Solution. Let us write s for the policy year. Then the mortality rate during year s is q 30+s 1. q 30+s 1

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1 Solutions to the May 213 Course MLC Examination by Krzysztof Ostaszewski, Copyright 213 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://wwwactexmadrivercom) and the Actuarial Bookstore (http://wwwactuarialbookstorecom) Exam MLC seminar at Illinois State University: May 213 Course MLC Examination, Problem No 1 For a fully discrete whole life insurance of 1 on (3), you are given: (i) Mortality follows the Illustrative Life Table (ii) i 6 (iii) The premium is the benefit premium Calculate the first year for which the expected present value at issue of that year s premium is less than the expected present value at issue of that year s benefit A 11 B 15 C 19 D 23 E 27 Let us write s for the policy year Then the mortality rate during year s is q 3+s 1 We are looking for the smallest value of s such that the expected present value at issue of that year s premium is less than the expected present value at issue of that year s benefit, or 16 ( s 1) s 1 p 3 1P < 3 16 s s 1 p 3 q 3+s 1 1 expected present value at issue of year s premium expected present value at issue of year s benefit Since the interest rate is 6%, and mortality follows the Illustrative Life Table, we can get the value of 1P 3 from values given in the table 1P 3 1A a Hence we have this inequality 16 ( s 1) s 1 p < 16 s s 1 p 3 q 3+s 1 1, or 1q 3+s 1 > The first age in the table where 1 times mortality rate exceeds is age 52 We set 3 + s 1 52, and conclude that s 23 Answer D

2 May 213 Course MLC Examination, Problem No 2 P&C Insurance Company is pricing a special fully discrete 3-year term insurance policy on (7) The policy will pay a benefit if and only if the insured dies as a result of an automobile accident You are given: (i) x ( τ ) l x ( 1) d x ( 2) d x ( 3) d Benefit x , , ( 1) where d x ( 2 represents deaths from cancer, d x represents deaths from automobile ( 3) accidents, and d x represents deaths from all other causes (ii) i 6 (iii) Level premiums are determined using the equivalence principle Calculate the annual premium A 122 B 133 C 144 D 155 E 166 The expected present value of benefits is Let P be the annual premium sought The expected present value of premiums is P + P P P Since the premium is set by the equivalence principle, those two quantities must be equal, and therefore P Answer A May 213 Course MLC Examination, Problem No 3 For a special fully discrete 2-year endowment insurance on (4), you are given: (i) The only death benefit is the return of annual benefit premiums accumulated with interest at 6% to the end of the year of death (ii) The endowment benefit is 1, (iii) Mortality follows the Illustrative Life Table (iv) i 6 Calculate the annual benefit premium A 2365 B 2465 C 2565 D 2665 E 2765

3 Let us write P for the annual benefit premium sought Note that in case of death, the death benefit is the accumulated value of all premiums paid up to that point, with interest, at the end of the year of death, so all dying policyholders simply get their own premiums with interest back, and any remaining premiums are those for surviving policyholders On the other hand, reserve is only calculated for surviving policyholders This is the central idea of the most efficient solution At policy duration 2, the benefit reserve equals, prospectively, the actuarial present value of future benefits minus the actuarial present value of future premiums, and since there are no future premiums, it simply equals 2V 1, On the other hand, retrospectively, the reserve equals the actuarial accumulated value of past premiums minus the accumulated actuarial value of past benefits But the past benefits were paid to those policyholders who died by a return of their own premiums with interest, which means that any past premiums left are the ones paid by policyholders surviving till policy duration 2, and since the reserve is calculated only for surviving policyholders, the reserve equals 2 V P s 2 6% Equating the two formulas for the same reserve at policy duration 2, we obtain 1, P s 2 6% Based on this, Answer C P 1, s 2 6% 2,56458 May 213 Course MLC Examination, Problem No 4 Employment for Joe is modeled according to a two-state homogeneous Markov model with states: Actuary (Ac) and Professional Hockey Player (H) You are given: (i) Transitions occur December 31 of each year The one-year transition probabilities are: Ac H Ac 4 6 H 8 2 Ac (ii) Mortality for Joe depends on his employment: q 35+k k for k, 1, 2 H q 35+k 1 + 5k for k, 1, 2, (iii) i 8 On January 1, 213, Joe turned 35 years old and was employed as an actuary On that date, he purchased a 3-year pure endowment of 1, Calculate the expected present value at issue of the pure endowment A 32,51 B 36,43 C 4,35 D 44,47 E 48,58 The specific mortalities given by the formulas in (ii) are: q Ac 35 1, q Ac 36 15, q Ac 37 2, q H 35 25, q H 36 3, q H Over the next three years, Joe has the following paths to collection of the endowment benefit at the end of that period (with

