hw6sol - UCSD ECE153 Handout #32 Prof. Young-Han Kim...

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Unformatted text preview: UCSD ECE153 Handout #32 Prof. Young-Han Kim Thursday, May 26, 2011 Solutions to Homework Set #6 (Prepared by TA Lele Wang) 1. Covariance matrices. Which of the following matrices can be a covariance matrix? Justify your answer either by constructing a random vector X , as a function of the i.i.d zero mean unit variance random variables Z 1 ,Z 2 , and Z 3 , with the given covariance matrix, or by establishing a contradiction. (a) bracketleftbigg 1 2 0 2 bracketrightbigg (b) bracketleftbigg 2 1 1 2 bracketrightbigg (c) 1 1 1 1 2 2 1 2 3 (d) 1 1 2 1 2 3 2 3 3 Solution: (a) This cannot be a covariance matrix because it is not symmetric. (b) This is a covariance matrix for X 1 = Z 1 + Z 2 and X 2 = Z 1 + Z 3 . (c) This is a covariance matrix for X 1 = Z 1 , X 2 = Z 1 + Z 2 , and X 3 = Z 1 + Z 2 + Z 3 . (d) This cannot be a covariance matrix. Suppose it is, then 2 23 = 9 > 22 33 = 6, which contradicts the Schwartz inequality. You can also verify this by showing that the matrix is not positive semidefinite. For example, the determinant is 2. Also one of the eigenvalues is negative ( 1 = . 8056). Alternatively, we can directly show that this matrix does not satisfy the definition of positive semidefiniteness by bracketleftbig 2 0 1 bracketrightbig 1 1 2 1 2 3 2 3 3 2 1 = 1 < . 2. Gaussian random vector. Given a Gaussian random vector X N ( , ), where = (152) T and = 1 1 0 1 4 0 0 0 9 . (a) Find the pdfs of i. X 1 , ii. X 2 + X 3 , iii. 2 X 1 + X 2 + X 3 , iv. X 3 given ( X 1 ,X 2 ), and v. ( X 2 ,X 3 ) given X 1 . (b) What is P { 2 X 1 + X 2 X 3 < } ? Express your answer using the Q function. 1 (c) Find the joint pdf on Y = A X , where A = bracketleftbigg 2 1 1 1 1 1 bracketrightbigg . Solution: (a) i. The marginal pdfs of a jointly Gaussian pdf are Gaussian. Therefore X 1 N (1 , 1). ii. 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). iii. 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). iv. 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)....
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hw6sol - UCSD ECE153 Handout #32 Prof. Young-Han Kim...

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