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Unformatted text preview: Stat 351 Fall 2007 Assignment #8 Solutions Exercise 7.1, page 134 : By Definition I, we know that X and Y X are normally distributed. Therefore, by Theorem 7.1, X and Y X are independent if and only if cov( X,Y X ) = 0. We compute cov( X,Y X ) = cov( X,Y ) cov( X,X ) = cov( X,Y ) var( X ) = SD( X )SD( Y ) var( X ) = var( X ) var( X ) = 0 since SD( X ) SD( Y ) = SD( X ) SD( X ) = var( X ) by the assumption that var( X ) = var( Y ). Hence, X and Y X are in fact independent. Problem #9, page 144 : Note that by Theorem 7.1, in order to show X 1 , X 2 , and X 3 are independent, it is enough to show that cov( X 1 ,X 2 ) = cov( X 1 ,X 3 ) = cov( X 2 ,X 3 ) = 0. Thus, if X 1 and X 2 + X 3 are independent, then cov( X 1 ,X 2 + X 3 ) = cov( X 1 ,X 2 ) + cov( X 1 ,X 3 ) = 0 and so cov( X 1 ,X 2 ) = cov( X 1 ,X 3 ) . (1) If X 2 and X 1 + X 3 are independent, then cov( X 2 ,X 1 + X 3 ) = cov( X 2 ,X 1 ) + cov( X 2 ,X 3 ) = 0 and so cov( X 2 ,X 1 ) = cov( X 2 ,X 3 ) . (2) Finally, if X 3 and X 1 + X 2 are independent, then cov( X 3 ,X 1 + X 2 ) = cov( X 3 ,X 1 )+cov( X 3 ,X 2 ) = 0 and so...
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 Fall '08
 MichaelKozdron
 Probability

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