Solutions to the May 2012 Course MLC Examination by Krzysztof Ostaszewski, http://www.krzysio.net, krzysio@krzysio.net Copyright 2012 by Krzysztof Ostaszewski All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner. Dr. Ostaszewski s manual for Course MLC is available from Actex Publications (http://www.actexmadriver.com) and the Actuarial Bookstore (http://www.actuarialbookstore.com) Exam MLC seminar at Illinois State University: http://math.illinoisstate.edu/actuary/exams/prep_courses.shtml May 2012 Course MLC Examination, Problem No. 1 For a 2-year select and ultimate mortality model, you are given: (i) q = 0.95q. x +1 x+1 (ii) l 76 = 98,153. (iii) l 77 = 96,124. Calculate l. 75 +1 A. 96,150 B. 96,780 C. 97,420 D. 98,050 E. 98,690 We first note that (i) says that 1 l x+2 = 1 l x +1 l x +2 We let x = 75 and obtain 1 96,124 l 75 +1 The solution is = 1 p l = q = 0.95q = 0.95 1 p x x +1 +1 x +1 x+1 x+1 = 1 l 77 l 75 +1 = 0.95 1 l 77 96,124 = 0.95 1 98,153. l = 96,124 75 +1 1 0.95 1 96,124 98,049.5178. 98,153 Answer D. l 76 = 0.95 1 l x+2 l x+1. May 2012 Course MLC Examination, Problem No. 2 You are given: (i) p x = 0.97. (ii) p x+1 = 0.95.
(iii) e x+1.75 = 18.5. (iv) Deaths are uniformly distributed between ages x and x +1. (v) The force of mortality is constant between ages x +1 and x + 2. Calculate e x+0.75. A. 18.6 B. 18.8 C. 19.0 D. 19.2 E. 19.4 The unusual twist in this problem is that the curtate life expectancy is calculated at fractional age. But we can just apply the standard formula for curtate life expectancy and see how things work out: e x+0.75 = 1 p x+0.75 + 2 p x+0.75 + 3 p x+0.75 +... = = 1 p x+0.75 + 1 p x+0.75 1 p x+1.75 + 1 p x+0.75 2 p x+1.75 +... = = 1 p x+0.75 1+ 1 p x+1.75 + 2 p x+1.75 +.... = 1 p x+0.75 1+ e x+1.75 Of course you could just memorize the formula e y = 1 p y 1+ e y+1 y = x + 0.75. Using the value given, we get e x+0.75 = 1 p x+0.75 1+ e x+1.75 = 1 p x+0.75 ( 1+18.5) = 19.5 1 p x+0.75. and apply it for Thus we need to calculate the probability 1 p x+0.75. Note that of the year between the age x + 0.75 and x + 1.75, a quarter of a year is between age x and age x + 1, under the UDD assumption, and the remaining three quarters are between age x + 1 and age x + 2, under the constant force assumption. We observe that p = p 1 x+0.75 0.25 x+0.75 p 0.75 x+1. UDD assumption applies CF assumption applies We use the UDD assumption to calculate 0.25 p x+0.75 : 0.25 p x+0.75 = 1 0.25 q x+0.75 = 1 0.25d x+0.75 l x+0.75 = UDD 1 0.25d x l x+0.75 = 1 d 0.25 x l x l x+0.75 l x = 1 0.25q x 0.75 p x = 1 0.25q x = 1 0.25q x = 1 0.25 1 p x 0.25 0.03 = 1 1 0.75 q UDD x 1 0.75q x 1 0.75( 1 p x ) 1 0.75 0.03 0.9923. Again, it is a good idea to just memorize the formula used in the above calculation: p = 1 sq x. s x+t UDD 1 tq x Now we will use the constant force assumption to calculate 0.75 p x+1. Let us denote that constant force between ages x + 1 and x + 2 by µ and observe that p x+1 = 0.95 = e µ, so that µ = ln0.95. Based on this: p = 0.75 x+1 e 0.75µ 0.75 ln0.95 = e = 0.95 0.75 0.9623. =
Combining these results we obtain e x+0.75 = 19.5 1 p x+0.75 = 19.5 0.25 p x+0.75 0.75 p x+1 19.5 0.9923 0.9623 18.6201. Answer A. May 2012 Course MLC Examination, Problem No. 3 For a fully discrete 3-year term insurance of 10,000 on (40), you are given: (i) µ 40+t, t 0, is a force of mortality consistent with the Illustrative Life Table. (ii) µ 40+t + 0.02, t = 0, 1, 2, is the force of mortality for this insured. (iii) i = 0.06. Calculate the annual benefit premium for this insurance. A. 196 B. 214 C. 232 D. 250 E. 268 Let us denote probabilities referring to the Illustrative Life Table by a superscript (ILT), and probabilities referring to this insured by a superscript (I). We have ds t µ ( ILT) p ( I) 40+s +0.02 t 40 = e 0 Therefore, ( I) = 1+ vp 40 + v 2 ( I) ( I) p p41 40 = and I a 40:3 ( ILT) = t p 40 e 0.02t. = 1+1.06 1 ( ILT) p 40 e 0.02 +1.06 2 ( ILT) p 40 e 0.02 ( ILT) p 41 e 0.02 = = 1+1.06 1 ( ILT) ( 1 q 40 ) e 0.02 +1.06 2 ( ILT) ( ILT) ( 1 q 40 ) ( 1 q 41 ) e 0.04 = = 1+1.06 1 ( 1 0.00278) e 0.02 +1.06 2 ( 1 0.00278) ( 1 0.00298) e 0.04 ( I) A 40:3 Furthermore, Therefore, 2.7723 ( I) = 1 d a 40:3 I E ( I) p 3 40 = 40 1 I A 40:3 ( I) p41 ( I) p42 1 0.06 2.7723 0.8431. 1.06 ( ILT) ( ILT) p41 ( ILT) p42 e 0.06 = p 40 = 1.06 3 1.06 3 = 1 q ( ILT) ( ILT) ( ILT) ( 40 ) ( 1 q 41 ) ( 1 q 42 ) e 0.06 = 1.06 3 ( 1 0.00278) ( 1 0.00298) ( 1 0.00320) e 0.06 = 0.7837. 1.06 3 = ( I) A40:3 The premium sought is ( I) 3 E 40 0.8431 0.7837 = 0.0594.
