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Dr. Hackney STA Solutions pg 54

# Dr. Hackney STA Solutions pg 54 - 4-10 Solutions Manual for...

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4-10 Solutions Manual for Statistical Inference the Poisson( ) pmf. Thus Y | Λ Poisson( ). Now calculations like those in a) yield the pmf of Y , for y = 0 , 1 , . . . , is f Y ( y ) = 1 Γ( α ) y !( ) α Γ( y + α ) 1+ y + α . Again, if α is a positive integer, Y negative binomial( α, 1 / (1 + )). 4.33 We can show that H has a negative binomial distribution by computing the mgf of H . E e Ht = EE ( e Ht N ) = EE e ( X 1 + ··· + X N ) t N = E E ( e X 1 t N ) N , because, by Theorem 4.6.7, the mgf of a sum of independent random variables is equal to the product of the individual mgfs. Now, E e X 1 t = x 1 =1 e x 1 t - 1 log p (1 - p ) x 1 x 1 = - 1 log p x 1 =1 ( e t (1 - p )) x 1 x 1 = - 1 log p ( - log 1 - e t (1 - p ) ) . Then E log { 1 - e t (1 - p ) } log p N = n =0 log { 1 - e t (1 - p ) } log p n e - λ λ n n ! (since N Poisson) = e - λ e λ log(1 - e t (1 - p )) log p n =0 e - λ log(1 - e t (1 - p )) log p λ log(1 - e t (1 - p )) log p n n ! . The sum equals 1. It is the sum of a Poisson [ λ log(1 - e t (1 - p ))] / [log p ] pmf. Therefore, E( e Ht
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