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.
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Therefore, the correlation between X and Y is E X Y = 1 2 = 2
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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
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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.
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 Fall '08
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 Normal Distribution, Variance, Probability theory, Cumulative distribution function, School of Electrical and Computer Engineering, fx 1Cx1D

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