# May 2012 Course MLC Examination, Problem No. 1 For a 2-year select and ultimate mortality model, you are given:

Save this PDF as:

Size: px
Start display at page:

Download "May 2012 Course MLC Examination, Problem No. 1 For a 2-year select and ultimate mortality model, you are given:"

## Transcription

1 Solutions to the May 2012 Course MLC Examination by Krzysztof Ostaszewski, Copyright 2012 by Krzysztof Ostaszewski All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner. Dr. Ostaszewski s manual for Course MLC is available from Actex Publications ( and the Actuarial Bookstore ( Exam MLC seminar at Illinois State University: May 2012 Course MLC Examination, Problem No. 1 For a 2-year select and ultimate mortality model, you are given: (i) q = 0.95q. x +1 x+1 (ii) l 76 = 98,153. (iii) l 77 = 96,124. Calculate l A. 96,150 B. 96,780 C. 97,420 D. 98,050 E. 98,690 We first note that (i) says that 1 l x+2 = 1 l x +1 l x +2 We let x = 75 and obtain 1 96,124 l The solution is = 1 p l = q = 0.95q = p x x x +1 x+1 x+1 = 1 l 77 l = l 77 96,124 = ,153. l = 96, ,124 98, ,153 Answer D. l 76 = l x+2 l x+1. May 2012 Course MLC Examination, Problem No. 2 You are given: (i) p x = (ii) p x+1 = 0.95.

2 (iii) e x+1.75 = (iv) Deaths are uniformly distributed between ages x and x +1. (v) The force of mortality is constant between ages x +1 and x + 2. Calculate e x A B C D E The unusual twist in this problem is that the curtate life expectancy is calculated at fractional age. But we can just apply the standard formula for curtate life expectancy and see how things work out: e x+0.75 = 1 p x p x p x = = 1 p x p x p x p x p x = = 1 p x p x p x = 1 p x e x+1.75 Of course you could just memorize the formula e y = 1 p y 1+ e y+1 y = x Using the value given, we get e x+0.75 = 1 p x e x+1.75 = 1 p x+0.75 ( ) = p x and apply it for Thus we need to calculate the probability 1 p x Note that of the year between the age x and x , a quarter of a year is between age x and age x + 1, under the UDD assumption, and the remaining three quarters are between age x + 1 and age x + 2, under the constant force assumption. We observe that p = p 1 x x+0.75 p 0.75 x+1. UDD assumption applies CF assumption applies We use the UDD assumption to calculate 0.25 p x+0.75 : 0.25 p x+0.75 = q x+0.75 = d x+0.75 l x+0.75 = UDD d x l x+0.75 = 1 d 0.25 x l x l x+0.75 l x = q x 0.75 p x = q x = q x = p x = q UDD x q x ( 1 p x ) Again, it is a good idea to just memorize the formula used in the above calculation: p = 1 sq x. s x+t UDD 1 tq x Now we will use the constant force assumption to calculate 0.75 p x+1. Let us denote that constant force between ages x + 1 and x + 2 by µ and observe that p x+1 = 0.95 = e µ, so that µ = ln0.95. Based on this: p = 0.75 x+1 e 0.75µ 0.75 ln0.95 = e = =

3 Combining these results we obtain e x+0.75 = p x+0.75 = p x p x Answer A. May 2012 Course MLC Examination, Problem No. 3 For a fully discrete 3-year term insurance of 10,000 on (40), you are given: (i) µ 40+t, t 0, is a force of mortality consistent with the Illustrative Life Table. (ii) µ 40+t , t = 0, 1, 2, is the force of mortality for this insured. (iii) i = Calculate the annual benefit premium for this insurance. A. 196 B. 214 C. 232 D. 250 E. 268 Let us denote probabilities referring to the Illustrative Life Table by a superscript (ILT), and probabilities referring to this insured by a superscript (I). We have ds t µ ( ILT) p ( I) 40+s t 40 = e 0 Therefore, ( I) = 1+ vp 40 + v 2 ( I) ( I) p p41 40 = and I a 40:3 ( ILT) = t p 40 e 0.02t. = ( ILT) p 40 e ( ILT) p 40 e 0.02 ( ILT) p 41 e 0.02 = = ( ILT) ( 1 q 40 ) e ( ILT) ( ILT) ( 1 q 40 ) ( 1 q 41 ) e 0.04 = = ( ) e ( ) ( ) e 0.04 ( I) A 40:3 Furthermore, Therefore, ( I) = 1 d a 40:3 I E ( I) p 3 40 = 40 1 I A 40:3 ( I) p41 ( I) p ( ILT) ( ILT) p41 ( ILT) p42 e 0.06 = p 40 = = 1 q ( ILT) ( ILT) ( ILT) ( 40 ) ( 1 q 41 ) ( 1 q 42 ) e 0.06 = ( ) ( ) ( ) e 0.06 = = ( I) A40:3 The premium sought is ( I) 3 E =

4 Answer B. 1 ( I) 10,000 A 40:3 I a 40:3 10, May 2012 Course MLC Examination, Problem No. 4 For a 3-year term insurance of 1000 on [75], you are given: (i) Death benefits are payable at the end of the year of death. (ii) Level premiums are payable at the beginning of each quarter. (iii) Mortality follows a select and ultimate life table with a two-year select period: x l l l x + 2 x x +1 x ,930 15,668 15, ,508 15,224 14, ,050 14,744 14, (iv) Deaths are uniformly distributed over each year of age. (v) i = Calculate the amount of each quarterly benefit premium. A. 5.3 B. 5.5 C. 5.7 D. 5.9 E. 6.1 The single benefit premium is 1 A 75 :3 = vq 75 + v 2 p 75 q v3 p 75 p q = = = The annuity factor is a = 1+ vp 75 + v 2 p : p = Furthermore, 3E 75 [ ] [ ]+3 l 75 p [ ] = = l 75 3 [ ] = l 78 3 l 75 [ ] = = and β ( 4) = from the Illustrative Life We obtain the values α 4 Table, although we should note that we can use the Table because the interest rate used, i = 6%, is the one used in the Table as well. Otherwise we would have to struggle with these formulas: α ( 4) = id and β 4 i ( 4) d 4 = i i 4 i 4 d 4. Based on the UDD assumption, we

5 calculate 4 a 75 [ ]:5 β ( 4) 1 3 E = a UDD [ 75]:5 α 4 [ ] The annual level benefit premium for the policy (remember that this policy has 1000 benefit, not unit benefit) is A [ 75]: ( 4) = a [ 75]: But the question asks about the quarterly premium, which is a quarter of this, i.e., Answer C. May 2012 Course MLC Examination, Problem No. 5 For a whole life insurance on (80): (i) Level premiums of 900 are payable at the start of each year. (ii) Commission expenses are 50% of premium in the first year and 10% of premium in subsequent years, payable at the start of the year. (iii) Maintenance expenses are 3 per month, payable at the start of each month. (iv) δ = (v) a 80 = (vi) µ 80 = Using the first 3 terms of Woolhouse s formula, calculate the expected value of the present-value-of-expenses random variable at age 80. A. 920 B. 940 C D E Recall that the Woolhouse's formula with three terms is ( m a ) x a x m 1 2m m2 1 ( δ + µ 12m 2 x ). You must memorize this formula for the test. While the premium is payable annually, the maintenance expenses are paid monthly, and we will use this Woolhouse s formula for the calculation of their present value. For the maintenance expenses, m = 12, a x = a 80 = 6.000, µ 80 = 0.033, δ = , so that ( 12 a ) ( ) The first year commission is 50% of 900, i.e., 450. The renewal commissions are 10% of 900, i.e., 90 each. Thus the expected present value of expenses as calculated at age 80 is ( 12) a a 80 = Answer D. Actuarial present value of commissions Actuarial present value of maintenance expenses ( 12 = a a ) 80 =

