SOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE SOLUTIONS. Copyright 2016 by the Society of Actuaries

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1 SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE SOLUTIONS Copyright 6 by the Society of Actuaries The solutions in this study note were previously presented in study note MLC-9-8 and MLC-9-. They have been edited for use under the 4 learning objectives and tetbook. Most solutions are mathematically the same as those in MLC-9-. Solutions whose wording has changed are identified with a * before the question number. The only question which has changed mathematically is 3. Questions 3-3 are new. Some of the questions are taken from past SOA eaminations. No questions from published eams after 5 are included ecept 3-33, which come from eams of or 3. Recent MLC eams are available at The average time allotted per multiple choice question will be shorter beginning with the Spring 4 eamination. Some of the questions here would be too long for the new format. However, the calculations, principles, and concepts they use are still covered by the learning objectives. They could appear in shorter multiple choice questions, perhaps with intermediate results given, or in written answer questions. Some of the questions here would still be appropriate as multiple choice questions in the new format. The weight of topics in these sample questions is not representative of the weight of topics on the eam. The syllabus indicates the eam weights by topic. September 6 changes: Questions 39-3 were added. MLC-9-6

2 Question # Answer: E q p p 3:34 3:34 3 3:34 p p p p :34 3:34 p p p p :34 3 3:34 ( )( ) ( )( ) ( )( ) (.7)(.7).54 (.)(.3).6 (.54)(.6) q :34.44 Alternatively, q q + q q 3: : 34 p q + p q p p : 34 3: 36 b g (.9)(.8)(.3) + (.5)(.4)(.7) (.9)(.8)(.5)(.4) [-(.7)(.3)] (.79).44 Alternatively, q q q q q 3: ( 3 p3 )( 3 p34 ) ( p3 )( p34 ) (.54)(.6) (.7)(.).44 (see first solution for p 3, p 34, 3 p 3, 3 p 34 ) MLC-9-6

3 Question # Answer: E A A + : A.4t.6t.4.6.5t.7t e e (.6) dt + e e e e (.7) dt.t.t.6 e dt + e (.7) e dt.6 + (.7).t.t e e e e + e ( ) Because this is a timed eam, many candidates will know common results for constant force and constant interest without integration. ( E ) For eample A : µ + δ E ( µ + δ) e A µ µ µ + δ With those relationships, the solution becomes A A E : A (.6+.4) (.6+.4).7 ( e ) + e (.6)( e ) e MLC-9-6

4 Question #3 Answer: D t.6 t.8 t.5 t E[ Z ] b tv t p µ + t dt e e e dt.7t 5 e 7 7 t ( ).t.6t.5t.9t E Z b tv t pµ + tdt e e e dt e dt.9t 5 e 9 9 [ ] Var Z Question #4 Answer: C Let ns nonsmoker and s smoker k b g ns q + k b g ns p + k b g s q + k ( s) p + k ( ns) ( ns) ( ns) ( ns) A v q + v p q + : ( ) ( s) ( s) ( s) ( s) A v q : + v p q + (.) ( ) : A weighted average (.75)(.43) + (.5)(.7).73 MLC-9-6 3

5 Question #5 Answer: B t ( τ ) ( ) ( ) ( 3) µ µ µ µ ( τ ) p e.45t δ t ( τ ) ( ) APV Benefits e,, p µ dt + + t ( ) ( ) e δ τ 5, p dt ( ) ( 3) e δτ τ, p dt ( ) µ t µ t t,,.645t 5,.645t 5,.645t,, e dt + e dt e dt 5, +, *Question #6 Answer: B 4: 4: EPV Benefits A + E vq k k k k EPV Premiums π a + E vq 4 4+ k 4 4+ k Net premiums Equivalence principle + 4: k 4 4+ k π + 4: k 4 4+ k k A E vq a E vq 4: π A / a 4: ( )( ) ( )( ) While this solution above recognized that π P and was structured to take 4: advantage of that, it wasn t necessary, nor would it save much time. Instead, you could do: MLC-9-6 4

6 EPV Benefits A EPV Premiums π a + E E vq Question #7 Answer: C 4: 4: 4 k 6 6+ k k π a + E A 4 6 ( )( ) ( )( ) π π π π ( ) ( ) ln.6 A7 δ A i.6 A7.547 a d.6/.6.97 a 69 + vp69a 7 + ( ) ( ) a a α ( ) β( ) ( )( ) ( m) ( m ) Note that the approimation a a works well (is closest to the eact m answer, only off by less than.). Since m, this estimate becomes Question #8 - Removed Question #9 - Removed MLC-9-6 5

7 Question # Answer: E d.5 v.95 At issue ( ) ( ) ( ) ( ) 49 k+ 4 k k A v q. v v.v v / d.3576 and a 4 A4 / d.3576 / A so P4 7.3 a Revised Revised ( 4 ) ( )( ) E LK A P a where Revised ( ) ( ) ( ) ( ) 4 Revised k+ Revised 5 k 5 k 5 5 Revised a 5 A5 d A v q.4 v v.4v v / d.5498 and /.5498 / *Question # 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 ( XY) + Var( E XY ) Let Y if smoker; Y if non-smoker S S A E( a Y T ) a δ Similarly E( a Y ) 7.4 T. E E a Y E E a Prob Y + E E a Prob Y ( ( )) ( ) ( ) ( ) ( ( )) ( ) T T T ( 7.4)(.7) ( 5.56)(.3) MLC-9-6 6

8 ( ( )) ( 7.4 )(.7 ) ( 5.56 )(.3 T ) E E a Y ( ( T )) ( ) Var E a Y ( Var ) ( 8.53 )(.7 ) ( 8.88 )(.3 T ) E a Y + ( ) 8.6 Var a T Alternatively, here is a solution based on Var( Y) E Y E Y, a formula for the variance of any random variable. This can be ( ) ( ) transformed into E( Y ) Var ( Y) E( Y) form (( ) NS ) Var ( NS ) + ( NS ) T T T E a a E a ( ) ( ) MLC which we will use in its conditional Var a E T a E a T T E a E a S Prob S E a NS Prob NS T + T T S NS.3a +.7a [ ] [ ] S NS ( A ) ( A ) (.444) +.7(.86) (.3)( 5.56) (.7)( 7.4) ( ) S Prob[ S] + NS Prob[ NS T T T ].3( Var ( a S) + ( E a S T T ) +.7( Var ( a NS) + E( a NS T T ) ) E a E a E a ( ) ( ) ( ) Var a T

9 Alternatively, here is a solution based on Var ( a ) T T v Var δ δ T ( v ) ( A) MLC a T v δ T v Var since Var + constant Var δ ( X ) ( X) Var δ since Var constant constant Var A δ ( X) ( X) A a A T This could be transformed into δ Var ( ) NS S and A A. T A E v A Var T ( ) E v ( ) T E v NS Prob NS + S Prob S ( a ) ( A ) ( ) NS δ Var NS + Prob NS T ( a ) ( A ) ( ) S + δ Var S + Prob S T ( )( ) ( )( ) (.6683)(.7) (.853)(.3).38 T E v T ( ) E v ( ) T E vns Prob NS S + Prob S + (.86)(.7) (.444)(.3).3334 ( a ) T A δ ( A ) T +, which we will use to get

