HW__8_SOlutions_to_POST - 1) {X 1 ,...., X n } can be...

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Unformatted text preview: 1) {X 1 ,...., X n } can be modeled by a Gamma distribution ( 29 ( 29 ( 29 1 1 exp / X f x x x- - = - with a fixed value of the shape parameter .V Thus you only need to calculate the value of . [In environmental applications when n is small, may be computed as an average from many samples to increase its precision.] a) What is the MLE of in this case? We know so we only need to estimate . Assuming that the observations are iid, we can construct the likelihood function as: 1 1 1 exp( / ) ( ) n i i i L x x - = = - We can simplify the likelihood function to obtain: 1 1 1 1 1 1 1 1 1 exp [( )] exp ( ) n n n n n n n i i i i i i i i L x x x x ---- = = = = =- = + - The MLE of will be the value which maximizes the likelihood function. NOTE: It is not necessary to compute the log of the likelihood function, but it is generally easier to find the optimal value of using the log of the likelihood function. 1 1 1 ln( ) ln( ) ln[ ( )] ( 1) ln( ) n n i i i i L n n x x = = = -- - +- Now to find the MLE of we take the derivative of the log of the likelihood function with respect to , set it equal to zero, and solve for . 2 1 ln( ) 1 n i i L n x =- = + = FINAL ANSWER 1 1 n i i X x n = = = This is our MLE for . Notice that it is only a function of observable/known values. (Not Required : You should also check the second order derivative, to ensure that this is in fact a maximum. ( 29 ln 2 2 < L ( 29 ( 29 2 1 2 1 2 ln 2 2 2 3 2 2 2 <- =- = - =- = n n n x n x n L i i Because for gamma distribution, we have > 0. b) What is the MM estimator of in this case? For the method-of-moments (MOM) estimator of when is not known, we simply equate the sample mean to the expected value of X, and solve for . We know that X ~ Gamma(,), therefore, FINAL ANSWER [ ] X E X X = = = (Not Required) : If were not known, then X = , and thus the estimator for is a function of the estimator for . But we really need an estimator for which is only based on observable values. However, we know that for a Gamma distribution Var[X] = 2 , and using MOM we can equate Var[X] to the sample variance. Therefore, using MOM we have two equations: 2 2 X S = = So we have two equations which we can use to solve for the two unknowns and ....
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This note was uploaded on 02/02/2008 for the course CEE 3040 taught by Professor Stedinger during the Fall '08 term at Cornell University (Engineering School).

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HW__8_SOlutions_to_POST - 1) {X 1 ,...., X n } can be...

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