MLC Fall 2015 Written Answer Questions. Model Solutions

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1 MLC Fall 215 Written Answer Questions Model Solutions 1

2 MLC Fall 215 Question 1 Model Solution Learning Outcome: 1(a) Chapter References: AMLCR Overall, this question was done reasonably well, although few candidates received maximum credit. (a) (i) For any age x, the survival function S x (t) must satisfy the following 1. S x () = 1 2. lim S x (t) = t 3. S x (t) must be a non-increasing function of t (ii) b = -.2: 1. S 4 () = lim t S 4 (t) = lim t.75e.2(t 25) = For the third criterion, we show that S 4 (t) is non-increasing before age 65, after age 65 and at age 65: d dt S 4(t) =.8t < for t < 25 d dt S 4(t) =.2.75 e.2(t 25) < for t 25 and lim S 4(t) =.75 = lim S 4(25) t 25 t 25 + Hence, b =.2 is a valid parameter. b=.: lim S 4(t) = lim.75 t t Hence, b =. is not a valid parameter. b=.2: lim S 4(t) = lim.75e.2(t 25) = t t Hence, b =.2 is not a valid parameter. 2

3 For full credit, candidates were expected to verify all three criteria from (a)(i), for b =.2. Few candidates verified, for example, that the function is non-increasing at t = 25. For the invalid values of b, candidates could justify the conclusion by indicating any criterion that is not satisfied. (b) (i) ( ) 1 d µ 4+t = S 4 (t) dt S 4(t) µ 6 = 1 ( ) = Alternative: µ 4+t = d dt log S 4(t) = d dt log ( 1 (.2t) 2) for t 25 µ 6 = 2(.2) ((.2)(2)) 2 =.195 (ii) µ 4+t = d dt log S 4(t) = d dt µ 7 = d (.1t) =.1 dt ( log.75e log e.1t) for t > 25 Alternative: ( ) 1 d µ 4+t = S 4 (t) dt S 4(t) (.75).1 e.1(t 25) = =.1 for t > e.1(t 25) µ 7 =.1 3

4 (iii) e 4:35 = = S 4 (t) dt (1.4t 2 )dt + S 4 (t) dt e.1(t 25) dt (1.4t 2 )dt = [ t.4t 3 /3 ] 25 = e.1(t 25) dt = 1 so e 4:35 = = e.1u du = Most candidates who attempted this part correctly calculated the first integral on the right hand side above. Only the stronger candidates successfully set up and evaluated the second integral. 4

5 MLC Fall 215 Question 2 Model Solution Learning Outcomes: 1(b), 1(d), 2(a), 4(a) Chapter References: AMLCR , 9.6 This proved to be one of the most challenging questions on the paper. (a) The future lifetimes of (x) and (y) are dependent, because the force of mortality for each is different depending on whether the other is alive or not, as µ 1 x+t:y+t µ 23 x+t, and µ 2 x+t:y+t µ 13 y+t. That means that the distribution of the time to death of (x) is different if (y) dies early than if (y) dies later (and vice versa), which means the future lifetime random variables T x and T y are dependent. The key point, which around one-third of the candidates identified, is that the force of mortality for (x) from state 2 is different than the force from state, and the force of mortality for (y) is different from state 1 than from state, which indicates dependence. The fact that there is no common shock transition does not imply independence. When justifying dependence, candidates were expected to compare appropriate pairs (i.e. µ 1 x+t:y+t and µ 23 x+t:y+t, which both concern the death of (x), and µ 2 x+t:y+t and µ 13 x+t:y+t, which both concern the death of (y)). (b) (i) (ii) d ( ) dt t p xy = t p xy µ 1 x+t: y+t + µ 2 x+t: y+t Boundary condition: p xy = 1. d ( ) dr r p xy = r p xy µ 1 x+r: y+r + µ 2 x+r: y+r 1 rp xy d dr r p xy = ( ) µ 1 x+r: y+r + µ 2 x+r: y+r d dr log rp xy = ( µ 1 x+r: y+r + µ 2 x+r: y+r ) 5

