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Unformatted text preview: STAT 410 Fall 2011 Homework #6 (due Friday, October 14, by 3:00 p.m.) 1  4. Let X 1 , X 2 , … , X n be a random sample from the distribution with probability density function ( )( )( ) g 1 1 g g ; X x x f + = ⋅ , 0 < x < 1, θ > – 1. 1. a) Obtain the method of moments estimator of θ , g ~ . E ( X ) = ( )( ) g + ⋅ ⋅ 1 g 1 1 g dx x x y = 1 – x dy = – dx = ( ) ( ) g ⋅ ⋅ + 1 g 1 1 g dy y y = ( ) ( ) g g + ⋅ ⋅ + + 1 1 1 g g 1 1 g g dy y dy y = 2 g 1 g 1 + + = 2 g 1 + . OR E ( X ) = ( )( ) g + ⋅ ⋅ 1 g 1 1 g dx x x u = x dv = ( )( ) g 1 1 g x + ⋅ dx du = dx v = ( ) 1 g 1 + x = ( ) ( ) g + + + 1 1 1 1 g g 1 1 dx x x x = ( ) g + 1 1 g 1 dx x = 2 g 1 + . OR Beta distribution, α = 1, β = θ + 1. G E ( X ) = G ¡ ¡ + = 2 g 1 + . 2 g ~ 1 X X 1 1 + = = ¡ = ⋅ n i i n G 2 X 1 X X 2 1 g ~ = = . b) Obtain the maximum likelihood estimator of θ , g ˆ . Likelihood function: L ( θ ) = ( ) ( ) ( ) ( ) g g 1 1 X 1 1 X 1 1 g g ¢ ¢ £ ¤ ¥ ¥ ¦ § + + ∏ ⋅ ⋅ = = = ∏ n i i n n i i . ln L ( λ ) = ( ) ( ) ¡ = + + ⋅ ⋅ n i i n 1 X 1 g 1 ln g ln . () ( ) ( ) ¡ = + + = n i d d i n 1 X 1 1 g ˆ g ˆ L g ln ln = 0. G ( ) ¡ = = n i i n 1 X 1 1 ln g ˆ . 2. a) What is the probability distribution of W = – ln ( 1 – X ) ? Let W = – ln ( 1 – X ) . F X ( x ) = ( )( ) g + ⋅ x dy y g 1 1 g = ( ) 1 1 g x y + = ( ) 1 g 1 1 + x , 0 < x < 1. F W ( w ) = P ( W ≤ w ) = P ( – ln ( 1 – X ) ≤ w ) = P ( X ≤ 1 – e – w ) = 1 – e – ( θ +1 ) w , 0 < w < ∞ . OR X = 1 – e – W = g – 1 ( W ) d x / d w = e – w f W ( w ) = f X ( g – 1 ( y ) ) w x d d = ( ) ( ) 1 g g w w e e × + ⋅ = ( θ + 1 ) e – ( θ +1 ) w , 0 < w < ∞ . G The distribution of W is Exponential with mean 1 g 1 + . b) What is the probability distribution of Y = ( ) g = n i i 1 X 1 ln = g = n i i 1 W ? The distribution of W is Exponential with mean 1 g 1 + . G Y = g = n i i 1 W has Gamma ( α = 2 n , “usual θ ” = 1 g 1 + ) distribution. c) Is the maximum likelihood estimator of θ , g ˆ , an unbiased estimator of θ ? If g ˆ is not an unbiased estimator of θ , construct an unbiased estimator g ˆˆ of θ based on g ˆ . E ( Y 1 ) = ( ) () ( ) ¡ ∞ Γ + + ⋅ 1 1 g 1 1 g dx x n x x n n e = ( ) ( ) ( ) ( ) ( ) ¡ ∞ Γ + + + ⋅ 1 2 1 g 1 1 1 1 g g dx x n n x n n e = 1 1 g + n . Therefore, E ( g ˆ ) = ( ) 1 1 1 g + + ⋅...
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This note was uploaded on 01/26/2012 for the course STAT 410 taught by Professor Monrad during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Monrad
 Statistics

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