6 May 2012 Course MLC Examination, Problem No. 6 Two whales, Hannibal and Jack, occupy the Ocean World aquarium. Both are currently age 6 with independent future lifetimes. Hannibal arrived at Ocean World at age 4 and Jack at age 6. The standard mortality for whales is as follows: x q x Whales arriving at Ocean World have mortality according to the following: 150% of standard mortality in year 1 130% of standard mortality in year 2 110% of standard mortality in year 3 100% of standard mortality in years 4 and later. As whales can get lonely on their own, a fund is set up today to provide 100,000 to replace a whale at the end of 3 years if exactly one survives. You are given that i = Calculate the initial value of the fund using the equivalence principle. A. 24,500 B. 29,200 C. 29,300 D. 30,900 E. 31,900 Hannibal arrived at Ocean World at age 4, and is now 6, thus Hannibal s future mortality rates are q 6 H = = 0.066, q 7 H = = 0.07, q 8 H = = Jack arrived at Ocean World at age 6, and is now 6, thus Jack s future mortality rates are q 6 J = = 0.09, q 7 J = = 0.091, q 8 J = = Based on this, p 6 H = 0.934, p 7 H = 0.93, p 8 H = 0.92, p 6 J = 0.91, p 7 J = 0.909, q 8 J = 0.912, p H 3 6 = p H 6 p H 7 p H 8 = , p J 3 6 = p J 6 p J 7 p J 8 = , and, thanks to independence of lives, p = p H 3 6: p J The probability of exactly one surviving three years, given that the lifetimes are independent, is p p = p H 3 6:6 3 6: p J 6 3 p 6:6 p = p H 3 6: p J p 6:6 = 3 p 6:6 by Poohsticks The initial value of the fund using the equivalence principle is simply the actuarial present value of the payment of 100,000 at time 3, i.e., 100, , Answer B.

7 May 2012 Course MLC Examination, Problem No. 7 For a single premium product replacement insurance on a computer, you are given: (i) The insurance pays a percentage of the cost of a new computer at the end of the year of failure within the first 5 years. (ii) The insurance will pay only once. (iii) The insurance is bought at the time of the purchase of the computer. (iv) The cost of a new computer increases by 2% each year, compounded annually. The insurance pays based on the increased cost. (v) The insurance benefit and probabilities of failure are as follows: Year of failure k Percentage of cost of new computer covered Probability of computer failure k 1 q % % % % % 0.02 (vi) The commission is c% of the gross premium. (vii) Other expenses are 5% of the gross premium. (viii) The gross premium, calculated using the equivalence principle, is 10% of the original cost of the computer. (ix) i = Calculate c%. A. 40% B. 45% C. 50% D. 55% E. 60% Note that the last column, the probability of computer failure, is the probability expressed as a fraction of the initial computer population at policy inception (because this is deferred probability). Let us write G for the gross single premium for this policy. Let us also write c for the commission, written as a fraction (instead of c%, we will switch back to percentages at the end of the solution). Let us also write K for the current cost of the computer. We use the equivalence principle and equate, at time 0, the actuarial present value of the premium (left-hand side) with the actuarial present value of benefits and expenses (right-hand side): G = 0.1K = ( c)g + K K K K K This means that K K = ( c) 0.1K K K K K We divide both sides by K and solve for c:

8 c = Answer D May 2012 Course MLC Examination, Problem No. 8 For a fully discrete whole life insurance of 1000 on (80): (i) i = (ii) a 80 = (iii) a 90 = (iv) q 80 = Calculate 10 V FTP, the full preliminary term reserve for this policy at the end of year 10. A. 340 B. 350 C. 360 D. 370 E. 380 Note that 5.89 = a 80 = p 80 a 81 = 1+ Therefore, 1.06 a 81 = We have and = α FPT = = , β FPT = 1000 P 81 = 1000 A 81 = d a = a 81 = a 81 a 81 a = We do not really need the value of α FPT but calculating it is a good exercise. We then conclude that Answer B. a V FTP = 1000 A β FPT a 90 = d a β FPT a 90 = =

9 May 2012 Course MLC Examination, Problem No. 9 A special fully discrete 3-year endowment insurance on (x) pays death benefits as follows: Year of Death Death Benefit 1 30, , ,000 You are given: (i) The maturity benefit is 50,000. (ii) Annual benefit premiums increase at 10% per year, compounded annually. (iii) i = (iv) q x = 0.09, q x+1 = 0.12, q x+2 = Calculate 2 V, the benefit reserve at the end of year 2. A. 28,000 B. 29,000 C. 30,000 D. 31,000 E. 32,000 Let us write P for the first year premium. The benefit reserve at the end of year 2 is the actuarial present value of future benefits minus the actuarial present value of future premiums, i.e., 2 V = P We do not know the value of P but we can find it. The use of the expressions benefit reserve and benefit premiums implies that we are using the equivalence principle. Therefore, the actuarial present value of future benefits is equal to the actuarial present value of future premiums at policy inception, resulting in: 30,000 q x ,000 p x q x ,000 p x p x = = P +1.1 P p x P p x p x Therefore, P = 30,000 q x ,000 p x q x ,000 p x p x p x p x p x = = 30, , , , We conclude that 2 V = P , , Answer C. May 2012 Course MLC Examination, Problem No. 10 For a universal life insurance policy on Julie, you are given: (i) At t = 0, Julie will pay a premium of 100. (ii) At t =1, Julie will pay a premium of 50 with a probability of 0.8 or will pay nothing otherwise. (iii) At t = 2, Julie will pay a premium of 50 with probability of 0.9 if she paid the

11 (iv) The policy remains in force for months 12 and 13, with the monthly premiums of 100 being paid at the start of each month. Calculate the cash surrender value at the end of month 13. A B C D E The credited interest rate is per month. We are given that the actual cash surrender value at the end of month 11 is 1000, and that the surrender charge during the first year is 300, so that we can conclude that the account value at the end of month 11 is the sum of these two values, = The cost of insurance in month 12 is of 10000, i.e., 10, if paid at the end of the month. Therefore, the account value at the end of the 12-th month is AV 12 months ( ( 1 0.3) 5) = The cost of insurance in month 13 is of 10,000, i.e., 20, if paid at the end of the month. Therefore, the account value at the end of the 13-th month is AV 13 months ( 5) = 1, The cash surrender value is the account value after subtraction of the surrender charge of 100, i.e., 1, Note that while we subtracted the cost of insurance in both months, there were no subtractions for death benefits in either, because we know that the policy remained in force till the end of the 13-th month. Answer C. May 2012 Course MLC Examination, Problem No. 12 Employees in Company ABC can be in: State 0: Non-executive employee State 1: Executive employee State 2: Terminated from employment John joins Company ABC as a non-executive employee at age 30. You are given: (i) µ 01 = 0.01 for all years of service. (ii) µ 02 = for all years of service. (iii) µ 12 = for all years of service. (iv) Executive employees never return to the non-executive employee state. (v) Employees terminated from employment never get rehired. (vi) The probability that John lives to age 65 is 0.9, regardless of state. Calculate the probability that John will be an executive employee of Company ABC at age 65. A B C D E Let us consider survival in terms of the years of future service. In order to be an executive employee at age 65, John, who is 30 years old now, has to survive 35 years as an

