Dr. Hackney STA Solutions pg 43

Dr. Hackney STA Solutions pg 43 - 3-16 Solutions Manual for...

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Unformatted text preview: 3-16 Solutions Manual for Statistical Inference To evaluate the second term, let u = obtain t x 1+x2 , dv = xe-x 2 /2 dx, v = -e-x 2 /2 , du = 1-x2 (1+x2 )2 , to x2 -x2 /2 e dx = 1 + x2 = 2 x (-e-x /2 ) 2 1+x 2 t e-t /2 + 2 1+t - t t 2 1 - x2 (-e-x /2 )dx 2 )2 (1 + x t 1 - x2 -x2 /2 e dx. (1 + x2 )2 Therefore, P (Z t) = = 3.48 For the negative binomial P (X = x + 1) = For the hypergeometric (M -x)(k-x+x+1)(x+1) P (X=x) M N -M P (X = x + 1) = (x+1)(k-x-1) (N ) k 0 3.49 a. 2 2 t e-t /2 + 1 + t2 2 2 t e-t /2 + 2 1+t 2 2 t e-t /2 . 1 + t2 2 2 t t 1 1 - x2 + 1 + x2 (1 + x2 )2 2 (1 + x2 )2 e-x 2 e-x 2 /2 dx /2 dx r+x+1-1 r p (1 - p)x+1 = x+1 r+x x+1 (1 - p)P (X = x). if x < k, x < M , x M - (N - k) if x = M - (N - k) - 1 otherwise. E(g(X)(X - )) = 0 g(x)(x - ) 1 x-1 e-x/ dx. () Let u = g(x), du = g (x), dv = (x - )x-1 e-x/ , v = -x e-x/ . Then Eg(X)(X - ) = 1 -g(x)x e-x/ () + 0 0 g (x)x e-x/ dx . Assuming g(x) to be differentiable, E|Xg (X)| < and limx g(x)x e-x/ = 0, the first term is zero, and the second term is E(Xg (X)). b. E g(X) -(-1) 1-X x = (+) ()() 1 g(x) - ( - 1) 0 1-x x x-1 (1 - x)-1 dx. Let u = g(x) and dv = ( - ( - 1) 1-x )x-1 (1 - x) . The expectation is x ( + ) g(x)x-1 (1 - x) ()() 1 0 1 + 0 (1 - x)g (x)x-1 (1 - x)-1 dx = E((1 - X)g (X)), assuming the first term is zero and the integral exists. ...
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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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