EX 29 - ⋅ ⋅ 1 1 1 θ θ 1 1 θ θ dy y dy y = 2...

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STAT 400 Examples for 11/10/2011 (3) Fall 2011 1. Let X 1 , X 2 , … , X n be a random sample from the distribution with probability density function ( ) ( ) ( ) θ 1 1 θ θ ; X x x f - + = , 0 < x < 1, θ > – 1. a) Obtain the maximum likelihood estimator of θ , θ ˆ . Likelihood function: L ( θ ) = ( ) ( ) ( ) ( ) θ θ 1 1 X 1 1 X 1 1 θ θ - + - + = = = n i i n n i i . ln L ( θ ) = ( ) ( ) = - + + n i i n 1 X 1 θ 1 ln θ ln . ( ) ( ) ( ) = - + + = n i d d i n 1 X 1 1 θ ˆ θ ˆ L θ ln ln = 0. ( ) = - - - = n i i n 1 X 1 1 ln θ ˆ . b) Obtain the method of moments estimator of θ , θ ~ . E ( X ) = ( ) ( ) - + 1 0 θ 1 1 θ dx x x y = 1 – x dy = – dx = ( ) ( ) + - - 0 1 θ 1 1 θ dy y y = ( ) ( )
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Unformatted text preview: ∫ ∫ + ⋅ ⋅ +-+ 1 1 1 θ θ 1 1 θ θ dy y dy y = 2 θ 1 θ 1 + +-= 2 θ 1 + . OR E ( X ) = ( ) ( ) ∫-+ ⋅ ⋅ 1 θ 1 1 θ dx x x u = x dv = ( ) ( ) θ 1 1 θ x-+ ⋅ dx du = dx v = ( ) 1 θ 1 +--x = ( ) ( ) ∫ + +-+--1 1 1 1 θ θ 1 1 dx x x x = ( ) ∫ +-1 1 θ 1 dx x = 2 θ 1 + . OR Beta distribution, α = 1, β = θ + 1. ⇒ E ( X ) = β α α + = 2 θ 1 + . 2 θ ~ 1 X X 1 1 + = = ∑ = ⋅ n i i n ⇒ 2 X 1 X X 2 1 θ ~--= = ....
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This note was uploaded on 02/15/2012 for the course STAT 400 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

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EX 29 - ⋅ ⋅ 1 1 1 θ θ 1 1 θ θ dy y dy y = 2...

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