

Test Question: 7 Key: E EN EN E X VarbNg Var X ES b gbg b gb g b g b gb g = 8. +. =. = 8. +. 4 = 6. = 6.. = 6. = 7 + = 7 b g = EX EX = b7 +, g 7 = 78, = ENEX =. b7g = 4 b g EN b g EX b g b g b g b g = 98, 344 = 336. Var S = Var X + Var N =. 78, + 7 6. = 98, 344 Std dev S So B = 4 + 34 = 58 Test Question: 8 Key: D Let π = benefit premium Actuarial present value of benefits = = b. 3gb, gv+ b. 97gb. 6gb5, gv + b. 97gb. 94gb. 9gb, gv = 566. 38 + 7769. 67 + 689. 8 =, 33. Actuarial present value of benefit premiums = a π x: 3 b gb g = +. 97v+. 97. 94 v =. 766π, 33. π = = 745. 55. 766 745. 55 6.,. 3 V = 3. = 958. 46 π b gb g b gb g Initial reserve, year = V + π = 958.56 + 745.55 = 94. 3 Test Question: 9 Key: A

Let π denote the premium. L = b v πa = + i v πa T T = π a T T b g EL = π a x = π = ax a T δ a L = π a = = T a v = T b δ a δ a x x g x T T x v A = A x x T d δ a v x T T i Test Question: Key: D (, ) (,.9) (.5,.8775) (,.885) tp t p = (. ) = 9. p =. 9. 5 = 855. b gb g since uniform, 5. p =. 9 + 855. / b = 8775. g

e5 :. = Area between t = and t = 5. F+ 9. I. 9 + 8775. = 5. HG K J bg F I + HG K Jb g =. 95+. 444 = 394. Alternatively, z z 5. e 5 :. = t pdt z 5. = tpdt + p x pdx z z 5. 5.. t 5. x = t + 9. x b g b g =. tdt+ 9. 5. xdx =. 95 +. 444 = 394. Test Question: Key: A, A63b. g = 533 A63 =. 467 Axb + ig qx Ax + = p b x gb g. 467 5.. 788 A64 =. 788 =. 483 b. 483gb5. g. 95 A65 =. 95 =. 4955 Single contract premium at 65 = (.) (,) (.4955) = 555 b g. 555 555 + i = i = 533 533 = 984

Test Question: Key: B Original Calculation (assuming independence): µ x = 6. µ y = 6. µ xy = 6. + 6. =. µ x Ax = x + = 6. =. 54545 µ δ 6. + 5. µ y Ay = y + = 6. =. 54545 µ δ 6. + 5. µ xy Axy = xy + =. =. 7588 µ δ. + 5. A = A + A A =. 54545 +. 54545. 7588 =. 385 xy x y xy Revised Calculation (common shock model): µ = 6., µ = 4. x T* x x T* y y µ = 6., µ = 4. y bg bg bg bg 4 4 T* x T* y Z µ xy = µ x + µ y + µ +. +. +. =. µ x Ax = x + = 6. =. 54545 µ δ 6. + 5. µ y 6. Ay = = =. 54545 µ y+ δ 6. + 5. µ xy Axy = xy + =. =. 66667 µ δ. + 5. A = A + A A =. 54545 +. 54545. 66667 =. 443 xy x y xy Difference =. 443. 385 =. 39

Test Question: 3 Key: E Treat as three independent Poisson variables, corresponding to, or 3 claimants. rate rate rate Var Var Var 6 3 3 L NM = = = 4 = = 6 = 6 = 4 = 8 total Var = 6+ 6+ 8= 4, since independent. Alternatively, E X 3 d i= + + = 3 6 3 O QP For compound Poisson, Var S = E N E X F = bg H G I K J = 3 4 Test Question: 4 Key: C z3 λbtdt g = 6 so Nb3g is Poisson with λ = 6. b g g P is Poisson with mean 3 (with mean 3 since Prob y i < 5 =. 5 P and Q are independent, so the mean of P is 3, no matter what the value of Q is.

