SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE SOLUTIONS The following questions or solutions have been modified since this document was prepared to use with the syllabus effective spring 2012, Prior to March 1, 2012: Questions: 151, 181, 289, 300 Solutions: 2, 284, 289, 290, 295, 300 Changed on March 19, 2012: Questions: 20, 158, 199 (all are minor edits) Changed on April 24, 2012: Solution: 292 Changed on August 20, 2012: Questions and Solutions 38, 54, 89, 180, 217 and 218 were restored from MLC-09-08 and reworded to conform to the current presentation. Question 288 was reworded to resolve an ambiguity. A typo in Question 122 was corrected. Questions and Solutions 301-309 are new questions Changed on August 23, 2012: Solution 47, initial formula corrected; Solution 72, minus signs added in the first integral Changed on December 11, 2012: Question 300 deleted Copyright 2011 by the Society of Actuaries MLC-09-11 PRINTED IN U.S.A. Question #1 Answer: E q  p  p 2 30:34 2 30:34 3 30:34 p 0.90.80.72 2 30 p 0.50.40.20 2 34 p 0.720.200.144 2 30:34 p 0.720.200.1440.776 2 30:34 p 0.720.70.504 3 30 p 0.200.30.06 3 34 p 0.5040.060.03024 3 30:34 p 0.5040.060.03024 3 30:34 0.53376 q 0.7760.53376 2 30:34 0.24224 Alternatively, q  q  q  q 2 30:34 2 30 2 34 2 30:34 b g  p q  p q  p 1 p 2 30 32 2 34 36 2 30:34 32:36 = (0.9)(0.8)(0.3) + (0.5)(0.4)(0.7) – (0.9)(0.8)(0.5)(0.4) [1-(0.7)(0.3)] = 0.216 + 0.140 – 0.144(0.79) = 0.24224 Alternatively, q  q  q  q  q 2 30:34 3 30 3 34 2 30 2 34 1 p 1 p 1 p 1 p  3 30 3 34 2 30 2 34 10.50410.0610.7210.20 0.24224 (see first solution for p , p , p , p ) 2 30 2 34 3 30 3 34 MLC‐09‐11 1 Question #2 Answer: E 1000A 1000A1  A  x  x:10 10 x 100010e0.04te0.06t(0.06)dte0.4e0.6e0.05te0.07t(0.07)dt  0 0  10000.0610e0.1tdte1(0.07)e0.12tdt  0 0   10  1000 0.06e0.10t e1(0.07)e0.12t   0.10   0.12    0 0  10000.061e1 0.07e1 0.10  0.12  10000.379270.21460593.87 Because this is a timed exam, many candidates will know common results for constant force and constant interest without integration.  For example A1  1 E  x:10  10 x E e10 10 x  A  x  With those relationships, the solution becomes 1000A 1000A1  E A  x  x:10 10 x x10 1000 0.06 1e0.060.0410e0.060.0410 0.07      0.060.04 0.070.05 10000.601e10.5833e1   593.86 MLC‐09‐11 2 Question #3 Answer: D   1 EZ bvt p  dt  e0.06te0.08te0.05t dt t t x xt 0 0 20 1 100  5  e0.07t     20 7  0 7 EZ2  bvt2 p  dt e0.12te0.16te0.05t 1 dt  1 e0.09tdt   t t x xt 0 0 20 20 0 1 100  5  e0.09t     20 9  0 9 2 5 5 VarZ  0.04535   9 7 Question #4 Answer: C Let ns = nonsmoker and s = smoker b g b g b g k = q ns pns q s ps xk xk xk xk 0 .05 0.95 0.10 0.90 1 .10 0.90 0.20 0.80 2 .15 0.85 0.30 0.70 A1ns v qns v2 pns qns x:2 x x x1 1 1 0.05 0.950.10 0.1403 1.02 1.022 A1s v qs  v2 ps qs x:2 x x x1 1 1 0.10 0.900.20 0.2710 1.02 1.022 A1  weighted average = (0.75)(0.1403) + (0.25)(0.2710) x:2 = 0.1730 MLC‐09‐11 3 Question #5 Answer: B  1 2 3     0.0001045 x x x x p e0.0001045t t x APV Benefitset1,000,000 p 1dt t x x 0 et500,000 p 2dt t x x 0 e200,000 p 3dt t x x 0 1,000,000  500,000  250,000    e0.0601045tdt  e0.0601045tdt  e0.0601045tdt 2,000,000 0 250,000 0 10,000 0 27.516.6377457.54 Question #6 Answer: B  EPV Benefits1000A1   E 1000vq 40:20 k 40 40k k20  EPV Premiumsa   E 1000vq 40:20 k 40 40k k20 Benefit premiums Equivalence principle    1000A1   E 1000vq a  E 1000vq 40:20 