Solutions to the November 202 Course MLC Examination by Krzysztof Ostaszewski, http://www.krzysio.net, krzysio@krzysio.net Copyright 202 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 November 202 Course MLC Examination, Problem No. For two lives, (80) and (90), with independent future lifetimes, you are given: k p 80+k p 90+k 0 0.9 0.6 0.8 0.5 2 0.7 0.4 Calculate the probability that the last survivor will die in the third year. A. 0.20 B. 0.2 C. 0.22 D. 0.23 E. 0.24 The probability sought is 2 p 80:90 3 p 80:90 Using Poohsticks 2 p 80 + 2 p 90 2 p 80 2 p 90 ( p + p p 2 80 2 90 2 80:90 ) 3 p 80 + 3 p 90 3 p 80:90 ( 3 p 80 + 3 p 90 3 p 80 3 p 90 ) ( p 80 p 8 p 82 + p 90 p 9 p 92 p 80 p 8 p 82 p 90 p 9 p 92 ) ( 0.9 0.8 0.7 + 0.6 0.5 0.4 0.9 0.8 0.7 0.6 0.5 0.4) ( 0.504 + 0.2 0.06048) 0.24048. p 80 p 8 + p 90 p 9 p 80 p 8 p 90 p 9 0.9 0.8 + 0.6 0.5 0.9 0.8 0.6 0.5 0.72 + 0.3 0.26 Answer E. November 202 Course MLC Examination, Problem No. 2 You are given: (i) An excerpt from a select and ultimate life table with a select period of 3 years: x l [ x] l [ x]+ l [ x]+2 l x+3 x + 3 60 80,000 79,000 77,000 74,000 63 6 78,000 76,000 73,000 70,000 64 62 75,000 72,000 69,000 67,000 65 63 7,000 68,000 66,000 65,000 66 (ii) Deaths follow a constant force of mortality over each year of age. Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
Calculate 000 2 3 q [ 60]+0.75. A. 04 B. 7 C. 22 D. 35 E. 42 Under constant force over each year of age, for 0 k, and x an integer, so that l x+k l x We have l x+k l x k p x e kµ x e µ x k ( l x+ ) k. k l x+ 000 2 3 q [ 60]+0.75 000 l [ l 60]+2.75 [ 60]+5.75 000 l [ 60]+0.75 l x k, l ( l [ 60]+2 )+0.75 65.75 l [ 60]+0.75 Answer B. 000 ( l [ 60]+2 ) 0.25 l 63 ( l [ 60] ) 0.25 l 60 0.75 l 0.25 0.75 65 l 66 ( [ ]+ ) 0.75 000 770000.25 74000 0.75 67000 0.25 65000 0.75 80000 0.25 79000 0.75 6.70. November 202 Course MLC Examination, Problem No. 3 You are given: (i) S ( 0 t ) t 4 ω, for 0 t ω. (ii) µ 65 80. Calculate e 06, the curtate expectation of life at age 06. A. 2.2 B. 2.5 C. 2.7 D. 3.0 E. 3.2 We have µ t d ( lns 0 ) d ln t dt dt ω Therefore, 4 d dt Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without 4 ln t ω 4 t ω 4 ω t. ω
80 µ 65 4 ω 65, so that 45 ω 65, and ω 0. Based on this Therefore, Answer B. p S 06 + t 0 t 06 S 0 06 06 + t 0 06 4 0 e 06 p 06 + 2 p 06 + 3 p 06 + 4 p +... 06 0 4 4 t 4 4. 4 4 4 2 4 4 3 4 4 + 4 + 4 2.4786. November 202 Course MLC Examination, Problem No. 4 For a special fully discrete whole life insurance on (40), you are given: (i) The death benefit is 50,000 in the first 20 years and 00,000 thereafter. (ii) Level benefit premiums of 6 are payable for 20 years. (iii) Mortality follows the Illustrative Life Table. (iv) i 0.06. Calculate 0 V, the benefit reserve at the end of year 0 for this insurance. A. 3,340 B. 3,370 C. 3,400 D. 3,430 E. 3,460 Prospectively, we calculate the reserve as ( +00,000 0 A 50 ) 6 a 50:0 V 50,000A 0 50:0 3,428. Answer D. 6( a 50 0 E 50 a 60 ) 6 ( 3.2668 0.508.45) 50,000 A 50 + 0 E 50 A 60 50,000 0.24905 + 0.508 0.3693 November 202 Course MLC Examination, Problem No. 5 A special fully discrete 2-year endowment insurance with a maturity value of 2000 is issued to (x). The death benefit is 2000 plus the benefit reserve at the end of the year of death. For year 2, the benefit reserve is the benefit reserve just before the maturity benefit is paid. You are given: (i) i 0.0. Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
(ii) q x 0.50 and q x+ 0.65. Calculate the level annual benefit premium. A. 070 B. 0 C. 50 D. 90 E. 230 The initial reserve at time 0 is 0 V 0. The terminal reserve at time 2 is 2 V 2000, as the policy endows at that time. Let us write P for the level annual benefit premium sought. We use the recursive reserve formula twice to find it: ( V + P) ( + i) q 0 x ( 2000 + V ) + ( q x ) V, ( V + P) ( + i) q x+ ( 2000 + 2 V ) + ( q x+ ) 2 V. Substituting known values, we obtain ( 0 + P). 0.50 ( 2000 + V ) + 0.850 V, ( V + P ). 0.65 ( 2000 + 2000) + 0.835 2000. From the first equation, V.P 300, and after we put this in the second equation (.P 300 + P). 0.65 4000 + 0.835 2000, or P 2660 2.3 5.52. Answer C. November 202 Course MLC Examination, Problem No. 6 For a special fully continuous 0-year increasing term insurance, you are given: (i) The death benefit is payable at the moment of death and increases linearly from 0,000 to 0,000. (ii) µ 0.0. (iii) δ 0.05. (iv) The annual premium rate is 450. (v) Premium-related expenses equal 2% of premium, incurred continuously. (vi) Claims-related expenses equal 200 at the moment of death. (vii) t V denotes the gross premium reserve at time t for this insurance. (viii) You estimate 9.6 V using Euler s method with step size 0.2 and the derivative of t V at time 9.6. (ix) Your estimate of 9.8 V is 26.68. Calculate the estimate of 9.6 V. A. 230 B. 250 C. 270 D. 280 E. 290 Using Euler s method with step size 0.2 and the derivative of t V at time 9.6, we obtain Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
