Let X be a random variable with pdf given by $\beta^2xe^{-\frac{1}{2}\beta^2x^2}$ Let X be a random variable with pdf given by $\beta^2xe^{-\frac{1}{2}\beta^2x^2}$ where $\beta>0, x >0$
Find $E[X]$.
I believe this is a function from the gamma distribution. Here is what I have thus far
$$\int_0^\infty\beta^2xe^{-\frac{1}{2}\beta^2x^2}dx$$
$$\beta^2\int_0^\infty xe^{-\frac{1}{2}\beta^2x^2}dx$$
$$u=-\frac{1}{2}\beta^2x^2$$ $$du=-\frac{1}{2}\beta^2x^2dx$$ $$dx=-\frac{du}{\beta^2x}$$
$$=\beta^2\int_0^\infty xe^{u}*-\frac{du}{\beta^2x}$$
$$=-\int_0^\infty e^{u}du$$
I just wanted to know if my steps are correct thus far. I haven't done integration in a while, so I am not sure if my steps are correct/incorrect.
 A: No, it is not a gamma but a reileigh with expectation
$$\mathbb{E}(X)=\frac{1}{\beta}\sqrt{\frac{\pi}{2}}$$
How to calculate it:
$$\mathbb{E}(X)=\int_0^{\infty}\beta^2x^2e^{-(\beta^2x^2)/2}dx$$
set $y=\beta x$ and the result is immediate using standard gaussian's second moment
In fact you get
$$\mathbb{E}(X)=\frac{1}{\beta}\int_0^{\infty}y^2e^{-y^2/2}dy=\frac{\sqrt{2\pi}}{\beta}\int_0^{\infty}\frac{1}{\sqrt{2\pi}}y^2e^{-y^2/2}dy=$$
$$=\frac{\sqrt{2\pi}}{2\beta}\underbrace{\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}y^2e^{-y^2/2}dy}_{=1}=  \frac{1}{\beta}\sqrt{\frac{\pi}{2}}    $$
I can say that
$$\int_{-\infty}^{\infty}f(y)dy=2\int_{0}^{\infty}f(y)dy$$
because $f(y)$ is even.
thus the latest integral is $E(Y^2)=1$ because $Y\sim N(0;1)$
A: There is an $$e^{-\frac{\beta^2}{2}x^2}$$ in the pdf: looks like we may be able to link that to a Gaussian distribution with mean $0$ and variance $\sigma^2 = 1/\beta^2$. Let's do that.
The pdf of that Gaussian would be
$$
f(x) = \frac{\beta}{\sqrt{2\pi}} e^{-\frac{\beta^2}{2}x^2}, \qquad x\in\mathbb{R}
$$
and if $Y$ is distributed according to this, then $\mathrm{Var}[Y] = \mathbb{E}[Y^2]$ (the mean being zero), so
$$
\frac{1}{\beta^2} = \mathbb{E}[Y^2] = \int_{-\infty}^\infty y^2 f(y)dy
= \frac{\beta}{\sqrt{2\pi}} \int_{-\infty}^\infty y^2 e^{-\frac{\beta^2}{2}y^2}dy
= \frac{2\beta}{\sqrt{2\pi}} \int_{0}^\infty y^2 e^{-\frac{\beta^2}{2}y^2}dy \tag{1}
$$
Interesting. what we want to compute, with that other probability distribution we had for some r.v. $X$, was
$$
\mathbb{E}[X] = \int_{0}^\infty x\cdot \beta^2 x e^{-\frac{\beta^2}{2}x^2} dx
= \beta^2\int_{0}^\infty x^2 e^{-\frac{\beta^2}{2}x^2} dx \tag{2}
$$
Great! We are done. Combining (1) and (2), we have
$$
\mathbb{E}[X] = \beta^2 \cdot \left(\frac{2\beta}{\sqrt{2\pi}}\right)^{-1}\mathbb{E}[Y^2]
=  \beta^2 \cdot \frac{\sqrt{2\pi}}{2\beta} \cdot \beta^{-2}
=  \boxed{\frac{1}{\beta}\sqrt{\frac{\pi}{2}}}
$$
and we never needed to know what the distribution of $X$ was named: just basic properties of gaussians.
