2 next we nd the 2nd order moments which are the same

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Unformatted text preview: ally distributed. 2 Next, we nd the 2nd-order moments which are the same as their variances E X = E Y = E Z = 3 + 3 =12 = 3. 2 By linearity of the E  operator and the independence between X and Y , we nd 6 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  1 6 1 6 1 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 : 1 Is ^ an unbiased estimator for ? Explain why or why not. 10 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 . 2 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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