Answer B. 1 ( I) 10,000 A 40:3 I a 40:3 10,000 0.0594 2.7723 214.3183. May 2012 Course MLC Examination, Problem No. 4 For a 3-year term insurance of 1000 on [75], you are given: (i) Death benefits are payable at the end of the year of death. (ii) Level premiums are payable at the beginning of each quarter. (iii) Mortality follows a select and ultimate life table with a two-year select period: x l l l x + 2 x x +1 x+2 75 15,930 15,668 15,286 77 76 15,508 15,224 14,816 78 77 15,050 14,744 14,310 79 (iv) Deaths are uniformly distributed over each year of age. (v) i = 0.06. Calculate the amount of each quarterly benefit premium. A. 5.3 B. 5.5 C. 5.7 D. 5.9 E. 6.1 The single benefit premium is 1 A 75 :3 = vq 75 + v 2 p 75 q + 75 +1 v3 p 75 p q = 75 +1 77 = 1 15668 15930 + 1 1.06 15930 1.06 15668 15286 15668 + 2 15930 15668 + 1 1.06 15668 2 15930 15286 14816 15286 15668 15286 = 1 1.06 262 15930 + 1 1.06 362 2 15930 + 1 1.06 470 2 15930 0.0620. The annuity factor is a = 1+ vp 75 + v 2 p :3 75 75 p = 1+ 1 75 +1 1.06 15668 15930 + 1 1.06 15668 2 15930 15286 15668 2.7819. Furthermore, 3E 75 [ ] [ ]+3 l 75 p [ ] = 3 75 1.06 = l 75 3 [ ] 1 1.06 = l 78 3 l 75 [ ] 1 1.06 = 14816 3 15930 1 1.06 0.7809. 3 = 1.00027 and β ( 4) = 0.38424 from the Illustrative Life We obtain the values α 4 Table, although we should note that we can use the Table because the interest rate used, i = 6%, is the one used in the Table as well. Otherwise we would have to struggle with these formulas: α ( 4) = id and β 4 i ( 4) d 4 = i i 4 i 4 d 4. Based on the UDD assumption, we
calculate 4 a 75 [ ]:5 β ( 4) 1 3 E 75 2.6985. = a UDD [ 75]:5 α 4 [ ] 2.7819 1.00027 0.38424 1 0.7809 The annual level benefit premium for the policy (remember that this policy has 1000 benefit, not unit benefit) is 1 1000 A [ 75]:3 1000 0.0620 ( 4) = 22.84. a [ 75]:5 2.6985 But the question asks about the quarterly premium, which is a quarter of this, i.e., 5.71. Answer C. May 2012 Course MLC Examination, Problem No. 5 For a whole life insurance on (80): (i) Level premiums of 900 are payable at the start of each year. (ii) Commission expenses are 50% of premium in the first year and 10% of premium in subsequent years, payable at the start of the year. (iii) Maintenance expenses are 3 per month, payable at the start of each month. (iv) δ = 0.0488. (v) a 80 = 6.000. (vi) µ 80 = 0.033. Using the first 3 terms of Woolhouse s formula, calculate the expected value of the present-value-of-expenses random variable at age 80. A. 920 B. 940 C. 1010 D. 1100 E. 1120 Recall that the Woolhouse's formula with three terms is ( m a ) x a x m 1 2m m2 1 ( δ + µ 12m 2 x ). You must memorize this formula for the test. While the premium is payable annually, the maintenance expenses are paid monthly, and we will use this Woolhouse s formula for the calculation of their present value. For the maintenance expenses, m = 12, a x = a 80 = 6.000, µ 80 = 0.033, δ = 0.0488, so that ( 12 a ) 12 1 80 6 2 12 122 1 ( 0.0488 + 0.033) 5.5349. 12 12 2 The first year commission is 50% of 900, i.e., 450. The renewal commissions are 10% of 900, i.e., 90 each. Thus the expected present value of expenses as calculated at age 80 is ( 12) 450 + 90a 80 + 3 12 a 80 = Answer D. Actuarial present value of commissions Actuarial present value of maintenance expenses ( 12 = 360 + 90 a 80 + 36 a ) 80 = 360 + 540 + 36 5.5349 1099.
May 2012 Course MLC Examination, Problem No. 6 Two whales, Hannibal and Jack, occupy the Ocean World aquarium. Both are currently age 6 with independent future lifetimes. Hannibal arrived at Ocean World at age 4 and Jack at age 6. The standard mortality for whales is as follows: x q x 6 0.06 7 0.07 8 0.08 9 0.09 Whales arriving at Ocean World have mortality according to the following: 150% of standard mortality in year 1 130% of standard mortality in year 2 110% of standard mortality in year 3 100% of standard mortality in years 4 and later. As whales can get lonely on their own, a fund is set up today to provide 100,000 to replace a whale at the end of 3 years if exactly one survives. You are given that i = 0.06. Calculate the initial value of the fund using the equivalence principle. A. 24,500 B. 29,200 C. 29,300 D. 30,900 E. 31,900 Hannibal arrived at Ocean World at age 4, and is now 6, thus Hannibal s future mortality rates are q 6 H = 1.1 0.06 = 0.066, q 7 H = 1 0.07 = 0.07, q 8 H = 1 0.08 = 0.08. Jack arrived at Ocean World at age 6, and is now 6, thus Jack s future mortality rates are q 6 J = 1.5 0.06 = 0.09, q 7 J = 1.30 0.07 = 0.091, q 8 J = 1.5 0.08 = 0.088. Based on this, p 6 H = 0.934, p 7 H = 0.93, p 8 H = 0.92, p 6 J = 0.91, p 7 J = 0.909, q 8 J = 0.912, p H 3 6 = p H 6 p H 7 p H 8 = 0.934 0.93 0.92 0.7991304, p J 3 6 = p J 6 p J 7 p J 8 = 0.91 0.909 0.912 0.75439728, and, thanks to independence of lives, p = p H 3 6:6 3 6 3 p J 6 0.7991304 0.75439728 0.6028618. The probability of exactly one surviving three years, given that the lifetimes are independent, is p p = p H 3 6:6 3 6:6 3 6 + 3 p J 6 3 p 6:6 p = p H 3 6:6 3 6 + 3 p J 6 2 3 p 6:6 = 3 p 6:6 by Poohsticks 0.7991304 + 0.75439728 2 0.6028618 0.34780408. The initial value of the fund using the equivalence principle is simply the actuarial present value of the payment of 100,000 at time 3, i.e., 100,000 0.34780408 1.06 3 29,202.30. Answer B.
May 2012 Course MLC Examination, Problem No. 7 For a single premium product replacement insurance on a computer, you are given: (i) The insurance pays a percentage of the cost of a new computer at the end of the year of failure within the first 5 years. (ii) The insurance will pay only once. (iii) The insurance is bought at the time of the purchase of the computer. (iv) The cost of a new computer increases by 2% each year, compounded annually. The insurance pays based on the increased cost. (v) The insurance benefit and probabilities of failure are as follows: Year of failure k Percentage of cost of new computer covered Probability of computer failure k 1 q 0 1 100% 0.01 2 80% 0.01 3 60% 0.02 4 40% 0.02 5 20% 0.02 (vi) The commission is c% of the gross premium. (vii) Other expenses are 5% of the gross premium. (viii) The gross premium, calculated using the equivalence principle, is 10% of the original cost of the computer. (ix) i = 0.04. Calculate c%. A. 40% B. 45% C. 50% D. 55% E. 60% Note that the last column, the probability of computer failure, is the probability expressed as a fraction of the initial computer population at policy inception (because this is deferred probability). Let us write G for the gross single premium for this policy. Let us also write c for the commission, written as a fraction (instead of c%, we will switch back to percentages at the end of the solution). Let us also write K for the current cost of the computer. We use the equivalence principle and equate, at time 0, the actuarial present value of the premium (left-hand side) with the actuarial present value of benefits and expenses (right-hand side): G = 0.1K = ( 0.05 + c)g + K 1.02 0.01 + 0.8 K 1.022 0.01 + 1.04 1.04 2 + 0.6 K 1.023 0.02 + 0.4 K 1.024 0.02 + 0.2 K 1.025 0.02. 1.04 3 1.04 4 1.04 5 This means that K 1.02 0.01 0.1K = ( 0.05 + c) 0.1K + + 0.8 K 1.022 0.01 + 1.04 1.04 2 + 0.6 K 1.023 0.02 + 0.4 K 1.024 0.02 + 0.2 K 1.025 0.02 1.04 3 1.04 4 1.04 5 We divide both sides by K and solve for c:
c = 0.1 0.005 0.1 0.6 1.023 0.02 1.04 3 0.1 Answer D. 1.02 0.01 1.04 0.1 0.8 1.022 0.01 1.04 2 0.1 0.4 1.024 0.02 1.04 4 0.1 0.2 1.025 0.02 1.04 5 0.1 0.55144085. May 2012 Course MLC Examination, Problem No. 8 For a fully discrete whole life insurance of 1000 on (80): (i) i = 0.06. (ii) a 80 = 5.89. (iii) a 90 = 3.65. (iv) q 80 = 0.077. Calculate 10 V FTP, the full preliminary term reserve for this policy at the end of year 10. A. 340 B. 350 C. 360 D. 370 E. 380 Note that 5.89 = a 80 = 1+1.06 1 p 80 a 81 = 1+ Therefore, 1.06 a 81 = 4.89 1 0.077 We have and = 4.89 1.06 0.923 1000α FPT = 1000 0.077 1.06 = 77 1.06, 1 0.077 1.06 5.6158. 1000β FPT = 1000 P 81 = 1000 A 81 = 1000 1 d a 1 0.06 81 = 1000 1.06 a 81 = a 81 a 81 a 81 1 0.06 4.89 1.06 = 1000 1.06 0.923 121.4647. 4.89 1.06 0.923 We do not really need the value of α FPT but calculating it is a good exercise. We then conclude that Answer B. a 81. 10V FTP = 1000 A 90 1000β FPT a 90 = 1000 1 d a 90 1000β FPT a 90 = = 1000 1 0.06 1.06 3.65 121.4647 3.65 350.0502.
