S09 Sol - IE 336: Operations Research - Stochastic Models...

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IE 336: Operations Research - Stochastic Models Exam #1 Solutions Spring 2009 1. (a) c Z 2 0 Z 2 - 2 x 2 + 2 xy + y 2 dxdy = c [ Z 2 0 Z 2 - 2 x 2 dxdy + Z 2 0 Z 2 - 2 y 2 dxdy ] * = c [ c Z 2 0 16 3 dy + Z 2 0 4 y 2 dy ] = c ( 32 3 + 4 3 [ y 3 ] 2 0 ) = c ( 32 3 + 32 3 ) = 1 Therefore, c = 3 64 *Since the random variable X is symmetric, R X R y 2 xydxdy will be 0. Thus, we don’t need to calculate it. (b) f x ( x ) = Z 2 0 3 64 x 2 + 2 xy + y 2 dy = 3 64 (2 x 2 + 4 x + 8 3 ) f y ( y ) = Z 2 - 2 3 64 x 2 + 2 xy + y 2 dx = 1 4 + 3 16 y 2 Since f ( x, y ) 6 = f x ( x ) f y ( y ), X and Y are not independent. (c) E ( Y ) = Z 2 0 1 4 y + 3 16 y 3 dy = 1 2 + 3 64 [ y 4 ] 2 0 = 5 4 1
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E ( XY ) = 3 64 Z 2 0 Z 2 - 2 xy ( x 2 + 2 xy + y 2 ) dxdy = 3 64 Z 2 0 Z 2 - 2 x 3 y + 2 x 2 y 2 + xy 3 dxdy * = 3 64 Z 2 0 [ 2 3 x 3 y 2 ] 2 x = - 2 dy = 3 64 · 32 3 Z 2 0 y 2 dy = 4 3 * Similarly, with the symmetric characteristic of X , x 3 y and xy 3 will be 0. 2. (a) Transition diagram is omitted.
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This note was uploaded on 05/05/2011 for the course IE 336 taught by Professor Bruce,s during the Spring '08 term at Purdue University-West Lafayette.

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S09 Sol - IE 336: Operations Research - Stochastic Models...

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