Fall 2003 Society of Actuaries Course 3 Solutions = (0.9)(0.8)(0.3) + (0.5)(0.4)(0.7) (0.9)(0.8)(0.5)(0.4) [1-(0.7)(0.3)] = (0.

Transcription

1 Fall 3 Society of Actuaries Course 3 Solutions 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 =.9.8 =.7 =.5.4 =. =.7. =.44 = =.776 =.7.7 =.54 =..3 =.6 =.54.6 =.34 = = q = :34 =.44 Alternatively, q = q + q q = p q + p q p p 3: : : 34 3: 36 b = (.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 ) g

2 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 =

3 Question #3 Key: A c 4 < 4 B = 4 = E B = c 4 ce X 4 i 3 = ci4 ci = c 4 3i 7 c = = Question #4 Key: C Let N = # of computers in department Let X = cost of a maintenance call Let S = aggregate cost ( X) ( X) λ λ Var = Standard Deviation = = 4, E( X ) = Var( X) + E( X) = 4, + 8 = 46, 4 E S = N E X = N 3 8 = 4N Var S = N E X = N 3 46, 4 = 39, N We want. Pr >. ( S E( S) ) S E S.E S. 4N Pr >.8 =Φ.9 39, N 39, N 373. N N =

4 Question #5 Key: B If you happen to remember this distribution from the Simulation tet (eample 4d in third edition), you could use: ( u) log log.95 n= Int + = Int + = + = log q log. For mere mortals, you get the simulated value of N from the definition of the inverse transformation method: f() = F() = so n = ( v ) = log = log.7 = λ. he amount of total claims during the year = 35.67

5 Question #6 Key: D ( τ ) µ =. time of death = 5ln (.35) = 5.49 ( adb µ ) ( τ ) =.5 Prob( non adb) =.75 µ.775 > Prob( non adb) so death is accidental and benefit is (see below for more detail) δ L= e.5a δ t -δ e =e.5 δ (.5)( 5.9) (.5)( 5.49) e = e.5.5 =.449 (.5)( 8.55) =.39 Another way of seeing that Prob(non adb) =.75. t ADB deaths in time.5 τ µ ( r t = e ) dr t =.5.e =.5 More detail on why benefit is : F t µ µ ( τ ) ( r) dr τ µ r e ( τ ) ( r) dr =.5 all deaths in time t Benefit (b) f(b) F(b) ().775 F < < so benefit =

6 Question #7 Key: B t ( τ ) ( 3) = + + =.45 ( τ ).45t p = e µ µ µ µ 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 t,,.645t 5,.645t 5,.645t =,, e dt+ e dt e dt 5, +, = =

7 Question #8 Key: B 4: 4: APV Benefits = A + E vq k k= k k= APV Premiums = π a + E vq 4 4+ k 4 4+ k Benefit premiums Equivalence principle + 4: k 4 4+ k = π + 4: k 4 4+ k k= A E vq a E vq π = A / a 4: 4: = = 5. 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 =.76π π +.9 = π = = 5..76

8 Question #9 Key: C ln.6 A7 = δ A7 =.53 =.547 i.6 A7.547 a 7 = = = d.6/.6.97 a 69 = + vp69a 7 = + ( ) = a 69 = α a 69 β = (.)( ).5739 = 8.59 ( m) ( m ) Note that the approimation a a works well (is closest to the eact answer, only m off by less than.). Since m =, this estimate becomes =

9 Question # Key: C he 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) =.. u k + i = u k 3. It is not E. he recursion for a pure endowment is simpler: 4. hus, it must be C. More rigorously, transform the recursion to its backward equivalent, u( k ) 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 in terms of : his is the form of (a), (b) and (c) on page 9 of Bowers with = k. hus, the recursion could be: 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 A u k = fails at = 69 since it is not true that A vq vp () 69 = 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.

