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Unformatted text preview: EE 278 November 6, 2009 Statistical Signal Processing Handout #14 Homework #6 Solutions 1. Gaussian random vector Suppose X N ( , ) is a Gaussian random vector with = 1 5 2 and = 1 1 1 4 9 . a. Find the pdf of X 1 . b. Find the pdf of X 2 + X 3 . c. Find the pdf of 2 X 1 + X 2 + X 3 . d. Find the pdf of X 3 given ( X 1 ,X 2 ). e. Find the pdf of ( X 2 ,X 3 ) given X 1 . f. Find the pdf of X 1 given ( X 2 ,X 3 ). g. Find P { 2 X 1 + X 2 + X 3 < } . h. Find the joint pdf of Y = A X , where A = bracketleftbigg 2 1 1 1 1 1 bracketrightbigg . Solution (40 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 ( 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 2 X 1 + X 2 + X 3 is a linear transformation of a Gaussian random vector, 2 X 1 + X 2 + X 3 = bracketleftbig 2 1 1 bracketrightbig X 1 X 2 X 3 , it is a Gaussian random vector with mean and variance = bracketleftbig 2 1 1 bracketrightbig 1 5 2 = 9 and 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. Since 13 = 0, X 3 and X 1 are uncorrelated and hence independent since they are jointly Gaussian; similarly, since 23 = 0, X 3 and X 2 are independent. Therefore the conditional pdf of X 3 given ( 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 ( 21  1 11 ( x 1 ) + 2 , 22 21  1 11 12 ) In the case of ( X 2 ,X 3 )  X 1 , 11 = bracketleftbig 1 bracketrightbig , 21 = bracketleftbigg 1 bracketrightbigg , 22 = bracketleftbigg 4 0 0 9 bracketrightbigg . Therefore the mean and variance of ( X 2 ,X 3 ) given X 1 = x 1 are ( X 2 ,X 3 )  X 1 = bracketleftbigg 1 bracketrightbigg bracketleftbig 1 bracketrightbig 1 bracketleftbig x 1 1 bracketrightbig + bracketleftbigg 5 2 bracketrightbigg = bracketleftbigg x 1 + 4 2 bracketrightbigg ( 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 densities are X 2 { X 1 = x 1 } N ( x 1 + 4 , 3) and X 3 { X 1 = x } N (2 , 9). f. In the case of X 1  ( X 2 ,X 3 ), 11 = bracketleftbigg 4 0 0 9 bracketrightbigg , 21 = bracketleftbig 1 0 bracketrightbig , 22 = bracketleftbig 1 bracketrightbig So the mean and variance of X 1 { X 2 = x 2 ,X 3 = x 3 } are X 1  X 2 ,X 3 = bracketleftbig 1 0 bracketrightbig bracketleftbigg 1 4 1 9 bracketrightbiggparenleftbiggbracketleftbigg...
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This note was uploaded on 11/28/2009 for the course EE 278 taught by Professor Balajiprabhakar during the Fall '09 term at Stanford.
 Fall '09
 BalajiPrabhakar
 Signal Processing

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