Long-Term Actuarial Mathematics Solutions to Sample Multiple Choice Questions

Transcription

1 Long-Term Actuarial Mathematics Solutions to Sample Multiple Choice Questions October, 8 Versions: July, 8 Original Set of Questions Published July 4, 8 Correction to question 6.5 August, 8 Correction to question S4., S4.3, S4.4, and S4.5 September 7, 8 Added 7 questions from the 6 and 7 multiple choice MLC eams. Corrected solutions to questions 6.5 and 9.7. October, 8 Corrected solution to question 7.7 October, 8 Page

2 Question. Answer C Answer C is false. If the purchaser of a single premium immediate annuity has higher mortality than epected, this reduces the number of payments that will be paid. Therefore, the Actuarial Present Value will be less and the insurance company will benefit. Therefore, single premium life annuities do not need to be underwritten. The other items are true. A Life insurance is typically underwritten to prevent adverse selection as higher mortality than epected will result in the Actuarial Present Value of the benefits being higher than epected. B In some cases such as direct marketed products for low face amounts, there may be very limited underwriting. The actuary would assume that mortality will be higher than normal, but the epenses related to selling the business will be low and partially offset the etra mortality. D If the insured s occupation or hobby is hazardous, then the insured life may be rated. E If the purchaser of the pure endowment has higher mortality than epected, this reduces the number of endowments that will be paid. Therefore, the Actuarial Present Value will be less and the insurance company will benefit. Therefore, pure endowments do not need to be underwritten. Question. Answer E Insurers have an increased interest in combining savings and insurance products so Item E is false. The other items are all true. October, 8 Page

3 Question #. Answer: B 4 t t Since S( t) F( t), we have ln[ S( t)] ln. 4 d Then t log S( t), and 65. dt 4 t e p, since p 6 t t t p 6 t S (6 t) 4 t (6) 4 6 /4 S 6 /4 /4 e i4 p 5 t 5.5 i October, 8 Page 3

4 Question #. Answer: D This is a mied distribution for the population, since the vaccine will apply to all once available. S = # of survivors Available? (A) Pr( A ) p A E( S A ) Var( S A ) E S ( A ) Yes ,,89 9,4,883,9 No ,4 3,83 9,3,685,43 ES ( ) ES ( ) As an eample, the formulas for the No row are Pr(No) =. =.8 96,36 9,6,54,98 Var( S ) 57,85 SD( S ) 397 p given No = (.98 during year )(.98 during year ) =.964. E( S No), Var( S No) and E( S No) are just binomial, n,; p(success) =.964 E( S), E( S ) are weighted averages, Var( S) E( S ) E( S) Or, by the conditional variance formula: Var( S) Var[ E( S A)] E[ Var( S A)] StdDev( S) 397.(.8)(97, 96, 4).(,89).8(3,83) 53, 664 3, 6 57, 85 October, 8 Page 4

5 Question #.3 Answer: A B t d d c c ln c ft St e dt dt B t c c ln c B t e c c ln c ln c B t c c ln c t e Bc t.7. e.7 ln. t.. f ln. ().7. e.4839 Alternative Solution: f ( t) p t t Then we can use the formulas given for Makeham with A, B.7 and c ln. 5 f ( t) e October, 8 Page 5

6 Question #.4 Answer: E t t t p t e75: t p75dt where t p for t t p 75 5t t 75 dt 3 t t 4375t 75t t Question #.5 Answer: B e e p e 4 4: (5) 38 e e p e 4 4: 4 4 e e e p : 4 4 e4 p4 p Question #.6 Answer: C d d lns ln d 3 d October, 8 Page 6

7 Therefore, 35 () Question #.7 Answer: B q 3 S 3 S 5 S Question #.8 Answer: C The -year female survival probability = e 3 The -year male survival probability = e We want -year female survival = e Suppose that there were M males and 3M females initially. After years, there are epected 3 to be Me and 3Me survivors, respectively. At that time we have: 3Me e 3 Me e October, 8 Page 7

8 Question # 3. Answer: B k k Under constant force over each year of age, lk ( l) ( l ) for an integer and k. 3 q [6].75 l l [6].75 [6] 5.75 l [6].75 l l l [6].75 [6].75 [6] (8, ) (79, ) 79, 49 (77, ).5.75 (74, ) 74, (67, ) (65, ) 65, q [6].75 l[6].75 l[6] , , l 79, 49 [6].75 q [6].75 Question # 3. Answer: D l l e p dt p p dt p p e [65] t [65] [65] t 66 [65] e () (98 935) e e[66] t p[66] dt p [66] e67 e e [66] Note that because deaths are uniformly distributed over each year of age,.5 t pdt q. October, 8 Page 8

9 Question # 3.3 Answer: E q. [5].5 [5] [5].5 l.5l.5l.5(97, ).5(93, ) 95, [5].5 [5] [5] l.3l.7l.3(89, ).7(83, ) 84, q. [5].5, l 95, 84, , q.5, 74. [5] Question # 3.4 Answer: B l l Let S denote the number of survivors. This is a binomial random variable with n 4 and success probability, ,87. ES ( ) 4,(.6) The variance is Var( S) (.6)(.6)(4,) StdDev( S) The 9% percentile of the standard normal is.8 Let S* denote the normal distribution with mean and standard deviation Since S is discrete and integer-valued, for any integer s, Pr(S ³ s) = Pr(S > s -.5)» Pr(S* > s -.5) æ = Pr S * > s ö ç è ø æ = Pr Z > s ö ç è ø For this probability to be at least 9%, we must have s 85.6 s So s 85 is the largest integer that works. October, 8 Page 9

10 Question # 3.5 Answer: E Using UDD l l (.6)66,666 (.4)(55,555) 6,.6 (.)(44, 444) (.9)(33,333) 34, 444. q 6 l63.4 l65.9 6,.6 34, (a) l 99,999 6 Using constant force.4 l l l l l l , ,555 6,977. l 65.9 l 65 l 66 44, , , q 6 6, , (b),( ab),( ) 6 Question # 3.6 Answer: D e[6] e [6]:3 3 p[6] ( e64) p [6] p [6] p 3 [6] e [6]:3.9,.9(.88).79,.79(.86).68 3 p k k [6] e e (5.) [6] 64 October, 8 Page

11 Question # 3.7 Answer: B.4 p[5] p[5] p5 q[5] q p p / p.5 [5].4.5 [5].4.9 [5] [5] /.9.4 q[5] q[5] q5 q[5].9.4 / /.5.64 q Question # 3.8 Answer: B 85, 3.5 6,84.9 E( N) 4 p35 4 p , , 33.9 Var( N) p p p p Since N 55 October, 8 Page

12 Question # 3.9 Answer: E From the SULT, we have: 99, p,. 9, p , The epected number of survivors from the sons is with variance The epected number of survivors from fathers is with variance The total epected number of survivors is therefore 38.. The standard deviation of the total epected number of survivors is therefore The 99 th percentile equals 38. (.36)(.96) 385 Question # 3. Answer: C The number of left-handed members after each year k is: L k Lk k k Similarly, the number of right-handed members after each year k is: R k Rk k k At the end of year 5, the number of left-handed members is epected to be , and the number of right-handed members is epected to be The proportion of left-handed members at the end of year 5 is therefore October, 8 Page

