# November 2012 Course MLC Examination, Problem No. 1 For two lives, (80) and (90), with independent future lifetimes, you are given: k p 80+k

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2 Calculate q [ 60] 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 q [ 60] l [ l 60]+2.75 [ 60] l [ 60]+0.75 l x k, l ( l [ 60]+2 ) l [ 60]+0.75 Answer B. 000 ( l [ 60]+2 ) 0.25 l 63 ( l [ 60] ) 0.25 l l l 66 ( [ ]+ ) November 202 Course MLC Examination, Problem No. 3 You are given: (i) S ( 0 t ) t 4 ω, for 0 t ω. (ii) µ 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

3 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 t e 06 p p p p t 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 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 ( ) 50,000 A E 50 A 60 50, 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

5 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 ,000, E , so that d t V dt t V ( 06, V ) V. Substituting into the Euler s formula, we obtain 9.6V 9.8 V 0.2 d tv This results in dt t V Answer B ( V ) 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 and q x 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 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

6 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 reduced by surrender charge of 20, i.e., 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 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% % (ii) The credited interest rate is i 2) (iii) The cash surrender value at the end of month is (iv) The cash surrender value at the end of month 3 is 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 Note that the cash surrender value equals the account value (i.e., reserve) minus the surrender charge. We are given that CSV Based on this, 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

7 AV Similarly, AV CSV + SC 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 ( AV ( 0.5) 0) , AV 2 ( ( W ) 0) From the first equation AV and substituting this into the second equation, we obtain W Answer A ( 0.5) , 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 (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 and 000q (vii) i 7%. Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without

8 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 ( ) Following into the second year, ( e 2 )( + i 2 ) q 6 B AV 2 AV + G 2 r 2 ( ) 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 ), so that the present value at time 0 of the expected surrender benefits in year 2 is ( 0.06) ( ) ( ).07 2 Account value available for surrenders 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 , k 0,, 2. (iii) p y+k , k 0,, 2. (iv) p x+k:y+k , 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

9 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 p p p x:y x:y x+:y The expected present value we are looking for is therefore 00, q + 00, x:y p x:yq x+:y+ + Present value of death benefit payment made at time + 00, 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 Probability that both ( x) and ( y) die during second year , 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

10 q 70 q 7 000A p p 70 p 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 , q , A , and therefore p , p , so that 000A Answer C 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 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 µ 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 B C D E 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 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 t e e µ 0 0 ( t +0.5µ t )dt ln2 ln2 e.5( t )dt e ln ( t ) 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

11 s e 0.05, and n p x e 0.05n e 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 ln 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, ,045, , , , (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, , q 50 3 l 50 τ d ( 3) 50,00 00, , l 50 τ ( τ q ) ( τ 50 p ) We calculate therefore q 50 ( 3 q 5 ) ( 3 q 50 ) ( 3 p 50 ) ( τ ) p 50 ( 3 This means that p 5 ) and q 5 ( p 5 ) ( τ ) p 5 From this we calculate ( 3) ( τ ) q 5 ( 3) ( τ ) q 50 80,000 90, ( 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

12 80,000 ( 3 q ) 90, 365 ln ln 80, , 365 This means that ( 3 d ) ( 3 5 q ) ( τ 5 l ) , , 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 ,8, ( q ) 5 d 5 ( τ ),8 l 5 90, , and we also have ( τ p ) 5 80,000 90, , ( τ q ) ( τ 5 p ) 5 80,000 90, 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 ( p 5 ) 0, 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 (v) x A x 20 E x (vi) E Z Calculate the standard deviation of Z. A B C D E 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

13 2v K+, K < 20, Z v K+, K 20. Therefore, E Z We conclude that the variance of Z is 2A E 40 A Var Z E Z 2 and the standard deviation of Z is σ Z Answer A. Var Z ( E( Z )) , 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 (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 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 , 0.5q x p x+0.5 p 0.5 x , 0.5q x+ 0.5 p x+ e 0.5µ x+ p 0.5 x , 0.5q x p x+.5 e 0.5µ x+ p 0.5 x 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 , q x 0.5 p x 0.5 q x , q p q 0.5 x x 0.5 x+ ( 0.6) , q p p q 0.5 x x 0.5 x+ 0.5 x+.5 ( 0.6) 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

14 300, , with probability 0.5 q x ,.9 330, , with probability.9 Z q x , 360, , with probability.9 q , x 390, , with probability.9 q x Also, Z 0 otherwise. We conclude that Pr Z > 277, Answer D q x 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 the following constant transition intensities: (i) (ii) (iii) (ii) 0 µ µ (iii) µ 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 B C D E 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) (B) 0.05 (C) (D) λ 0 λ 02 λ 0 State 2 Liquidated λ 0 µ , λ 02 µ , λ 0 λ 0 + λ , Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without

15 λ 0 λ 2 λ λ 20 λ 2 λ 2 λ 0 µ , λ 2 µ 2 0.0, t λ λ 0 + λ , λ 20 µ , λ 2 µ , λ 2 λ 20 + λ 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 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 d d d. dr r p 02 dr r p 2 dr r p 22 ( t 0.02 r p ) ( t r p r p ) r p 0 0. r p 0 t ( 0.02 r p ) ( t r p 0.02 r p ) r p 0. r p. ( t 0.02 r p ) ( t r p r p ) 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 r p ) r p Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without

17 ( 0.25V + 0) q p V, ( 0.50V + 60 ( 0.0 )) q p V. Now we substitute the following known data: V , 0.75 we obtain 0.25q d l l l l , 0.25 p l l , d 0.25q l l l l , 0.25 p l l , V V, 0.50 ( 0.50V + 54) From the second equation V Using this in the first equation, we obtain V 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 (vii) E 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

20 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) A ( 2) 62 A , d ( d ( 2) ) E( X ) ,936.2, 200Var( X ) , ,235,38.7. Pr 0.90 Pr 200Π > S 200Π E S Var( S) , > 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 Π E( S) 200Π 366,936.2 Var S 85,235,38.7, resulting in Π Answer E. 366, ,235, 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 (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 B C D E Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without

22 Also, P 00A x:y a x:y. a x:y Therefore, Answer A. + e 0.05t 4 e 0.0t e 0.03t dt e 0.06t dt e 0.08t dt , A x:y Poohsticks A x + A y A x:y A x + A y δ a x:y ( ) P 00A x:y a x:y 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 (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 , Copyright 202 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without

23 and Therefore, Answer D. 000A 80:2 ( IA) 80:2 P a 80:2 IA q p q q p 80q 8 000A 80:2 80: , 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 V Expected deaths are q 47 from the Illustrative Life Table, i.e., , times the number of alive policyholders at the beginning of 20, which is 0, , which gives The actual number of deaths in 20 is 8. Hence, the mortality gain equals ( ) , 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

24 (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 Z L D 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 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 H à L H à Z à D H à L à D H à H à Z H à H à Z The total actuarial present value of the benefits at policy issue is the sum of all numbers in the last column, which is approximately 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

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