6 Integrate both sides from to t t d dr log rp xy dr = t log t p xy log p xy = µ 1 x+r: y+r + µ 2 t x+r: y+r dr µ 1 x+r: y+r + µ 2 x+r: y+r dr from the boundary condition log p xy =, so log t p xy = t t p xy = exp µ 1 x+r: y+r + µ 2 ( t x+r: y+r dr ) µ 1 x+r: y+r + µ 2 x+r: y+r dr Most candidates scored partial credit for this part. Some candidates wrote down a few steps, but missed the key parts of the proof. An acceptable alternative approach was to plug the given solution into the Kolmogorov equation and demonstrate that the integral equation for t p xy satisfies the Kolmogorov differential equation and the boundary condition. This approach was awarded full credit if done correctly. (c) (i) At time 1, (x) is age 6 and (y) is age 65. The value at time 1 of the joint and reversionary annuities is 5, ā 6: , ā 1 6: , ā 2 6: 65 = 5, (8.8219) + 3, (1.3768) + 3, (3.175) = 572, 924 (ii) The EPV of the deferred annuity for the case where both lives survive 1 years uses the result from (i): 1p 5: 55 v 1 572, 924 = , 924 = 32, 627 The EPV of the deferred annuity for the case where (y) survives 1 years but (x) does not, is 1p 1 5: 55 v 1 (3, ā 11 65) = , = 978 6

7 The EPV of the deferred annuity for the case where (x) survives 1 years but (y) does not, is 1p 2 5: 55 v 1 (3, ā 22 6) = , = 18, 799 Let P denote the premium. Then the EPV of the benefit paid on second death during the deferred period is 3P Ā3 5: 55: 1 = 3P (.3421) = (.1263) P Putting these together gives the premium P = 32, , , P P = 3354 = 333, Only the top few candidates achieved full credit for this part. Many candidates did not allow for the possibility that only one life would survive the deferred period. The lower scoring candidates used combinations of probabilities and annuities that indicated less than adequate understanding of multiple state models and notation. For example, the expression ( 1 p 1 5:55 ā 1 6:65) is meaningless, as it requires the lives to be in state and also in state 1 at time 1. The second superscript of t p ij in this type of combination must be the same as the first superscript of ā jk. (d) (i) 1V () = 572, 924 from (c)(i) 1V (1) = 3, ā = 3, = 35, 844 1V (2) = 3, ā 22 6 = 3, = 354, 96 (ii) For t 1, and δ = log(1.5), d dt t V () = δ t V () 5, µ 1 5+t: 55+t (tv (1) t V ()) µ 2 5+t: 55+t (tv (2) t V ()) 7

8 (iii) We need µ 1 6:65 = A+Bc 6 =.976 and µ 2 6:65 = A+Bc 65 = Then t+hv () t V () + h ( δ t V () 5, µ 1 5+t: 55+t (tv (1) t V ()) µ 2 5+t: 55+t (tv (2) t V ())) { 572, (572, 924) 5, 564, (35, , 924) }.15919(354, , 924) As in (c), a few of the the very best candidates achieved full credit for this part. Many students correctly calculated the three reserves in (i). Some did not understand that, for example, t V (1) is the reserve assuming the policy is in state 1 at t, that is that (5) has already died, so the correct annuity factor must be a 11 65, not a 1 6:65. Common errors in (ii) and (iii) included omitting the release of reserve terms involving t V () on the right hand side, omitting the annuity paid in state, including the annuities paid in the other states (these are implicitly valued in t V (1) and t V (2) ), and getting one or more signs wrong. 8

9 MLC Fall 215 Question 3 Model Solution Learning Outcomes: 3(c), 4(c), 4(g) Chapter References: AMLCR Overall, this question was done well, with many candidates achieving at least 9/1 points. (a) P ä 5 = 1, A ä 5 ä 5 = A 5 = , P = = + 5 = This part was very well done by most candidates. (b) The profit test table for t =, 1, 2, 3 using notation from the textbook, is as follows. t t 1V P t E t I t EDB t ECV t E t V P r t * Explanation and useful numbers I t = i t ( t 1 V + P t E t ), i t =.5 for t = 1, i t =.7 for t = 2, 3 EDB t = q 5+t 1 1, ECV t = p 5+t 1.5 CV t E t V = p 5+t 1.95 t V q 5 =.592 q 51 =.642 q 52 =.697 p 5 =.9948 p 51 = p 52 =.9933 Most candidates received partial credit for this part. Other than calculation errors, the most common mistakes were (i) Missing or incorrect use of surrender probabilities when calculating the expected surrender values and/or reserves, and (ii) Using the previous year profit instead of the reserve when calculating the interest income and annual profit. 9

