hw9_soln

hw9_soln - Homework 9 Solutions Problem Solutions : Yates...

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Unformatted text preview: Homework 9 Solutions Problem Solutions : Yates and Goodman, 5.7.1 5.7.2 6.1.1 6.1.4 6.3.2 6.4.2 and 6.4.4 Problem 5.7.1 Solution (a) From Theorem 5.12, the correlation matrix of X is R X = C X + X X (1) = 4- 2 1- 2 4- 2 1- 2 4 + 4 8 6 4 8 6 (2) = 4- 2 1- 2 4- 2 1- 2 4 + 16 32 24 32 64 48 24 48 36 = 20 30 25 30 68 46 25 46 40 (3) (b) Let Y = X 1 X 2 . Since Y is a subset of the components of X , it is a Gaussian random vector with expected velue vector Y = E [ X 1 ] E [ X 2 ] = 4 8 . (4) and covariance matrix C Y = Var[ X 1 ] Cov [ X 1 , X 2 ] C X 1 X 2 Var[ X 2 ] = 4- 2- 2 4 (5) We note that det( C Y ) = 12 and that C- 1 Y = 1 12 4 2 2 4 = 1 / 3 1 / 6 1 / 6 1 / 3 . (6) This implies that ( y- Y ) C- 1 Y ( y- Y ) = y 1- 4 y 2- 8 1 / 3 1 / 6 1 / 6 1 / 3 y 1- 4 y 2- 8 (7) = y 1- 4 y 2- 8 y 1 / 3 + y 2 / 6- 8 / 3 y 1 / 6 + y 2 / 3- 10 / 3 (8) = y 2 1 3 + y 1 y 2 3- 16 y 1 3- 20 y 2 3 + y 2 2 3 + 112 3 (9) The PDF of Y is f Y ( y ) = 1 2 12 e- ( y- Y ) C- 1 Y ( y- Y ) / 2 (10) = 1 48 2 e- ( y 2 1 + y 1 y 2- 16 y 1- 20 y 2 + y 2 2 +112) / 6 (11) 1 Since Y = X 1 , X 2 , the PDF of X 1 and X 2 is simply f X 1 ,X 2 ( x 1 , x 2 ) = f Y 1 ,Y 2 ( x 1 , x 2 ) = 1 48 2 e- ( x 2 1 + x 1 x 2- 16 x 1- 20 x 2 + x 2 2 +112) / 6 (12) (c) We can observe directly from X and C X that X 1 is a Gaussian (4 , 2) random variable. Thus, P [ X 1 > 8] = P X 1- 4 2 > 8- 4 2 = Q (2) = 0 . 0228 (13) Problem 5.7.2 Solution We are given that X is a Gaussian random vector with X = 4 8 6 C X = 4- 2 1- 2 4- 2 1- 2 4 . (1) We are also given that Y = AX + b where A = 1 1 / 2 2 / 3 1- 1 / 2 2 / 3 b =- 4...
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hw9_soln - Homework 9 Solutions Problem Solutions : Yates...

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