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hw6sol

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

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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 σ 2 23 = 9 > σ 22 σ 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 ( λ 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 ( μ , Σ), where μ = (152) T and Σ = 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 ( 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). 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 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). iv. Since σ 13 = 0, X 3 and X 1 are uncorrelated and hence independent since they are jointly Gaussian; similarly, since σ 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 Thursday...

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