10 Question # - Removed Question #3 Answer: D Let NS denote non-smokers, S denote smokers. ( T < t) ( T < t ) ( ) + ( T < t ) ( ) Prob Prob NS Prob NS P rob S P rob S.t.t ( e ) ( e ) t.t.7e.3e.t.t S ( t).3e +.7e Want ˆt such that.75 S( tˆ) or.5 S( tˆ) ( ).tˆ.tˆ.tˆ.tˆ.5.3e +.7e.3 e +.7e Substitute: let e.t ˆ This is quadratic, so ( )( ) ( ).7 ± e.347.tˆ ˆ.347 so t.56 MLC-9-6 9

11 *Question #4 Answer: A At a constant force of mortality, the net premium equals the force of mortality and so µ.3. µ.3 A. δ + µ δ +.3 δ.6 ( L) ( A) ( ).6 ( ) A. 3 ( δ a ).9 Var. where µ.3 A a µ + δ.9 3 µ + δ.9 Question #5 - Removed *Question #6 Answer: A P A a V a.454 ( V + 5 P4 )( + i) 5q6 V P6 ( (5)(.89)).6 5(.376) a [Note: For this insurance, V V 4 because retrospectively, this is identical to whole life] Though it would have taken much longer, you can do this as a prospective reserve. The prospective solution is included for educational purposes, not to suggest it would be suitable under eam time constraints. P 4.89 as above A + 4 E A P + 5P E a + π E E a where 4 4 6: : A A 6:5 6 5 E6 A MLC-9-6

12 a a 4: 4 E4 a 6.76 a a E a : ( ) ( )( )( ) (.89)(.76) ( 5)(.89)(.744)( 4.347) π (.744)(.68756)( ) π Having struggled to solve for π, you could calculate V prospectively then (as above) calculate V recursively. V 4A + A 6:5 6 5P4 a π 6:5 5 E6 a 65 ( 4)(.6674) ( 5)(.89)( 4.347) (.3)(.68756)( ) (minor rounding difference from V ) Or we can continue to V prospectively 4 V 5A + 6:4 4 E6 A65 5P4 a π 6:4 4 E6 a where E v ( ) 4 6 l 7,533, l 8,75, :4 A A6 4 E6 A a a E a : ( )( ) ( )( )( ) ( 5)(.89)( 3.595).3(.73898)( ) V Finally. A moral victory. Under eam conditions since prospective net premium reserves must equal retrospective net premium reserves, calculate whichever is simpler. MLC-9-6

13 Question #7 Answer: C ( ) ( ) Var Z A A 4 4 A A.8 A ( vq + vp A ) ( ) A (.8/ /.5 A ) 4 4 A.65 4 ( ) ( ) A4 A4.433 A4 v q4 + v p4 A4 A4 A4 (.8/ /.5 ) A ( Z ) Var Question #8 - Removed Question #9 - Removed MLC-9-6

14 *Question # Answer: D ( τ ) ( ) ( ) + µ µ µ + t + t ( τ ) ( ) µ + µ. + t + t ( ) ( τ ) µ.8µ + t + t ' ( ) '. ( ).k t dt q p e e 3.4 ( ) ( ) k 3 ln.4 /..4 k.63 t + t t + t ( ) ( τ) ( ) ( τ) ( τ) q p µ dt.8 p µ dt ( ) ( τ) ( τ).8 q.8 p ( ) p e ( τ ) τ µ + t dt kt dt e 8k 3 e ( 8) (.63) e ( ) q.8(.9538).644 k MLC-9-6 3

15 Question # Answer: A k min(,3) k f(k) f ( k) ( min( k,3) ) ( ) ( ) f k min k,3. (.9)(.) (.7)(.3) E[ min( K,3) ].4 E{ min ( K,3) } 5.58 Var [ min( K,3)] Note that E[ min( K,3)] is the temporary curtate life epectancy, age. e if the life is :3 Question # Answer: B (.) ( 6) (.8) ( 6) (6) e + e S.5354 (.)( 6) (.8)( 6) (6) e + e S q MLC-9-6 4

16 *Question #3 Answer: D Let q 64 for Michel equal the standard q 64 plus c. We need to solve for c. Recursion formula for a standard insurance: ( )(.3) ( ) V V + P q V Recursion formula for Michel s insurance ( )( ) ( ) V V + P +..3 q + c ( V ) The values of 9 V 45 and V 45 are the same in the two equations because we are told Michel s net premium reserves are the same as for a standard insurance. Subtract the second equation from the first to get: ( ) ( ) ( V ).3 (.) + c( V ) (.3). c MLC-9-6 5

17 Question #4 Answer: B K is the curtate future lifetime for one insured. L is the loss random variable for one insurance. L AGG is the aggregate loss random variables for the individual insurances. σ AGG is the standard deviation of L AGG. M is the number of policies. K+ π K+ L v π a v π K + + d d ( A ) E[ L] ( A πa ) A π d π.5 Var[ L] + ( A A) + (.9476 (.495) ).6834 d.5664 [ AGG ] [ ] [ ] [ ] E L ME L.868M Var L M Var L M(.6834) σ.6833 M AGG AGG [ ] ( ) L E L E L Pr[ LAGG > ] > σ AGG σ.868m Pr N(,) > M (.6833).868 M M 6.97 AGG AGG AGG minimum number needed 7 AGG MLC-9-6 6

18 Question #5 Answer: D Annuity benefit: Death benefit: New benefit: Z K + v, for K,,,... d K Z Bv + for K,,,... K + v K + Z Z+ Z, + Bv d,, K + B v d d + K + ( ), Var( Z) B Var v d, Var ( Z) if B 5,..8 In the first formula for Var ( Z ), we used the formula, valid for any constants a and b and random variable X, Var ( a + bx ) b Var ( X ) Question #6 Answer: A µ + t: y+ t µ + t+ µ y+ t A µ / µ + δ.574 ( ) + t + t + / ( + ).4 + : + ( + : + ) ( + t: y+ t ) A µ µ + δ y y t y t A µ / µ + δ.6667 y ty t ty t a / µ + δ y A A + A A y y y Premium.3476/ MLC-9-6 7

19 *Question #7 Answer: B P A / a.63 / P A / a.7636/ a45 a E 3 LK4 3 A45 P4 P4 a ( )( ) Many similar formulas would work equally well. One possibility would be 3V4 + ( P4 P4 ), because prospectively after duration 3, this differs from the normal net premium reserve in that in the net year you collect P 4 instead of P 4. *Question #8 Answer: E ( T ) ( ) E min, e w 4 4 ( ) tf t dt 54 w 4 w ( ) ( ) 3 t f t dt + 4f t dt 4 w w ( ) t f ( t) dt 4 (.6) t f t dt + 4 w 86 ( t ) ( ) S ( ) 4 ( ) tf t dt ( ) 4 f t dt MLC-9-6 8

