Find Maximum Likelihood Estimator of Gamma Distribution

[Ceci020]

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) * (x₁ * x₂ * … *x_n) * e^ [(-1/ β) * (x₁ + x₂ +…+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 * (x₁ + x₂+ … + 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

 

[mmwave]

.1/ Yes, you are correct. The given function is indeed a gamma distribution with alpha = 2.

2/ Yes, you are correct till this point. Your method for finding the MLE is correct.

3/ No, you cannot assume that X must be 2 β. The given information only tells you the expected value of X, not the actual value of X. The MLE for β is the sample mean divided by 2, but this does not mean that the population mean is necessarily equal to 2 β. It is possible that the population mean is different from 2 β.

4/ To find the variance of the MLE, you can use the fact that the variance of a sample mean is equal to the population variance divided by the sample size, which in this case is n. So, the variance of the MLE, denoted as Var(β''), would be Var(β'') = Var(X/2) = Var(X)/4 = (2β^2)/4 = (β^2)/2. This means that the MLE is unbiased, as the expected value of the MLE is equal to the true parameter β.

You do not need to use integration to find the variance of the MLE. You can simply use the properties of variance and the fact that the sample mean is a consistent estimator of the population mean.