Post a New Question

Statistics

posted by .

Find Maximum Likelihood Estimator of Gamma Distribution

Given:
f(x; β) = [ 1/( β^2) ] * x * e^(-x/ β) for 0 < x < infinity

EX = 2β and VarX = 2(β^2)

Questions:
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 fn 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. If I let this MLE of β being β’, then for variance, 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.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question