SPRING 2007 EXAM MLC SOLUTIONS

Transcription

1 SPRING 7 EXAM MLC SOLUTIONS Question # Key: E p p p μ 7 d p7 e e p Question # Key: B A μ/ μ+ δ μ/ μ μ.4 A μ.79 μ+ δ Var a T A A δ

2 Question # 3 Key: D [ 6] A[ 6] / d.359 /.6/ P[ 6] A[ 6] / [ 6] 359/ V[ 6] A65 P[ 6] && a Alternatively, A65 A[ 6] 5V [ 6] A [ 6]

3 Question # 4 Key: E Let K curtate future lifetime random variable σ Var L L K && K + L v Pa P K + P 5 + v d d P K + P Var ( L) Var 5 + v d d P 5, + d Var v because Var ( ax b) a Var ( ) K + + if a and b are constants P is a constant, a numeric value, not a random variable. P There are various ways to evaluate + d. One possibility is d P a + && d d d da && A P K Var ( L ) 5, + + Var ( v ) d ( 5,)(.699) ( A A ) ( 5,755, 4,)( ) 58, 478,9 Standard deviation 6,77

4 Question # 5 Key: C Number of claims in months is Poisson with λ n P(N n) Pr( N 3) Question # 6 Key: A Donations compound Poisson λ Withdrawals compound Poisson λ 7. 4 epected donations where 5 mean epected withdrawals where 4 mean epected net Variance of donations 56 ( ) 6, ,86 Variance of net change 6,8 + 9,86 46,66 Variance of withdrawals ( 6 8) Pr[ S > 6] Pr N(,) > Pr N(,) > ,66

5 Question # 7 Key: E All have the same benefits, so retrospectively the one with the highest accumulated value of premiums wil have the highest reserve. All have total premiums of, so the one with the premiums earliest will have the highest accumulated value. That is E. Alternatively, all have the same benefits, so all have the same actuarial present value of premiums. E has the highest APV of the first 5 premiums, so it has the lowest APV (at both time and time 5) of premiums for years 6 and later. Thus E has the highest prospective reserve. Question #8 Key: E Let t p probability Kevin still there p probability Kira still there t y t t y.7t.6t ( e ) e dt.6t.3t e dt e dt Kira s epected playing time ( p ) pdt Alternatively and equivalently ey ey Alternatively, Kira has 7 chance of surviving Kevin, 3 and (memoryless).667 future lifetime then if she does.6 7 (.667) o o

6 Question # 9 Key: E For decrement in its associated single decrement table: () t p 5. t () () ( ) Thus. p 5.98 ;.6 p 5.94 ;.4 p 5..94/ For decrement in its associated single decrement table: No one age 5 dies before age %. of the lives which reach 5. die at %. of the lives age 5 die at /.9.33 is the probability a life which reaches 5.6 will Thus die at 5.6. For a life age 5 in the double decrement table: ( ) Probability of surviving until t. is.98, because. p 5.98 and no one dies from cause () before.. Probability of dying at 5. from cause () is , since 9% of those who reach 5. die then. Probability of surviving beyond ( τ ) ( ) Probability of surviving to t.6. p5.4 p5., since no one dies from cause () between t. and t.6 ; probability Probability of dying at 5.6 from cause () is , since 3.3% of those who reach 5.6 die then. Total probability of dying from cause ()

7 Question # Key: C Let A be values at 6% all years. Let A* be values at % for one year, then 6%. * A vq vp A..988A A (.988)( A ). 67 A *. (.99)(.3) A Question # Key: A +.4 && : 4: 4: 4:9 4:9 G A Ga G Ga a && && 4: 4:.G+.5Ga 5+ 5a A4: + + 5a 4:9 G.9. 4: A E A E.9 E. ( ) ( ) ( ) + +.9( ( 3.668) )

8 Question # Key: D Possible outcomes: Result L Probability K(55), J L v K(55), J L v K(55), J L v (.8).764 L v a K(55), J 5&& (.6).573 K(55) > L (.93).896 F(-97.7).896 F(79.83) F(893.4) which is > 95% Question # 3 Key: C v. The shortest solution uses Hattendorf. Here s one Hattendorf solution. ( ) Var 3L K 55 3 since no coverage there. ( 55 ) ( 3 3 ) ( 3 ( 55) 3) Var L K v b V p q v p Var L K ( ) (.5)(.5) v (.5)( ) v + 6,6 ( ) ( ) Var L K 55 v b V p q + v p Var L K 55 (.833) (.6)(.4) ( v )(.6)( 6,6) v + 53,39 +, ,756 Without Hattendorf, you need to use the general formula for variance of any random variable, applied here to the contingent distribution: Var( ) ( ) L K E L K E L K Before we can evaluate L for any outcome, we must know the benefit premium P.

