Trouble deriving sum of squared normals is Exponential with mean $2$ Box-Muller method hinges on the fact that $R = Z_1^2 + Z_2^2$ is Exponential with mean 2, where $Z_1, Z_2$ are independent standard normals. I want to derive this fact but am getting stuck. I proceed as follows:
\begin{split}
P(R = r) 
&= 
\int_{-\infty}^\infty P(R = r\mid Z_1 = z_1)P(Z_1 = z_1)
\\
&= 
\int_{-\sqrt{r}}^{\sqrt{r}} P(R = r\mid Z_1 = z_1)P(Z_1 = z_1)
\\
&= 
\int_{-\sqrt{r}}^{\sqrt{r}} P\left(Z_2 = \pm\sqrt{r - z_1^2}\right)P(Z_1 = z_1)
\\
&= 
\int_{-\sqrt{r}}^{\sqrt{r}} P\left(Z_2 = \pm\sqrt{r - z_1^2}\right)P(Z_1 = z_1)
\\
&= 
\int_{-\sqrt{r}}^{\sqrt{r}} \frac{e^{-r/2}}{\pi}dz_1
\\
&= 
\frac{e^{-r/2}}{\pi}2\sqrt{r}
\\
\end{split}
However, this is not Exponential with mean 2. Can someone please point out what is wrong with above?
 A: I wasn't making a careful distinction between probability of a continuous r.v. and its density; hence all the confusion. A simple way around is to  work with CDF:
\begin{split}
P(R \leq u) 
&= 
\int_{x^2 + y^2 \leq u} \frac{e^{-x^2/2}}{\sqrt{2\pi}}\frac{e^{-y^2/2}}{\sqrt{2\pi}} \, dx \, dy
\\
&= 
\int_{x^2 + y^2 \leq u} \frac{e^{-(x^2 + y^2)/2}}{2\pi} \,dx \,dy
\\
&= 
\int_{r=0}^\sqrt{u} \left( \int_{\theta=0}^{2\pi} \frac{e^{-r^2/2}}{2\pi}  d\theta\right) r dr
\\
&= 
\int_{r=0}^\sqrt{u}e^{-r^2/2} r\,dr
\\
&= 
1 - e^{-u/2}
\\
\end{split}
This is the CDF of an exponential with mean $2.$
A: I had great difficulty interpreting your work because your use of notation is a bit unclear.
First, let's figure out what the distribution of $S = X^2$ is, where $X \sim \mathrm{Normal}(0,1)$.  We note $$\Pr[S \le s] = \Pr[-\sqrt{s} \le X \le \sqrt{s}] = F_X(\sqrt{s}) - F_X(-\sqrt{s}) = 1 - 2 F_X(-\sqrt{s})$$ from the symmetry of the standard normal distribution.  It follows that $$f_S(s) = \frac{d}{ds}\left[1 - 2F_X(-\sqrt{s})\right] = \frac{1}{\sqrt{s}} f_X(-\sqrt{s}) = \frac{1}{\sqrt{2\pi s}} e^{-s/2}, \quad s > 0.$$  Then we use convolution to obtain the density of $R = S + T$ where $T \sim S$:  $$f_R(r) = \int_{t=0}^r f_S(r-t)f_T(t) \, dt = \frac{e^{-r/2}}{2\pi} \int_{t=0}^r \frac{dt}{\sqrt{(r-t)t}}.$$  This last integral is not difficult to evaluate and is left as an exercise.  Hint:  try the substitution $t = r \sin^2 \theta$.
