Dr. Hackney STA Solutions pg 57

Dr. Hackney STA Solutions pg 57 - Second Edition 4-13 r = =...

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Unformatted text preview: Second Edition 4-13 r = = 1 (r)(1 - r) 0 1 x x u 1 u e-(u+x/) du xr-1 e-x/ , (r)r xr-1 e-x/ r (r)(1 - r) r e-u du = 0 since the integral is equal to (1 - r) if r < 1. b. Use the transformation t = / to get p ()d = 0 1 (r)(1 - r) r-1 ( - )-r d = 0 1 (r)(1 - r) 1 tr-1 (1 - t)-r dt = 1, 0 c. since this is a beta(r, 1 - r). d d 1 r-1 1 log f (x) = log +(r - 1) log x - x/ = - >0 dx dx (r)r x for some x, if r > 1. But, d log dx 0 - e-x/ q ()d = 1 -x/ e q ()d 0 2 1 -x/ e q ()d 0 < 0 x. 4.39 a. Without loss of generality lets assume that i < j. From the discussion in the text we have that f (x1 , . . . , xj-1 , xj+1 , . . . , xn |xj ) (m - xj )! = x1 ! xj-1 !xj+1 ! xn ! Then, f (xi |xj ) = (x1 ,...,xi-1 ,xi+1 ,...,xj-1 ,xj+1 ,...,xn ) p1 1 - pj x1 pj-1 1 - pj xj-1 pj+1 1 - pj xj+1 pn 1 - pj xn . f (x1 , . . . , xj-1 , xj+1 , . . . , xn |xj ) (m - xj )! x1 ! xj-1 !xj+1 ! xn ! = (xk =xi ,xj ) ( p1 x1 pj-1 xj-1 pj+1 xj+1 pn xn ) ( ) ( ) ( ) 1 - pj 1 - pj 1 - pj 1 - pj pi 1-pj pi 1-pj m-xi -xj m-xi -xj m-x -x (m - xi - xj )! 1 - (m - xi - xj )! 1 - = i j (m - xj )! pi xi pi ( ) 1- xi !(m - xi - xj )! 1 - pj 1 - pj (m - xi - xj )! x1 ! xi-1 !, xi+1 ! xj-1 !, xj+1 ! xn ! (xk =xi ,xj ) ( p1 pi-1 pi+1 )x1 ( )xi-1 ( )xi+1 1 - pj - p i 1 - p j - pi 1 - pj - pi ...
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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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