410Hw06ans.pdf - STAT 410 Spring 2019 A Stepanov Homework#6(due Friday April 12 by 4:00 p.m Please include your name with your last name underlined your

# 410Hw06ans.pdf - STAT 410 Spring 2019 A Stepanov...

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STAT 410 Homework #6 (due Friday, April 12, by 4:00 p.m.) Spring 2019 A. Stepanov Please include your name ( with your last name underlined ) , your NetID, and your section number at the top of the first page. No credit will be given without supporting work. 1. Let > 0. Let X 1 , X 2 , … , X n be a random sample from the distribution with probability density function 1 2 λ ln λ λ ; x x x f , 0 < x < 1, zero otherwise. Note: Since 0 < x < 1, ln x < 0. A better way to write this density function would be 1 2 λ ln λ λ ; x x x f = 2 1 λ ln λ x x , 0 < x < 1. a) Obtain a method of moments estimator for , λ . 1 = 1 0 1 2 λ ln λ dx x x . 1 0 1 λ ln dx x x = 2 λ 1 , > 0. E ( X k ) = 1 0 1 2 λ ln λ dx x x x k = 1 0 1 2 λ ln λ dx x x k = 2 2 λ λ k , k > – . = E ( X ) = E ( X 1 ) = 2 2 1 λ λ . OR = E ( X ) = 1 0 1 2 λ ln λ dx x x x = … by parts …
X = 2 2 1 λ ~ λ ~ . X = 1 λ ~ λ ~ . λ ~ = X 1 X . b) Suppose n = 4, and x 1 = 0.4, x 2 = 0.7, x 3 = 0.8, x 4 = 0.9. Obtain a method of moments estimate for , λ . x = 0.70 . λ ~ = 0.70 1 0.70 5.1222 . c) Show that λ is a consistent estimator of . ( NOT enough to say “because it is a method of moments estimator” ) By WLLN, μ X P = 2 2 1 λ λ . Since g ( x ) = x x 1 is continuous at , λ ~ = X 1 X = g ( X ) P g ( ) = 1 μ μ = 1 1 1 λ λ λ λ = . λ ~ is a consistent estimator of .
d) Obtain the maximum likelihood estimator for , ˆ λ .

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