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# 409Hw01ans - STAT 409 Homework#1 due Friday September 2 by...

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STAT 409 Homework #1 Fall 2011 ( due Friday, September 2, by 4:00 p.m. ) 1. Let X 1 , X 2 , … , X n be a random sample from the distribution with probability density function ( ) ( ) ( ) ( ) 1 θ 1 2 X X θ θ θ ; x x x f x f - + = = - , 0 < x < 1, θ > 0. a) Obtain the method of moments estimator of θ , θ ~ . E ( X ) = ( ) ( ) - + - 1 0 1 2 1 θ θ θ dx x x x = ( ) + - + 1 0 1 2 θ θ θ θ dx x x = ( ) 0 1 2 1 1 1 1 2 1 θ θ θ θ θ θ + + + - + + x x = ( ) ( ) ( ) 2 1 1 θ θ θ θ + + + = 2 θ θ + . OR Beta distribution, α = θ , β = 2. E ( X ) = 2 θ θ + . X 2 θ ~ θ ~ = + ( ) 2 X θ ~ θ ~ + = X 2 X θ ~ θ ~ = - Answer: θ ~ = X 1 X 2 - , where = = n i i n 1 X 1 X .

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b) Is θ ~ an unbiased estimator of θ ? Justify your answer. Consider g ( x ) = x x - 1 2 . Then g ( X ) = θ ~ , g ( 2 θ θ + ) = θ . Also g " ( x ) = ( ) 3 1 4 x - > 0 for 0 < x < 1, i.e., g ( x ) is strictly convex. By Jensen’s Inequality, E ( θ ~ ) = E [ g ( X ) ] > g ( E ( X ) ) = g ( μ X ) = g ( 2 θ θ + ) = θ . Therefore, θ ~ is NOT an unbiased estimator of θ . c) Is θ ~ a consistent estimator of θ ? Justify your answer. By WLLN, ( ) X E X P = 2 θ θ + . ( Var ( X ) = ( )( ) 2 2 3 2 θ θ θ + + < ) Consider g ( x ) = x x - 1 2 . Then g ( x ) is continuous at 2 θ θ + . g ( X ) = θ ~ g ( 2 θ θ + ) = θ . a P n X , g is continuous at a ( ) ( ) X a g g P n θ θ ~ P . θ ~ is a consistent estimator of θ .
2. Let X 1 , X 2 , … , X n be a random sample of size n from a distribution with probability density function ( ) ( ) 2 1 θ X X θ ; x x f x f + = = , – 1 < x < 1, – 1 < θ < 1. a) Obtain the method of moments estimator of θ , θ ~ . μ = E ( X ) = - + 1 1 2 1 θ dx x x = 1 1 3 2 6 4 θ - + x x = 3 θ . 3 X θ ~ = θ ~ = 3 X .

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409Hw01ans - STAT 409 Homework#1 due Friday September 2 by...

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