Independence of Gaussian variables

Solution 1:

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 $$F_{R,\Theta}(r,\theta) = -\frac{\theta}{2\pi}\exp\left(-\frac{r^2}{2\sigma^2}\right) + \frac{\theta}{2\pi}.$$