9.6V 9.8 V 0.2 d tv dt t9.6. Recall the Thiele s differential equation when expenses are included: d dt tv δ V + P e S t t t t ( + E V t t t )µ x+t. In this case, t 9.6, δ 9.6 δ 0.05, t V is denoted by 9.6 V and it is unknown, P t is a gross premium with the annual rate of 450, e t are premium related expenses at 2% of premium, S t is constant with S 9.6 06,000, E 9.6 200, so that d t V dt t9.6 0.05 9.6 V + 450 0.02 450 ( 06,000 + 200 9.6 V ) 0.0 62+ 0.06 9.6 V. Substituting into the Euler s formula, we obtain 9.6V 9.8 V 0.2 d tv This results in dt t9.6 9.6V 250.88.02 247.9. Answer B. 26.68 0.2 ( 62 + 0.06 9.6 V ) 250.88 0.02 9.6 V. November 202 Course MLC Examination, Problem No. 7 You are calculating asset shares for a universal life insurance policy with a death benefit of 000 on (x) with death benefits payable at the end of the year of death. You are given: (i) The account value at the end of year 4 is 30. (ii) The asset share at the end of year 4 is 20. (iii) At the beginning of year 5: a. A premium of 20 is paid. b. Annual cost of insurance charges of 2 and annual expense charges of 7 are deducted from the account value. (iv) At the beginning of year 5, the insurer incurs expenses of 2. (v) All withdrawals occur at the end of the policy year; the withdrawal benefit is the account value less a surrender charge of 20. d w (vi) q x+4 0.00 and q x+4 0.050 are the probabilities of death and withdrawal, respectively. (vii) The annual interest rate for asset shares is 0.08; the annual interest rate credited to the universal life insurance policy is 0.06. Calculate the asset share at the end of year 5. A. 39 B. 40 C. 4 D. 42 E. 43 The account value (a.k.a., reserve) at the end of year 5 is Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
30 + 20 9.06 43.46. Account value at Premium Cost of insurance of 2 Accumulated at the end of year 4 plus annual expense the interest rate charge of 7 credited (i.e., 6%) The surrender value at the end of year 5 is therefore this account value of 43.46 reduced by surrender charge of 20, i.e., 43.46 20 23.46. The asset share at the end of year 5 is Expenses Accumulated Expected payout AS 4 Premium incurred at asset share rate Expected death benefit payout at withdrawal 20 + 20 2.08 0.00 000 0.05 23.46 40.96. 0.00 0.05 Answer C. Probability of policy continuing November 202 Course MLC Examination, Problem No. 8 For a universal life insurance policy with a death benefit of 50,000 plus the account value, you are given: (i) Policy Year Monthly Premium Percent of Premium Charge Monthly Cost of Insurance Rate per 000 Monthly Expense Charge 300 W% 2 0 500 2 300 5% 3 0 25 (ii) The credited interest rate is i 2) 0.054. (iii) The cash surrender value at the end of month is 200.00. (iv) The cash surrender value at the end of month 3 is 802.94. Calculate W%, the percent of premium charge in policy year. A. 25% B. 30% C. 35% D. 40% E. 45% Surrender Charge Let us write CSV k for the cash surrender value at time k, with time counted in months, and SC k for the surrender charge at time k. Let be COI k be the cost of insurance for month k. Since i ( 2) 0.054, the effective monthly interest rate is j i ( 2) 2 0.054 0.0045 0.45%. 2 Note that the cash surrender value equals the account value (i.e., reserve) minus the surrender charge. We are given that CSV 3 802.94. Based on this, 802.94 CSV 3 AV 3 SC 3 AV 3 25, so that Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
AV 3 802.94 +25 927.94. Similarly, AV CSV + SC 200 + 500 700. Now we use the recursive formula for the account value (reserve) for universal life with increasing total benefit and fixed ADB ( AV t + G t ( r t ) e t )( + i t ) q x+t ( B + AV t+ ) + p x+t AV t q x+t B + AV t. The cost of insurance in this formula is COI t q x+t B, so that AV t ( AV t + G t ( r t ) e t )( + i t ) COI t. In this case ( e 3 )( + i 3 ) COI 3, e 2 COI 2. AV 3 AV 2 + G 3 r 3 + i 2 AV 2 AV + G 2 r 2 Substituting known values (note that the cost of insurance figures are given per thousand), we obtain 927.94 ( AV 2 + 300( 0.5) 0).0045 50 3, AV 2 ( 700 + 300( W ) 0).0045 50 2. From the first equation 927.94 + 50 3 AV 2.0045 and substituting this into the second equation, we obtain 823.636 + 50 2 700 +0 W.0045 0.25. 300 Answer A. +0 300( 0.5) 823.636, November 202 Course MLC Examination, Problem No. 9 You are given the following about a universal life insurance policy on (60): (i) The death benefit equals the account value plus 200,000. (ii) Age x Annual Premium Annual Cost of Insurance Rate per 000 Annual Expense Charges 60 5000 5.40 00 6 5000 6.00 00 (iii) Interest is credited at 6%. (iv) Surrender value equals 93% of account value during the first two years. Surrenders occur at the ends of policy years. (v) Surrenders are 6% per year of those who survive. (vi) Mortality rates are 000q 60 3.40 and 000q 6 3.80. (vii) i 7%. Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
