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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 ( ) for >- 1 is given by L ( ) = n Y i =1 f X ( x i ; ) = ( + 1) n n Y i =1 x i ! . So the log-likelihood function is log L ( ) = n log( + 1) + log n Y i =1 x i ! = n log( + 1) + 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 ( ) d = n + 1 + n X i =1 log x i = 0 , which yields =- n n i =1 log x i- 1 . Since d 2 log L ( ) d 2 =- n/ ( + 1) 2 < 0, we conclude that the computed value of maximizes log L ( ). So the MLE for is 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 ( + 1) x dx = ( + 1) Z 1 x +1 dx = + 1 + 2 . To compute the method of moments estimator for , we equate this expected value with the first sample moment...
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