Let X be a random variable with pdf given by $\beta^2xe^{-\frac{1}{2}\beta^2x^2}$

Question

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.

Answer 1

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 $Y\sim N(0;1)$

Answer 2

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.