Find Maximum Likelihood Estimator of Gamma Distribution

  1. 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(xi; β)
    - After doing all arithmetic, I get
    L(β) = β^(-2n) * (x1 * x2 * … *xn) * e^ [(-1/ β) * (x1 + x2 +…+xn)]
    - 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 * (x1 + x2+ … + xn)) / 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
     
  2. jcsd
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