(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Find Maximum Likelihood Estimator of Gamma Distribution

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