**BEGINNING OF EXAMINATION**

November 00 Course 3 Society of Actuaries **BEGINNING OF EXAMINATION**. You are given: R = S T µ x 0. 04, 0 < x < 40 0. 05, x > 40 Calculate e o 5: 5. (A) 4.0 (B) 4.4 (C) 4.8 (D) 5. (E) 5.6 Course 3: November 00 -- GO ON TO NEXT PAGE

. 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 60 0.09 0. 0.3 0.5 63 6 0.0 0. 0.4 0.6 64 6 0. 0.3 0.5 0.7 65 63 0. 0.4 0.6 0.8 66 64 0.3 0.5 0.7 0.9 67 (ii) White was a newly selected life on 0/0/000. (iii) White s age on 0/0/00 is 6. (iv) P is the probability on 0/0/00 that White will be alive on 0/0/006. Calculate P. (A) 0 P < 0.43 (B) 0.43 P < 0.45 (C) 0.45 P < 0.47 (D) 0.47 P < 0.49 (E) 0.49 P.00 Course 3: November 00 -- GO ON TO NEXT PAGE

3. For a continuous whole life annuity of on ( x ) : (i) (ii) T( x) is the future lifetime random variable for ( x ). The force of interest and force of mortality are constant and equal. (iii) a x = 50. Calculate the standard deviation of a T ( x ). (A).67 (B).50 (C).89 (D) 6.5 (E) 7. Course 3: November 00-3- GO ON TO NEXT PAGE

4. For a special fully discrete whole life insurance on (x): (i) (ii) The death benefit is 0 in the first year and 5000 thereafter. Level benefit premiums are payable for life. (iii) q x = 0. 05 (iv) v = 090. (v) a&& x = 500. (vi) 0V x = 0. 0 (vii) 0V is the benefit reserve at the end of year 0 for this insurance. Calculate 0 V. (A) 795 (B) 000 (C) 090 (D) 80 (E) 5 Course 3: November 00-4- GO ON TO NEXT PAGE

5. For a fully discrete -year term insurance of on (x): (i) 0.95 is the lowest premium such that there is a 0% chance of loss in year. (ii) p x = 0.75 (iii) p x+ = 0.80 (iv) Z is the random variable for the present value at issue of future benefits. Calculate VarbZg. (A) 0.5 (B) 0.7 (C) 0.9 (D) 0. (E) 0.3 Course 3: November 00-5- GO ON TO NEXT PAGE

6. A group dental policy has a negative binomial claim count distribution with mean 300 and variance 800. Ground-up severity is given by the following table: Severity Probability 40 0.5 80 0.5 0 0.5 00 0.5 You expect severity to increase 50% with no change in frequency. You decide to impose a per claim deductible of 00. Calculate the expected total claim payment after these changes. (A) Less than 8,000 (B) At least 8,000, but less than 0,000 (C) At least 0,000, but less than,000 (D) At least,000, but less than 4,000 (E) At least 4,000 Course 3: November 00-6- GO ON TO NEXT PAGE

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: 50 Variance of the number of boxes dropped per month: 00 Expected value per box: 00 Variance of the value per box: 400 You pay your employees a bonus if the value of light bulbs destroyed in a month is less than 8000. Assuming independence and using the normal approximation, calculate the probability that you will pay your employees a bonus next month. (A) 0.6 (B) 0.9 (C) 0.3 (D) 0.7 (E) 0.3 Course 3: November 00-7- GO ON TO NEXT PAGE

8. Each of 00 independent lives purchase a single premium 5-year deferred whole life insurance of 0 payable at the moment of death. You are given: (i) µ = 0. 04 (ii) δ = 0. 06 (iii) F is the aggregate amount the insurer receives from the 00 lives. Using the normal approximation, calculate F such that the probability the insurer has sufficient funds to pay all claims is 0.95. (A) 80 (B) 390 (C) 500 (D) 60 (E) 70 Course 3: November 00-8- GO ON TO NEXT PAGE

9. For a select-and-ultimate table with a -year select period: x p x p x + p x+ x+ 48 0.9865 0.984 0.973 50 49 0.9858 0.983 0.9698 5 50 0.9849 0.989 0.968 5 5 0.9838 0.9803 0.9664 53 Keith and Clive are independent lives, both age 50. Keith was selected at age 45 and Clive was selected at age 50. Calculate the probability that exactly one will be alive at the end of three years. (A) Less than 0.5 (B) At least 0.5, but less than 0.5 (C) At least 0.5, but less than 0.35 (D) At least 0.35, but less than 0.45 (E) At least 0.45 Course 3: November 00-9- GO ON TO NEXT PAGE

