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

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.

The inverse linear transformation isThe positive support $x>0,y>0$ is exactly $u>0$ and $u<v<2u$. The independence and rate-one exponential distributions give $f_{X,Y}(x,y)=e^{-x-y}$ there. Since $x+y=u$, the change of variables for probability densities yieldsAlthough the density's value depends only on $u$, its support does not factor as a product of fixed intervals. Thus this expression does not imply independence of $U$ and $V$.

Integrating the joint probability density across $u<v<2u$ gives the marginal densityFor every $u>0$, the conditional probability density is consequentlyHenceThis also follows from two independent exponential variables conditioned on their sum: given $X+Y=u$, the variable $Y$ is uniform on $(0,u)$, and $V=u+Y$. Conditioning on $U=u$ here is defined through the conditional probability density, rather than division by $\mathbb{P}(U=u)$, which is zero. The endpoint $u=0$ lies outside the positive marginal support and needs no specified conditional law.

For fixed $v>0$, the support inequalities become $v/2<u<v$. The marginal density is thereforeIt is nonnegative and integrates to $2-1=1$. Another interpretation is that $V=X+2Y$ sums independent exponential random variables of rates $1$ and $1/2$, so it has a hypoexponential distribution. The mean is $1+2=3$, which agrees with ${\int }_0^{\infty }v(e^{-v/2}-e^{-v})dv=4-1=3$.