HW7sol - ORIE 3500/5500 Fall Term 2008 Assignment 7...

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ORIE 3500/5500 Fall Term 2008 Assignment 7 Solution 1. (a) We have M Y ( s ) = E ( e sY ) = Z 0 e s · min( x, 1) λe - λx dx = Z 1 0 e sx λe - λx dx + Z 1 e s λe - λx dx = λ λ - s (1 - e - ( λ - s ) ) + e s e - λ = λ - se s e - λ λ - s (b) E ( Y ) = M 0 Y (0). M 0 Y ( s ) = - e - λ ( s + 1) e s λ - s + λ - se s e - λ ( λ - s ) 2 So E ( Y ) = (1 - e - λ ) . 2. M X ( s ) = ( c/ 3)(1 + 2 e 3 s )(1 - e s / 3) - 1 = ( c/ 3)(1 + 2 e 3 s ) X k =0 ( e s / 3) k = ( c/ 3) " X k =0 (1 / 3) k e sk + 2 X k =0 (1 / 3) k e ( k +3) s # If P ( X = k ) = p k for k 0, then M X ( s ) = k =0 p k e sk . Comparing the coefficients, p k = ± ( c/ 3)(1 / 3) k ,k < 3 ( c/ 3)((1 / 3) k + 2(1 / 3) k - 3 ) ,k 3 . Then 1 = X k p k = ( c/ 3)( X k 0 (1 / 3) k + X k 3 2(1 / 3) k - 3 ) = ( c/ 3)((3 / 2)+2(3 / 2)) = 3 c/ 2 . 1
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So c = 2 / 3. (a) p X (3) = ( c/ 3)((1 / 3) 3 + 2) = 110 / 243. (b) E ( X ) = M 0 X (0).
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This note was uploaded on 12/04/2008 for the course STSCI 6000 taught by Professor Turnbull during the Fall '08 term at Cornell University (Engineering School).

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HW7sol - ORIE 3500/5500 Fall Term 2008 Assignment 7...

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