13 Question # 3. Answer: B.5q5 q5 p5.5q Question # 3. Answer: C 3.5 p [6] p [6]+ =.5 p 64 ( 3 p [6] - 3 p [6]+ ) æ = l ö 65 ç è ø l 64.5 æ = 46 ö è ç 5737 ø =.5466 æ ç è.5 l 64 l [6] - l 64 l [6]+ ö ø æ ö è ç 96ø Question # 3.3 Answer: B eé ë 58 ù û + = e [58]+ +.5 e [58]+ = p [58]+ (+ e 6 ) = p [58]+ + e 6 = l 6 l [58]+ e6 p 6 = 3548 é ê ë ù -ú p 6 û ( / 394) = =.338 eé ë 58 ù û + = =.6 October, 8 Page 3

14 Question # 4. Answer: A E[ Z] A E A ()(.36987) (.576)(.6567).489 EZ [ ].4954 which is given in the problem. Var Z E Z E Z ( ) [ ] ( [ ]) SD( Z) An alternative way to obtain the mean is E[ Z] A A Had the problem asked 4: for the evaluation of the second moment, a formula is 4 6 E[ Z ] ( ) A ( v ) p A 4: 4. Question # 4. Answer: D Half-year 3 4 PV of Benefit.5 3, v (3,)(.9) 75, 9 33, v (33,)(.9) 77, , v (36,)(.9) 77, , v (39,)(.9) 76, 86 PV 77, if and only if () dies in the nd or 3 rd half years..5 Under CF assumption,.5 p.5 p.5 (.84).965 and.5.5 p.5 p.5 (.77).8775 Then the probability of dying in the nd or 3 rd halfyears is.5 p.5 p.5 p.5 p (.965)(.835) (.84)(.5).794 October, 8 Page 4

15 Question # 4.3 Answer: D A q v q q v q q v 3 6 ( 6) 6:3 6 ( 6)( 6) q..99 A q v ( q ) v : ( q6) v ( q6) v when v = /.5. The primes indicate calculations at 4.5% interest. A q v ( q ) q v ( q )( q ) v 6: (.7).99(.983) Question # 4.4 Answer: A Var( Z) E( Z ) E( Z) E( Z) E (. T )(. T) E (. T) 4 4 f () T t dt dt (. t) 4.t 4 ln (. t) ln(9) ( ) {(. ) [(. ) ] } [(. ) ] E Z E T T E T Var Z 4 (. t) ft ( t) 4. (. t) ( ). (.7465) October, 8 Page 5

16 Question # 4.5 Answer: B Outcome (z) Prob (p) p z p z v ,63.9 E( Z) F E Z Fv..Fv.( Fv ).76 35, F Var( Z) 35, F ( F) 34, F.44F Take derivative with respect to F. Derivative.848 F 4.43 Set = and solve; get F = 4.43 /.848 = 5 It should be obvious that this is a minimum rather than a maimum; you could prove it by noting that the second derivative =.848 >. Note: the result does not depend on v. If you carry v symbolically through all steps, all instances cancel. Question # 4.6 Answer: B Time Age SULT q Improvement q factor 7.43.% % %.86 v / EPV v v v 3, [ (.87) (.98893)(.86) ] 9.85 October, 8 Page 6

17 Question # 4.7 Answer: C We need to determine q q 9 l 93 l 93.5 l 93 l 95.5 l 93 ( l 95.5 d 95) 85 [6.5(4)].345 l l l, where l9,, l93 85, l l 95 l p d , l q l , p., and d 95 l 95 l Question # 4.8 Answer: C SULT Let A 5 designate A 5 using the Standard Ultimate Life Table at 5%. SULT APV (insurance) q5 p5 A Question # 4.9 Answer: D A A A A 35 35:5 35: A 5.7 A5.5.4 October, 8 Page 7

18 Question # 4. Answer: D Drawing the benefit payment pattern: 3 E[ Z] E A E A E A 3 3 Question # 4. Answer: A Var Z A A 5, : n : n A A A A : n : n : n n A A A : : : A n n n n () A : n A n () A A A A : n : n : n : n A : () A : A n n n () : n n : n () A : A n n V Z A A A 5, Var Z ()(36) (9) (58)(9) Therefore, Var Z 5, 36, 43,68,74 43,385. October, 8 Page 8

19 Question # 4. Answer: C so that Var Z3 4Var Z Var Z 4 CovZ, Z Z Z Z 3 Cov Z, Z E Z Z E Z E Z (.65)(.75) where 3 Var Z 4(46.75) (.65)(.75) Question # 4.3 Answer: C A v p q payment year 3 Lives years Die year 3 4 v 3 p65 q65 3 payment year 4 Lives 3 years Die year 4 3 (.9) (.9) (.) (.9)(.9)(.88)(.4) The actuarial present value of this insurance is therefore October, 8 Page 9

20 Question # 4.4 Answer: E Out of 4 lives initially, we epect l 6, p survivors l 6 96, 634. The standard deviation of the number of survivors is p p To ensure 86% funding, using the normal distribution table, we plan for The initial fund must therefore be F (64)(5) 389, Question # 4.5 Answer: E..6.4 t t t t t t t E Z b v p dt e e e.4 dt.4.8.8t.4 e dt t.4t.t.4.4 E Z b.4 t v t p t dt e e e dt. 3 Var Z October, 8 Page

21 Question # 4.6 Answer: D 3 A vq v p q v p p q :3 where: v.4 q 5 p q p q q q So: A 5 :3.6 Question # 4.7 Answer: A The median of K 48 is the integer m for which P( K m).5 and P( K m) This is equivalent to finding m for which l l.5 and.5. l 48m 48m l Based on the SULT and l 48 (.5) (98,783.9)(.5) 49,39.95, we have m 4 since l 49, and l 49, APV 5 A 5 E A 5 A 5 E E A So: (.343) October, 8 Page

22 Question # 4.8 Answer: A The present value random variable PV =,,e -.5T, T is a decreasing function of T so that its 9 th percentile is,,e -.5 p where p is the solution to.4t - dt =. p ò. p æ ò.4t - dt = -.4 t - ö è ç -ø p = 4 p æ =.4 - ö è ç pø =.,,e = 8,873.8 Question # 4.9 Answer: B Ming Ming q.8q SULT.664 p Ming Ming Ming A vq vp A SULT (.5) (.664) (.5) ( )(.6984).595 Ming, A 59, 5 8 October, 8 Page

23 Question # 4. Answer: B Var( Z). E[ Z] v p ( p ).v p (.57). ( i) ( i) ( i) 4.3 i.6. Question # 4. Answer: C The earlier the death (before year ), the larger the loss. Since we are looking for the 95 th percentile of the present value of benefits random variable, we must find the time at which 5% of the insureds have died. The present value of the death benefit for that insured is what is being asked for. l l l , l 94, 8. 94, ,.3 So, the time is between ages 65 and 66, i.e. time and time. l l l 94, , l 94, , t / The time just before the last 5% of deaths is epected to occur is: =.8893 The present value of death benefits at this time is:.8893(.5), e 35,88 October, 8 Page 3

