COURSE 3 MAY 2001 MULTIPLE-CHOICE ANSWER KEY

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1 COURSE 3 MAY 00 MULTIPLE-CHOICE ANSWER KEY E 6 B A 7 D 3 C 8 D 4 E 9 E 5 D 30 D 6 B 3 D 7 C 3 A 8 D 33 D 9 C 34 A 0 A 35 B B 36 E A 37 C 3 E 38 A 4 B 39 E 5 A 40 B 6 B 7 A 8 E 9 C 0 C E D 3 A 4 B 5 C COURSE 3: MAY 00-4 STOP

2 SOLUTIONS FOR MAY 00 COURSE 3 EXAM Test Question: Key: E For de Moivre s law, z o ω 30 t e 30 F 0 ω 30 L N M HG t t ω 30 b O gq P 0 I K J dt ω 30 ω Prior to medical breakthrough ω 00 e o After medical breakthrough e 30 e o ω so 30 e ω 08 o o

3 Test Question: Key: A 0. 5 L 00, 000v 4000a&& 77,079

4 Test Question: 3 Key: C Ε Ν Ε Ε Ν Λ Ε Λ Λ Var Ν Ε Var Ν Λ + Var E N Λ Λ Ε Λ + Var Λ + 4 Λ Λ Λ Distribution is negative binomial (Loss Models, 3.3.) Per supplied tables mean rβ b g Var rβ + β 4 b + β β rβ r g From tables 3 3 rbr + gb r + gβ bgb 3gb4g 4 p3 05. r ! b + βg 3! p 3 5 Λ

5 Test Question: 4 Key: E E N b gb g b. gb g b. gb g b. gb g. b gb g b gb g b gb g Var N E X E X Var X For any compound distribution, per Loss Models Var S E N Var X + Var N ce X h (30) (.64) + (30) d36. i 768 For specifically Compound Poisson, per Probability Models Var S λt E X (60) (0.5) (5.6) 768 Alternatively, consider this as 3 Compound Poisson processes (coins worth ; worth 5; worth b g 0, thus for each VarbSg 0), where for each Var X Processes are independent, so total Var is b g Var N E X b gb gb g b gb gb g b gb gb gb g Var

6 Test Question: 5 Key: D 0 d ib g b g V + 0P. q V 000A+ 0 p+ 9 b gb. 06g b 000g a 0. && / 06. so b A P a&& g

7 Test Question: 6 Key: B k k bτg k p e e k F e z0 b g b g bg b g τ µ t dt µ 0 0 z z p 0 60 p 60 HG b p µ g b g t dt b g t dt I KJ where p is from Illustrative Life Table, since µ follows I.L.T. k k 6, 66, ,, 074 6, 396, ,, 074 b g b g b g τ τ τ 0 q60 0p60 p60 b 0 p60g b p60g from I.L.T b g

8 Test Question: 7 Key: C State : State : light Training heavy Training P P P P P L N M O Q P π 0. π π π 08. π+ 0. 6π π + π 8 π 0. b π g π π. π 08. π 3 Note: the notation in Probability Models would label the states 0 and, and would label the top row and left column of the matri P with subscript 0. The underlying calculations are the same. The matri P would look different, but the result would be the same, if you chose to make heavy the lower-numbered state.

9 Test Question: 8 Key: D σ Ybg is normal b0. 0, g corresponds to - standard deviation Y () 0.0-()(0.0) -0.0 Y( ) Y( ) is normal (0.0, ) corresponds to +.5 standard deviation Y( ) Y( ) ( 5. )( 00. ) b g b g F 00e 00e G 00e 00e G F 405. Y 0. 0 Y 0. 03

10 Test Question: 9 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

11 Test Question: 0 Key: A bτ g z bµ gt 0 0 d 000 e µ dt b µ e µ e g e µ ln µ bg z t 3 3 b b g b gj d 000 e dt e b b gb g b gb g e e g g j

12 Test Question: Key: B b g τ p b g τ p70 F( 0) 00. b g b g b g 0 70 d00 i b979g b943g+ b g b00 g b038g ( 979) + b g d i b g b g b g 70. good; must have bmaimum possibleg F( 000v) F 943 F 0 + q F v F F q F v F F q F 00v F 068 F q 00 F F(943) < random number < F(979), so choose 979

