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Unformatted text preview: Statistics 351 Fall 2007 Midterm #2 – Solutions 1. (a) Let B = 1 1 1 1 and b = 2 so that B X + b = 1 1 1 1 X 1 X 2 + 2 = X 1 X 2 2 X 1 + X 2 = Y 1 Y 2 = Y . By Theorem V.3.1, we conclude that Y ∈ N ( Bμ + b,B Λ B ) where E ( Y ) = Bμ + b = 1 1 1 1 1 1 + 2 = and cov( Y ) = B Λ B = 1 1 1 1 2 2 2 3 1 1 1 1 = 1 4 5 1 1 1 1 = 1 1 1 9 . 1. (b) From (a) , we can read that Y 1 ∈ N (0 , 1), Y 2 ∈ N (0 , 9), and cov( Y 1 ,Y 2 ) = 1. Hence, ρ = corr( Y 1 ,Y 2 ) = 1 / 3 so that the density of Y is given by f Y 1 ,Y 2 ( y 1 ,y 2 ) = 1 2 π √ 8 exp 1 2 9 y 2 1 8 + 2 y 1 y 2 8 + y 2 2 8 . 1. (c) We now find f Y 2  Y 1 =0 ( y 2 ) = f Y (0 ,y 2 ) f Y 1 (0) = 1 2 π √ 8 exp n 1 2 9(0) 2 8 + 2(0) y 2 8 + y 2 2 8 o . 1 √ 2 π exp 1 2 (0) = 1 √ 8 √ 2 π exp 1 16 y 2 2 In other words, the distribution of Y 2  Y 1 = 0 is N (0 , 8). 2. If Y = ( Y 1 ,Y 2 ) , then since Y 1 and Y 2 are linear combinations of the components of X , we conclude from Definition I that Y...
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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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