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


 Lenard Whitehead
 2 years ago
 Views:
Transcription
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 3year 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 2year 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 twostate homogeneous Markov model with states: Actuary (Ac) and Professional Hockey Player (H) You are given: (i) Transitions occur December 31 of each year The oneyear 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 3year 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 righthand 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 multistate A multistate 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 oneyear 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 threeyear 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 threeyear 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 midyear 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 twostep 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 twoyear endowment insurance policies to smokers (S) and nonsmokers (NS) age x You are given: (i) The death benefit is 1, The maturity benefit is 3, (ii) The level annual premium for nonsmoker policies is determined by the equivalence principle
17 (iii) The annual premium for smoker policies is twice the nonsmoker 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 nonsmokers 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 nonsmokers, 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 nonsmokers 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 oneyear 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 twoterm Woolhouse formula for annuities A 95 B 98 C 11 D 111 E 114
May 2012 Course MLC Examination, Problem No. 1 For a 2year select and ultimate mortality model, you are given:
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
More informationNovember 2012 Course MLC Examination, Problem No. 1 For two lives, (80) and (90), with independent future lifetimes, you are given: k p 80+k
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
More informationSOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS The following questions or solutions have been modified since this document was prepared to use with the syllabus effective
More informationSOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS The following questions or solutions have been modified since this document was prepared to use with the syllabus effective
More informationChapter 2. 1. You are given: 1 t. Calculate: f. Pr[ T0
Chapter 2 1. You are given: 1 5 t F0 ( t) 1 1,0 t 125 125 Calculate: a. S () t 0 b. Pr[ T0 t] c. Pr[ T0 t] d. S () t e. Probability that a newborn will live to age 25. f. Probability that a person age
More informationSOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE WRITTENANSWER QUESTIONS AND SOLUTIONS
SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE WRITTENANSWER QUESTIONS AND SOLUTIONS Questions February 12, 2015 In Questions 12, 13, and 19, the wording was changed slightly
More informationINSTRUCTIONS TO CANDIDATES
Society of Actuaries Canadian Institute of Actuaries Exam MLC Models for Life Contingencies Friday, October 31, 2014 8:30 a.m. 12:45 p.m. MLC General Instructions 1. Write your candidate number here. Your
More informationJANUARY 2016 EXAMINATIONS. Life Insurance I
PAPER CODE NO. MATH 273 EXAMINER: Dr. C. BoadoPenas TEL.NO. 44026 DEPARTMENT: Mathematical Sciences JANUARY 2016 EXAMINATIONS Life Insurance I Time allowed: Two and a half hours INSTRUCTIONS TO CANDIDATES:
More informationPremium Calculation. Lecture: Weeks 1214. Lecture: Weeks 1214 (STT 455) Premium Calculation Fall 2014  Valdez 1 / 31
Premium Calculation Lecture: Weeks 1214 Lecture: Weeks 1214 (STT 455) Premium Calculation Fall 2014  Valdez 1 / 31 Preliminaries Preliminaries An insurance policy (life insurance or life annuity) is
More informationManual for SOA Exam MLC.
Chapter 6. Benefit premiums Extract from: Arcones Fall 2010 Edition, available at http://www.actexmadriver.com/ 1/90 (#4, Exam M, Spring 2005) For a fully discrete whole life insurance of 100,000 on (35)
More informationManual for SOA Exam MLC.
