November 2000 Course 3

Transcription

1 November Course 3 Society of Actuaries/Casualty Actuarial Society November - - GO ON TO NEXT PAGE

2 Questions through 36 are each worth points; questions 37 through 44 are each worth point.. 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 November - - GO ON TO NEXT PAGE

3 . 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. November GO ON TO NEXT PAGE

4 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 November GO ON TO NEXT PAGE

5 4. 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). November GO ON TO NEXT PAGE

6 5. 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) (C) (D) (E) November GO ON TO NEXT PAGE

7 6. An insurance company has two insurance portfolios. Claims in Portfolio P occur in accordance with a Poisson process with mean 3 per year. Claims in portfolio Q occur in accordance with a Poisson process with mean 5 per year. The two processes are independent. Calculate the probability that 3 claims occur in Portfolio P before 3 claims occur in Portfolio Q. (A).8 (B).33 (C).38 (D).43 (E).48 November GO ON TO NEXT PAGE

8 7. 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). November GO ON TO NEXT PAGE

9 8. The number of claims, N, made on an insurance portfolio follows the following distribution: n Pr(N=n) 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). November GO ON TO NEXT PAGE

10 9. 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 b g b g = x t 8., t b g b g = (v) µ x t., t, where µ x accident. b g is the force of decrement due to death by Calculate the single benefit premium for this insurance. (A),765 (B),95 (C),6 (D) 3,44 (E) 3,35 November - - GO ON TO NEXT PAGE

11 . 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, ,37 78,4 77, ,575 76,77 75, Assume that deaths are uniformly distributed between integral ages. Calculate. 9 q (A). (B).3 (C).4 (D).5 (E).6 November - - GO ON TO NEXT PAGE

12 . Customers arrive at a service center in accordance with a Poisson process with mean per hour. There are two servers, and the service time for each server is exponentially distributed with a mean of minutes. If both servers are busy, customers wait and are served by the first available server. Customers always wait for service, regardless of how many other customers are in line. Calculate the percent of the time, on average, when there are no customers being served. (A).8% (B) 3.9% (C) 7.% (D) 9.% (E).7% November - - GO ON TO NEXT PAGE

13 . For a special single premium -year term insurance on (7): (i) The death benefit, payable at the end of the year of death, is equal to plus the benefit reserve. (ii) q7+ t = 3., t (iii) i =. 7 Calculate the single benefit premium for this insurance. (A) 6 (B) 67 (C) 38 (D) 369 (E) 4 November GO ON TO NEXT PAGE

14 3. 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 November GO ON TO NEXT PAGE

15 4. For a double decrement table with l 4 bτg = : x qbg x qbg x q bg x.5. q bg x y y τ Calculate l 4 b g. (A) 8 (B) 8 (C) 84 (D) 86 (E) 88 November GO ON TO NEXT PAGE

16 5. You own one share of a stock. The price is 8. The stock price changes according to a standard Brownian motion process, with time measured in months. Calculate the probability that the stock reaches a price of at some time within the next 4 months. (A).668 (B).735 (C).885 (D).96 (E).336 November GO ON TO NEXT PAGE

17 6. 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. November GO ON TO NEXT PAGE

18 7. You are such a fan of a weekly television drama about a big city hospital that you begin tracking the arrival and departure of patients on the show. Over the course of a season, you determine the following: (i) (ii) (iii) (iv) Patients arrive according to a Poisson process with mean 5 per hour. The time from arrival to departure for each patient is exponentially distributed with mean 3 minutes. There is no limit to the number of doctors available to treat patients. Patients who do not depart the show by the end of an episode appear on the following week s show without a loss of time. Calculate the limiting probability that there are fewer than three patients on the show at any given time. (A).544 (B).597 (C).65 (D).74 (E).758 November GO ON TO NEXT PAGE

19 8. For a fully discrete whole life insurance of on (4), you are given: (i) (ii) (ii) The contract premium is equal to the benefit premium. Λ is the loss variable for the one-year term insurance in the eleventh contract year with a benefit equal to the amount at risk on the whole life insurance. Mortality follows the Illustrative Life Table. (iii) i =. 6 (iv) K 4 b g is the curtate-future-lifetime random variable. Calculate the conditional variance of Λ, given that Kb4g. (A) 39 (B) 4 (C) 4 (D) 4 (E) 43 November GO ON TO NEXT PAGE

