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Unformatted text preview: IE 336: Operations Research  Stochastic Models Homework Set #3 Solutions Spring 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 ) = integraldisplay x xf X ( x ) dx = integraldisplay ∞ x ( 2 e x 2 e 2 x ) dx = integraldisplay ∞ 2 xe x dx integraldisplay ∞ 2 xe 2 x dx = 2 bracketleftBigg e x ( 1) 2 ( 1 x 1) bracketrightBigg ∞ 2 bracketleftBigg e 2 x ( 2) 2 ( 2 x 1) bracketrightBigg ∞ = 2 bracketleftbig e + 1 bracketrightbig 2 bracketleftbigg 4 e + 1 4 e bracketrightbigg 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 integraltext ∞ y xe x dx = ( y + 1) and f ( y ) = 2 e 2 y Then, E ( X ) = integraldisplay y E ( X  y ) f ( y ) dy = integraldisplay ∞ ( y + 1) 2 e 2 y dy = bracketleftbig ye 2 y + 0 . 5 e 2 y + 0 . 5 e 2 y + 1 bracketrightbig ∞ = 1 . 5 1 For E(Y  x): E ( Y  x ) = integraldisplay y yf ( y  x ) dy * = integraldisplay x y e y 1 e x dy = 1 1 e x integraldisplay x ye y dy =...
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This note was uploaded on 03/29/2010 for the course IE 383 taught by Professor Leyla,o during the Spring '08 term at Purdue.
 Spring '08
 Leyla,O
 Operations Research

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