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410Hw12ans - STAT 410 Homework#12(due Thursday November 19...

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STAT 410 Homework #12 Fall 2009 (due Thursday, November 19, by 4:00 p.m.) 1. a) 7.2.4 b) 7.2.7 a) 7.2.4 f ( x ; θ ) = ( 1 – θ ) x θ , x = 0, 1, 2, … , 0 < θ < 1, zero elsewhere. f ( x 1 ; θ ) f ( x 2 ; θ ) f ( x n ; θ ) = ( 1 – θ ) Σ x i θ n . By Factorization Theorem, = n i i 1 X is a sufficient statistic for θ . b) 7.2.7 f ( x ; θ ) = ( 29 6 1 6 1 θ θ θ x e x - - Γ , 0 < x < . f ( x 1 ; θ ) f ( x 2 ; θ ) f ( x n ; θ ) = ( 29 6 1 1 6 1 θ θ θ i x n i i n e x Σ Γ - - = By Factorization Theorem, = n i i 1 X is a sufficient statistic for θ .
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2. a) 7.2.6 b) 7.2.8 a) 7.2.6 f ( x ; θ ) = ( ( 29 ( 29 ( 29 1 2 1 1 2 2 θ θ θ - - - Γ Γ Γ x x = ( 29 ( 29 1 1 1 θ θ θ x x - + - , 0 < x < 1. f ( x 1 ; θ ) f ( x 2 ; θ ) f ( x n ; θ ) = ( 29 [ ] ( 29 - + = - = n i i n i i n x x 1 1 1 1 1 θ θ θ By Factorization Theorem, = n i i 1 X is a sufficient statistic for θ . a) 7.2.8 f ( x ; θ ) = ( ( 29 ( 29 ( 29 1 1 θ θ 1 θ θ θ θ - - - Γ Γ Γ x x = ( ( 29 ( 29 [ ] 1 2 θ 1 θ θ 2 - - Γ Γ x x , 0 < x < 1. f ( x 1 ; θ ) f ( x 2 ; θ ) f ( x n ; θ ) = ( 29 ( 29 ( 29 [ ] 1 1 2 θ 1 θ θ 2 - = - Γ Γ n i i i n x x By Factorization Theorem, ( 29 [ ] = - n i i i 1 X 1 X is a sufficient statistic for θ .
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3. Let X 1 , X 2 , … , X n be a random sample of size n from a N ( 0 , σ 2 ) distribution. a) Find the sufficient statistic Y for σ 2 . ( See 7.2.1 . ) ( = - = n i i n x x x x 1 2 2 2 2 1 2 2 1 2 1 ,..., , ; σ exp σ π σ L = - = 2 1 2 1 1 2 2 2 σ exp σ π n i i n x . By Factorization Theorem, Y = = n i i 1 2 X is a sufficient statistic for σ 2 . OR f ( x ) = ( - - 2 2 1 2 1 2 1 2 2 2 2 σ π σ exp σ π ln x , < x < . K ( x ) = x 2 . Y = = n i i 1 2 X is a sufficient statistic for σ 2 . b) Show that the maximum likelihood estimator for σ 2 is a function of Y . ln ( 2 σ L = ( 29 ( = - - - n i i x n n 1 2 2 2 2 1 2 2 2 σ σ π ln ln . ( 2 2 σ L σ d d ln = ( 29 = + - n i i x n 1 2 2 2 2 2 1 2 σ σ . ( 2 2 σ ˆ L σ d d = 0 = = n i i n 1 2 2 X 1 σ ˆ = n Y .
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OR ln ( σ L = ( 29 ( 29 = - - - n i i x n n 1 2 2 2 1 2 2 σ σ π ln ln .
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