20 9. A community is able to obtain plasma at the continuous rate of units per day. The daily demand for plasma is modeled by a compound Poisson process where the number of people needing plasma has mean and the number of units needed by each person is approximated by an exponential distribution with mean. Assume all plasma can be used without spoiling. At the beginning of the period there are units available. Calculate the probability that there will not be enough plasma at some time. (A). (B). (C).3 (D).4 (E).5 November - - GO ON TO NEXT PAGE

21 . Y is the present-value random variable for a special 3-year temporary life annuity-due on (x). You are given: (i) t px = 9., t 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 November - - GO ON TO NEXT PAGE

22 . 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),, November - - GO ON TO NEXT PAGE

23 . 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). November GO ON TO NEXT PAGE

24 3. 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 November GO ON TO NEXT PAGE

25 4. 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 (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 November GO ON TO NEXT PAGE

26 5. Given: (i) (ii) µ Superscripts M and N identify two forces of mortality and the curtate expectations of life calculated from them. N 5 btg = R S T (iii) e5 M =. µ µ M 5 M 5 b g b g t +. * t t btg t > Calculate e N 5. (A) 9. (B) 9.3 (C) 9.4 (D) 9.5 (E) 9.6 November GO ON TO NEXT PAGE

27 6. 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). November GO ON TO NEXT PAGE

28 7. 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 November GO ON TO NEXT PAGE

29 8. 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 November GO ON TO NEXT PAGE

30 9. 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 November GO ON TO NEXT PAGE

31 3. For independent lives (5) and (6): b g = µ x, x x < Calculate e o 5: 6. (A) 3 (B) 3 (C) 3 (D) 33 (E) 34 November GO ON TO NEXT PAGE

32 3. 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, , ,8 53, , ,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 November GO ON TO NEXT PAGE

33 3. 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. b b b b b (A) Φ 33. (B) Φ 66. g g g g g (C) Φ. 33 (D) Φ. 66 (E) Φ 333. November GO ON TO NEXT PAGE

34 33. 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, November GO ON TO NEXT PAGE

35 34. The Town Council purchases weather insurance for their annual July 4 picnic. (i) (ii) For each of the next three years, the insurance pays on July 4 if it rains on that day. Weather is modeled by a Markov chain, with two states, as follows: The probability that it rains on a day is.5 if it rained on the prior day. The probability that it rains on a day is. if it did not rain on the prior day. (iii) i =. Calculate the single benefit premium for this insurance purchased one year before the first scheduled picnic. (A) 34 (B) 4 (C) 54 (D) 7 (E) 76 November GO ON TO NEXT PAGE

36 35. 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 = 9. 9 (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 November GO ON TO NEXT PAGE

37 36. Given: x b g = + (i) µ x F e, x (ii). 4 p = 5. Calculate F. (A) -. (B) -.9 (C). (D).9 (E). November GO ON TO NEXT PAGE

38 Questions through 36 are each worth points; questions 37 through 44 are each worth point Use the following information for questions 37 and 38. A complaint department has one worker. Customers arrive according to a nonhomogeneous Poisson process with intensity function λbtg = 5 t, for t < 5. Customers always wait to be served. (i) Arrival times are simulated using the thinning algorithm with λ = 5. (ii) Random numbers are from the uniform distribution on [,]. (iii) (iv) (v) Use the following random numbers to simulate the potential arrival times, where low random numbers correspond to longer times:.,.5,.4,. Use the following random numbers to determine acceptance or rejection where low numbers indicate acceptance:.5,.94,.6,.47 Simulated service times are:.,.8, Calculate the arrival time of the third person. (A).6 (B).7 (C).9 (D). (E).3 November GO ON TO NEXT PAGE