12 employee, and at some point between now and the end of those 35 years has to become an executive employee and not get terminated for the remaining period during the 35 years. Note that he also has to live for those 35 years, i.e., be alive at age 65, but we are told that the probability of that is 0.9 and is independent of the state he is in at age 65, so we can simply calculate the probability of him being an executive employee at age 65 assuming he is alive, and then multiply that by 0.9, the probability that he will be alive, to get the desired answer. In formula terms, we are simply looking for p = 0.9 t p µ t 35 t p 0 dt = 0 Survive as a non-executive employee for t years Become an executive employee at time t Survive as a executive employee for the rest of 35 years Answer A. t ds µ s +µ s µ 12 s ds 35 0 = 0.9 e µ 01 t t e dt = 0.9 e µ s ( +µ s )t µ 01 t e µ 12( s 35 t ) dt = 0 35 = 0.9 e ( )t 0.01 e 0.002( 35 t) dt = 0.009e 0.07 e 0.014t dt = 0 0 ( = 0.009e e 0.49 ) =a 35 δ =14% May 2012 Course MLC Examination, Problem No. 13 Lorie s Lorries rents lavender limousines. On January 1 of each year they purchase 30 limousines for their existing fleet; of these, 20 are new and 10 are one-year old. Vehicles are retired according to the following 2-year select-and-ultimate table, where selection is age at purchase: Limousine age (x) q [ x] q [ x]+1 q x+2 x Lorie s Lorries has rented lavender limousines for the past ten years and has always purchased its limousines on the above schedule. Calculate the expected number of limousines in the Lorie s Lorries fleet immediately after the purchase of this year s limousines. A. 93 B. 94 C. 95 D. 96 E. 97 For the limousines that were purchased new: - Expected number of limousines just purchased (this number is certain) = 20.

13 - Expected number of limousines purchased one year ago = = = 18. = 20 1 p [ 0] = 20 p [ 0] = 20 1 q [ 0] - Expected number of limousines purchased two years ago = ( 1 q 0 ) = = = 20 2 p [ 0] = 20 p [ 0] p [ 0]+1 = 20 1 q [ 0] [ ]+1 - Expected number of limousines purchased three years ago = = 20 3 p [ 0] = 20 p [ 0] p [ 0]+1 p [ 0]+2 = 20 1 q 0 ( [ ] ) 1 q 0 ( [ ]+1 ) 1 q 2 = = Expected number of limousines purchased four years ago = = 20 4 p [ 0] = 20 p [ 0] p [ 0]+1 p [ 0]+2 p [ 0]+3 = 20 1 q 0 ( [ ] ) 1 q 0 ( [ ]+1 ) 1 q 2 = Expected number of limousines purchased five years ago = = 20 5 p [ 0] = 20 p [ 0] p [ 0]+1 p [ 0]+2 p [ 0]+3 p [ 0]+4 = ( 1 q [ 0]+1 ) 1 q 2 = 20 1 q [ 0] ( 1 q 3 )( 1 q 4 ) = ( 1 q 3 ) = = = 0. We also see that limousines purchased new any further back in the past are already out of service. For the limousines that were purchased used: - Expected number of limousines just purchased (this number is certain) = Expected number of limousines purchased one year ago = = = 9. = 10 1 p [ 1] = 10 p [ 1] = 10 1 q [ 1] - Expected number of limousines purchased two years ago = ( 1 q 1 ) = = = 10 2 p [ 1] = 10 p [ 1] p [ 1]+1 = 10 1 q [ 1] [ ]+1 - Expected number of limousines purchased three years ago = = 10 3 p [ 1] = 10 p [ 1] p [ 1]+1 p [ 1]+2 = 10 1 q 1 ( [ ] ) 1 q 1 ( [ ]+1 ) 1 q 3 = = = Expected number of limousines purchased four years ago = = 10 4 p [ 1] = 10 p [ 1] p [ 1]+1 p [ 1]+2 p [ 1]+3 = 10 1 q 1 ( [ ] ) 1 q 1 ( [ ]+1 ) 1 q 3 ( 1 q 4 ) = = = 0. We also see that limousines purchased used any further back in the past are already out of service. We conclude that the expected number of limousines in the Lorie s Lorries fleet immediately after the purchase of this year s limousines is: Answer D. May 2012 Course MLC Examination, Problem No. 14 XYZ Insurance Company sells a one-year term insurance product with a gross premium equal to 125% of the benefit premium. You are given: (i) The death benefit is payable at the end of the year of death, and the amount depends

14 on whether the cause of death is accidental or not. (ii) Death benefit amounts, together with the associated one-year probabilities of death, are as follows: Cause of Death Death Benefit Probabilities of Death Non-accidental Accidental (iii) i = 0%. Using the normal approximation without continuity correction, calculate the smallest number of policies that XYZ must sell so that the amount of gross premium equals or exceeds the 95th percentile of the distribution of the total present value of death benefits. A. 378 B. 431 C. 484 D. 537 E. 590 Let n be the number of policies that is sought in this problem. Let X i denote the claim amount for policy i, with i = 1, 2,, n. Let S = X i be the aggregate claim. We have = = 0.6, E X i = = 3.5, 2 = E( X i ) E( X i ) 2 E X i 2 Var X i for every i = 1, 2,, n. Therefore, E S = ne( X i ) = 0.6n, = nvar( X i ) = 3.14n. n i=1 = = 3.14 Var S Since the 95-th percentile of the standard normal distribution is 1.645, using the normal approximation without continuity correction for S gives its 95-th percentile as 0.6n n 0.6n n. The benefit premium for one policy is 0.6 and for all policies combined it is 0.6n. The gross premium for one policy is = 0.75, and for all policies combined it is 0.75n. We want 0.75n 0.6n n, or n , equivalent to 2 n Since the answer has to be a whole number, n 378. Answer A. May 2012 Course MLC Examination, Problem No. 15 For a 20-year temporary life annuity-due of 100 per year on (65), you are given:

15 (i) µ x = 0.001x, x 65. (ii) i = (iii) Y is the present-value random variable for this annuity. Calculate the probability that Y is less than A B C D E We are looking for the probability = Pr 100 a K65 +1 < 1000 Pr Y < 1000 = Pr a K65 +1 < 10 If we set up an equation a k = 10, we obtain ln k = ln1.05. is This means that the largest integer value of k = K such that a K65 +1 < 10 k = K = 13. Therefore, Pr( Y < 1000) = Pr a < 10 K65 +1 = Pr K = Pr( K 65 12) = Pr T 65 < 13. We can calculate that probability from the given formula for the force of mortality Answer C. Pr( T 65 < 13) = 13 q 65 = 1 13 p 65 = 1 e 78 µ x dx = 1 e = 1 e x dx 65 x=78 = 1 e x2 x=65 = May 2012 Course MLC Examination, Problem No. 16 For a special continuous joint life annuity on (x) and (y), you are given: (i) The annuity payments are 25,000 per year while both are alive and 15,000 per year when only one is alive. (ii) The annuity also pays a death benefit of 30,000 upon the first death. (iii) i = (iv) a xy = 8. (v) a xy = 10. Calculate the actuarial present value of this special annuity. A. 239,000 B. 246,000 C. 287,000 D. 354,000 E. 366,000 The annuity portion only of this special continuous joint life annuity on (x) and (y) can be thought of as a continuous payment of 10,000 per year until the first death plus a continuous payment of 15,000 per year until the second death. Therefore, the present value of the annuity portion only is