Calculate the present value at issue of the insurer s expected surrender benefits paid in the second year. A. 380 B. 390 C. 400 D. 40 E. 420 The account value at time 0, just before the first premium is paid is zero. Then, since death benefit equals fixed ADB plus account value, AV ( AV 0 + G ( r ) e )( + i ) q 60 B ( 0 + 5000 00).06 200 5.40 44. Following into the second year, ( e 2 )( + i 2 ) q 6 B AV 2 AV + G 2 r 2 ( 44 + 5000 00).06 200 6.00 8354.84. If a policy is surrendered in year 2 (and for a policy in existence at the end of year this happens with probability 0.06), then it was not surrendered in year (and for a policy issued at time 0 this happens with probability 0.06 0.94), so that the present value at time 0 of the expected surrender benefits in year 2 is 8354.84 0.93 0.06 ( 0.06) ( 0.0034) ( 0.0038).07 2 Account value available for surrenders 380.0. Answer A. Adjusted for surrender charge Fraction that surrenders in year 2 Fraction that did not surrender in year q 60 q 6 Present value factor November 202 Course MLC Examination, Problem No. 0 For a special 3-year term life insurance policy on (x) and (y) with dependent future lifetimes, you are given: (i) A death benefit of 00,000 is paid at the end of the year of death if both (x) and (y) die within the same year. No death benefits are payable otherwise. (ii) p x+k 0.84366, k 0,, 2. (iii) p y+k 0.86936, k 0,, 2. (iv) p x+k:y+k 0.7705, k 0,, 2. (v) Maturity (in years) Annual Effective Spot Rate 3% 2 8% 3 0% Calculate the expected present value of the death benefit. A. 9,500 B. 0,00 C. 2,00 D. 2,500 E. 4,00 The death benefit of $00,000 is paid at the end of year if both (x) and (y) die during the first year, or it is paid at the end of year 2 if both (x) and (y) are alive at time and Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
both die during the second year, or it is paid at the end of year 3 if both (x) and (y) are alive at time 2 and both die during the third year. Let us note the following, true for k 0,, or 2: Also, q x+k:y+k p x+k:y+k Poohsticks p x+k + p y+k p x+k:y+k 0.05803. 0.84366 + 0.86936 0.7705 p p p 0.7705 0.7705 2 x:y x:y x+:y+ 0.77052. The expected present value we are looking for is therefore 00,000.03 q + 00,000.08 2 x:y p x:yq x+:y+ + Present value of death benefit payment made at time + 00,000.0 3 Present value of death benefit payment made at time 2 00,000 Answer C. Probability that both ( x) and ( y) die during first year p 2 x:y q x+2:y+2 Probability that both ( x) and ( y) die during second year Present value of death benefit payment made at time 2 0.05803 0.7705 0.05803 +.03 Probability that both ( x) and ( y) die during second year + 0.77052 0.05803.08 2.0 3 2,062.09. November 202 Course MLC Examination, Problem No. For a whole life insurance of 000 on (70), you are given: (i) Death benefits are payable at the end of the year of death. (ii) Mortality follows the Illustrative Life Table. (iii) Maturity (in years) Annual Effective Spot Rate.6% 2 2.6% (iv) For the year starting at time k and ending at time k, k 3, 4, 5,..., the one-year forward rate is 6%. Calculate the expected present value of the death benefits. A. 520 B. 530 C. 550 D. 570 E. 600 It seems like it will be a lot of calculations to go from age 70 to the limiting age if we do every year calculation separately. But the Illustrative Life Table of course gives us actuarial present values of whole life insurance death benefits, except that it does so for the valuation interest rate of 6%. Here, the spot rates for the first two years are well below 6%, but beyond that all one-year forward rates are 6% (so that all rates for all periods starting at time 2 or later are also 6%). Thus starting from age 72, we can determine actuarial present values from the Illustrative Life Table. We have Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
