Dr. Hackney STA Solutions pg 106

Dr. Hackney STA Solutions pg 106 - Second Edition 7-9 7.21...

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Unformatted text preview: Second Edition 7-9 7.21 a. E b. Var 1 n Yi 1 = 2 xi n Var Yi 2 = 2 x2 n i 1 . x2 i 1 n 1 Yi = xi n 1 E Yi = xi n xi = . xi i i i i i i Using Example 4.7.8 with ai = 1/x2 i we obtain 1 n 1 x2 i n 2. i xi 1 1 = Var x2 n i Yi . xi i Thus, ^ Var = 2 2 2 2 n i xi i i Because g(u) = 1/u2 is convex, using Jensen's Inequality we have 1 1 x2 n Thus, Var 7.22 a. 2 2 n -n(-)2 /(22 ) 1 f (, ) = f (|)() = x x e x e-(-) /2 . 2 2 b. Factor the exponent in part (a) as 1 1 1 -n ( - )2 - 2 ( - )2 = - 2 ( - (x))2 - 2 x ( - )2 , x 2 2 2 2v + 2 /n where (x) = ( 2 x + ( 2 /n))/( 2 + 2 /n) and v = ( 2 2 /n) ( + 2 /n). Let n(a, b) denote the pdf of a normal distribution with mean a and variance b. The above factorization shows that f (x, ) = n(, 2 /n) n(, 2 ) = n((x), v 2 ) n(, 2 + 2 /n), where the marginal distribution of X is n(, 2 + 2 /n) and the posterior distribution of |x is n((x), v 2 ). This also completes part (c). 7.23 Let t = s2 and = 2 . Because (n - 1)S 2 / 2 2 , we have n-1 f (t|) = 1 ((n - 1)/2) 2(n-1)/2 n-1 t [(n-1)/2]-1 i i 1 . x2 i 1 1 2 = Var n xi Yi . xi Yi i xi = 2 2 2 n2 x n i i e-(n-1)t/2 n-1 . With () as given, we have (ignoring terms that do not depend on ) (|t) 1 1 ((n-1)/2)-1 e-(n-1)t/2 ((n-1)/2)++1 1 1 -1/ e +1 , exp - 1 (n - 1)t 1 + 2 ...
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