SOCIETY OF ACTUARIES EXAM MLC ACTUARIAL MODELS EXAM MLC SAMPLE SOLUTIONS

Transcription

1 SOCIETY OF ACTUARIES EXAM MLC ACTUARIAL MODELS EXAM MLC SAMPLE SOLUTIONS Copyright 8 by the Society of Actuaries Some of the questions in this study note are taken from past SOA eaminations. MLC-9-8 PRINTED IN U.S.A.

2 Question # Key: E q p p 3:34 3:34 3 3:34 p p p :34 p 3:34 p p p p :34 3 3:34 ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) q :34.44 Alternatively, q q + q q p q + p q p p 3: : : 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

3 Question # Key: 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) e e e...t.t.6.7. 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 Question #3 Key: D t.6 t.8 t.5 t E[ Z] bv ( ) t t p μ + t dt e e e dt.7t 5 e 7 7 MLC

4 ( t ) t μ ( ) t.t.6t.5t.9t E Z b v p + t dt e e e dt e dt.9t 5 e 9 9 [ ] Var Z Question #4 Key: C Let ns nonsmoker and s smoker k bg ns q + k bg ns p + k bg 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 Question #5 Key: B t ( τ ) ( ) ( ) ( 3) ( τ ).45t p e μ μ μ μ MLC

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

6 Question #7 Key: C ( ) ( ) ln.6 A7 δ A i.6 A7.547 a&& d.6 /.6.97 a&& 69 + vp69a&& 7 + ( ) ( ) a&& 69 α( ) a&& 69 β( ) (.)( ) ( m) ( m ) Note that the approimation a&& a&& works well (is closest to the eact answer, m only off by less than.). Since m, this estimate becomes Question #8 Key: C The following steps would do in this multiple-choice contet:. From the answer choices, this is a recursion for an insurance or pure endowment.. Only C and E would satisfy u(7).. + i u k u k 3. It is not E. The recursion for a pure endowment is simpler: ( ) ( ) 4. Thus, it must be C. More rigorously, transform the recursion to its backward equivalent, u k in terms of u k : ( ) ( ) q + i u k pk pk p u k q i u k u k vq vp u k k ( ) + u( k ) k ( ) k + ( + ) ( ) ( ) + ( ) k k pk This is the form of (a), (b) and (c) on page 9 of Bowers with k. Thus, the recursion could be: MLC

7 or or A vq + vp A + y : + : y A vq vp A + A vq + vp A + y : : y Condition (iii) forces it to be answer choice C ( ) u k A fails at 69 since it is not true that ( ) A vq vp ( )() u k A : y fails at 69 since it is not true that 69: A vq + vp ( ) : ( )() u k A is OK at 69 since 69: y A vq + vp ( )() Note: While writing recursion in backward form gave us something eactly like page 9 of Bowers, in its original forward form it is comparable to problem 8.7 on page 5. Reasoning from that formula, with π h and bh+, should also lead to the correct answer. MLC

8 Question #9 Key: A You arrive first if both (A) the first train to arrive is a local and (B) no epress arrives in the minutes after the local arrives. PA ( ).75 Epresses arrive at Poisson rate of (.5)( ) 5 per hour, hence per minutes. e f ( ).368! A and B are independent, so P A and B ( ) ( )( ) Question # Key: E d.5 v.95 At issue ( ) ( ) ( ) ( ) 49 k+ 4 k k A v q. v v.v v / d.3576 and a&& A / d.3576 / A so P4 7.3 a&& Revised Revised ( ( ) ) 5 4&& 5 ( )( ) E L K 4 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 / MLC

9 Question # Key: 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 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) E E a Y ( ( )) ( 7.4 )(.7 ) ( 5.56 )(.3 T ) ( ( T )) ( ) Var E a Y ( Var ) ( 8.53 )(.7 ) ( 8.88 )(.3 T ) E a Y 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) E( ( a ) NS ) Var ( a NS ) + E( a NS ) T T T ( ) ( ) + which we will use in its conditional form Var a E T a E a T T E a E a S Prob S E a NS Prob NS T + T T [ ] [ ] MLC

10 S.3a +.7a NS S NS ( A ) ( A ) (.444) +.7(.86) (.3)( 5.56) (.7)( 7.4) ( ) S Prob[ S] + NS Prob[ NS] E a E a E a T T T ( ( a ) ( E a T T ) ( ( a T ) E( a T ) ).3 Var S + S +.7 Var NS + NS ( ) ( ) ( ) Var a T v Alternatively, here is a solution based on a T δ T v Var ( a ) Var T δ δ T v Var since Var ( X + constant) Var ( X) δ T Var ( v ) since Var constant constant Var δ A ( A) which is Bowers formula 5..9 δ ( X ) ( X ) A a A T This could be transformed into δ Var ( ) NS S and A A. T +, which we will use to get MLC

11 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 ) Question # Key: A To be a density function, the integral of f must be (i.e., everyone dies eventually). The solution is written for the general case, with upper limit. Given the distribution of f t, we could have used upper limit here. () Preliminary calculations from the Illustrative Life Table: l l l l MLC

12 5 () () () f t dt k f t dt +. f t dt T 5 5 (). () ( ) k f t dt+ f t dt 5 ( 5). ( ) ( 5) ( 5 p ).(.5) (.895).6 kf + F F k + k +.6 k For 5, F ( ) 3.83 f ( t) dt 3.83F ( ) F F T T T l l 4 ( ) ( ) l l FT ( 5).4 p4.83 F ( ) ( ) T ( ) Question #3 Key: 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.. ().3 t t +.7 S t e e Want ˆt such that.75 S( tˆ) or.5 S( tˆ) ( ) tˆ.tˆ.tˆ.tˆ.5.3e +.7e.3 e +.7e ˆ Substitute: let e.t MLC

