# 09_09_09ans - STAT 409 1. Fall 2009 Examples for 09/09/09...

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STAT 409 Examples for 09/09/09 Fall 2009 1. Let X 1 , X 2 , … , X n be a random sample of size n from a Poisson distribution with mean λ . That is, f ( k ; λ ) = P ( X 1 = k ) = ! λ λ k e k - , k = 0, 1, 2, 3, … . a) Use Factorization Theorem to find Y = u ( X 1 , X 2 , , X n ), a sufficient statistic for λ . f ( x 1 ; λ ) f ( x 2 ; λ ) f ( x n ; λ ) = = - n i i x x e i 1 ! λ λ = = - = n i i n x x e n i i 1 ! λ 1 1 λ . By Factorization Theorem, Y = = n i i 1 X is a sufficient statistic for λ . [ X is also a sufficient statistic for λ . ] b) Show that P ( X 1 = x 1 , X 2 = x 2 , … , X n = x n | Y = y ) does not depend on λ . Since Y = = n i i 1 X has a Poisson distribution with mean n λ , if = n i i x 1 = y , P ( X 1 = x 1 , X 2 = x 2 , … , X n = x n | Y = y ) = = ( 29 ! λ ! λ ! λ ! λ 2 1 λ λ ... λ λ 2 1 y n x x x n y n x x x e e e e n - - - - = y n n x x x y 1 ... ! ! ! ! 2 1 does not depend on λ . [ P ( X 1 = x 1 , X 1 = x 1 , … , X n = x n | Y = y ) = 0 if = n i i x 1 y . ]

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1¼. Let X 1 , X 2 , … , X n be a random sample of size n from an Exponential distribution with probability with mean θ . ( 29 - = = = n i i n n i i x x f 1 1 1 1 θ exp θ θ ; . By Factorization Theorem, = n i i 1 X is a sufficient statistic for θ . OR f ( x ; θ ) = θ 1 θ x e - = exp [ θ 1 x ln θ ]. K ( x ) = x . Y =
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## This note was uploaded on 10/12/2011 for the course STATISTICS stat 410 taught by Professor Stepanov during the Spring '11 term at University of Illinois, Urbana Champaign.

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09_09_09ans - STAT 409 1. Fall 2009 Examples for 09/09/09...

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