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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. a. For the uniform r.v. on [-b,b], the mean μ = 0 and the variance σ 2 = b 2 / 3 . The exact P [ | X - μ | > c ] = (1 - ( c/b ) , for 0 c b ; = 0 , for c > b. CB gives P [ | X - μ | > c ] σ 2 c 2 = b 2 3 c 2 . b. For the Laplacian r.v., μ = 0 and σ 2 = 2 2 . The exact P [ | X - μ | > c ] = P [ | X | > c ] = exp ( - αc ) . CB gives P [ | X - μ | > c ] 2 α 2 c 2 . c. For the Gaussian r.v., the mean is μ = 0 and the variance is σ 2 . The exact P [ | X - μ | > c ] = 2 Q ( c/σ ) . CB gives P [ | X - μ | > c ] σ 2 c 2 . 5. a. V ( t ) = X cos(2 πft ) + Y sin(2 πft ) = X 2 + Y 2 X X 2 + Y 2 cos(2 πft ) + Y X 2 + Y 2 sin(2 πft ) · = X 2 + Y 2 (cos(Θ) cos(2 πft ) + sin(Θ) sin(2 πft )) = R cos(2 πft - Θ) , where R = X 2 + Y 2 and Θ = tan - 1 ( Y/X

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