4 probabilities for each step indicated below): 1/1/13 12/31/13 1/1/14 12/31/14 1/1/15 12/31/15 Actuary Alive Actuary Alive Actuary Alive /1/13 12/31/13 1/1/14 12/31/14 1/1/15 12/31/15 Actuary Alive Actuary Alive Hockey Alive /1/13 12/31/13 1/1/14 12/31/14 1/1/15 12/31/15 Actuary Alive Hockey Alive Hockey Alive /1/13 12/31/13 1/1/14 12/31/14 1/1/15 12/31/15 Actuary Alive Hockey Alive Actuary Alive Therefore, the probability of collecting the 1, endowment benefit is The expected present value at issue of the pure endowment is , ,3528 Answer C May 213 Course MLC Examination, Problem No 5 5 The The joint joint lifetime lifetime of Kevin, of Kevin, age age 65, 65, and and Kira, Kira, age age 6, 6, is modeled is modeled as: as: State Kevin alive Kira alive 2 µ 1 µ State 1 3 µ Kevin alive Kira dead 13 µ You are given the following constant transition intensities: You are given the following constant transition intensities: (i) µ 1 4 (ii) µ 2 5 (i) State 2 Kevin dead Kira alive 1 µ 4 23 µ State 3 Kevin dead Kira dead 2 (ii) µ 5 Copyright 213 by Krzysztof 3 Ostaszewski All rights reserved No reproduction in any form is permitted without explicit permission (iii) of the copyright µ 1 owner (iv) (v) 13 µ 23 µ 1 8

5 (iii) µ 3 1 (iv) µ 13 1 (v) µ Calculate 1 p 65:6 A 46 B 48 C 5 D 52 E 54 2 The probability 1 p 65:6 is the probability of transition from state to state 2 in such a way that the joint life status (65:6) is in state at time and is in state 2 at time 1 The subscript 65:6 indicates that the probability refers to the joint life status (65:6) The probability sought equals (remember that under constant forces of transition, the probability of remaining in state j over period of time t equals the expression of the form exp( t times the sum of forces of transition out of state j) ) Answer A p :6 1 p µ 2 t 65:6 dt p 22 1 t 65+t:6+t 1 Remain in state for t years ( )t Transition from state to state 2 in period ( t,t+dt) Remain in state 2 til the end of 1 years e µ1 +µ 2 +µ 3 µ 2 e µ23( 1 t ) dt e 1t 8 1 t 5 e ( ) dt 1 5 e 1t e 8 e 8t dt 5 e e 1t e 8t dt e e 2t dt 5 a 8 e 8 1 2% 5 e e May 213 Course MLC Examination, Problem No 6 For a wife and husband ages 5 and 55, with independent future lifetimes, you are given: 1 (i) The force of mortality on (5) is µ 5+t, for t < 5 5 t (ii) The force of mortality on (55) is µ 55+t 4, for t > (iii) For a single premium of 6, an insurer issues a policy that pays 1 at the moment of the first death of (5) and (55) (iv) δ 5 Calculate the probability that the insurer sustains a positive loss on the policy A 45 B 47 C 49 D 51 E 53 The force of mortality on (5) represents De Moivre s Law, so that t p 5 5 t 5 1 t 5