q 70 q 7 000A 70 000.06 + p 70.026 + p 70 p 7 2.026 A 2 72, where all values on the right-hand side can now be looked up in the Illustrative Life Table. From the table: q 70 0.0338, q 7 0.03626, A 72 0.54560, and therefore p 70 0.96682, p 7 0.96374, so that 000A 70 000 0.0338.06 548.89. Answer C. 0.03626 0.96682 0.96374 + 0.96682 +.026 2 0.54560.026 2 November 202 Course MLC Examination, Problem No. 2 A party of scientists arrives at a remote island. Unknown to them, a hungry tyrannosaur lives on the island. You model the future lifetimes of the scientists as a three-state model, where: State 0: no scientists have been eaten. State : exactly one scientist has been eaten. State 2: at least two scientists have been eaten. You are given: (i) Until a scientist is eaten, they suspect nothing, so µ t 0 0.0+ 0.02 2 t, t > 0. (ii) Until a scientist is eaten, they suspect nothing, so the tyrannosaur may come across two together and eat both, with µ t 02 0.5µ t 0, t > 0. (iii) After the first death, scientists become much more careful, so µ t 2 0.0, t > 0. Calculate the probability that no scientists are eaten in the first year. A. 0.928 B. 0.943 C. 0.95 D. 0.956 E. 0.962 Note that because it is impossible to return to state 0, the probability we are looking for, p 0 00 is the same as p 0 00. We calculate p 0 00 p 0 00 e e 0.05t+ We could also note that µ 0 02 ( t +µ t )dt 0 t 0.03 ln2 2t t0 µ t 0 + µ t 02 0.0+ 0.02 2 t e e 0.05+ µ 0 0 ( t +0.5µ t )dt 0 0.06 0.03 0 ln2 ln2 e.5( 0.0+0.02 2 t )dt 0 0.05 0.03 e ln2 0.94338496. + 0.5( 0.0+ 0.02 2 t ) 0.05 + 0.03 2 t, and this is the Makeham s Law with A 0.05, B 0.03, c 2. Recall that under Makeham s Law, n p x s n ( g cx c n ) B, where g e lnc and s e A. Here, g e 0.03 ln2, Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
s e 0.05, and n p x e 0.05n e 0.03 2 x ( 2 n ) ln2, so that the quantity sought, i.e., p 00 0 p 00 0, is the same as p 0 under this Makeham model, i.e., p 00 0 p 00 0 e 0.05 e Answer B. 0.03 2 ln2 0.94338496. November 202 Course MLC Examination, Problem No. 3 You are given: (i) The following excerpt from a triple decrement table: x ( τ ) l x d x 50 00,000 490 8,045,00 5 90,365 -- 8,200 -- 52 80,000 -- -- -- (ii) All decrements are uniformly distributed over each year of age in the triple decrement table. 3 (iii) q x is the same for all ages.. Calculate 0,000 q 5 ( 2) d x A. 30 B. 33 C. 36 D. 38 E. 4 Recall the key formula for UDD over each year of age in the multiple decrement table: We have q x ( j p ) ( τ ) x p x p 50 τ ( j) q ( τ ) x. l ( τ ) 5 90,365 00,000 0.90365, q 50 3 l 50 τ d ( 3) 50,00 00,000 0.0, l 50 τ ( τ q ) ( τ 50 p ) 50 0.90365 0.09635. We calculate therefore q 50 ( 3 q 5 ) ( 3 q 50 ) ( 3 p 50 ) ( τ ) p 50 ( 3 This means that p 5 ) 0.9885 and q 5 ( 3 0.9855 p 5 ) ( τ ) p 5 From this we calculate ( 3) ( τ ) q 5 ( 3) ( τ ) q 50 80,000 90, 365 0.0 0.09635 0.90365 0.05. ( 3) q 5 80,000 90,365. ( 3) d x Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
80,000 ( 3 q ) 90, 365 ln0.9855 5 ln 80,000 0.0088985. 90, 365 This means that ( 3 d ) ( 3 5 q ) ( τ 5 l ) 5 0.0088985 90,365 984.058, d 5 2 ( d ) ( τ 5 d ) ( 2 5 d ) ( 3 5 d ) ( τ 5 l ) ( τ ) 50 l 5 90,365 80,000 d 5 3 8200 984,8, ( q ) 5 d 5 ( τ ),8 l 5 90,365 0.0306922, and we also have ( τ p ) 5 80,000 90,365 0.8852985, ( τ q ) ( τ 5 p ) 5 80,000 90,365 0.47049. We therefore conclude that q 5 ( p 5 ) ( τ ) p 5 ( τ ) q 5 The quantity sought is ( 0,000 q 5 ) 0,000 Answer D. 0.0306922 0.47049 0.8852985 0.986244. ( p 5 ) 0,000 0.986244 38. November 202 Course MLC Examination, Problem No. 4 For a special whole life insurance policy issued on (40), you are given: (i) Death benefits are payable at the end of the year of death. (ii) The amount of benefit is 2 if death occurs within the first 20 years and is thereafter. (iii) Z is the present value random variable for the payments under this insurance. (iv) i 0.03. (v) x A x 20 E x 40 0.36987 0.5276 60 0.62567 0.7878 (vi) E Z 2 0.24954. Calculate the standard deviation of Z. A. 0.27 B. 0.32 C. 0.37 D. 0.42 E. 0.47 The present value of death benefit random variable is Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
2v K+, K < 20, Z v K+, K 20. Therefore, E Z We conclude that the variance of Z is 2A 40 20 E 40 A 60 2 0.36987 0.5276 0.62567 0.4892. Var Z E Z 2 and the standard deviation of Z is σ Z Answer A. Var Z ( E( Z )) 2 0.24954 0.4892 2 0.07405, 0.07405 0.2722. November 202 Course MLC Examination, Problem No. 5 For a special 2-year term insurance policy on (x), you are given: (i) Death benefits are payable at the end of the half-year of death. (ii) The amount of the death benefit is 300,000 for the first half-year and increases by 30,000 per half-year thereafter. (iii) q x 0.6 and q x+ 0.23. (iv) i ( 2) 0.8. (v) Deaths are assumed to follow a constant force of mortality between integral ages. (vi) Z is the present value random variable for this insurance. Calculate Pr(Z > 277,000). A. 0.08 B. 0. C. 0.4 D. 0.8 E. 0.2 Note that the effective