24 Question # 4. Answer: C t d Ia s Ia 4 : s sp t 4v ds t t p4v 4 : t dt At t.5,.5 E.5 p p v (.5.9)( ) 6.39 E Question # 5. Answer: A p v () () E( Y ) a e e a.6 e.7.5 e T e Y E( Y ) T (.) Pr[ Y E( Y)] Pr( T 34.9) e.75 t Question # 5. Answer: B A E n : n A A E A n : n n.3 A (.35)(.4) A.6 : n : n A A E a : n : n n n : An :.5.9 d (.5 /.5) a a E : n : n n October, 8 Page 4

25 Question # 5.3 Answer: C a y a a y a y Ay a y Ay 9.4 Ay.34.7 A A A y y y.34 A.9 y Ay.5 Question # 5.4 Answer: A e 4 5 So receive K for 5 years guaranteed and for life thereafter., K a a a 5 5(.) 5 t e e e ( )5.5 5 a4 5 E4a45 e e 7.44, K Question # 5.5 Answer: D Under the Equivalence Principle 6 Pa 5, a a P ( IA) 6: 6: 6: where ( IA) A A (.9) So 6: 6: 6: k k 5, a6 a6: 5, ( ) P 34, 687 a ( IA) : 6: October, 8 Page 5

26 Question # 5.6 Answer: D Let Y i be the present value random variable of the payment to life i. A A A..45 E[ Yi ] a.55 Var[ Yi ] d d.5 /.5 Then Y Yi is the present value of the aggregate payments. i E[ Y] E[ Y ] 55 and Var[ Y] Var[ Y ] i F55 F55 Pr[ Y F] Pr Z F Question # 5.7 Answer: C ( m ) a a v p m a () 3 35:3 35: :3 Since A A E 35:3 35:3 d d A35 3 E35 A65 3 E35 d E v p a 35: , then 3 3 A35 v 3 p 35 A65 v 3 p 35 d (.88 (.7)(.498)).7 (.4 /.4) i () a (.7) :3 4 () a ,38 35:3 October, 8 Page 6

27 Question # 5.8 Answer: B a a v p a 5:6: 5:6 5:6 7:8 a v p p a 5: :8 a E 8 a 5:6 5 7:8 6 75, , Question # 5.9 Answer: C a vp a k vp a k a : : : n n n : n Therefore, we have a [ ]: n.67 k.5 a.854 n : Question # 5. Answer: C The epected present value is: a 5 E 5 55a The probability that the sum of the undiscounted payments will eceed the epected present value is the probability that at least 7 payments will be made. This will occur if (55) survives to age 7. The probability is therefore: 9, p , October, 8 Page 7

28 Question # 5. Answer: A a vp a S S SULT S SULT SULT 45 tdt ( 45 t.5) dt ( 45 t ) dt (.5) dt S SULT.5 98, p e e e e p e e 99, a S S SULT 45 vp45a46 (.5) ( )(7. 676) 7. a 7 S 45 Question # 6. Answer: D The equation of value is given by Actuarial Present Value of Premiums = Actuarial Present Value of Death Benefits. The death benefit in the first year is P. The death benefit in the second year is P. The formula is Pa A P( IA). Solving for P we obtain 8: 8: 8: A P a ( IA ) 8: 8: 8: a p 8: 8v (.96734)(.3667) A 8: vq8 v p8q (.96734)(.3667) ( IA) vq 8: 8 v p8q8 () P D October, 8 Page 8

29 Question # 6. Answer: E Ga, A G( IA).45G.5Ga a : : : : : (, )(.794) (6.8865) G (.5)(6.8865) Question # 6.3 Answer: C Let C be the annual contribution, then C E a Let K 65 be the curtate future lifetime of (65). The required probability is a 45: Ca 45: E45 a a 65 45: Pr a Pr a Pr a a Pr a E a E 65 K65 K65 K65 K : 45 Thus, since a 3.46 and a 3.8 we have a K65 K l86 57, Pr Pr p l 94, October, 8 Page 9

30 Question # 6.4 Answer: E Let X be the present value of a life annuity of / per month on life i for i,,...,. i Let S i X i be the present value of all the annuity payments. () () A6 6 ().475 E[ X i ] a.967 d.583 Var X i A6 A6.5 (.475) d (.583) ES [ ] ()(8)(.967) 366,936. Var( S) ()(8) (3.555) 85, 8,54 With the normal approimation, for Pr( S M).9 M E[ S].8 Var( S) 366, ,8,54 378, , So Question # 6.5 Answer: D Let k be the policy year, so that the mortality rate during that year is q3 k. The objective is to determine the smallest value of k such that k k v p ( P ) v p q () P k 3 3 k 3 3k 3 3k.7698 q9k q.47 9k vq 9 k 6 k 3 Therefore, the smallest value that meets the condition is 33. October, 8 Page 3

31 Question # 6.6 Answer: A Epected Present Value of Benefits: 5( /) /.6 75(5 /) /.6, (8 /) / Epected Present Value of Premiums: [ (87 /) /.6 (7/) /.6 ] P.4446 P. The annual premium is P = 98.4/.4446 = Question # 6.7 Answer: C 3 There are four ways to approach this problem. In all cases, let π denote the net premium. The first approach is an intuitive result. The key is that in addition to the pure endowment, there is a benefit equal in value to a temporary interest only annuity due with annual payment π. However, if the insured survives the years, the value of the annuity is not received. a, E a p a. 4: 4 4: 4 6% Based on this equation,, E4, v,, 88 p a a s The second approach is also intuitive. If you set an equation of value at the end of years, the present value of benefits is, for all the people who are alive at that time. The people who have died have had their premiums returned with interest. Therefore, the premiums plus interest that the company has are only the premiums for those alive at the end of years. The people who are alive have paid premiums. Therefore s,. The third approach uses random variables to derive the epected present value of the return of premium benefit. Let K be the curtate future lifetime of (4). The present value random variable is then K s v, K K Y, K a, K K, K a, K K. a a, K October, 8 Page 3

32 The first term is the random variable that corresponds to a -year temporary annuity. The second term is the random variable that corresponds to a payment with a present value of a contingent on surviving years. The epected present value is then a p4 a The fourth approach takes the most steps.. 4: 9 9 k k k ( i), 4: 4 4, k k 4 k 4 k k d a E v s q E v q,,, E4 q4 da v 4: p4 d v, E4 a 4: p4 d, E a p a. 9 k E4 q4 v q k k 4 E4 q4 A4: d k d 4 4: 4 6% Question # 6.8 Answer: B a a 6: 6: Annual level amount Question # 6.9 Answer: D 4 5a 5a 6: 6: a a : A A E 5: 5: a.848 5: APV of Premiums = APV Death Benefit + APV Commission and Taes + APV Maintenance Ga, A.Ga.3G 5a 5 5: 5: 5: 5: 8.55G 4.666G G G October, 8 Page 3