13 This is quadratic, so ( )( ) ( ).7 ± e.347.tˆ ˆ.347 so t.56 Question #4 Key: A ( ) μ.3 P A μ.3 A. δ + μ δ +.3 δ.6 ( L) ( A) ( δ a ) ( 3 ).6 ( ) A. Var..9 μ.3 where A a μ + δ.9 3 μ + δ.9 Question #5 Key: C Let N number of sales on that day S aggregate prospective loss at issue on those sales K curtate future lifetime N~Poisson (. 5) E[ N] [ N] Var K + L,v 5a&& E L,A 5a&& [ ] K K L, + v Var[ L], + A65 ( A65) d d d S L+ L L N [ ] [ ] [ ] [ ] [ ] [ ] + ( [ ]) Var[ ] E S E N E L Var S Var L E N E L N MLC

14 Pr( S < ) Pr Z < [ ] [ ] E S Var S Substituting d.6/(+.6), A , A and a&& yields E [ L] [ L] [ ] [ S] Var 5,, E S Var 54,5, Std Dev (S),46 S Pr( S < ) Pr <,46,46 Pr Z < ( ) With the answer choices, it was sufficient to recognize that:.6554 Φ.4 < Φ.443 < Φ ( ) ( ) ( ) By interpolation, Φ(.443) (.43) Φ(.5) + (.57) Φ(.4) (.43)(.695) + (.57)(.6554).679 MLC

15 Question #6 Key: A A P4.89 a&& a&&.454 V + 5 P ( + i) 5q 6 V a&& V ( ) 4 6 P 6 ( (5)(.89)).6 5(.376) [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&& E E a&& where π 6:5 6: A A 6:5 6 5 E6 A 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 ) 4 MLC

16 Or we can continue to V prospectively 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, : A A E A.5779 a&& a&& 6:4 6 4 E6 a&& ( )( ) ( )( )( ) ( 5)(.89)( 3.595).3(.73898)( ) V Finally. A moral victory. Under eam conditions since prospective benefit reserves must equal retrospective benefit reserves, calculate whichever is simpler. Question #7 Key: C ( ) ( ) Var Z A4 A4 A4 A4.8 A4 ( vq4 + vp4a4 ) A4 (.8/.5 + (.997 /.5) A4) A.65 4 ( ) ( ) A4 A4.433 A4 v q4 + v p4 A4 A4 + A4 A4.793 ( Z ) Var (.8/.5.997/.5 ) MLC

17 Question #8 Key: D This solution looks imposing because there is no standard notation. Try to focus on the big picture ideas rather than starting with the details of the formulas. Big picture ideas:. We can epress the present values of the perpetuity recursively.. Because the interest rates follow a Markov process, the present value (at time t ) of the future payments at time t depends only on the state you are in at time t, not how you got there. 3. Because the interest rates follow a Markov process, the present value of the future payments at times t and t are equal if you are in the same state at times t and t. Method : Attack without considering the special characteristics of this transition matri. Let k s state you are in at time k ( thus s, or ) k Let Y k present value, at time k, of the future payments. Y k is a random variable because its value depends on the pattern of discount factors, which are random. The epected value of Y k is not constant; it depends on what state we are in at time k. Recursively we can write ( ) Yk v Y k + +, where it would be better to have notation that indicates the v s are not constant, but are realizations of a random variable, where the random variable itself has different distributions depending on what state we re in. However, that would make the notation so comple as to mask the simplicity of the relationship. Every time we are in state we have E Yk sk.95 + E Yk+ sk ( ) ( {( E Yk+ sk+ ) ) ( k+ sk ]) + ( E Yk+ sk+ ) Prob( sk+ sk ] ) ( E Yk+ sk+ ) Prob sk+ sk ].95 + Pr ob s ( E Yk+ sk+ ).95 + ( )} MLC

18 That last step follows because from the transition matri if we are in state, we always move to state one period later. Similarly, every time we are in state we have E Yk sk.93 ( + E Yk+ sk ).93 ( + E Yk+ sk+ ) That last step follows because from the transition matri if we are in state, we always move to state one period later. Finally, every time we are in state we have E Yk sk.94 ( + E Yk+ sk ).94 + E Yk+ sk+ Pr sk+ sk + E Yk+ sk+ Pr sk+ sk } ( { ) ( { E Yk+ sk+ E Yk+ sk+ }) Those last two steps follow from the fact that from state we always go to either state (with probability.9) or state (with probability.). Now let s write those last three paragraphs using this shorter notation: n E Yk sk n. We can do this because (big picture idea #3), the conditional epected value is only a function of the state we are in, not when we are in it or how we got there. ( ) ( ) ( ) That s three equations in three unknowns. Solve (by substituting the first and third into the second) to get 6.8. That s the answer to the question, the epected present value of the future payments given in state. The solution above is almost eactly what we would have to do with any 3 3 transition matri. As we worked through, we put only the non-zero entries into our formulas. But if for eample the top row of the transition matri had been (.4.5. ), then the first of our three equations would have become.95( ), similar in structure to our actual equation for. We would still have ended up with three linear equations in three unknowns, just more tedious ones to solve. Method : Recognize the patterns of changes for this particular transition matri. MLC

