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Unformatted text preview: ally distributed.
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Next, we nd the 2ndorder moments which are the same as their variances E X =
E Y = E Z = 3 + 3 =12 = 3.
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By linearity of the E operator and the independence between X and Y , we nd
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E X Y +Z =E X Y +E Z =E X Y =E X E Y =33= 9:
2 2 2 2 2 2 2 2 2 2 2 2 b Because X and Y are independent, we have 5 fXY x; y = fX xfY y = 0; = ; ,3 x; y: 3;
otherwise
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36 4. Suppose that fXigN are i.i.d. random variables with mean 0 and variance . An estii
mator for is de ned as
PN
i
^ = N ,Xi :
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Is ^ an unbiased estimator for ? Explain why or why not.
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2 =1 2 =1 2 2 2 2 Solution: The mean of ^ is
2 E^ =
2 E Xi = N
N ,1
N ,1 PN 2 i=1 2 6= :
2 8 Therefore the estimator is biased .
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5. Suppose that random signal X is Gaussian distributed with mean 5 and variance 1; i.e.,
X N 5; 1, and additive noise N is Gaussian distributed with mean 0 and variance 1;
i.e., N N 0; 1. We observe the noise corrupted signal Y = X + N . If a value y = 1 is
observed, what is the best estimate of X in the sense of maximizing f...
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 Fall '08
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