13 Test Question: Key: A Let Z i be random variable indicating death; W i be random variable indicating lapse for policy. Let U denote the random number used. policy # : q from Illustrative Life Table U 03. < Z W 0 policy # : q from Illustrative Life Table U 05. > Z 0 net checking lapse U 0. < 05. (surrender rate) W policy # 3 q U 04. > Z 3 0 net checking lapse U 08. > 05. W3 0 total Death and Surrender Benefits

14 Test Question: 3 Key: E 3 p p Use l (arbitrary, doesn t affect solution) so l+ 07. l By hyperbolic l l l l p. 5q Prob (all 3 failed) b 3 g

15 Test Question: 4 Key: B This is a graph of l µbg. b g would be increasing in the interval 80, 00 µ b g. The graphs of l p, l and l would be decreasing everywhere. The graph shown is comparable to Figure 3.3. on page 65 of Actuarial Mathematics

16 Test Question: 5 Key: A Using the conditional mean and variance formulas: E N E Λ N Λ c h Λd c hj Λd c h Var N Var E N Λ + E Var N Λ Since N, given lambda, is just a Poisson distribution, this simplifies to: E N E Λ Λ b g ΛbΛg Var N Var + E Λ bλg We are given that E N 0. and Var N 04., subtraction gives Var Λ j b g 0.

17 Test Question: 6 Key: B N number of salmon X eggs from one salmon S total eggs. E(N) 00t Var(N) 900t b g b g b g b g b g b g b g b g F > , b, g > HG 3, 000, E S E N E X 500t Var S E N Var X + E X Var N 00t t 3, 000t P S P S t t t 3 000t 0, t t 50 t 40 t d i t t t ± t. 4 round up to 3 t I KJ. 95

18 Test Question: 7 Key: A + A P V ( s benefits) v k b k + k p q k v v v , 89 + k 3 b g b gb g b gb gb g

19 Test Question: 8 Key: E π denotes benefit premium 9 V APV future benefits - APV future premiums 0. 6 π π b 0V + πgb08. g bq65gb0g V p 65 b gb08. g b00. gb0g

20 Test Question: 9 Key: C X losses on one life E X b gb g b gb g b gb g S total losses E S 3E X 3 b g+ c sb g E S E S F 0 b gc b gd E S f h sb gh i

21 Test Question: 0 Key: C M r E e b g r r r 3r e + e + e e + e + e M b05. g p E X b g b g b g λ M r cr Since λ and r 0.5, M c c c c 7. 8 premium rate per period

22 Test Question: Key: E Simple s surplus at the end of each year follows a Markov process with four states: State 0: out of business State : ending surplus State : ending surplus State 3: ending surplus 3 (after dividend, if any) State 0 is absorbing (recurrent). All the other states are transient states. Thus eventually Simple must reach state 0.

23 Test Question: Key: D (See solution to problem for definition of states). t L N M O P Q P at t t L N M O P Q P at t Epected dividend at the end of the third year 3 k 0 (probability in state k at t ) (epected dividend if in state k) 0.0* * (0* *0.5) *(0*0.6 +*0.5 + *0.5)

24 Test Question: 3 Key: A 80 70a + 50a 0a a a 30: 40 8 a a + a a a : 40 30: : 40 30: 40 b gb g b gb g 30: 40

25 Test Question: 4 Key: B a a f t dt zo b g t z 0. 05t e o 005 z. Γb g b L NM b g L F H G I K J N 0 05M. 05. te t te 005. o t te dt 05. t F HG O Q P i dt t t t + e I K J 05. t 05 e O QP 0 0, , 88

26 Test Question: 5 Key: C b g b g p k k p k L 0+ O k p b k NM QP g Thus an (a, b, 0) distribution with a 0, b. Thus Poisson with λ. b g p 4 4 e 4! 0. 09

27 Test Question: 6 Key: B By the memoryless property, the distribution of amounts paid in ecess of 00 is still eponential with mean 00. With the deductible, the probability that the amount paid is 0 is F Thus the average amount paid per loss is (0.393) (0) + (0.607) (00).4 The epected number of losses is (0) (0.8) 6. The epected amount paid is (6) (.4) / 00 b g 00 e

28 Test Question: 7 Key: D l96 + l97 From UDD l l l Likewise, from l 360 and l. 88,we get l For constant force, e l µ p l l µ e b0. 6g b g b gb g l

29 Test Question: 8 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 f Mbµ gdµ 0 z z µ µ e t dt d µ 0 0 z µ + + e du e e 0 s d i d i d i