39 37-38 (Repeated for convenience.) Use the following information for questions 37 and 38. A complaint department has one worker. Customers arrive according to a nonhomogeneous Poisson process with intensity function λbtg = 5 t, for t < 5. Customers always wait to be served. (vi) Arrival times are simulated using the thinning algorithm with λ = 5. (vii) Random numbers are from the uniform distribution on [,]. (viii) (ix) (x) Use the following random numbers to simulate the potential arrival times, where low random numbers correspond to longer times:.,.5,.4,. Use the following random numbers to determine acceptance or rejection where low numbers indicate acceptance:.5,.94,.6,.47 Simulated service times are:.,.8, Determine the status of the system at time. (A) (B) (C) (D) (E) have completed service, is being served, are waiting for service have completed service, is being served, is waiting for service has completed service, is being served, are waiting for service has completed service, is being served, is waiting for service have completed service, is being served, are waiting for service November GO ON TO NEXT PAGE

40 39-4 Use the following information for questions 39 and 4. For a new car dealer commencing business: (i) (ii) The dealer has neither assets nor liabilities. In exchange for continuous payments at the rate of expected sales, the car dealer receives cars on demand from the manufacturer. (iii) Car sales occur in accordance with a compound Poisson process at the rate of cars per month. (iv) (v) The price of each car sold is either, or 3,, with equal probability. The dealer is in a positive position when cumulative sales exceed cumulative payments to the car manufacturer. 39. Calculate the probability that the dealer will ever be in a positive position. (A). (B).5 (C).5 (D).75 (E). November GO ON TO NEXT PAGE

41 39-4 (Repeated for convenience.) Use the following information for questions 39 and 4. For a new car dealer commencing business: (vi) (vii) The dealer has neither assets nor liabilities. In exchange for continuous payments at the rate of expected sales, the car dealer receives cars on demand from the manufacturer. (viii) Car sales occur in accordance with a compound Poisson process at the rate of cars per month. (ix) (x) The price of each car sold is either, or 3,, with equal probability. The dealer is in a positive position when cumulative sales exceed cumulative payments to the car manufacturer. 4. Given that the dealer will attain a positive position, calculate the conditional expected value of the amount of the first positive position. (A), (B) 3, (C) 5, (D) 7, (E), November GO ON TO NEXT PAGE

42 4-4 Use the following information for questions 4 and 4. 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 b g = F x 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. 4. Calculate C. (A),, (B) 3,3, (C) 4,4, (D) 5,5, (E) 6,6, November GO ON TO NEXT PAGE

43 4-4 (Repeated for convenience.) Use the following information for questions 4 and 4. An insurer has excess-of-loss reinsurance on auto insurance. You are given: (vii) Total expected losses in the year are,,. (viii) In the year, individual losses have a Pareto distribution with b g = F x F I x HG x + K J, >. (ix) Reinsurance will pay the excess of each loss over 3. (x) (xi) (xii) 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. 4. Calculate C / C. (A).4 (B).5 (C).6 (D).7 (E).8 November GO ON TO NEXT PAGE

44 43-44 Use the following information for questions 43 and 44. Lucky Tom finds coins on his 6 minute walk to work at a Poisson rate of per minute. 6% of the coins are worth each; % are worth 5 each; % are worth each. The denominations of the coins found are independent. Two actuaries are simulating Tom s 6 minute walk to work. (i) (ii) The first actuary begins by simulating the number of coins found, using the procedure of repeatedly simulating the time until the next coin is found, until the length of the walk has been exceeded. For each coin found, he simulates its denomination, using the inverse transform algorithm. The expected number of random numbers he needs for one simulation of the walk is F. The second actuary uses the same algorithm for simulating the times between events. However, she first simulates finding coins worth, then simulates finding coins worth 5, then simulates finding coins worth, in each case simulating until the 6 minutes are exceeded. The expected number of random numbers she needs for one complete simulation of Tom s walk is G. 43. Which of the following statements is true? (A) (B) (C) (D) (E) Neither is a valid method for simulating the process. The first actuary s method is valid; the second actuary s is not. The second actuary s method is valid; the first actuary s is not. Both methods are valid, but they may produce different results from the same sequence of random numbers. Both methods are valid, and will produce identical results from the same sequence of random numbers. November GO ON TO NEXT PAGE