Chapter 4. Life insurance. Extract from: Arcones Fall 2009 Edition, available at http://www.actexmadriver.com/ (#1, Exam M, Fall 2005) For a special whole life insurance on (x), you are given: (i) Z is
More informationPremium Calculation. Lecture: Weeks 1214. Lecture: Weeks 1214 (Math 3630) Annuities Fall 2015  Valdez 1 / 32
Premium Calculation Lecture: Weeks 1214 Lecture: Weeks 1214 (Math 3630) Annuities Fall 2015  Valdez 1 / 32 Preliminaries Preliminaries An insurance policy (life insurance or life annuity) is funded
More informationInsurance Benefits. Lecture: Weeks 68. Lecture: Weeks 68 (STT 455) Insurance Benefits Fall 2014  Valdez 1 / 36
Insurance Benefits Lecture: Weeks 68 Lecture: Weeks 68 (STT 455) Insurance Benefits Fall 2014  Valdez 1 / 36 An introduction An introduction Central theme: to quantify the value today of a (random)
More informationPlease write your name and student number at the spaces provided:
MATH 3630 Actuarial Mathematics I Final Examination  sec 001 Monday, 10 December 2012 Time Allowed: 2 hours (6:008:00 pm) Room: MSB 411 Total Marks: 120 points Please write your name and student number
More informationTABLE OF CONTENTS. 4. Daniel Markov 1 173
TABLE OF CONTENTS 1. Survival A. Time of Death for a Person Aged x 1 B. Force of Mortality 7 C. Life Tables and the Deterministic Survivorship Group 19 D. Life Table Characteristics: Expectation of Life
More informationPremium Calculation  continued
Premium Calculation  continued Lecture: Weeks 12 Lecture: Weeks 12 (STT 456) Premium Calculation Spring 2015  Valdez 1 / 16 Recall some preliminaries Recall some preliminaries An insurance policy (life
More informationPractice Exam 1. x l x d x 50 1000 20 51 52 35 53 37
Practice Eam. You are given: (i) The following life table. (ii) 2q 52.758. l d 5 2 5 52 35 53 37 Determine d 5. (A) 2 (B) 2 (C) 22 (D) 24 (E) 26 2. For a Continuing Care Retirement Community, you are given
More informationACTUARIAL MATHEMATICS FOR LIFE CONTINGENT RISKS
ACTUARIAL MATHEMATICS FOR LIFE CONTINGENT RISKS DAVID C. M. DICKSON University of Melbourne MARY R. HARDY University of Waterloo, Ontario V HOWARD R. WATERS HeriotWatt University, Edinburgh CAMBRIDGE
More informationSOA EXAM MLC & CAS EXAM 3L STUDY SUPPLEMENT
SOA EXAM MLC & CAS EXAM 3L STUDY SUPPLEMENT by Paul H. Johnson, Jr., PhD. Last Modified: October 2012 A document prepared by the author as study materials for the Midwestern Actuarial Forum s Exam Preparation
More informationAnnuities. Lecture: Weeks 911. Lecture: Weeks 911 (STT 455) Annuities Fall 2014  Valdez 1 / 43
Annuities Lecture: Weeks 911 Lecture: Weeks 911 (STT 455) Annuities Fall 2014  Valdez 1 / 43 What are annuities? What are annuities? An annuity is a series of payments that could vary according to:
More informationHeriotWatt University. BSc in Actuarial Mathematics and Statistics. Life Insurance Mathematics I. Extra Problems: Multiple Choice
HeriotWatt University BSc in Actuarial Mathematics and Statistics Life Insurance Mathematics I Extra Problems: Multiple Choice These problems have been taken from Faculty and Institute of Actuaries exams.
More informationEXAMINATION. 6 April 2005 (pm) Subject CT5 Contingencies Core Technical. Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE
Faculty of Actuaries Institute of Actuaries EXAMINATION 6 April 2005 (pm) Subject CT5 Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Enter all the candidate and
More informationINSTITUTE OF ACTUARIES OF INDIA
INSTITUTE OF ACTUARIES OF INDIA EXAMINATIONS 17 th November 2011 Subject CT5 General Insurance, Life and Health Contingencies Time allowed: Three Hours (10.00 13.00 Hrs) Total Marks: 100 INSTRUCTIONS TO
More informationManual for SOA Exam MLC.
Chapter 4. Life Insurance. Extract from: Arcones Manual for the SOA Exam MLC. Fall 2009 Edition. available at http://www.actexmadriver.com/ 1/14 Level benefit insurance in the continuous case In this chapter,
More informationManual for SOA Exam MLC.
Chapter 5. Life annuities Extract from: Arcones Fall 2009 Edition, available at http://www.actexmadriver.com/ 1/60 (#24, Exam M, Fall 2005) For a special increasing whole life annuitydue on (40), you
More information**BEGINNING OF EXAMINATION**
November 00 Course 3 Society of Actuaries **BEGINNING OF EXAMINATION**. You are given: R = S T µ x 0. 04, 0 < x < 40 0. 05, x > 40 Calculate e o 5: 5. (A) 4.0 (B) 4.4 (C) 4.8 (D) 5. (E) 5.6 Course 3: November
More informationO MIA009 (F2F) : GENERAL INSURANCE, LIFE AND
No. of Printed Pages : 11 MIA009 (F2F) kr) ki) M.Sc. ACTUARIAL SCIENCE (MSCAS) N December, 2012 0 O MIA009 (F2F) : GENERAL INSURANCE, LIFE AND HEALTH CONTINGENCIES Time : 3 hours Maximum Marks : 100
More informationManual for SOA Exam MLC.
