Dr. Hackney STA Solutions pg 54

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

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Unformatted text preview: 4-10 Solutions Manual for Statistical Inference the Poisson(p) pmf. Thus Y | Poisson(p). Now calculations like those in a) yield the pmf of Y , for y = 0, 1, . . ., is fY (y) = 1 (y + ) ()y!(p) p 1+p y+ . Again, if is a positive integer, Y negative binomial(, 1/(1 + p)). 4.33 We can show that H has a negative binomial distribution by computing the mgf of H. EeHt = EE eHt N = EE e(X1 ++XN )t N = E E eX1 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, EeX1 t = x1 =1 ex1 t -1 (1 - p) logp x1 x1 = -1 logp x 1 (et (1 - p)) x1 =1 x1 = -1 - log 1-et (1 - p) logp . Then E log {1-et (1 - p)} logp N = n=0 log {1-et (1 - p)} logp log(1-et (1-p)) logp n e- n n! (since N Poisson) log(1-et (1-p)) logp n = e- e e -log(1-et (1-p)) logp n=0 t n! . The sum equals 1. It is the sum of a Poisson [log(1 - e (1 - p))]/[logp] pmf. Therefore, E(e Ht ) = e- elog(1-e = t (1-p)) / log p = . elogp -/ logp 1 1-et (1 - p) -/ log p p 1-et (1 - p) -/ logp This is the mgf of a negative binomial(r, p), with r = -/ log p, if r is an integer. 4.34 a. 1 P (Y = y) = 0 1 P (Y = y|p)fp (p)dp n y 1 p (1 - p)n-y p-1 (1 - p)-1 dp y B(, ) = 0 = = b. 1 n (+) py+-1 (1 - p)n+-y-1 dp y ()() 0 n (+) (y+)(n+-y) , y = 0, 1, . . . , n. y ()() (+n+) 1 P (X = x) = 0 1 P (X = x|p)fP (p)dp r+x-1 r ( + ) -1 p (1 - p)x p (1 - p)-1 dp x ()() = 0 ...
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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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