6 for t 5 The force of mortality on (55) is constant, so that t p 55 e 4t for t Since the future lifetimes are independent, t p 5:55 1 t 5 e 4t for t 5, but for t > 5, t p 5:55 because (5) will be dead with certainty for t > 5 Let us write (using two possible notations for the future lifespan) T 5:55 T ( 5 :55) min( T 5,T 55 ) min( T ( 5),T ( 55) ) for the future lifespan of the joint life status (5:55) The loss of the insurer on this policy is given by the formula L 1e 5T 5:55 6 We want to know the probability that L > We have Answer B ( ) Pr 1e 5T 5:55 6 > Pr L > ( ) Pr 1e 5T 5:55 ( > 6 ) ( ) Pr lne 5T 5:55 ( > ln6 ) Pr e 5T 5:55 > 6 ( ) Pr T 5:55 < ln6 Pr 5T 5:55 > ln6 ln6 1 Pr T 5:55 ln e 4 ln ln6 25 ( e ln6 ) ln May 213 Course MLC Examination, Problem No 7 You are given: (i) q 6 1 (ii) Using i 5, A 6: Using i 45 calculate A 6:3 A 866 B 87 C 874 D 878 E 882 Using i 5, A 6:3 q 6 ( q 6 )q 61 ( + 1 q 6 )( 1 q 61 ) But we are given that q 6 1, and using that, we obtain so that ( ) q q 61, q ( 1 q 61 ),

7 q Now we calculate using i 45 A 6:3 q 6 ( q 6 )q 61 ( + 1 q 6 )( 1 q 61 ) Answer D May 213 Course MLC Examination, Problem No 8 For a special increasing whole life insurance on (4), payable at the moment of death, you are given: (i) The death benefit at time t is b t 1+ 2t, t (ii) The interest discount factor at time t is v t ( ) ( 1+ 2t) 2, t 25, t < 4 (iii) t p 4 µ 4+t, otherwise (iv) Z is the present value random variable for this insurance Calculate Var(Z) A 36 B 38 C 4 D 42 E 44 We have Also, ( ) E b T v( T ) E Z Therefore, E Z 2 ( ) E 1+ 2T f T ( t) 1+ 2t dt t dt 125 ln9 ln1 ( ) E ( b T v( T )) 2 Var Z Answer A (( )( 1+ 2T ) 2 ) E ( 1+ 2T ) 1 ( ) ( ) E 1+ 2T ( ) ( ) dt 2 ( ) 1 ln ( 1+ 2t ) t4 4 2 t ((( ) ( 1+ 2T ) 2 ) ) f T t 1 ( 1+ 2t 1+ 2t 4 ) 2 dt t E( ( 1+ 2T ) ) 2 ( ) ( ) E Z 2 ( ) E( Z 2 ) t4 ( ) t

8 May 213 Course MLC Examination, Problem No 9 For a fully discrete whole life insurance of 1, on (x), you are given: (i) Deaths are uniformly distributed over each year of age (ii) The benefit premium is (iii) The benefit reserve at the end of year 4 is 1458 (iv) q x (v) i 3 Calculate the benefit reserve at the end of 45 years A 157 B 168 C 175 D 183 E 19 Let us write t V for the benefit reserve at policy duration t, and P for the bene We use a slightly modified standard recursive reserve formula ( 4V + P) ( 1+ i) 5 5 p x+4 45 V + 5q x+4 1, ( 1+ i) 5 Reserve from policy duration 4 with premium accumulated with interest pays for the items on the right-hand side Reserve at policy duration 45, of course only held for surviving policyholders Death benefits for those who die by policy duration 45, note that we have to discount for half a year, as the death benefit is paid at the end of the year We substitute known values and obtain ( ) p x+4 45 V + 5 q x+4 1, 13 5 We are not given 5 q x+4 but we know q x and we know that the UDD assumption applies Hence q 5q x+4 x+4 UDD and p 1 q x+4 5 x+4 Substituting this, we obtain ( ) V , 13 5, so that Answer E 45V ( ) , ,9177 May 213 Course MLC Examination, Problem No 1 1 A multi-state A multi-state model model is being is being used to used value to value sickness sickness benefit benefit insurance: insurance: healthy (h) sick (s) dead (d) For a policy on (x) you are given: For a (i) policy Premiums on (x) are you payable are given: continuously at the rate of P per year while the policyholder is healthy (i) Premiums are payable continuously at the rate of P per year while the policyholder is healthy (ii) (iii) Sickness benefits are payable continuously at the rate of B per year while the policyholder is sick There are no death benefits