20 *Question #9 Answer: B d.5 v.95 Step Determine p from Kevin s work: vp vq v p( p+ q+ ) ( ) p ( )( p) ( ) p( ) p ( p ) p p 34/38.9 Step Calculate P :, as Kira did: P.95.9 : + [ ] P :.855 ( )( ) ( )( ) The first line of Kira s solution is that the epected present value of Kevin s net premiums is equal to the epected present value of Kira s, since each must equal the epected present value of benefits. The epected present value of benefits would also have been easy to calculate as ( )( )( ) ( )( )( ) Question #3 Answer: E Because no premiums are paid after year for (), V A + ( V + )( + i) b q One of the recursive reserve formulas is h+ V p+ h ( 3,535 +,78) (.5),. V 35, ( 35, ) (.5),. V 36,657.3 A h π h h+ + h MLC-9-6 9

21 Question #3 Answer: B The survival function is () t S t ω : Then, ω t e and t p ω 5 45 e e t 4 t e 45:65 t p 45:65 dt dt 6 4 F HG t t t 3 3 I K J e45: 65 e45+ e65 e45: In the integral for e 45:65, the upper limit is 4 since 65 (and thus the joint status also) can survive a maimum of 4 years. Question #3 Answer: E µ S (4) / S (4) 4 d 4 e / 4 e / i 4 e / 4 e / 4 e e MLC-9-6

22 Question # 33 Answer: A q b g i q b g b g τ L NM τ µ q µ p b g i bτg ln p ln big bτg O P Q q P b g τ L NM b g b g b g b g τ 3 µ µ + µ + µ b g τ b g µ τ q e e.7769 b g q. 5 µ big ln e µ b g ln e - τ 5. b. 7769gµ b g b5. gb. 7769g µ b τ g O QP Question # 34 Answer: D 6 3 A v p q B B B + pay at end live then die of year 3 years in year v p q pay at end live then die of year 4 3 years in year ( ) ( ) ( ) ( ).3 (.3) MLC-9-6 ( ) ( ) ( ) ( )

23 Question # 35 Answer: B a a + E a + a 5 : 5 :5 5 5 b g. 7 5 e b g. 7 5 E e. 75, where. 7 µ + δ for t < 5 a , where. 8 µ + δ for t 5. 8 b gb g a Question #36 Answer: D ( τ ) ( ) ( ) p p' p' ( ) ( ) ( ) ln p' ( ) ( τ ) q q since UDD in double decrement table ( τ ) ln ( p ) ln (.8).44 ln (.56).693 ( ) ( ).3q.3q+..53 ( ).q τ To elaborate on the last step: ( ) q.3 +. Number dying from cause between +. and +.4 Number alive at +. Since UDD in double decrement, ( τ ) ( ) ( l ).3 q ( τ) ( τ) l.q ( ) MLC-9-6

24 *Question #37 Answer: E The net premium is δ a T o L v ( ) a a T T T T v v δ T v + +. δ δ.4693 Var L Var v +.(4.73).473 δ T ( o ) ( ) Question #38 Answer: D T T Actuarial present value (APV) prem 8( + (.7 +.) + (.5 +.3)),96 APV claim 5( ) + 3( ) 8 Difference 6 Question # 39 - Removed MLC-9-6 3

25 Question # 4 Answer: D Use Mod to designate values unique to this insured. b g b g b g b g b g a A / d /. 6 / P A / a / d i b gb g Mod Mod Mod A6 v q6 + p6 A Mod Mod d 6 i b g a A / d. 444 /. 6 / Mod Mod Mod E L A P a d i b g *Question # 4 Answer: D The prospective reserve at age 6 per of insurance is A 6, since there will be no future premiums. Equating that to the retrospective reserve per of coverage, we have: A P s 4: Mod P5 s k 5: 4 E A 6 5 A a a A 4 4: Mod 5: : + P5 a E E E E P Mod b gb g P Mod P Mod 5 MLC-9-6 4

26 Alternatively, you could equate the retrospective and prospective reserves at age 5. Your equation would be: A a A Mod 4 4: : A5 P5 a 5: a E E where A 4 : 4 4 A4 E4 A5 4 4 b gb g d ib g P Mod bgb4437 P Mod. g Alternatively, you could set the epected present value of benefits at age 4 to the epected present value of net premiums. The change at age 5 did not change the benefits, only the pattern of paying for them. Mod 4 4 4: 5 4 5: A P a + P E a b g d ib gb g F H G I K J + P Mod bgb 7748 P Mod. g Question # 4 Answer: A b g b g b g τ d q l 4 d b g b g bg q l b g d b g db g. τ p b g Note: The UDD assumption was not critical ecept to have all deaths during the year so that - 8 lives are subject to decrement. MLC-9-6 5

27 Question #43 Answer: D Use age subscripts for years completed in program. E.g., p applies to a person newly hired ( age ). Let decrement fail, resign, 3 other. Then q b g, q b g, q b g q bg 5 q, b g 3, q b g 8 q b3g 3 q 3, b g 9, q b g 4 bτg b gb gb g bτg b gb gb g bτg b gb gb g b τ g bτg τ b g τ τ log / log This gives p / 4 / 5 / 54. p / 5 / 3 / p / 3 / 8 / b g b g So,. 54 8, and b g b g b g b g q p p q b g c h b g b. 45/ 86. gb56. g log 3 / log q. 76 bg b g b τ g d l q b gb g Question #44 - Removed MLC-9-6 6

28 Question #45 Answer: E ω For the given life table function: e k q ω ω ω k+ k+ k k b ω k b A v q v a ω A ω A a d e5 5 ω for typical annuitants ey 5 y Assumed age 7 A a a ba b 5, 97 7 Question #46 Answer: B 3: d 3 id 4 ib g E3 E4 i E p p v p v p v + i b gb gb g b gb gb g The above is only one of many possible ways to evaluate p 3 p 4 v, all of which should give.564 a a E a 3: 4: 3: 4 3: 4 3+ : 4+ b g b gb g a 3: a 4: 5 b3. 68g b. 564gb4784. g MLC-9-6 7

29 *Question #47 Answer: A Equivalence Principle, where π is annual net premium, gives A + ( IA) π a π π d A a IA ( ) b g i We obtained a 35 from a 35 A d Question #48 - Removed Question #49 Answer: C µ y µ + µ y 4. µ A Ay µ δ µ y Ay ay y µ δ µ y + δ. and Ay A + Ay Ay b5833. g P. 87 a a 563. y y 563. MLC-9-6 8

30 Question #5 Answer: E ( )( ) ( ) V + P + i q V V 4 b gb g b g b + ig b. gb. g i ( )( ) ( ) V + P + i q V V 4 b gb g 4b g q q Question #5 Answer: E P A / a b 6 6 6g b 6 6g b 6 6 6g b 6 6g v q + p A / + p v a q + p A / 6. + p a c b gb gh c b gb gh / Question #5 - Removed MLC-9-6 9