10 (c) (i) NP V () = 1 NP V (1) = v 1% = NP V (2) = p (τ) 5 v 2 1% 1p (τ) 5 = = NP V (2) = 39.3 So the third year is the first year with Pr t > and NP V (t 1) >. (ii) The insurer pays out 85% of the profit in year 3, which is = per policy in force at t = 2 Only policies in force at t = 3 participate. The probability that a policy in force at t = 2 is still in force at t = 3 is 1p (τ) 52 =.95p 52 = Hence the projected dividend payment per policy in force at t = 3 is 684.4/ = Most candidates received partial credit for this part. The most common mistake was ignoring or using incorrect probabilities when discounting the profit vector. Also, some candidates multiplied the distributed profit by 1 p (τ) 52, instead of dividing by it. (d) Possible Advantages: Maintain more assets under management if policyholders choose to convert, which gives potential for future profits. Reduce liquidity requirements. Increase attractiveness of the contract. More tax-efficient distribution than cash dividends Less expensive to operate than cash dividends Possible Disadvantages: Adverse selection sicker lives will choose the option, which will then cost more than anticipated. More complex for policyholders to understand Extra cost of administration, acquisition and maintenance expenses. Candidates were only required to mention one advantage and one disadvantage for full credit. Partial credit was awarded for some answers where the candidate showed some understanding, but did not complete the explanation. Some candidates discussed advantages and disadvantages of offering a participating policy, instead of discussing the dividend option. These answers received no credit. 1

11 MLC Fall 215 Question 4 Model Solution Learning Outcomes: 3(a) Chapter References: AMLCR 6.4, 6.8 This question was the highest scoring on the paper, with a large number of candidates achieving full credit. (a) Death in Year 1: L Event = (1 + G(1 + i))v G = 1v = Probability =.6 Withdrawal in Year 1: L Event = G Probability =.4 Death in Year 2: L Event = (1 + G((1 + i) + (1 + i) 2 ))v 2 G(1 + v) = 1v 2 = 89. Probability =.9.12 =.18 Survival in force to end of Year 2: L Event = G(1 + v) = G Probability =.9.88 =.792 In table form: Event Value of L given Event Probability of Event Death in Year Withdrawal Year 1 G.4 Death in Year Neither death nor withdrawal G.792 Most candidates did well in this part. Some candidates confused dependent and independent rates of mortality and withdrawal. A few calculated an equivalence principle premium, even though it was not required or relevant. Amongst candidates who did not achieve full marks, the most common problem was determining the amount of the return of premiums benefit. 11

12 (b) (i) E[L ] = 943.4(.6) G(.4) + 89.(.18) G(.792) = G a = b = (ii) E[L 2 ] = (.6) + G 2 (.4) (.18) + (1.9434G) 2 (.792) = G 2 E[L ] 2 = ( G) 2 = G G 2 V [L ] = E[L 2 ] E[L ] 2 = 115, G +.538G 2 c =.538 d = e = 115, 63 The table in part (a) was used by stronger candidates to answer part (b), as the examiners intended. Standard variance formulas for level benefit term insurance do not work in this case, and candidates who tried to use memorized formulas received little or no credit. (c) For each policy E[L ] = a bg = V [L ] = cg 2 + dg + e = 187, 48 So for the aggregate loss E[L agg ] = 2 ( 52.57) = 1514 V [L agg ] = 2 187, 48 = = ( ) ( 1, 514) Pr[L agg > ] = 1 Φ = 1 Φ(1.72) = =.427 Some candidates used the wrong tail of the Normal distribution for this part. Otherwise, this was done well. 12

13 MLC Fall 215 Question 5 Model Solution Learning Outcomes: 3(c), 4(e) Chapter References: AMLCR 13.4 Many candidates scored full marks for parts (a), (b), and (c)(i). Only the best candidates managed to score full marks for (c)(ii) and (d). (a) P 2 = 1: AV 2 = (AV 1 +.9P 2 1 2v) (1.6) = ( P 2 1) (1.6) 2 = Generally well done. Errors included incorrect discounting of the CoI, and incorrect treatment of the surrender charge. (b) AV 2 = ( P 2 1) (1.6) 2 =.954P AV 3 = (AV 2 +.9P 3 1) (1.6) 3 Generally well done. = (.954P P 3 1) (1.6) 3 = 1.112P P a = b =.954 c = (c) (i) Calculate the different possible values for AV 3, using the result from (b), AV 3 = 1.112P P as follows: P 2 P 3 Prob. AV

