# lec12 - Lecture 12 Let us give several more examples of...

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Unformatted text preview: Lecture 12 Let us give several more examples of finding sufficient statistics. Example 1. Poisson Distribution Π( λ ) has p.f. f ( x | λ ) = λ x x ! e- λ for x = 0 , 1 , 2 , . . . and the joint p.f. is f ( x 1 , ··· , x n | λ ) = λ P x i Q n i =1 x i ! e- nλ = 1 Q n i =1 X i ! e- nλ λ P X i . Therefore we can take u ( x 1 , . . . , x n ) = 1 Q n i =1 X i ! , T ( x 1 , . . . , x n ) = n X i =1 x i and v ( T, λ ) = e- nλ λ T . Therefore, by Neyman-Fisher factorization criterion T = ∑ n i =1 X i is a sufficient statis- tics. Example 2. Consider a family of normal distributions N ( α, σ 2 ) and assume that σ 2 is a given known parameter and α is the only unknown parameter of the family. The p.d.f. is given by f ( x | α ) = 1 √ 2 πσ e- ( x- α ) 2 2 σ 2 and the joint p.d.f. is f ( x 1 , . . . , x n | α ) = 1 ( √ 2 πσ ) n exp n- n X i =1 ( x i- α ) 2 2 σ 2 o = 1 ( √ 2 πσ ) n exp n- ∑ x 2 i 2 σ 2 + ∑ x i α σ 2- nα 2 2 σ 2 o = 1 ( √ 2 πσ ) n exp n- ∑ x 2 i 2 σ 2 o exp n X x i α σ 2- nα 2 2 σ 2 o . 45 LECTURE 12. 46 If we take T = ∑ n i =1 X i , u ( x 1 , . . . , x n ) = 1 ( √ 2 πσ ) n exp n- ∑ x 2 i 2 σ 2 o and v ( T, α ) = exp n T α σ 2- nα 2 2 σ 2 o , then Neyman-Fisher criterion proves that T is a sufficient statistics....
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## This note was uploaded on 10/11/2009 for the course STATISTICS 18.443 taught by Professor Dmitrypanchenko during the Spring '09 term at MIT.

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lec12 - Lecture 12 Let us give several more examples of...

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