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2006solutions2 - Statistics 351 Fall 2006(Kozdron Midterm#2...

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Statistics 351 Fall 2006 (Kozdron) Midterm #2 — Solutions 1. (a) Recall that a square matrix is strictly positive definite if and only if the determinants of all of its upper block diagonal matrices are strictly positive. Since Λ = 2 - 2 - 2 3 we see that det(Λ 1 ) = det(2) = 2 > 0 and det(Λ 2 ) = det(Λ) = 6 - 4 = 2 > 0, and so we conclude that Λ is, in fact, strictly positive definite. 1. (b) Since X is multivariate normal, we conclude that its characteristic function is given by ϕ X ( t ) = exp it 1 + 2 it 2 - 1 2 (2 t 2 1 - 4 t 1 t 2 + 3 t 2 2 ) . 1. (c) If Y 1 = X 1 - 2 X 2 and Y 2 = X 1 + X 2 , then by Definition I, Y 1 is normal with mean E ( Y 1 ) = E ( X 1 ) - 2 E ( X 2 ) = - 3 and variance var( Y 1 ) = var( X 1 ) + 4 var( X 2 ) - 4 cov( X 1 , X 2 ) = 22, and Y 2 is normal with mean E ( Y 2 ) = E ( X 1 ) + E ( X 2 ) = 3 and variance var( Y 1 ) = var( X 1 ) + var( X 2 ) - 2 cov( X 1 , X 2 ) = 1. Since cov( Y 1 , Y 2 ) = cov( X 1 - 2 X 2 , X 1 + X 2 ) = var( X 1 ) - cov( X 1 , X 2 ) - 2 var( X 2 ) = - 2 , we conclude Y = ( Y 1 , Y 2 ) N - 3 3 , 22 - 2 - 2 1 . 2. By Definition I, we see that X 1 and X 1 + X 2 are each normally distributed random variables. Therefore, by Theorem V.7.1, X 1 and X 1
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