{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Dr. Hackney STA Solutions pg 69

# Dr. Hackney STA Solutions pg 69 - Second Edition 5-3 Hence...

This preview shows page 1. Sign up to view the full content.

Second Edition 5-3 Hence, Var( S 2 n +1 ) = 1 4 n 2 ( n + 1) 2 [Var( A ) + 4Var( B ) + 4Cov( A, B )] = 1 n + 1 θ 4 - n - 2 n θ 2 2 , establishing the induction and verifying the result. c. Again assume that θ 1 = 0. Then Cov( ¯ X, S 2 ) = 1 2 n 2 ( n - 1) E n k =1 X k n i =1 n j =1 ( X i - X j ) 2 . The double sum over i and j has n ( n - 1) nonzero terms. For each of these, the entire expectation is nonzero for only two values of k (when k matches either i or j ). Thus Cov( ¯ X, S 2 ) = 2 n ( n - 1) 2 n 2 ( n - 1) E X i ( X i - X j ) 2 = 1 n θ 3 , and ¯ X and S 2 are uncorrelated if θ 3 = 0. 5.9 To establish the Lagrange Identity consider the case when n = 2, ( a 1 b 2 - a 2 b 1 ) 2 = a 2 1 b 2 2 + a 2 2 b 2 1 - 2 a 1 b 2 a 2 b 1 = a 2 1 b 2 2 + a 2 2 b 2 1 - 2 a 1 b 2 a 2 b 1 + a 2 1 b 2 1 + a 2 2 b 2 2 - a 2 1 b 2 1 - a 2 2 b 2 2 = ( a 2 1 + a 2 2 )( b 2 1 + b 2 2 ) - ( a 1 b 1 + a 2 b 2 ) 2 . Assume that is true for n , then n +1 i =1 a 2 i n +1 i =1 b 2 i - n +1 i =1 a i b i 2 = n i =1 a 2 i + a 2 n +1 n i =1 b 2 i + b 2 n +1 - n i =1 a i b i + a n +1 b n +1 2 = n i =1 a 2 i n i =1 b 2 i - n i =1 a i b i 2 + n i =1 a 2 i b 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}