16 10,000a xy +15,000a xy = 10, , = 230,000. But the special annuity also includes a death benefit of 30,000A xy = 30,000( 1 δ a xy ) = 30,000 1 ( ln1.06) 8 The total of the two is 230, , = 246, Answer B. 16, May 2012 Course MLC Examination, Problem No. 17 Your company issues fully discrete whole life policies to a group of lives age 40. For each policy, you are given: (i) The death benefit is 50,000. (ii) Assumed mortality and interest are the Illustrative Life Table at 6%. (iii) Annual gross premium equals 125% of benefit premium. (iv) Assumed expenses are 5% of gross premium, payable at the beginning of each year, and 300 to process each death claim, payable at the end of the year of death. (v) Profits are based on gross premium reserves. During year 11, actual experience is as follows: (a) There are 1000 lives inforce at the beginning of the year. (b) There are five deaths. (c) Interest earned equals 6%. (d) Expenses equal 6% of gross premium and 100 to process each death claim. For year 11, you calculate the gain due to mortality and then the gain due to expenses. Calculate the gain due to expenses during year 11. A B C D E Let us write Ga for the total actuarial gain, Ga M for the actuarial gain due to expenses, Ga E for the actuarial gain due to mortality, Ga I for the actuarial gain due to interest, and Ga W for the actuarial gain due to withdrawals. In general, the formula for the actuarial gain, assuming unit death benefit, is: Ga = k+1as k+1 AS = ( k AS + G) ( îk+1 i ) + ( Gc + e ) 1+ i k k ( Gĉ k + ê k )( 1+ ) îk+1 + ˆq x+k Ga M Ga I ( 1) 1 + q x+k 1 k+1 AS 1 k+1as + q ( 2) x+k Ga E ( CV AS) ( 2) k+1 k+1 ˆq CV k+1as x+k k+1. In this problem, we are told that the experience and assumed interest rate are equal, so that Ga I = 0. Furthermore, the problem has this peculiar structure where it has no reference to withdrawals, so we assume that withdrawals and cash value do not exist and do not enter into the picture, so that Ga W = 0. This leads to ( 1+ i) ( Gĉ k + ê k ) 1+ îk+1 Ga = Gc k + e k Ga E + q ( 1) x+k Ga W ( 1 k+1 AS) ˆq ( 1 ) x+k 1 k+1as. Ga M

17 Additionally, the problem does not give us any information about asset shares, so we cannot calculate the second term, and expenses are also assessed on payment of death benefit, resulting in interaction of mortality and expenses gains, which is not assumed in the above standard formula. We resolve the first issue by noting that information to calculate the reserve is given, so that we can calculate the gains based solely on reserves, and assuming that the assets portion is irrelevant. We resolve the second issue by following the order of calculations suggested by the wording of the problem: First calculate the mortality gain, then calculate the expenses gain. But how do we get the expenses gain from mortality gain? We subtract the mortality gain from the total gain, so we will have to find the total actuarial gain, in aggregate, without referring to its sources, to solve this problem. Let us also write G for the level annual gross premium, which is equal to 125% of level annual benefit premium. The level annual benefit premium is 50,000P 40 = 50,000A = a ILT Therefore, the gross premium is ,000P The expense-loaded policy reserve at policy duration 10 is V = 50, Benefit paid at the end of the year of death Expenses paid at the end of the year of death A G a = 50 Fraction of premium after expenses = 50, ILT Also, the expense-loaded policy reserve at policy duration 11 is V = 50, Benefit paid at the end of the year of death Expenses paid at the end of the year of death ILT A G a = 51 Fraction of premium after expenses = 50, ILT Of course, you can also find the second reserve from the first using the standard recursive formula 11V p 50 = ( 10 V + G 0.95) ,300q 50, so that ( ) , V = This is actually a slightly different value, but this error does not matter than much. If the actuarial assumptions are realized perfectly, the actuarial gain is zero. But if we substitute the actual experience for actuarial assumptions, we obtain the total actuarial gain as ILT

19 Illustrative Life Table at 6%. (iii) Expected expenses equal 40 plus 5% of gross premium, payable at the beginning of each year. (iv) Expected mortality equals 70% of the Illustrative Life Table. (v) The expected interest rate is 7%. Calculate the expected profit in the eleventh policy year, for a policy in force at the beginning of that year. A. 683 B. 719 C. 756 D. 792 E. 829 Based on the data from the Illustrative Life Table, the gross premium for this policy is G = ,000P 50 = , , Also, this policy s reserve at policy duration 10 is V = 100,000 A 10 ( P a ) = 36,913 1, , The reserve at policy duration 11 is 11 V = 100,000 A 61 P 50 a 61 ILT = ILT 38,729 1, , The problem does not clearly say this, but while we are supposed to calculate the expected profit for the 11-th policy year for a policy in force at the beginning of the 11-th year, the value of the profit is supposed to be established as of the end of that policy year, i.e., at duration 11. Note also that the calculation is on expected value basis, not based on experience, and therefore it represents expected values based on the original design of the policy. The expected value of the profit is 10V + G 0.05G q ,000 ( 1 0.7q 60 ) 11 V = Expenses Expected mortality = ( 15, , ) , Answer C. May 2012 Course MLC Examination, Problem No You You are are pricing pricing disability disability insurance using the following model: Healthy State 0 Disabled State 1 Dead State 2 You assume the following constant forces of transition: You assume the following constant forces of transition: Copyright 2012 by Krzysztof 01 Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission (i) of the µ copyright = 0.06owner. (ii) 10 µ = 0.03

20 (i) µ 01 = (ii) µ 10 = (iii) µ 02 = (iv) µ 12 = Calculate the probability that a disabled life on July 1, 2012 will become healthy at some time before July 1, 2017 but will not then remain continuously healthy until July 1, A B C D E The probability sought is Answer D t p µ 10 x 1 00 x+t 5 t p x+t dt = 0 A life disabled now will Then this life becomes be disabled continuously healthy at time t Then this life does not until time t remain continuously healthy for the rest of 5 years t 5 5 ( µ 10 +µ 12 )ds ( µ 01 +µ 02 )ds 0 = e µ 10 t 1 e dt = 0 t 5 5 ( )ds ( )ds 0 t = e e dt = 0 5 = e 0.07t t 0.03 ( 1 e 5 )dt = 0.03 ( e 0.07t e 0.35 )dt = 0 = 0.03a 5 δ =7% 0.15e May 2012 Course MLC Examination, Problem No. 20 Jenny joins XYZ Corporation today as an actuary at age 60. Her starting annual salary is 225,000 and will increase by 4% each year on her birthday. Assume that retirement takes place on a birthday immediately following the salary increase. XYZ offers a plan to its employees with the following benefits: A single sum retirement benefit equal to 20% of the final salary at time of retirement for each year of service. Retirement is compulsory at age 65; however, early retirement is permitted at ages 63 and 64, but with the retirement benefit reduced by 40% and 20%, respectively. The retirement benefit is paid on the date of retirement. A death benefit, payable at the end of the year of death, equal to a single sum of 100% of the annual salary rate at the time of death, provided death occurs while the employee is still employed. You are given that δ = 5% and the following multiple decrement table (w =withdrawal; r = retirement; and d = death):

