hwsol4 - EE 278 Handout#13 Statistical Signal Processing...

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Unformatted text preview: EE 278 Handout #13 Statistical Signal Processing Monday, August 3, 2009 Homework #4 Solutions 1. (10 points) (a) The marginal pdfs of a jointly Gaussian pdf are Gaussian. Therefore X 1 ∼ N (1 , 1). (b) Since X 2 and X 3 are independent ( σ 23 = 0), the variance of the sum is the sum of the variances. Also the sum of two jointly Gaussian random variables is also Gaussian. Therefore X 2 + X 3 ∼ N (7 , 13). (c) Since it is a linear combination of GRV’s, 2 X 1 + X 2 + X 3 = bracketleftbig 2 1 1 bracketrightbig   X 1 X 2 X 3   is a Gaussian random vector with mean and variance μ = bracketleftbig 2 1 1 bracketrightbig   1 5 2   = 9 and σ 2 = bracketleftbig 2 1 1 bracketrightbig   1 1 0 1 4 0 0 0 9     2 1 1   = 21 . Thus 2 X 1 + X 2 + X 3 ∼ N (9 , 21). (d) X 3 and X 1 are uncorrelated and therefore are independent since they are jointly Gaussian, and similarly for X 3 and X 2 . Therefore the conditional pdf of X 3 | ( X 1 ,X 2 ) is the same as the pdf of X 3 , which is N (2 , 9). (e) We use the general formula for the conditional Gaussian pdf: X 2 |{ X 1 = x 1 } ∼ N ( Σ 21 Σ − 1 11 ( x − μ 1 ) + μ 2 , Σ 22 − Σ 21 Σ − 1 11 Σ 12 ) In the case of ( X 2 ,X 3 ) | X 1 , Σ 11 = bracketleftbig 1 bracketrightbig , Σ 21 = bracketleftbigg 1 bracketrightbigg , Σ 22 = bracketleftbigg 4 0 0 9 bracketrightbigg So the mean and variance of ( X 2 ,X 3 ) given X 1 = x 1 are μ ( X 2 ,X 3 ) | X 1 = bracketleftbigg 1 bracketrightbigg bracketleftbig 1 − 1 bracketrightbigbracketleftbig x 1 − 1 bracketrightbig + bracketleftbigg 5 2 bracketrightbigg = bracketleftbigg x 1 + 4 2 bracketrightbigg Σ ( X 2 ,X 3 ) | X 1 = bracketleftbigg 4 0 0 9 bracketrightbigg − bracketleftbigg 1 bracketrightbigg bracketleftbig 1 0 bracketrightbig = bracketleftbigg 4 0 0 9 bracketrightbigg − bracketleftbigg 1 0 0 0 bracketrightbigg = bracketleftbigg 3 0 0 9 bracketrightbigg Thus X 2 and X 3 are conditionally independent given X 1 . The conditional densi- ties are X 2 |{ X 1 = x 1 } ∼ N ( x 1 + 4 , 3) and X 3 |{ X 1 = x } ∼ N (2 , 9). 1 (f) In the case of X 1 | ( X 2 ,X 3 ), Σ 11 = bracketleftbigg 4 0 0 9 bracketrightbigg , Σ 21 = bracketleftbig 1 0 bracketrightbig , Σ 22 = bracketleftbig 1 bracketrightbig So the mean and variance of X 1 |{ X 2 = x 2 ,X 3 = x 3 } are μ X 1 | X 2 ,X 3 = bracketleftbig 1 0 bracketrightbig bracketleftbigg 1 4 1 9 bracketrightbiggparenleftbiggbracketleftbigg x 2 x 3 bracketrightbigg − bracketleftbigg 5 9 bracketrightbiggparenrightbigg + 1 = bracketleftbig 1 4 bracketrightbig bracketleftbigg x 2 − 5 x 3 − 9 bracketrightbigg + 1 = 1 4 x 2 + 4 9 σ 2 X 1 | X 2 ,X 3 = σ 22 − bracketleftbig 1 0 bracketrightbig bracketleftbigg 1 4 1 9 bracketrightbiggbracketleftbigg 1 bracketrightbigg = 1 − 1 4 = 3 4 The conditional mean does not depend on x 3 since X 1 and X 3 are independent....
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This note was uploaded on 10/26/2011 for the course ECE 180 taught by Professor Eggers during the Spring '11 term at Aarhus Universitet.

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hwsol4 - EE 278 Handout#13 Statistical Signal Processing...

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