Past exam of the probability course of the University of Oxford / 2015 / 2 / a / i / Solution

The joint probability density of the two independent standard normal random variables is $(2\pi {)}^{-1}{e}^{-(x^2+y^2)/2}$. The polar coordinates map has absolute Jacobian determinant $r$, so the change of variables for probability densities givesAt the origin the angle may be assigned arbitrarily: that point has probability zero. The joint probability density factorizes as a Rayleigh distribution density $r{e}^{-r^2/2}$ and a continuous uniform distribution density $1/(2\pi )$. Hence $R$ and $\Theta$ are independent.

For $u>0$, substitute $s=r^2/2$ to findTherefore $R^2\sim \mathrm{Exp}(1/2)$, equivalently ${\chi }_2^2$, and applying a measurable function to $R$ preserves its independence from $\Theta$. Explicitly, with $S=R^2$,Thus $\Theta \sim \mathrm{Unif}[0,2\pi )$ and $S\sim \mathrm{Exp}(1/2)$ independently. This polar decomposition of a planar standard normal vector separates the uniformly distributed direction from its random squared distance.