21 Age x ( τ ) l x ( w) d x Calculate the expected present value of Jenny s total benefits. A. 85,000 B. 92,250 C. 99,500 D. 106,750 E. 113,750 In order to receive a benefit, Jenny can retire at ages 63, 64, or 65, or die at ages 60, 61, 62, 63 (without having retired previously), or 64 (without having retired previously). The following table shows the calculation of the actuarial present value of retirement benefits (note that the actuarial present value is calculated as salary times discount factor times probability times (number of years of service times 20%), corrected for reduction, and note that the benefit is paid at the beginning of the year of retirement (as opposed to the death benefit, which is paid at the end of the year of death): Age at retirement Salary Discount for present value Probability ( r) d x No. of years of service times 20% ( d) d x Reductio n Actuarial present value , e q ( r ) 3 60 = 6 60% 40% , e q ( r ) 4 60 = 5 80% 20% , e q ( r ) 5 60 = % 0% The sum of these actuarial present values is = 103,275. For the death benefits, the actuarial present values are calculated as follows (death benefit, which is salary at death, times the discount factor, times probability) Age at death Salary Discount for present value calculation Probability ,000 e 0.05 q ( d ) 0 60 = , e q ( d ) 1 60 = , e q ( r ) 2 60 = , e ( r 3 q ) 60 = , e q ( r ) 4 60 = Actuarial present value

22 The total actuarial present value of death benefits is = 10,474. The total actuarial present value of all benefits is 103, ,474 = 113,749. Answer E. May 2012 Course MLC Examination, Problem No. 21 For a fully continuous whole life insurance of 1000 on (x), you are given: (i) Benefit premiums are 10 per year. (ii) δ = (iii) µ x+20.2 = (iv) t V denotes the benefit reserve at time t for this insurance. at t = 20.2 is equal to (v) d t dt V Calculate 20.2 V. A. 480 B E. 730 Recall the Thiele s differential equation d t V dt Change in reserve = δ t V + Interest on reserve P ( S t V )µ x+t Premium (paid continuously) Net amount at risk payout at death We substitute numerical values at t = 20.2 and obtain 20.5 = V V This is a simple linear equation with 20.2 V being the unknown. We solve it and obtain 20.2 V Answer A.. May 2012 Course MLC Examination, Problem No. 22 For a fully discrete 10-year deferred whole life insurance of 100,000 on (30), you are given: (i) Mortality follows the Illustrative Life Table. (ii) i = 6%. (iii) The benefit premium is payable for 10 years. (iv) The gross premium is 120% of the benefit premium. (v) Expenses, all incurred at the beginning of the year, are as follows: Policy Year Percent of gross premium Per policy 1 10% % and later 0% 40 (vi) L 0 is the present value of future losses at issue random value.

23 Calculate E( L 0 ). A. 334 B. 496 C. 658 D. 820 E. 982 The gross premium is G = A 30 a 30:10 = A E 30 a E 30 a 40 = ILT = ILT Therefore, E( L 0 ) = G a 30: A G G a 30: a 30 = Answer C A a 30 = G a 30:10 ( ) + G May 2012 Course MLC Examination, Problem No. 23 On January 1 an insurer issues 10 one-year term life insurance policies to lives age x with independent future lifetimes. You are given: (i) Each policy pays a benefit of 1000 at the end of the year if that policyholder dies during the year. (ii) Each policyholder pays a single premium of 90. (iii) q x is the same for every policyholder. With probability 0.30, q x = 0.0 for every policyholder. With probability 0.70, q x = 0.2 for every policyholder. (iv) i = Calculate the variance of the present value of future losses at issue random variable for the entire portfolio. A. 800,000 B. 900,000 C. 1,000,000 D. 1,400,000 E. 1,800,000 Let Q be the random probability of death for a policyholder. Then, for each of the 10 policyholders, 0 with probability 0.3, Q = 0.2 with probability 0.7. Note that the problem says that the probability of dying q x is the same, so that once a value of Q is given, that value is q x, and it is the probability of dying for all policyholders. Let L k be the present value of future losses random variable for policyholder number k, where k = 1, 2,, 10. We have, for every k,

24 L k = 90 with probability 1 Q, with probability Q, = with probability 1 Q, 1 with probability Q. This effectively says that L k equals a constant of 90 added to a multiple (by the factor of 1000/1.04) of a Bernoulli Trial with probability of success of Q. Let us write L 1 = = 90 + = = X 10, 1.04 X, L X,, L 2 10 where each X k for k = 1, 2,, 10, is a Bernoulli Trial with probability of success Q. Let S be the random present value of future losses at issue random variable for the entire portfolio. We have S = L k = = X k. k= X k k= k=1 We observe that for Q = 0, each X k is equal to 900 with probability 1, while for Q = 0.2, We have Var S Note now that and Therefore, 10 X k is binomial with n = 10 and p = 0.2. We need to find the variance of S. k=1 ( ) + Var E( S Q) = E Var S Q E( S Q) = Var( S Q) = Var S Answer E when Q = 0,which happens with probability 0.3, when Q = 0.2,which happens with probability 0.7, when Q = 0,which happens with probability 0.3, = E Var S Q when Q = 0.2,which happens with probability 0.7. ( ) + Var E( S Q) = = ,812, Because Var( S Q) equals to a Bernoulli Trial Because E( S Q) is a constant of 900 with probability of success 0.7 multiplied by the number plus multiplied by a Bernoulli 1.04 Trial with probability of success of 0.7

25 May 2012 Course MLC Examination, Problem No. 24 For a three-year term insurance of 10,000 on (65), payable at the end of the year of death, you are given: (i) x q x (ii) Forward interest rates at the date of issue of the contract, expressed as annual rates, are as follows: Start time End time Annual forward rate Calculate the expected present value of this insurance. A. 105 B. 110 C. 113 D. 115 E. 120 We will write s k for the annual effective spot rate for maturity k, and f k,l for the effective annual forward rate from time k to time l. We are given f 0,3 = s 3 = 0.05, f 1,3 = 0.07, and f 2,3 = But ( 1+ s 3 ) 3 = ( 1+ f 0,1 )( 1+ f 1.3 ) 2 = ( 1+ s 1 )( 1+ f 1.3 ) 2, so that 3 = , 2 3 = ( 1+ s 2 ) 2 1+ f 2.3 ( ( 1+ s 2 ) 2 = 1+ s 3) 3 = s 1 = 1+ s 3 1+ f 1.3 as well as 1+ s 3, so that 1+ f 2.3 The expected present value of this insurance is q s 1 ( 1+ s 2 ) p 2 65 q s 3 = p 65 p 66 q 67 = ( ) ( ) ( ) Answer B.

26 May 2012 Course MLC Examination, Problem No. 25 An insurer issues fully discrete whole life insurance policies of 100,000 to insureds age 40 with independent future lifetimes. You are given: (i) Issue expenses are incurred upon policy issuance and are 20% of the first year premium. (ii) Renewal expenses are 6% of each premium, incurred on the premium due date, starting in the second year. (iii) The annual premium is calculated using the percentile premium principle and the normal approximation, such that the probability of a loss on the portfolio is 5%. (iv) A 40 = (v) 2 A 40 = (vi) a 40 = (vii) i = (viii) The annual premium per policy for a portfolio of 2000 policies is Calculate the difference in annual premium per policy for a portfolio of 2000 policies and for a portfolio of 40,000 policies. A. 37 B. 42 C. 48 D. 52 E. 56 As the annual premium per policy for a portfolio of 2000 policies is given, we simply need to find the annual premium per policy for a portfolio of 40,000 policies, call it G, and then calculate the difference. Let us write L i for the loss at issue for policy i. We have L i = 100, K ( )G ( )G a = K40 +1 = 100, K G 0.94G K ( 100, G) 1.06 K G. Therefore, E( L i ) = E( ( 100, G) 1.06 K G) = ( 100, G) A G = = 100, ( )G = 16, G, Var( L i ) = Var ( 100, G) 1.06 K G = 2 2 ( A 40 A 40 ) = 2 = 100, G = ( 100, G) , G Let L denote the total loss for the entire portfolio of 40,000 policies. Then (note that we are using independence of policies for the variance calculation):

