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Find Maximum Likelihood Estimator of Gamma Distribution 
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#1
Feb2512, 06:45 PM

P: 11

Given
f(x; β) = [ 1/( β^2) ] * x * e^(x/ β) for 0 < x < infinity EX = 2β and VarX = 2(β^2) Questions: Find the Maximum likelihood estimator of β (I call it β''), then find Bias and variance of this β'' 1/ First, I believe this is a gamma distribution with alpha = 2. Is that right? 2/ Find Maximum likelihood estimator of β  First, I get the likelihood function L(β) = product of all the f(x_{i}; β)  After doing all arithmetic, I get L(β) = β^(2n) * (x_{1} * x_{2} * … *x_{n}) * e^ [(1/ β) * (x_{1} + x_{2} +…+x_{n})]  Then I take the log function of L(β), I call it l(β), find derivative of this l(β), equate the derivative to 0 and solve. I get the MLE is (1/n * (x_{1} + x_{2}+ … + x_{n})) / 2 = (sample mean )/2. Am I correct till this point? 3/ If the given said EX = 2 β, can I assume that X must be 2 β, and thus for the population, β is X / 2 ?? 4/ I think this MLE is not bias, but I get confused when I try to find the Variance of the MLE. For β'', I believe that I should find E(β’' ^ 2) – [ E(β’') ] ^2. But from here, my question is that should I use the fact that β’' is (sample mean) / 2, then just plug in and solve ? Or should I do integration? Because I think since this distribution is continuous, isn’t it that E(anything) is integration of that “anything” with f(x)dx ?? Please help me, thanks in advance 


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