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March 27, 2015

March 27, 2015

Posted by **Cecile** on Saturday, February 25, 2012 at 7:51pm.

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.

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