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HW__8_SOlutions_to_POST

# HW__8_SOlutions_to_POST - 1{X1 Xn can be modeled by a Gamma...

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1) {X 1 , .... , X n } can be modeled by a Gamma distribution ( 29 ( 29 ( 29 1 1 exp / X f x x x - α α- = β Γ α - β with a fixed value of the shape parameter α .V Thus you only need to calculate the value of β . [In environmental applications when n is small, α may be computed as an average from many samples to increase its precision.] a) What is the MLE of β in this case? We know α so we only need to estimate β. Assuming that the observations are iid, we can construct the likelihood function as: 1 1 1 exp( / ) ( ) n i i i L x x α α β β α - = = Π - Γ 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 α α α α β α β α β β - - - - = = = = = - Π = + Γ - Π Γ The MLE of β 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 β 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 α β α α β = = = - - Γ - + - Now to find the MLE of β we take the derivative of the log of the likelihood function with respect to β, set it equal to zero, and solve for β. 2 1 ln( ) 1 0 n i i L n x α β β β = - = + = FINAL ANSWER 1 1 ˆ n i i X x n β α α = = = This is our MLE for β. 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 0 ˆ ln 2 2 < β β L ( 29 ( 29 0 ˆ 2 1 2 1 2 ˆ ln 2 2 2 3 2 2 2 < - = - = - = - = β α α α β β α β β β α β β n n n x n x n L i i Because for gamma distribution, we have α > 0. b) What is the MM estimator of β in this case? For the method-of-moments (MOM) estimator of β when α is not known, we simply equate the sample mean to the expected value of X, and solve for β. We know that X ~ Gamma(α,β), therefore, FINAL ANSWER ˆ [ ] X E X X αβ αβ β α = =≈ = (Not Required) : If α were not known, then ˆ ˆ X β α = , and thus the estimator for β is a function of the estimator for α. But we really need an estimator for β which is only based on observable values. However, we know that for a Gamma distribution Var[X] = αβ 2 ,

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and using MOM we can equate Var[X] to the sample variance. Therefore, using MOM we have two equations: 2 2 X S αβ αβ = = So we have two equations which we can use to solve for the two unknowns ˆ α and ˆ β .
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HW__8_SOlutions_to_POST - 1{X1 Xn can be modeled by a Gamma...

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