I am confused on how to get the cumulative distribution function, mean and variance for the continuous random variable below: Given the condition below

Integrating it by parts makes me confused because of the denominator R^2.

Hope you can help me.

Thank you,

  • 1
    $\begingroup$ It seem to me there is a small mistake there (in the picture). Yes $r$ is taken to be positive, but there should be written $x>0$ otherwise $0$. $\endgroup$
    – kolobokish
    Nov 19, 2016 at 9:07
  • $\begingroup$ Unfortunately, that's the only thing given on the problem. No condition for x $\endgroup$
    – J. Pagana
    Nov 19, 2016 at 9:24
  • $\begingroup$ Then it wouldn't be distribution. What I mean is that $\int_{-\infty}^{\infty}f(x)dx = 2$ (for any $r>0$). I meant in the previous comment, that the function is generally defined for some support. So I think there is mistake. $\endgroup$
    – kolobokish
    Nov 19, 2016 at 9:29

1 Answer 1


So as I mentioned in comment $x>0$. Than I am sure you could do the following by yourself, but anyway I'll write. We have $F(y) = 0 $ while $y\leq 0$. And $F(y) = \int_{0}^{y}\frac{x}{r^{2}}e^{-\frac{x^{2}}{2r^{2}}}dx =\int_{0}^{y}e^{-\frac{x^{2}}{2r^{2}}}d\frac{x^{2}}{2r^{2}} = 1-e^{-\frac{y^{2}}{2r^{2}}} $.

For the mean we would need gamma function. We will make change of variable like this $\frac{x^{2}}{2r^{2}}= t$ $$E(X) = \int_{0}^{\infty}\frac{x^{2}}{r^{2}}e^{-\frac{x^{2}}{2r^{2}}}dx=\int_{0}^{\infty}\sqrt{2}t e^{-t}t^{-\frac{1}{2}}rdt = \sqrt{2}r\int_{0}^{\infty}t^{\frac{3}{2}-1}e^{-t}dt =\sqrt{2}r \Gamma(\frac{3}{2}) = \frac{r}{\sqrt{2}}\Gamma(\frac{1}{2}) = \frac{r}{\sqrt{2}} \sqrt{\pi}$$


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