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,
Best Answer
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 $\frac{x^{2}}{2r^{2}}= t$ $$E(X) = \int_{0}^{\infty}\frac{x^{2}}{r^{2}}e^{-\frac{x^{2}}{2r^{2}}}dx=\int_{0}^{\infty}\sqrt{2}t e^{-t}t^{-\frac{1}{2}}rdt = \sqrt{2}r\int_{0}^{\infty}t^{\frac{3}{2}-1}e^{-t}dt =\sqrt{2}r \Gamma(\frac{3}{2}) = \frac{r}{\sqrt{2}}\Gamma(\frac{1}{2}) = \frac{r}{\sqrt{2}} \sqrt{\pi}$$