10 Question # 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 = =.76

11 Question # Key: E (his is covered in Probability Models, section 4.6 in the eighth edition). Let states be,, 3 for acutely ill, in remission, cured/dead. he transition matri is: States and are transient. he matri for transitions from transient states to transient states is. P.6.3 = I P = where I =..5 is the identity matri S I P = = In that matri, the entry s ij is the epected time in state j, given that you started in state i. In this problem, we started in state, so the relevant epected times are s and s. otal cost = ( epected time in state ) + ( epected time in state ) = s + s.5.3 = ( ) = 37.84

12 Question #3 Key: C Since all claims are for, claims rate = λ = 8. With relative security loading.5, premium rate = (8)(.5) =. Per period f() F() -F() Ruin if or more claims by t = or 3 or more claims by t = or more claims by t = is ruin since we must pay or more when we started with and have collected less than in premium. 3 or more by t = is similar: we can t pay 3 or more with the initial and less than in premium. Evaluate the probability of ruin by summing the probabilities of three separate cases:. wo or more claims before time.. No claims by time, and three or more between and. 3. Eactly one claim by time, and two more between and. P(ruin) = -F() =.9 + f()(-f()) (.4493)(.474) =.3 + f()(-f()) (.3595)(.9) =.687.8

13 Question #4 Key: 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 /.5 =.9848 A so P4 = = = 7.3 a Revised Revised ( ) E L K 4 = A P a = = 35.6 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 /

14 Question #5 Key: E Let NS denote non-smokers and S denote smokers. he 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 = ) = a = δ.444 = = Similarly E( a Y = ) = = 7.4. E E a Y = E E a Prob Y= + E E a Prob Y= ( ) ( ) ( 7.4)(.7) ( 5.56)(.3) = + = 6.67 E E a Y = + = ( ) ( 7.4 )(.7 ) ( 5.56 )(.3 ) ( ( )) Var E a Y = =.47 ( Var ) ( 8.53 )(.7 ) ( 8.88 )(.3 ) E a Y = + = 8.6 Var a = = 9.7 Alternatively, here is a solution based on Var( Y) E( Y ) E( Y) transformed into E( Y ) Var ( Y) E( Y) E( ( a ) NS ) = Var ( a NS ) + E( a NS ) =, a formula for the variance of any random variable. his can be = + which we will use in its conditional form Var a E = a E a E a E a S Prob S E a NS Prob NS = + [ ] [ ]

15 S =.3a +.7a NS S NS ( A ) ( A ).3.7 = +...3(.444) +.7(.86) = = +. = = 6.67 (.3)( 5.56) (.7)( 7.4) = S Prob[ S] + NS Prob[ NS] E a E a E a ( ( a ) ( E a ) ( ( a ) E( a ) ) =.3 Var S + S +.7 Var NS + NS = = Var a = = 9. v Alternatively, here is a solution based on a = δ v Var ( a ) = Var δ δ v = Var since Var ( X + constant) = Var ( X) δ Var ( v ) = since Var constant = constant Var δ A ( A) = which is Bowers formula 5..9 δ ( X ) ( X ) A a A his could be transformed into δ Var NS S and A A. = +, which we will use to get

16 = A E v A Var E v = E v NS Prob NS + S Prob S ( a ) ( A ) NS = δ Var NS + Prob NS ( a ) ( A ) S + δ Var S + Prob S = = + (.6683)(.7) (.853)(.3) =.38 = E v E v = E vns Prob NS S + Prob S = + (.86)(.7) (.444)(.3) =.3334 ( a ) = A δ ( A ) = = 9..

17 Question #6 Key: E p p 3 =.7 =.4 Since hyperbolic, = (.5) + (.75) s(.5) s( 3) s =.5 / /.7 = p s.5 = = Alternatively, since hyperbolic.4 p = = p.5 p = (.5) q.5743 = =.84.5 p = (.7)(.84) = No matter which way we got.5 p (ignoring rounding in last digit).5q = =.453 Probe not transmitting means all three failed Prob = (.453) 3 =.69

18 Question #7 Key: A o be a density function, the integral of f must be (i.e., everyone dies eventually). he 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 able: l l 5 l l 4 =.895 = () () () = f t dt = k f t dt +. f t dt 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 l 4 = 3.83 = =.6 l 4 l 5 = 3.83 = =.4 l F ( 5).4 p4 = = =.83 F 4.6 5

19 Question #8 Key: D Let NS denote non-smokers, S denote smokers. ( < 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 = his is quadratic, so.7 ± =.3 e =.347.tˆ ˆ =.347 so t =.56

20 Question #9 Key: D he modified severity, X*, represents the conditional payment amount given that a payment occurs. Given that a payment is required (X > d), the payment must be uniformly distributed between and c ( b d). he modified frequency, N*, represents the number of losses that result in a payment. he b d deductible eliminates payments for losses below d, so only F ( d) = of losses will b require payments. herefore, the Poisson parameter for the modified frequency distribution is b d λ. (Reimbursing c% after the deductible affects only the payment amount and not the b frequency of payments).

