V in a b has mean a b2 hence for y u 1 3 we have that

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Unformatted text preview: nt, we have that E X Y = E X E Y . Since X is an exponential r.v. with parameter  = 1, its mean is E X = 1= = 1. 3 A uniform r.v. in a; b has mean a + b=2; hence for Y  U 1; 3 , we have that E Y = 2. 3 Therefore, the correlation between X and Y is E X Y = 1  2 = 2 2 e Due to the independence between X and Y , we have that Z31 E X=Y = E X E 1=Y = 1  y 1 dy = 1 lnyj3=1 = ln3=2 = 0.5493 y 2 2 1 5 Note: it is wrong to write E X=Y = E X =E Y = 1=2. f First, we need to nd out f xjy. Since X and Y are independent, we have that f xjy = fX x, 8y. Hence, to maximize f xjy over all x, we only need to maximize fX x = e,x over x  0. Obviously, the maximum is achieved at x = 0 . Hence, the best estimate of X is x = 0. Note that due to the independence, the best estimate of X is una ected what value of Y is observed. 5 3...
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