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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 loglikelihood 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 loglikelihood 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 loglikelihood 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.
 Spring '07
 Polonik
 Statistics

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