19 This particular transition matri has a recurring pattern that leads to a much quicker solution. We are starting in state and are guaranteed to be back in state two steps later, with the same prospective value then as we have now. Thus, EY EY first move is to Pr first move is to + EY first move is to Pr first move is to [ ] [ ] [ ] ( ( EY [ ]) ( ( EY [ ]) (Note that the equation above is eactly what you get when you substitute and into the formula for in Method.) [ ] ( ) [ ] EY [ ] EY EY 6.8 Question #9 Key: E The number of problems solved in minutes is Poisson with mean. If she solves eactly one, there is /3 probability that it is #3. If she solves eactly two, there is a /3 probability that she solved #3. If she solves #3 or more, she got #3. f().353 f().77 f().77 P (.77) + (.77) + ( ) MLC

20 Question # Key: D ( τ ) ( ) () t + μ μ μ ( ) () t ( τ ) (.μ () ) t + μ () t ( μ ) () t.8 μ ( τ ) () t k. ' () ' ().kt dt q p e e 3.4 ( ) ( ) k 3 ln.4 /..4 k.63 ( ) ( ) ( ) ( ) ( q p τ μ dt.8 p τ μ τ) () t dt t t ( ) ( τ) ( τ).8 q.8 p ( τ ) μ () p e e 8k 3 t dt kt dt e ( 8) (.63) e ( ) q.8(.9538).644 Question # Key: A k f(k) ( ) ( ) ( ) ( ) k min(,3) f k min( k,3) f k min k,3. (.9)(.) (.7)(.3) MLC

21 E[ min( K,3) ].4 E{ min ( K,3) } 5.58 Var[ min( K,3)] e if the life is age. :3 Problem 3.7 in Bowers, pages 86 and 87, gives an alternative formula for the variance, Note that E[ min( K,3)] is the temporary curtate life epectancy, basing the calculation on k p rather than Question # Key: B k q. (.)( 6) (.8)( 6) e + e s( 6).5354 (.)( 6) (.8)( 6) e + e s( 6) q Question #3 Key: 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 benefit reserves are the same as for a standard insurance. Subtract the second equation from the first to get: MLC

22 ( ) ( ) ( V ).3 (.) + c( V45 ) (.3). c Question #4 Key: B K is the curtate future lifetime for one insured. L is the loss random variable for one insurance. L is the aggregate loss random variables for the individual insurances. AGG σ AGG is the standard deviation of AGG M is the number of policies. L. 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) (.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

23 Question #5 Key: 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 Key: A () t () t + () t μ ()/( μ () + δ).574 μ ()/( μ () + δ).4 μ ()/( μ () + δ).6667 / μ () t + δ μ μ μ y y A t t A t t y y y A t t a y y y y ( y ) A A + A A y y y Premium.3476/ MLC

24 Question #7 Key: B P A / a&&.63/ P A / a&&.7636/ a45 a&& ( ) E 3L K 4 3 A45 P4 P4a ( )( ) Many similar formulas would work equally well. One possibility would be 3V4 + ( P4 P4), because prospectively after duration 3, this differs from the normal benefit reserve in that in the net year you collect P 4 instead of P 4. MLC

25 Question #8 Key: E ( T ) ( ) E min, e w 4 4 () tf t dt 54 w 4 w () () 3 tf t dt+ 4 f t dt 4 w w () tf() t dt 4 (.6) tf t dt ( t ) ( ) s( ) w 4 () tf t dt ( ) 4 f t dt Question #9 Key: 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 actuarial present value of Kevin s benefit premiums is equal to the actuarial present value of Kira s, since each must equal the actuarial present value of benefits. The actuarial present value of benefits would also have been easy to calculate as ( )( )( ) ( )( )( ) MLC

26 Question #3 Key: E Because no premiums are paid after year for (), V A + ( V + )( + i) b q Rearranging 8.3. from Bowers, we get h+ V p ( ) ( ) h π h h+ + h 3,535 +,78.5,. V 35, ( 35, ) (.5),. V 36,657.3 A h Question #3 Key: B s For De Moivre s law where ( ) ω t e and t p ω 5 45 e e65 ω : t 4 t e 45:65 t p 45:65 dt dt 6 4 F HG t t t 3 3 I K J o o o o e45: 65 e45+ e65 e45: In the integral for e o 45:65, the upper limit is 4 since 65 (and thus the joint status also) can survive a maimum of 4 years. MLC

27 Question #3 Answer: E b g b g b g μ 4 s' 4 / s 4 d 4 e / 4 e / 4 e / 4 e / 4 e e i Question # 33 Answer: A q bg i q bg bg τ L NM τ μ q μ ln p ln p bg i bg τ bg i bg τ O P Q q P bg τ L NM bg bg bg bg τ 3 μ μ + μ + μ bg τ bg μ τ q e e.7769 bg q 5. bg i μ ln e μbg ln e - τ 5. b. 7769gμbg b5. gb. 7769g μbg τ O QP MLC

28 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 b g b gb gb g b g b gb gb gb g 9. Question # 35 Answer: B a a + E a + a 5 : 5 :5 5 5 bg 7. 5 e bg 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 MLC

29 Question #36 Key: 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

30 Question #37 Key: E P( A ) δ a L L+ E o e o T ( ) o ( ) v P A a + c + g e a T T T T v v v P( A) + co + ( g e) δ δ ( ) ( g e) P( A) ( g e) P A T v + + co + δ δ δ δ T ( ) ( ) P A T g e Var ( ole) Var ( v ) + δ δ Above step is because for any random variable X and constants a and b, Var a X + b a Var X. ( ) ( ) Apply that formula with X Plugging in, Var ( L ) ( ) o e (.)(.735) T v. (.3.66) Question #38 Key: D T T MLC