9 The shortest way is to calculate P from the benefit reserves. E.g.,.833.5v P; P Or (much more time consuming), you could calculate P as the actuarial present value of benefits divided by the actuarial present value of an annuity-due. t q p t p && :3 a v p t t t t :3 + t A v q t P :3 Whichever way you solved for P, you can now finish the variance (.6698) K L Conditional ( L ) Prob.4 33,6 P P.3 35, P.3 45, ( ) ( ) ,6.3 35, ,884 E L K (the benefit reserve) E L K ,947 ( ( 55) ) 64, Var L K 55,763

10 Question # 4 Key: E The intent of this problem, not the same as the wording used on the eam, is that the future lifetimes of (3) and (35) are independent and follow the same mortality table. p a 5 3 q3:35 b ω 35 [ ] t t p3 t p35 μ35 () t dt 5 p3 t p35 t p35 μ35 ( t) dt Pr 3 dies within 5 years after (35) () t t p35 μ35 t p3 5 p3 t p35 dt 3:35 35:35 q a q ( 3:35 ) q a (equally likely to die first since same age) ( b) a

11 Question # 5 Key: D Present Value of Future Benefits Year Benefit Paid Pattern Necessary Probability Yr.3 Yr (.6)(.4) Yr 3 (.6)(.4)(.) PVFB.3(.8)( ) +.4(.8) ( ) +.4(.8) 3 ( ) Present Value of Future Premiums Year Premium Paid Pattern Necessary Probability Yr P.6 P Yr Yr 3 (.6)(.4) P PVFP P +.8.6P +.8.4P.6336 P PVFB PVFP P P Question # 6 Key: B Find, then find (.6)(.3).8 Find, then find Find, then find.3.4. Total

12 Question # 7 Key: D K K Q Q Q K K K PV of option (i) ( v v v 3 ) PV of option (ii).8 vr.455r So R 8. / Question # 8 Key: C 3 q[ + ] q[ ] q[ + ] + q 5 + l69 77 l[ 67] p[ 67] p[ 67] ( [ ] ) p[ 67 ] p[ 67 ] p[ 67 ] + q[ 67 ] q q[ 67] + q q[ 67] q[ 66] + q p[ 67] l[ 67]

13 Question # 9 Key: D V V vp5 + v p5 + v 3 p vp5 + ( vp5) + ( vp5 ) 5 * + vp + ( vp ) ( vp5) Numerator is a geometric series, so 3 ( vp5 ) vp5 3 ( vp5 ) vp5 vp5 p *In a multiple choice test with numeric answers, you could have tried each of the answer choices and discovered.954 works. Alternatively, consider the recursive reserve formula: ( n V + π )( + i) q nv q The only difference in the recursion for V 4, V 4 and 3 V 4 is the initial reserve. If V 4 were larger than V 4, then V 4 would be even larger, and 3 V 4 even larger still. If V 4 were smaller than V 4, then V 4 would be even smaller, and 3 V 4 even smaller still. Therefore V 4 V 4 V 4 3 V vp + vp vp 5.9 p v

14 Question # Key: C Let μ be the force of all non-crash deaths. ( τ ) ( ) μ μ μ...8 In year, μ DA : In year, μ DA : p : e e.996 DA Question # Key: B For a model of the form l ( ω ) α o we have e ( ω ) /( α + ) This can be verified by integration. For the De Moivre model e ( ω )/. o Consequently [ ] ( ω 6 )/ ( 3/ ) ω 9 Hence ( ω 7 )/ 4/3 ω 3 / α + / ω 3 / / α + α.5

15 Question # Key: C.T In first years, Pr( Z > 7) Pr( e > 7) ( T ) ( T ) Pr. > ln.7 Pr < 3.57.T After years Pr( Z > 7) Pr( 5e > 7) ( T ) ( T ) Pr. > ln.8 Pr <.73 Pr Z > 7 q + q Question # 3 Key: B ( ) ( τ ) ( ) μ. p q e. e.7369 ( ) ln q ln μ + + ( τ ) q+ tp+ μ+ dt () μ+ + μ + e μ dt ( + ) t e.877dt t ( e ) ( e.4377 ) ( τ ) q p q+ (.7369)(.33).7

16 Question # 4 Key: A S S S e () μ tdt.( 75) 75 e ( 76) 76 e..643.( 77) S( 77) e : Alternatively, 76 μ() tdt 75 p e e 75..t 76 e (same as similarly for p 75) Question # 5 Key: B 5 t 5 m( 5) λ ( t) dt.5dt dt.dt + + t N is Poisson with mean p 4 e.75 /4! p 4.68

17 Question # 6 Key: A There are two ways to attack this problem. One approach focuses on the distribution of T. T is the sum of 89 independent eponential random variables, each with mean 4 and therefore variance 4. ET Var T Using normal approimation T Pr( T > 68) Pr > Φ.84 The other approach looks at the distribution of N, the number of cell divisions that occur in 68 days. N is Poisson with mean variance ( T > ) ( N < ) ( N < ) Pr 68 Pr 89 Pr 88.5 Φ (.46).84 N Pr < In evalutaing the normal approimation to N, we needed a continuity correction since N has a discrete distribution. We didn t need one in evaluating the normal approimation to T, a continuous distribution.

18 Question # 7 Key: A t δt μt t( μ δ ) E Z e e e μdt μe dt + + t μ+ δ+ μ μ μ e.3636 μ.4 μ+ δ + μ+ δ μ t δ μt μ.4 E( Z ) e e e μdt Var Z μ δ Question # 8 Key: C ,.5 AS CV CV CV 9.4 CV

19 Question # 9 Key: A For an annuity benefit of : Single Benefit Premium NSP + NSP A 3 3 3:3 E E + NSP A E E A (.54733)(.744)(.454) NSP.48 (.54733)(.744)(.3693) NSP.673 NSP For benefit of, Question # 3 Key: D a ( τ ) δ + μ () () ( ) δt () 3 μ + μ δt μ μ + μ μ A e e dt+ e e dt () 3μ μ ( τ) ( τ) ( τ) ( τ) δ + μ δ + μ δ + μ δ + μ P A. ( τ ) δ + μ. ( τ ) δ + μ