# Hw5sol - UCSD ECE 153 Handout#19 Prof Young-Han Kim Thursday Solutions to Homework Set#5(Prepared by TA Halyun Jeong 1 Work on the midterm problems

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Unformatted text preview: UCSD ECE 153 Handout #19 Prof. Young-Han Kim Thursday, November 20, 2008 Solutions to Homework Set #5 (Prepared by TA Halyun Jeong) 1. Work on the midterm problems to make sure you understand everything clearly. 2. Read Sections 6.1–6.5 in the text. Try to work on all examples. 3. 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 can be a covariance matrix for X 1 = Z 1 + Z 2 and X 2 = Z 1 + Z 3 . (c) This can be 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 nonnegative definite. 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 < . 4. Given a Gaussian random vector X ∼ N ( μ , Σ), where μ = (1 5 2) T and Σ =   1 1 0 1 4 0 0 0 9   . (a) What is P (2 X 1 + X 2 + X 3 < 0)? 1 (b) Find the joint pdf on Y = A X , where A = bracketleftbigg 2 1 1 1- 1 1 bracketrightbigg . Solution: (a) Let Y = X 1 + X 2 + 2 X 3 . Now Y ∼ N (9 , 21), so P( X 1 + X 2 + 2 X 3 < 0) = P( Y < 0) = Φ parenleftbigg- μ Y σ y parenrightbigg = Φ parenleftbigg- 9 √ 21 parenrightbigg = Φ (- 1 . 96) = Q (1 . 96) = 2 . 48 × 10- 2 . (b) μ Y = Aμ X = bracketleftbigg 1 1- 1 2- 1 1 bracketrightbigg   1 2 3   = bracketleftbigg 3 bracketrightbigg , Σ Y = A Σ X A T = bracketleftbigg 2 1 1 1- 1 1 bracketrightbigg   1 1 0 1 4 0 0 0 9     2 1 1- 1 1 1   = bracketleftbigg 21 6 6 12 bracketrightbigg . Thus Y ∼ N parenleftbiggbracketleftbigg 9- 2 bracketrightbigg , bracketleftbigg 21 6 6 12 bracketrightbiggparenrightbigg ....
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## This note was uploaded on 10/26/2011 for the course MATH 180C taught by Professor Eggers during the Winter '09 term at Aarhus Universitet.

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Hw5sol - UCSD ECE 153 Handout#19 Prof Young-Han Kim Thursday Solutions to Homework Set#5(Prepared by TA Halyun Jeong 1 Work on the midterm problems

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