hw6-sol

# hw6-sol - EE 278 November 6, 2009 Statistical Signal...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 11/28/2009 for the course EE 278 taught by Professor Balajiprabhakar during the Fall '09 term at Stanford.

### Page1 / 13

hw6-sol - EE 278 November 6, 2009 Statistical Signal...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online