# 10_26ans - STAT 410 Examples for Fall 2011 Theorem 1...

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STAT 410 Examples for 10/26/2011 Fall 2011 Theorem 1 ( Factorization Theorem ) : Let X 1 , X 2 , … , X n denote random variables with joint p.d.f. or p.m.f. f ( x 1 , x 2 , , x n ; θ ) , which depends on the parameter θ . The statistic Y = u ( X 1 , X 2 , , X n ) is sufficient for θ if and only if f ( x 1 , x 2 , , x n ; θ ) = φ [ u ( x 1 , x 2 , , x n ) ; θ ] h ( x 1 , x 2 , , x n ) , where depends on x 1 , x 2 , , x n only through u ( x 1 , x 2 , , x n ) and h ( x 1 , x 2 , , x n ) does not depend on θ . ½ . Let X 1 , X 2 , … , X n be a random sample of size n from the distribution with probability density function ( ) ( ) ( ) θ 2 X X ln 1 θ θ ; x x x f x f - = = , x > 1, θ > 1. Find a sufficient statistic Y = u ( X 1 , X 2 , … , X n ) for θ . f ( x 1 ; θ ) f ( x 2 ; θ ) f ( x n ; θ ) = ( ) = - n i i i x x 1 θ 2 ln 1 θ = ( ) = - = - n i i n i i n x x 1 θ 1 2 ln 1 θ . Y 1 = = n i i 1 X is a sufficient statistic for θ . Y 2 = ln Y 1 = ln = n i i 1 X = = n i i 1 X ln is also a sufficient statistic for θ .

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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 ) = = ( ) ! λ
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10_26ans - STAT 410 Examples for Fall 2011 Theorem 1...

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