20082ee131A_1_HW8SO

20082ee131A_1_HW8SO - EE 131A Homework#8 Solution Spring...

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Unformatted text preview: EE 131A Homework #8 Solution Spring 2008 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. E { XY } = Z ∞ { Z x 2 ye- 2 x dy } dx = Γ(3) / 8 = 1 / 4 . f X ( x ) = Z x (2 /x ) e- 2 x dy = 2 e- 2 x , ≤ x < ∞ ; E { X } = Z ∞ xf X ( x ) dx = 1 / 2 . f Y ( y ) = Z ∞ y (2 /x ) e- 2 x dx, ≤ y < ∞ ; E { Y } = Z ∞ o yf Y ( y ) dy = Γ(2) / 4 = 1 / 4 . cov ( X,Y ) = E { XY } - E { X } E { Y } = 1 / 4- (1 / 2)(1 / 4) = 1 / 8 . 3. (a) f N ( n ) = P ( N = n ) = 1 / 3 for n = 1 , 2 , 3 X | N is uniform. so f ( x | n ) = 1 /n for 0 ≤ x ≤ n ; n = 1 , 2 , 3 f ( x,n ) = f ( n ) f ( x | n ) = 1 / (3 n ) for 0 ≤ x ≤ n ; n = 1 , 2 , 3 f ( x ) = 3 ∑ n =1 f ( n,x ) But the number of non-zero terms in the sum depends on x . Case 1. 0 ≤ x ≤ 1 f ( x ) = f (1 ,x ) + f (2 ,x ) + f (3 ,x ) = 1 / 3 + 1 / (3 · 2) + 1 / (3 · 3) = 11 / 18 Case 2. 1 < x ≤ 2 f ( x ) = f (2 ,x ) + f (3 ,x ) = 1 / (3 · 2) + 1 / (3 · 3) = 5 /...
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20082ee131A_1_HW8SO - EE 131A Homework#8 Solution Spring...

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