# Finding estimator for a one-parameter Weibull distribution

I'm doing some practice problems on methods of moments from a textbook. I am stuck on the following question:

The pdf of a one-parameter Weibull distribution is given by:

$f(x) = \begin{cases} 2\alpha xe^{-\alpha x^2}, &\mbox{if }\: x > 0 \\ 0 & \mbox{otherwise}. \end{cases}$

Using a random sample of size $n$, obtain a moment estimator for $\alpha$.

My Question:

I understand that we need to match population moments with sample moments to get the estimator for $\alpha$. In trying to find the first population moment E[X], I set up the following integral:

$E[X]=\int^{\infty}_0x2\alpha xe^{-\alpha x^2}$

$E[X]=2 \alpha\int^{\infty}_0x^2e^{-\alpha x^2}$.

My question is: how would I best evaluate this integral? Do I need to do integration by parts from here? Or is there an easier "trick" I can use to evaluate this?

Let $y=x^2$, so $dy = 2x dx$ and $x=y^\frac{1}{2}$, leaving you with an integral of a gamma density and some constants.
• After doing the recommended substitution, I get $E(X)=\int^{\infty}_0 \alpha y^{1/2} e^{-\alpha y} dy$. If we take a gamma density such that $\Gamma (t)=\int^{\infty}_0 z^{t-1} e^{-z} dz$, for some variable $z$, does that mean $E(X)$ evaluates to a gamma density with $t=\frac{3}{2}$ and $z=\alpha y$? I am unsure how to evaluate the gamma density now. Do I pull out the $\alpha$ and proceed from there? Apologies if I am missing something very basic here. – Uday Pramod Mar 19 '14 at 9:03