45 43-44 (Repeated for convenience). Use the following information for questions 43 and 44. Lucky Tom finds coins on his 6 minute walk to work at a Poisson rate of per minute. 6% of the coins are worth each; % are worth 5 each; % are worth each. The denominations of the coins found are independent. Two actuaries are simulating Tom s 6 minute walk to work. (i) (ii) The first actuary begins by simulating the number of coins found, using the procedure of repeatedly simulating the time until the next coin is found, until the length of the walk has been exceeded. For each coin found, he simulates its denomination, using the inverse transform algorithm. The expected number of random numbers he needs for one simulation of the walk is F. The second actuary uses the same algorithm for simulating the times between events. However, she first simulates finding coins worth, then simulates finding coins worth 5, then simulates finding coins worth, in each case simulating until the 6 minutes are exceeded. The expected number of random numbers she needs for one complete simulation of Tom s walk is G. 44. Determine which of the following ranges contains the ratio F/G. (A).<F/G.4 (B).4<F/G.8 (C).8<F/G. (D).<F/G.6 (E).6<F/G ** END OF EXAMINATION ** November STOP

46 Solutions for November Course 3 Exam Test Question: Key: B p p q x y = =. 965 = 75.. =. 95 = b gb g b. 75 xy. 75 xy = b. 75 pxgd 75. pyi p gb g = -b.965gb95. g = 97. since independent Test Question: 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 = = 89 Standard deviation (S) = 43.5 Pr(S>9.5) = Pr b g Var(N) = FS > I HG K J = Φb68. g Course 3: November Course 3 November

47 Test Question: 3 Key: A The person receives K per year guaranteed for years Ka&& = K The person receives K per years alive starting years from now &&a 4 K *Hence we have = E 4 a&& 5 K Derive E 4 : b A = A + E A E : = A 4 A 4 A 5 b g 4 5 : = = g Derive a&& 5 A5 35. = = = 69. d Plug in values:, = hk c = K K = b gb g Test Question: 4 Key: D STANDARD: e F I HG K J = = z o 5 :. t t = dt t ds. MODIFIED: p = e z = e = Difference =.847 z F HG o t e 5 : = t p5dt + p5 dt 74 z z z F HG t.. t = e dt + e dt 74. F HG = e. t + e t. 74 I K J I K J I KJ g = = b Course 3 November

48 Test Question: 5 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 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: 6 Key: A For each claim, probability it is from portfolio P = 3 8. Prob (3 from P before 3 from Q) = Prob (3 or more of first 5 are from P) d ic h c h d ic h c h d 8 5ic 8 h c h = = = 3, 768 = Test Question: 7 Key: C Since only decrements () and () occur during the year, probability of reaching the end of the year is bg b g 6 6 b gb g p p =.. 5 = 945. Probability of remaining through the year is bg b g b 3 g b gb gb g p p p =.. 5. = Probability of exiting at the end of the year is q 6 3 b g = = Course 3 November 3

49 Test Question: 8 Key: E b g =.7 b g = =. b g = b g =. 4 = 6 b g =. 7 = 4. bsg bng b Xg b Xg bng = = 6. 4 b g = 4 bsg Standard DevbSg bs > 9 4g = bs = g E N Var N E X Var X E S Var = E Var + E Var = Standard Dev S E + = = 9. 4 Since there are no possible values of S between and, Pr. Pr 3 = =. Test Question: 9 Key: D z APV of regular death benefit = e z = -.6t -.8t bgde ib. 8gd e idt =. 8 / = 764,. 7 b g - t - t b gd ib.8gde i δ µ dt 3 b gd -δ t ib gd -µ t 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. = 7937,. Total APV = = 344 Course 3 November 4

