Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 33 / 5 / b / Solution

First fix a deterministic point $y$ different from the initial point. By translation, work with $y=0$ and $B_0=x\ne 0$. For $R>|x|$, hitting $0$ before $T_R$ requires hitting every circle of radius $r\in (0,|x|)$ before $T_R$. Part (a) givesAny finite-time hit of $0$ lies before exit from some sufficiently large integer-radius circle, since a continuous path on a finite interval is bounded. A countable union over such radii therefore shows that this fixed point is never hit. This is the polar point for planar Brownian motion property; it is a statement about each fixed point, not about avoiding all points simultaneously.

For the initial point $x$, fix a deterministic $\epsilon >0$. The Gaussian distribution of $B_{\epsilon }$ has a density, so $B_{\epsilon }\ne x$ almost surely. Conditional on ${\mathcal{F}}_{\epsilon }$, the future is Brownian motion starting from $B_{\epsilon }$, and the preceding fixed-point result shows that it never hits $x$. Taking $\epsilon =1/m$ and a countable union proves there is no return to the starting point at any positive time.

In contrast, neighborhoods are revisited arbitrarily late. Choose $r>0$ so that the closed disk centered at $x$ of radius $r$ lies in the given open set $U$. From any point $z$ outside this disk, let $R\to \infty$ in part (a), centered at $x$. It givesAt each deterministic integer time $n$, condition on $B_n$. If it is in the disk there is already a visit to $U$ at time $n$; if it is outside, the Markov property and the last calculation ensure a later visit to that disk, hence to $U$. With probability one this holds for all $n$ simultaneously. Thus the set of visits to $U$ is unbounded. The recurrence of planar Brownian motion concerns neighborhoods even though every fixed point is polar.