Chapter 6. Benefit premiums. Extract from: Arcones Fall 2010 Edition, available at http://www.actexmadriver.com/ 1/77 Fully discrete benefit premiums In this section, we will consider the funding of insurance
More informationPremium calculation. summer semester 2013/2014. Technical University of Ostrava Faculty of Economics department of Finance
Technical University of Ostrava Faculty of Economics department of Finance summer semester 2013/2014 Content 1 Fundamentals Insurer s expenses 2 Equivalence principles Calculation principles 3 Equivalence
More informationMath 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 2 Solutions
Math 70/408, Spring 2008 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 2 Solutions About this problem set: These are problems from Course /P actuarial exams that I have collected over the years,
More informationTABLE OF CONTENTS. GENERAL AND HISTORICAL PREFACE iii SIXTH EDITION PREFACE v PART ONE: REVIEW AND BACKGROUND MATERIAL
TABLE OF CONTENTS GENERAL AND HISTORICAL PREFACE iii SIXTH EDITION PREFACE v PART ONE: REVIEW AND BACKGROUND MATERIAL CHAPTER ONE: REVIEW OF INTEREST THEORY 3 1.1 Interest Measures 3 1.2 Level Annuity
More informationSome Observations on Variance and Risk
Some Observations on Variance and Risk 1 Introduction By K.K.Dharni Pradip Kumar 1.1 In most actuarial contexts some or all of the cash flows in a contract are uncertain and depend on the death or survival
More informationFurther Topics in Actuarial Mathematics: Premium Reserves. Matthew Mikola
Further Topics in Actuarial Mathematics: Premium Reserves Matthew Mikola April 26, 2007 Contents 1 Introduction 1 1.1 Expected Loss...................................... 2 1.2 An Overview of the Project...............................
More informationManual for SOA Exam MLC.
Chapter 5. Life annuities. Extract from: Arcones Manual for the SOA Exam MLC. Spring 2010 Edition. available at http://www.actexmadriver.com/ 1/114 Whole life annuity A whole life annuity is a series of
More informationb g is the future lifetime random variable.
**BEGINNING OF EXAMINATION** 1. Given: (i) e o 0 = 5 (ii) l = ω, 0 ω (iii) is the future lifetime random variable. T Calculate Var Tb10g. (A) 65 (B) 93 (C) 133 (D) 178 (E) 333 COURSE/EXAM 3: MAY 0001
More informationMATH 3630 Actuarial Mathematics I Class Test 2 Wednesday, 17 November 2010 Time Allowed: 1 hour Total Marks: 100 points
MATH 3630 Actuarial Mathematics I Class Test 2 Wednesday, 17 November 2010 Time Allowed: 1 hour Total Marks: 100 points Please write your name and student number at the spaces provided: Name: Student ID:
More information1. Datsenka Dog Insurance Company has developed the following mortality table for dogs:
1 Datsenka Dog Insurance Company has developed the following mortality table for dogs: Age l Age l 0 2000 5 1200 1 1950 6 1000 2 1850 7 700 3 1600 8 300 4 1400 9 0 Datsenka sells an whole life annuity
More informationManual for SOA Exam MLC.
Chapter 4. Life Insurance. c 29. Miguel A. Arcones. All rights reserved. Extract from: Arcones Manual for the SOA Exam MLC. Fall 29 Edition. available at http://www.actexmadriver.com/ c 29. Miguel A. Arcones.
