20061ee131A_1_131ahw8sol1

20061ee131A_1_131ahw8sol1 - EE 131A 1. Homework #8 Solution...

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EE 131A Homework #8 Winter 2006 Solution K. Yao 1. E { X | Y = 1 } = 3; E { X | Y = 2 } = 5+ E { X } ; E { X | Y = 3 } = 7+ E { X } . Thus, E { X } = 1 3 (3 + 5 + E { X } + 7 + E { X } ) or E { X } = 15 . 2. f X ( x ) = Z x 0 (2 /x ) e - 2 x dy = 2 e - 2 x , 0 x < ; E { X } = Z 0 xf X ( x ) dx = 1 / 2 . f Y ( y ) = Z x = y (2 /x ) e - 2 x dx, 0 y < E { Y } = R y =0 y " R x = y (2 /x ) e - 2 x dx # dy = R x =0 (2 /x ) e - 2 x " x R y =0 ydy # dx = R x =0 (2 /x ) e - 2 x ( y 2 / 2) | x y =0 dx = R x =0 xe - 2 x dx = 1 / 4 E { XY } = R x =0 x R y =0 2 ye - 2 x dxdy = R x =0 x 2 e - 2 x dx = (1 / 8) R u =0 u 2 e - u du = (1 / 8)Γ(3) = 1 / 4 cov ( X,Y ) = E { XY } - E { X } E { Y } = 1 / 4 - (1 / 2)(1 / 4) = 1 / 8 . 3. From Chebyshev Inequality, we have P {| X - 50 | ≥ 10 } ≤ σ 2 10 2 = 1 / 4 . Then P {| X - 50 | < 10 } ≥ (1 - 1 4 ) = 3 / 4 . Thus, the probability this week’s production will be between 40 and 60 is at least 0.75. 4.
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This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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20061ee131A_1_131ahw8sol1 - EE 131A 1. Homework #8 Solution...

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