EXAM P SAMPLE SOLUTIONS

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122. Solution: D x 6 e − The marginal distribution of Y is given by f2(y) = e e dx = 6 e dx –2y –y –2y –2y –3y = −6 e e + 6e = 6 e – 6 e , 0 < y < ∞ ∞ −2y −3y Therefore, E(Y) = ∫ y f2(y) dy = ∫ (6ye − 6 ye ) dy = 6 ∞ 6 –2y 6 –3y 2 2 ∫ ye dy − 3 3 ∫ y e dy 0 0 ∞ 0 ∞ 0 y ∫ 0 –x –2y –2y ∞ ∫ 0 ye −2 y y ∫ 0 ∞ ∫ –3y dy – 6 y e dy = ∞ ∞ But –2y 2 y e dy and –3y 3y e dy are equivalent to the means of exponential random ∫ 0 ∫ 0 ∞ –2y variables with parameters 1/2 and 1/3, respectively. In other words, 2 y e dy = 1/2 ∞ –3y and 3y e dy = 1/3 . We conclude that E(Y) = (6/2) (1/2) – (6/3) (1/3) = 3/2 – 2/3 = ∫ 0 9/6 − 4/6 = 5/6 = 0.83 . -------------------------------------------------------------------------------------------------------- 123. Solution: C Observe Pr 4< S < 8 = Pr⎡ ⎣4< S < 8 N = 1⎤ ⎦Pr N = 1 + Pr⎡ ⎣4< S < 8 N > 1⎤ ⎦Pr 1 −4 −8 1 −1 5 5 2 −1 = 3( e − e ) + 6( e −e ) * = 0.122 N > 1 *Uses that if X has an exponential distribution with mean μ a b 1 t 1 t Pr ( a X b) Pr ( X a) Pr ( X b) e dt e dt e e μ ∞ ∞ − − − μ − μ μ ≤ ≤ = ≥ − ≥ = ∫ − = − μ ∫ μ [ ] [ ] [ ] Page 54 of 55 a b ∫ 0 0
124. Solution: A − x −2 y Because f(x,y) can be written as f ( x) f ( y) = e 2e and the support of f(x,y) is a cross product, X and Y are independent. Thus, the condition on X can be ignored and it suffices −2 y to just consider f ( y) = 2e . Because of the memoryless property of the exponential distribution, the conditional density of Y is the same as the unconditional density of Y+3. Because a location shift does not affect the variance, the conditional variance of Y is equal to the unconditional variance of Y. Because the mean of Y is 0.5 and the variance of an exponential distribution is always equal to the square of its mean, the requested variance is 0.25. Page 55 of 55
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SOCIETY OF ACTUARIES/CASUALTY ACTUA
Page 3 and 4: 4. Solution: A For i = 1, 2, let Ri
Page 5 and 6: 9. Solution: B Let M = event that c
Page 7 and 8: 14. Solution: A 1 11 1 1 1 ⎛1⎞
Page 9 and 10: 19. Solution: B Apply Bayes’ Form
Page 11 and 12: 25. Solution: B Let Y = positive te
Page 13 and 14: 31. Solution: D Let X denote the nu
Page 15 and 16: 36. Solution: B To determine k, not
Page 17 and 18: 41. Solution: E Let X = number of g
Page 19 and 20: 44. Solution: C If k is the number
Page 21 and 22: 5 ( ) = ∑ ( ) Pr[ = ] E⎡⎣f X
Page 23 and 24: 54. Solution: B Let Y denote the cl
Page 25 and 26: 59. Solution: B The distribution fu
Page 27 and 28: 64. Solution: A Let X denote claim
Page 29 and 30: 68. Solution: C Note that X has an
Page 31 and 32: 74. Solution: E First note R = 10/T
Page 33 and 34: ( X < ) ∪ ( Y < ) Pr ⎡⎣ 1 1
Page 35 and 36: 40 − 3n −2≥ n To find such an
Page 37 and 38: -----------------------------------
Page 39 and 40: 92. Solution: B Let X and Y denote
Page 41 and 42: Consequently, 1 2 1 2 ( , ) 34 f t
Page 43 and 44: 99. Solution: C We use the relation
Page 45 and 46: 105. Solution: A The calculation re
Page 47 and 48: 2 [ X Y] E⎡( X Y) ⎤ ( E[ X Y] )
Page 49 and 50: 111. Solution: E 3 f Pr ⎡ ⎣1< Y
Page 51 and 52: 116. Solution: D Denote the number
Page 53: 120. Solution: A We are given that
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EXAM P SAMPLE SOLUTIONS

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