9 (ii) Sickness benefits are payable continuously at the rate of B per year while the policyholder is sick (iii) There are no death benefits ij (iv) µ x+t denotes the intensity rate for transition from i to j, where i, j s, h or d (v) δ is the force of interest (vi) t V ( i) is the reserve at time t for an insured in state i where i s, h or d Which of the following gives Thiele s differential equation for the reserve that the insurance company needs to hold while the policyholder is sick? A d dt t δ t sh + B µ x+t B d dt t δ t sh B µ x+t C d dt t δ t sh + B µ x+t D d dt t δ t sh B µ x+t E d dt t δ t sh B µ x+t ( V ( h) t ) t ( V ( h) t ) t V ( h) sd ( t ) µ t x+t V ( h) sd ( t ) µ t x+t V ( h) sd ( t ) + µ t x+t t t t The rate of change of reserve held while the policyholder is sick, d dt t, has the following components: Instantaneous interest on the current reserve: δ t, Rate of premiums received while in state s: (premium is paid only while the policyholder is healthy), Rate of benefits paid (outgo, so a negative component) while in state s: B, Transition intensity for transition to state h, times the change in reserve upon transition (as a result of such transition, the insurer must hold reserve for h and release reserve for s): µ sh x+t ( tv ( h) t ), Transition intensity for transition to state d, times the change in reserve upon transition (as a result of such transition, the insurer releases reserve for s, and there is no reserve for state d, because the policy ends): µ sd x+t ( t ) By adding all of those terms, we obtain d dt t δ t sh B µ V ( h) sd x+t ( t ) + µ t x+t t Answer E May 213 Course MLC Examination, Problem No 11 For a one-year term insurance on (45), whose mortality follows a double decrement

10 model, you are given: (i) The death benefit for cause (1) is 1 and for cause (2) is F (ii) Death benefits are payable at the end of the year of death ( 1 (iii) q ) ( and q ) 45 2 (iv) i 6 (v) Z is the present value random variable for this insurance Calculate the value of F that minimizes Var(Z) A B 5 C 167 D 2 E 5 Z can have the following values: with probability 4, F 16 1 with probability 2, with probability Therefore, the mean of Z is F 16 1, while the second moment of Z is ( ) ( F 16 1 ) The variance of Z is Var Z ( ) ( ) ( F 16 1 ) 2 ( F 16 1 ) ( )2 + 2 F F F F 2 16F This numerator of this expression is a quadratic function of F, which is minimized for F Answer B May 213 Course MLC Examination, Problem No 12 Russell entered a defined benefit pension plan on January 1, 2, with a starting salary of 5, You are given: (i) The annual retirement benefit is 17% of the final three-year average salary for each year of service (ii) His normal retirement date is December 31, 229 (iii) The reduction in the benefit for early retirement is 5% for each year prior to his normal retirement date (iv) Every January 1, each employee receives a 4% increase in salary (v) Russell retires on December 31, 226

11 Calculate Russell s annual retirement benefit A 49, B 52, C 55, D 58, E 61, The final three-year average salary is 5, ,362 3 Based on this, the annual retirement benefit is approximately , ,3 Answer B 17% for each year of service Number of years of service Final average salary 5% reduction for each of three years of early retirement May 213 Course MLC Examination, Problem No 13 An automobile insurance company classifies its insured drivers into three risk categories The risk categories and expected annual claim costs are as follows: Risk Category Expected Annual Claim Cost Low 1 Medium 3 High 6 The pricing model assumes: At the end of each year, 75% of insured drivers in each risk category will renew their insurance i 6 All claim costs are incurred mid-year For those renewing, 7% of Low Risk drivers remain Low Risk, and 3% become Medium Risk 4% of Medium Risk drivers remain Medium Risk, 2% become Low Risk, and 4% become High Risk All High Risk drivers remain High Risk Today the Company requires that all new insured drivers be Low Risk The present value of expected claim costs for the first three years for a Low Risk driver is 317 Next year the company will allow 1% of new insured drivers to be Medium Risk Calculate the percentage increase in the present value of expected claim costs for the first three years per new insured driver due to the change A 14% B 16% C 19% D 21% E 23% There are three states: Low Risk, Medium Risk and High Risk The transition matrix is 7 3 Q and the two-step transition matrix is

