A3-S-431F08

# A3-S-431F08 - SOLUTIONS TO ASSIGNMENT 3 ACTSC 431/831 FALL...

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SOLUTIONS TO ASSIGNMENT 3 – ACTSC 431/831, FALL 2008 1. (a) We have 1 = R -∞ f X ( x ) dx = 0 . 06 × 5 + 0 . 1 × 5 + a 3000 , which gives a = 600. Then E ( X ) = R -∞ xf X ( x ) dx = 7 . 5 . Furthermore, we have E ( X 2 ) = R -∞ x 2 f X ( x ) dx = 91 . 6667 . Hence, V ar ( X ) = 35 . 42. (b) We have F X ( x ) = Z x -∞ f X ( t ) dt = 0 , x < 0 , R x 0 0 . 06 dt, 0 x < 5 , R 5 0 0 . 06 dt + R x 5 0 . 1 dt, 5 x < 10 , R 5 0 0 . 06 dt + R 10 5 0 . 1 dt + R x 10 600 t 4 dt, x 10 , = 0 , x < 0 , 0 . 06 x, 0 x < 5 , 0 . 1 x - 0 . 2 , 5 x < 10 , 1 - 200 x 3 , x 10 . (c) The probability is Pr { 4 < X < 8 } = F X (8) - F X (4) = 0 . 36. 2. Let X have the Pareto (3 , 50) distribution, p 1 = Pr { 5 < X < 10 } = 0 . 172611 and p 2 = Pr { X > E ( X ) } = 0 . 296296. (a) We have V ar ( N 1 ) = E ( V ar ( N 1 | N )) + V ar ( E ( N 1 | N )) = E ( Np 1 (1 - p 1 )) + V ar ( Np 1 ) = 39 . 67 . (b) We know that

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A3-S-431F08 - SOLUTIONS TO ASSIGNMENT 3 ACTSC 431/831 FALL...

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