Expectation of random variable. Problem with gausian error function.

Question

Let $c>0$ and $f_c: (0,\infty) \to \mathbb R$ defined by $f_c(x)=\frac{2x}{c}e^{-x^2/c}$. I have shown that $f_c$ is a density. Let be $X_c$ a related random variable. Now I want to calculate the expectation of $X_c$. My try: $E[X_c]= \int_0^\infty \! xf_c(x) \, \mathrm{d}x$. How can one calculate this without any software? I did not get it without any because of the relation to gausian error function. I guess it is not possible to calculate this so easily. Are there any hidden tricks? Any help is much appreciated!

Answer 1

You cannot find this integral without using some knowledge either on gaussian distribution or on Euler gamma-function. $$\mathbb E[X_c]= \int_0^\infty x\frac{2x}{c}e^{-x^2/c} \, dx = -\int_0^\infty x\,de^{-x^2/c} =\int_0^\infty e^{-x^2/c}\, dx = \dfrac{\sqrt{\pi c}}{2}$$ since $\frac{1}{\sqrt{\pi c}}e^{-x^2/c}$ is the pdf of normal distribution $N(0,c/2)$ and is integrated to $1$ over $\mathbb R$.

The last integral can be found directly usual way as Gaussian integral: let $I=\int_0^\infty e^{-x^2/c}\, dx$ , then $$ I^2=\int\limits_0^\infty e^{-x^2/c}\, dx \int\limits_0^\infty e^{-y^2/c}\, dy = \int\limits_0^\infty \int\limits_0^\infty e^{-(x^2+y^2)/c}\, dx dy = \int\limits_0^{\pi/2}\int\limits_0^\infty e^{-r^2/c}r\,drd\varphi = \frac{\pi c}{4}, $$ so $I=\frac{\sqrt{\pi c}}{2}.$

Or you can change variable in original integral $y=x^2/c$ to get gamma-function $$\mathbb E[X_c]= \int\limits_0^\infty x\frac{2x}{c}e^{-x^2/c} \, dx =\sqrt{c}\int\limits_0^\infty \sqrt{y}\,e^{-y}\,dy = \sqrt{c}\,\Gamma(1.5)=\sqrt{c}\,\frac12\Gamma(0.5)=\frac{\sqrt{\pi c}}2,$$ since $\Gamma(x+1)=x\Gamma(x)$ for $x>0$ and $\Gamma(0.5)=\sqrt{\pi}$. The last equality is the same Gaussian integral again.