21 Question # Key: C Let N = number of sales on that day S = aggregate prospective loss at issue on those sales K = curtate future lifetime [ ] [ ] K + [ ] N Poisson.*5 E N = Var N = L=,v 5a E L =,A 5a K K L=, + v Var[ L] =, + A65 ( A65) d d d = E S E N E L S L L... L N [ ] = [ ] [ ] [ ] = [ ] [ ] + ( [ ]) Var[ ] Var S Var L E N E L N Pr( S < ) = Pr Z < [ ] [ ] E S Var S Substituting d =.6/(+.6), A 65 =.363, A65 =.4398 and a 65 = yields E [ L] [ L] [ ] [ S] = Var = 5,, E S = Var = 54,5, Std Dev (S) =,46 S Pr( S < ) = Pr <,46,46 = Pr Z <.443 =.67 With the answer choices, it was sufficient to recognize that:.6554 = φ.4 < φ.443 < φ.5 =.695 By interpolation, φ(.443) (.43) φ(.5) + (.57) φ(.4) = (.43)(.695) + (.57)(.6554) =.679

22 Question # Key: A A P4 = = =.89 a a V 4 = = = a V = V + 5 P ( + i) 5q 4 6 P 6 ( (5)(.89)).6 5(.376) = = [Note: For this insurance, V = V 4 because retrospectively, this is identical to whole life] hough it would have taken much longer, you can do this as a prospective reserve. he 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 A65 =.6674 a = a 4: 4 E4 a 6 =.76 a = a E a = : = = (.89)(.76) + ( 5)(.89)(.744)( 4.347) + π (.744)(.68756)( ) π = =.3 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

23 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 7,533,964 = = l = 8,75, :4 A = A6 4 E6 A65 = =.5779 a = a 6:4 6 4 E6 a 65 = = ( 5)(.89)( 3.595).3(.73898)( ) V = = 55 Finally. A moral victory. Under eam conditions since prospective benefit reserves must equal retrospective benefit reserves, calculate whichever is simpler. Question # 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 = (.8/ /.5 ) A 4 =.793 ( Z ) = Var =.544

24 Question #3 Key: A Let L, be the amount by which surplus first drops below 4, given that it does drop below 4. [ 4 < 35] = [ 5 < ] = PR L PR L f y dy 5 L = P y 5 p dy he last step above is equation in Bowers (page 46) For claim size distributed uniformly over (, ) p y < and P( y) = y/ < y< y > = 5 y PR = ( y /) dy =.5 5 = 5 5 5

25 Question #4 Key: D his solution looks imposing because there is no standard notation. ry 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. he 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 + = hat last step follows because from the transition matri if we are in state, we always move to state one period later.

26 Similarly, every time we are in state we have E Yk sk = =.93 ( + E Yk+ sk = ) =.93 ( + E Yk+ sk+ = ) hat 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+ }). hose last two steps follow =.94 + =.9 + =. 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. ( ) ( ) =.95 + = =.93 + hat s three equations in three unknowns. Solve (by substituting the first and third into the second) to get = 6.8. hat s the answer to the question, the epected present value of the future payments given in state. he 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. his 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.

27 hus, 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

28 Question #5 Key: A Let state s = number of stocks with market price > strike price. s =,, ransition probability matri: π = limiting probability of stocks with strike price < market price. π = limiting probability of stock with strike price < market price. π = limiting probability of stocks with strike price < market price. 9 π = π + π+ π π = π + π+ π 8 8 π + π + π = Solving algebraically the three equations in three unknowns gives us π =. 7

29 Question #6 Key: E he 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) + ( ) =

30 Question #7 Key: D ( τ ) ( = ) () t + µ µ µ () t ( τ ) ( =.µ () ) t + µ () t ( ) () t =.8 µ 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 3 =.9538 q =.8(.9538) =.644

