This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 methodofmoments (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 ....
View
Full
Document
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).
 Fall '08
 Stedinger

Click to edit the document details