k 40 40k 40:20 k 40 40k k20 20 1000A1 /a 40:20 40:20 161.320.27414369.13  14.81660.2741411.1454 5.11 While this solution above recognized that 1000P1 and was structured to take 40:20 advantage of that, it wasn’t necessary, nor would it save much time. Instead, you could do: MLC‐09‐11 4 EPV Benefits 1000A 161.32 40  EPV Premiums =a  E  E 1000vq 40:20 20 40 k 60 60k k0 a  E 1000A 40:20 20 40 60 14.81660.2741411.14540.27414369.13   11.7612101.19 11.7612101.19161.32 161.32101.19  5.11 11.7612 Question #7 Answer: C ln1.06  A  A  0.530.5147 70 i 70 0.06 1 A 10.5147 a  70  8.5736 70 d 0.06/1.06 a 1vp a 10.978.57368.8457 69 69 70 1.06 a2 2a 21.000218.84570.25739 69 69 8.5902 m1 Note that the approximation am a  works well (is closest to the exact x x 2m answer, only off by less than 0.01). Since m = 2, this estimate becomes 1 8.8457 8.5957 4 Question #8 - Removed Question #9 - Removed MLC‐09‐11 5 Question #10 Answer: E d = 0.05  v = 0.95 At issue 49 A  vk1 q 0.02v1...v500.02v1v50/d 0.35076 40 k 40 k0 and a 1 A /d 10.35076/0.0512.9848 40 40 1000A 350.76 so P  40  27.013 40 a 12.9848 40 E L K 10 1000ARevised P aRevised 549.1827.0139.0164305.62 10 40 50 40 50 where 24 ARevised  vk1 qRevised 0.04v1...v250.04v1v25/d 0.54918 50 k 50 k0 and aRevised 1 ARevised/d 10.54918/0.059.0164 50 50 Question #11 Answer: E Let NS denote non-smokers and S denote smokers. The shortest solution is based on the conditional variance formula VarX EVarX YVarEX Y Let Y = 1 if smoker; Y = 0 if non-smoker 1 AS Ea Y 1aS  x T x  10.444  5.56 0.1 10.286   Similarly E a Y 0  7.14 T 0.1 EEa Y EEa 0ProbY=0EEa 1ProbY=1 T T T 7.140.705.560.30 6.67 MLC‐09‐11 6 EEa Y2 7.1420.705.5620.30  T  44.96 VarEa Y44.966.672 0.47 T EVara Y8.5030.708.8180.30 T 8.60   Var a 8.600.479.07 T Alternatively, here is a solution based on Var(Y) EY2EY2, a formula for the variance of any random variable.   This can be transformed into EY2VarYEY2 which we will use in its conditional   form Ea 2 NSVara NSEa NS2 T T  T   2  2 Vara   E a  Ea   T   T   T  Ea   Ea SProbSEa NSProbNS       T T T 0.30a S 0.70a NS x x 0.301 AS 0.701 ANS x x   0.1 0.1 0.3010.4440.7010.286  0.305.560.707.14 0.1 1.675.006.67 Ea 2  Ea 2 SProbSEa 2 NSProbNS  T   T   T      2 0.30 Var a S  Ea S  T T       2 0.70 Var a NS E a NS T T 0.308.8185.5620.708.5037.142     11.919 + 41.638 = 53.557 Vara  53.5576.672 9.1   T MLC‐09‐11 7 1vT Alternatively, here is a solution based on a  T   1 vT    Var a Var   T     vT  Var  since VarX constantVarX    VarvT  since VarconstantXconstant2VarX 2 2A A 2 x x  which is Bowers formula 5.2.9 2 This could be transformed into 2A 2Vara  A2, which we will use to get x T x 2A NSand 2A S. x x 2A  Ev2T x    Ev2T NSProbNSEv2T SProbS      2Vara NSA NS2ProbNS  T x  2Vara SA S2ProbS  T x   0.018.5030.28620.70   0.018.8180.44420.30   0.166830.700.285320.30 0.20238 A  EvT x    EvT NSProbNSEvT SProbS     0.2860.700.4440.30 0.3334 MLC‐09‐11 8 2A A 2   x x Var a  T 2 0.202380.33342  9.12 0.01 Question #12 - Removed Question #13 Answer: D Let NS denote non-smokers, S denote smokers. ProbT tProbT t NSProbNSProbT t SProbS 1e0.1t0.71e0.2t0.3 10.7e0.1t 0.3e0.2t S (t)0.3e0.2t 0.7e0.1t 0 Want tˆ such that 0.751S tˆ or 0.25 S tˆ 0 0 0.250.3e2tˆ 0.7e0.1tˆ 0.3e0.1tˆ20.7e0.1tˆ Substitute: let xe0.1tˆ 0.3x20.7x0.250 0.7 0.490.30.254 This is quadratic, so x 20.3 x0.3147 e0.1tˆ 0.3147 so tˆ11.56 MLC‐09‐11 9