411_hw03_soln

411_hw03_soln - Stat 411 Homework 03 1. Problem 6.1.3....

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Stat 411 – Homework 03 Solutions 1. Problem 6.1.3 . (a) Let X 1 ,...,X n be iid with PMF f θ ( x ) = e - θ θ x /x !, x = 0 , 1 , 2 ,... (a Poisson distribution). Then the likelihood function is L x ( θ ) = n Y i =1 e - θ θ x i x i ! = const · e - θ x 1 + ··· + x n . Taking log and then derivative gives 0 set = ∂θ log L x ( θ ) = ∂θ ± const - + ( x 1 + ··· + x n ) log θ ² = - n + n x θ . From here it’s clear that the MLE is ˆ θ = x . (b) Let X 1 ,...,X n be iid with PDF f θ ( x ) = θx θ - 1 , x (0 , 1) (a beta distribution). The likelihood function is L x ( θ ) = n Y i =1 θx θ - 1 i = θ n ³ n Y i =1 x i ´ θ - 1 . Taking log and then derivative gives 0 set = ∂θ log L x ( θ ) = ∂θ h n log θ + ( θ - 1) n X i =1 log x i i = n θ + n X i =1 log x i . Therefore, the MLE is ˆ θ = - n ( n i =1 log x i ) - 1 . (c) Let X 1 ,...,X n be iid with PDF f θ ( x ) = (1 ) e - x/θ , x > 0 (an exponential distribution with mean θ ). The likelihood equation is L x ( θ ) = n Y i =1 1 θ e - x i = 1 θ n e - ( x 1 + ··· + x n ) . Taking log and then derivative gives 0 set = ∂θ log L x ( θ ) = ∂θ ± - n log θ - n x/θ ² = - n θ + n x θ 2 . Then the MLE is clearly ˆ θ = x . (d) Let X 1 ,...,X n be iid with PDF f θ ( x ) = e - ( x - θ ) I [ θ, ) ( x ) (a shifted exponential distribution). Here, like in the uniform problem from class, we use the indicator function because the support set depends on θ . The likelihood function is L x ( θ ) = n Y i =1 e - ( x i - θ ) I [ θ, ) ( x
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This note was uploaded on 03/12/2012 for the course STAT 411 taught by Professor Staff during the Spring '08 term at Ill. Chicago.

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411_hw03_soln - Stat 411 Homework 03 1. Problem 6.1.3....

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