31 Actuarial present value (A.P.V.) prem 8( + (.7 +.) + (.5 +.3)),96 A.P.V. claim 5( ) + 3( ) 8 Difference 6 MLC

32 Question # 39 Answer: D b gb gb g Per minutes, find coins worth eactly at Poisson rate 5.. b g Per minutes, f F f g Fbg fbg 839. Fbg. 997 f F b g b g b g Let Period first minutes; period net. Method, succeed with 3 or more in period ; or eactly, then one or more in period P c Fbg h + fbg c Fbg h b. 997g+ b839. gb. 3679g 965. Method : fail in period if < ; Pr ob F fail in period if eactly in period, then ; Pr ob fbg fbg b839. gb. 3679g. 677 Succeed if fail neither period; Pr ob b g (Method is attacking the problem as a stochastic process model; method attacks it as a ruin model.) 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 MLC

33 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 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 : A4 E4 A b gb g MLC

34 d ib g P Mod bgb4437 P Mod. g Alternatively, you could set the actuarial present value of benefits at age 4 to the actuarial present value of benefit 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&& F H G I K J b g + dp ib gb g Mod b gb g 7748 P Mod Question # 4 Answer: A bg bg bg τ d q l 4 bg d bg b g q l bg d bg dbg. τ bg p Note: The UDD assumption was not critical ecept to have all deaths during the year so that - 8 lives are subject to decrement. MLC

35 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 bg, q bg, q bg q bg q 5, bg 3, q bg 8 q bg, q bg, q bg This gives p / 4 / 5 / 54. p / 5 / 3 / p / 3 / 8 / bg b g So,. 54 8, and bg τ b gb gb g bg τ b gb gb g bg τ b gb gb g bg τ bg τ τ b g τ τ bg bg bg bg q log p / log p q bg ch b g q log 3 / log b. 45/ 86. gb56. g. 76 bg bg bg τ d l q b gb g Question #44 Answer: C Let: N number X profit S aggregate profit subscripts G good, B bad, AB accepted bad c hb g λ G MLC

36 c hc hb g (If you have trouble accepting this, think instead of a heads-tails λ AB 3 6 rule, that the application is accepted if the applicant s government-issued identification number, e.g. U.S. Social Security Number, is odd. It is not the same as saying he automatically alternates accepting and rejecting.) b Gg b Gg b Gg b Gg b Gg Var S E N Var X + Var N E X b4gb, g + b4gd3 i 4,, b ABg b ABg b ABg + b ABg b ABg bgb9, g + bgb g,, Var S E N Var X Var N E X S G and S AB are independent, so bg b g b g Var S Var SG + Var SAB 4,, +,, 5,, If you don t treat it as three streams ( goods, accepted bads, rejected bads ), you can compute the mean and variance of the profit per bad received. λ B c 3hb6g If all bads were accepted, we would have E X Var X + E X d Bi b Bg b Bg b g 9, +, Since the probability a bad will be accepted is only 5%, E XB Prob accepted E XB accepted + Prob not accepted E XB not accepted b g b g c h b g c h b gb g b gb g d i b5. gb, g + b5. gbg 5, E X B Likewise, Now Var S E N Var X + Var N E X b Bg b Bg b Bg b Bg b Bg b gb g b gd i 47, 5 + 5,, S G and S B are independent, so Varbg S VarbSGg+ VarbSBg 4,, +,, 5,, Question #45 Key: E MLC

37 For De Moivre s Law: o e k q ω ω ω ω k+ k+ k k b ω k b A v q v aω A ω A a&& d o e5 5 ω for typical annuitants o ey 5 y Assumed age 7 A a&& a ba&& b 5, 97 Question #46 Answer: B 3: d 3 id 4 ib g b E3gb E4gb ig E p p v p v p v + i + b gb gb7985. g 564. 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

38 Question #47 Answer: A Equivalence Principle, where π is annual benefit premium, gives d b g A + IA π a&& π i π d A a&& IA ( ) b g i We obtained &&a 35 from a&& 35 A d MLC

39 Question #48 Answer: C Time until arrival waiting time plus travel time. Waiting time is eponentially distributed with mean λ. The time you may already have been waiting is irrelevant: eponential is memoryless. You: E (wait) hour 3 minutes E (travel) b. 5gb6g + b. 75gb8g 5 minutes E (total) 8 minutes Co-worker: E (wait) hour minutes 5 E (travel) 6 minutes E (total) 8 minutes Question #49 Answer: C μ y μ + μ y 4. μ A Ay μ δ μ y Ay ay μ δ μ + δ. and y Ay A + A A P a a y y y y b g y 563. MLC

40 Question #5 Answer: E b gb g b g b gb i g b g b + ig b. gb. g. V + P + i q V V b gb g b g b gb g q 4b g V + P + i q V V q Question #5 Answer: E P A / a&& b 6 6 6g b 6 6g b 6 6 6g b 6 6g vq + p A / + p va&& q + p A / 6. + p a&& c b gb gh c b gb gh / MLC

41 Question #5 Key: D Since the rate of depletion is constant there are only ways the reservoir can be empty sometime within the net days. Way #: There is no rainfall within the net 5 days Way # There is one rainfall in the net 5 days And it is a normal rainfall And there are no further rainfalls for the net five days Prob (Way #) Prob( in 5 days) ep(-.*5).3679 Prob (Way #) Prob( in 5 days).8 Prob( in 5 days) 5*. ep(-.* 5)*.8 * ep(-.* 5) ep(-) *.8 * ep(-).83 Hence Prob empty at some time Question #53 Key: E.96 ( λ ) + e μ ( ) μ + λ ln μ.48 λ Similarly ( ) μ ln.97 λ μ μ + μ + λ y ( 5)(.68) py e e.736 Question #54 Answer: B Transform these scenarios into a four-state Markov chain, where the final disposition of rates in any scenario is that they decrease, rather than if rates increase, as what is given. MLC