31 Question #53 Answer: E g ln(.96).48 µ + ty : + t + ty : + t h ln(.97).346 µ ( µ µ µ ) p p ep + + dt e (.68) 5 y 5 y ty : t ty : t ty : t Question #54 Answer: B Transform these scenarios into a four-state Markov chain, as shown below. State from year t 3 to year t from year t to year t Probability that year t will decrease from year t - Decrease (D) Decrease (I).8 Increase Decrease.6 Decrease Increase.75 3 Increase Increase The transition probability matri is Note for eample that the value of. in the first row is a transition from DD to DI, that is, the probability of an increase in year three after two successive years of decrease. The requested probability is that of starting in state and then being in either state or after two years. That requires the first row of the square of the transition probability matri. It can be obtained by multiplying the first row of the matri by each column. The result is [ ]. The two required probabilities are.64 and.5 for a total of.79. The last two values in the vector do not need to be calculated, but doing so provides a check (in that the four values must sum to one). Alternatively, the required probabilities are for the successive values DDID and DDDD. The first case is transitions from state to and then to for a MLC-9-6 3

32 probability of.(.75).5. The second case is transitions from to and then to for a probability of.8(.8).64 for a total of.79. MLC-9-6 3

33 Question #55 Answer: B l ω 5 p l / l (6 t) / 6 t t 45 Let K be the curtate future lifetime of (45). Then the sum of the payments is if K 9 and is K 9 if K. a 45 b 6 K F 6 KI HG 6 K J g b4gb4g b6g Hence, ProbcK 9 > 366. h ProbcK > h b g b g Prob K 33 since K is an integer Prob T 33 l78 33p45 l MLC-9-6 3

34 Question #56 Answer: C µ A. 5 µ. 4 µ + δ µ A µ + δ diai z s z s E A ds d. s e z F e b. 4g HG. 4 A ds ib. 4g ds. s. I. 4 4 KJ. Alternatively, using a more fundamental formula but requiring more difficult integration. c h z z b g b δ t t. 4t. 6t g z. t. 4 t e dt IA t p µ t e dt t e. 4 e dt (integration by parts, not shown) F t HG I. t. 4 K J e MLC

35 Question #57 Answer: E Subscripts A and B here distinguish between the tools and do not represent ages. ο We have to find e AB z t e ο F t A dt t 5 5 HG I K J F I 7 HG K J F ti t dt οhg K JF I HG K J z 7 7F + HG t e ο t z7 B dt t 7 4 e ο AB z t t t t t t t e ο AB e ο A+ e ο B e ο AB I KJ dt Question #58 Answer: A ( τ ) µ t t p b g. 4 τ e t Epected present value (EPV) EPV for cause + EPV for cause. z5 z. 4 t. 4 t 5. 4 t. 4 t b g b g e e. dt + 5, e e. 4 dt ( ) 5.44t.44( 5) ( (.) + 5, (.4) ) e dt e MLC

36 Question #59 Answer: A R p q µ + + k dt µ + tdt k dt ( k ) ( t ) S p e since e e e So S. 75R p e. 75q k µ + tdt k dt k q e. 75 p k p q e. 75q. 75q L q k ln. 75q NM O QP e MLC

37 Question #6 Answer: C A 6 A 6 and.3693 d A6 A6.86 π π + d d Epected Loss on one policy is E L( π ), A6 π Variance on one policy is Var L( π ), + ( A6 A6 ) On the lives, E S,E L π d [ ] ( ) and Var[ S], Var L( π ) The π is such that E[ S] / Var[ S].36 since ( ) π π,, + A6 d d.36 π, + A6 A6 d Φ π π, + (.3693) d d.36 π, + (.86) d π d.479 π, + d π π d d π d π 5976 d 3379 MLC

38 Question #6 Answer: C ( π )( ) ( ) V V + + i + V V q 75.5π q 75 Similarly, V ( V + π ).5 q V V + π.5 q 76 ( ) 3 77 ( π π π) + ( q75 + q76 + q77 ) 3 3V q75 q76 q *.5.5 π ( ) ( ) ( ) * This equation is an algebraic manipulation of the three equations in three V, V, π. One method usually effective in problems where benefit unknowns ( ) stated amount plus reserve, is to multiply the V equation by.5, the V equation by.5, and add those two to the 3 V equation: in the result, you can cancel out the V, and V terms. Or you can substitute the V equation into the V equation, giving V in terms of π, and then substitute that into the 3 V equation. Question #6 Answer: D 8: z δ t 7 A e dt d i b g δ e. 6 since δ ln δ F 7 a + v : H G I 7 K J V 5, A 6643 a : 8: MLC

39 Question #63 Answer: D Let A and a be calculated with µ + t and δ. 6 * * Let A and a be the corresponding values with µ + t increased by.3 and δ decreased by.3 a a * A δ. 6 a t * ( µ + s+.3) ds.3t Proof: a e e dt * * t sds.3t.3t e µ + e e dt t µ + sds.6t e e dt a ] A. 3a. 3a b gb g MLC

40 Question #64 Answer: A bulb ages Year 3 # replaced The diagonals represent bulbs that don t burn out. E.g., of the initial,, (,) (-.) 9 reach year. (9) (-.3) 63 of those reach year. Replacement bulbs are new, so they start at age. At the end of year, that s (,) (.) At the end of, it s (9) (.3) + () (.) 7 + At the end of 3, it s (8) (.) + (9) (.3) + (63) (.5) Epected present value Question #65 Answer: E z z z 5 HG KJ z. 6. 6L. 5 d i d i e 5 : 5 t p 5 dt+ p t p dt t F. 4ds. 5t e dt + e z I e dt e e e NM O QP MLC

41 Question #66 Answer: C p e q 6 + je q 6 + jb q63gb q64gb q65g b. 89gb. 87gb. 85gb. 84gb. 83g Question # 67 Answer: E. 5 a µ + δ µ δ µ δ A A µ +. 5 µ δ µ µ + δ 3 e T j Var a A A δ S.D MLC-9-6 4

42 *Question # 68 Answer: D v. 9 d. A da b gb g 5A 5vq Net premium π a a+ 4 A b gb g b gb g a Net premium reserve at the end of year a a da V 5A π a b gb g b gb g b gb g Question #69 Answer: D v is the lowest premium to ensure a zero % chance of loss in year (The present value of the payment upon death is v, so you must collect at least v to avoid a loss should death occur). Thus v.95. E Z vq + v p q b g + b g d i + b g b g E Z v q + v p q b g d i c b gh b g Var Z E Z E Z MLC-9-6 4

43 Question #7 Answer: D Epected present value (EPV) of future benefits ( )/.6 + (.5.4)( ) / EPV of future premiums + ( ) (.99)( 5) 95.5 E LK /.6 5 Question #7 - Removed Question #7 Answer: A Let Z be the present value random variable for one life. Let S be the present value random variable for the lives. δ t µ t E( Z) e e µ dt E Z 5 µ e δ + µ.46 ( δ + µ )5 d i F µ I e b HG δ + µ KJ F H G I K J d i b g d i c b gh δ + µ e Var Z E Z E Z b g b g E S b g E Z 4. 6 b g Var S Var Z F F MLC g

44 Question #73 Answer: D Prob{only survives} -Prob{both survive}-prob{neither survives} b ge j b gb gb gb gb gb g b gb g p p p p Question # 74 - Removed Question #75 - Removed Question # 76 Answer: C This solution applies the equivalence principle to each life. Applying the equivalence principle to the life group just multiplies both sides of the first equation by, producing the same result for P. ( Prems) ( Benefits) + + EPV P EPV q v p q v Pp p v bgb. 338g bgb. 338gb. 366g Pb. 338gb. 366g + P P. 678 P 3.. (EPV above means Epected Present Value). 8. MLC