27 = 40,000( 16, G), 40,000 ( 100, G) E L Var L Because of independence of the policies, and identical distributions of individual policies losses, L is approximately normal, and L E( L) 0.95 = Pr( L < 0) = Pr Var( L) < E( L) = Φ E( L) Var( L) Var( L) = 40,000 16, G = Φ 40,000 ( 100, G) The 95-th percentile of the standard normal distribution is 1.645, so that 40,000( 16, G) 40, , G = 1.645, and from this we calculate G The difference sought is = 37. Answer A. May 2012 Course MLC Examination, Problem No. 26 For a special fully discrete whole life insurance on (35) you are given: (i) Mortality follows the Illustrative Life Table. (ii) i = (iii) Initial annual premiums are level for the first 30 years; thereafter, annual premiums are one-third of the initial annual premium. (iv) The death benefit is 60,000 during the first 30 years and 15,000 thereafter. (v) Expenses are 40% of the first year s premium and 5% of all subsequent premiums. (vi) Expenses are payable at the beginning of the year. Calculate the initial annual premium using the equivalence principle. A. 290 B. 310 C. 330 D. 350 E. 370 From the Illustrative Life Table 30 E 35 = 20 E E 55 = , a 35:30 = a E 35 a , 1 A 35:30 = A E 35 A Let us denote the initial annual premium by G. We then calculate the actuarial present value of premiums as G 2 3 a :30 3 a 35 G G, the actuarial present value of benefits as 3 60,000 4 A :30 4 A , ,

28 and the actuarial present value of expenses as 2 G a :30 3 a 35 G G. Using the Equivalence Principle, we obtain the equation G = 4, G, resulting in G Answer E. May 2012 Course MLC Examination, Problem No. 27 For a universal life insurance policy with death benefit of 100,000 on (40), you are given: (i) The account value at the end of year 4 is (ii) A premium of 200 is paid at the start of year 5. (iii) Expense charges in renewal years are 40 per year plus 10% of premium. (iv) The cost of insurance charge for year 5 is 400. (v) Expense and cost of insurance charges are payable at the start of the year. (vi) Under a no lapse guarantee, after the premium at the start of year 5 is paid, the insurance is guaranteed to continue until the insured reaches age 49. (vii) If the expected present value of the guaranteed insurance coverage is greater than the account value, the company holds a reserve for the no lapse guarantee equal to the difference. The expected present value is based on the Illustrative Life Table at 6% interest and no expenses. Calculate the reserve for the no lapse guarantee, immediately after the premium and charges have been accounted for at the start of year 5. A. 0 B. 10 C. 20 D. 30 E. 40 The account value (reserve) at policy duration 5 (which is the start of policy year 5) is = The expected present value of the guaranteed coverage is the actuarial present value of term insurance to age 49, which is 1 100,000A 44:5 = 100,000( A 44 5 E 44 A 49 ) = = 100, ILT Note what happens here: The account value (i.e., retrospective reserve for a universal life policy is 1769, while the company guarantees that life insurance coverage will continue for the next five years, and that guarantee has actuarial present value of 1799, and that extra cost needs to be reserved for). Therefore, the no lapse reserve is = 30. Answer D. ILT 1799.

29 May 2012 Course MLC Examination, Problem No. 28 You are using Euler s method to calculate estimates of probabilities for a multiple state model with states {0,1, 2}. You are given: (i) The only possible transitions between states are: 0 to 1 1 to 0 1 to 2 (ii) For all x, µ x 01 = 0.3, µ x 10 = 0.1, µ x 12 = 0.1. (iii) Your step size is 0.1. (iv) You have calculated that (a) 0.6 p x 00 = (b) 0.6 p x 01 = (c) 0.6 p x 02 = Calculate the estimate of 0.8 p x 01 using the specified procedure. A B C D E The probability we are looking for is 0.8 p x 01, but in notation of Models for Quantifying ( 0 Risk, it is 0.8 p ) 01, if we consider age x to be time 0. Recall the Kolmogorov Forward Equation (we will continue using notation from Models for Quantifying Risk): d dr p ( t) r ij = p ( t ) λkj t + r r ik r p ( t ) λ t + r ( ij ) = p ( t) jk λkj t + r r ik k j k j p ( t) r ij λ j ( t + r). In this case, we have s = 0.3, λ 10 s = 0.1, λ 12 s = 0.1, λ 01 = µ x+s ( s) = µ x+s = µ x+s = µ x+s = µ x+s = µ x+s λ = 0, λ 20 s 20 = 0, λ 21 s 21 = 0. We also have the following specific Kolmogorov forward equations d dr p ( 0) r 01 = ( 0) r p λ01 00 ( r) r p ( 0 ) λ10 r ( 0) ( 01 ) + r p λ21 02 ( r) r p ( 0 ) λ12 r 01 ( 0) = 0.3 r p 00 ( 0) 0.1r p 01 ( 0) 0.1r p 01 = ( 0) 0.3r p ( 0) r p 01, ( ) = 0 + ( r p 02 ) = d dr p ( 0) r 00 = ( 0) r p λ10 01 ( r) r p ( 0 ) λ01 r 00 λ20 ( r) r p ( 0 ) λ02 r 00 ( 0) = 0.1 r p 01 ( 0) 0.3r p 00. The Euler step method replaces the derivative on the left-hand side of the equation by a difference quotient: ( 0) ( 0) 0.6 p 01 = ( 0) 0.7 p p 01 ( 0) so that 0.7 p Also, = , ( 0) p ( 0) 0.6 p 01

### 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

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

### 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

Solutions to the May 213 Course MLC Examination by Krzysztof Ostaszewski, http://wwwkrzysionet, krzysio@krzysionet Copyright 213 by Krzysztof Ostaszewski All rights reserved No reproduction in any form

### Chapter 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

### SOCIETY 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

### SOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE WRITTEN-ANSWER QUESTIONS AND SOLUTIONS

SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE WRITTEN-ANSWER QUESTIONS AND SOLUTIONS Questions February 12, 2015 In Questions 12, 13, and 19, the wording was changed slightly

### INSTRUCTIONS 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

### Premium Calculation. Lecture: Weeks 12-14. Lecture: Weeks 12-14 (STT 455) Premium Calculation Fall 2014 - Valdez 1 / 31

Premium Calculation Lecture: Weeks 12-14 Lecture: Weeks 12-14 (STT 455) Premium Calculation Fall 2014 - Valdez 1 / 31 Preliminaries Preliminaries An insurance policy (life insurance or life annuity) is

### ACTUARIAL 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 Heriot-Watt University, Edinburgh CAMBRIDGE

### Contents. 3 Survival Distributions: Force of Mortality 37 Exercises Solutions... 51

Contents 1 Probability Review 1 1.1 Functions and moments...................................... 1 1.2 Probability distributions...................................... 2 1.2.1 Bernoulli distribution...................................