May 2012 Course MLC Examination, Problem No. 9 A special fully discrete 3-year endowment insurance on (x) pays death benefits as follows: Year of Death Death Benefit 1 30,000 2 40,000 3 50,000 You are given: (i) The maturity benefit is 50,000. (ii) Annual benefit premiums increase at 10% per year, compounded annually. (iii) i = 0.05. (iv) q x = 0.09, q x+1 = 0.12, q x+2 = 0.15. Calculate 2 V, the benefit reserve at the end of year 2. A. 28,000 B. 29,000 C. 30,000 D. 31,000 E. 32,000 Let us write P for the first year premium. The benefit reserve at the end of year 2 is the actuarial present value of future benefits minus the actuarial present value of future premiums, i.e., 2 V = 50000 1.05 1 P 1.1 2. We do not know the value of P but we can find it. The use of the expressions benefit reserve and benefit premiums implies that we are using the equivalence principle. Therefore, the actuarial present value of future benefits is equal to the actuarial present value of future premiums at policy inception, resulting in: 30,000 q x 1.05 1 + 40,000 p x q x+1 1.05 2 + 50,000 p x p x+1 1.05 3 = = P +1.1 P p x 1.05 1 +1.1 2 P p x p x+1 1.05 2. Therefore, P = 30,000 q x 1.05 1 + 40,000 p x q x+1 1.05 2 + 50,000 p x p x+1 1.05 3 1+1.1 p x 1.05 1 +1.1 2 p x p x+1 1.05 2 = = 30,000 0.09 1.05 1 + 40,000 0.91 0.12 1.05 2 + 50,000 0.91 0.88 1.05 3 1+ 0.91 1.05 1 + 0.91 0.88 1.05 2 14,519.1584. We conclude that 2 V = 50000 1.05 1 P 1.1 2 50000 1.05 1 14,519.1584 1.1 2 30,050.8660. Answer C. May 2012 Course MLC Examination, Problem No. 10 For a universal life insurance policy on Julie, you are given: (i) At t = 0, Julie will pay a premium of 100. (ii) At t =1, Julie will pay a premium of 50 with a probability of 0.8 or will pay nothing otherwise. (iii) At t = 2, Julie will pay a premium of 50 with probability of 0.9 if she paid the
premium at t = 1 or will pay nothing otherwise. (iv) During the first three years, Julie will pay no other premiums. (v) i = 0.04. (vi) Assume no deaths or surrenders in the first three years. Calculate the standard deviation of the present value at issue of premiums paid during the first three years. A. 30 B. 32 C. 34 D. 36 E. 38 Let us write Y for the random present value at issue of premiums paid during the first three years. The premiums can be paid only at time 0, or only at times 0 and 1, or only at times 0, 1, and 2. Y is a discrete random variable with the following description: Values of Y Circumstances producing Probability of the value the value 100 No premium paid at time t = 1 100 + 50 1.04 148.08 Premium paid at time t = 1 but not paid at time t = 2 100 + 50 1.04 + 50 1.04 194.31 Premiums paid at all 2 possible times: t = 0, 1, 2 Based on this, and = 100 0.20 + 100 + 50 E Y E Y 2 = 100 2 0.20 + 100 + 50 σ Y = Answer E. 30,938.80, Var( Y ) = E Y 2 0.20 0.80 0.10 = 0.08 0.80 0.90 = 0.72 1.04 0.08 + 100 + 50 1.04 + 50 1.04 2 0.72 171.7496, 1.04 2 0.08 + 100 + 50 1.04 + 50 1.04 2 2 0.72 ( E( Y )) 2 30938.80 171.7496 2 37.96. May 2012 Course MLC Examination, Problem No. 11 For a universal life insurance policy with death benefit of 10,000 plus account value, you are given: (i) Policy Monthly Percent of Cost of Monthly Surrender Year Premium Premium Insurance Rate Expense Charge Charge Per Month Charge 1 100 30% 0.001 5 300 2 100 10% 0.002 5 100 (ii) The credited interest rate is i 12) = 0.048. (iii) The actual cash surrender value at the end of month 11 is 1000.
