410Hw06ans - STAT 410 Fall 2011 Homework #6 (due Friday,...

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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, > 1. 1. a) Obtain the method of moments estimator of , 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, = 1, = + 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 , g . Likelihood function: L ( ) = ( ) ( ) ( ) ( ) g g 1 1 X 1 1 X 1 1 g g - +- + = = = n i i n n i i . ln L ( ) = ( ) ( ) =- + + n i i n 1 X 1 g 1 ln g ln . () ( ) ( ) =- + + = n i d d i n 1 X 1 1 g g L g ln ln = 0. G ( ) =--- = n i i n 1 X 1 1 ln g . 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 ( +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-- + = ( + 1 ) e ( +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 ( = 2 n , usual = 1 g 1 + ) distribution. c) Is the maximum likelihood estimator of , g , an unbiased estimator of ? If g is not an unbiased estimator of , construct an unbiased estimator g of based on g . E ( Y 1 ) = ( ) () ( ) + +-- 1 1 g 1 1 g dx x n x x n n e = ( ) ( ) ( ) ( ) ( ) - +- + +--- 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 ) = ( ) 1 1 1 g +- +-...
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410Hw06ans - STAT 410 Fall 2011 Homework #6 (due Friday,...

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