50 Test Question: Key: B l l b gb g b gb g =. 6 79, , 65 = 8, 4. b gb g b gb g =. 5 79, , 839 = 79, , q + = 8, 4. = 3. Test Question: Key: D λ i = i customers arrive at rate of /hr. µ = µ = 6 if customer is being served, customer leaves at 6/hr. µ i = i λ ipi = µ i+ Pi + 6P = P Pi + Pi i P = i i= P = 5 3 P P = P = P P = i c h 5 i 6 53 P P = P P i 3 i= = P + 5F H I K P = c h i 6 P P = = 9. 99% Course 3 November 5

51 Test Question: Key: C (Per instead of per ) vh+ V = hv + π h vqx+ h for all h; π h = for h > h+ h h h+ h + h h x+ h v V = v V + v π v q h+ h h h+ h + h h x+ h v V v V v π = v q 9 h= h+ h+ h h ( v V v V v π ) = v q h h 9 h= h+ x+ h 9 h= h+ π = v q x+ h 9 + π =. 3 v h h= =.3a =.378 π = 37.8 Test Question: 3 Key: A b g z λ λ 5 5d i 5d i P = e dλ = e = e = P e d e e 5 b g z = 5 = 6e 99 d i = 5d i λ λ λ λ λ λ =. 5 PbN g = =. 694 Course 3 November 6

52 Test Question: 4 Key: A bτg b 4 4 g b 4 g = p b g 4 p b g 4 b g q = q + q = =. 75p 4 b g b 4 g p = q =. = y b g. 4 bτg 4 q = y = 4 q l b g τ 4 b gb g b gb g = =. 39 = = 83 Test Question: 5 Key: E For standard Brownian motion: µ = Var = per month = 4 per 4 months Thus distribution of stock price at time 4 is normal with mean 8 and variance 4. Prob Price at time 4 > L NM = Prob Price-8 8 > = Φb.5g =.933 =.668 = O QP Prob Maximum price during,4 = Prob Price at 4 > = b gb.668 =.336 g > Course 3 November 7

53 Test Question: 6 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 = Since L is a decreasing function of K, to have Pr L π > < π b g.5 means we must have L b g for K 47. bπ g for K 47 is at K = bπ g at = 47 =, π && Highest value of L L K v a = π b g b g L π π π > = Test Question: 7 Key: A This is an M / M / queuing model. λ n = λ, n, µ n = nµ, n, since patients recover individually and number of doctors is unlimited. λ P = µ P λ P = µ P λ P = 3µ P3 M λ P nµ P n = n Also, P = P + P = P = l n n x e j n c h n x n n! x λ Pn = λ µ n P = P µ x = 5!! where x = so for this problem = 5. + x n n! = = e e x Pr # patients < = P + P + P q 3 = P + x + x x e = e + x + = for x =.5 x j Course 3 November 8

54 Test Question: 8 Key: C You may find equation and exercise 8.3 helpful in conjunction with this solution. The random variable Λ in question 8 is Λ in those references. For Kb4g, the loss variable Λ = R S T b 4g b g b g V v π a&&, K 4 = where π is the benefit premium for the -year term insurance. b g, but you don't need its value. π = V 4 vq 5 π a&&, K 4 > Since Var(X + c) = Var(X), where c is any constant, the π &&a term can be ignored for computing VarbΛg. The conditional probabilities, given K are Pr K = K = q ; Pr K > K = p c h 5 c h 5 c b g h = b V 4gv p5 q5 = L a&& 5 O NM e e a jj 6QP b gb g && 4... = e c hi 6. b. 5885g = 888 b gb. g 6. b. 5885g The variance of a binomial distribution with parameters q, (also called a Bernoulli distribution) is m q(-q), so Var Λ K 4 = 4, 83 Test Question: 9 Key: E L N θ u Ψbug = exp + M θ b + θg p + θ = θ =. u = p = Ψbug = e. = 5.. O Q P Course 3 November 9

55 Test Question: Key: B b g x b g x x b g x b g = =. 475 d i = =. VARbYg = = Pr K = = p =. Pr K = = p p = =. 9 Pr K > = p = 8. E Y E Y Test Question: Key: D Let X be the occurrence amount, Y = max(x-, ) be the amount paid. E[X] =, Var[X] = b, g P(X>) = exp(-/,) = 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 E Y = , = E Y = , =, 89, 675 b Var Y = 89,, = 99, 944 g b g 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,, Course 3 November

