Question: Assume I have two random independent standard normal variables $X,Y\sim N(0,1)$, How can I find the distribution of $Z=X^2+Y^2$? 

I thought integrating the convolution, i.e $F\_Z(z)=\displaystyle\int\_{-\infty}^{\infty}\int\_{-\infty}^{\infty}f\_{X^2}(t)f\_{Y^2}(x-t)dt$. If X is a normal variable, does also $X^2$ is a normal variable (then I could find the density by simply substituting)?

Answer: 
**IF** ${X},{Y}\overset{\tt iid}{\sim}\mathcal{N} ({0},{1} )$, **THEN** ${R}=\sqrt{{X}^2+{Y}^2}\sim{Ray}$ where a $\tt Rayleigh\,\,distribution$ is just the square root of and ${Exp}({\lambda}=\frac{1}{2})$

So ${Z}={R}^2\sim{Exp}(\frac{1}{2})$

${Exp}({\lambda}=\frac{1}{2})={gamma}({r}={1},{\lambda}=\frac{1}{2})={gamma}(\frac{n}{2}=\frac{2}{2},\frac{1}{2})={chi-square}({n}={2})$

I did this quite fast. Let me know if I made any mistakes.

Thanks,   

Jason  

Math Undergrad   

University of California, Berkeley
First, both $X^2$ and $Y^2$ are known distributions, namely a chi-squared distribution. See <https://en.wikipedia.org/wiki/Chi-squared_distribution>.

If they are independent, to add them just convolve the pdf's. See: <https://en.wikipedia.org/wiki/Convolution_of_probability_distributions>