12 Q We know that the present value of expected claim costs for the first three years for a Low Risk driver is 317 The present value of expected claim costs for the first three years for a Medium Risk driver is ( ) Probability of renewing + ( ) Probability of renewing twice Therefore, the present value of costs for the new portfolio is The rate of increase is approximately Answer A May 213 Course MLC Examination, Problem No 14 For a universal life insurance policy with a death benefit of 15,, you are given: (i) Policy Monthly Percent of Monthly Cost Monthly Year Premium Premium of Insurance Expense Charge Rate per 1 Charge % 1 5 (ii) i 12) 6 (iii) The account value at the end of month 11 is 25, Calculate the account value at the end of month 12 A 26,83 B 26,85 C 26,87 D 26,89 E 26,91 We have the standard recursive formula (we have only one interest rate here, no difference between credited rate and rate used for discounting) AV End ( AV Start + P( 1 f ) e COI )( 1+ i), where COI DB AV End End ( COI rate) 1+ i Let us write AV t for the account value at the end of month t Then ( ( ) e COI )( 1+ i) AV 12 AV 11 + P 1 f ( 25, + 2 ( 1 35) 5 COI ) COI

13 Also, so that ( ) 15 15, COI COI COI, COI We conclude that AV COI Answer D May 213 Course MLC Examination, Problem No 15 For fully discrete whole life insurance policies of 1, issued on 6 lives with independent future lifetimes, each age 62, you are given: (i) Mortality follows the Illustrative Life Table (ii) i 6 (iii) Expenses of 5% of the first year gross premium are incurred at issue (iv) Expenses of 5 per policy are incurred at the beginning of each policy year (v) The gross premium is 12% of the benefit premium (vi) L is the aggregate present value of future loss at issue random variable Calculate Pr( L < 6,), using the normal approximation A 74 B 78 C 82 D 86 E 9 The aggregate present value of future loss random variable is the sum of 6 individual policies present value of future loss random variables, which are independent and identically distributed, thus by the Central Limit Theorem the aggregate present value of future loss random variable can be approximated by a normal random variable with the same mean and variance Let us begin by finding the mean and variance of L We will need to know the gross premium for this calculation The benefit premium is 1,A 62 a ILT 62 Therefore, the gross premium is G Let us denote by L Single the present value of future loss at issue random variable for a single policy Then

14 ( ) L Single 1 16 ( K+1) G 5 a + 5G K+1 Gross premium after recurrent expense of 5 a K+1 1 vk+1 d One time expense of 5% of gross premium at issue, it contributes to the loss 1 16 ( K+1) ( K+1 ) ( K+1) ( K+1) The expected value of L Single is ( ) A E L Single ILT The variance of L Single Var L Single is ( ) ( A 62 A 62 ) ILT ( ) 11,61,2929, ILT ILT so that the standard deviation is approximately 11,61,2929 3,47388 The aggregate loss present value of future loss at issue random variable L is the sum of six hundred independent identically distributed random variables with the mean and variance we have just calculated Therefore, E L and ( ) 6 ( ) , Var( L) 6 3, , Therefore, if we write Z for a standard normal random variable, and Φ for its cumulative distribution function, we obtain Pr( L < 6,) Pr L ( ) 6, < ( ) 83, , ( ) 83, , Pr Z < ( ) 7792 Φ Answer B Pr ( Z < ) May 213 Course MLC Examination, Problem No 16 For a fully discrete whole life insurance policy of 2 on (45), you are given: (i) The gross premium is calculated using the equivalence principle

