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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. P = 4 7 8 4 q/ 2 r 0 . 5 7 0 . 3 p 0 . 4 7 p 0 . 5 q = 4 7 8 4 0 . 1 0 . 4 0 . 5 7 0 . 3 0 . 3 0 . 4 8 0 . 3 0 . 5 0 . 3 p = 0 . 3 q = 0 . 2 , r = 0 . 4. (b) P ( X 1 = 7 | X 0 = 4) = 0 . 4 P ( X 5 = 4 | X 4 = 8 , X 2 = 7) = P ( X 5 = 4 | X 4 = 8) = P 84 = 0 . 3 P ( X 8 = 4 | X 5 = 4 , X 3 = 7) = P ( X 8 = 4 | X 5 = 4) = P (3) 44 = 0 . 244 (c) P 7847844847 = P 78 P 84 P 47 P 78 P
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