Practice Midterm I S

Practice Midterm I S - Solution to Practice Exercises...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution to Practice Exercises STA131C (Spring 07). Problem 1. a) First we find the likelihood function, ` ( x ; α,λ ) = n Y i =1 f ( X i ; α,λ ) = n Y i =1 1 Γ( α ) λ α X α- 1 i e- λX i = 1 Γ( α ) n λ αn n Y i =1 X α- 1 i e- λX i Hence, the log-likelihood is given by log ` ( x ; α,λ ) =- n log Γ( α ) + αn log λ + ( α- 1) n X i =1 log X i- λ n X i =1 X i . This log-likelihood is differentiable in λ and hence we can find the MLE by differentiat- ing. We have d dλ log ` ( x ; α,λ ) = α n 1 λ- n X i =1 X i . Equating this to zero and solving for λ we obtain b λ = α 1 n n X i =1 X i !- 1 = α ( X )- 1 . Since the second derivative of the log-likelihood is given by d 2 dλ 2 log ` ( x ; α,λ ) =- α n 1 λ 2 < 0 it follows that b λ is a maximum and hence the MLE. b) If a MLE is sufficient, then it is automatically minimal sufficient. Hence, we need to check sufficiency of X. To this end, we see that the joint pdf of the X i ,i = 1 ,...,n can be written as 1 Γ( α ) n λ αn n Y i =1 x α- 1 i e- λx i = 1 Γ( α ) n λ αn exp ( α- 1) n X i =1 log x i- λ n X i =1 x i ....
View Full Document

This note was uploaded on 04/20/2008 for the course STATS 131C taught by Professor Polonik during the Spring '07 term at UC Davis.

Page1 / 4

Practice Midterm I S - Solution to Practice Exercises...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online