Finding the expectation of an pdf to the exponential family

Question

Hi i've seen this question in a book i am using to learn some statistics. I don't have answers, hence I don't know what I'm looking for. The other expectation questions were straightforward but I struggled with this one.

$$ f(x)=\begin{cases} \frac{x}{\theta^{2}}exp\left(\frac{-x^{2}}{2\theta^{2}}\right) & x>0\\ 0 & otherwise \end{cases}$$

Now i understand that the integral would be $$\int_{0}^{\infty}\frac{x^{2}}{\theta^{2}}exp\left(\frac{-x^{2}}{2\theta^{2}}\right)$$

But i was quite sure how I could evaluate this. Can you use a substitution like $u=\frac{x^{2}}{\theta^{2}}$ and they it will almost look like gamma variable.

Any help is appreciated

thank you

Answer 1

$u = \frac {x^2}{2\theta^2}\\ du = \frac {x}{\theta^2}\ dx\\ x = \theta \sqrt {2u}$

$\theta \sqrt 2\int_0^{\infty} u^{\frac 12} e^{-u} \ du\\ \theta \sqrt 2\Gamma(\frac 32) = \theta \sqrt \frac {\pi}{2}$

Answer 2

This is actually an example of a pretty famous class of integrals called Gaussian integrals. They all involve functions of the form $e^{-x^2}$, so you're in luck.

To do this one, use the $u$-substitution $$x = \sqrt{2}\theta u \quad dx = \sqrt{2}\theta du \quad u^2 = \frac{x^2}{2\theta^2}$$ to obtain $$2\sqrt{2}\theta\int_0^\infty u^2 e^{-u^2} du$$ The integral might not look much easier to evaluate, but it's actually not too bad; see this Mathworld link for details. The upshot is that it evaluates to $$2\sqrt{2}\theta\int_0^\infty u^2 e^{-u^2} du = 2\sqrt{2}\theta \frac{\sqrt{\pi}}{4} = \theta\sqrt{\frac{\pi}{2}}$$ and that's your expectation value.