42 P from year t 3 from year t Probability that year t will State to year t to year t decrease from year t - Decrease Decrease.8 Increase Decrease.6 Decrease Increase.75 3 Increase Increase.9 Transition matri is + P 8. * * L N M For this problem, you don t need the full transition matri. There are two cases to consider. Case : decrease in 3, then decrease in 4; Case : increase in 3, then decrease in 4. For Case : decrease in 3 (following decreases) is.8; decrease in 4 (following decreases is.8. Prob(both) For Case : increase in 3 (following decreases) is.; decrease in 4 (following a decrease, then increase) is.75. Prob(both) Combined probability of Case and Case is O Q P MLC

43 Question #55 Answer: B l ω 5 tp45 l45+ t / l45 6 t/ 6 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 b gb g b g Hence, ProbcK 9 > h ProbcK > h b g b g Prob K 33 since K is an integer Prob T 33 l78 33p45 l MLC

44 Question #56 Answer: C μ A 5. μ 4. μ+ δ A diai μ μ+ δ 4. z s z se Ads d. s e z F e b4. ghg A ds ib4. g ds. s. I 4. 4 KJ. Alternatively, using a more fundamental formula but requiring more difficult integration. c h z z bg δ t t 4. t 6. t b g z. t 4. te dt IA t p μ t e dt te 4. e dt (integration by parts, not shown) F t HG I. t 4. K J e MLC

45 Question #57 Answer: E Subscripts A and B here just distinguish between the tools and do not represent ages. ο We have to find e AB e ο A F I HG K J z t t dt t 5 5 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 MLC

46 Question #58 Answer: A bg btg μ τ t p bg 4. τ e t Actuarial present value (APV) APV for cause + APV for cause. z5 z 4. t. 4 t 5. 4 t. 4 t z5. 44t.,. b g b g e e. dt + 5, e e. 4 dt c b g b gh e dt e bgj e. 784 Question #59 Answer: A R p q S p e since e e b g z k μ t + k dt μ t dt k dt So S 75. R p e 75. q e e k k e z z d bg i bg k bg μ tdt q 75. p p q 75. q 75. q L NM q k ln 75. q O QP e z z kdt MLC

47 Question #6 Key: C A 6 A 6 and.3693 d A6 A6.86 π π E L π, + A d d Epected Loss on one policy is ( ) 6 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

48 Question #6 Key: C ( π )( ) ( ) V V + + i + V V q.5π q Similarly, V ( V + π ).5 q V V + π.5 q 76 ( ) V + + q75 q76 q77 ( π π π) + ( q75 + q76 + q77 ) *.5.5 π ( ) ( ) ( ) * This equation is algebraic manipulation of the three equations in three unknowns ( V, V, π ). One method usually effective in problems where benefit 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 3V 5, A 6643 a&& 8: 8: 87 MLC

49 Question #63 Answer: D b g and δ 6 b g Let A and a be calculated with μ t. * * 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 L N M * * t * μ s + 3. ds 3. t Proof: a e e dt z z z a z z z d t μ sds 3. t 3. t e e e dt t μ A 3. a 3. a Question #64 Answer: A bg bg bg sds 6. t e e dt b gb g i 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. MLC

50 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) Actuarial present value Question #65 Key: E o 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+ ez I e dt e e e NM O QP Question #66 Key: C p e q 6 + je q 6 + jb q63gb q64gb q65g b. 89gb. 87gb. 85gb. 84gb. 83g MLC

51 Question # 67 Key: E. 5 a μ + δ μ δ μ δ A A μ + 5. μ δ μ μ+ δ 3 Var a e T j A A δ S.D Question # 68 Key: D v 9. d. A da&& b gb g 5A 5vq Benefit premium π a&& V a&& a&&. A + a&& 5 && + a+ b gb g b gb g da&& V 5A π a&& b gb g b gb g b gb g MLC

52 Question #69 Key: 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. Ebg Z vq + v.95.5 b.95g pq E Z vq + vpq d i + b g b g.3478 b g d i c b gh b g Var Z E Z E Z Question #7 Key: D Actuarial present value (APV) of future benefits / /.6 ( ) ( )( ) APV of future premiums + ( ) (.99)( 5) 95.5 E L K( 55) /.6 5 Question #7 Key: A This is a nonhomogeneous Poisson process with intensity function λ (t) 3+3t, t, where t is time after noon MLC

53 () λ t dt Average λ ( 3 3tdt ) + f ( ) 7.5 e ! 3t 3t Question #7 Key: A Let Z be the present value random variable for one life. Let S be the present value random variable for the lives. δt μt EZ bg z e e μ dt 5 μ e bδ+ μ g5 δ + μ EZ d i F μ I eb δ+ μ g HG δ + μkj F.4 H G I.8 K J de i b g d i c b gh VarZ EZ EZ b g b g ES EZ 4. 6 Varbg S Varbg Z F F MLC

