Dr. Hackney STA Solutions pg 184

# Dr. Hackney STA Solutions pg 184 - 11-12 Solutions Manual...

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Unformatted text preview: 11-12 Solutions Manual for Statistical Inference b. Note that for ci = 1 n - x(xi -) x , (xi -)2 x ci = 1 and ci Yi = ci xi = 0. Then ci ( + xi = , c2 , i E^ Var^ and c2 i = = ^ c. Write = 1 + n = = E c2 VarYi = 2 i x(xi - x) 1 - n (xi - x)2 x2 (xi - x)2 = 2 = x2 i . nSxx x2 (xi - x)2 1 + 2 n2 ( (xi - x)2 ) (cross term = 0) di yi , where di = xi - x . (xi - x)2 From Exercise 11.11, Cov(^ , ) ^ = Cov ci Yi , di Yi = 2 ci di = - 2 x . (xi - x)2 = 2 11.31 The fact that x(xi - x) 1 - n (xi - x)2 ^i = i (xi - x) (xi - x)2 [ij - (cj + dj xi )]Yj i ci Yi , follows directly from (11.3.27) and the definition of cj and dj . Since = ^ 11.3.2 Cov(^i , ) = 2 ^ j from Lemma cj [ij - (cj + dj xi )] cj (cj + dj xi ) j = 2 ci - = 2 ci - j c2 - xi j j cj dj . Substituting for cj and dj gives ci c2 j j = = 1 (xi - x) x - n Sxx 1 x2 + n Sxx xi x , Sxx xi j cj dj = - ^ and substituting these values shows Cov(^i , ) = 0. Similarly, for , ^ ^ Cov(^i , ) = 2 di - j cj dj - xi j d2 j ...
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## This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.

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