### JANUARY 2016 EXAMINATIONS. Life Insurance I

PAPER CODE NO. MATH 273 EXAMINER: Dr. C. Boado-Penas TEL.NO. 44026 DEPARTMENT: Mathematical Sciences JANUARY 2016 EXAMINATIONS Life Insurance I Time allowed: Two and a half hours INSTRUCTIONS TO CANDIDATES:

### Premium Calculation. Lecture: Weeks 12-14. Lecture: Weeks 12-14 (Math 3630) Annuities Fall 2015 - Valdez 1 / 32

Premium Calculation Lecture: Weeks 12-14 Lecture: Weeks 12-14 (Math 3630) Annuities Fall 2015 - Valdez 1 / 32 Preliminaries Preliminaries An insurance policy (life insurance or life annuity) is funded

### Premium Calculation - continued

Premium Calculation - continued Lecture: Weeks 1-2 Lecture: Weeks 1-2 (STT 456) Premium Calculation Spring 2015 - Valdez 1 / 16 Recall some preliminaries Recall some preliminaries An insurance policy (life

### INSTITUTE 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

### Manual 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)

### SOCIETY 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

### Annuities. Lecture: Weeks 9-11. Lecture: Weeks 9-11 (STT 455) Annuities Fall 2014 - Valdez 1 / 43

Annuities Lecture: Weeks 9-11 Lecture: Weeks 9-11 (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:

### SOA 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

### TABLE 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

### Heriot-Watt University. BSc in Actuarial Mathematics and Statistics. Life Insurance Mathematics I. Extra Problems: Multiple Choice

Heriot-Watt 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.

### Further 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...............................

### Please 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 (6:00-8:00 pm) Room: MSB 411 Total Marks: 120 points Please write your name and student number

### Practice 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

### O MIA-009 (F2F) : GENERAL INSURANCE, LIFE AND

No. of Printed Pages : 11 MIA-009 (F2F) kr) ki) M.Sc. ACTUARIAL SCIENCE (MSCAS) N December, 2012 0 O MIA-009 (F2F) : GENERAL INSURANCE, LIFE AND HEALTH CONTINGENCIES Time : 3 hours Maximum Marks : 100

### Insurance Benefits. Lecture: Weeks 6-8. Lecture: Weeks 6-8 (STT 455) Insurance Benefits Fall 2014 - Valdez 1 / 36

Insurance Benefits Lecture: Weeks 6-8 Lecture: Weeks 6-8 (STT 455) Insurance Benefits Fall 2014 - Valdez 1 / 36 An introduction An introduction Central theme: to quantify the value today of a (random)

### Manual 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

### INSTITUTE 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

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

### MATH 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:

### Manual 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 annuity-due on (40), you

### EXAMINATION. 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

### Institute of Actuaries of India

Institute of Actuaries of India Subject CT5 General Insurance, Life and Health Contingencies May 2008 Examination INDICATIVE SOLUTION Introduction The indicative solution has been written by the Examiners

### Some 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

### **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

### INSURANCE 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

### INSTITUTE 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

### 1 Cash-flows, discounting, interest rate models

Assignment 1 BS4a Actuarial Science Oxford MT 2014 1 1 Cash-flows, discounting, interest rate models Please hand in your answers to questions 3, 4, 5 and 8 for marking. The rest are for further practice.

### Premium 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

### 1. 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

### Math 630 Problem Set 2

Math 63 Problem Set 2 1. AN n-year 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,

### 2 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

### PREMIUM 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

### THE 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

### ACTS 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

### b 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 000-1

### CHAPTER 6 DISCOUNTED CASH FLOW VALUATION

CHAPTER 6 DISCOUNTED CASH FLOW VALUATION Answers to Concepts Review and Critical Thinking Questions 1. The four pieces are the present value (PV), the periodic cash flow (C), the discount rate (r), and

### Manual 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

### Glossary of insurance terms

Glossary of insurance terms I. Insurance Products Annuity is a life insurance policy where an insurance company pays an income stream to an individual, usually until death, in exchange for the payment

### APPENDIX 4 - LIFE INSURANCE POLICIES PREMIUMS, RESERVES AND TAX TREATMENT

APPENDIX 4 - LIFE INSURANCE POLICIES PREMIUMS, RESERVES AND TAX TREATMENT Topics in this section include: 1.0 Types of Life Insurance 1.1 Term insurance 1.2 Type of protection 1.3 Premium calculation for

### For educational use by Illinois State University students only, do not redistribute - 71 -

Valuation of life annuities A life annuity is a financial contract in the form of an insurance policy in which a seller (issuer), a life insurance company, makes a series of future payments to a buyer

### GLOSSARY. A contract that provides for periodic payments to an annuitant for a specified period of time, often until the annuitant s death.

The glossary contains explanations of certain terms and definitions used in this prospectus in connection with the Group and its business. The terms and their meanings may not correspond to standard industry

### Finance 350: Problem Set 6 Alternative Solutions

Finance 350: Problem Set 6 Alternative Solutions Note: Where appropriate, the final answer for each problem is given in bold italics for those not interested in the discussion of the solution. I. Formulas

### CHAPTER 5 INTRODUCTION TO VALUATION: THE TIME VALUE OF MONEY

CHAPTER 5 INTRODUCTION TO VALUATION: THE TIME VALUE OF MONEY 1. The simple interest per year is: \$5,000.08 = \$400 So after 10 years you will have: \$400 10 = \$4,000 in interest. The total balance will be

### The Lafayette Life Insurance Company Agents Products Quiz The Marquis Series of Products and Other Annuities

There are different types of annuity products that can serve different needs. Even within a particular type of annuity product category, for example a fixed indexed annuity, the benefits and features can

### November 2000 Course 3

November Course 3 Society of Actuaries/Casualty Actuarial Society November - - GO ON TO NEXT PAGE Questions through 36 are each worth points; questions 37 through 44 are each worth point.. For independent

### EDUCATION 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,

### Heriot-Watt University. M.Sc. in Actuarial Science. Life Insurance Mathematics I. Tutorial 5

1 Heriot-Watt University M.Sc. in Actuarial Science Life Insurance Mathematics I Tutorial 5 1. Consider the illness-death model in Figure 1. A life age takes out a policy with a term of n years that pays

### Math 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,

### INSTITUTE 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

### Solution. 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

### Your guide to participating life insurance SUN PAR PROTECTOR SUN PAR ACCUMULATOR

Your guide to participating life insurance SUN PAR PROTECTOR SUN PAR ACCUMULATOR Participate in your brighter future with Sun Life Financial. Participating life insurance is a powerful tool that protects

### VERMONT DEPARTMENT OF BANKING AND INSURANCE REVISED REGULATION 77-2 VERMONT LIFE INSURANCE SOLICITATION REGULATION

VERMONT DEPARTMENT OF BANKING AND INSURANCE REVISED REGULATION 77-2 VERMONT LIFE INSURANCE SOLICITATION REGULATION Section 1. AUTHORITY This rule is adopted and promulgated by the Commissioner of Banking

### WHAT IS LIFE INSURANCE?

UNDERSTANDING LIFE INSURANCE Presented by The Kansas Insurance Department WHAT IS LIFE INSURANCE? a. Insurance Contract issued by an Insurance Company. b. Premiums paid under the contract provide for a

### EXAMINATIONS. 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

### SECTION 99.1 Purposes. The purposes of this Part are:

INSURANCE DEPARTMENT OF THE STATE OF NEW YORK REGULATION NO. 151 (11 NYCRR 99) VALUATION OF ANNUITY, SINGLE PREMIUM LIFE INSURANCE, GUARANTEED INTEREST CONTRACT AND OTHER DEPOSIT RESERVES I, Neil D. Levin,