54 Question #73 Key: D Prob{only survives} -Prob{both survive}-prob{neither survives} p p p p b ge j b gb gb gb gb gb g b gb g Question # 74 Key: C The tyrannosaur dies at the end of the first day if it eats no scientists that day. It dies at the end of the second day if it eats eactly one the first day and none the second day. If it does not die by the end of the second day, it will have at least, calories then, and will survive beyond.5. b g b g b g b gb g Prob (dies) f + f f e since fbg. 368! e fbg. 368! Question #75 Key: B Let X epected scientists eaten. For each period, EX EXdead Prob already dead + EXalive Prob alive Probbdeadg + EX alive Probbaliveg Day, E X b g fbg b g Prob (dead at end of day ).53 [per problem ] e Prob dead at end of day. 368! Day, E X b g Day.5, E X b g b g MLC

55 where EX 5. alive 5. since only day in period. E X E X + E X + E X E, X 8, 8 Question # 76 Key: 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. b g b g bgb. 338g bgb. 338gb. 366g Pb. 338gb. 366g APV Prems P APV Benefits q v + p q v + Pp p v P P. 678 P 3.. (APV above means Actuarial Present Value) Question #77 Key: E Level benefit 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 MLC

56 Another approach is to think in terms of retrospective reserves. Here is one such solution: V P P && s n n : n : P P P P e e e e P e P P n : n : n : n : j j j j j a&& n : ne a&& n : a&& n : n : P P /. 864 P n : e n : j b gb g Question #78 Key: A b g δ ln zω z t bg δt A p μ t e dt ω e δt dt for DeMoivre ω a ω ω From here, many formulas for VcA 4 hcould be used. One approach is: Since MLC

57 so P A A A 5 4 c 4h a A so a δ a A so a δ V A A P A a F HG F HG I KJ I KJ c h c h b gb g Question #79 Key: D A E v E v NS Prob NS + E v S Prob S T T T F b g F bg bg bg 3. I HG + K J F Similarly, 3. A I 7 HG 3+ 6 K J. +.. F H Var a bg T I K A A δ I HG + K J F 6. I HG 6+ 6 K J b g Question #8 Key: B q q + q q 8: : (.6) +..5 (.).36 Using new p 8 value of (.3) +..5 (.) MLC

58 .68 Change Alternatively, p p8 p8.6. p p84 p84..3 p p8 + p84 8:84 p8 p84 since independent. +.3 (.)(.3).4 p p + p p p 3 8: ( )( ) q p p 8:84 8:84 3 8: Revised 3p p :84 ( )( ).68 q :84 change Question #8 Key: D Poisson processes are separable. The aggregate claims process is therefore equivalent to two independent processes, one for Type I claims with epected F frequency 3 HG I 3K J b g and one for Type II claims. Let S I aggregate Type I claims. N I number of Type I claims. X I severity of a Type I claim (here ). MLC

59 b g b g Since X, aconstant, E X ;Var X. I I I b Ig b Ig b Ig b Ig b Ig Var S E N Var X + Var N E X bgbg bgbg +, bg b g b g Var S Var SI + Var SII since independent,,, + VarbSIIg Var S,, b IIg Question #8 Key: A bg bg bg τ 5p5 5p5 5p5 F HG b gb I K J e g b gb g Similarly 6 5 pbg τ F I 5 e b. gb g HG 5 K J b gb bg τ bg τ bg τ 55q5 5p5 p g Question #83 Key: C Only decrement operates before t.7 bg 7q4 b7gqbg 4 b7gbg 7 since UDD..... Probability of reaching t.7 is Decrement operates only at t.7, eliminating.5 of those who reached.7 bg b gb g q MLC

60 Question #84 Key: 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 Question #85 Key: E At issue, actuarial present value (APV) of benefits t bv p μ t dt z z b g t t t 4. t de ide it p t dt 65 μ65bg z t 65 65b g 65 p μ t dt q I APV of premiums π π a HG + K J π Benefit premium π / z z u + u u 67 μ65 4. b+ ug 4. u u z 8. u V b v p + u du πa b g e e p μ + u du F e p + u du 67 I K J b g b gb g q μ b g MLC

61 Question #86 Key: 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 Key: A 3λ ( λ /) e λγ() ( λ ) λ e λ p p( λ ) f ( λ) dλ dλ! λe [Integrate by parts; not shown] dλ 4 λe e λ λ MLC

62 Question #88 Key: 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 Question #89 Key: E M T Initial state matri One year transition matri L N M O Q P cb b M T g h M T T g M T T T Probability of being in state F after three years d ib g Actuarial present value. 468v 5 7 Note: Only the first entry of the last matri need be calculated (verifying that the four sum to is useful quality control. ) MLC

63 Question #9 Key: B Let Y i be the number of claims in the ith envelope. Let Xbg 3 be the aggregate number of claims received in 3 weeks. EY EY b g b g b g b g b g b g b g b g bg b g i i EX Var X ProbmXbg 3 Zr. 9 Φb8. g R X 3 65 S b g U Prob 8. T V 468 W X bg Note: The formula for Var Xbg 3 took advantage of the frequency s being Poisson. The more general formula for the variance of a compound distribution, Var S E N Var X + Var N E X, would give the same result. bg bg bg bgbg Question #9 Key: E μ t t μ M p p F b g b g M 65 F 6 6 ω ω 85 ω t t 5 Let denote the male and y denote the female. MLC

64 o e o ey o ey 5. 5 z z F HG c c F HG F HG mean for uniform distribution over, mean for uniform distribution over,5 I K JF HG t t 5 7 t t I K J I KJ I dt dt 7 t 7 t t KJ b b gh gh o o o o + e e e e y + y y Question #9 Key: B A A 3 μ μ+ δ μ μ+ δ 5 PcA h μ 4. Var L F P A bg 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 45 A A I KJ j Question #93 Key: A MLC