More informationINSTITUTE AND FACULTY OF ACTUARIES EXAMINATION
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION 8 October 2015 (pm) Subject CT5 Contingencies Core Technical
More information1. A survey of a group s viewing habits over the last year revealed the following
1. A survey of a group s viewing habits over the last year revealed the following information: (i) 8% watched gymnastics (ii) 9% watched baseball (iii) 19% watched soccer (iv) 14% watched gymnastics and
More informationMultistate transition models with actuarial applications c
Multistate transition models with actuarial applications c by James W. Daniel c Copyright 2004 by James W. Daniel Reprinted by the Casualty Actuarial Society and the Society of Actuaries by permission
More information2 Policy Values and Reserves
2 Policy Values and Reserves [Handout to accompany this section: reprint of Life Insurance, from The Encyclopaedia of Actuarial Science, John Wiley, Chichester, 2004.] 2.1 The Life Office s Balance Sheet
More informationHeriotWatt University. M.Sc. in Actuarial Science. Life Insurance Mathematics I. Tutorial 5
1 HeriotWatt University M.Sc. in Actuarial Science Life Insurance Mathematics I Tutorial 5 1. Consider the illnessdeath model in Figure 1. A life age takes out a policy with a term of n years that pays
More information1 Cashflows, discounting, interest rate models
Assignment 1 BS4a Actuarial Science Oxford MT 2014 1 1 Cashflows, discounting, interest rate models Please hand in your answers to questions 3, 4, 5 and 8 for marking. The rest are for further practice.
More information4. Life Insurance. 4.1 Survival Distribution And Life Tables. Introduction. X, Ageatdeath. T (x), timeuntildeath
4. Life Insurance 4.1 Survival Distribution And Life Tables Introduction X, Ageatdeath T (x), timeuntildeath Life Table Engineers use life tables to study the reliability of complex mechanical and
More informationSOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS Copyright 5 by the Society of Actuaries Some of the questions in this study note are taken from past SOA examinations. M95 PRINTED
More informationEDUCATION COMMITTEE OF THE SOCIETY OF ACTUARIES MLC STUDY NOTE SUPPLEMENTARY NOTES FOR ACTUARIAL MATHEMATICS FOR LIFE CONTINGENT RISKS VERSION 2.
EDUCATION COMMITTEE OF THE SOCIETY OF ACTUARIES MLC STUDY NOTE SUPPLEMENTARY NOTES FOR ACTUARIAL MATHEMATICS FOR LIFE CONTINGENT RISKS VERSION 2.0 by Mary R. Hardy, PhD, FIA, FSA, CERA David C. M. Dickson,
More informationMath 370, Spring 2008 Prof. A.J. Hildebrand. Practice Test 2 Solutions
Math 370, Spring 008 Prof. A.J. Hildebrand Practice Test Solutions About this test. This is a practice test made up of a random collection of 5 problems from past Course /P actuarial exams. Most of the
More informationSafety margins for unsystematic biometric risk in life and health insurance
Safety margins for unsystematic biometric risk in life and health insurance Marcus C. Christiansen June 1, 2012 7th Conference in Actuarial Science & Finance on Samos Seite 2 Safety margins for unsystematic
More informationSOCIETY OF ACTUARIES EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS Copyright 015 by the Society of Actuaries Some of the questions in this study note are taken from past examinations. Some of the questions
More informationMath 630 Problem Set 2
Math 63 Problem Set 2 1. AN nyear term insurance payable at the moment of death has an actuarial present value (i.e. EPV) of.572. Given µ x+t =.7 and δ =.5, find n. (Answer: 11) 2. Given: Ā 1 x:n =.4275,
More informationACTS 4301 FORMULA SUMMARY Lesson 1: Probability Review. Name f(x) F (x) E[X] Var(X) Name f(x) E[X] Var(X) p x (1 p) m x mp mp(1 p)
ACTS 431 FORMULA SUMMARY Lesson 1: Probability Review 1. VarX)= E[X 2 ] E[X] 2 2. V arax + by ) = a 2 V arx) + 2abCovX, Y ) + b 2 V ary ) 3. V ar X) = V arx) n 4. E X [X] = E Y [E X [X Y ]] Double expectation
More informationSOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS Copyright 005 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study
More informationEXAMINATIONS. 18 April 2000 (am) Subject 105 Actuarial Mathematics 1. Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE
Faculty of Actuaries Institute of Actuaries EXAMINATIONS 18 April 2000 (am) Subject 105 Actuarial Mathematics 1 Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Write your surname in full, the
More informationINSTITUTE AND FACULTY OF ACTUARIES EXAMINATION
INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION 27 April 2015 (pm) Subject CT5 Contingencies Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Enter all the candidate and examination
More informationActuarial Science with
Actuarial Science with 1. life insurance & actuarial notations Arthur Charpentier joint work with Christophe Dutang & Vincent Goulet and Giorgio Alfredo Spedicato s lifecontingencies package Meielisalp
More informationValuation Report on Prudential Annuities Limited as at 31 December 2003. The investigation relates to 31 December 2003.
