# 409Hw02ans - STAT 409 Homework#2 due Friday September 9 by...

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STAT 409 Homework #2 Fall 2011 ( due Friday, September 9, by 4:00 p.m. ) 1 – 2. Let X 1 , X 2 , … , X n be a random sample of size n from the distribution with probability density function ( ) < < - = otherwise 0 1 0 ; 1 θ θ θ x x x f 0 < θ < . 1. a) Find the probability distribution of = - n i i 1 X ln . Let Y = ln X. F X ( x ) = x θ , 0 < x < 1. Then F Y ( y ) = P ( Y y ) = P ( X e y ) = 1 – e θ y , y > 0. Y has Exponential distribution with mean θ 1 . OR Let Y = ln X = u ( X ) x = u 1 ( y ) = v ( y ) = e y v ' ( y ) = e y f Y ( y ) = f X ( v ( y ) ) | v ' ( y ) | = θ e y ( θ – 1 ) e y = θ e θ y , y > 0. Y has Exponential distribution with mean θ 1 . OR Let Y = ln X M Y ( t ) = E ( e Y t ) = E ( e ln X t ) = E ( X t ) = ( ) - - 1 0 1 θ θ dx x x t = - - 1 0 1 θ θ dx x t = t - θ θ = t θ 1 1 1 - , t < θ . Y has Exponential distribution with mean θ 1 .

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Suppose X and Y are independent , X is Gamma ( α 1 , θ ) , Y is Gamma ( α 2 , θ ) . If random variables X and Y are independent, then M X + Y ( t ) = M X ( t ) M Y ( t ) . M X + Y ( t ) = ( ) ( ) 2 1 α α θ θ 1 1 1 1 t t - - = ( ) 2 1 α α θ 1 1 + - t , t < θ 1 . X + Y is Gamma ( α 1 + α 2 , θ ) ; = - n i i 1 X ln = = n i i 1 Y has Gamma ( α = n , β = “usual θ ” = θ 1 ) distribution. b) Recall that the maximum likelihood estimator of θ is = = - = - = n i i i i n n n 1 1 X X θ ˆ ln ln . Is θ ˆ an unbiased estimator of θ ? If θ ˆ is not an unbiased estimator of θ , construct an unbiased estimator θ ˆ ˆ of θ based on = n i i 1 X ln . Let Y i = – ln X i , i = 1, 2, … , n . Then θ ˆ = Y 1 . Since g ( x ) = x 1 is strictly convex for x > 0, by Jensen’s Inequality E ( θ ˆ ) = E ( Y 1 ) > ( ) Y E 1 = θ . θ ˆ is NOT an unbiased estimator for θ . Since Y 1 , Y 2 , … , Y n are i.i.d. Exponential with mean θ 1 ,
Y M ( t ) = Y E t e = ( ) + + + n t n e 2 1 Y ... Y Y E = n n t Y M = n t n θ 1 1 1 - . Y has Gamma distribution with α = n and β = “usual θ ” = θ 1 n . E ( θ ˆ ) = E ( Y 1 ) = ( ) ( ) Γ - - 0 1 θ θ 1 dx x n n x x n n n e = ( ) ( ) ( ) ( ) - Γ - - - - 0 2 1 θ 1 1 θ θ dx x n n n n x n n n e = 1 θ - n n = 1 θ θ - + n . θ ˆ ˆ = θ ˆ 1 - n n = = - - n i i n 1 X 1 ln is an unbiased estimator for θ . Note also that = n i i 1 X ln is sufficient for θ .

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