Finding distribution of $X^2+Y^2$ where $X,Y\sim N(0,1)$ 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)?
 A: 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
A: Your distribution is the $\chi^2$ distribution with $2$ degrees of freedom, aka the exponential distribution with mean $2$.  You should be able to calculate the cdf of $X^2+Y^2$ directly by switching to polar coordinates.
Or else you can use your convolution idea. The random variable $X^2$ is not normally distributed, but you can find its distribution in various ways, such as the method of transformations. Or else you can express the probability that $X^2\le w$ as an integral of the normal density function, and then differentiate under the integral sign (fundamental theorem of calculus).
A: 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