PRUDENTIAL ANNUITIES LIMITED Returns for the year ended 31 December 2003 SCHEDULE 4 Valuation Report on Prudential Annuities Limited as at 31 December 2003 1. Date of investigation The investigation relates
More informationCOURSE 3 SAMPLE EXAM
COURSE 3 SAMPLE EXAM Although a multiple choice format is not provided for some questions on this sample examination, the initial Course 3 examination will consist entirely of multiple choice type questions.
More informationStochastic Analysis of LongTerm MultipleDecrement Contracts
Stochastic Analysis of LongTerm MultipleDecrement Contracts Matthew Clark, FSA, MAAA, and Chad Runchey, FSA, MAAA Ernst & Young LLP Published in the July 2008 issue of the Actuarial Practice Forum Copyright
More informationTABLE OF CONTENTS. Practice Exam 1 3 Practice Exam 2 15 Practice Exam 3 27 Practice Exam 4 41 Practice Exam 5 53 Practice Exam 6 65
TABLE OF CONTENTS Practice Exam 1 3 Practice Exam 2 15 Practice Exam 3 27 Practice Exam 4 41 Practice Exam 5 53 Practice Exam 6 65 Solutions to Practice Exam 1 79 Solutions to Practice Exam 2 89 Solutions
More informationManual for SOA Exam MLC.
Chapter 5 Life annuities Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at http://wwwactexmadrivercom/ 1/94 Due n year temporary annuity Definition 1 A due n year term annuity
More informationINSURANCE DEPARTMENT OF THE STATE OF NEW YORK REGULATION NO. 147 (11 NYCRR 98) VALUATION OF LIFE INSURANCE RESERVES
INSURANCE DEPARTMENT OF THE STATE OF NEW YORK REGULATION NO. 147 (11 NYCRR 98) VALUATION OF LIFE INSURANCE RESERVES I, Gregory V. Serio, Superintendent of Insurance of the State of New York, pursuant to
More informationMath 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 5 Solutions
Math 370/408, Spring 2008 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 5 Solutions About this problem set: These are problems from Course 1/P actuarial exams that I have collected over the
More informationSOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS Copyright 007 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study
More information1. Revision 2. Revision pv 3.  note that there are other equivalent formulae! 1 pv 16.5 4. A x A 1 x:n A 1
Tutorial 1 1. Revision 2. Revision pv 3.  note that there are other equivalent formulae! 1 pv 16.5 4. A x A 1 x:n A 1 x:n a x a x:n n a x 5. K x = int[t x ]  or, as an approximation: T x K x + 1 2 6.
More informationCHAPTER 450412 VALUATION OF LIFE INSURANCE POLICIES
CHAPTER 450412 VALUATION OF LIFE INSURANCE POLICIES Section 45041201 Applicability 45041202 Definitions 45041203 General Calculation Requirements for Basic Reserves and Premium Deficiency Reserves
More informationINSTITUTE OF ACTUARIES OF INDIA
INSTITUTE OF ACTUARIES OF INDIA EXAMINATIONS 13 th May 2015 Subject CT5 General Insurance, Life and Health Contingencies Time allowed: Three Hours (10.30 13.30 Hrs) Total Marks: 100 INSTRUCTIONS TO THE
More informationMaking use of netting effects when composing life insurance contracts
Making use of netting effects when composing life insurance contracts Marcus Christiansen Preprint Series: 2113 Fakultät für Mathematik und Wirtschaftswissenschaften UNIVERSITÄT ULM Making use of netting
More information( ) = 1 x. ! 2x = 2. The region where that joint density is positive is indicated with dotted lines in the graph below. y = x
Errata for the ASM Study Manual for Exam P, Eleventh Edition By Dr. Krzysztof M. Ostaszewski, FSA, CERA, FSAS, CFA, MAAA Web site: http://www.krzysio.net Email: krzysio@krzysio.net Posted September 21,
More informationManual for SOA Exam MLC.