0-. Use the following information for questions 0 and. For a tyrannosaur with 0,000 calories stored: (i) (ii) (iii) (iv) The tyrannosaur uses calories uniformly at a rate of 0,000 per day. If his stored calories reach 0, he dies. The tyrannosaur eats scientists (0,000 calories each) at a Poisson rate of per day. The tyrannosaur eats only scientists. The tyrannosaur can store calories without limit until needed. 0. Calculate the probability that the tyrannosaur dies within the next.5 days. (A) 0.30 (B) 0.40 (C) 0.50 (D) 0.60 (E) 0.70 Course 3: November 00-0- GO ON TO NEXT PAGE

0-. (Repeated for convenience) Use the following information for questions 0 and. For a tyrannosaur with 0,000 calories stored: (i) (ii) (iii) (iv) The tyrannosaur uses calories uniformly at a rate of 0,000 per day. If his stored calories reach 0, he dies. The tyrannosaur eats scientists (0,000 calories each) at a Poisson rate of per day. The tyrannosaur eats only scientists. The tyrannosaur can store calories without limit until needed.. Calculate the expected calories eaten in the next.5 days. (A) 7,800 (B) 8,800 (C) 9,800 (D) 0,800 (E),800 Course 3: November 00 -- GO ON TO NEXT PAGE

-3. Use the following information for questions and 3. A fund is established by collecting an amount P from each of 00 independent lives age 70. The fund will pay the following benefits: 0, 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: (i) Mortality follows the Illustrative Life Table. (ii) i = 0.08. Calculate P, using the equivalence principle. (A).33 (B).38 (C) 3.0 (D) 3.07 (E) 3.55 Course 3: November 00 -- GO ON TO NEXT PAGE

-3. (Repeated for convenience) Use the following information for questions and 3. A fund is established by collecting an amount P from each of 00 independent lives age 70. The fund will pay the following benefits: 0, 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: (i) Mortality follows the Illustrative Life Table. (ii) i = 0.08 3. For this question only, you are also given: The number of claims in the first year is simulated from the binomial distribution using the inverse transform method (where smaller random numbers correspond to fewer deaths). The random number for the first trial, generated using the uniform distribution on [0, ], is 0.8. Calculate the simulated claim amount. (A) 0 (B) 0 (C) 0 (D) 30 (E) 40 Course 3: November 00-3- GO ON TO NEXT PAGE

4. You are simulating a continuous surplus process, where claims occur according to a Poisson process with frequency, and severity is given by a Pareto distribution with parameters α = and θ = 000. The initial surplus is 000, and the relative security loading is 0.. Premium is collected continuously, and the process terminates if surplus is ever negative. You simulate the time between claims using the inverse transform method (where small numbers correspond to small times between claims) using the following values from the uniform distribution on [0,]: 0.83, 0.54, 0.48, 0.4. You simulate the severities of the claims using the inverse transform method (where small numbers correspond to small claim sizes) using the following values from the uniform distribution on [0,]: 0.89, 0.36, 0.70, 0.6. Calculate the simulated surplus at time. (A) 09 (B) 935 (C) 85 (D) 400 (E) Surplus becomes negative at some time in [0,]. Course 3: November 00-4- GO ON TO NEXT PAGE

5. You are given: (i) P x = 0. 090 (ii) (iii) nv x = 0563. P x: n =. 0 00864 Calculate P x: n. (A) 0.008 (B) 0.04 (C) 0.040 (D) 0.065 (E) 0.085 Course 3: November 00-5- GO ON TO NEXT PAGE

6. You are given: (i) Mortality follows De Moivre s law with ω = 00. (ii) i = 0. 05 (iii) The following annuity-certain values: a a a 40 50 60 = 758. = 8. 7 = 9. 40 Calculate 0 VcA 40 h. (A) 0.075 (B) 0.077 (C) 0.079 (D) 0.08 (E) 0.083 Course 3: November 00-6- GO ON TO NEXT PAGE

7. For a group of individuals all age x, you are given: (i) 30% are smokers and 70% are non-smokers. (ii) The constant force of mortality for smokers is 0.06. (iii) The constant force of mortality for non-smokers is 0.03. (iv) δ = 0. 08 F H Calculate Var a T x I K for an individual chosen at random from this group. (A) 3.0 (B) 3.3 (C) 3.8 (D) 4. (E) 4.6 Course 3: November 00-7- GO ON TO NEXT PAGE

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).00 (B).36 (C).45 (D).8 (E).96 Course 3: November 00-8- GO ON TO NEXT PAGE

9. A Poisson claims process has two types of claims, Type I and Type II. (i) The expected number of claims is 3000. (ii) The probability that a claim is Type I is /3. (iii) Type I claim amounts are exactly 0 each. (iv) The variance of aggregate claims is,00,000. Calculate the variance of aggregate claims with Type I claims excluded. (A),700,000 (B),800,000 (C),900,000 (D),000,000 (E),00,000 Course 3: November 00-9- GO ON TO NEXT PAGE