(iv) The policy remains in force for months 12 and 13, with the monthly premiums of 100 being paid at the start of each month. Calculate the cash surrender value at the end of month 13. A. 1130 B. 1230 C. 1330 D. 1400 E. 1460 The credited interest rate is 0.004 per month. We are given that the actual cash surrender value at the end of month 11 is 1000, and that the surrender charge during the first year is 300, so that we can conclude that the account value at the end of month 11 is the sum of these two values, 1000 + 300 = 1300. The cost of insurance in month 12 is 0.001 of 10000, i.e., 10, if paid at the end of the month. Therefore, the account value at the end of the 12-th month is AV 12 months ( 1300 +100 ( 1 0.3) 5) 1.004 10 = 1360.46. The cost of insurance in month 13 is 0.002 of 10,000, i.e., 20, if paid at the end of the month. Therefore, the account value at the end of the 13-th month is AV 13 months 1360.46.+100 1 0.1 ( 5) 1.004 20 = 1,431.24. The cash surrender value is the account value after subtraction of the surrender charge of 100, i.e., 1,331.24. Note that while we subtracted the cost of insurance in both months, there were no subtractions for death benefits in either, because we know that the policy remained in force till the end of the 13-th month. Answer C. May 2012 Course MLC Examination, Problem No. 12 Employees in Company ABC can be in: State 0: Non-executive employee State 1: Executive employee State 2: Terminated from employment John joins Company ABC as a non-executive employee at age 30. You are given: (i) µ 01 = 0.01 for all years of service. (ii) µ 02 = 0.006 for all years of service. (iii) µ 12 = 0.002 for all years of service. (iv) Executive employees never return to the non-executive employee state. (v) Employees terminated from employment never get rehired. (vi) The probability that John lives to age 65 is 0.9, regardless of state. Calculate the probability that John will be an executive employee of Company ABC at age 65. A. 0.232 B. 0.245 C. 0.258 D. 0.271 E. 0.284 Let us consider survival in terms of the years of future service. In order to be an executive employee at age 65, John, who is 30 years old now, has to survive 35 years as an
employee, and at some point between now and the end of those 35 years has to become an executive employee and not get terminated for the remaining period during the 35 years. Note that he also has to live for those 35 years, i.e., be alive at age 65, but we are told that the probability of that is 0.9 and is independent of the state he is in at age 65, so we can simply calculate the probability of him being an executive employee at age 65 assuming he is alive, and then multiply that by 0.9, the probability that he will be alive, to get the desired answer. In formula terms, we are simply looking for 35 0.9 35 p 01 00 0 = 0.9 t p µ 01 0 11 t 35 t p 0 dt = 0 Survive as a non-executive employee for t years Become an executive employee at time t Survive as a executive employee for the rest of 35 years Answer A. t ds 35 35 µ 01 02 s +µ s µ 12 s ds 35 0 = 0.9 e µ 01 t t e dt = 0.9 e µ s 01 02 ( +µ s )t µ 01 t e µ 12( s 35 t ) dt = 0 35 = 0.9 e ( 0.01+0.006)t 0.01 e 0.002( 35 t) dt = 0.009e 0.07 e 0.014t dt = 0 0 ( = 0.009e 0.07 1 e 0.49 ) 0.2320. 0.014 0 35 =a 35 δ =14% May 2012 Course MLC Examination, Problem No. 13 Lorie s Lorries rents lavender limousines. On January 1 of each year they purchase 30 limousines for their existing fleet; of these, 20 are new and 10 are one-year old. Vehicles are retired according to the following 2-year select-and-ultimate table, where selection is age at purchase: Limousine age (x) q [ x] q [ x]+1 q x+2 x + 2 0 0.100 0.167 0.333 2 1 0.100 0.333 0.500 3 2 0.150 0.400 1.000 4 3 0.250 0.750 1.000 5 4 0.500 1.000 1.000 6 5 1.000 1.000 1.000 7 Lorie s Lorries has rented lavender limousines for the past ten years and has always purchased its limousines on the above schedule. Calculate the expected number of limousines in the Lorie s Lorries fleet immediately after the purchase of this year s limousines. A. 93 B. 94 C. 95 D. 96 E. 97 For the limousines that were purchased new: - Expected number of limousines just purchased (this number is certain) = 20.
- Expected number of limousines purchased one year ago = = 20 0.9 = 18. = 20 1 p [ 0] = 20 p [ 0] = 20 1 q [ 0] - Expected number of limousines purchased two years ago = ( 1 q 0 ) = 20 0.9 0.833 = 14.994. = 20 2 p [ 0] = 20 p [ 0] p [ 0]+1 = 20 1 q [ 0] [ ]+1 - Expected number of limousines purchased three years ago = = 20 3 p [ 0] = 20 p [ 0] p [ 0]+1 p [ 0]+2 = 20 1 q 0 ( [ ] ) 1 q 0 ( [ ]+1 ) 1 q 2 = = 20 0.9 0.833 0.667 10.0010. - Expected number of limousines purchased four years ago = = 20 4 p [ 0] = 20 p [ 0] p [ 0]+1 p [ 0]+2 p [ 0]+3 = 20 1 q 0 ( [ ] ) 1 q 0 ( [ ]+1 ) 1 q 2 = 20 0.9 0.833 0.667 0.5 5.0005. - Expected number of limousines purchased five years ago = = 20 5 p [ 0] = 20 p [ 0] p [ 0]+1 p [ 0]+2 p [ 0]+3 p [ 0]+4 = ( 1 q [ 0]+1 ) 1 q 2 = 20 1 q [ 0] ( 1 q 3 )( 1 q 4 ) = ( 1 q 3 ) = = 20 0.9 0.833 0.667 0.5 0 = 0. We also see that limousines purchased new any further back in the past are already out of service. For the limousines that were purchased used: - Expected number of limousines just purchased (this number is certain) = 10. - Expected number of limousines purchased one year ago = = 10 0.9 = 9. = 10 1 p [ 1] = 10 p [ 1] = 10 1 q [ 1] - Expected number of limousines purchased two years ago = ( 1 q 1 ) = 10 0.9 0.667 = 6.003. = 10 2 p [ 1] = 10 p [ 1] p [ 1]+1 = 10 1 q [ 1] [ ]+1 - Expected number of limousines purchased three years ago = = 10 3 p [ 1] = 10 p [ 1] p [ 1]+1 p [ 1]+2 = 10 1 q 1 ( [ ] ) 1 q 1 ( [ ]+1 ) 1 q 3 = = 10 0.9 0.667 0.5 = 3.0015. - Expected number of limousines purchased four years ago = = 10 4 p [ 1] = 10 p [ 1] p [ 1]+1 p [ 1]+2 p [ 1]+3 = 10 1 q 1 ( [ ] ) 1 q 1 ( [ ]+1 ) 1 q 3 ( 1 q 4 ) = = 10 0.9 0.667 0.5 0 = 0. We also see that limousines purchased used any further back in the past are already out of service. We conclude that the expected number of limousines in the Lorie s Lorries fleet immediately after the purchase of this year s limousines is: 20 + 18 + 14.994 + 10.001 + 5.0005 + 0 + 10 + 9 + 6.003 + 3.0015 + 0 96. Answer D. May 2012 Course MLC Examination, Problem No. 14 XYZ Insurance Company sells a one-year term insurance product with a gross premium equal to 125% of the benefit premium. You are given: (i) The death benefit is payable at the end of the year of death, and the amount depends
on whether the cause of death is accidental or not. (ii) Death benefit amounts, together with the associated one-year probabilities of death, are as follows: Cause of Death Death Benefit Probabilities of Death Non-accidental 5 0.10 Accidental 10 0.01 (iii) i = 0%. Using the normal approximation without continuity correction, calculate the smallest number of policies that XYZ must sell so that the amount of gross premium equals or exceeds the 95th percentile of the distribution of the total present value of death benefits. A. 378 B. 431 C. 484 D. 537 E. 590 Let n be the number of policies that is sought in this problem. Let X i denote the claim amount for policy i, with i = 1, 2,, n. Let S = X i be the aggregate claim. We have = 5 0.1+10 0.01 = 0.6, E X i = 5 2 0.1+10 2 0.01 = 3.5, 2 = E( X i ) E( X i ) 2 E X i 2 Var X i for every i = 1, 2,, n. Therefore, E S = ne( X i ) = 0.6n, = nvar( X i ) = 3.14n. n i=1 = 3.5 0.6 2 = 3.14 Var S Since the 95-th percentile of the standard normal distribution is 1.645, using the normal approximation without continuity correction for S gives its 95-th percentile as 0.6n +1.645 3.14n 0.6n + 2.9149 n. The benefit premium for one policy is 0.6 and for all policies combined it is 0.6n. The gross premium for one policy is 1.25 0.6 = 0.75, and for all policies combined it is 0.75n. We want 0.75n 0.6n + 2.9149 n, or n 2.9149 0.15 19.4330, equivalent to 2 n 2.9149 0.15 377.6408. Since the answer has to be a whole number, n 378. Answer A. May 2012 Course MLC Examination, Problem No. 15 For a 20-year temporary life annuity-due of 100 per year on (65), you are given:
