pracprelim3soln

pracprelim3soln - ORIE 3500/5500 Fall ’10 Solutions for...

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Unformatted text preview: ORIE 3500/5500, Fall ’10 Solutions for Practice Problems Solutions for Selected Practice Problems Problem 1 (a) The likelihood function L ( θ ) for θ >- 1 is given by L ( θ ) = n Y i =1 f X ( x i ; θ ) = ( θ + 1) n n Y i =1 x i ! θ . So the log-likelihood function is log L ( θ ) = n log( θ + 1) + θ log n Y i =1 x i ! = n log( θ + 1) + θ n X i =1 log x i . Taking the derivative of the log-likelihood function and setting it equal to zero, we obtain d log L ( θ ) dθ = n θ + 1 + n X i =1 log x i = 0 , which yields θ =- n ∑ n i =1 log x i- 1 . Since d 2 log L ( θ ) dβ 2 =- n/ ( θ + 1) 2 < 0, we conclude that the computed value of θ maximizes log L ( θ ). So the MLE for θ is ˆ θ MLE =- n ∑ n i =1 log X i- 1 . (b) If X is a random variable with pdf f X as given in the problem, then E ( X ) = Z 1 x ( θ + 1) x θ dx = ( θ + 1) Z 1 x θ +1 dx = θ + 1 θ + 2 . To compute the method of moments estimator for θ, we equate this expected value with the first sample moment...
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pracprelim3soln - ORIE 3500/5500 Fall ’10 Solutions for...

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