Dr. Hackney STA Solutions pg 87

# Dr. Hackney STA Solutions pg 87 - i =1 w i ( ) n X j =1 t i...

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Chapter 6 Principles of Data Reduction 6.1 By the Factorization Theorem, | X | is suﬃcient because the pdf of X is f ( x | σ 2 ) = 1 2 πσ e - x 2 / 2 σ 2 = 1 2 πσ e -| x | 2 / 2 σ 2 = g ( | x || σ 2 ) · 1 | {z } h ( x ) . 6.2 By the Factorization Theorem, T ( X ) = min i ( X i /i ) is suﬃcient because the joint pdf is f ( x 1 ,...,x n | θ ) = n Y i =1 e - x i I ( iθ, + ) ( x i ) = e inθ I ( θ, + ) ( T ( x )) | {z } g ( T ( x ) | θ ) · e - Σ i x i | {z } h ( x ) . Notice, we use the fact that i > 0, and the fact that all x i s > iθ if and only if min i ( x i /i ) > θ . 6.3 Let x (1) = min i x i . Then the joint pdf is f ( x 1 ,...,x n | μ,σ ) = n Y i =1 1 σ e - ( x i - μ ) I ( μ, ) ( x i ) = ± e μ/σ σ ² n e - Σ i x i I ( μ, ) ( x (1) ) | {z } g ( x (1) , Σ i x i | μ,σ ) · 1 | {z } h ( x ) . Thus, by the Factorization Theorem, ( X (1) , i X i ) is a suﬃcient statistic for ( μ,σ ). 6.4 The joint pdf is n Y j =1 ³ h ( x j ) c ( θ ) exp ´ k X i =1 w i ( θ ) t i ( x j ) = c ( θ ) n exp k X
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Unformatted text preview: i =1 w i ( ) n X j =1 t i ( x j ) | {z } g ( T ( x ) | ) n Y j =1 h ( x j ) | {z } h ( x ) . By the Factorization Theorem, n j =1 t 1 ( X j ) ,..., n j =1 t k ( X j ) is a sucient statistic for . 6.5 The sample density is given by n Y i =1 f ( x i | ) = n Y i =1 1 2 i I (-i ( -1) x i i ( + 1)) = 1 2 n n Y i =1 1 i ! I min x i i -( -1) I max x i i + 1 . Thus (min X i /i, max X i /i ) is sucient for ....
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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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