Dr. Hackney STA Solutions pg 38

Dr. Hackney STA Solutions pg 38 - Second Edition 3-11 (iii)...

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Unformatted text preview: Second Edition 3-11 (iii) , unknown, 1 h(x) = I[0,1] (x), c(, ) = B(,) , w1 () = - 1, t1 (x) = log x, w2 () = - 1, t2 (x) = log(1 - x). 1 d. h(x) = x! I{0,1,2,...} (x), c() = e- , w1 () = log , t1 (x) = x. e. h(x) = r - 1 I{r,r+1,...} (x), c(p) = x-1 p 1-p r , w1 (p) = log(1 - p), t1 (x) = x. 2 2 3.29 a. For the n(, 2 ) f (x) = 1 2 e- /2 e-x 2 /2 2 +x/ 2 , so the natural parameter is (1 , 2 ) = (-1/2 2 , / 2 ) with natural parameter space {(1 ,2 ):1 < 0, - < 2 < }. b. For the gamma(, ), 1 f (x) = e(-1) log x-x/ , () so the natural parameter is (1 , 2 ) = ( - 1, -1/) with natural parameter space {(1 ,2 ):1 > -1,2 < 0}. c. For the beta(, ), f (x) = (+) ()() e(-1) log x+(-1) log(1-x) , so the natural parameter is (1 , 2 ) = ( - 1, - 1) and the natural parameter space is {(1 ,2 ):1 > -1,2 > -1}. d. For the Poisson 1 f (x) = e- exlog x! so the natural parameter is = log and the natural parameter space is {:- < < }. e. For the negative binomial(r, p), r known, P (X = x) = r+x-1 x (pr ) ex log (1-p) , so the natural parameter is = log(1 - p) with natural parameter space {: < 0}. 3.31 a. 0 = k h(x)c() exp i=1 k wi ()ti (x) dx = h(x)c () exp i=1 k wi ()ti (x) dx k + h(x)c() exp i=1 wi ()ti (x) i=1 k wi () ti (x) dx j k = h(x) logc() c() exp j k wi ()ti (x) dx + E i=1 i=1 wi () ti (x) j = Therefore E logc() + E j k wi () i=1 j ti (x) i=1 wi () ti (x) j = - j logc(). ...
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