### 4. Life Insurance. 4.1 Survival Distribution And Life Tables. Introduction. X, Age-at-death. T (x), time-until-death

4. Life Insurance 4.1 Survival Distribution And Life Tables Introduction X, Age-at-death T (x), time-until-death Life Table Engineers use life tables to study the reliability of complex mechanical and

### Stochastic Analysis of Long-Term Multiple-Decrement Contracts

Stochastic Analysis of Long-Term Multiple-Decrement 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

### 58-58-50. Standard Valuation Law.

58-58-50. Standard Valuation Law. (a) This section shall be known as the Standard Valuation Law. (b) Each year the Commissioner shall value or cause to be valued the reserve liabilities ("reserves") for

### (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

### Be it enacted by the People of the State of Illinois,

AN ACT concerning insurance. Be it enacted by the People of the State of Illinois, represented in the General Assembly: Section 5. The Illinois Insurance Code is amended by changing Sections 223 and 229.2

### The Basics of Interest Theory

Contents Preface 3 The Basics of Interest Theory 9 1 The Meaning of Interest................................... 10 2 Accumulation and Amount Functions............................ 14 3 Effective Interest

### NORTH CAROLINA GENERAL ASSEMBLY 1981 SESSION CHAPTER 761 SENATE BILL 623

NORTH CAROLINA GENERAL ASSEMBLY 1981 SESSION CHAPTER 761 SENATE BILL 623 AN ACT TO AMEND CHAPTER 58, ARTICLE 22, OF THE GENERAL STATUTES RELATING TO NONFORFEITURE BENEFITS OF LIFE INSURANCE POLICIES AND

### 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

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

### LEVELING THE NET SINGLE PREMIUM. - Review the assumptions used in calculating the net single premium.

LEVELING THE NET SINGLE PREMIUM CHAPTER OBJECTIVES - Review the assumptions used in calculating the net single premium. - Describe what is meant by the present value of future benefits. - Explain the mathematics

### 001. Statutory authority. This regulation is promulgated under the authority of NEB. REV. STAT. 44-101.01 and is operative October 1, 2008.

Title 210 - NEBRASKA DEPARTMENT OF INSURANCE Chapter 40 - UNIVERSAL LIFE INSURANCE 001. Statutory authority. This regulation is promulgated under the authority of NEB. REV. STAT. 44-101.01 and is operative

### GLOSSARY. A contract that provides for periodic payments to an annuitant for a specified period of time, often until the annuitant s death.

The glossary contains explanations of certain terms and definitions used in this prospectus in connection with us and our business. The terms and their meanings may not correspond to standard industry

### Universal Life Insurance

Universal Life Insurance Lecture: Weeks 11-12 Thanks to my friend J. Dhaene, KU Leuven, for ideas here drawn from his notes. Lecture: Weeks 11-12 (STT 456) Universal Life Insurance Spring 2015 - Valdez

### SOCIETY 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. M-9-5 PRINTED

### Chapter 16 Fundamentals of Life Insurance

Chapter 16 Fundamentals of Life Insurance Overview This chapter begins a block of material on several important personal risks: premature death, poor health, and excessive longevity. This chapter examines

### Manual 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,

### 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

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.

### QUICK NOTES SUPPLEMENTAL STUDY GUIDE NEW JERSEY

QUICK NOTES SUPPLEMENTAL STUDY GUIDE NEW JERSEY A REVIEW SUPPLEMENT FOR THE NEW JERSEY LIFE, ACCIDENT & HEALTH STATE LICENSING EXAM (April 2016 Edition) What is Insurance Schools Quick Notes Supplemental

### EDUCATION AND EXAMINATION COMMITTEE SOCIETY OF ACTUARIES RISK AND INSURANCE. Copyright 2005 by the Society of Actuaries

EDUCATION AND EXAMINATION COMMITTEE OF THE SOCIET OF ACTUARIES RISK AND INSURANCE by Judy Feldman Anderson, FSA and Robert L. Brown, FSA Copyright 25 by the Society of Actuaries The Education and Examination

### Alamere Indexed Universal Life Product

Alamere Indexed Universal Life Product V1.0 Document Part No. E55321-01 June 2014 Alamere Indexed Universal Life Product 1 of 18 Table of Contents I. REVISION HISTORY 5 II. PRODUCT DESCRIPTION & POLICY

### CHAPTER 5 LIFE INSURANCE POLICY OPTIONS AND RIDERS

CHAPTER 5 LIFE INSURANCE POLICY OPTIONS AND RIDERS There are a number of policy options and riders that the purchasers of insurance need to understand. Obviously, life and disability agents need to have

### RULES OF THE DEPARTMENT OF INSURANCE DIVISION OF INSURANCE CHAPTER 0780-1-40 RELATING TO LIFE INSURANCE SOLICITATION TABLE OF CONTENTS

RULES OF THE DEPARTMENT OF INSURANCE DIVISION OF INSURANCE CHAPTER 0780-1-40 RELATING TO LIFE INSURANCE SOLICITATION TABLE OF CONTENTS 0780-1-40-.01 Purpose 0780-1-40-.05 General Rules 0780-1-40-.02 Scope

### Profit 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.

### CHAPTER 57 - LIFE AND HEALTH REINSURANCE AGREEMENTS

TITLE 210 - NEBRASKA DEPARTMENT OF INSURANCE CHAPTER 57 - LIFE AND HEALTH REINSURANCE AGREEMENTS 001. Authority. This regulation is adopted and promulgated by the Director of Insurance of the State of

### COURSE 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.

### MATHEMATICS OF FINANCE AND INVESTMENT

MATHEMATICS OF FINANCE AND INVESTMENT G. I. FALIN Department of Probability Theory Faculty of Mechanics & Mathematics Moscow State Lomonosov University Moscow 119992 g.falin@mail.ru 2 G.I.Falin. Mathematics

### CHAPTER 5 INTRODUCTION TO VALUATION: THE TIME VALUE OF MONEY

CHAPTER 5 INTRODUCTION TO VALUATION: THE TIME VALUE OF MONEY Answers to Concepts Review and Critical Thinking Questions 1. The four parts are the present value (PV), the future value (FV), the discount

### Effective 5/10/ A Standard Nonforfeiture Law for Life Insurance.

Effective 5/10/2016 31A-22-408 Standard Nonforfeiture Law for Life Insurance. (1) (a) This section is known as the "Standard Nonforfeiture Law for Life Insurance." (b) This section does not apply to group

### A Technical Guide for Individuals. The Whole Story. Understanding the features and benefits of whole life insurance. Insurance Strategies

A Technical Guide for Individuals The Whole Story Understanding the features and benefits of whole life insurance Insurance Strategies Contents 1 Insurance for Your Lifetime 3 How Does Whole Life Insurance

### KCI Tax Interpretation

KCI Koehler Consulting, Inc. KCI Tax Interpretation Sample March 10, 2006 This document contains a sample sub-section from the KCI Tax Interpretation. This is for evaluation purposes, only. The table of

### ACTUARIAL GUIDELINE XXXIII DETERMINING CARVM RESERVES FOR ANNUITY CONTRACTS WITH ELECTIVE BENEFITS

ACTUARIAL GUIDELINE XXXIII DETERMINING CARVM RESERVES FOR ANNUITY CONTRACTS WITH ELECTIVE BENEFITS Background Information 1. Introduction The Standard Valuation Law (SVL) defines the methods and assumptions

### INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION

INSTITUTE AND FACULTY OF ACTUARIES EXAMINATION 30 April 2015 (am) Subject SA2 Life Insurance Specialist Applications Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1. Enter all the candidate and