45 *Question #77 Answer: E Level net premiums can be split into two pieces: one piece to provide term insurance for n years; one to fund the reserve for those who survive. Then, P P + P V n : n : n And plug in to get. 9 P P : n : n. 85 b gb g Another approach is to think in terms of retrospective reserves. Here is one such solution: ( n : ) a ( P P n : ) V P P s n n : ( P P n : ) ( P P n : ) ( P n : ) n P n : E a n : n : a n : P /. 864 P : n e : n j b gb g MLC

46 Question #78 Answer: A b g δ ln ω δt A tpµ + te dt ω e ω a ω ω δt dt for the given mortality function From here, many formulas for the reserve could be used. One approach is: Since a A 5 A5.374 so a δ a A4 A4.333 so a δ.333 P( A4 ) reserve A5 P( A4 ) a (.33)(.83).75. Question #79 Answer: D ( ) Prob( S) A T T Prob NS T E v E v NS + E v S F Similarly,. 3 A I 7 HG K J. +.. F H Var a b g T I K A A δ F. 6 I HG K J MLC

47 Question #8 Answer: B q q + q q 8: : (.6) +..5 (.) (.6.).36 Using new p 8 value of (.3) +..5 (.) (.3.).68 Change Alternatively, p p8 p8.6. p p84 p84..3 p p8 + p84 8:84 p8 p84 since independent. +.3 (.)(.3).4 3 p 3 p8 + 3 p84 8:84 3 p8 3 p (.)(.3).64 q p p 8:84 8:84 3 8: Revised 3 p p :84 (.6)(.3).68 q :84 change Question #8 - Removed MLC

48 Question #8 Answer: A b g b g b g τ 5 p5 5p5 5 p5 F HG b gb 55 5 I K J e g b gb g Similarly 6 5 pbτg F I 5 e b. gb g HG 5 K J b gb bτg bτg bτg 5 5q5 5p5 p g. 57 Question #83 Answer: C Only decrement operates before t.7 b g 7q4 b 7gq b g 4 b 7gbg 7 since UDD..... Probability of reaching t.7 is Decrement operates only at t.7, eliminating.5 of those who reached.7 b g b gb g q MLC

49 Question #84 Answer: C e j π π π 8 vq8 v p q + p v A8 + + F F H G b g b g b7468. g πb. 756g b6754. g π. 839 I π HG 6 K J π π π Where p 3, 84, ,, 839 b gb g Or p I K J *Question #85 Answer: E At issue, epected present value (EPV) of benefits.4.4 ( )( ) t t t t t 65 µ t 65 µ 65+ t 65+ t t p65 µ 65+ tdt q65 b v p dt e e p dt EPV of premiums t.4t.t t πa65 π v p dt π e e dt π 6.667π.6 Net premium π / u V b + u v up67 µ udu πa67.4( + u).4u e e up67 µ udu (6)(6.667).8 e u p67 µ udu 83.9 q MLC

50 Question #86 Answer: B () a a + E : : A: () a : d (3) A A + A : : : + : + (4) A A E A :. 8 A A : 8. b gb g Now plug into (3): A : Now plug into ():. 43 a :. 5 / 5. b g 97. Now plug into (): a : Question #87 - Removed Question #88 Answer: B e 8.83 e p+ pe + p e+ 9.9 a + vp + v p +... : a + v+ v p +... a a vq : v(.9548).459 v.969 i.789 v MLC

51 Question #89 Answer: E One approach is to enumerate the possible paths ending in F and add the probabilities: FFFF. 3.8 FFGF.(.8)(.5).8 FGFF.8(.5)(.).8 FGHF.8(.5)(.75).3 The total is.468. An alternative is to use matri multiplication. The desired probability is the value in the upper left corner of the transition probability matri raised to the third power. Only the first row needs to be evaluated. The first row of the matri squared is [ ], obtained by multiplying the first row by each column, in turn. The first row of the matri cubed is obtained by multiplying the first row of the squared matri by each column. The result is [ ]. Note that only the first of the four calculations in necessary, though doing the other three and observing that the sum is provides a check. Either way, the required probability is.468 and the actuarial present value is v (.468) 5(.9) (.468) 7. Question #9 Removed MLC-9-6 5

52 Question #9 Answer: E M µ F 3 µ 6 ω 85 ω M t t p65 t p F 6 t 5 Let denote the male and y denote the female. e 5 mean for uniform distribution over, e e y y.5 ( ( )) mean for uniform distribution over (,5) ( ) t t dt 5 7 t t + dt t 7 t t e e e e y + y y MLC-9-6 5

53 *Question #9 Answer: B A A µ µ + δ 3 µ µ + δ 5 Let PA ( ) denote the net premium. c h P A Var L µ. 4 F P A b g c hi + H G K J e δ F I F + HG K J F HG H G I. 4 K J F H G 3I K J F H G 4 I K J 5 *Question #93 Answer: A 45 A A MLC I KJ j Let π be the net premium Let k V denote the net premium reserve at the end of year k. For any n, ( nv + π )( + i) ( q5+ n n+ V + p5+ n n+ V) n+ V V V + π + i Thus ( )( ) ( )( ) ( ( ) )( ) ( )( ) ( )( ) V V + π + i π + i + π + i π s V V + π + i πs + π + i π s 3 3 By induction (proof omitted) nv π s n For n 35, n V a 6 (epected present value of future benefits; there are no future premiums) a π s 6 35 a π s 6 35 For n, V π s a s 6 s 35

54 Alternatively, as above V + + i V ( n π )( ) n+ Write those equations, for n to n 34 : V + π + i V ( )( ) ( π )( ) ( π )( ) : V + + i V : V + + i V 3 ( π )( ) 34 : V + + i V Multiply equation k by ( ) 34 k + i and sum the results: ( V π)( i) ( V π)( i) ( V π)( i) ( V π)( i) ( ) ( ) ( ) ( ) V + i + V + i + V + i + + V + i + V For,,, 34, k the ( ) 35 k V i k + terms in both sides cancel, leaving ( ) 35 π ( ) 35 ( ) 34 ( ) V + i + + i + + i i 35V Since V π s V a 6 (see above for remainder of solution) MLC

55 *Question #94 Answer: B Note: The symbol µ + ty : + can be ambiguous. Here it means that both lives were t alive at ages and y and the last survivor status is intact at time t (that is, at least one of the two lives is alive). This meaning matches the wording of the question, which does not use the symbol. Then, tqy tpµ + t+ tq tpyµ y+ t µ + ty : + t q p + p q + p p For () (y) (5) t t y t t y t t y (.5q5 )( p5 ) q6 ( q )( p ) ( p ) (.95)(.9478)(.376)( ) ( )( )( ) ( ) µ : where p.5 5 q ( l + l ) ( 8,88,74 + 8,75, 43) p l 8,95,9 p.95 8,88, ,95,9 ( 5.5) ( ) p µ + p q since UDD Alternatively, ( t ) p5 p5 p t ( + t) p5:5 ( p5 ) ( t p6 ) ( + t) p p 5:5 5 t p6 ( p5 ) ( t p6 ) p5 ( tq6 ) ( p5 ) ( tq6 ) Derivative p q + ( p ) ( tq ) q since UDD Derivative at + t.5 is ( )( ) ( ) ( ( )( ))( ) p p 5:5 ( p ) (.9848) (.9848) dp (.3) µ (for any sort of lifetime) dt.3 p.996 MLC