0. Don, age 50, 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 = 00. Decrement is leaving academic employment, with btg µ 50 = 0. 05, t 0 Calculate the probability that Don remains an actuarial science professor for at least five but less than ten years. (A) 0. (B) 0.5 (C) 0.8 (D) 0.3 (E) 0.34 Course 3: November 00-0- GO ON TO NEXT PAGE

. For a double decrement model: (i) In the single decrement table associated with cause (), = 000 are uniformly distributed over the year. (ii) In the single decrement table associated with cause (), = 05 decrements occur at time 0.7. q 40 q 40. and decrements. and all Calculate qbg 40. (A) 0.4 (B) 0.5 (C) 0.6 (D) 0.7 (E) 0.8 Course 3: November 00 -- GO ON TO NEXT PAGE

. A taxi driver provides service in city R and city S only. If the taxi driver is in city R, the probability that he has to drive passengers to city S is 0.8. If he is in city S, the probability that he has to drive passengers to city R is 0.3. The expected profit for each trip is as follows: a trip within city R:.00 a trip within city S:.0 a trip between city R and city S:.00 Calculate the long-run expected profit per trip for this taxi driver. (A).44 (B).54 (C).58 (D).70 (E).80 Course 3: November 00 -- GO ON TO NEXT PAGE

3. The Simple Insurance Company starts at time 0 with a surplus of 3. At the beginning of every year, it collects a premium of. Every year, it pays a random claim amount as shown: Claim Amount Probability of Claim Amount 0 0.5 0.5 0.40 4 0.0 If, at the end of the year, Simple s surplus is more than 3, it pays a dividend equal to the amount of surplus in excess of 3. If Simple is unable to pay its claims, or if its surplus drops to 0, it goes out of business. Simple has no administration expenses and interest is equal to 0. Calculate the probability that Simple will be in business at the end of three years. (A) 0.00 (B) 0.49 (C) 0.59 (D) 0.90 (E).00 Course 3: November 00-3- GO ON TO NEXT PAGE

4. For a special -payment whole life insurance on (80): (i) Premiums of π are paid at the beginning of years and 3. (ii) (iii) The death benefit is paid at the end of the year of death. There is a partial refund of premium feature: If (80) dies in either year or year 3, the death benefit is 000 + π. Otherwise, the death benefit is 000. (iv) Mortality follows the Illustrative Life Table. (v) i = 0.06 Calculate π, using the equivalence principle. (A) 369 (B) 38 (C) 397 (D) 409 (E) 45 Course 3: November 00-4- GO ON TO NEXT PAGE

5. For a special fully continuous whole life insurance on (65): (i) The death benefit at time t is b = 000 e, t 0. t 0. 04t (ii) Level benefit premiums are payable for life. = (iii) µ 65 t 0. 0, t 0 (iv) δ = 0. 04 Calculate V, the benefit reserve at the end of year. (A) 0 (B) 9 (C) 37 (D) 6 (E) 83 Course 3: November 00-5- GO ON TO NEXT PAGE

6. You are given: (i) A x = 0. 8 (ii) A x+ 0 = 0. 40 (iii) A x: = 0. 5 0 (iv) i = 0. 05 Calculate a x:0. (A).0 (B). (C).7 (D).0 (E).3 Course 3: November 00-6- GO ON TO NEXT PAGE

7. 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) 0. (B) 0.4 (C) 0.6 (D) 0.8 (E) 0.30 Course 3: November 00-7- GO ON TO NEXT PAGE

8. The unlimited severity distribution for claim amounts under an auto liability insurance policy is given by the cumulative distribution: 0. 0x 0. 00 x = F x 0. 8e 0. e, x 0 The insurance policy pays amounts up to a limit of 000 per claim. Calculate the expected payment under this policy for one claim. (A) 57 (B) 08 (C) 66 (D) 05 (E) 40 Course 3: November 00-8- GO ON TO NEXT PAGE

9. 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 0.0 0.80 0.00 0.00 G 0.50 0.00 0.50 0.00 H 0.75 0.00 0.00 0.5 I.00 0.00 0.00 0.00 At time 0, the machine is in State F. A salvage company will pay 500 at the end of 3 years if the machine is in State F. Assuming v = 0. 90, calculate the actuarial present value at time 0 of this payment. (A) 50 (B) 55 (C) 60 (D) 65 (E) 70 Course 3: November 00-9- GO ON TO NEXT PAGE

30. The claims department of an insurance company receives envelopes with claims for insurance coverage at a Poisson rate of λ = 50 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 0.0 0.5 3 0.40 4 0.5 Using the normal approximation, calculate the 90 th percentile of the number of claims received in 3 weeks. (A) 690 (B) 70 (C) 730 (D) 750 (E) 770 Course 3: November 00-30- GO ON TO NEXT PAGE