65 Let π be the benefit premium Let k V denote the benefit reserve a the end of year k. For any n, ( nv + π )( + i) ( q5+ n n+ V + p5+ n n+ V) V Thus V ( V + π )( + i) n+ ( )( ) ( ( ) )( ) ( )( ) ( && )( ) 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 (actuarial present value of future benefits; there are no future premiums) a&& 6 π && s 35 a6 π && && s 35 For n, V π && s a&& && 6 s && s 35 Alternatively, as above ( nv + π )( + i) n+ V Write those equations, for n to n 34 : V + π + i V ( )( ) ( π )( ) ( π )( ) : V + + i V : V + + i V 3 M ( π )( ) 34 : V + + i V Multiply equation k by ( i) 34 k + and sum the results: ( V π)( i) ( V π)( i) ( V π)( i) L ( 34V π)( i) V( + i) + V( + i) + V( + i) + L+ V( + i) + V MLC

66 k L the ( ) 35 ( ) 35 π ( ) 35 ( ) 34 L ( ) For,,, 34, k V i k + terms in both sides cancel, leaving V + i + + i + + i i 35V Since V π && s 35 35V a&& 6 (see above for remainder of solution) Question #94 Key: B μ y () t ( + ) + μ( + ) q p μ t q p y t t y t t t y q p + p q + p p t t y t t y t t y For () (y) (5) μ 5:5 ( ) (.5q5 )( p5 ) q6 ( q )( p ) ( p ) (.95)(.9478)(.376)( ) ( )( )( ) ( ) 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 t Alternatively, ( t ) p5 p5 p6 ( + t) p5:5 ( p5 ) ( t p6 ) ( + t) p p 5:5 5 t p6 ( p5 ) ( t p6 ) ( ) ( ) ( ) p5 tq6 p5 tq6 since UDD ( ) ( ) Derivative p q + p tq q Derivative at + t.5 is ( )( ) ( ) ( ( )( ))( ) MLC

67 ( ) p p p.5 5: ( ) ( ) μ (for any sort of lifetime) dp dt (.3).3 p.996 Question #95 Key: D ( τ ) () () ( t () t + ) () t μ μ μ t ( τ) ( ) t ( τ) ( () ) t ( τ) ( τ P P v p μ t dt+ 5, v p μ () t dt+ 5, v p μ ) () t dt t t t.t.3t.t.3t.t.3t P P e e.9dt + 5, e e.dt + 5, e e.3dt.4 ( ).4 ( ).4 ( ) e e e P P,94 Question #96 Key: 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 vbe g K J. b g Let π benefit premium. MLC

68 e π v+ 98. v 87 j. 858π 87 π 84 Question #97 Key B && 3 : 3 b g3 : b ge 3j πa A + P IA + π A A3 π a IA A b g && 3 : 3 : 3 b g b g Test Question: 98 Key: E For de Moivre s law, z o ω 3 t e 3 F ω 3 L N M HG t t ω 3 b O gq P I K J dt ω 3 ω 3 3 Prior to medical breakthrough ω e o 3 35 After medical breakthrough e 3 e o ω so 3 e 3 39 ω 8 o o MLC

69 Test Question: 99 Key: A 5. L, v 4a&& 77,79 Question # Key: D ( accid ) μ. ( total) μ. ( other) μ...9 Actuarial present value.5 t 5,. t(.9) e e dt ( ).4t.5t.t + 5, e e e. dt.9. 5, +,.6. MLC

70 Test Question: Key: E b gb g b gb g b gb g b gb g b gb g b gb g b gb g E N Var N E X E X Var X For any compound distribution Var S E N Var X + Var N ce X h (3) (.64) + (3) d36. i 768 For specifically Compound Poisson Var S λt E X (6) (.5) (5.6) 768 Alternatively, consider this as 3 Compound Poisson processes (coins worth ; worth 5; worth ), where for each Varb Xg, thus for each VarbSg VarbNgE X. Processes are independent, so total Var is Var bgbgbg bgbgbg bgbgbgbg Test Question: Key: D d ib g b g 6 9V+ P. q+ 9 V A+ p+ 9 b gb6. g. 54bg a. && / 6. so b A P a&& g MLC

71 Test Question: 3 Key: B k k bg k p e e F e z bg bg bg bg τ τ μ tdt μ z z p 6 p 6 HG b p k μ g bg tdt bg tdt I KJ where p is from Illustrative Life Table, since μ follows I.L.T. k k 6, 66, , 88, 74 6, 396, , 88, 74 bg bg bg τ τ τ q6 p6 p6 b p6g b p6g from I.L.T bg Test Question: 4 Key: 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

72 Test Question: 5 Key: A τ z μ+ 4. t bg b g d e μ +. 4 dt b e μ+ 4. e 48 g μ 4. e μ +. 4 ln μ. 9 bg z t 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 Key: B This is a graph of lμbg. μbg would be increasing in the interval b8, g. The graphs of lp, l and l would be decreasing everywhere. Question #7 Key: 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

73 Question #8 Key: E + i q + p p () ( ) ( ) A A + V V i q + p p ( ) ( ) ( ) B B B V V π () ( ) ( ) ( ) A B A B B + i V V V V π ( ) p + (.6) Test Question: 9 Key: A k + A P V ( s benefits) v bk+ kpq k 3v. + 35v v , 89 + k b g b gb g 3b gb gb g MLC