Chapter 6. Benefit premiums. Extract from: Arcones Fall 2010 Edition, available at http://www.actexmadriver.com/ 1/24 Nonlevel premiums and/or benefits. Let b k be the benefit paid by an insurance company
More informationProfit Measures in Life Insurance
Profit Measures in Life Insurance Shelly Matushevski Honors Project Spring 2011 The University of Akron 2 Table of Contents I. Introduction... 3 II. Loss Function... 5 III. Equivalence Principle... 7 IV.
More information[Source: Added at 17 Ok Reg 1659, eff 71400]
TITLE 365. INSURANCE DEPARTMENT CHAPTER 10. LIFE, ACCIDENT AND HEALTH SUBCHAPTER 17. VALUATION OF LIFE INSURANCE POLICIES REGULATION INCLUDING THE INTRODUCTION AND USE OF NEW SELECT MORTALITY FACTORS)
More informationSOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM M ACTUARIAL MODELS EXAM M SAMPLE QUESTIONS Copyright 005 by the Society of Actuaries Some of the questions in this study note are taken from past SOA examinations. M0905 PRINTED
More informationManual for SOA Exam MLC.
Chapter 4. Life Insurance. Extract from: Arcones Manual for the SOA Exam MLC. Fall 2009 Edition. available at http://www.actexmadriver.com/ 1/44 Properties of the APV for continuous insurance The following
More information( ) is proportional to ( 10 + x)!2. Calculate the
PRACTICE EXAMINATION NUMBER 6. An insurance company eamines its pool of auto insurance customers and gathers the following information: i) All customers insure at least one car. ii) 64 of the customers
More informationUniversal Life Insurance
Universal Life Insurance Lecture: Weeks 1112 Thanks to my friend J. Dhaene, KU Leuven, for ideas here drawn from his notes. Lecture: Weeks 1112 (STT 456) Universal Life Insurance Spring 2015  Valdez
More informationLIFE INSURANCE AND PENSIONS by Peter Tryfos York University
LIFE INSURANCE AND PENSIONS by Peter Tryfos York University Introduction Life insurance is the business of insuring human life: in return for a premium payable in one sum or installments, an insurance
More informationFundamentals of Actuarial Mathematics
Fundamentals of Actuarial Mathematics S. David Promislow York University, Toronto, Canada John Wiley & Sons, Ltd Contents Preface Notation index xiii xvii PARTI THE DETERMINISTIC MODEL 1 1 Introduction
More informationYanyun Zhu. Actuarial Model: Life Insurance & Annuity. Series in Actuarial Science. Volume I. ir* International Press. www.intlpress.
Yanyun Zhu Actuarial Model: Life Insurance & Annuity Series in Actuarial Science Volume I ir* International Press www.intlpress.com Contents Preface v 1 Interest and AnnuityCertain 1 1.1 Introduction
More informationManual for SOA Exam MLC.
Chapter 5 Life annuities Extract from: Arcones Manual for the SOA Exam MLC Fall 2009 Edition available at http://wwwactexmadrivercom/ 1/70 Due n year deferred annuity Definition 1 A due n year deferred
More informationShould I Buy an Income Annuity?
The purchase of any financial product involves a trade off. For example when saving for retirement, you are often faced with making a trade off between how much you want to protect your investments from
More informationThe Fair Valuation of Life Insurance Participating Policies: The Mortality Risk Role
The Fair Valuation of Life Insurance Participating Policies: The Mortality Risk Role Massimiliano Politano Department of Mathematics and Statistics University of Naples Federico II Via Cinthia, Monte S.Angelo
More informationEXAM 3, FALL 003 Please note: On a onetime basis, the CAS is releasing annotated solutions to Fall 003 Examination 3 as a study aid to candidates. It is anticipated that for future sittings, only the
More informationSolution. Because you are not doing business in the state of New York, you only need to do the calculations at the end of each policy year.
Exercises in life and annuity valuation. You are the valuation actuary for the Hard Knocks Life Insurance Company offering life annuities and life insurance to guinea pigs. The valuation interest rate
More informationShould I Buy an Income Annuity?