3. An insurer s losses are given by a Poisson process with mean per year. Severity is constant at. Annual premiums of 8 are collected. Initial surplus is 6. Which of the following best describes the insurer s future? (A) (B) (C) (D) (E) Ruin is certain within 3 years. Ruin is certain, but not necessarily within 3 years. Ruin is possible, but not certain. Ruin is possible, but can be avoided with certainty by increasing the initial surplus. The current surplus is adequate so that ruin will not occur. Course 3: November 00-3- GO ON TO NEXT PAGE

3. An actuary is evaluating two methods for simulating T(xy), the future lifetime of the joint-life status of independent lives (x) and (y): (i) Mortality for (x) and (y) follows De Moivre s law with ω = 00. (ii) 0 < x y < 00 (iii) (iv) (v) Both methods select random numbers R and R independently from the uniform distribution on [0, ]. Method sets: = b00 g = smaller of Tbxg and Tb yg (a) T x = 00 x R (b) T y y R (c) T xy Method first determines which lifetime is shorter: (a) If R 050., it chooses that x T xy = T x = ( 00 x) R. (b) If R > 050., it chooses that y T xy = T y = 00 y R. is the first to die, and sets is the first to die, and sets Which of the following is correct? (A) Method is valid for x = y but not for x < y; Method is never valid. (B) Method is valid for x = y but not for x < y; Method is valid for x = y but not for x < y. (C) Method is valid for x = y but not for x < y; Method is valid for all x and y. (D) Method is valid for all x and y; Method is never valid. (E) Method is valid for all x and y; Method is valid for x = y but not for x < y. Course 3: November 00-3- GO ON TO NEXT PAGE

33. You are given: (i) (ii) (iii) x The survival function for males is sbxg =, 0 < x < 75. 75 Female mortality follows De Moivre s law. At age 60, the female force of mortality is 60% of the male force of mortality. For two independent lives, a male age 65 and a female age 60, calculate the expected time until the second death. (A) 4.33 (B) 5.63 (C) 7.3 (D).88 (E) 3.7 Course 3: November 00-33- GO ON TO NEXT PAGE

34. For a fully continuous whole life insurance of : (i) µ = 0. 04 (ii) δ = 0. 08 (iii) L is the loss-at-issue random variable based on the benefit premium. Calculate Var (L). (A) (B) (C) (D) (E) 0 5 4 3 Course 3: November 00-34- GO ON TO NEXT PAGE

35. The random variable for a loss, X, has the following characteristics: x Eb X xg F x 0 0.0 0 00 0. 9 00 0.6 53 000.0 33 Calculate the mean excess loss for a deductible of 00. (A) 50 (B) 300 (C) 350 (D) 400 (E) 450 Course 3: November 00-35- GO ON TO NEXT PAGE

36. 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) 0 0.4 0.3 0. 3 0. (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 0% of the expected claims. All other expenses are 5% of revenues. Calculate the expected dividend. (A) 0.43 (B) 0.47 (C) 0.5 (D) 0.55 (E) 0.59 Course 3: November 00-36- GO ON TO NEXT PAGE

37. 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) 0.44 (B) 0.50 (C) 0.56 (D) 0.6 (E) 0.67 Course 3: November 00-37- GO ON TO NEXT PAGE

38. For a triple decrement model: (i) x q x q bg x q b3g x 63 0.0000 0.03000 0.5000 64 0.0500 0.03500 0.0000 65 0.03000 0.04000 0.5000 (ii) q 65 =. 0 076 (iii) Each decrement has a constant force over each year of age. Calculate 64 q. (A) 0.048 (B) 0.0448 (C) 0.0434 (D) 0.04480 (E) 0.04546 Course 3: November 00-38- GO ON TO NEXT PAGE

39. For a special 3-year deferred whole life annuity-due on (x): (i) i = 0. 04 (ii) The first annual payment is 000. (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 =05. is the curtate expectation of life for (x). (vii) k 3 k px 0.99 0.98 0.97 Calculate the annual benefit premium. (A) 65 (B) 85 (C) 305 (D) 35 (E) 345 Course 3: November 00-39- GO ON TO NEXT PAGE

40. For a special fully discrete 0-payment whole life insurance on (30) with level annual benefit premium π : (i) The death benefit is equal to 000 plus the refund, without interest, of the benefit premiums paid. (ii) A 30 = 00. (iii) 0 A 30 = 0. 088 (iv) biag 30 0 = 0 078 :. (v) a&& 30 0 = 7. 747 : Calculate π. (A) 4.9 (B) 5.0 (C) 5. (D) 5. (E) 5.3 **END OF EXAMINATION** Course 3: November 00-40- STOP

