Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 201 / 5

Planar Brownian motion and the initial law ${\nu }_{0,R}$ are invariant under every rotation about the origin. The hitting time ${\tau }_{0,r}$ is also rotation invariant, so the law of $B_{{\tau }_{0,r}}$ is invariant under every rotation of the circle of radius $r$. Planar Brownian motion hits that circle almost surely by recurrence of planar Brownian motion. The unique rotation-invariant probability measure on the circle is its uniform measure, and therefore

Reflection in $L$ interchanges $x$ and $y$. Relabel the two points if necessary so that $x$ lies on the side of $L$ containing the center $a$; the absolute difference is symmetric in $x$ and $y$. Couple a Brownian motion $B$ started at $x$ to one $B^{\prime }$ started at $y$ by setting $B_t^{\prime }=\varphi (B_t)$ before $T_L$ and $B_t^{\prime }=B_t$ afterwards. The Brownian reflection coupling has the correct marginal laws because reflection is an isometry and the Strong Markov property applies at $T_L$. Once the paths meet on $L$, they agree forever.

If the path from $x$ does not hit the target circle before $T_L$, the coupled paths meet before the relevant uncoupled hitting outcomes can differ. The coupling inequality for total variation therefore gives

WriteThis is the harmonic measure of $A$ viewed from $z$, and is a bounded harmonic function outside the closed target disc. The disc is contained in $B(0,r_2)$. In the half-plane cut out by a line $L$ through the origin, the probability of reaching $B(0,r_2)$ before $L$, from a point of modulus $R$, is $O(r_2/R)$ uniformly in the direction. One sees this by mapping the half-plane outside the disc with $z\mapsto \log (z/r_2)$ to a half-strip and solving the corresponding Dirichlet problem by a sine series.

Part (b) consequently shows that the angular oscillation of $H_A$ on a large circle tends to zero, uniformly in the Borel set $A$. The same half-strip estimate in the annulus $R-|a|\le |z|\le R+|a|$ shows that shifting the center of that circle by the fixed vector $a$ changes the average of $H_A$ by $o(1)$, uniformly in $A$. Hence, for every $\epsilon >0$, all sufficiently large $R$ satisfyfor every Borel subset $A$ of the target disc.

By part (a), translated by $a$, starting with $B_0\sim {\nu }_{a,R}$ givesPart (c) says that the same hitting law from ${\nu }_{0,R}$ tends in total variation distance to ${\nu }_{a,r_1}$. On the other hand, every path from the circle of radius $R>r_2$ to the target disc must first hit the circle of radius $r_2$. Part (a) and the Strong Markov property show that its position there has law ${\nu }_{0,r_2}$ and thatindependently of $R$. Letting $R\to \infty$ in part (c) proveswhen $B_0\sim {\nu }_{0,r_2}$.