Originally Posted by

**reflex** Hello,

I think I'm missing something silly here, but I'm stuck in a rut - a push would be appreciated.

I'm working with this distribution $\displaystyle f(x\mid\theta) = 2\theta^{-2}x^3exp(-x^2/\theta) (x > 0)$

I want to find the mean and variance. Often it's easy to convert $\displaystyle \int\f(x)dx$ into something that looks like $\displaystyle moment*\int\tilde{f(x)}$ where $\displaystyle \tilde{f(x)}dx$ is some pdf and therefore integrates to 1.

I didn't see a solution like this, so I went brute force and just tried to integrate it figuring it's probably not the most elegant way, but it should work.

using 'by parts':

$\displaystyle \int_{0}^\infty x\cdot2\theta^{-2}x^3exp(-x^2/\theta)dx$

$\displaystyle 2\theta^{-2}\int_{0}^\infty x^4exp(-x^2/\theta)dx$

$\displaystyle 2\theta^{-2}[-\frac{\theta}{2}x^3exp(-x^2/\theta) + \int_{0}^\infty2x^2\theta exp(-x^2/\theta)dx]$

I then try to do by parts again, but it's already clear that when I later go to evaluate the result between 0 and $\displaystyle \infty$ the first x^3 term will blow up and there aren't any other convenient x^3 terms to cancel it out later.

Thanks!