56 Question #95 Answer: D ( τ ) ( ) ( ) µ µ µ + t + t + t t ( τ) ( ) t ( τ) ( ) t ( τ) ( τ) 5, 5, t µ + t + t µ + t + t µ + t.t.3t.t.3t.t.3t P P v p dt v p dt v p dt P P e e.9dt + 5, e e.dt + 5, e e.3dt (.4) (.4) (.4) e e e P P,94 *Question #96 Answer: B e p + p + 3p Annuity v p + v p k 3 v b 4. 3 k k 3 g p k 3 k v k p 3 3 F H G I v be g K J. Let π net premium. e π +. 99v +. 98v 87 j. 858π 87 π 84 b g MLC

57 Question #97 Answer: B 3 : 3 b g3 : b ge 3j π a A + P IA + π A A3 π a IA A b g 3 : 3 : 3 b g b g Question #98 Answer: E For the general survival function ( ) t S, t t ω, ω z ω 3 t e 3 F I dt 3 K J ω L M HG O gq P ω 3 t t N bω 3 ω 3 3 Prior to medical breakthrough ω e 3 35 After medical breakthrough e 3 e ω so 3 e 3 39 ω 8 Question #99 Answer: A.5 L, v 4a 77,79 MLC

58 Question # Answer: D ( accid ) µ. ( total) µ. ( other) µ...9 Epected present value.5 t 5,. t(.9) Question # Removed e e dt ( ).4t.5t.t + 5, e e e. dt.9. 5, +,.6. Question # Answer: D + ( V + P )( ) q ( ) V A + p so a ( )(.6).54( ) (.6 /.6) A P a MLC

59 Question #3 Answer: B k ( τ ) k ( τ k ) ( ) µ + tdt µ + tdt p e e k ( ) µ + tdt e ( p ) k k p µ p 6 p 6 where is from Illustrative Life Table, since follows I.L.T. 6, 66, , 88, 74 6, 396, , 88, 74 b g b g b g τ τ τ q6 p6 p6 b p6g b p6g from I.L.T ( ) Question #4 Answer: C P s d a, where s can stand for any of the statuses under consideration. s a a a s y P + d s a y a + a a + a y y y a y P y MLC

60 Question #5 Answer: A bτ g z bµ +. 4gt d e µ +. 4 dt b e µ +. 4 e 48 g µ. 4 e µ +. 4 ln µ. 9 bg z 4. 49t 3 3 b b g b gj d e. 9 dt e b b gb g b gb g e e g g j Question #6 Answer: B This is a graph of l µ. µ would be increasing in the interval b8, g. The graphs of l p, l and l would be decreasing everywhere. Question #7 Answer: B Variance v p q Epected value v p q v p v q.65 q.357 v 5 5 p Since µ is constant 5 ( p ) q 5 p 5 ( ( p) ) q.5 MLC

61 *Question #8 Answer: E ( ) ( ) ( + i V V ) q + I I + p+ p+ ( ) ( ) ( + i V V ) q + II II II π + p+ p+ ( ) ( ) ( ) ( ) I II I II II + i V V V V π ( ) p + (.6) Question #9 Answer: A k + EPV ( s benefits) v bk + k p q k + k 3 b g b gb g b gb gb g 3v. + 35v v , 89 *Question # Answer: E π denotes the net premium V EPV 9 future benefits - EPV future premiums. 6 π π bv + πgb8. g bq65gbg V p65 b gb8. g b. gbg. 5.8 MLC-9-6 6

62 Question # Answer: A Epected present value Benefits (.8) (.8)(.9),39.75 P P(.3955) P (.8)(.)(,) (.8)(.9)(.97)( 9,) +.6.6, Question # Answer: A 8 7a3 + 5a4 a3: 4 8 b7gbg + b5gbg a3: 4 a3: 4 8 a a a4 a3 4 + : : a 4 3: 4 Question #3 Answer: B.5t e t a a f () t dt te dt t.5 Γ() z b L NM b g L F H G I K J 5NM. 5. te t te. 5. 5t F HG O QP i dt t t t + e I K J. 5t e 5 O QP, , 88 MLC-9-6 6

63 Question #4 Answer: C Event Prob Present Value.5 5 ( ) ( )( ) ( )( ) / / / E[ X ] (.5)( 5) + (.95)( 33.87) + (.855)( 56.) 5.95 ( )( ) ( )( ) ( )( ) + + E X [ ] ( ) ( ) ( ) Var X E X E X Question #5 Answer: B Let K be the curtate future lifetime of ( + k) K + k L v P :3 a K + When (as given in the problem), () dies in the second year from issue, the curtate + is, so future lifetime of ( ) L v P a : The premium came from A :3 P :3 a A P :3 da :3 :3 da :3 79. :3 a a :3 :3 d MLC-9-6 6

64 Question #6 Answer: D Let M the force of mortality of an individual drawn at random; and T future lifetime of the individual. n Pr T E Pr T M z Pr T M µ f Mbµ g dµ z z µ µ e t dt d µ z µ + + e du e e s d i d i d i Question #7 Answer: E For this model: () / 6 () µ 4+ t ; µ 4+ / 4.5 t/ 6 6 t () / 4 () µ 4+ t ; µ 4+ /.5 t/ 4 4 t ( τ ) µ MLC

65 *Question #8 Answer: D Let π net premium Epected present value of benefits b. 3gb, gv + b. 97gb. 6gb5, gv + b. 97gb. 94gb. 9gb, gv , 33. Epected present value of net premiums a π : 3 b gb g +. 97v v. 766π, 33. π ,. 3 V π b gb g b gb g Initial reserve, year V + π Question #9 Answer: A Let π denote the premium. L b v π a + i v π a T T π a T T b g E L π a π a a T δ a d v L π a T a δ a T T v b δ ag v A δ a A T T T T i MLC

66 Question # Answer: D (, ) (,.9) (.5,.8775) (,.885) tp t p (. ). 9 p b gb g since uniform,. 5 p / b e:.5 area between t and t () + (.5) g MLC

67 Alternatively,.5.5 :.5 t t + t e p dt p dt p p dt.5 (. t) dt +.9 (.5 t) dt.5.t.5t t +.9 t Question # Answer: A, A63b. g 533 A Ab + ig q A + p b. 467gb5. g. 788 A b. 483gb5. g. 95 A Single gross premium at 65 (.) (,) (.4955) 555 b g i i MLC

68 Question #A Answer: C Because your original survival function for () was correct, you must have 3 µ.6 µ + µ µ +. µ + t + t: y+ t + t: y+ t + t: y+ t + ty : + t.4 Similarly, for (y) 3 µ.6 µ + µ µ +. µ y+ t + t: y+ t + t: y+ t + t: y+ t + ty : + t.4 The first-to-die insurance pays as soon as State is left, regardless of which state is net. The force of transition from State is 3 µ + ty : + t+ µ + ty : + t+ µ + ty : + t With a constant force of transition, the epected present value is δt 3.5t.t. e ( tpy µ + ty : + t+ µ + ty : + t+ µ + ty : + t) dt e e (.) dt.5 MLC