Preliminary November 00 Course 3 Exam Answer Key Exam Test Key E C 3 E 4 D 5 D 6 D 7 A 8 A 9 D 0 C B C 3 C 4 C 5 E 6 A 7 D 8 B 9 D 0 A C B 3 D 4 C 5 E 6 B 7 A 8 C 9 E 30 B 3 C 3 D 33 E 34 B 35 B 36 E 37 A 38 B 39 B 40 B Course 3: November 00-4- STOP

November, 00 Society of Actuaries Question # Key: E o z z z 5 0 0 HG KJ z0. 60. 60L. 50 d i d i e 5 : 5 = t p dt p p dt 5 + 0 5 5 0 t 40 5 5. 04t F. 04 ds 0 0. 05t = e dt + e z I e dt e e e 04 = +.. 05 = 797. + 4387. = 560. NM O QP Question # Key: C p = 5 60 + e q 60 + je q q 60 + jb 63gb q64gb q65g = b0. 89gb0. 87gb0. 85gb0. 84gb0. 83g = 0. 4589

Question # 3 Key: E 50. = a x = 0. 08 0. 04 + µ + δ µ δ = µ = δ = A A x x µ = µ + δ = 0. 5 µ = = µ + δ 3 Var a e T j = Ax A δ x = 3 4 = 5. 083 0. 006 S.D.= 5. 083 = 7. 7

Question # 4 Key: D v = 090. d = 00. A = da&& = 00. 5 = 0. 5 x x b g 5000Ax 5000vqx Benefit premium π = a&& x 5000 0. 5 5000 0. 90 0. 05 = 5 0 V x a&& = a&& 0. A x+ 0 x a&& 5 a&& x+ 0 = x+ 0 = = da&& = 00. 4 = 0. 6 x+ 0 x+ 0 4 = 455 V = 5000A π a&& = 5000 0. 6 455 4 = 80 0 x+ 0 x+ 0 b g b g

Question #5 Key: D 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 = 0.95. x x x+ E Z = vq + v p q = 0.95 0.5 + 0.95 0.75 0. = 0.379 4 d i x x x+ 4 E Z = v q + v p q = 0.95 0.5 + 0.95 0.75 0. = 0.3478 d i c h Var Z = E Z E Z = 0.3478 0.379 = 0. Question # 6 Key: D Severity after increase Severity after increase and deductible 60 0 0 0 80 80 300 00 Expected payment per loss = 0. 5 0 + 0. 5 0 + 0. 5 80 + 0. 5 00 = 75 Expected payments = Expected number of losses Expected payment per loss = 75 300 =,500

Question # 7 Key: A E (S) = E (N) E (X) = 50 00 = 0,000 Var S = E N Var X + E X Var N b = b50gb400g + d00 ib00g = 4, 00, 000 g F HG b 8, 000 0, 000 Pr S < 8, 000 = Pr Z < 4, 00, 000 = Pr Z < 0.998 6% g I KJ Question #8 Key: A Let Z be the present value random variable for one life. Let S be the present value random variable for the 00 lives. δ t µ t EbZg = 0z e e µ dt E Z 5 µ = 0 e δ + µ =. 46 b g δ + µ 5 d i F µ I = 0 e b HG δ + µ KJ F = H G I K J d i = = d i c h δ + µ 5 0.04 0.8 0 e 33. 0.6 Var Z E Z E Z E S = 33.. 46 = 5348. = 00 E Z = 46. Var S = 00 Var Z = 5348. F 4. 6 = 645. F = 8 534. 8 g

Question #9 Key: D Prob{only survives} = -Prob{both survive}-prob{neither survives} e j = p p p p 3 50 3 50 3 50 3 50 = 0.973 0.9698 0.968 0.9849 0.989 0.968 0.90 0.9363 4444 4444 34444 4444 3 = 0.4046 = 0.90 0.93630 Question # 0 Key: C 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 0,000 calories then, and will survive beyond.5. Prob (ruin) = f 0 + f f 0 = 0. 368 + 0. 368 0368. = 0. 503 0 e since fb0g = = 0368. 0! e fbg = = 0368.!

Question # Key: B Let X = expected scientists eaten. For each period, E X = E X dead Prob already dead + E X alive Prob alive Day, E X = fb0g b E X = 0 Prob dead + alive Prob alive e 0 Prob dead at end of day = = = 0. 368 0! Day, E X = 0 0. 368 + 0. 368g = 0. 63 Prob (dead at end of day ) = 0.503 [per problem 0] Day.5, E X. 5 = 0 0. 503+ 0. 5 0. 503 = 0. 49 where E X. 5 alive = 05. since only day in period. E X = E X + E X + E X. 5 = + 0. 63 + 0. 49 = 88. E 0, 000X = 8, 80 b g Question # Key: C This solution applies the equivalence principle to each life. Applying the equivalence principle to the 00 life group just multiplies both sides of the first equation by 00, producing the same result for P. 70 70 7 70 7 b0gb0. 0338g b0gb 0. 0338gb0. 0366g Pb 0. 0338gb 0. 0366g APV Prems = P = APV Benefits = 0q v + 0p q v + Pp p v P = + 08. 08. = 0. 307 + 0. 3006 + 0. 7988P 0. 6078 P = = 3. 0 0. 0 (APV above means Actuarial Present Value). + 08.