56 Test Question: Key: B P In general Var L = + A A Here P A b g b g c δ x x h c xh = = = a δ 8 x 5.. b g F c xh.. So Var L = + Ax A. 8 I HG K J = b * g F bgi = + c x x HG h. 8 KJ 4 5 b g + = 8 b g b 565 g = x x xbδ g b g VarbL g 75 and Var L A A So Var L *.. E L * = A. 5a = a +. 5 = 5. 3 =. 5 E L * + * =. Test Question: 3 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 bλ = 4g. Plug in and try d i too high, so try 3 months bλ = 6g d / i., okay. The answer is 3 months. e + 4e + 4 / e =. 38, 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]. Course 3 November

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

58 Test Question: 5 Key: D e = p + e e p N M 6 = e6 N µ 5 t +. t dt 5 = e z b M g since same µ b g c h M = e z µ 5btgdt z = e. t dt z M 5 b g µ t dt. t dt e z c h c h = p M 5 e L NM F H. t t IO KQP = e. 5 M p 5 b N N e = p + e. 5 M 5 6 = e p + e M 5 b g = 95. e = 95.. = 95. g b gb g Course 3 November 3

59 Test Question: 6 Key: D E Y = E Y =, a σ σ AGG b g x b g F I H c h K Ax =, G J =,, δ b g c h = Var Y =, A A δ b, g = b5. g 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 F = 8. σ + E Y b AGG b AGG AGG AGG g g AGG F = 8. 5, +,, =, 64, O Q P Test Question: 7 Key: C Expected claims under current distribution = 5 θ = parameter of new distribution X = claims E X b g = θ bonus =.5 5 X 5 F HG θ E (claims + bonus) = θ+.5 5 θ θ F b g b g θ θ H G 5 I + θ K J = θ θ 5θ = θ θ + θ 5θ = θ θ = 5 5 = 354 F HG I K JI = KJ Course 3 November 4

60 Test Question: 8 Key: E b g eb g DA = vq + vp DA 8: : b g b g.. 8 q 8 =. 3= + DA 89 : b DA g b6. g 4 = 89 = 5. :. 8 q 8 =. DA = v. + v : b g b gb g b = = g j Test Question: 9 Key: B Let T denote the random variable of time until the college graduate finds a job Let N t, 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 b gr Type II - - reject with probability b pg Nbtg T m r m b gr is Poisson process with λ = λ b p 8 g b 8 g F w. 4. b w 4g > I HG K J = Φbg.. 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 = = Pr T > 3 = F 3 b g b g 3 = e3.5 =.386 Course 3 November 5

61 Test Question: 3 Key: A t p x o e5: 6 L NM = exp zt ds x s O 5 ti 4 t HG 5 K J 4 t = t t b g t x x s QP = = exp ln x o o o o e5: 6 = e5 + e6 e5: 6 z o t t e5 dt L = = t N M O Q P = z o t t e6 dt L = = t N M O Q P = z o 4F F z e5: 6 = d i F HG = = dt 9t t dt I HG K J = + 3 4I 467. KJ = t Course 3 November 6

62 Test Question: 3 Key: A b gb g b gb g UDD l =. 8 53, , 384 = 46, 67. z z t o b g b g b g Mrl = e = p = areas = 44. Sb Xig Sbg. S + t dt S = Ο Ο Ο3 Ο x i Test: 3 Key: Question D µ = E N = 5 µ µ n x N E X = Var X = 675 = Var N = 5 = 5 E S = E X E N = 5 5 = 5 Var S = E N Var X + Var N E X = = 79, 375 Standard Deviation S = 79, 375 = 874. Pr S > = Pr S 5 / 874. > 5 / = Φ. 66 b g b g b g b g Course 3 November 7