Test Question: 5 Key: A At age x: Actuarial Present value (APV) of future benefits = APV of future premiums = F HG 4 5 a x I K J π F A x I HG 5 K J A 5 A 4 a 4 = π a by equivalence principle 5 865. = π π = = 58. 4 6. 4 5 5 5 5 V = APV (Future benefits) APV (Future benefit premiums) 4 = A35 π a 35 5 5 = 5 8 7 4 b. g b58. gb5396. g 5 =. 5 Test Question: 6 Key: E Let EY Var Y Y = present value random variable for payments on one life S = Y = present value random variable for all payments = a = 4866. 4 d A A 4 4 = d =. 4863 63. 6. /. 6 = 7555. ES = EY = 4, 86. 6 Var S = Var Y = 7, 555 Standard deviation S = 7, 555 = 656. i d ib g By normal approximation, need E [S] +.645 Standard deviations = 4,86.6 + (.645) (65.6) = 5,54

Test Question: 7 Key: B 5A 4 A Initial Benefit Prem = 5a 4a e 3 3: 3: 35 3: b. g b. g j 5 48 4 933 = 5b4. 835g 4b959. g 54. 73.. 3958 = = =. 5 7475. 47. 836 6. 339 Where A = A A 337 9374 933 3: e =.. =. 3: 3: j and A3 :. 337 a = = = 959. 3: d F 6. H G I 6. K J Comment: the numerator could equally well have been calculated as A + 4 E A =.48 + (4) (.9374) (.495) =.395 3 3 5 Test Question: 8 75. 75. p p q x y = 75. 5. =. 965 =. 75. =. 95 = b gb g b 75. xy 75. xy = b75. pxgd75. pyi p gb g = -b.965gb. 95g = 97. Key: B since independent Test Question: 9 Key: A N = number of physicians E(N) = 3 Var (N) = X = visits per physician E(X) = 3 Var (X) = 3

S = total visits E(S) = E (N) E(X) = 9 Var(S) = E(N) Var(X) + E X = 3 3 + 9 = 89 Standard deviation (S) = 43.5 F HG b g Var(N) = Pr(S>9.5) = Pr S 9 9. 5 9 > 435. 435. I K J = Test Question: 3 Key: A Φb68. gcourse 3: November The person receives K per year guaranteed for years Ka = 8. 4353 K The person receives K per years alive starting years from now a 4 K *Hence we have = 8. 4353+ E 4 a 5 K Derive E 4 : Derive a 5 A = A + E A E 4 4 4 : = A A b 4 A 4 5 b g 4 5 A5 35. = = = 6. 9 d. 4 4. Plug in values:, = c8. 4353+ b. 6gb6. 9ghK = 8. 5753K K = 538. 35 : 3. 9. = = 6. 35. g Test Question: 3 Key: D STANDARD: e F I HG K J = = z 5 :. t t = dt t 933 75 75. ds. MODIFIED: p = ez = e =. 9484 5

z z t e 5 : = t p dt p dt 5 + 5 74 z F HG t.. t = e dt + e dt 74. Difference =.847 F HG = e. t + e t. 74 z I K J F I HG K J I KJ b g =. 9563 +. 9484 9. 343 = 9. 3886 Test Question: 3 Key: B Comparing B & D: Prospectively at time, they have the same future benefits. At issue, B has the lower benefit premium. Thus, by formula 7.., B has the higher reserve. Comparing A to B: use formula 7.3.5. At issue, B has the higher benefit premium. Until time, they have had the same benefits, so B has the higher reserve. Comparing B to C: Visualize a graph C* that matches graph B on one side of t= and matches graph C on the other side. By using the logic of the two preceding paragraphs, C s reserve is lower than C* s which is lower than B s. Comparing B to E: Reserves on E are constant at. Test Question: 33 Key: C Since only decrements () and () occur during the year, probability of reaching the end of the year is bg bg 6 6 b gb g p p =.. 5 =. 945 Probability of remaining through the year is bg bg bg 3 6 6 6 b gb gb g p p p =.. 5. =. 84645 Probability of exiting at the end of the year is qbg 3 6 =. 945. 84645 =. 945