74 Test Question: Key: E π denotes benefit premium 9 V APV future benefits - APV future premiums 6. π π bv + πgb8. g bq65gbg V p 65 b gb8. g b. gbg 5.8. Question # Key: A Actuarial present value Benefits (.8) (.8)(.9),39.75 P P(.3955) P (.8)(.)(,) (.8)(.9)(.97)( 9,) +.6.6, Test Question: Key: A 8 7a + 5a a a a 3: 4 8 a a + a a a : 4 3: : 4 3: 4 b gb g b gb g 3: 4 Test Question: 3 Key: B MLC

75 a a f t dt zo bg t z 5. t e o 5 z. Γb g b L NM b g L F H G I K J 5NM. 5. te t te 5. o t te dt 5. t F HG O QP i dt t t t+ e I K J 5. t e 5 O QP, , 88 Question #4 Key: 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 Key: B Let K be the curtate future lifetime of ( + k) K + k L v P :3 ak + && When (as given in the problem), () dies in the second year from issue, the curtate + is, so future lifetime of ( ) MLC

76 L v P a&& : The premium came from A :3 P :3 a && A P :3 da&& :3 :3 da&& 79. :3 :3 a&& a&& :3 :3 d Test Question: 6 Key: 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 fmbμg dμ z z μ t μe dt dμ z μ + + e du e e s d i d i d i Question #7 Key: E Note that above 4, decrement is DeMoivre with omega ; decrement is DeMoivre with omega 8. That means μ / 4.5; μ /.5 ( τ ) ( ) () ( ) 4 4 μ Or from basic definition of μ, ( ) ( ) ( ) 6 t 4 t 4 t t t p τ MLC

77 ( τ ) d( t p4 )/ dt ( + t) / 4 at t gives 6 / 4.5 ( τ ) p4 ( /3 )*( /) /3 ( τ ) ( τ) ( τ) μ4 ( ) d( t p4 )/ dt / p4.5/ ( / 3).75 Test Question: 8 Key: D Let π benefit premium Actuarial present value of benefits b. 3gb, gv+ b. 97gb. 6gb5, gv + b. 97gb. 94gb. 9gb, gv , 33. Actuarial present value of benefit 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 Test Question: 9 Key: A Let π denote the premium. L b v πa + i v πa T T π a T T b g EL π a π a MLC T T T

78 a T δ a L π a T a v T b δ a δ a g T v A A d δ a v T i Test Question: Key: D (, ) (,.9) (.5,.8775) (,.885) tp t p (. ) 9. p b gb g since uniform, 5. p / b g o e5 :. Area between t and t 5. F+ 9. I HG K J bg F I + HG K Jb g Alternatively, MLC

79 z z o 5. e 5 :. t pdt z 5. tpdt + p pd z z t 5. t + 9. b g b g. tdt d Test Question: Key: A, A63b. g 533 A Ab + ig q A + p b gb g A b. 483gb5. g. 95 A Single contract premium at 65 (.) (,) (.4955) 555 b g i i MLC

80 Test Question: Key: B Original Calculation (assuming independence): μ 6. μ y 6. μ y μ A μ δ μ y Ay y μ δ μ y Ay y μ δ A A + A A y y y Revised Calculation (common shock model): μ 6., μ 4. μ y bg bg T* T* y y 6., μ 4. bg bg 4 4 T* T* y Z μy μ + μy + μ μ A μ δ μ y 6. Ay μ y+ δ μ y Ay y μ δ A A + A A y y y Difference MLC

81 Question #3 Key: B q q + q q 5p35q4 + 5p45q5 5p35: 45q4: 5 5p35q4 + 5p45q5 5p35 5p45b p4 : 5g 5p35q4 + 5p45q5 5p35 5p45b p4 p5g : : 45 Alternatively, b gb g b gb g b gb g b gb g ( )( ) ( )( ) p p p p p p q p p 5 35: : :45 ( 5 p p45 5 p35:45 ) ( 6 p p45 6 p35:45 ) ( 5p35+ 5p45+ 5p35 5p45) ( 6p35+ 6p45 6p35 6p45) ( ) ( ) Test Question: 4 Key: C z3 λbtdt g 6 so Nb3g is Poisson with λ 6. b g g P is Poisson with mean 3 (with mean 3 since Prob y i < 5. 5 P and Q are independent, so the mean of P is 3, no matter what the value of Q is. MLC

82 Test Question: 5 Key: A At age : Actuarial Present value (APV) of future benefits APV of future premiums F HG 4 5 &&a I K J π F A I HG 5 K J A 5 A 4 a&& 4 π a&& by equivalence principle π π V APV (Future benefits) APV (Future benefit premiums) 4 A35 π a&& b. g b58. gb5396. g 5. 5 Test Question: 6 Key: E Let EY Var Y 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 / ES EY 4, Var S Var Y 7, 555 Standard deviation S 7, i d ib g By normal approimation, need E [S] Standard deviations 4, (.645) (65.6) 5,54 MLC

83 Test Question: 7 5A 4 A Initial Benefit Prem 5a&& 4a&& e 3 3: Key: B 3: 35 3: b. g b. g j b4. 835g 4b959. g Where A A A : e... 3: 3: j and A3 :. 337 a&& : d F 6. H G I 6. K J Comment: the numerator could equally well have been calculated as A + 4 E A.48 + (4) (.9374) (.495) Test Question: p p q y b gb g b 75. y 75. y b75. pgd75. pyi p gb g -b.965gb. 95g 97. Key: B since independent Question #9 Key: D Let G be the epense-loaded premium. Actuarial present value (APV) of benefits,a 35 APV of premiums Ga&& 35 APV of epenses.g+ 5 + (.5)( ) a && 35 Equivalence principle: MCL