hw1_514_11

# hw1_514_11 - STAT 514(Spring 11 Homework 1 6.1 By the...

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STAT 514 (Spring 11) Homework 1 6.1 By the Factorization Theorem, | X | is sufficient because the pdf of X is f ( x | σ 2 ) = 1 2 πσ 2 e - x 2 / 2 σ 2 = 1 2 πσ 2 e -| x | 2 / 2 σ 2 = g ( | x || σ 2 ) × 1 where h ( x ) = 1 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 )) × e - Σ i x i where g ( T ( x ) | θ ) = e inθ I ( θ, ) ( T ( x )) and h ( x ) = e - Σ i x i when i > 0 and min i ( x i /i ) > θ for all x i > iθ . 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 ) I ( μ,σ ) ( x i ) = e μ/σ σ ! n e - i x i I ( μ, ) ( x (1) ) × 1 where g ( x (1) , i x i | μ, σ ) = e μ/σ σ n e - i x i I ( μ, ) ( x (1) ) and h ( x ) = 1 . 6.4 The joint pdf is n Y j =1 ( h ( x j ) c ( θ ) exp k X i =1 w i ( θ ) t i ( x j ) !) = c ( θ ) exp k X i =1 w i ( θ ) n X j =1 t i ( x j ) × n Y j =1 h ( x j ) By the Factorization theorem, n j =1 t 1 ( X j ) , · · · , n j =1 t k ( X j ) !

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• Spring '15
• Michael Parzen
• Algebra, Statistical theory, i=1, max Xi

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