Test Question: 34 Key: E b g =.7 b g = 4. + 9.. 49=. b g = b g =. 4 = 6 b g =. 7= 4. bg S bg N bg X bg X bg N =. 7 6+ 4. = 64. b g = 4 bsg Standard DevbSg bs > 94g = bs = g E N Var N E X Var X E S Var = E Var + E Var = Standard Dev S E + = 4. + 4= 94. Since there are no possible values of S between and, Pr. Pr 3 =. 7.. 8.. 8 =. Test Question: 35 Key: D z APV of regular death benefit = z = -.6t -.8t bgde ib. 8gde idt =. 8 /. 6 +. 8 =, 764. 7 b g -δ t -µ t b gde ib.8gde i =z APV of accidental death benefit e. e dt =z -.6t -.8t e. e dt 3 b gd ib gd i = e -.4 /. 68 =, 79. 37 Total APV = 765 + 79 = 344 dt 3 b gd -δ t ib gd -µ t i

Test Question: 36 Key: B l l 6 +. 6 6 +. 5 bgb gbgb g =. 6 79, 954 +. 4 8, 65 = 8,. 4 bgb g bgb g =. 5 79, 954 +. 5 78, 839 = 79, 396. 5 8. 4 79, 396. 5 q + = 8,. 4 =. 3 9. 6. 6 P = = 9. 99% Test Question: 37 Key: A 5 5 5 bg z λ λ 5 5 d i 5 d i P = e dλ = e = e = 987. 5 5 5 Pbg z λ λ λ = 5 λe dλ = e e 6e 99 d λ i = 5 d i=. 5 PN b g =. 987. 99 =. 694 Test Question: 38 Key: A bg bg bg 4 = pbg 4 pbg 4 bg τ 4 4 q = q + q = 34. 34. = 75. p 4 bg bg 4 p = 88. 4 q =. = y bg. 4 q = y = 4

q τ 4 l bg τ 4 bg b gb g b gb g =. 8. 4 =. 39 =. 34. 39 = 83 Test Question: 39 Key: C Pr Lbπ' g > <.5 K+ Pr, v π' a > <.5 K + From Illustrative Life Table, 47 p 3 =. 586 and 48 p 3 =. 4768 Since L is a decreasing function of K, to have Pr L π > < π Highest value of L π 47+ Lbg π at K = 47 =, v π a 47+ = 69. 98 6. 589π L bπ g b69. 98 6. 589π g 69. 98 π > = 36. 77 6. 589 b g.5 means we must have L b g for K 47. b g for K 47 is at K = 47. Test Question: 4 Key: B b g x b g x x b g x b g =. +.. +.. =. d i =. +.. +.. =. VARbYg = = Pr K = = p =. Pr K = = p p = 9. 8. = 9. Pr K > = p = 8. EY EY 9 87 8 7 475 9 87 8 7 6 47 6. 47. 475. 99 Test Question: 4 Key: D Let X be the occurrence amount, Y = max(x-, ) be the amount paid. E[X] =, Var[X] = b, g P(X>) = exp(-/,) =.94837

The distribution of Y given that X>, is also exponential with mean, (memoryless property). So Y is R S T with prob.9563 exponential mean with prob.94837 E Y =. 9563 +. 94837, = 94. 837 b g E Y =. 9563 +. 94837, = 89,, 675 b g Var Y =, 89, 675 94. 837 = 99, 944 Alternatively, think of this as a compound distribution whose frequency is Bernoulli with p =.94837, and severity is exponential with mean,. b gb g b g Var = Var N E X + Var X E N = p p,, + p,,

