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Unformatted text preview: 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 = 2 2 2 3 we see that det( 1 ) = det(2) = 2 > 0 and det( 2 ) = det() = 6 4 = 2 > 0, and so we conclude that is, in fact, strictly positive definite. 1. (b) Since X is multivariate normal, we conclude that its characteristic function is given by 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 )...
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This note was uploaded on 03/26/2012 for the course STAT 351 taught by Professor Michaelkozdron during the Fall '08 term at University of Regina.
 Fall '08
 MichaelKozdron
 Statistics, Probability

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