half a year interest rate is 9%. Also, under the constant force of mortality between integral ages assumption, if we write µ x for the constant force between ages x and x +, and µ x+ for the constant force between ages x + and x + 2, then 0.5q x 0.5 p x e 0.5µ x p 0.5 x 0.6 0.083485, 0.5q x+0.5 0.5 p x+0.5 p 0.5 x 0.6 0.083485, 0.5q x+ 0.5 p x+ e 0.5µ x+ p 0.5 x+ 0.23 0.22504, 0.5q x+.5 0.5 p x+.5 e 0.5µ x+ p 0.5 x+ 0.23 0.22504. Based on the above, probabilities of the insured dying in each of the four half-years are: 0 0.5q x 0.5 q x 0.083485, 0.5 0.5q x 0.5 p x 0.5 q x+0.5 0.9655 0.083485 0.07655, q p q 0.5 x x 0.5 x+ ( 0.6) 0.22504 0.02903, q p p q 0.5 x x 0.5 x+ 0.5 x+.5 ( 0.6) 0.877496 0.22504 0.090297. The random present value of the death benefit is Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
300,000 275,229.36 with probability 0.5 q x 0.083485,.9 330,000 277,754.40 with probability.9 Z 2 0.5 0.5q x 0.07655, 360,000 277,986.05 with probability.9 q 0.02903, 3 0.5 x 390,000 276,285.83 with probability.9 q 0.090297. 4 0.5 x Also, Z 0 otherwise. We conclude that Pr Z > 277,000 0.07655 + 0.02903 0.7948. Answer D. 0.5 0.5 q x + 0.5 q x You are evaluating November the financial 202 Course strength MLC of companies Examination, based on Problem the following No. 6 multiple You are evaluating the financial strength of companies based on the following multiple state model: state model: State 0 Solvent State Bankrupt For each company, you assume the following constant transition intensities: For each company, (i) µ you 0 assume 0.02. the following constant transition intensities: (i) (ii) (iii) (ii) 0 µ µ 0 0.02 0.06. (iii) µ 2 0.0. 0 Using µ 0.06 Kolmogorov s forward equations with step h, calculate the probability that a 2 company 2 µ 0.0currently Bankrupt will be Solvent at the end of one year. A. 0.048 B. 0.05 C. 0.054 D. 0.057 E. 0.060 Using Kolmogorov s forward equations with step h /2, calculate the probability that a company currently We have, Bankrupt using will the be Markov Solvent process at the end notation: of one year. t (A) 0.048 (B) 0.05 (C) 0.054 (D) 0.057 λ 0 λ 02 λ 0 State 2 Liquidated λ 0 µ 0 0.02, λ 02 µ 02 0.00, λ 0 λ 0 + λ 02 0.02 + 0.00 0.02, Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
λ 0 λ 2 λ λ 20 λ 2 λ 2 λ 0 µ 0 0.06, λ 2 µ 2 0.0, t λ λ 0 + λ 2 0.06 + 0.0 0.6, λ 20 µ 20 0.00, λ 2 µ 2 0.00, λ 2 λ 20 + λ 2 0 + 0 0. This tells us that the generator matrix is: λ 0 ( t + t) λ 0 ( t + r) λ 02 ( t + r) Q t+r λ 0 ( t + r) λ ( t + t) λ 2 ( t + r) λ 20 ( t + r) λ 2 ( t + r) λ 2 ( t + t) For 0.02 0.02 0 0.06 0.6 0. 0 0 0 r 00 r 0 r 02 P r t, r 0 r r 2 r 20 r 2 r 22 the resulting Kolmogorov Forward Equation, in matrix form, is d d t d P dr r t d r p 00 r p 0 r p p 02 dr r 00 p dr r 0 r dr d 0 r r 2 dr d r 0 dr p t r r p 20 r p 2 r p 22 d dr d r 20 dr p t r 2 r 00 r p 0 r 20 r 0 r p r 2 r 02 r p 2 r 22 0.02 0.02 0 0.06 0.6 0. 0 0 0 d d d. dr r p 02 dr r p 2 dr r p 22 ( t 0.02 r p ) ( t 00 + 0.06 r p 0 0.02 r p ) 00 0.6 r p 0 0. r p 0 t ( 0.02 r p ) ( t 0 + 0.06 r p 0.02 r p ) 0 0.6 r p 0. r p. ( t 0.02 r p ) ( t 20 + 0.06 r p 2 0.02 r p ) 20 0.6 r p 2 0. r p 2 ( 0 We are looking for an approximate value of the probability p ) 0. We can actually remove the superscript related to the starting point in time, because due to constant intensities of ( 0 transition, the starting time does not affect probabilities, so p ) 0 p 0, and similarly we will drop the superscript for other probabilities. The equation d dr p ( t ) ( t r 0 0.02 r p ) 0 + 0.06 r p Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
gives us this approximation or p 0 0.5 p 0 0.5 0.02 0.5 p 0 + 0.06 0.5 p, p 0.99 p + 0.03 p. 0 0.5 0 0.5 Additionally d dr p ( t ) ( t r 0 0.02 r p ) t 0 + 0.06 r p give us approximations or 0.5 p 0 0 p 0 0.5 d dr p ( t ) ( t r 0.02 r p ) 0 0.6 r p 0.02 0 p 0 + 0.06 0 p, 0.5 p 0 p 0.5 p 0.99 p + 0.03 p, p 0.0 p + 0.92 p. 0.5 0 0 0 0 0.5 0 0 0 But 0 p 0 0, 0 p, so that p 0.99 p + 0.03 p 0.5 0 0 0 0 0.99 0 + 0.03 0.03, p 0.0 p + 0.92 p 0.5 0 0 0 0.0 0 + 0.92 0.92. We conclude that p 0.99 p + 0.03 p 0 0.5 0 0.5 0.99 0.03+ 0.03 0.92 0.0573. Answer D. 0.02 0 p 0 0.6 0 p, November 202 Course MLC Examination, Problem No. 7 For a whole life insurance of 000 with semi-annual premiums on (80), you are given: (i) A gross premium of 60 is payable every 6 months starting at age 80. (ii) Commissions of 0% are paid each time a premium is paid. (iii) Death benefits are paid at the end of the quarter of death. (iv) t V denotes the gross premium reserve at time t. (v) 0.75 V 753.72. (vi) t l 90+t 0 000 0.25 898 0.50 800 0.75 706 (vii) i ( 4) 0.08. Calculate 0.25 V. A. 680 B. 690 C. 700 D. 70 E. 730 We can use the recursive reserve formula twice, in quarterly steps. Note that the effective quarterly interest rate is 2%, and that when going from policy duration 0.25 to 0.75, we only have a premium payment at policy duration 0.50. Thus Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
