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Unformatted text preview: STAT 409 Homework #3 Fall 2008 (due Friday, September 19, by 4:00 p.m.) 1. Let X 1 , X 2 , … , X n be a random sample from the distribution with probability density function … a) ( ) 3 2 & 3 & x e x x f = x > 0 θ > 0. Find the sufficient statistic Y = u ( X 1 , X 2 , … , X n ) for θ . f ( x 1 , x 2 , … x n ; θ ) = f ( x 1 ; θ ) f ( x 2 ; θ ) … f ( x n ; θ ) = & & ¡ ¢ £ £ ¤ ¥ ¦ ¦ § ¨ © © ª « ∏ = ¬ = n i i x n n x n i i e 1 2 1 3 & 3 & . By Factorization Theorem, Y = ¬ = n i i 1 3 X is a sufficient statistic for θ . OR f ( x ; λ ) = { } 3 ln ln ln exp 2 3 & & x x + + + ⋅ . K ( x ) = x 3 . Y = ¬ = n i i 1 3 X is a sufficient statistic for λ . b) ( ) ( ) ( ) 1 1 2 & & & x x x f + = 0 < x < 1 θ > 0. Find the sufficient statistic Y = u ( X 1 , X 2 , … , X n ) for θ . f ( x 1 , x 2 , … x n ; θ ) = f ( x 1 ; θ ) f ( x 2 ; θ ) … f ( x n ; θ ) = ( ) ( ) ∏ ∏ = = & & ¡ ¢ £ £ ¤ ¥ + n i i n i i n x x 1 1 1 2 1 & & & . By Factorization Theorem, Y = ∏ = n i i 1 X is a sufficient statistic for θ . OR f ( x ; λ ) = ( ) ( ) ( ) { } 2 1 1 ln & & ln ln & exp x x + + + ⋅ . & K ( x ) = x ln . & Y = ( ) ¡ = n i i 1 X K = ¡ = n i i 1 X ln is a sufficient statistic for λ . 2. Let X 1 , X 2 , … , X n be a random sample of size n from the distribution with probability density function ( ) ( ) ( ) & 2 X X ln 1 & & ; x x x f x f ⋅ = = , x > 1, θ > 1....
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This note was uploaded on 10/11/2010 for the course STATS 410 taught by Professor Alexey during the Spring '10 term at University of Illinois at Urbana–Champaign.
 Spring '10
 Alexey
 Probability

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