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Unformatted text preview: Stat 411 – Homework 05 Solutions 1. (a) The likelihood function looks like L ( θ 1 ,θ 2 ) = n Y i =1 f θ ( X i ) ∝ θ n/ 2 2 e (1 / 2 θ 2 ) ∑ n i =1 (log X i θ 1 ) 2 . Taking a natural logarithm gives ‘ ( θ 1 ,θ 2 ) = const n log θ 2 2 1 2 θ 2 n X i =1 (log X i θ 1 ) 2 . Differentiating with respect to θ 1 and θ 2 separately and setting equal to zero gives the likelihood equations: 0 = ∂ ∂θ 1 ‘ ( θ 1 ,θ 2 ) = 1 θ 2 n X i =1 (log X i θ 1 ) , 0 = ∂ ∂θ 2 ‘ ( θ 1 ,θ 2 ) = n 2 θ 2 + 1 2 θ 2 2 n X i =1 (log X i θ 1 ) 2 . The solution to the first equation is ˆ θ 1 = 1 n ∑ n i =1 log X i , irrespective of the value of θ 2 . With the solution ˆ θ 1 plugged in to the second equation, the solution for θ 2 is clearly ˆ θ 2 = 1 n ∑ n i =1 (log X i ˆ θ 1 ) 2 . This is just like if we had originally made a transformation Y = log X and did the calculations with Y 1 ,...,Y n iid ∼ N ( θ 1 ,θ 2 ); that is, the MLEs would be ˆ θ 1 = Y and ˆ θ 2 = 1 n n X i =1 ( Y i Y ) 2 , Y i = log X i , i = 1 ,...,n. (b) Towards the Fisher information matrix, we need various derivatives of log f θ ( x ): ∂ ∂θ 1 log f θ ( x ) = log x θ 1 θ 2 ∂ ∂θ 2 log f θ ( x ) = 1 2 θ 2 + (log x θ 1 ) 2 2 θ 2 2 ∂ 2 ∂θ 2 1 log f θ ( x ) = 1 θ 2 ∂ 2 ∂θ 2 2 log f θ ( x ) = 1 2 θ 2 2 (log x θ 1 ) 2 θ 3 2 ∂ 2 ∂θ 1 ∂θ 2 log f θ ( x ) = log x θ 1 θ 2 2 ....
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 Spring '08
 STAFF
 Derivative, Conditional Probability, Maximum likelihood, Francis Ysidro Edgeworth, Fisher Information

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