 Solution

The Brownian reflection principle gives, for $t>0$,For a standard normal random variable $N^{\prime }$, this is also $\mathbb{P}(|N^{\prime }|\ge x/\sqrt{t})=\mathbb{P}(x^2/(N^{\prime }{)}^2\le t)$. Thus the inverse-square Gaussian law of Brownian first passage isIn particular, ${\tau }_x$ is finite almost surely.

The Brownian motions $X,Y$ are independent, so conditional on ${\tau }_x=s$, the ordinate $Y_{{\tau }_x}$ has the same distribution as $\sqrt{s}N$, with $N$ a standard normal independent of the random time. By Brownian scaling and the hitting-time identity,The last equality uses symmetry of $N$ and independence of $N,N^{\prime }$: inserting the independent sign of $N^{\prime }$ does not change the distribution of the numerator. ThereforeFor completeness, the ratio has density ${\int }_{\mathbb{R}}|v|\phi (cv)\phi (v)dv=1/(\pi (1+c^2))$, where $\phi$ is the standard normal density. The scaled exit ordinate has density $x/(\pi (x^2+y^2))$.