33 Question # 6. Answer: D Actuarial PV of the benefit 5.85 a.77 :3 Level Annual Premium 56.5 a : p.77.6 (.6) p.93 Actuarial PV of the benefit =.5.975(.93).975(.93) 5.85,.6 (.6) (.6) 3 q q.9 p.9 Question # 6. Answer: C For calculating P A5 vq 5 vp 5 A5 v(.48) v(.48)(.39788) a A / d P A / a For this particular life, A5 vq5 vp5 A5 v(.48) (.48)(.39788).437 a A / d Epected PV of loss = A5 Pa (5.33).48 October, 8 Page 33

34 Question # 6. Answer: E, in the solution is the, death benefit plus the death benefit claim epense. A da d(.).3755 Ga, A.65G.Ga 8 a, A 8 a, (.3755) 8(.) G a.65. a..65.(.) K Let Z v denote the present value random variable for a whole life insurance of on (). Let Y denote the present value random variable for a life annuity-due of on (). ak L, Z.65G.GY 8 Y GY, Z (.9 G) Y.65G 8 K K v, v (.9 G).65G 8 d.9g.9g d d K, v.65g 8.9G Var( L) A ( A), d.376(,394,6) ( ), 88,86 Question # 6.3 Answer: D If T45.5, then K45 and K45..9(35.386) d K 45 K 45 L, v G(.) a G(.8.), v.9ga.7g 4953,(.58468).9 G(8.773).7G G ( ) / (7.4956) 5. E( L), A (.) Ga (.8.) G (, )(.56) (.9)(5.)(7.86) (.7)(5.) E( L) 4.87 October, 8 Page 34

35 Question # 6.4 Answer: D, A P[ a.5 a ] 4 4: 4:, A4, (.6),6 P a.5 a (4.97) : 4: where a 4: E 4 a : There are several other ways to write the right hand side of the first equation. Question # 6.5 Answer: B Woolhouse: W a (4) 3 UDD: UDD a (4) (4) a (4).9(3.46) and A d a (.476)(3.46).8358 ( W ) (.8358) P ( UDD) (.8358) P P P ( UDD) ( W ) October, 8 Page 35

36 Question # 6.6 Answer: A P, a d, a a 3: 3: 3: 3: A d a.5(4).3 3: 3: Ga, A ( 5 a ) (.33G.6 G a ) 3: 3: 3: 3: 4G,(.3) [ 5(4)] (.33G.84 G).83 G 3,9 G 48 Question # 6.7 Answer: E Future epenses at.8g 5 Epense load at P 3.64 (.8G 5) P e P 58.8 e e P :3 Question # 6.8 Answer: D P 3, a PA 4 4: P 3, a 4 / A4: 3, (5.4649) / (.46346) 66,363 October, 8 Page 36

37 Question # 6.9 Answer: C Let be the annual premium, so that a 5 A5.a 5.9 A a k k Loss at issue: L v. a v / d.9 k. Var L A A d 5 5 (.5467)( ) (.5467)(.54).33 Question # 6. Answer: B EPV(premiums) = EPV(benefits) 3 P vp v p P vq v p q v p q P P P.9588 P 56.3 P 459 Question # 6. Answer: C A P a A P A P a A 75:5 5 75:5 75:5 75:5 5 75:5 75:5 A A A :5 75:5 75:5 A75:5 a (.7) / :5 d 59 So P (.) October, 8 Page 37

38 Question # 6. Answer: C Let the monthly net premium ()., A45 ().4665 () a 45: i.48 i, A45, A45 (.48)(5,6) 5, a () a () E () 45: 45: (.35994).643 5, Question # 6.3 Answer: D Ga :3 APVBenefits + epenses APV gross premium :3 :5 FA a G a a FA 3 3a G a.6. a. a :3 :3 :5 FA 3 3(5.396) (4.45).(.39) FA FA FA h 44.7 October, 8 Page 38

39 Question # 6.4 Answer: E In general, the loss at issue random variable can be epressed as: Z P P L Z P Y Z P Z Using actuarial equivalence to determine the premium rate: A.3 P.3 a.3 /.7 P.3 Var( L) Var Z Var Z Var Z P *.6 Var( L*) Var Z Question # 6.5 Answer: C Need EPV(Ben Ep) EPV(Prem) 8 EPV(Prem) Ga 8.9G 55: EPV(Ben Ep)=, a 3a () 55 55, E a 3a () m, E55 a65 3a55 m, (.5934) (6.599) 4 98, 4.83 Therefore, 98, G 8 G,36 October, 8 Page 39

40 Question # 6.6 Answer: D EPV (Premiums) = Pa9 P( a 9 ) (4.835) P EPV(Benefits) A9 (.7537) Therefore, P Question # 6.7 Answer: D EPV(Premiums) EPV(Benefits) EPV(Premiums) 3P a P E a ( ) P e 3P.8 3P Pe P EPV(Benefits),, A 5, E A ( ),, 5,e.3.8.3,,.7 5,e.9 35, P 35, ,783.5 P 9.66 P,39.6 Question # 6.8 Answer: B Ga A.5G.5Ga 5 5a 4:5 4 4:5 4:5 a a E a 4 4: (.783)(7.86) (4.54) G (4.54) October, 8 Page 4

41 Question # 6.9 Answer: B Per equivalence Principle: Ga,A.4G 5.Ga 5a a, da.4(77) 5.(77) a 5a a35, 78 5 a ,.35 Solving for a 35, we have,858,858 a Question # 6.3 Answer: A The loss at issue is given by: L v.5 G.5 G a G a K K K K K v v.5g.95g d.95g K G v.5g.95 d d Thus, the variance is Var L A A / / October, 8 Page 4

42 Question # 6.3 Answer: D A35 e e A a P A A a The annual net premium for this policy is therefore,.3654,36.54 Question # 6.3 Answer: C Assuming UDD Let P = monthly net premium EPV(premiums) Pa P a EPV(benefits),A P P i i, A, da.6.6, 9.9 log P October, 8 Page 4

43 Question # 6.33 Answer: B The probability that the endowment payment will be made for a given contract is: 5 p 5 ep.t dt ep.t 5 ep.(5).54 Because the premium is set by the equivalence principle, we have E L. Further, 5 L v 5 p 5 p Var 5,,94,39, Then, using the normal approimation, the approimate probability that the aggregate losses eceed 5, is 5, P L 5, PZ P( Z.3).3,94,39, Question # 6.34 Answer: A Let B be the amount of death benefit. EPV(Premiums) 5a 5(4.649) EPV(Benefits) B A (.343)B 6 6 EPV(Epenses) (.)(5) (.3)(5) a (.)(5) (.3)(734.55) EPV(Premiums) EPV(Benefits) EPV(Epenses) (.343)B (.343)B B 3, 94 6 October, 8 Page 43

44 Question # 6.35 Answer: D Let G be the annual gross premium. By the equivalence principle, we have Ga, A.5G.4Ga so that G , A, a Question # 6.36 Answer: B By the equivalence principle, 45a,A Ra : : : where ( ).4 (.) A : e e.33. a : ( ) (.) e e Solving for R, we have.33 R 45, October, 8 Page 44

