Independence of Gaussian variables I want to solve the following problem:
Suppose that $X$ and $Y$ are two independent, Gaussian variables, with mean zero and common variance $\sigma^2$. Let $R^2 = X^2+ Y^2$ and $Z = \arctan (X/Y)$. I know that in this case $R^2$ has an exponential distribution and $Z$ has a uniform distribution. However, are $R$ and $Z$ independent from each other?
 A: Yes, let $X=R\cos\Theta$ and $Y = R\sin\Theta$ where $R \in [0,\infty)$ and $\Theta \in [0,2\pi)$. Then, for $r \in [0,\infty), \theta \in [0,2\pi)$,
\begin{align}
F_{R,\Theta}(r,\theta) & = P\{R \leq r, \Theta\leq \theta\}\\
&= P\{(X,Y) \in ~\text{sector of circle of radius $r$ and angle $\theta$}\}\\
&= \iint_{\text{sector}} \frac{1}{2\pi\sigma^2}\exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)\,\mathrm dx\, \mathrm dy
\end{align}
Change to polar coordinates $(\rho,\psi)$ and evaluate the integral to get
\begin{align}F_{R,\Theta}(r,\theta) &= \int_0^r\int_0^\theta \frac{1}{2\pi\sigma^2}\exp\left(-\frac{\rho^2}{2\sigma^2}\right) \,\rho\,\mathrm d\psi \,\mathrm d\rho
\\
&=\int_0^r \frac{\rho}{\sigma^2}\exp\left(-\frac{\rho^2}{2\sigma^2}\right) \,\mathrm d\rho\int_0^\theta \frac{1}{2\pi}\,\mathrm d\psi \\
&= \left[-\exp\left(-\frac{\rho^2}{2\sigma^2}\right)\bigg|_0^r\right] \cdot \left[\frac{\psi}{2\pi}\bigg|_0^\theta\right]\\
&=\left[1 -\exp\left(-\frac{r^2}{2\sigma^2}\right)\right]\frac{\theta}{2\pi} \tag{1}\\
&= F_R(r)\cdot F_\Theta(\theta)\end{align}
which shows that $R$ and $\Theta$ are independent random variables, and also that $R$ is a Rayleigh random variable while $\Theta$ is uniformly distributed on $[0,2\pi]$.
Comment: I wish to thank @Titus for pointing out the error in what I had written in the first version of this answer and also for insisting that his calculation of $F_{R,\Theta}(r,\theta)$ was correct by gently asking where the error was in his calculation. Upon multiplying out the right hand side of $(1)$ and re-arranging terms, we do get his result
$$F_{R,\Theta}(r,\theta) = -\frac{\theta}{2\pi}\exp\left(-\frac{r^2}{2\sigma^2}\right) + \frac{\theta}{2\pi}.$$
