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

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 18

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

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online