15 (ii) Expenses, payable at the beginning of the year, are: % of Premium Per 1 Per Policy First year 25% 15 3 Renewal years 5% 5 1 (iii) Mortality follows the Illustrative Life Table (iv) i 6 Calculate the expense reserve at the end of policy year 1 A 2 B 1 C 14 D 19 E 27 While this is not stated clearly, it is implied by the structure of expenses that all premiums are level annual premiums The expense reserve equals the actuarial present value of future expenses minus the actuarial present value of future expense premiums Or, equivalently, it equals the actuarial accumulated value of past expense premiums minus the actuarial accumulated value of past expenses In order to know the expense premiums, we must know the gross and net (ie, benefit) premium The benefit premium is simpler to calculate: 2A a ILT Since the gross premium is calculated using the equivalence principle, it must satisfy the equation G a 45 2A a 45 + First year extra expenses, per thousand and per policy + 2G + 5G a 45 First year percentage of premium expenses in excess of such expenses in all renewal years Percentage of premium expenses that are the same in all years Recurring expenses, per thousand and per policy From this, we can solve for G and obtain G a 45 2A a G + 5G a 45, and G 2A a a 45 2 ILT Based on this, the expense premium is approximately At policy duration 1, the expense reserve equals a + 5G a a ( ) a 55 ILT ( ) ILT Of course, you should know that there is nothing strange about having a negative expense reserve Answer E

16 May 213 Course MLC Examination, Problem No 17 You are profit testing a fully discrete whole life insurance of 1, on (7) You are given: (i) Reserves are benefit reserves based on the Illustrative Life Table and 6% interest (ii) The gross premium is 8 (iii) The only expenses are commissions, which are a percentage of gross premiums (iv) There are no withdrawal benefits (v) ( death) ( withdrawal) Policy Year k q 7+k 1 q 7+k 1 Commission Rate Interest Rate Calculate the expected profit in policy year 2 for a policy in force at the start of year 2 A 18 B 19 C 2 D 21 E 22 The benefit premium (used for benefit reserves calculations) is 1P 7 1,A a ILT The benefit reserve at policy duration 1 is V 1 A 1 7 ( P a ) ILT The benefit reserve at policy duration 2 is 2V 7 1A 72 P a 71 ILT The general recursive formula, which shows emergence of the expected profit in policy year t is: ( )( 1+ i) Sq x+t 1 + t Vp x+t 1 + Pr t P + t 1 V E t The version of it that applies to policy year 2 in this problem is: ( death) ( P + 1 V E 1 )( 1+ i) Sq 71 + ( withdrawal) q7+k Vp 71 + Pr 2 Therefore, ( death) Pr 2 ( P + 1 V E 1 )( 1+ i) Sq 71 ( death) 2 V 1 q 71 ( withdrawal ) ( q7+k 1 ) ( ) 17 1, Answer E ( ) May 213 Course MLC Examination, Problem No 18 An insurance company sells special fully discrete two-year endowment insurance policies to smokers (S) and non-smokers (NS) age x You are given: (i) The death benefit is 1, The maturity benefit is 3, (ii) The level annual premium for non-smoker policies is determined by the equivalence principle

17 (iii) The annual premium for smoker policies is twice the non-smoker annual premium (iv) µ NS x+t 1, t > S (v) q x+k 15q NS x+ j, for k,1 (vi) i 8 Calculate the expected present value of the loss at issue random variable on a smoker policy A 3, B 29, C 28, D 27, E 26, Given that the force of mortality for non-smokers is constant, q NS x q NS x+1 1 e And, of course, S q x+k 15q NS x+ j Since we are not told that the premium for smokers is determined by the equivalence principle, while it is the case for non-smokers, we must conclude that the premium for smokers is not derived from the equivalence principle, so the only way to find it is by finding the premium for non-smokers and doubling it Let us find the premium for nonsmokers, P NS, which is determined by the equivalence principle: so that ( ) P NS a x:2 P NS p x NS 1, 18 1 q NS x +1, 18 2 p NS x q NS x+1 + 3, 18 2 p NS x p NS x+1, P NS 1, 18 1 q NS x +1, 18 2 p NS x q NS x+1 + 3, 18 2 p NS NS x q x NS p x We calculate the numerator of the above as 1, 1, ( ) , ( ) We calculate the denominator as Therefore, P NS ,26948, and P S 2P NS 2 2, ,53896 The expected present value of the loss at issue random variable on a smoker policy is therefore, approximately,