( 0.25V + 0).02 0.25 q 90.25 000 + 0.25 p 90.25 0.50 V, ( 0.50V + 60 ( 0.0 )).02 0.25 q 90.50 000 + 0.25 p 90.50 0.75 V. Now we substitute the following known data: V 753.72, 0.75 we obtain 0.25q 90.25 0.25 d 90.25 l 90.25 l 90.25 l 90.50 l 90.25 898 800 898 98 898, 0.25 p 90.25 l 90.50 800 l 90.25 898, d 0.25q 90.50 0.25 90.50 l 90.50 l 90.75 800 706 94 l 90.50 l 90.50 800 800, 0.25 p 90.50 l 90.75 706 l 90.50 800, 98 800 0.25V.02 000 + 898 898 V, 0.50 ( 0.50V + 54).02 94 800 000 + 706 800 753.72. From the second equation 94 706 000 + 0.50V 800 800 753.72 54 753.37.02 Using this in the first equation, we obtain 98 800 000 + V 898 898 753.37 0.25 729.9984..02 Answer E. November 202 Course MLC Examination, Problem No. 8 For a special fully discrete whole life insurance on (40), you are given: (i) The death benefit is 000 during the first years and 5000 thereafter. (ii) Expenses, payable at the beginning of the year, are 00 in year and 0 in years 2 and later. (iii) Π is the level annual premium, determined using the equivalence principle. (iv) G.02Π is the level annual gross premium. (v) Mortality follows the Illustrative Life Table. (vi) i 0.06. (vii) E 40 0.50330. Calculate the gross premium reserve at the end of year for this insurance. A. 70 B. 60 C. 50 D. 40 E. 30 Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
We begin by finding Π. The actuarial present value of benefits at issue is 000A 40 + 4 E 40 000A 5 6.32 + 4 0.50330 259.6 683.9669. The actuarial present value of expenses at issue is 00 +0a 40 00 +0( a 40 ) 00 +0 ( 4.866 ) 238.7. All the values used in calculations above came from the Illustrative Life Table. The premium is paid as a life annuity due on (40), hence 683.97 + 238.7 683.97 + 238.7 Π 62.2365. a 40 4.866 Therefore, G.02Π 63.482. If we calculated the reserve retrospectively, we would obtain G 000q 40.06 00 E 63.482 2.78.06 00 4.6056. 40 ( 0.00278).06 But prospectively, we obtain ( 000A 4 + 4 0 E 4 000A 5 ) + 0 a 63.482 a 4 4 Actuarial present value of future benefits Actuarial present value of future expenses Actuarial present value of future gross premiums 68.69 + 4 0.53499 259.6+0 4.6864 63.48 4.6864 6.202. The difference of these two is ( 4.6056) ( 6.202) 9.60. Note that the actuarial present value at policy inception of the profit portion of the gross premium is 0.02 Π a 40 0.02 62.2365 4.866 8.4427. Its actuarial accumulated value at policy duration is approximately 8.4427 8.4427 9.60. E 40 ( 0.00278).06 Let us write the actuarial equivalence for gross premium valuation at policy inception: Actuarial Present Value of Future Gross Premiums Actuarial Present Value of Future Benefits and Expenses + + Actuarial Present Value of Future Profits. Note that at policy inception, the actuarial present value of future benefits and expenses minus the actuarial present value of future gross premiums equals the opposite (i.e., minus) of the actuarial present value of future profits. Now consider a policy duration k, where k is an integer. Let us write APV for actuarial present value, and AAV for actuarial accumulated value. We have at policy inception: APV (Future Gross Premiums) APV (Future Gross Premiums up to time k) + + APV (Future Gross Premiums from time k on) APV (Future Benefits and Expenses) + APV (Future Profits) APV (Future Benefits and Expenses up to time k) + + APV (Future Benefits and Expenses from time k on) + + APV (Future Profits). Now we accumulate all of these amounts to policy duration k Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
AAV (Past Gross Premiums up to time k) + + APV (Future Gross Premiums from time k on) AAV (Past Benefits and Expenses up to time k) + + APV (Future Benefits and Expenses from time k on) + + Actuarial Value (Past and Future Profits). Therefore, at policy duration k, Retrospective Reserve AAV (Past Gross Premiums) AAV (Past Benefits and Expenses) APV (Future Benefits and Expenses) APV (Future Gross Premiums) + + Actuarial Value (Past and Future Profits) Prospective Reserve + Actuarial Value (Past and Future Profits). If you recall that the purpose of the reserve is to provide funds for benefits and expenses payment that are not paid by remaining premiums, you now should be able to clearly see that the prospective reserve calculation is the correct one, and if you do the calculation retrospectively, you must subtraction the actuarial value of all profits at the moment of