Practice Midterm I S

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

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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 ....
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## 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.

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

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