30 Test Question: 9 Key: E E N E N E X VarbNg Var X E S b gb g b gb g b g b gb g b g E X E X b , 000g , E N E X. b70g 04 b g E N b g E X b g b g b g b g 98, Var S Var X + Var N. 78, , 344 Std dev S So B

31 Test Question: 30 Key: D f f f s s s b000g b08. gb0. g + b0. gbgb 0. gb0. g b00g b0. gbgb07. gb 0. g 008. b000g b0. gb0. g b 00g b0. 088gb800g b0. 08gb900g b0. 00gb800g E S With 75% relative security loading, cost (.75) (99.) 7.8 Alternatively, f F f 0 F s s sb g b gb g b gb gb gb g s b g b g b gb g b gb g b g E S 04 [from problem 9] b g+ b g+ b gc sb gh E S b00gc Fsb0gh b00gc Fsb00g E S 00 E S F cost (.75) (99.) 7.8 b gb g b gb g h

32 Test Question: 3 Key: D Let π benefit premium Actuarial present value of benefits , 000 v , 000 v , 000 v b gb g b gb gb g b gb gb gb g , 303. Actuarial present value of benefit premiums a&& π : v v. 766π b gb g 0, 303. π b gb 06. g b00, 000gb0. 03g V Initial reserve, year V + π π 3

33 Test Question: 3 Key: A Let π denote the premium. L b v a i v a a T T T T T T T + π π π b g E L a a 0 π π L a a a a v a v a a v A A T T T T T π δ δ δ δ d i b g

34 Test Question: 33 Key: D (0, ) (, 0.9) (.5, ) (, 0.885) tp t p ( 0. ) 0. 9 b gb p since uniform, 5. p / g b g o e 5 :. Area between t 0 and t 5. F+ 0. 9I HG K J b g F I + HG K Jb g Alternatively, z z o 5. e 5 :. t pdt 0 z 0. 5 tpdt+ p p d 0 0 z z t 0. t b g b g 0. t dt d

35 Test Question: 34 Key: A 0, 000A63b. g 533 A A + A b A + i q p b g gb g b gb g A Single contract premium at 65 (.) (0,000) (0.4955) 5550 b g i i

36 Test Question: 35 Key: B Original Calculation (assuming independence): µ µ µ A A A y y y y µ µ δ µ y µ δ y µ y µ δ y A A + A A y y y Revised Calculation (common shock model): µ 006., µ T* µ 006., µ 004. y b g b g T* y y µ µ + µ + µ A A A y y y b g b g T* T* y Z y µ µ δ µ y µ y+ δ µ y µ δ y A A + A A y y y Difference

37 Test Question: 36 Key: E Treat as three independent Poisson variables, corresponding to, or 3 claimants. rate rate rate Var Var Var L NM total Var , since independent. Alternatively, d i + + E X O QP For compound Poisson, Var S E N E X b gf H G I K J

38 Test Question: 37 Key: C z3 λbtgdt 6 so Nb3g is Poisson with λ 6. 0 b g g P is Poisson with mean 3 (with mean 3 since Prob y i < P and Q are independent, so the mean of P is 3, no matter what the value of Q is.

39 Test Question: 38 Key: A At age : Actuarial Present value (APV) of future benefits 4 APV of future premiums F I 5 &&a K J π HG F HG 5 A I K J A A 4 a&& 4 π a&& by equivalence principle π π V APV (Future benefits) APV (Future benefit premiums) A35 π a&& b. g b58. gb5396. g 5 05.

40 Test Question: 39 Key: E Let E Y Var Y E S Var Y present value random variable for payments on one life S Y present value random variable for all payments 0a&& d A d / A40 i d ib g 00E Y 4, S 00Var Y 70, 555 Standard deviation S 70, By normal approimation, need E [S] Standard deviations 4, (.645) (65.6) 5,54

41 Test Question: 40 Key: B 5A 4 A Initial Benefit Prem 5a&& 4a&& Where e e 30 30: 0 30: 35 30: 0 b. g b. g b g b g j j A30 : 0 A30 : A 0 30: and A30: a&& : 0 d F H G I 06. K J Comment: the numerator could equally well have been calculated as A E A (4) (0.9374) (0.4905)

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

SOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE SOLUTIONS. Copyright 2016 by the Society of Actuaries SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE SOLUTIONS Copyright 6 by the Society of Actuaries The solutions in this study note were previously presented in study note MLC-9-8