hw3 - solutions - f08

# hw3 - solutions - f08 - IE 336 Operations Research...

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IE 336: Operations Research - Stochastic Models Homework Set #3 Solutions Fall 2008 1. From HW#1 we know that f X ( x ) = 2 e x - 2 e - 2 x and f Y ( y ) = 2 e - 2 y For E(X): E ( X ) = Z x xf X ( x ) dx = Z 0 x ( 2 e - x - 2 e - 2 x ) dx = Z 0 2 xe - x dx - Z 0 2 xe - 2 x dx = 2 " e - x ( - 1) 2 ( - 1 x - 1) # 0 - 2 " e - 2 x ( - 2) 2 ( - 2 x - 1) # 0 = 2 e - 0 + 1 - 2 0 4 e 0 + 1 4 e 0 E ( X ) = 1 . 5 Note: It may be tempting to think that the interval for E(X) is from y to infinity, but you are asked for the E(X) (which is independent of y) so the interval is from 0 to infinity. An alternative point of view would be to look the problem as if you did not know the joint distribution, but were given the marginal of X (which is the pdf of X, independent of Y) and the interval 0 x ≤ ∞ . If you were asked for E(X | y) then the interval would be from y to infinity. This distinction is the focus of an alternative solution. Alternative solution: E ( X | y ) = e y R y xe - x dx = ( y + 1) and f ( y ) = 2 e - 2 y Then, E ( X ) = Z y E ( X | y ) f ( y ) dy = Z 0 ( y + 1) 2 e - 2 y dy = - ye - 2 y + 0 . 5 e - 2 y + 0 . 5 - e - 2 y + 1 0 = 1 . 5 1

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For E(Y | x): E ( Y | x ) = Z y yf ( y | x ) dy * = Z x 0 y e - y 1 - e - x dy = 1 1 - e - x Z x 0 ye - y dy = 1 1 - e - x e - y ( - y - 1) x 0 = - xe - x - e - x + 1 1 - e - x = - xe - x 1 - e - x + 1 - e - x 1 - e - x E ( Y | x ) = - xe - x 1 - e - x + 1 * Note: f ( y | x ) was one of the calculations required for HW#1 2. Given the probabilities of fishing day-to-day, the transition diagram is as follows:
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