Question #3 Key: C For a binomial random variable with n = 00 and p = q = 0. 0338, simulate number of deaths: 00 i = 0: p = 0. 0344 = f 0 = F 0 Since 08. > Fb0g, continue i = : f = f 0 n p / p = b0. 0344gb00gb 0. 0338g / b0. 9668g = 075. F = F 0 + f = 0. 0344 + 075. = 0575. Since 08. > F, continue i = : f = f n / p / p = b0. 75gb 99 / gb0. 0338 / 0. 9668g = 0996. F = F + f = 0575. + 0996. = 0. 3537 Since 08. < F, number of claims =, so claim amount = 0. 70

Question #4 Key: C For the Pareto parameters given, the mean is θ α = 000,so the annual premium is 000. = 00. b g Simulating the times of claims, we use the formula t = 0. 5 lnb xg to get the times between claims to be 0.886, 0.388, 0.37, so claims occur at times 0.886,.74, and.60, so we only have to worry about the first claim. For the Pareto, F x = F HG 000 x + 000 I K J, so inverting gives x = 000 F u 000. Using this on the first value provided gives a severity of 000 089. 000 = 05. Clearly this will not cause ruin, since more than 5 of premium has been collected by time 0.886, so final surplus = initial surplus + premium losses = 000 + 00 05 = 85.

Question #5 Key: E 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 x: n x: n n x And plug in to get 0. 090 = P + 0. 00864 0. 563 P x: n x: n = 0. 085 Another approach is to think in terms of retrospective reserves. Here is one such solution: V = P P && s n x x x: n x: n = P P x = P P = e e e e x P x e x: n x: n x: n P P x: n j j j j j a&& n P x: n Ex a&& x: n a&& x: n x: n 0. 563 = 0. 090 P / 0. 00864 P x: n e = 0. 090 0. 00864 0. 563 = 0. 085 x: n j

Question #6 Key: A b g δ = ln 05. = 0. 04879 z z µ ω x δt A = p t e dt x t x x 0 ω x = x e 0 ω = x a ω x ω δt dt for DeMoivre From here, many formulas for 0 Vc A 40 h could be used. One approach is: Since so P A A A 50 40 c 40h a 50 8. 7 A50 = = = 0. 374 so a50 =. 83 50 50 δ a 60 9. 40 A40 = = = 0. 333 so a40 = 387. 60 60 δ 0. 333 = = 0. 033 387. V A = A P A a = 0374. 0. 033. 83 = 0. 075. 0 40 50 40 50 F HG F HG I = KJ I = KJ c h c h

Question #7 Key: D A = E v = E v NS Prob NS + E v S Prob S x T x T x T x F F 0. 03 I = HG + K J 0. 70 + 0. 03 0. 08 = 0395. F Similarly, 0. 03 A x I = 0 70 HG 0 03+ 06 K J. +.. F H Var a T x I K = x Ax A δ 093. 0. 395 = 0. 08 0. 6 I HG + K J 030. 0. 06 0. 08 F 0. 06 I HG 0 06 + 06 K J =.. = 4.. 0. 30 093.. Question #8 Key: B Let S denote aggregate losses before deductible. E S = = 4, since mean severity is. 0 = = e f S 0 0! f Sbg F = H G I K J F H G I K J = e! 0353., since must have 0 number to get aggregate losses = 0. 3 0. 090, since must have loss whose size is to get aggregate losses =. S S S S E S = 0 f 0 + f + f 0 f b g = 0 0353. + 0. 090 + 0353. 0. 090 = 639. E S = E S E S + = 4 639. =. 3608 c b h g

Question #9 Key: D Poisson processes are separable. The aggregate claims process is therefore equivalent to two independent F processes, one for Type I claims with expected frequency b3000g 000 and 3 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 = 0). Since X = 0, a constant, E X = 0; Var X = 0. I I I b Ig b Ig b Ig b Ig b Ig Var S = E N Var X + Var N E X b000gb0g b000gb0g = + = 00, 000 Var S = Var S + Var S since independent,00,000 = 00,000 + Var S Var S b Ig b IIg b IIg b IIg =, 000, 000 HG I K J =

Question #0 Key: A τ 5 p50 = 5p50 5 p50 = F HG b 00 55 00 50 I K J 0. 05 5 = 09. 0. 7788 = 0. 7009 Similarly τ 00 60 0 p50 = F 00 50 e g b g b0. 05gb0g HG b I K J e g = 0. 8 0. 6065 = 0. 485 bτg bτg bτg q 5 5 50 = 5p50 0p50 = 0. 7009 0. 485 = 0. 57 Question # Key: C Only decrement operates before t = 0.7 0 7q40 b0 7g q 40 b0 7gb00g 0 07 since UDD. =. =.. =. Probability of reaching t = 0.7 is -0.07 = 0.93 Decrement operates only at t = 0.7, eliminating 0.5 of those who reached 0.7 q 40 = 093. 05. = 065.