45 Question # 6.37 Answer: D By the equivalence principle, we have Ga 5,A a A 35: so 5, A35 a35 5, G a : Question # 6.38 Answer: B Let P be the annual net premium P A (.9) n : a a : n : n where An a (.5) A A d (.5) i A : A n : n ne.5.9 A : n :.956 : : n : n : n A n.5 a n : (.956.7) Therefore, we have.9 P October, 8 Page 45

46 Question # 6.39 Answer: A Premium at issue for (4): Premium at issue for (8): A a A a Lives in force after ten years: 98,576.4 Issued at age 4:, p4, , ,84., p8, Issued at age 8: 75, 657. The total number of lives after ten years is therefore: , The average premium after ten years is therefore: ( ) ( ) 9.3 5, October, 8 Page 46

47 Question # 6.4 Answer: C * Let P be the annual net premium at +. Also, let Ay be the epected present value for the special insurance described in the problem issued to ( y ). k k *.3 k k Pa v q A We are given k k * k k a.3 v q A Which implies that * vpa vq vp A.3.3 Solving for A, we get * A * v v.843.3v.95 Thus, we have.843 P Question # 6.4 Answer: B Let P be the net premium for year. Then: P.Pvp, vq (.)(, ) v p q.. (.)(.99)(.) P.99, P (.5) October, 8 Page 47

48 Question # 6.4 Answer: D The policy is fully discrete, so all cash flows occur at the start or end of a year. Die Year ==> L v Die Year ==> L v 35.8( v) Survive Year ==> L v 35.8( v v ) 58.3 There is a loss if death occurs in year or year, otherwise the policy was profitable. Pr death in year or e.3 Question # 6.43 Answer: C epenses :4 3:9 APV G Ga a.g 4.5G a 4a 3:5 3: G a.g 4.5G a 4a, A 3:5 3:5 3: 3:, A 4 4a G.85a. 3: 3: 3:5, A, 3: A 3: E 3,( ) G October, 8 Page 48

49 Question # 6.44 Answer: D Let P be the premium per of insurance. Pa P I A E A 5: 5: 5 6 a a E a : A da P a I A E A 5: 5: 5 6 E A P a I A 5 6 5: 5: P.8 October, 8 Page 49

50 Question # 6.45 Answer: E 56 56, T T L v 56a, e T Since L is a decreasing function of T, the 5 th percentile of L is L (t) where t is such that Pr[T 35 t] =.5 or Pr[T 35 > t] = t , , t 35 74, t (8 35) s s ( s) 8s , ,86.3s 75,657.( s) s.454 t L (45.454), e ln(.5) ln(.5) 56 ln(.5) Question # 6.46 Answer: E Let P be the premium per of insurance. Pa.53P v p a 55: a a v p a v p a : P.53P P P 8. October, 8 Page 5

51 Question # 6.47 Answer: D Ga, E a.5ga.7g 7: 7 8 7: G (, )(.5994)(8.5484).5 G(7.649).7G G 66,383.54, G 66, Question # 6.48 Answer: A Actuarial present value of insured benefits: , 5, P 5, P 3, Question # 6.49 Answer: C i Ga, A.4Ga () () 4: 4 4: a () a ( E ) () () 4: 4: (.36663) (,)(.48)(.6)+ G G/ = 86.6 October, 8 Page 5

52 Question # 6.5 Answer: A A a 35, P, Benefits paid during July 8:,, q, Premiums payable during July 8:, ( q ) , , Cash flow during July 8: 39 5,858 46,948 Question # 7. Answer: C : V 5, A E A (875) a 5, [.893 (.68)(.98)] 875[8.55],5 October, 8 Page 5

53 Question # 7. Answer: C V V Year : V P i q V p V P(.).5( V ).85( V ).P3 V Year : V P i q ( V ) p () (.P 3 P)(.).65( ).835().3P P 5.3 October, 8 Page 53

54 Question # 7.3 Answer: E i (4).8 means an interest rate of j. per quarter. This problem can be done with two quarterly recursions or a single calculation. Using two recursions: [.5V 6(.)](.) () V [ V 54](.) V V [.5V ](.) () 898 [.5V ](.) V 73. Using a single step, V is the EPV of cash flows through time.75 plus.5 the EPV of cash flows thereafter (that is, V )..75 E times V () (6)(.) ( (.) 898(.) 898(.) 898(.).7) 73.5 October, 8 Page 54

55 Question # 7.4 Answer: B EPV of benefits at issue A4 4 E4( A5).6 (4)(.57949)(97.8) EPV of epenses at issue ( a4 ) (7.4578) ( ) / a 854.3/ G. 47. EPV of benefits at time A4 4 E 4 A (4)(.6879)(97.8) 68.3 EPV of epenses at time ( a 4) (8.343) 83.4 Gross Prem Reserve Ga (8.343) Question # 7.5 Answer: A Net Amount at Risk = V Epected Deaths = (, ) q 999(.96) Actual Deaths = 6 Mortality Gain/Loss = (Epected Deaths (9.5 6)(996.5) 339 Actual Deaths)(Net Amount At Risk) October, 8 Page 55

56 Question # 7.6 Answer: B d ( t V ) G E S 9.6 V ( ) where G, E, S and are evaluated at t 9.6 and dt t9.6 where S includes claims-related epenses. d ( t V ) 45 (.)(45) (6, )(.) 9.6 V (..5) V dt t V V d V V V 9.8. ( t ) 6.68 (.)[ ] dt t V Question # 7.7 Answer: E q p V v p V v q b, where b, is the death benefit during year q 4.5(.456).334 q.5(.456) ( 4V P)(.3) q 4b V p 4 (, )(.3).456(, ) V, V (.3) (.97666)(, 737.5) (.3) (.334) (, ), ,9 4.5V can also be calculated recursively:.5 p 4.5(.456) (, )(.3).8(, ) / (.3) 4.5V,9.8 The interest adjustment on the death benefit term is needed because the death benefit will not be paid for another one-half year. October, 8 Page 56

57 Question # 7.8 Answer: B Net Premium, A / a, (.3495) / G 8.93(.3) Let * L be the present value of future loss at issue for one policy. L, v ( G 5) a.5g * K K v d K K, v (5.5 5).5(5.5) K, v K 4, 63.5v 469. E( L ) 4, 63.5 A , 63.5(.3495) * 6 Var( L ) (4, 63.5) A A (4, 63.5) , 536, 763 * 6 6 Let L be the aggregate loss for 6 such policies. E L E L * ( ) 6 ( ) 6(.34) 684 Var L Var L * ( ) 6 ( ) 6(5,536, 763) 3,3, 57,8 StdDev.5 ( L) 3,3, 57,8 57, 637 4, 684 Pr L 4, , 637 Question # 7.9 Answer: E Gross premium = G Ga A.5 a.g.5ga a. G A a October, 8 Page 57