31 Question #8 Key: A k 3 k f(k) f ( k) ( k 3) f ( k) ( k 3). (.9)(.) = (.7)(.3) = = ( K ) E K 3 =.4 E 3 = 5.58 Var( K 3) = =.7 Note that E[ K 3] is the temporary curtate life epectancy, e if the life is age. :3 Problem 3.7 in Bowers, pages 86 and 87, gives an alternative formula for the variance, basing the calculation on k p rather than k q. Question #9 Key: E f( ) =., 8 =..5( 8) =.3.5, 8 < 8 E =.d+.3.5 d = + 3 = = EX ( ) = EX f d ( f d) + ( ) = = (.8) = Loss Elimination Ratio = =

32 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 ) he 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: ( V ) =.3 (.) + c( V45 ) (.3). c = 45.3 =.47 =.8

33 Question #3 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.7595 = = 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.645 =.6833 M = 6.97 AGG AGG AGG minimum number needed = 7 AGG

34 Question #3 Key: D Annuity benefit: Death benefit: New benefit: K + v Z =, 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)

35 Question #33 Key: B First restate the table to be CAC s cost, after the % payment by the auto owner: owing Cost, p() 7 5% 9 4% 44 % hen E( X ) =.5*7 +.4*9 +.*44 = 86.4 E X =.5*7 +.4*9 +.*44 = Var ( X ) = = Because Poisson, E( N) = Var ( N) = E( S) = E( X) E( N) = 86.4* = 86,4 Var( S) = E( N)Var( X) + E( X) Var( N) = * * = 7,95,6 S E( S) 9, 86, 4 Pr( S > 9,) + Pr > = Pr( Z >.8) = Φ(.8) =. Var( S) 7,95,6 Since the frequency is Poisson, you could also have used Var S = λ E X = = 7,95,6 hat way, you would not need to have calculated Var ( X ).

36 Question #34 Key: C ( d) E( X) d / θ ( e ) E X θ LER = = θ d / θ = e Last year d / θ.7 = e d = θ log.3 Net year: dnew = θ log( LER new ) 4 Hence θ log( LER new ) = dnew = θ log.3 3 log LER =.653 ( new ).653 e LER new = =. LER =.8 new Question #35 Key: E E( X) = e( d) S( d) + E( X d) [Klugman Study Note, formula 3.] = e p + E 4 e4 6 = (.6) + 4 (.5)(4 ) 4 e e4 = ( 6 3) = = 5.6 he first equation, in the notation of Bowers, is e = e 4 4p + e :4. he corresponding formula, with i >, is a very commonly used one: a = a + E a + n : n n

37 Question #36 Key: C n e = n : t p dt q 9 =.8877 = so =.8877 =.956 For U p t q e t 9:.8877 t 9 For C t p = ( p) so e9: = = =.93 log p9 log(.83) p p9.83 For H p e ( p ) t Alternatively For U, ( z) q = so = log = log.83 =.899 p + tq q : 9 9 q µ + = is increasing for < t < tq µ + t = p is constant for < t < For C, log q For H, ( t) ( ) µ + = is decreasing for < t < t q With p the same for all three, we must have and U C H t 9 > t 9 > t 9 p dt p dt p dt U C H t p t p t p > > for <t< Alternatively (not rigorously) U.5 p9 C.5 p9 H.5 p9 =.956 =.97 =.8958 hat should strongly suggest that U C H t 9 > t 9 > t 9 p dt p dt p dt You could compare additional values of p for greater comfort.

38 Question #37 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 he 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. he actuarial present value of benefits would also have been easy to calculate as = 97.5

39 Question #38 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 + h 3,535 +,78.5,. V = = 35, ( 35, ) (.5),. V = = 36,657.3 = A Question #39 Key: B For De Moivre s law where s( ) ω t e = and t p = ω 5 45 e45 = = e65 = = = ω : t 4 t e45:65 = t p45:65dt = dt 6 4 F HG + = t t t 3 3 I K J 4 = 556. e45: 65 = e45+ e65 e45: 65 = = 34 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.

40 Question #4 Key: B F() =.8 F(t) =.8 +.5(t-), t 5.75 found since F ( ) found since F ( ) =.85 Average of those two outcomes is.