18 Answer A 1, 18 1 q S x +1, 18 2 p NS x q NS x+1 + 3, 18 2 p NS x q NS x+1 P S P S 18 1 p x NS 1, ( ) + 3, + 1, , , ( ) ( ) 3,1743 May 213 Course MLC Examination, Problem No 19 You are given: (i) The following extract from a mortality table with a one-year select period: x l [ x] d [ x] l x+1 x (ii) Deaths are uniformly distributed over each year of age o (iii) e 65 [ ] 15 Calculate e 66 o [ ] A 141 B 143 C 145 D 147 E 149 We note immediately that l 66 l 65 [ ] , and l 67 l 66 [ ] , so that d 66 l 66 l We also have o 15 e [ 65] UDD 1 t p [ 65] dt + p [ 65] t p 66 dt + p [ 65] p 66 e 67 1 o UDD 1 1 o ( 1 tq [ 65] )dt + p [ 65] ( 1 tq 66 )dt + p [ 65] p 66 e q [ 65 ] + p [ 65] q o 66 + p [ 65] p 66 e e o 67 Therefore, o e We also have

19 o e [ 66] 1 t p [ 66] dt + p [ 66] e 67 o UDD 1 o ( 1 tq [ 66] )dt + p [ 66] e q o [ 66 ] + p [ 66] e Answer D May 213 Course MLC Examination, Problem No 2 Scientists are searching for a vaccine for a disease You are given: (i) 1, lives age x are exposed to the disease (ii) Future lifetimes are independent, except that the vaccine, if available, will be given to all at the end of year 1 (iii) The probability that the vaccine will be available is 2 (iv) For each life during year 1, q x 2 (v) For each life during year 2, q x+1 1, if the vaccine has been given, and q x+1 2, if it has not been given Calculate the standard deviation of the number of survivors at the end of year 2 A 1 B 2 C 3 D 4 E 5 The probability distribution of the random number of survivors is mixed, with probability 2 of having lower mortality due to the vaccine, and with probability 8 of having the same mortality as the first year Let A be the event that the vaccine becomes available Let N be the random number of survivors at the end of year 2 If event A happens, for each of the original 1, lives, that person is still alive at the end of year 2 with probability This means that for each person, the number of survivors for that person is 1 with probability 98 99, and with probability , so it is a Bernoulli Trial with p The number of survivors for all 1, lives insured is binomial with n 1, and p This means that its first moment is 1, , and the second moment is 1, ( ) Variance Square of the first monent If event A does not happen, the number of survivors for all 1, lives insured is binomial with n 1, and p 98 98, with mean 1, , and the second moment 1, ( ) This implies that E N Variance ( ) , E N 2 and therefore Square of the first monent ( ) ,

20 Var( N ) E N 2 and the standard deviation of N is Answer D Var N ( ) ( E( N )) , ( ) May 213 Course MLC Examination, Problem No 21 You are given: (i) δ t 6, t (ii) µ x ( t) 1, t (iii) Y is the present value random variable for a continuous annuity of 1 per year, payable for the lifetime of (x) with 1 years certain Calculate Pr( Y > E( Y )) A 75 B 71 C 715 D 72 E 725 We have Therefore, E( Y ) a 1 + e 6 1 e 1 1 a x+1 Pr( Y > E( Y )) Pr Y > e ( ) Pr ( ) 6 ln Pr T > e Answer A 1 + e e 6T 6 > Pr ( T > ) May 213 Course MLC Examination, Problem No 22 For a whole life insurance of 1, on (x), you are given: (i) Death benefits are payable at the end of the year of death (ii) A premium of 3 is payable at the start of each month (iii) Commissions are 5% of each premium (iv) Expenses of 1 are payable at the start of each year (v) i 5 (vi) 1A x+1 4 (vii) 1 V is the gross premium reserve at the end of year 1 for this insurance Calculate 1 V using the two-term Woolhouse formula for annuities A 95 B 98 C 11 D 111 E 114

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