58 A45 a45 (.56) (7.86) G a..95(7.86). 45 There are two ways to proceed. The first is to calculate the epense-augmented reserve and the net premium reserve and take the difference. A45 (.56) The net premium is 7. a The net premium reserve is A55 7.a55 (.354) 7.(6.599) 97.4 The epense-augmented reserve, which is the gross premium reserve, is A [.5(3.6).5( /) ] a 3.6a (.354) ( )(6.599) 7.77 Epense reserve is = 5 The second is to calculate the epense reserve directly based on the pattern of epenses. The first step is to determine the epense premium. The present value of epenses is [.5G.5( /) ] a.g.( /).558(7.86) The epense premium is 5.97/7.86=4.4 The epense reserve is the epected present value of future epenses less future epense premiums, that is, [.5G.5( /) ] a 4.4a.58(6.599) There is a shortcut with the second approach based on recognizing that epenses that are level throughout create no epense reserve (the level epense premium equals the actual epenses). Therefore, the epense reserve in this case is created entirely from the etra first year epenses. They occur only at issue so the epected present value is.(3.6)+.(/)+ = 8.3. The epense premium for those epenses is then 8.3/7.86 =.585 and the epense reserve is the present value of future non-level epenses () less the present value of those future epense premiums, which is.585(6.599) = 5 for a reserve of 5. October, 8 Page 58

59 Question # 7. Answer: A NS NS. q q e.95 Then the annual premium for the non-smoker policies is P NS, where P vp, vq, v p q 3, v p p NS NS NS NS NS NS NS P P NS NS, (.96)(.95), (.857)(.95)(.95) 3, (.857)(.95) (.96)(.95), 5 And then S P 4,5. q q e S S..5( ).43 EPV ( L ), vq, v p q 3, v p p P P vp S S S S S S S S S, (.96)(.43), (.857)(.857)(.43) 3, (.857)(.857) 4,5 4,5(.96)(.857) 3, 7 Question # 7. Answer: D a ( A ) / d (.4) / (.5 /.5).6 a () ().6 / 4.4 V, A a a (3)(.5) V, (.4) (.6) (.4)(8.5) V 7 October, 8 Page 59

60 Question # 7. Answer: C Simplest solution is recursive: V since the reserves are net premium reserves. q 7 (.7)(.43).789 V ( 35.68)(.5) ()(.789) Prospectively, q (.8)(.67).9336; q 7 7 (.9)(.38).773 A (.9336) v (.9336)(.773) v (.9336)(.773)(.4758) v.4497 a A / d (.4497) / (.5 /.5).786 V ()(.4497) (.786)(35.68) 9.85 Question # 7.3 Answer: A Let P =.53 be the monthly net premium per of insurance. i () V, A A Pa 55: 55: 55:,.48(.47).5934 ()(.53)(7.83) 38, Where A A E A a 55: 55: 55: 55: E () a () a () 55: 55: E 55.(8.9).4665(.5934) 7.83 October, 8 Page 6

61 Question # 7.4 Answer: C Use superscript g for gross premiums and gross premium reserves. Use superscript n (representing net ) for net premiums and net premium reserves. Use superscript e for epense premiums and epense reserves. g P (given) g 45 g.58p 45.P 5 a e P a 45.58(977.6) 45 [.(977.6) 5] Alternatively,, A n 45 e g n P P P P 6.63 a 45 e g e V.P 5 a P a [.(977.6) 5](7.45) 6.64(7.45) 97 Alternatively, n n V, A P a 5 5 5, (.893) 85.97(7.45) V, A 5.P P a g g g e g n 5 5 5, (.893) [5.(977.6) 977.6](7.45) V V V 97 Question # 7.5 Answer: B, K 45, K 45 L v Pa v Pa 445,(.58468) 8.77P P (5, , 45) / A da (.5 /.5)(3.45) V, A Pa (, )(.3693) (6.5)(3.45), October, 8 Page 6

62 Question # 7.6 Answer: B T T L A v.a v T 6 6 T T Var[ LA] Var[ v ].455 Var[ v ] T 6 T 6 LB v.6a v T 6 6 T 6 6 Var[ LB ] Var[ v ] (.6398) Question # 7.7 Answer: E In the final year: ( 4V P)( i) b5( q 68) ( p 68) Since b5, this reduces to ( 4V P)( i) (.6 P)(.4) P.3654 Looking back to the th year: ( V P)( i) b( q 55) V ( p55) (5.3654)(.4) 4(.5) V(.85) V 4.89 Question # 7.8 Answer: A This first solution recognizes that the full preliminary term reserve at the end of year for a 3 year endowment insurance on (4) is the same as the net premium reserve at the end of year 9 for a 9 year endowment insurance on (4). Then, using superscripts of FPT and NLP to distinguish the reserves, we have V V ( A P a ) or FPT NLP 9 5: 4:9 5: [ (.848)] 8 a 5: a :9 October, 8 Page 6

63 where a a E a A 4:9 4: E4 v l 4 99, 85.9 P 4: (.876)(.83) d(5.664).5495 l 9, 8.4 (.4946) FPT Alternatively, working from the definition of full preliminary term reserves as having V and the discussion of modified reserves in the Notation and Terminology Study Note, let be the valuation premium in year and be the valuation premium thereafter. Then (with some of the values taken from above), vq.59 4 APV (valuation premiums) APV (benefits) E ( a ) A 4 4:9 4: (5.664) Where E 4 4:3 (.57) v.9588 A A E ( E )( A ) (.57864)(.488).437 V A a (.38844) 6.(.847) 8 FPT 5: 5: October, 8 Page 63

64 Question # 7.9 Answer: E No cash flow beginning of year, the one item earning interest is the reserve at the end of the previous year Gain due to interest = (reserves at the beginning of year)(actual interest anticipated interest) (899.8)(.4.3) 899. Question # 7. Answer: A V.96G 5 (.5) q (, ) p V G 5 (.5) (.9)(, ) (.9)(7) (.5)(.96) G (.5)(.96) G 5.4 G 97.8 Question # 7. Answer: E V i p V q V.4 (.5) (49.78).4553() V 5.9 Question # 7. Answer: D 3 APV future benefits.4v.5.96v ( ) v 63.5 APV future premiums 3(.96 v) E[ 3L] October, 8 Page 64

65 Question # 7.3 Answer: D p V L A A d V L p A A d A vq vp A (.973) (.67) (.973) (.67)(.5536).5969 A v q v p A (.973)(.67) (.973)(.67)(.3783).99 k Var k L (.99) (.5969).34.8 (.3783) (.5536).383 L Var Question # 7.4 Answer: A V A P a da P a a V a a P d a a Since a vpa substituting we get a a V a V vp a vp Solving for a, we get a V vp (.) October, 8 Page 65

66 Question # 7.5 Answer: D Let G be the annual gross premium. Using the equivalence principle,.9ga 4.4G, A4 3, (.6) 3 So G (8.4578).4 The gross premium reserve after the first year and immediately after the second premium and associated epenses are paid is,a.9g a 4 4,665.9( )(7.343) 73 Question # 7.6 Answer: C The epected end of the year profit B 7 V G WG 6.8 (.4736)(5) (.4736)( V ) Plug in given values, we have W 7 4, , , W Solving W, W % October, 8 Page 66