calculation from the retrospective calculation result. In this problem, the reserve at policy duration is: 6.20 4.60 9.60. Interestingly, 40 was one of the answer choices, nice little trap waiting for the candidates taking the exam. The correct answer choice is 60, the one closest to 6.20. Answer B. November 202 Course MLC Examination, Problem No. 9 For whole life annuities-due of 5 per month on each of 200 lives age 62 with independent future lifetimes, you are given: (i) i 0.06. ( 2 (ii) A ) 62 0.4075 and 2 ( 2 A ) 62 0.205. (iii) Π is the single premium to be paid by each of the 200 lives. (iv) S is the present value random variable at time 0 of total payments made to the 200 lives. Using the normal approximation, calculate Π such that Pr(200Π > S) 0.90. A. 850 B. 860 C. 870 D. 880 E. 890 Let X, X 2,, X 200, be the random present values of life annuities-due considered. These are independent identically distributed random variables, and S X + X 2 +...+X 200. This implies that S is approximately normal and E S E( X + X 2 +...+X 200 ) 200E( X ) as well as Var( S) Var( X + X 2 +...+X 200 ) 200Var( X ). Recall that d ( 2) 2 + i 2. We have Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
and Based on this, E S and Var S We have Since S E S Var S 2 E( X ) 5 2 a 62 Var( X ) 2 2 A ( 5 2) 2 62 5 2 A ( 2) 62 A 62 2 0.4075 5 2 834.7545, 2.06 2 d 2 2 80 ( d ( 2) ) 2 2 200E( X ) 200 834.7545 366,936.2, 200Var( X ) 200 426,76.9086 85,235,38.7. Pr 0.90 Pr 200Π > S 200Π E S Var( S) 0.205 0.4075 2 2.06 426,76.9086. 2 2 > S E S Var S. 200Π E S is approximately standard normal, Var S must be the 90-th percentile of the standard normal distribution, i.e.,.282. This gives the equation.282 200Π E( S) 200Π 366,936.2 Var S 85,235,38.7, resulting in Π Answer E. 366,936.2 +.282 85,235,38.7 200 893.8597. November 202 Course MLC Examination, Problem No. 20 Stuart, now age 65, purchased a 20-year deferred whole life annuity-due of per year at age 45. You are given: (i) Equal annual premiums, determined using the equivalence principle, were paid at the beginning of each year during the deferral period. (ii) Mortality at ages 65 and older follows the Illustrative Life Table. (iii) i 0.06. (iv) Y is the present value random variable at age 65 for Stuart s annuity benefits. Calculate the probability that Y is less than the actuarial accumulated value of Stuart s premiums. A. 0.40 B. 0.42 C. 0.44 D. 0.46 E. 0.48 Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
Let us write P for the annual premium that Stuart has paid. Since it was calculated using the equivalence principle, we have P 20 E 45 a 65 a 45:20. Let us write K 65 for the curtate future lifetime of (65) (we could also write K(65), but we need to get used to the new notation, so this is a good time to practice). The probability we are looking for is (note that a 65 9.8969 from the Illustrative Life Table) P a Pr 45:20 > a 20 E K65 + 45 Pr 20 E 45 a 65 a 45:20 > a a 45:20 20 E K65 + 45 Pr( a 65 > a K65 + ) Pr( 9.8969 > a K65 + ). If we use BA II Plus Pro, set BGN mode by pushing 2ND, BGN if BGN is not displayed, and enter 0 FV, PMT, 9.8969 PV, 6 I/Y, CPT N, we obtain 4.0973895. This means that the probability sought is Pr K 65 + < 4.0973895 Pr( K 65 + 4) 4 p 65 l 79 422563 l 65 7533964 0.439733. For a moment, when you see the formula for P you might worry about working with mortality from before age 65, when we are told that mortality follows the Illustrative Life Table only from age 65 on, but as you can see, items related to mortality from before age 65 cancel nicely in the work involved in finding the probability sought. Answer C. November 202 Course MLC Examination, Problem No. 2 For a fully continuous whole life insurance issued on (x) and (y), you are given: (i) The death benefit of 00 is payable at the second death. (ii) Premiums are payable until the first death. (iii) The future lifetimes of (x) and (y) are dependent. (iv) t p xy 4 e 0.0t + 3 4 e 0.03t, t 0. (v) t p x e 0.0t, t 0. (vi) t p y e 0.02t, t 0. (vii) δ 0.05. Calculate the annual benefit premium rate for this insurance. A. 0.96 B..43 C..9 D. 2.39 E. 2.86 Note that the future lifetime of (x) is ruled by constant force of mortality of %, and the future lifetime of (y) is rules by constant force of mortality of 2%. Let us write P for the annual benefit premium rate sought. We have Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