69 Question #B Answer: E Because () is to have a constant force of.6 regardless of (y) s status (and viceversa) it must be that µ µ t y+ t There are three mutually eclusive ways in which both will be dead by the end of year 3: : Transition from State directly to State 3 within 3 years. The probability of this is t..t.3 t py µ + t: y+ tdt e.dt e.( e ).58. : Transition from State to State and then to State 3 all within 3 years. The probability of this is t.6(3 t) tpyµ + t: y+ t 3 tp+ tdt e.4( e ) dt t.8.4t.4.t.4e.4t.4 e e e e e ( e ) e ( e ).9 3: Transition from State to State and then to State 3 all within 3 years. By symmetry, this probability is.9. The answer is then.58 + (.9).7. MLC

70 Question #C Answer: D Because the original survival function continues to hold for the individual lives, with a constant force of mortality of.6 and a constant force of interest of.5, the epected present values of the individual insurances are.6 A Ay.54545, Then, A A + A A y y y Alternatively, the answer can be obtained be using the three mutually eclusive outcomes used in the solution to Question B. :. e p + + dt e e.dt t 3.5t.t t y µ t: y t.5t.5r 3 t yµ + t: y+ t r + t: y+ tµ ++ t r: y++ t r.5t.t.5r.6r e e e e drdt e p e p drdt and 3: The solution is (.4545).443. The fact that the double integral factors into two components is due to the memoryless property of the eponential transition distributions. MLC

71 Question #3 Answer: B q q + q q 5 35: : 45 p q + p q p q b g b g b gb g b gb g b gb g b gb g : 45 4: 5 p q + p q p p p p q + p q p p p p : Alternatively, ( )( ) ( )( ) p p p p p p q p p 5 35: : :45 ( 5 p p45 5 p35:45 ) ( 6 p p45 6 p35:45 ) ( 5 p p p35 5 p45 ) ( 6 p p45 6 p35 6 p45 ) ( ) ( ) Question #4 Removed Question #5 - Removed MLC-9-6 7

72 Question #6 Answer: E Let E Y Var Y E S Var Y present value random variable for payments on one life S Y present value random variable for all payments a d A A 4 4 d / E Y 4, S i d ib g Var Y 7, 555 Standard deviation S 7, By normal approimation, need E [S] Standard deviations 4, (.645) (65.6) 5,54 *Question #7 Answer: B 5A3 4 A 3: Initial net premium 5a 4a 3: 35 3: 5b g 5b4. 835g 4b959. g Where A A A : e... 3: 3: j and A3 :. 337 a : d F. 6 H G I 6. K J e Comment: the numerator could equally well have been calculated as A + 4 E A.48 + (4) (.9374) (.495) j MLC-9-6 7

73 Question #8 Answer: B p p. 75. q y. 95 p b gb g b gb g. 75 y. 75 y b. 75 pgd. 75 pyi -b.965gb. 95g 97. since independent *Question #9 Answer: D Let G be the gross premium. Epected present value (EPV) of benefits,a 35 EPV of premiums Ga 35 EPV of epenses.g+ 5 + (.5)( ) a 35 Equivalence principle: Ga 35,A35 + (.G ) a 35 A35 G, +.G+ 75 a 35.9G,P ( )( 8.36) + 75 G.9 34 MLC-9-6 7

74 Question #3 Answer: A The person receives K per year guaranteed for years Ka K The person receives K per years alive starting years from now Hence we have E 4 a 5 K b g a 4 K Derive Derive E 4: a 5 A A + E A E : A 4 A 4 A 5 b g 4 5 A d : Plug in values:, hk c K K b gb g MLC

75 Question #3 Answer: D STANDARD: MODIFIED: t t e5: dt t ds. p e z e t e5: t p5 dt p + 5 dt 74 Difference.847 z z F HG. t. t e dt + e dt 74. F HG e. t + e t. 74 I K J I KJ g b *Question #3 Answer: B Comparing B & D: Prospectively at time, they have the same future benefits. At issue, B has the lower net premium. Thus, B has the higher reserve. Comparing A to B: Consider them retrospectively. At issue, B has the higher net premium. Until time, they have had the same benefits, so B has the higher reserve. Comparing B to C: Visualize a graph C* that matches graph B on one side of t and matches graph C on the other side. By using the logic of the two preceding paragraphs, C s reserve is lower than C* s which is lower than B s. Comparing B to E: Reserves on E are constant at. MLC

76 Question #33 Answer: C Since only decrements () and () occur during the year, probability of reaching the end of the year is p p b g b g 6 6 b gb g Probability of remaining through the year is p p p b g b g b g b gb gb g Probability of eiting at the end of the year is qb3g Question #34 - Removed Question #35 Answer: D z EPV of regular death benefit -δ t -µ t bgde ib.8gde idt z -.6t -.8t bgde ib. 8gde idt. 8 / b g, z3 EPV of accidental death benefit b gd -δ t e ib gd -µ t. e i dt z3 b gd -.6t ib gd -.8t e. e i dt e -.4 /. 68, Total EPV MLC

77 Question #36 Answer: B l l b gb g b gb g. 6 79, , 65 8, , , 839 b gb g b gb g 79, , q + 8, Question #37 - Removed Question #38 Answer: A b g b g b g τ bg bg q q + q. 34 p p p b g b g b g 4 p q. y b g. 4 bτg 4 q y 4 q τ 4 l b g b gb g b gb g MLC

78 Question #39 Answer: C Pr b g > < L π'.5 K + Pr, v π' a > <.5 K + From Illustrative Life Table, 47 p and 48 p Since L is a decreasing function of K, to have Pr L bπ g > <.5 means we must have L bπ g for K 47. Highest value of L bπ g for K 47 is at K L π K, v π a b g at π L bπ g b π g π > Question #4 Answer: B b b b g g g Pr K p. Pr K p p Pr K > p. 8 EbYg E Y d i VARbYg MLC

79 Question #4 Answer: E [ ] ba E Z since constant force A µ /( µ + δ) ( ) ( ) bµ b. E(Z) b /3 µ + δ.6 T Τ [ Z] bv b v b ( A A ) Var Var Var Var Z µ µ b µ + δ µ + δ 4 b b 9 45 ( ) E( Z) 4 b b b b *Question #4 Answer: B In general, for any premium P and corresponding loss at issue random variable L, P Var L + A A Here So and So b g b δg c h P π δ.8. a 5 b g F I c HG K J h b * g F bgi + c HG h. 8 KJ 4 5 b g + 8 b g b 565 g bδ g b g VarbL g 75. Var L + A A. 8 Var L A A Var L *.. E L * A. 5a a E L * + *. MLC

80 Question #43 - Removed Question #44 Answer: B b g l τ Let number of students entering year superscript (f) denote academic failure superscript (w) denote withdrawal subscript is age at start of year; equals year - bτg p b g b g b g b g τ τ f f l l q q. q q q l q. 4 l q q q q b g b g b g w τ f b g b g b g b g b g τ f τ f w b g f f b g f e e b g. 4 q b g b g b g τ f w b p q q b g b g b g b g b g b g b g w w τ w τ τ w 3q q + p q + p p q b gb g b gb gb g j g j MLC