67 Question # 7.7 Answer: E We assume that we have assets equal to reserves at the beginning of the year. We collect premiums, earn interest, and pay the claims which gives us the assets at the end of the year: j (98) , 66 8,6 j Actual total reserve at the end of the year = (98 7)(7.77) 7,. Since there is no gain or loss that means that the assets must equal the reserves, so: 7, 8.7 4, 66 j 8.5% 8,6 Question # 7.8 Answer: A If G denotes the gross premium, then A35 3a35 7 (.9653) 3(8.978) 7 G 5..96a.6.96(8.978).6 So that, 35 R A (3.96 G) a (.) (3.965.)(8.8788) 77.3 Note that S as per definition of FPT reserve. October, 8 Page 67

68 Question # 7.9 Answer: D 55: 55: a a IA V a64 A a 64: 64: 94, ,8.5.5,866 Question # 7.3 Answer: C.5(.6) P V L# B A A.55 8 A A.55 d.6 A.55 A.67.5(.6) (.6).875(.6) V L # A A October, 8 Page 68

69 Question # 7.3 Answer: D We have Present Value of Modified Premiums = Present Value of level net premiums 5 vq a P E a Pa 5: 45: 5:4 5 P a P E a vq 5:4 45: P a vq 5: a a We are given that P.6 5: 5:.6(.87) (.4) For insurance of,, 33. Question # 7.3 Answer: C P P P where P g n e e is the epense loading P n A.893 a ,,,,, e g n P P P,8, 68 October, 8 Page 69

70 Question # 7.33 Answer: B V, A, P a FPT , A, A a58 a ,.7.4 d d Question # 7.34 Answer: D V V P i 5, V V q 5, P i q P i q P P V V P i 5, V V q P( i) (5, ) q 5, 5, 5,.5 5,.5.5 5,.5 5, Solving for P, we get 6,437.5 P 8, October, 8 Page 7

71 Question # 7.35 Answer: C 4 V ,q53 q The easiest way to calculate the profit is to calculate the assets at the end of the period using actual eperience and the reserves at the end of the period using the reserve assumptions. The difference is the profit. Ending assets 4885(55 34)(.6) 4(,) 3,58,67. Ending reserves (4885 4)(666.87) 3,6, Profit = 3,58,67. 3, 6, ,73.33 Alternatively, we can calculate the profit by source of profit. For eample, if we calculated the gain by source calculation, done in the order of interest, epense, and mortality, the profit for policy year 4 is (4885)[(55 3)(.) (3 34)(.6) (, )(.68 4 / 4885)] 68,73.37 Question # 7.36 Answer: B Since G is determined using the equivalence principle, V e P G87.5G.3 e Then, V G G.97 October, 8 Page 7

72 Question # 7.37 Answer: D d V V P e ( S E V ) dt t t t t t t t t At t 3.5, 9.5 V (, V) V (.5.38) V 65 Question # 7.38 Answer: E G G HA 8 a a HA : 45.95a 45: ( ) A G H ( ) ( ) A45 g ( ) f ( ) 5.8 October, 8 Page 7

73 Question # 7.39 Answer: D æ G =.35G +. è ç.8.65g = G = 7.6 Question # 7.4 Answer: C ö.8 3 ø Epected epense financial impact = Epected epenses + Foregone interest +.7 =7 Actual epense financial impact = Actual epenses + Foregone interest = 8.5 Gain from epenses: = 6.75 October, 8 Page 73

74 Question # 7.4 Answer: A P A d da45 On a unit basis, Var( L ) A ( A ) A ( A ) ( 45) A45 ( A45) da45 ( da) da da A A The standard deviation of L.793 (,)(The standard deviation of L ) 5, 439 Question # 7.4 Answer: D V A ( P W ) a At issue, present value of benefits must equal present value of premium, so: A Pa W E a ( P W )(3.5498) P W P W PW( )(3.5498) ( W) W(.35994)(3.5498) W October, 8 Page 74

75 Question # 7.43 Answer: E a.4 V B B B a 4.8 Ga 5 5a B A, 9 9, A da G , G 96.8 g V 9,968.4 A 5a 96.8a A V g.4 da , V g, 7.8 Alternatively, the epense net premium is based on the etra epenses in year, so P e (3 5) / V e.6899(.4) 9.6 V V V g n e Question # 7.44 Answer: E L L 35 * L, A 965.3, L, * October, 8 Page 75

76 Question # 8. Answer: B Because it is impossible to return to state, p and p are the same, so t t.3( ) j t.5t s ds.5.3( ) dt ln j p p e e e t ep[ ( )] ep(.588).943 Note: The sum of.5.3( t t t ) is a form of Makeham s law and p could be calculated using the formula provided in the tables instead of integrating. Question # 8. Answer: D For t and h.5, p p.5 p p p Similarly.5 p.5.5 Then,.5 p For t.5 and h.5, p p.5 p p p [.3(.).9(.6) ].573 October, 8 Page 76

77 Question # 8.3 Answer: D possible transitions probability discounted benefits APV H Z.5 5v.9476 H L.4 5v H Z D.5(.7).35 v H L D.4(.6).4 v H H Z.9(.5).45 5v.48 H H L.9(.4).36 5v Sum of these gives total APV = October, 8 Page 77

78 Question # 8.4 Answer: D p l 9, , ( ) 5 5 ( ) l5 3 (3) ( ) p 5 d 5 / l 5 /, p 5 p 5 (3) (3) (3) q q p p p ln p ln(.5) d l p l p l (3) ( ) 3 ( ) 3 ( ) 5 ( ) l ( ) ( ) ln p5 l l 5 5 ln ( ) l 5 ln(.5) 9,365 ( 8, / 9,365) 984 ln(8, / 9,365) () ( ) ( ) () (3) d 5 l5 l5 d 5 d 5 9,365 8, () ( ) p 5 d 5 / l 5 8/9,365 p 5 p 5 () () 8,/9, q p p p 8, / 9,365.38, q, () 5 Note: This solution uses multi-state notation for dependent probabilities. There is alternative notation for these when the contet is strictly multiple decrement, as it is here. October, 8 Page 78

79 Question # 8.5 Answer: C There are four career paths Joe could follow. Each has probability of the form: p ( p )( p )(transition probability )(transition probability ) where the survival probabilities depend on each year s employment type. For eample, the first entry below corresponds to Joe being an actuary for all three years. The probability that Joe is alive on January, 6 is: (.9)(.85)(.8)(.4)(.4) (.9)(.85)(.65)(.4)(.6) (.9)(.7)(.8)(.6)(.8) (.9)(.7)(.65)(.6)(.).583 The epected present value of the endowment is 3 (,).583 / (.8) 4,35 Question # 8.6 Answer: E This is an application of Thiele s differential equation for a multi-state model. d ( s) The components of t V are dt ( s) Interest on the current reserve: V t Rate of premiums received while in state s: Rate of benefits paid while in state s: B Transition intensity for transition to state h, times the change in reserve upon transition (hold reserve ( ) ( ) for h and release reserve for s): sh ( V h V s ) t t t sd ( s) Similar to previous, noting that the reserve if dead is : ( ) Adding these terms yields the solution. t t V October, 8 Page 79