(i) µ x = 0.001x, x 65. (ii) i = 0.05. (iii) Y is the present-value random variable for this annuity. Calculate the probability that Y is less than 1000. A. 0.54 B. 0.57 C. 0.61 D. 0.64 E. 0.67 We are looking for the probability = Pr 100 a K65 +1 < 1000 Pr Y < 1000 = Pr a K65 +1 < 10 If we set up an equation a k = 10, we obtain ln 1 0.5 1.05 k = 13.2532. ln1.05. is This means that the largest integer value of k = K 65 +1 such that a K65 +1 < 10 k = K 65 +1 = 13. Therefore, Pr( Y < 1000) = Pr a < 10 K65 +1 = Pr K 65 +1 13 = Pr( K 65 12) = Pr T 65 < 13. We can calculate that probability from the given formula for the force of mortality Answer C. Pr( T 65 < 13) = 13 q 65 = 1 13 p 65 = 1 e 78 µ x dx = 1 e 0.0005 782 +0.0005 65 2 0.6052. 65 = 1 e 78 0.001x dx 65 x=78 = 1 e 0.0005 x2 x=65 = May 2012 Course MLC Examination, Problem No. 16 For a special continuous joint life annuity on (x) and (y), you are given: (i) The annuity payments are 25,000 per year while both are alive and 15,000 per year when only one is alive. (ii) The annuity also pays a death benefit of 30,000 upon the first death. (iii) i = 0.06. (iv) a xy = 8. (v) a xy = 10. Calculate the actuarial present value of this special annuity. A. 239,000 B. 246,000 C. 287,000 D. 354,000 E. 366,000 The annuity portion only of this special continuous joint life annuity on (x) and (y) can be thought of as a continuous payment of 10,000 per year until the first death plus a continuous payment of 15,000 per year until the second death. Therefore, the present value of the annuity portion only is
10,000a xy +15,000a xy = 10,000 8 +15,000 10 = 230,000. But the special annuity also includes a death benefit of 30,000A xy = 30,000( 1 δ a xy ) = 30,000 1 ( ln1.06) 8 The total of the two is 230,000 + 16,015.4621 = 246,015.4621. Answer B. 16,015.4621. May 2012 Course MLC Examination, Problem No. 17 Your company issues fully discrete whole life policies to a group of lives age 40. For each policy, you are given: (i) The death benefit is 50,000. (ii) Assumed mortality and interest are the Illustrative Life Table at 6%. (iii) Annual gross premium equals 125% of benefit premium. (iv) Assumed expenses are 5% of gross premium, payable at the beginning of each year, and 300 to process each death claim, payable at the end of the year of death. (v) Profits are based on gross premium reserves. During year 11, actual experience is as follows: (a) There are 1000 lives inforce at the beginning of the year. (b) There are five deaths. (c) Interest earned equals 6%. (d) Expenses equal 6% of gross premium and 100 to process each death claim. For year 11, you calculate the gain due to mortality and then the gain due to expenses. Calculate the gain due to expenses during year 11. A. 5900 B. 6200 C. 6400 D. 6700 E. 7000 Let us write Ga for the total actuarial gain, Ga M for the actuarial gain due to expenses, Ga E for the actuarial gain due to mortality, Ga I for the actuarial gain due to interest, and Ga W for the actuarial gain due to withdrawals. In general, the formula for the actuarial gain, assuming unit death benefit, is: Ga = k+1as k+1 AS = ( k AS + G) ( îk+1 i ) + ( Gc + e ) 1+ i k k ( Gĉ k + ê k )( 1+ ) îk+1 + ˆq x+k Ga M Ga I ( 1) 1 + q x+k 1 k+1 AS 1 k+1as + q ( 2) x+k Ga E ( CV AS) ( 2) k+1 k+1 ˆq CV k+1as x+k k+1. In this problem, we are told that the experience and assumed interest rate are equal, so that Ga I = 0. Furthermore, the problem has this peculiar structure where it has no reference to withdrawals, so we assume that withdrawals and cash value do not exist and do not enter into the picture, so that Ga W = 0. This leads to ( 1+ i) ( Gĉ k + ê k ) 1+ îk+1 Ga = Gc k + e k Ga E + q ( 1) x+k Ga W ( 1 k+1 AS) ˆq ( 1 ) x+k 1 k+1as. Ga M
Additionally, the problem does not give us any information about asset shares, so we cannot calculate the second term, and expenses are also assessed on payment of death benefit, resulting in interaction of mortality and expenses gains, which is not assumed in the above standard formula. We resolve the first issue by noting that information to calculate the reserve is given, so that we can calculate the gains based solely on reserves, and assuming that the assets portion is irrelevant. We resolve the second issue by following the order of calculations suggested by the wording of the problem: First calculate the mortality gain, then calculate the expenses gain. But how do we get the expenses gain from mortality gain? We subtract the mortality gain from the total gain, so we will have to find the total actuarial gain, in aggregate, without referring to its sources, to solve this problem. Let us also write G for the level annual gross premium, which is equal to 125% of level annual benefit premium. The level annual benefit premium is 50,000P 40 = 50,000A 40 50 161.32 = a ILT 40 14.8166 544.3894. Therefore, the gross premium is 1.25 50,000P 40 1.25 544.3894 680.4868. The expense-loaded policy reserve at policy duration 10 is V = 50,000 10 + 300 Benefit paid at the end of the year of death Expenses paid at the end of the year of death A 50 1 0.05 G a = 50 Fraction of premium after expenses = 50,300 0.24905 0.95 680.4868 13.2668 3950.7275. ILT Also, the expense-loaded policy reserve at policy duration 11 is V = 50,000 11 + 300 Benefit paid at the end of the year of death Expenses paid at the end of the year of death ILT A 51 1 0.05 G a = 51 Fraction of premium after expenses = 50,300 0.25961 0.95 680.4868 13.0803 4602.4607. ILT Of course, you can also find the second reserve from the first using the standard recursive formula 11V p 50 = ( 10 V + G 0.95) 1.06 50,300q 50, so that ( 3950.7275 + 680.4868 0.95) 1.06 50,300 0.00592 11V = 4602.4920. 0.99408 This is actually a slightly different value, but this error does not matter than much. If the actuarial assumptions are realized perfectly, the actuarial gain is zero. But if we substitute the actual experience for actuarial assumptions, we obtain the total actuarial gain as ILT
Ga = 3950.7275 + 0.94 680.4868 1.06 50,100 0.005 10V Actual % Gross premium Actual payment of premium expense upon death Actual accumulation with interets 4602.4607 0.995 35.8596983, Expected reserve, but for actual surviving policies held at the end of year 10 Actual probability of death, 5 out of 1000 per policy, and for one thousand polies the total actuarial gain 1000Ga 35,859.6983. Note that we disregarded the surplus (and thus we do not work with asset shares), and we basically calculated actual reserve at policy duration 11 based on experience, minus expected reserve for surviving policies at the same policy duration. Now we need to find the gain from mortality. It is found by using the actual mortality experience and assumed expenses, giving: Ga M = 3950.7275 + 0.95 680.4868 1.06 50,300 10V Assumed % of premium expense Gross premium 4602.4607 0.995 42.0729611 Expected reserve, but for actual surviving policies held at the end of year 10 Accumulation at assumed (and experience) interest rate of 6% 0.005 Death benefit plus assumed expenses at death Experience mortality per policy, or 42,072.9611 for 1000 policies. Based on the, the actuarial gain (for 1000 policies) due to expenses is 35,859.6983 42,072.9611 = 6213.2628. This problem gives you a good perspective on calculation of actuarial gain based solely on reserves. If interest, expenses, mortality and withdrawals interact with each other (and that is always the case in reality), allocation of the total actuarial gain to its sources will depend on the order of allocation. For example, if their order is: interest, expenses, mortality, withdrawals, then: (1) To compute the gain from interest, we assume that expenses, mortality and withdrawals follow assumptions. (2) Then to compute the gain from expenses, we assume interest follows experience, while mortality and withdrawals follows assumptions. (3) Then to compute the gain from mortality, we assume interest and expenses follow experience, while withdrawals follow assumptions. (4) Finally, to compute the gain from withdrawals, we assume that interest, expenses and mortality follow experience. Answer B. May 2012 Course MLC Examination, Problem No. 18 For a fully discrete whole life policy on (50) with death benefit 100,000, you are given: (i) Reserves equal benefit reserves calculated using the Illustrative Life Table at 6%. (ii) The gross premium equals 120% of the benefit premium calculated using the