Test Question: 4 Key: B bg b gc xh P In general Var L = + δ Ax A Here PcAxh= = = a δ 8 5.. x bg F I c x HG K J h=. 565 5. b * g F bgi = + c x x HG h. 8 KJ 4 5 b g = + 8 b g b 565 g = 744 + 8 x x xbδ g b g VarbL g 75. So Var L = + Ax A. 8 and Var L A A So Var L *.. E L* = A. 5a = a +. 5 = 5. 3 =. 5 E L* + * =. Test Question: 43 Key: C Serious claims are reported according to a Poisson process at an average rate of per month. The chance of seeing at least 3 claims is ( the chance of seeing,, or claims). b g. is the same as Pb,, g. is the same as Pb g Pb g Pb g P 3+ 9 λ λ λ. e + λe + λ / e d i + +. The expected value is per month, so we would expect it to be at least months λ = 4 b g. Plug in and try 4 4 4 e + 4e + 4 / e =. 38, 6 6 6 d i too high, so try 3 months bλ = 6g d / i., okay. The answer is 3 months. e + 6e + 6 e = 6 [While is a reasonable first guess, it was not critical to the solution. Wherever you start, you should conclude is too few, and 3 is enough].

Test Question: 44 Key: B Let lbg τ = number of students entering year superscript (f) denote academic failure superscript (w) denote withdrawal subscript is age at start of year; equals year - pbg τ =. 4. =. 4 bg bg bg bg τ τ f f l = l q q =. q = q q =. 6.. = 3. l q = 4. l q q q q bg bg bg w τ f bgbg bg bg bg τ f τ f w bg f f bg f e e bg = 4. q 3. 8. = =. 4. bg bg bg τ f w b p = q q =. 3. = 5. bg bg bgbg bgbgbg w w τ w τ τ w 3q = q + p q + p p q =. + 4. 3. + 4. 5. 3. = 38. b gb g b gb gb g j g j

Test Question: 45 Key: D e = p + e e p 5 5 6 N M 6 = e6 = ez b bg g since same µ M = e z µ 5 tdt z z c h M N µ 5 t +. t dt 5 bg. c tdt h = e M µ 5 tdt. tdt e z bg c h = p M 5 e L NM F H t t. IO KQP = e 5. M p 5 b N N 5 5 6 e = p + e 5. M 5 6 = e p + e M 5 b g = 95. e = 95.. = 95. g b gb g

Test Question: 46 Key: D E Y = E Y =, a σ σ AGG b g F HG x b g c h I KJ = Ax =,,, δ b g c h = Var Y =, A A δ b, g = b. 5g b6. g = 5, δ Y x x AGG = σ = 5, = 5, L N M F E YAGG 9. = Pr > σ F E Y 8. = σ Y AGG AGG AGG F = 8. σ + E Y b AGG b g g AGG F = 8. 5, +,, =, 64, O Q P Test Question: 47 Key: C Expected claims under current distribution = 5 θ = parameter of new distribution X = claims Eb Xg = θ bonus =.5 5 X 5 F F θ I E (claims + bonus) = θ+.5 5 θ + K J 5 5 θ F b g b g θ H θ G 5 I + θ K J = 5 5 5 + θ θ 5θ = 5 5 + θ I HG HG K J = θ + θ 5θ = 5 5 + 5θ θ = 5 5 = 354

Test Question: 48 Key: E b g eb g DA = vq + vp DA 8: 8 8 89 : b g b g.. 8 q 8 =. 3 = + DA 89 : 6. 6. 3 b DA g b6. g 4 = 89 =. 5 :. 8 q 8 =. DA = v. + v. 9. 5 8: b g b gb. b. g = + 9 5 =. 67 6. g j Test Question: 49 Key: B Let T denote the random variable of time until the college graduate finds a job Let Nt, t denote the job offer process m b g r Each offer can be classified as either RType I - - accept with probability p N t S m bgr Type II - - reject with probability b pg Nbtg T m r m b gr is Poisson process with λ λ b 8 g b 8 g F w. 4. b I w 4g > HG K J = Φ bg.. By proposition 5., N t p= Pr w>, = Pr ln w> ln, = Pr ln >. = Pr ln. = 587. λ = 587. =. 374 T has an exponential distribution with θ = = 35.. 374 Pr T > 3 = F 3 b g bg 3 = e3.5 =.386 = p