Prepared For: Valued Client Prepared By: CPS The purchase of any financial product involves a trade off. For example when saving for retirement, you are often faced with making a trade off between how
More informationTHE EFFECT OF A CHANGE IN THE INTEREST RATE ON LIFE ASSURANCE PREMIUMS
380 THE EFFECT OF A CHANGE IN THE INTEREST RATE ON LIFE ASSURANCE PREMIUMS By R. E. WHITE, F.I.A., AND B. T. HOLMES, A.I.A., F.A.S., F.A.I.A. of the Confederation Life Association THE recent paper by Dr
More informationMath 370, Actuarial Problemsolving Spring 2008 A.J. Hildebrand. Practice Test, 1/28/2008 (with solutions)
Math 370, Actuarial Problemsolving Spring 008 A.J. Hildebrand Practice Test, 1/8/008 (with solutions) About this test. This is a practice test made up of a random collection of 0 problems from past Course
More informationLloyd Spencer Lincoln Re
AN OVERVIEW OF THE PANJER METHOD FOR DERIVING THE AGGREGATE CLAIMS DISTRIBUTION Lloyd Spencer Lincoln Re Harry H. Panjer derives a recursive method for deriving the aggregate distribution of claims in
More informationMath 370, Spring 2008 Prof. A.J. Hildebrand. Practice Test 2
Math 370, Spring 2008 Prof. A.J. Hildebrand Practice Test 2 About this test. This is a practice test made up of a random collection of 15 problems from past Course 1/P actuarial exams. Most of the problems
More informationMath 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 3 Solutions
Math 37/48, Spring 28 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 3 Solutions About this problem set: These are problems from Course /P actuarial exams that I have collected over the years,
More information(Common) Shock models in Exam MLC
Making sense of... (Common) Shock models in Exam MLC James W. Daniel Austin Actuarial Seminars www.actuarialseminars.com June 26, 2008 c Copyright 2007 by James W. Daniel; reproduction in whole or in part
More informationChapter 2 Premiums and Reserves in Multiple Decrement Model
Chapter 2 Premiums and Reserves in Multiple Decrement Model 2.1 Introduction A guiding principle in the determination of premiums for a variety of life insurance products is: Expected present value of
More informationFundamentals of Actuarial Mathematics. 3rd Edition
Brochure More information from http://www.researchandmarkets.com/reports/2866022/ Fundamentals of Actuarial Mathematics. 3rd Edition Description:  Provides a comprehensive coverage of both the deterministic
More informationPREMIUM AND BONUS. MODULE  3 Practice of Life Insurance. Notes
4 PREMIUM AND BONUS 4.0 INTRODUCTION A insurance policy needs to be bought. This comes at a price which is known as premium. Premium is the consideration for covering of the risk of the insured. The insured
More informationMath 370, Actuarial Problemsolving Spring 2008 A.J. Hildebrand. Problem Set 1 (with solutions)
Math 370, Actuarial Problemsolving Spring 2008 A.J. Hildebrand Problem Set 1 (with solutions) About this problem set: These are problems from Course 1/P actuarial exams that I have collected over the years,
More informationMathematics of Risk. Introduction. Case Study #1 Personal Auto Insurance Pricing. Mathematical Concepts Illustrated. Background
Mathematics of Risk Introduction There are many mechanisms that individuals and organizations use to protect themselves against the risk of financial loss. Government organizations and public and private
More informationTraditional, investment, and risk management actuaries in the life insurance industry
Traditional, investment, and risk management actuaries in the life insurance industry Presentation at California Actuarial Student Conference University of California, Santa Barbara April 4, 2015 Frank
More informationTHE MATHEMATICS OF LIFE INSURANCE THE NET SINGLE PREMIUM.  Investigate the component parts of the life insurance premium.
THE MATHEMATICS OF LIFE INSURANCE THE NET SINGLE PREMIUM CHAPTER OBJECTIVES  Discuss the importance of life insurance mathematics.  Investigate the component parts of the life insurance premium.  Demonstrate
More informationDisability insurance: estimation and risk aggregation
Disability insurance: estimation and risk aggregation B. Löfdahl Department of Mathematics KTH, Royal Institute of Technology May 2015 Introduction New upcoming regulation for insurance industry: Solvency
More informationFINANCIAL ANALYSIS ****** UPDATED FOR 55% BENEFIT ****** ****** FOR ALL SURVIVORS ******
FINANCIAL ANALYSIS ****** UPDATED FOR 55% BENEFIT ****** ****** FOR ALL SURVIVORS ****** This fact sheet is designed to supplement the Department of Defense brochure: SBP SURVIVOR BENEFIT PLAN FOR THE
More information