indep_Xbar_S

# indep_Xbar_S - N ( , ) where = 1 /n ( n-1) /n-1 /n -1 /n-1...

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Statistics 351 Fall 2007 Independence of X and S 2 Theorem. Suppose that X 1 , . . . , X n are independent N (0 , 1) random variables. If X = 1 n n X i =1 X i and S 2 = 1 n - 1 n X i =1 ( X i - X ) 2 denote the sample mean and sample variance , respectively, then X and S 2 are independent. Our proof of this theorem will follow the outline proposed by Exercise V.8.2. Proof. Since X 1 , . . . , X n are i.i.d. N (0 , 1), we conclude (using, say moment generating func- tions) that X ∈ N (0 , 1 /n ). Similarly, we can show that X - X j = 1 n ( X 1 + ··· + X j - 1 + X j +1 + ··· + X n ) - n - 1 n X j ∈ N ± 0 , n - 1 n 2 + ( n - 1) 2 n 2 ² = N ± 0 , n - 1 n ² and so X j - X ∈ N ± 0 , n - 1 n ² as well. We also note that Cov( X j , X ) = Cov ³ X j , 1 n n X i =1 X i ! = 1 n n X i =1 Cov( X j , X i ) = 1 n Cov( X j , X j ) = 1 n , and so for j 6 = k it follows that Cov( X j - X, X k - X ) = Cov( X j , X k ) - Cov( X j , X ) - Cov( X, X k ) + Cov( X, X ) = 0 - 1 n - 1 n + 1 n = - 1 n using the fact that Cov( X, X ) = Var( X ) = 1 /n . Similarly, Cov( X, X j - X ) = Cov( X j , X ) - Cov( X, X ) = 1 n - n n 2 = 0 . Thus, we see that ( X, X 1 - X, . . . , X n - X )

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Unformatted text preview: N ( , ) where = 1 /n ( n-1) /n-1 /n -1 /n-1 /n ( n-1) /n -1 /n . . . . . . . . . . . . . . .-1 /n-1 /n ( n-1) /n . By a theorem from last lecture (Theorem V.7.2), we conclude from the form of that X and ( X 1-X, . . . , X n-X ) are independent normal random vectors. It now follows from the transformation theorem (Theorem I.2.1 on page 23) that since X = ( X 1-X, . . . , X n-X ) and X are independent, so too are X and X X = n X i =1 ( X i-X ) 2 . This now implies that X and S 2 are independent....
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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.

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indep_Xbar_S - N ( , ) where = 1 /n ( n-1) /n-1 /n -1 /n-1...

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