Test Question: 5 Key: A t p x L NM = exp zt ds x s O b g t x x s QP = = exp ln x t e = e + e e e e 5: 6 5 6 5: 6 z 5 e5 = e 6 5: 6 5: 6 z L N M L M 5 t t dt = t 5 5 5 P = 5 4 4 t t = dt = t 4 4 4 N Q P = z4f 5 ti 4 t 4 = dt 9t t dt HG 5 K JF I HG 4 K J z = d + i 3 F t 4I = t t 45 + 4. 67 3 HG = 5 + 4. 67 = 3. 33 O Q O 5 4 KJ =

Test Question: 5 Key: A b gb g b gb g UDD l =. 8 53, 488 +. 7, 384 = 46, 67. S t Mrlbg z z b + g = e = t p dt S = areas b g =. 75+ 8. +. 36+. 7 = 4. 4 Sb Xig Sb g..37573.56.8897 Ο Ο Ο3 Ο4 5 3 35 4 x i Test: 5 Key: Question D µ n = E N = 5 µ x = Var X = 675 µ N = Var N = 5 EX = 5 E S = E X E N = 5 5 = 5 Var S = E N Var X + Var N E X = 5 675 + 5 5 = 79, 375

Standard Deviation S = 79, 375 = 874. b g b g b g b g Pr S > = Pr S 5 / 874. > 5 / 874. = Φ. 66 Test Question: 53 Key: E b g = VarbΛg+ VarbΛg b g = vb b Vg p5q5 b, 3, 9g b. 83gb. 9968g Var L Var Λ Var Λ = = 358664. 9 v since VarbΛ g= 3. b g = vb b Vg p5q5p5 b, 6, 539g b. 9968gb. 9gb. 9989g = = 759. 3. 75. 9 b g = 358664. 9 + = Var L 3. 453937. 6 Alternative solution: π =, v V = 978. 74 6539 = 369. 74 L = R S T 5: 3, v π a = 6539 for K =, v π a = 378.8 for K = 3, v π a = 83.5 for K > 3

Pr K = = q =.83 Pr b g 5 b g 5 5 b gb g K = = p q =.9968.9 =.934 b g b g b g b g Pr K > = Pr K = Pr K = =.9865 π Var L = E L E L = E L since is benefit premium =.83 6539 +.93 378.8 +.9865 83.5 = 453, 895 [difference from the other solution is due to rounding] b g Test Question: 54 Key: C Let π denote the single benefit premium. π = a + π A 3 35 35: 3 3 a A A a 35 35: 3 35: 3 π = = A A 35: 3 e b 35: 3 j g 65..7 9.9 = b.7 g =.386.93 =.49 = Test Question: 55 Key: E 4. x F e dx p = = ez e + j 4.. 5. 5 =. 4F L e NM. 4F F H e x. 4 O QP I K e 8. = e. 4. 68 = F e b g F ln. 5 =. 4. 68. 693 =. 4F. 68 F =.

Test Question: 56 Key: C E X = b!/ g b! g = F E X = H G I K J L 3 N M b O gq P = 3 + 5 F I HG K J = = 3 5 So the fraction of the losses expected to be covered by the reinsurance is = 4.. The expected ceded losses are 4,, the ceded premium is 4,4,. Test Question: 57 Key: E X = 5. X F HG I K J L = N M L = NM b g P so: F x 5. x / 5. + x + O QP O Q This is just another Pareto distribution with α =, θ =. E X =. and L M F E X 3 = H G I K J L NM 3 = NM 5 F HG b O QP = 3 + 35 So the fraction of the losses expected to be covered by the reinsurance is 35 =. 4. IO g KJ QP

The total expected losses have increased to,5,, so C =.. 4, 5, = 4, 758, 6 And C C = 4, 758, 6 4, 4, =8.