Also, P 00A x:y a x:y. a x:y Therefore, Answer A. + e 0.05t 4 e 0.0t + 3 4 e 0.03t dt 0 4 + e 0.06t dt + 3 4 0 + e 0.08t dt 4 0.06 + 3 4 0.08 3.546667, A x:y Poohsticks A x + A y A x:y A x + A y δ a x:y 0 0.0 0.0+ 0.05 + 0.02 ( 0.05 3.546667 ) 0.02 + 0.05 0.2946429. P 00A x:y 00 0.2946429 0.95604396. a x:y 3.546667 November 202 Course MLC Examination, Problem No. 22 You are given the following information about a special fully discrete 2-payment, 2-year term insurance on (80): (i) Mortality follows the Illustrative Life Table. (ii) i 0.075. (iii) The death benefit is 000 plus a return of all premiums paid without interest. (iv) Level premiums are calculated using the equivalence principle. Calculate the benefit premium for this special insurance. A. 82 B. 86 C. 90 D. 94 E. 98 Let us write P for the level annual premium sought. Based on the equivalence principle, we write so that We calculate P a 80:2 000A 80:2 000A 80:2 P a 80:2 IA 80:2. + P ( IA) 80:2, a 80:2 + vp 80 + q 80 0.0803 +.075.075.90388206, Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
and Therefore, Answer D. 000A 80:2 ( IA) 80:2 P a 80:2 IA q 80 000.075 + p q 80 8.075 2 000 0.0803.075 q 80.075 + 2 p 80q 8 000A 80:2 80:2 + 0.997 0.08764.075 2.075 0.0803 2.075 56.772702, + 2 0.997 0.08764.075 2 0.23462648. 56.772702.90388206 0.23462648 93.977342. November 202 Course MLC Examination, Problem No. 23 A life insurance company issues fully discrete 20-year term insurance policies of 000. You are given: (i) Expected mortality follows the Illustrative Life Table. (ii) Death is the only decrement. (iii) 3 V, the reserve at the end of year 3, is 2.8. On January, 2009, the company sold 0,000 of these policies to lives all aged 45. You are also given: (i) During the first two years, there were 30 actual deaths from these policies. (ii) During 20, there were 8 actual deaths from these policies. Calculate the company s gain due to mortality for the year 20. A. 28,00 B. 28,300 C. 28,500 D. 28,700 E. 28,900 The mortality gain equals ( Expected deaths Actual deaths) Net Amount at Risk. In this case, the net amount at risk is 000 3 V 000 2.8 987.82. Expected deaths are q 47 from the Illustrative Life Table, i.e., 0.00466, times the number of alive policyholders at the beginning of 20, which is 0,000 30 9970, which gives 0.00466 9970 46.4602. The actual number of deaths in 20 is 8. Hence, the mortality gain equals ( 46.4602 8) 987.82 28,3.5548. Answer A. November 202 Course MLC Examination, Problem No. 24 An insurance company is designing a special 2-year term insurance. Transitions are modeled as a four-state homogeneous Markov model with states: Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
(H) Healthy (Z) infected with virus Zebra (L) infected with virus Lion (D) Death The annual transition probability matrix is given by: H Z L D H 0.90 0.05 0.04 0.0 Z 0.0 0.20 0.00 0.70 L 0.20 0.00 0.20 0.60 D 0.00 0.00 0.00.00 You are given: (i) Transitions occur only once per year. (ii) 250 is payable at the end of the year in which you become infected with either virus. (iii) For lives infected with either virus, 000 is payable at the end of the year of death. (iv) The policy is issued only on healthy lives. (v) i 0.05. Calculate the actuarial present value of the benefits at policy issue. A. 66 B. 75 C. 84 D. 93 E. 02 We list all possible cases of payouts, with their probabilities and actuarial present values (calculated as products of probabilities and discounted benefits) Possible Probability Discounted Actuarial transition benefit present value H à Z 0.05 250.05.904769 H à L 0.04 250.05 9.52380952 H à Z à D 0.05 0.7 000.05 2 3.746037 H à L à D 0.04 0.6 H à H à Z 0.9 0.05 H à H à Z 0.9 0.04 000.05 2 2.7687075 250.05 2 0.204086 250.05 2 8.632653 The total actuarial present value of the benefits at policy issue is the sum of all numbers in the last column, which is approximately 93.306576. Answer D. November 202 Course MLC Examination, Problem No. 25 For a fully discrete 0-year term life insurance policy on (x), you are given: Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without
(i) Death benefits are 00,000 plus the return of all gross premiums paid without interest. (ii) Expenses are 50% of the first year s gross premium, 5% of renewal gross premiums and 200 per policy expenses each year. (iii) Expenses are payable at the beginning of the year. (iv) A x:0 0.7094. x:0 (v) IA 0.96728. (vi) a x:0 6.8865. Calculate the gross premium using the equivalence principle. A. 3200 B. 3300 C. 3400 D. 3500 E. 3600 Let us write G for the gross premium sought. Using the equivalence principle, we write: G a x:0 00,000A x:0 APV of death benefit x:0 + G IA + 0.45G + 0.05G a x:0 + 200 a x:0. APV of return of premium benefit We solve for G and obtain G 00,000A + 200 a x:0 x:0 0.95 a x:0 IA 0.45 Answer E. x:0 % of premium expenses Per policy expenses 00,000 0.7094 + 200 6.8865 0.95 6.8865 0.96728 0.45 3604.2299. Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without