# 409Hw05ans - STAT 409 Homework#5 due Friday October 7 by...

This preview shows pages 1–5. Sign up to view the full content.

STAT 409 Homework #5 Fall 2011 ( due Friday, October 7, by 4:00 p.m. ) 1. Consider Gamma ( α , β = “usual θ ) distribution. That is, f ( x ; α , β ) = ( ) β α α 1 1 β α x e x - - Γ , 0 < x < . Suppose α is known. a) Determine the Fisher information I ( β ). ln f ( x ; β ) = ( ) ( ) β ln β ln ln 1 α α α x x - - + - Γ - . β ln f ( x ; β ) = 2 β β α x + - . 2 2 β ln f ( x ; β ) = 3 2 β β 2 α x - . I ( β ) = – E [ 2 2 β ln f ( X; β ) ] = – E [ 3 2 β β X 2 α - ] = ( ) 3 2 β β X E 2 α + - = 3 2 β β β α α 2 + - = 2 β α . OR I ( β ) = Var [ θ ln f ( X; β ) ] = Var [ 2 β β X α + - ] = ( ) 4 β X Var = 4 2 β β α = 2 β α . b) Find the maximum likelihood estimator β ˆ of β . L ( β ) = ( ) = n i i x f 1 β ; = ( ) - Γ = - = 1 α α 1 1 1 1 β exp β α n i i n i i n x x . ln L ( β ) = n ln Γ ( α ) – n α ln β + ( α – 1 ) = n i i x 1 ln = n i i x 1 1 β . β d d ln L ( β ) = β α n + = n i i x 1 2 1 β = 0. β ˆ = = n i i n 1 X 1 α = α X .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
c) Is β ˆ an unbiased estimator for β ? Justify your answer. E ( β ˆ ) = α μ = α α β = β . β ˆ is an unbiased estimator for β . d) Is β ˆ an efficient estimator for β ? Justify your answer. Var ( β ˆ ) = n 1 2 2 σ α = n 1 2 2 β α α = α 2 β n . Rao-Cramer Lower Bound: ( ) β I 1 n = α 2 β n . β ˆ is an efficient estimator for β . ( Variance of β ˆ attains Rao-Cramer Lower Bound )
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 < . Recall that = - n i i 1 X ln has Gamma ( α = n , β = “usual θ ” = θ 1 ) distribution. If n = 12, find c and d such that P ( c < – 2 θ = n i i 1 X ln < d ) = 0.95. Then - - = = X 2 , X 2 1 1 ln ln n i i n i i d c is a 95% confidence interval for θ . – 2 θ = n i i 1 X ln = = n i i 1 Y has χ 2 ( 2 n d.f. ) distribution. P ( c < – 2 θ = n i i 1 X ln < d ) = 0.95. P ( c < χ 2 ( 24 d.f. ) < d ) = 0.95. c = ( ) 24 2 975 . 0 χ = 12.40 . d = ( ) 24 2 025 . 0 χ = 39.36 . ( OR c = 0, d = 36.42 OR c = 13.85, d = )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. 6.4-6 ( ) Let X 1 , X 2 , … , X n be a random sample of size n from an exponential distribution with unknown mean of μ = θ . a)
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 15

409Hw05ans - STAT 409 Homework#5 due Friday October 7 by...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online