63 Test Question: 33 Key: E b g = VarbΛ g + VarbΛg b g = b g = vbb V g p5q5 b, 3, 9g b. 83gb. 9968g Var L Var Λ Var Λ = = v since Var Λ 3. b g = vbb V g p5q5p5 b, 6, 539g b. 9968gb. 9gb. 9989g = = b g = = Var L Alternative solution: π =, v V = = L = Pr Pr b R S T 5 : 3, v π a&& = 6539 for K =, v π a&& = for K = 3, v π a&& = 83.5 for K > g K = = q = b g 5 5 b gb g K = = p q = =.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 = = 453, 895 [difference from the other solution is due to rounding] b g Course 3 November 8

64 Test Question: 34 Key: D After 365 days, probabilities have converged to limiting probabilities. Let p = limiting probability of rain. 5p+. p = p p = 7 b g L F Benefit premium = H G I K JF H G I K J F + H G I K JF H G I K J F + H G I K JF H G 3 I K J O NM QP = 7 Test Question: 35 Key: C Let π denote the single benefit premium. π = a&& + π A : 3 3 a&& A A a&& 35 35: 3 35: 3 π = = A A 35: 3 e = b 35: 3 j g g b = =.49 = Test Question: 36 Key: E. 4 p. 5 = = e z. 5 b g. 4 F+ e e =. 4F L N M e. 4F F H e x x j. 4 O QP I K. 8 = e e = F e dx ln. 5 =. 4F =. 4F. 68 F =. Course 3 November 9

65 Test Question: 37 Key: E b g (the end of the interval) so continue first t = log. =. 46 < 5 5 Test:. 5 < λb. 46g / λ = b5. 46g / 5 =. 9 True, so first arrival time =.46 Next t =.46 /5 log.5 =.6 Test:. 94 < λb6. g / λ =. 79 False, so continue Next t =.6 /5 log.4 =.35 Test:. 6 < λb35. g / λ =. 73 True, so second arrival time =.35 Next t =.35 /5 log. =.7 Test:. 47 < λb. 7g / λ =. 55 True, so third arrival time =.7 Test Question: 38 Key: C First person arrives at t =.46. Service time =., so departs at t =.56 Second person arrives at t =.35. Service time =.8, so departs at t = =.36 No further arrivals before t =, so has been served, is being served. Course 3 November

66 Test Question: 39 Key: E Model is dealer financial position at t = sales through t payments through t b g N t = x j + θ λ p t j= L NM b b where g b g payment rate + θ = = expected sales rate λ = rate of sales = p = average sales price = 5, No term needed for dealer s starting position; it is. = + θ λp t g b g N t x j j= So final position > surplus process <. O P Q By formula 3.5., Ψ P = model for surplus process. b g = + = = θ ( Since θ ). Course 3 November

67 Test Question: 4 Key: B See solution to #39 for definition and values of θ, λ and p. N t Dealer s final position = x j + θ λ p t b g L NM b = + θ λp t b g g b g N t x j O P Q P surplus process. Dealer s first final position > first negative for surplus process (denoted L ) pdf b g P y of L = P y p = R S T b g, y <, 5.,, y < 3,., 3, y z z, 3, E L = y dy + y dy = + = 5 5,, 5, 5,, = 3, Test Question: 4 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,. Course 3 November

68 Test Question: 4 Key: E X = 5. X F I HG 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 E X =. =, θ =. and L NM F E X 3 = H G I K J L 3 = NM 5 F HG b O QP = IO gkj QP So the fraction of the losses expected to be covered by the reinsurance is 35 = 4.. The total expected losses have increased to,5,, so C =. 4., 5, = 4, 758, 6 And C C = 4, 758, 6 4, 4, = 8. Test Question: 43 Key: D Proposition 5.3 (page 66) of Probability Models makes it clear that the two views of the process are equivalent. The simulation procedures are consistent with Ross. It is obvious that the two methods can give different results. (E.g., let the first number be so extreme that actuary finds no coins; actuary is still likely to get some worth 5 or ) Course 3 November 3

69 Test Question: 44 Key: E For F: Expected coins = 6 = Need (expected) + random numbers (because you need the one that exceeds 6 minutes) Need (expected) to determine the denominations. Thus F = + + = 4 For G: 6. + = 73 for coins worth = 5 for coins worth = 5 for coins worth F/G =.96 3 Course 3 November 4