80 Question # 8.7 Answer: A Q..4.4 Q Let p =.75 be the probability of renewing. PV costs for Medium Risk: 3v (.) 3(.4) 6(.4) pv (.) 3(.) 6(.56) p v The present value of costs for the new portfolio is (.9)37 + (.)759 = 36.. The increase is (36./37) =.4, or 4%. Question # 8.8 Answer: B.5 Prob( H D in months) (.75..5)..75 You could do more etensive matri multiplication and also obtain the probability that it is H after or it is S after, but those aren t needed. Let D be the number of deaths within years out of lives Then D~binomial with n, p.75 PD ( 4) (.75) (.75).45 October, 8 Page 8

81 Question # 8.9 Answer: E Let A denote Alive, which is equivalent to not Dead. It is also equivalent to Healthy or Disabled. Let H denote Healthy. The conditional probability is: P( H and A) P( H ) P( H A), P( A) P( H ) P(Disabled) Where And (.5).5 ( ) ds ds P H p e e e.67 u ds ds u P(Disabled) p e e du.5u.5u.5 e (.) e du (.) e.5 du.5 (.) e. Then PH ( ).67 P( H A).83 P( H ) P(Disabled).67. Question # 8. Answer: A t p ep ds ep..s.5.5s ds t ep.5s.75s ep.5t.75t t t s s 3 p ep ep(.45).45 EPV, through time, g( t) dt, g n t n where g( t), t p t t p t e dt (3), e 3., 4 October, 8 Page 8

82 Question # 8. Answer: B Let be the person s age. 5 p 5 ep s s 5 ep (..5) ds ep[ 5(.5)] ep[.5].54 ds Question # 8. Answer: D Let P be the annual premium APV(premium) P v v.779p APV(benefits) v v v P Note: The term benefits v is the sum of 3 3 v for death benefits and v for maturity October, 8 Page 8

83 Question # 8.3 Answer: A By U.D.D.: () () (). q q q q..95 q q.5 () () () p p p (.9)(.5).755 ( ) () () q q q p q () ( ) () ( ) () Question # 8.4 Answer: C Path Probability SSS S.6 S S C S. S C S S. S C C S. Total across all paths:.5 Question # 8.5 Answer: D Intuitively: (A) (B) (C) (D) (E) A lower interest rate increases premium, but a higher recovery rate decreases premium, because there are lower projected benefits and more policyholders paying premiums. A lower death rate of healthy lives more pay premium lower premium A higher death rate of sick lives fewer benefits lower premium A lower recovery rate higher sickness benefits higher premium A lower death rate of sick lives higher sickness benefits higher premium A higher rate of interest lower premium A lower mortality rate for healthy lives may result in lower premium because more healthy lives are paying premium. October, 8 Page 83

84 Question # 8.6 Answer: C The desired probability is: 4 u ep ep ds 4.5u. e (.) e du 4..5u (.) e e du..5 e e.8..7 ds du The limits of the outer integral are to 4 because you must be disabled by 64 if you are to have been disabled for at least one year by 65. Question # 8.7 Answer: C The probabilities are: Sick t.5 Sick t (.95)(.5) (.5)(.6).3875 Sick t 3 (.95)(.95)(.5) (.95)(.5)(.6) (.5)(.6)(.6) (.5)(.3)(.5).46 EPV v v v 3, October, 8 Page 84

85 Question # 8.8 Answer: A The probability that Johnny will not have any accidents in the net year is: p e s s ds t t t, where t So that tln e dt ln and p e The probability Johnny will have at least one accident is therefore Note: The sum of is a form of Makeham s law and p could be calculated using the formula t t provided in the tables instead of integrating. Question # 8.9 Answer: A The probability of being fully functional after two years for a single television is: (.8)(.8) (.)(.6) (.8)(.).734. The number of the five televisions being fully functional has a binomial distribution with parameters of n 5 and p.734. The probability that there will be eactly two televisions that are fully functioning is therefore: 5 (.734).734 ()(.5364)(.963).79 3 October, 8 Page 85

86 Question # 8. Answer: E Probability 5 p 5 t t p dt 5 5.6t.8(5 t).6t.4.8t e.5e dt e.5e e dt 5.4.t.4.5. e (.5) e dt e e.76. Question # 8. Answer: C 5 p 5 = p ò m t +t 5-t p dt = e -(m +m 3 )t m e -(5-t )(m +m 3 ) ò dt 5 = e -(.+.)t -(5-t )(.3+.4) ò (.)e dt = ò e -(.3)t (.)e 5 5 -(5-t )(.7) = (.)e æ e ò.67(5) -ö e.67t dt = (.)e -3.5 è ç.67 ø = dt October, 8 Page 86

87 Question # 8. Answer: D Healthy lives probabilities: Probability of a Healthy life at time being Healthy at: Time :.9 Time : (.9)(.9) =.8 Probability of a Healthy life at time is Sick at time and then Healthy at: Time : (.5)(.3) =.5 Probability of being Healthy at time :.9 Probability of being Healthy at time : =.85 Sick lives probabilities: Probability of a Healthy life at time being Sick at: Time :.5 Probability of a Healthy life at time zero is Sick at time and then Sick at: Time : (.5)(.6) =.3 Probability of a Healthy life at time is Healthy at time and thensick at: Time : (.9)(.5) =.45 Probability of being Sick at time :.5 Probability of being Sick at time : =.75 APV (Healthy): 5(e e e ) = APV (Sick): 5(.5e e ) = Total APV: = 86.9 October, 8 Page 87

88 Question # 8.3 Answer: C q () = q () (t p ) () 53 q 54 () (t q 54 = q ) 54 - q () (t 54 = (- p ) 54 ) - q 54 (t p ) 53 = - q () 53 - q () 53 = =.945 q () = = Question # 8.4 Answer: A The probability of death by year 3: Epected number of deaths = æ () = ö è ç 5ø -.4 =.35 Variance of the number of deaths = æ Pr( X < 375) = Pr è ç X ö < ø = Pr(Z < -.) = F(-.) =.357 Question # 8.5 Answer: B d (3) (t 4 = l ) (3) 4 q 4 (t l ) (t 4 = l ) 4 (- q () 4 - q () 4 - q (3) 4 ) () (t If q 4 increases by., then the change in l ) (t ) 4 = -.l 4. The change in d (3) (t 4 = -.l ) 4 q (3) 4 = -. 5,. = -5 October, 8 Page 88

89 Question # 9. Answer: E Pr(last survivor dies in the third year) = p p 8:9 3 8:9 p8 p9 p8:9 3 p8 3 p 9 3 p 8:9 [(.9)(.8) (.6)(.5) (.9)(.8)(.6)(.5)] [(.9)(.8)(.7) (.6)(.5)(.4) (.9)(.8)(.7)(.6)(.5)(. 4)] (.7.3.6) ( ) Question # 9. Answer: E,, APV(benefits), A Ra.6Ra.7Ra where a a a :65, A R.3a.3a : : ,,, A R.3a.3a 65 65:65 65 a A a : ,53 R 68, Question # 9.3 Answer: A 65:6 t 65:6 t 65 t:6t p p p dt.t.8( t) e (.5) e dt.8.t.5e e dt e...8.5e.457 October, 8 Page 89