# Why is $S = X^2 + Y^2$ distributed $Exponential(\frac{1}{2})$?

Let $X$ and $Y$ be independent, standard normally distributed random variables ($\sim Normal(0, 1)$). Why is $S = X^2 + Y^2$ distributed $Exponential(\frac{1}{2})$?

I understand that an exponential random variable describes the time until a next event given a rate at which events occur. However, I do not understand what this has to do with $S = X^2 + Y^2$.

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Let $W=X^2+Y^2$. We find the cumulative distribution function $F_W(w)$ of $w$.
This is the probability that $W\le w$. Clearly $F_W(w)=0$ if $w\le 0$. Now we deal with $w\gt 0$.
So we want $\Pr(X^2+Y^2 \le w)$. This is the integral over the disk with centre the origin, and radius $\sqrt{w}$, of the joint density function of $X$ and $Y$. So for positive $w$, $$F_W(w)=\iint_D \frac{1}{2\pi}e^{-(x^2+y^2)/2}\,dx\,dy.$$ Change to polar coordinates. Then $dx\,dy=r\,dr\,d\theta$, and our integral is $$\int_0^{2\pi}\frac{1}{2\pi}\left(\int_0^{\sqrt{w}}re^{-r^2/2}\,dr \right)\,d\theta.$$ The integration is easy. For the inner integral, let $u=r^2$. The outer integral is even simpler. Quickly we get $\dfrac{1}{2}e^{-w/2}$, which is the cdf of the exponential with parameter $\dfrac{1}{2}$.