Illustrative Life Table at 6%. (iii) Expected expenses equal 40 plus 5% of gross premium, payable at the beginning of each year. (iv) Expected mortality equals 70% of the Illustrative Life Table. (v) The expected interest rate is 7%. Calculate the expected profit in the eleventh policy year, for a policy in force at the beginning of that year. A. 683 B. 719 C. 756 D. 792 E. 829 Based on the data from the Illustrative Life Table, the gross premium for this policy is G = 1.2 100,000P 50 = 1.2 24,905 13.2268 2,252.50. Also, this policy s reserve at policy duration 10 is V = 100,000 A 10 ( P a 60 50 60 ) = 36,913 1,877 11.1454 15,990.90. The reserve at policy duration 11 is 11 V = 100,000 A 61 P 50 a 61 ILT = ILT 38,729 1,877 10.9041 17,809.80. The problem does not clearly say this, but while we are supposed to calculate the expected profit for the 11-th policy year for a policy in force at the beginning of the 11-th year, the value of the profit is supposed to be established as of the end of that policy year, i.e., at duration 11. Note also that the calculation is on expected value basis, not based on experience, and therefore it represents expected values based on the original design of the policy. The expected value of the profit is 10V + G 0.05G + 40 1.07 0.7q 60 100,000 ( 1 0.7q 60 ) 11 V = Expenses Expected mortality = ( 15,990.90 + 2,252.50 112.65 40) 1.07 963.20 0.99037 17,809.80 756. Answer C. May 2012 Course MLC Examination, Problem No. 19 19. You You are are pricing pricing disability disability insurance using the following model: Healthy State 0 Disabled State 1 Dead State 2 You assume the following constant forces of transition: You assume the following constant forces of transition: Copyright 2012 by Krzysztof 01 Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission (i) of the µ copyright = 0.06owner. (ii) 10 µ = 0.03
(i) µ 01 = 0.06. (ii) µ 10 = 0.03. (iii) µ 02 = 0.01. (iv) µ 12 = 0.04. Calculate the probability that a disabled life on July 1, 2012 will become healthy at some time before July 1, 2017 but will not then remain continuously healthy until July 1, 2017. A. 0.012 B. 0.015 C. 0.018 D. 0.021 E. 0.024 The probability sought is Answer D. 5 11 t p µ 10 x 1 00 x+t 5 t p x+t dt = 0 A life disabled now will Then this life becomes be disabled continuously healthy at time t Then this life does not until time t remain continuously healthy for the rest of 5 years t 5 5 ( µ 10 +µ 12 )ds ( µ 01 +µ 02 )ds 0 = e µ 10 t 1 e dt = 0 t 5 5 ( 0.03+0.04)ds ( 0.06+0.01)ds 0 t = e 0.03 1 e dt = 0 5 = e 0.07t 0.07 5 t 0.03 ( 1 e 5 )dt = 0.03 ( e 0.07t e 0.35 )dt = 0 = 0.03a 5 δ =7% 0.15e 0.35 0.0209. 0 May 2012 Course MLC Examination, Problem No. 20 Jenny joins XYZ Corporation today as an actuary at age 60. Her starting annual salary is 225,000 and will increase by 4% each year on her birthday. Assume that retirement takes place on a birthday immediately following the salary increase. XYZ offers a plan to its employees with the following benefits: A single sum retirement benefit equal to 20% of the final salary at time of retirement for each year of service. Retirement is compulsory at age 65; however, early retirement is permitted at ages 63 and 64, but with the retirement benefit reduced by 40% and 20%, respectively. The retirement benefit is paid on the date of retirement. A death benefit, payable at the end of the year of death, equal to a single sum of 100% of the annual salary rate at the time of death, provided death occurs while the employee is still employed. You are given that δ = 5% and the following multiple decrement table (w =withdrawal; r = retirement; and d = death):
Age x ( τ ) l x ( w) d x 60 100 21 0 1 61 78 13 0 1 62 64 7 0 1 63 56 0 6 1 64 49 0 5 1 65 43 0 43 0 Calculate the expected present value of Jenny s total benefits. A. 85,000 B. 92,250 C. 99,500 D. 106,750 E. 113,750 In order to receive a benefit, Jenny can retire at ages 63, 64, or 65, or die at ages 60, 61, 62, 63 (without having retired previously), or 64 (without having retired previously). The following table shows the calculation of the actuarial present value of retirement benefits (note that the actuarial present value is calculated as salary times discount factor times probability times (number of years of service times 20%), corrected for reduction, and note that the benefit is paid at the beginning of the year of retirement (as opposed to the death benefit, which is paid at the end of the year of death): Age at retirement Salary Discount for present value Probability ( r) d x No. of years of service times 20% ( d) d x Reductio n Actuarial present value 63 225,000 1.04 3 e 3 0.05 q ( r ) 3 60 = 6 60% 40% 4705 100 64 225,000 1.04 4 e 4 0.05 q ( r ) 4 60 = 5 80% 20% 6896 100 65 225,000 1.04 5 e 5 0.05 q ( r ) 5 60 = 43 100% 0% 91674 100 The sum of these actuarial present values is 4705 + 6896 + 91674 = 103,275. For the death benefits, the actuarial present values are calculated as follows (death benefit, which is salary at death, times the discount factor, times probability) Age at death Salary Discount for present value calculation Probability 60 225,000 e 0.05 q ( d ) 0 60 = 1 100 61 225,000 1.04 e 2 0.05 q ( d ) 1 60 = 1 100 62 225,000 1.04 2 e 3 0.05 q ( r ) 2 60 = 1 100 63 225,000 1.04 3 e 4 0.05 ( r 3 q ) 60 = 1 100 64 225,000 1.04 4 e 5 0.05 q ( r ) 4 60 = 1 100 Actuarial present value 2140 2117 2095 2072 2050
The total actuarial present value of death benefits is 2140 + 2117 + 2095 + 2072 + 2050 = 10,474. The total actuarial present value of all benefits is 103,275 + 10,474 = 113,749. Answer E. May 2012 Course MLC Examination, Problem No. 21 For a fully continuous whole life insurance of 1000 on (x), you are given: (i) Benefit premiums are 10 per year. (ii) δ = 0.05. (iii) µ x+20.2 = 0.026. (iv) t V denotes the benefit reserve at time t for this insurance. at t = 20.2 is equal to 20.5. (v) d t dt V Calculate 20.2 V. A. 480 B. 540 3. 610 4. 670 E. 730 Recall the Thiele s differential equation d t V dt Change in reserve = δ t V + Interest on reserve P ( S t V )µ x+t Premium (paid continuously) Net amount at risk payout at death We substitute numerical values at t = 20.2 and obtain 20.5 = 0.05 20.2 V +10 1000 20.2 V 0.026. This is a simple linear equation with 20.2 V being the unknown. We solve it and obtain 20.2 V 480.26. Answer A.. May 2012 Course MLC Examination, Problem No. 22 For a fully discrete 10-year deferred whole life insurance of 100,000 on (30), you are given: (i) Mortality follows the Illustrative Life Table. (ii) i = 6%. (iii) The benefit premium is payable for 10 years. (iv) The gross premium is 120% of the benefit premium. (v) Expenses, all incurred at the beginning of the year, are as follows: Policy Year Percent of gross premium Per policy 1 10% 100 2-10 3% 40 11 and later 0% 40 (vi) L 0 is the present value of future losses at issue random value.