Question # Key: B Let π = long run probability of being in R at the start of a trip Let π = probability of being in S π π City R City S π City R 0. 0.8 π City S 0.3 0.7 3 0. π + 0. 3π = π 08. π = 03. π π = π 8 π + π = 3 π π 8 3 π π π 8 + = 8 = = = E(R) = expected profit per trip if in city R= 0. + 0.8 =.8 E(S) = expected profit per trip if in city S= 0.3 +. 0.7 =.44 3 8 In the long run, the profit per trip =.8 +.44 =.54

Question #3 Key: D Let states 0,,, 3 correspond to surplus of 0,,, 3. One year transition matrix T = T T 3 L N M L N M L N M = = 00. 0. 00 0. 00 0. 00 0. 0 0. 40 0. 5 05. 0. 0 0. 00 0. 40 0. 40 0. 00 0. 0 0. 00 080. O Q P O Q P 00. 0. 00 0. 00 0. 00 0. 33 09. 0. 0 0. 8 0. 8 0. 08 06. 0. 48 0. 04 0. 4 0. 05 0. 67 000. 0. 000 0. 000 0. 000 0. 408 03. 075. 0. 335 0. 38 08. 0. 084 0. 460 0. 098 0. 30 0. 080 059. 3 0 0 0 T = 0. 098 0. 30 0. 080 059. Prob of starting in 3 and ending in 0 = 0.098 0.098 = 0.90 or 0.30 + 0.080 + 0.59 = 0.90. O Q P If you anticipate how T 3 will be used, you see it is sufficient to calculate only the fourth row of it, or even only column of row 4. Alternatively, it is certainly correct, and probably shorter, to calculate 3 M T T T M T, M = 0 0 0. rather than where See problem 9 for an example of that method.

Question #4 Key: C e j 3 80 8 π π π 80 vq80 v p q + p v = 000A80 + + F F H G π 08390. I 0. 08030 0. 8390 0. 0956 + 665 75 π HG 06 K J =. + + 06 3.. 06. = + π. 74680 665. 75 π 0. 0756 π. 6754 = 665. 75 π = 397. 4 Where p = 3, 84, 54 80 3 94 365 = 0.,, 8390 Or p 80 = 0. 08030 0. 08764 = 08390. I K J

Question #5 Key: E At issue, actuarial present value (APV) of benefits = = z z t bt v t p65 µ t dt 0 65 0. 04t 0. 04t 000de ide it p65 µ 65btgdt 0 z 0 t 65 65 65 = 000 p µ t dt = 000 q = 000 I APV of premiums = π = π a HG + K J = π 65 6. 667 0. 04 0. 0 Benefit premium π = 000 / 6. 667 = 60 z z u 0 + u u 67 65 V = b v p µ + u du π a 0 b 0. 04 + u 0. 04u u z 0. 08 u 0 67 65 g b g F = 000e e p µ + u du 60 6. 667 67 65 = 000e p + u du 000 67 = 0839. q 000 = 0839. 000 = 839. 67 µ

Question #6 Key: B () a = a&& + E x: 0 x: 0 0 x Ax : 0 () a&& x : = 0 d (3) A = A + A x: 0 x: 0 x: 0 :0 0 0 (4) Ax = Ax + Ex Ax + x: 0 0. 8 = A + 0. 5 0. 40 A x: 0 = 08. Now plug into (3): A x: = 08. + 0. 5 = 0. 43 0 Now plug into (): 0. 43 a&& x : 0 = 0. 05 / 05. b g = 97. Now plug into (): a x: = 97. + 0. 5 =. 0

Question #7 Key: A E N = E E N Λ = 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. Per supplied tables: p rβ = = = 0. r+! + β 3 b g Λ Λ rβ = rβ + β = 6 + β = 3 β = and 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 θ α 4 θ = = = θα α = θα = θ e = θ α = 4 j

p p f d e e d e d e e 0 0 0 3 3 0 3 4 9 9 0 = = = = F HG I KJ = = z z z λ λ λ λ λ λ λ λ λ λ λ λ α λ λ c h! Integrate by parts; not shown /. / Γ

