Dr. Hackney STA Solutions pg 33

Dr. Hackney STA Solutions pg 33 - 3-6 Solutions Manual for...

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Unformatted text preview: 3-6 Solutions Manual for Statistical Inference 3.15 The moment generating function for the negative binomial is M (t) = the term r(1 - p)(e -1) (e -1) = (et - 1) t 1 1-(1 - p)e t t p t 1-(1 - p)e r = 1+ 1 r(1 - p)(e -1) r 1-(1 - p)et t r , as r , p 1 and r(p - 1) . Thus by Lemma 2.3.14, the negative binomial moment generating function converges to t e(e -1) , the Poisson moment generating function. 3.16 a. Using integration by parts with, u = t and dv = e-t dt, we obtain ( + 1) = 0 t(+1)-1 e-t dt = t (-e ) -t - 0 0 t-1 (-e-t )dt = 0 + () = (). b. Making the change of variable z = (1/2) = 0 t-1/2 e-t dt = 0 2t, i.e., t = z 2 /2, we obtain 2 2 -z2 /2 e zdz = 2 e-z /2 dz = 2 = . z 2 0 where the penultimate equality uses (3.3.14). 3.17 EX = = 1 x-1 e-x/ dx = () 0 (+) + (+) = . () () x 1 () x(+)-1 e-x/ dx 0 Note, this formula is valid for all > -. The expectation does not exist for -. 3.18 If Y negative binomial(r, p), its moment generating function is MY (t) = from Theorem 2.3.15, MpY (t) = lim p 1-(1-p)ept r p 1-(1-p)et r , and, . Now use L'H^pital's rule to calculate o 1 1 = , pt 1-t (p - 1)te +ept p0 p pt 1-(1 - p)e = lim p0 so the moment generating function converges to (1 - t)-r , the moment generating function of a gamma(r, 1). 3.19 Repeatedly apply the integration-by-parts formula 1 (n) z n-1 z -z dz = x xn-1 e-x 1 + (n - 1)! (n - 1) z n-2 z -z dz, x until the exponent on the second integral is zero. This will establish the formula. If X gamma(, 1) and Y Poisson(x). The probabilistic relationship is P (X x) = P (Y - 1). 3.21 The moment generating function would be defined by 0 1 etx dx. - 1+x2 On (0, ), etx > x, hence etx dx > 1+x2 0 x dx = , 1+x2 thus the moment generating function does not exist. ...
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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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