Calculate E( L 0 ). A. 334 B. 496 C. 658 D. 820 E. 982 The gross premium is G = 1.2 100000 10 A 30 a 30:10 = 1.2 100000 A 40 10 E 30 a 30 10 E 30 a 40 = ILT = 1.2 161.32 0.54733 ILT 15.8561 0.54733 14.8166 1367.77. Therefore, E( L 0 ) = G a 30:10 +100000 10 A 30 + 0.07G + 0.03G a 30:10 + 60 + 40 a 30 = Answer C. +100000 10 A 30 + 60 + 40 a 30 = G 0.07 0.97 a 30:10 ( ) + G 0.07 0.97 15.8561 0.54733 14.8166 +161.32 0.54733+ 60 + 40 15.8561 658.06. May 2012 Course MLC Examination, Problem No. 23 On January 1 an insurer issues 10 one-year term life insurance policies to lives age x with independent future lifetimes. You are given: (i) Each policy pays a benefit of 1000 at the end of the year if that policyholder dies during the year. (ii) Each policyholder pays a single premium of 90. (iii) q x is the same for every policyholder. With probability 0.30, q x = 0.0 for every policyholder. With probability 0.70, q x = 0.2 for every policyholder. (iv) i = 0.04. Calculate the variance of the present value of future losses at issue random variable for the entire portfolio. A. 800,000 B. 900,000 C. 1,000,000 D. 1,400,000 E. 1,800,000 Let Q be the random probability of death for a policyholder. Then, for each of the 10 policyholders, 0 with probability 0.3, Q = 0.2 with probability 0.7. Note that the problem says that the probability of dying q x is the same, so that once a value of Q is given, that value is q x, and it is the probability of dying for all policyholders. Let L k be the present value of future losses random variable for policyholder number k, where k = 1, 2,, 10. We have, for every k,
L k = 90 with probability 1 Q, 1000 1.04 90 with probability Q, = 90 + 1000 1.04 0 with probability 1 Q, 1 with probability Q. This effectively says that L k equals a constant of 90 added to a multiple (by the factor of 1000/1.04) of a Bernoulli Trial with probability of success of Q. Let us write L 1 = 90 + 1000 1000 = 90 + = = 90 + 1000 1.04 X 10, 1.04 X, L 1 2 1.04 X,, L 2 10 where each X k for k = 1, 2,, 10, is a Bernoulli Trial with probability of success Q. Let S be the random present value of future losses at issue random variable for the entire portfolio. We have 10 10 S = L k = 90 + 1000 = 900 + 1000 10 X k. k=1 1.04 X k k=1 1.04 k=1 We observe that for Q = 0, each X k is equal to 900 with probability 1, while for Q = 0.2, We have Var S Note now that and Therefore, 10 X k is binomial with n = 10 and p = 0.2. We need to find the variance of S. k=1 ( ) + Var E( S Q) = E Var S Q E( S Q) = Var( S Q) = Var S Answer E.. 900 when Q = 0,which happens with probability 0.3, 900 + 1000 2 when Q = 0.2,which happens with probability 0.7, 1.04 0 when Q = 0,which happens with probability 0.3, 1000 1.04 = E Var S Q 2 1.6 when Q = 0.2,which happens with probability 0.7. ( ) + Var E( S Q) = 2 2 1000 1000 = 0.3 0 + 0.7 1.04 1.6 + 1.04 2 0.3 0.7 1,812,130.18. Because Var( S Q) equals to a Bernoulli Trial Because E( S Q) is a constant of 900 with probability of success 0.7 multiplied by 2 1000 the number 1.04 1.6 plus 1000 2 multiplied by a Bernoulli 1.04 Trial with probability of success of 0.7
May 2012 Course MLC Examination, Problem No. 24 For a three-year term insurance of 10,000 on (65), payable at the end of the year of death, you are given: (i) x q x 65 0.00355 66 0.00397 67 0.00444 (ii) Forward interest rates at the date of issue of the contract, expressed as annual rates, are as follows: Start time End time Annual forward rate 0 3 0.050 1 3 0.070 2 3 0.091 Calculate the expected present value of this insurance. A. 105 B. 110 C. 113 D. 115 E. 120 We will write s k for the annual effective spot rate for maturity k, and f k,l for the effective annual forward rate from time k to time l. We are given f 0,3 = s 3 = 0.05, f 1,3 = 0.07, and f 2,3 = 0.091. But ( 1+ s 3 ) 3 = ( 1+ f 0,1 )( 1+ f 1.3 ) 2 = ( 1+ s 1 )( 1+ f 1.3 ) 2, so that 3 = 1.05 3 2 1.07, 2 3 = ( 1+ s 2 ) 2 1+ f 2.3 ( ( 1+ s 2 ) 2 = 1+ s 3) 3 = 1.053 1.091. 1+ s 1 = 1+ s 3 1+ f 1.3 as well as 1+ s 3, so that 1+ f 2.3 The expected present value of this insurance is 10000 q 65 + 10000 1+ s 1 ( 1+ s 2 ) p 2 65 q 66 + 10000 1+ s 3 = 10000 1.05 3 0.00355 + 10000 1.05 3 3 p 65 p 66 q 67 = ( 1 0.00355) 0.00397 + 1.07 2 1.091 + 10000 3 ( 1 0.00355) ( 1 0.00397) 0.00444 110.4586. 1.05 Answer B.