Question #8 Key: C Limited expected value = 000 z 000 z 000 0. 0 x 0. 00x 0. 0 x 0. 00x c Fbxghdx = d08. e + 0. e idx = d 40e 00e i = 40 + 6. 4 0 0 0 = 66.4 Question #9 Key: E M = Initial state matrix = 0 0 0 L0.0 0.80 0 0 T = One year transition matrix = M0.50 0 0.50 0 0.75 0 0 0.5 N M.00 0 0 0 O Q P cb b M T = g h M T T = g M T T T = 0. 0 0. 80 0 0 0. 44 06. 0. 40 0 0. 468 0. 35 0. 08 00. Probability of being in state F after three years = 0.468. 3 d i Actuarial present value = 0. 468v 500 = 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 #30 Key: B Let Y i be the number of claims in the ith envelope. Let Xb3g be the aggregate number of claims received in 3 weeks. E Yi = b 0. g + b 0. 5g + b3 0. 4g + b4 05. g =. 5 E Yi = b 0. g + b4 0. 5g + b9 0. 4g + b6 05. g = 7. E Xb3g = 50 3. 5 = 65 Var Xb3g = 50 3 7. = 4680 ProbmXb3g Zr = 0. 90 = Φb8. g R X 3 65 S U Prob 8. T V 4680 W Xb3g 7.7 Note: The formula for Var Xb3g 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. Question #3 Key: C 8 + θ = = 5. θ = 0. 5 θ > 0 so (C)

Question #3 Key: D No real issues with #. It matches the real survival process. # is never valid. Even if x = y, so that Prob (x dies first) = 0.50, the conditional future lifetime of x would not follow De Moivre. Question #33 Key: E µ t t µ p p M F M 65 F 60 60 = = = ω 60 75 60 5 60 3 = = = ω = 85 ω 60 5 5 5 t = 0 t = 5 Let x denote the male and y denote the female. o e = 5 mean for uniform distribution over 0,0 x o ey o e xy =. 5 = = = z z F HG 0 0 c c F HG F HG 0 0 mean for uniform distribution over 0,5 I K JF HG t t 0 5 7 t t + 50 50 3 I K J I KJ I 0 dt dt 7 t 7 000 t t + = 0 00 + 00 750KJ 0 00 750 4 = 0 7 + = 3 3 3 b b gh gh o o o o + e = e e e xy x + y xy = 5 + 5 3 30 75 6 = = 37. 3 6

Question #34 Key: B A A x x µ = µ + δ = 3 µ = = µ + δ 5 Pc A x h = µ = 0. 04 Var L F c xhi H G K J e F I F HG K J F HG H G I K J I 0 04. 0. 08 5 3 KJ P A = + δ = + F = H G I K J 3 F H G 4 I K J = 5 45 Ax Ax j Question #35 Key: B g 00 E X E X Mean excess loss = F 00 33 9 = = 300 0. 8 E X = E X 000 since F 000 = 0.

Question #36 Key: E Expected insurance benefits per factory = E X + = 0. + 0. = 0. 4. Insurance premium = (.) ( factories) (0.4 per factory) = 0.88. Let R = retained major repair costs, then f f f R R R 0 = 0. 4 = 06. = 0. 4 0. 6 = 0. 48 = 0. 6 = 0. 36 Dividend = 3 0. 88 05. 3, = 67. R, if positive E Dividend R if positive = 06. 67. 0 + 0. 48 67. + 0. 36 0 = 0. 5888 [The (0.36)(0) term in the last line represents that with probability 0.36, b67. Rg is negative so the dividend is 0.] b g Question #37 Key: A αθ E X = = 4α = 8 α α 4α = 8α 8 α = = F 4 F 6 H G θi 6 K J F = H G I α 6 K J = 0. 555 s 6 = F 6 = 0. 444

Question #38 Key: B 64 64 τ 64 65 q = q + p q bτg p 64 = 0. 05 0. 035 0. 00 = 0. 7570 q τ τ 64 = p64 = 0. 4730 = 0. 05 = 0. 975 p 64 q ln p64 64 = bτg ln p64 64 qbτ g lnb0. 975g 64 = 0. 473 = 0. 004 ln 0. 7570 = 0. 004 + 0. 7570 0. 076 = 0. 0448 q

Question #39 Key: B ex = px + px + 3px +... = 05. Annuity = v p 000 + v p 000. 04 +... = 3 4 3 x 4 k = 3 = 000v b 000 04. 3 k k = 3 g p x k 3 k v k p x x 3 3 F = = H G I 000v bex 0. 99 0. 98g 000 K J. = b g 04. 9 08 807 Let π = benefit premium. e π + 0. 99v + 0. 98v = 807 j. 8580π = 807 π = 84 Question #40 Key B && 300 : 30 30 : b 0 ge 0 30j π a = 000A + P IA + 0π A 000A30 π = a IA 0 A && 30 : 0 30 : 0 0 30 000 00. = 7. 747 0. 078 0 0. 088 = 0 6. 789 = 5. 04