81 Question #45 Answer: D e p ( + e ) e e N M 6 6 due to having the same µ N M M p5 ep µ 5+ t +.( t) dt p5e.5 e p ( + e ) e p ( + e ).95e 9.5 N N.5 M M Question #46 Answer: D EY [ AGG ] EY [ ] (, ) a A (, ),, δ σ σ b g c h Var Y, A A δ b, g b. 5g b6. g 5, δ Y AGG F E YAGG. 9 Pr > σ 8. σ 5, 5, L N M Y F E Y σ AGG F 8. σ + E Y b b AGG AGG AGG g g AGG F 8. 5, +,,, 64, O Q P MLC-9-6 8

82 Question #47 Answer: A : A vq v q v q (.99847) + 5 (.99847)(.99839) ( ) (.53 a + 3 ) (.979) 3: q (.979 )( ) Annualized premium Each semiannual premium.8 Question: #48 Answer: E b g eb g DA vq + vp DA 8: : b g b g.. 8 q DA 89 : b DA g b6. g :. 8 q 8. DA v. + v : b b g b gb g j g Question #49 - Removed MLC-9-6 8

83 Question #5 Answer: A t t t t p ep ds ep ln( s) s e5:6 e5 + e6 e5: t t 5 5 e dt t t t 4 e dt t t 4 t 4 9t+ t t e5:6 dt dt t 45t e 5: Question #5 Answer: C Ways to go in years ; p (.7)(.).7 ; p (.)(.5).5 ; p.. ( )( ) Total. Binomial m q. Var () (.) (.78) 7 MLC-9-6 8

84 Question #5 Answer: A For death occurring in year.3 EPV For death occurring in year 3, two cases: () State State State 4: (..). () State State State 4: (.5.3).5 Total.7.7 EPV Total. EPV Question #53 - Removed *Question #54 Answer: C Let π denote the single net premium. π a + π A : 3 3 a A A a 35 35: 3 35: 3 π A A 35: 3 e b 35: 3 j g b.7g MLC

85 Question #55 Answer: E. 4 p e z e F+ e d. 4F L e NM. 4F F H e j. 4 O QP I K e. 8 e F e b g F ln F. 68 F. *Question #56 Answer: C ( + )(.3) + + ( + ) + 9 ( ) + ( 343)(.3).94( 87) + V V P q b q V q b V V V b ( b V) + V P b d a V net premium reserve at the start of year ten P Question #57 Answer: B d.6 V.94 p Step Determine vp vq + v p( p+ + q+ ) p.94 p p p 94( p) p p 7/ MLC ( ) ( )( ) ( ) ( )

86 Step Determine P : P : [ ] P 479 :.8554 ( )( ) ( )( ) Question #58 Answer: D ( IA) v p ( IA) v p q + A ( ) [ ],, 4 4: 4: 9 4 5, see comment 4: 8,95, , 87, 64,.6736 (.59) (.766,) 5,53 Where 4: A A E A ( )( ) Comment: the first line comes from comparing the benefits of the two insurances. At each of age 4, 4, 4,,49 ( IA ) provides a death benefit greater than 4: ( ) IA. Hence the A term. But 4: 4: ( IA ) provides a death benefit at 5 of, 4: while ( IA ) provides. Hence a term involving 4: 9 q 4 9 p 4 q 5. The various v s and p s just get all epected present values at age 4. MLC

87 Question #59 Answer: A π ( + ) ( ) ( ) q ( ) V i q V q AS ( )( i) G epenses + q p ( (.4)( ) )(.) ( )(.5).5 6(.) Question #6 Answer: C At any age, ( ). p e.98 ( ) q.98.98, which is also end of the year. ( ) q, since decrement occurs only at the Epected present value (EPV) at the start of each year for that year s death benefits,*.98 v 88. ( ) p τ.98* ( τ ) E p v.94 v.94* EPV of death benefit for 3 years E4 *88. + E4 * E4 * MLC

88 Question #6 Answer: B e 3:4 4 t 4 p dt 3 ω 3 t dt ω 3 4 t t 3 ( ω ) 8 4 ω ω 95 Or, with a linear survival function, it may be simpler to draw a picture: p 4 p e 3:4 t p 3 area p ω ω 3 ω t 65 ( ) ( ( )) Var ET ET ( + p ) 4 3 MLC

89 Using a common formula for the second moment: ( ) Var T t p dt e t 65 t 65 t t dt dt ( ) ( ) * / Another way, easy to calculate for a linear survival function is ( ) ( ) µ µ Var T t p dt t p dt t + t t + t t dt t dt t t ( 3.5) 35.8 With a linear survival function and a maimum future lifetime of 65 years, you probably didn t need to integrate to get E( T ) 3 e Likewise, if you realize (after getting ω 95 ) that T 3 is uniformly distributed on (, 65), its variance is just the variance of a continuous uniform random variable: ( 65 ) Var 35.8 Question #6 Answer: E ( ) ( ) 8.5.6,. V ( )(.6) ( 9,)(.) V MLC

90 Question #63 Answer: D e e p y t k e p + p +... y y y ( )( ) ( ) ( ) ( ) e e + e e 8.56 y y y Question #64 - Removed Question #65 - Removed Question #66 Answer: E.8t.8t a e dt.5 3 A e (.3) dt t 3 A e (.3) dt σ ( a ) Var a A ( A ) (.375) 6.48 T T δ Pr a a σ ( a ) Pr a > > T T T T v.5t Pr > Pr.6754 > e.5 ln.6754 Pr T > Pr[ T > ] e.79 MLC

91 Question #67 Answer: A ( τ ) (.5)( 5).5 5 p5 e e.7788 ( ) 5 ( ) 5 (.3+.) t.5t 5q55 µ 55 (. /.5) + t e dt e.4( e.5 ).885 Probability of retiring before.7788* ( τ ) ( ) 6 p q MLC-9-6 9

92 Question #68 Answer: C Complete the table: l8 l[ 8] d[ 8] 9 l8 l[ 8] d[ 8] 83 (not really needed) e e + since UDD e [ ] e[ ] + l 8 + l8 + l83 + e[ ] + l[ 8] e[ 8] l[ 8] l8 + l8 + Call this equation (A) e[ 8] l[ 8] l8 + Formula like (A), one age later. Call this (B) Subtract equation (B) from equation (A) to get l 8 e [ 8] l [ 8] e[ 8] l[ 8 ] 9 [ 8.5.5] e[ 8] e[ 8] 8. 9 MLC-9-6 9

93 Alternatively, and more straightforward, 9 p[ 8].9 83 p[ 8] p8.9 9 e [ ] q[ 8] p + [ 8] + e 8 8 where q [ 8] contributes 8.5 (.9) + (.9) + e8 since UDD e8 8.9 e8 q8 + p8 + e8 e8 8.9 (.9) e8 e[ 8] q[ 8] p + [ 8] 8 + e ( ) ( )( ) Or, do all the recursions in terms of e, not e, starting with e [ 8] , then final step e [ ] e [ ] + MLC-9-6 9