Dr. Hackney STA Solutions pg 107

# Dr. Hackney STA Solutions pg 107 - 7-10 Solutions Manual...

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Unformatted text preview: 7-10 Solutions Manual for Statistical Inference which we recognize as the kernel of an inverted gamma pdf, IG(a, b), with a= n-1 + 2 and b= (n - 1)t 1 + 2 -1 . Direct calculation shows that the mean of an IG(a, b) is 1/((a - 1)b), so E(|t) = n-1 1 2 t+ n-1 2 + -1 = n-1 2 2 s n-1 2 + + 1 -1 . This is a Bayes estimator of 2 . 7.24 For n observations, Y = i Xi Poisson(n). a. The marginal pmf of Y is m(y) = 0 (n)y e-n 1 -1 e-/ d y! () 0 = Thus, ny y!() (y+)-1 e- /(n+1) d = ny (y + ) y!() n+1 y+ . (y+)-1 e- /(n+1) f (y|)() = (|y) = y+ gamma y + , m(y) n+1 (y+) n+1 b. E(|y) Var(|y) = = (y + ) (y + ) n+1 2 = 2. . 1 y+ (). n+1 n+1 (n+1) 7.25 a. We will use the results and notation from part (b) to do this special case. From part (b), the Xi s are independent and each Xi has marginal pdf m(x|, 2 , 2 ) = - f (x|, 2 )(|, 2 ) d = - 1 -(x-)2 /22 -(-)2 /2 2 e e d. 2 Complete the square in to write the sum of the two exponents as - - x 2 2 + 2 + 2 2 + 2 2 2 2 2 + 2 2 - (x - )2 . 2( 2 + 2 ) Only the first term involves ; call it -A(). Also, e-A() is the kernel of a normal pdf. Thus, e-A() d = - 2 2 , + 2 and the marginal pdf is m(x|, 2 , 2 ) = = a n(, 2 + 2 ) pdf. (x - )2 1 exp - 2 2( 2 + 2 ) 2 2 + 2 (x - )2 1 exp - , 2( 2 + 2 ) 2 2 + 2 ...
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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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