14 which gives E[DB 3 ] =.36 11, , , , 44.6 =1, (ii) E[CV 3 ] = = The main issue in this part was recognizing that CV 3 cannot be less than zero, so it is not appropriate simply to subtract 1 from E[AV 3 ]. (d) Let AV (1) 1 = 5114 denote the AV assuming all premiums paid; AV (2) 1 denotes the AV at time 1 assuming all premiums except the last are paid, and AV (3) 1 denotes the AV at time 1 assuming all premiums except the third are paid. Then AV (2) 1 = AV (1) 1 1(.95)(1.6) = 417. AV (3) 1 = AV (1) 1 1(.9)(1.6) 8 = E[AV 1 ] =.5 AV (2) AV (3) 1 = This part was done well by those who made a substantive attempt. omitted this part. Many candidates 14

15 MLC Fall 215 Question 6 Model Solution Learning Outcomes: 2(a), 2(b) Chapter References: AMLCR 4.4, 11.4 This was the lowest scoring question on the paper. There was some evidence that candidates were pressed for time, but also evidence that candidates struggled with some of the concepts covered. Generally, passing candidates understood what to do with a force of interest that varies with time (from Exam FM), and, further, understood that conditional variance was needed, and knew how to find it. (a) Let Y j denote the PV of benefits for the jth life, Y = 1 j=1 Y j. Let v(5) denote the discount factor for time 5. That is ( exp ) 5 v(5) =.3t 1 2 dt with prob..6 ( exp ) 5.2dt with prob t.5 dt =.2 t 3 2 ] 5 = e 5.3 t.5 dt = { with prob..6 v(5) =.9484 with prob..4 Also 5 p 85 = = E[Y j v(5)] = v(5) { with prob..6 E[Y j v(5)] =.4613 with prob..4 { with prob..6 E[Y v(5)] = with prob..4 E[Y ] = E[E[Y v(5)]] = The most common error was incorrect calculation of v(5). 15

16 (b) V [Y ] = E [V [Y v(5)]] + V [E[Y v(5)]] E[Y v(5)] = 1v(5) 5 p 65 V [Y v(5)] = 1 (v(5)) 2 5p 85 (1 5 p 85 ) so we have { with prob..6 E[Y v(5)] = with prob..4 { with prob..6 V [Y v(5)] = with prob..4 So E [ V [Y v(5)] ] = = V [ E[Y v(5)] ] = (35.891) (4.613) 2.4 (37.78) 2 = V [Y ] = = = So, the required probability is ( ) Pr[Y 3] = Φ 4.79 = Φ( 1.62) = 1 Φ(1.62) = =.526 The most common error for this part was omitting the second part of the conditional variance calculation. (c) (i) A risk is diversifiable if adding more units of risk reduces the standard deviation of the mean loss per contract, with a limit of zero as the number tends to infinity. Mathematically, a portfolio of risks X j is diversifiable if and only if lim N V [ N N j=1 X j ] = (ii) The risk is not diversifiable, because the uncertainty about the discount factor cannot be eliminated by increasing the number of policies. 16

17 Any verbal or mathematical explanation of diversification that incorporates the idea that for a diversifiable portfolio, adding policies reduces the variance or standard deviation of the mean loss, with a limit of, was given full credit. It was not sufficient however, to say that the variance of the portfolio goes to as N ; the variance of the portfolio as a whole tends to, it is the variance of the mean that tends to. For (ii) it was acceptable to say that the mortality risk is diversifiable, but the interest rate risk is not. 17

MLC Fall 2014 Written Answer Questions. Model Solutions

MLC Fall 2014 Written Answer Questions Model Solutions 1 MLC Fall 2014 Question 1 Model Solution Learning Objectives: 2(a), (a), 4(f) Textbook References: 5.4, 6.6, 1.5 (a) The EPV of the commission payments