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

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.

Set $\alpha =\arctan c\in (0,\pi /2)$. In polar coordinates, the event is $\sin \Theta >c|\cos \Theta |$, the angular sectorBecause $\Theta$ has the continuous uniform distribution on an interval of length $2\pi$, the probability is the sector's angular length divided by $2\pi$:The symmetry of the standard normal distribution in $Y$ makes the lower sector $Y<-c|X|$ equally likely. Their disjoint union is $|Y/X|>c$, apart from $X=0$, an event of probability zero. Therefore the requested cumulative distribution function isThis is the unit-scale half-Cauchy distribution. At $z=0$ there is no atom, and as $z\to \infty$ the cumulative distribution function tends to one.

The left panel uses $c=1$, giving a sector of angle $\pi /2$ and probability $1/4$. The right panel displays the domain $u<v<2u$ for the next change of variables for probability densities; a vertical section through it determines the conditional range.