# CAnswerskey - An Introductory Guide in the Construction of...

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An Introductory Guide in the Construction of Actuarial Models: A Preparation for the Actuarial Exam C/4 ANSWER KEY Marcel B. Finan Arkansas Tech University c All Rights Reserved November 17, 2014
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3 The answer key manuscript is to help the reader to check his/her answers against mine. I am not in favor of providing complete and detailed solutions to every single problem in the book. The worked out examples in the book are enough to provide the user with the skills needed to tackle the practice problems. This manuscript should not be made public or shared with others. Best of wishes. Marcel B. Finan Russellville, Arkansas July 2013
4 Section 1 1.1 Deterministic 1.2 Stochastic 1.3 Stochastic 1.4 Stochastic 1.5 Mostly stochastic Section 2 2.1 (a) A c = B, B c = A and C c = { 1 , 3 , 4 , 5 , 6 } (b) A [ B = { 1 , 2 , 3 , 4 , 5 , 6 } A [ C = { 2 , 4 , 6 } B [ C = { 1 , 2 , 3 , 5 } (c) A \ B = ; A \ C = { 2 } B \ C = ; (d) A and B are mutually exclusive as well as B and C 2.2 Note that Pr( E ) > 0 for any event E. Moreover, if S is the sample space then Pr( S ) = 1 X i =1 Pr( O i ) = 1 2 1 X i =0 1 2 i = 1 2 · 1 1 - 1 2 = 1 Now, if E 1 , E 2 , · · · is a sequence of mutually exclusive events then Pr( [ 1 n =1 E i ) = 1 X n =1 1 X j =1 Pr( O nj ) = 1 X n =1 Pr( E n )
5 where E n = { O n 1 , O n 2 , · · · } . Thus, Pr defines a probability function. 2.3 0.5 2.4 0.56 2.5 0.66 2.6 0.52 2.7 0.05 2.8 0.6 2.9 0.48 2.10 0.04 2.11 The probability is given is the figure below. The probability that the first ball is red and the second ball is blue is PR( RB ) = 0 . 3 . 2.12
6 The probability that the first ball is red and the second ball is blue is PR( RB ) = 6 / 25 . 2.13 0.173 2.14 0.467 2.15 0.1584 2.16 0.0141 2.17 0.29 2.18 0.42 2.19 0.22 2.20 0.657 Section 3 3.1 (a) Discrete (b) Discrete (c) Continuous (d) Mixed. 3.2 x 2 3 4 5 6 7 8 9 10 11 12 p(x) 1 36 2 36 3 36 4 36 5 36 6 36 5 36 4 36 3 36 2 36 1 36
7 3.3 p ( x ) = 8 < : p, x = 1 1 - p, x = 0 0 , x 6 = 0 , 1 . 3.4 1/9 3.5 0.469 3.6 0.132 3.7 0.3 3.8 = 1 . 3.9 F ( x ) = 8 > > > > < > > > > : 0 x < 1 0 . 25 1 x < 2 0 . 75 2 x < 3 0 . 875 3 x < 4 1 4 x 3.10 (a) F ( x ) = 0 , x < 0 1 - 1 (1+ x ) a - 1 , x 0 .
8 (b) F ( x ) = 0 , x < 0 1 - e - kx , x 0 . 3.11 F ( n ) = P ( X n ) = n X k =0 P ( X = k ) = n X k =0 1 3 2 3 k = 1 3 1 - ( 2 3 ) n +1 1 - 2 3 =1 - 2 3 n +1 3.12 (a) 0.135 (b) 0.233 (c) F ( x ) = 1 - e - x 5 x 0 0 x < 0 3.13 f ( x ) = F 0 ( x ) = e x (1+ e x ) 2 3.14 (a) We have that S (0) = 1 , S 0 ( x ) = - 1 20 (100 - x ) - 1 2 0 , s ( x ) is right continuous, and S (100) = 0 . Thus, S satisfies the properties of a survival function. (b) F ( x ) = 1 - S ( x ) = 1 - 1 10 (100 - x ) 1 2 . (c) 0.092 3.15 0.149 3.16 F ( x ) = 1 - S ( x ) = x 2 100 , x 0 3.17 (a) 0.3 (b) 0.3 3.18 h ( x ) = - S 0 ( x ) S ( x ) = 1 2 (100 - x ) - 1 3.19 S ( x ) = e - μx , F ( x ) = 1 - e - μx , and f ( x ) = F 0 ( x ) = μe - μx .
9 3.20 1/480 Section 4 4.1 (b) np (1 - p ) 4.2 (b) λ 4.3 (c) (1 - p ) p - 2 4.5 (c) 1 λ 2 4.6 (a) E ( X ) = 1 Γ ( ) Z 1 0 xe - x x - 1 dx = Γ ( ) Z 1 0 1 e - x x dx = Γ ( + 1) Γ ( ) = ↵✓ (b) E ( X 2 ) = 1 Γ ( ) Z 1 0 x 2 e - x x - 1 dx = 1 Γ ( ) Z 1 0 x +1 1 e - x dx = 2 Γ ( + 2) Γ ( ) Z 1 0 x +1 +2 Γ ( + 2) e - x dx = 2 Γ ( + 2) Γ ( ) where the last integral is the integral of the pdf of a Gamma random variable with parameters ( + 2 , ) . Thus, E ( X 2 ) = 2 Γ ( + 2) Γ ( ) = 2 ( + 1) Γ ( + 1) Γ ( ) = 2 ( + 1) .
10 Finally, V ar ( X ) = E ( X 2 ) - ( E ( X )) 2 = 2 2 ( + 1) - 2 2 = ↵✓ 2 4.7 4 4.8 1,417,708,752 4.9 730,182,499.20 4.10 0 4.11 9 4.12 0.3284 4.13 We have μ 0 n = Z 1 0 x n f ( x ) dx = A Z 1 0 x B + n e - Cx dx = A - x B + n e - Cx C 1 0 + B + n C Z 1 0 x B + n - 1 e - Cx dx = B + n C Z 1 0 Ax B + n - 1 e - Cx dx = B + n C E ( X n - 1 ) .
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