May 2012 Course MLC Examination, Problem No. 25 An insurer issues fully discrete whole life insurance policies of 100,000 to insureds age 40 with independent future lifetimes. You are given: (i) Issue expenses are incurred upon policy issuance and are 20% of the first year premium. (ii) Renewal expenses are 6% of each premium, incurred on the premium due date, starting in the second year. (iii) The annual premium is calculated using the percentile premium principle and the normal approximation, such that the probability of a loss on the portfolio is 5%. (iv) A 40 = 0.161. (v) 2 A 40 = 0.048. (vi) a 40 = 14.822. (vii) i = 0.06. (viii) The annual premium per policy for a portfolio of 2000 policies is 1215. Calculate the difference in annual premium per policy for a portfolio of 2000 policies and for a portfolio of 40,000 policies. A. 37 B. 42 C. 48 D. 52 E. 56 As the annual premium per policy for a portfolio of 2000 policies is given, we simply need to find the annual premium per policy for a portfolio of 40,000 policies, call it G, and then calculate the difference. Let us write L i for the loss at issue for policy i. We have L i = 100,000 1.06 K40+1 + ( 0.2 0.06)G ( 1 0.06)G a = K40 +1 = 100,000 1.06 K40+1 + 0.14G 0.94G 1 1.06K 40+1 0.06 1.06 ( 100,000 +16.6067 G) 1.06 K40+1 16.4667G. Therefore, E( L i ) = E( ( 100,000 +16.6067 G) 1.06 K40+1 16.4667G) = ( 100,000 + 16.6067 G) A 40 16.4667G = = 100,000 0.161+ ( 16.6067 0.161 16.4667)G = 16,100 13.7930G, Var( L i ) = Var ( 100,000 +16.6067 G) 1.06 K40+1 16.4667G = 2 2 ( A 40 A 40 ) = 2 = 100,000 + 16.6067 G = ( 100,000 + 16.6067 G) 2 0.048 0.161 2 100,000 +16.6067 G 2 0.02208. Let L denote the total loss for the entire portfolio of 40,000 policies. Then (note that we are using independence of policies for the variance calculation):
= 40,000( 16,100 13.7930G), 40,000 ( 100,000 +16.6067 G) 2 0.02208. E L Var L Because of independence of the policies, and identical distributions of individual policies losses, L is approximately normal, and L E( L) 0.95 = Pr( L < 0) = Pr Var( L) < E( L) = Φ E( L) Var( L) Var( L) = 40,000 16,100 13.7930G = Φ 40,000 ( 100,000 +16.6067 G) 0.02208. The 95-th percentile of the standard normal distribution is 1.645, so that 40,000( 16,100 13.7930G) 40,000 100,000 +16.6067 G = 1.645, 2 0.02208 and from this we calculate G 1178. The difference sought is 1215 1178 = 37. Answer A. May 2012 Course MLC Examination, Problem No. 26 For a special fully discrete whole life insurance on (35) you are given: (i) Mortality follows the Illustrative Life Table. (ii) i = 0.06. (iii) Initial annual premiums are level for the first 30 years; thereafter, annual premiums are one-third of the initial annual premium. (iv) The death benefit is 60,000 during the first 30 years and 15,000 thereafter. (v) Expenses are 40% of the first year s premium and 5% of all subsequent premiums. (vi) Expenses are payable at the beginning of the year. Calculate the initial annual premium using the equivalence principle. A. 290 B. 310 C. 330 D. 350 E. 370 From the Illustrative Life Table 30 E 35 = 20 E 35 10 E 55 = 0.28600 0.48686 0.13924, a 35:30 = a 35 30 E 35 a 65 15.3926 0.13924 9.8969 14.0146, 1 A 35:30 = A 35 30 E 35 A 65 0.12872 0.13924 0.43980 0.06748. Let us denote the initial annual premium by G. We then calculate the actuarial present value of premiums as G 2 3 a + 1 35:30 3 a 35 G 2 3 14.0146 + 1 3 15.3926 14.4734G, the actuarial present value of benefits as 3 60,000 4 A 1 + 1 35:30 4 A 3 35 60,000 4 0.06748 + 1 4 0.12872 4.967.40,
and the actuarial present value of expenses as 2 G 0.35 + 0.05 3 a + 1 35:30 3 a 35 G 0.35 +0.05 2 3 14.0146 + 1 3 15.3926 1.0737G. Using the Equivalence Principle, we obtain the equation 14.4734G = 4,967.40 +1.07370G, resulting in G 370.71. Answer E. May 2012 Course MLC Examination, Problem No. 27 For a universal life insurance policy with death benefit of 100,000 on (40), you are given: (i) The account value at the end of year 4 is 2029. (ii) A premium of 200 is paid at the start of year 5. (iii) Expense charges in renewal years are 40 per year plus 10% of premium. (iv) The cost of insurance charge for year 5 is 400. (v) Expense and cost of insurance charges are payable at the start of the year. (vi) Under a no lapse guarantee, after the premium at the start of year 5 is paid, the insurance is guaranteed to continue until the insured reaches age 49. (vii) If the expected present value of the guaranteed insurance coverage is greater than the account value, the company holds a reserve for the no lapse guarantee equal to the difference. The expected present value is based on the Illustrative Life Table at 6% interest and no expenses. Calculate the reserve for the no lapse guarantee, immediately after the premium and charges have been accounted for at the start of year 5. A. 0 B. 10 C. 20 D. 30 E. 40 The account value (reserve) at policy duration 5 (which is the start of policy year 5) is 2029 + 200 0.10 200 40 400 = 1769. The expected present value of the guaranteed coverage is the actuarial present value of term insurance to age 49, which is 1 100,000A 44:5 = 100,000( A 44 5 E 44 A 49 ) = = 100,000 0.19261 0.73117 0.23882 ILT Note what happens here: The account value (i.e., retrospective reserve for a universal life policy is 1769, while the company guarantees that life insurance coverage will continue for the next five years, and that guarantee has actuarial present value of 1799, and that extra cost needs to be reserved for). Therefore, the no lapse reserve is 1799 1769 = 30. Answer D. ILT 1799.
May 2012 Course MLC Examination, Problem No. 28 You are using Euler s method to calculate estimates of probabilities for a multiple state model with states {0,1, 2}. You are given: (i) The only possible transitions between states are: 0 to 1 1 to 0 1 to 2 (ii) For all x, µ x 01 = 0.3, µ x 10 = 0.1, µ x 12 = 0.1. (iii) Your step size is 0.1. (iv) You have calculated that (a) 0.6 p x 00 = 0.8370 (b) 0.6 p x 01 = 0.1588 (c) 0.6 p x 02 = 0.0042 Calculate the estimate of 0.8 p x 01 using the specified procedure. A. 0.20 B. 0.21 C. 0.22 D. 0.23 E. 0.24 The probability we are looking for is 0.8 p x 01, but in notation of Models for Quantifying ( 0 Risk, it is 0.8 p ) 01, if we consider age x to be time 0. Recall the Kolmogorov Forward Equation (we will continue using notation from Models for Quantifying Risk): d dr p ( t) r ij = p ( t ) λkj t + r r ik r p ( t ) λ t + r ( ij ) = p ( t) jk λkj t + r r ik k j k j p ( t) r ij λ j ( t + r). In this case, we have s 01 10 10 = 0.3, λ 10 s = 0.1, λ 12 s = 0.1, λ 01 = µ x+s ( s) = µ x+s = µ x+s = µ x+s = µ x+s = µ x+s λ 02 02 = 0, λ 20 s 20 = 0, λ 21 s 21 = 0. We also have the following specific Kolmogorov forward equations d dr p ( 0) r 01 = ( 0) r p λ01 00 ( r) r p ( 0 ) λ10 r ( 0) ( 01 ) + r p λ21 02 ( r) r p ( 0 ) λ12 r 01 ( 0) = 0.3 r p 00 ( 0) 0.1r p 01 ( 0) 0.1r p 01 = ( 0) 0.3r p 00 0.2 ( 0) r p 01, ( ) = 0 + ( r p 02 ) = d dr p ( 0) r 00 = ( 0) r p λ10 01 ( r) r p ( 0 ) λ01 r 00 λ20 ( r) r p ( 0 ) λ02 r 00 ( 0) = 0.1 r p 01 ( 0) 0.3r p 00. The Euler step method replaces the derivative on the left-hand side of the equation by a difference quotient: ( 0) ( 0) 0.6 p 01 = ( 0) 0.7 p 01 0.1588 0.7 p 01 ( 0) so that 0.7 p 01 0.180934. Also, = 0.03 0.8370 0.02 0.1588, ( 0) 0.1 0.3 0.6 p 00 0.2 ( 0) 0.6 p 01