Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 201 / 5

Let ${\tau }_r=\inf \{t\ge 0:|B_t|=r\}$, ${\tau }_R=\inf \{t\ge 0:|B_t|=R\}$, and $\tau ={\tau }_r\wedge {\tau }_R$. The function $u(y)=|y{|}^{-1}$ is harmonic away from the origin: its radial Laplacian in dimension three isThe Itô formula therefore makes $u(B_{t\wedge \tau })$ a local martingale, bounded between $1/R$ and $1/r$, hence a true martingale. The annulus exit time is finite almost surely. For instance, exit from the containing open ball is finite because from every point in the open ball a sufficiently large Gaussian vector increment has a uniform positive probability of leaving it in unit time; iterate the geometric tail bound from a uniform escape probability.

The optional stopping theorem at $t\wedge \tau$, followed by the dominated convergence theorem, gives $\mathbb{E}u(B_{\tau })=u(x)$. By continuity, there is no radial overshoot and the two exit spheres cannot be hit simultaneously. Writing $p={\mathbb{P}}_x({\tau }_r<{\tau }_R)$ givesAll harmonicity and boundedness statements apply only in the annulus, which stays away from the singularity at zero.

Take a sequence $R\uparrow \infty$ with $R>|x|$. The events $\{{\tau }_r<{\tau }_R\}$ increase, and their union is $\{{\tau }_r<\infty \}$. Indeed, if the sphere of radius $r$ is reached at a finite time, continuity makes the path bounded up to that time, so a sufficiently large outer sphere has not yet been hit. Conversely, each event in the union includes a finite inner hit.

Taking the limit of part (a)'s Brownian sphere-hitting probability in dimension three yields the sharper formulaThis establishes the required escape probability without presupposing the transience of Brownian motion in dimension at least three.

First suppose $|x+z|\le r$ for some lattice point $-z$. If $|x+z|=r$, the desired hit is already at time zero. If $|x+z|<r$, finite exit from that bounded open ball and continuity give the required sphere hit almost surely.

It remains to consider $|x+z|>r$ for every $z\in {\mathbb{Z}}^3$. At each integer time $n$, choose a nearest lattice point $-z_n$ to $B_n$ by rounding its three coordinates, with a fixed rule at ties. Then $|B_n+z_n|\le \sqrt{3}/2$. Put $\delta =\min (r/2,1/4)>0$. Conditional on the past, $B_{n+1}-B_n$ is a standard three-dimensional Gaussian vector. Its probability density function is bounded below on the open ball of radius $\delta$ centered at $-(B_n+z_n)$ byMultiplying by the volume of that open ball gives a constant $\epsilon >0$, independent of $n$ and the past, such thatThe probability of avoiding all lattice-centered radius-$r$ open balls at the first $N$ positive integer times is therefore at most $(1-\epsilon {)}^N$, by iterated conditional expectation. Eventually the path enters one such open ball almost surely. Since the initial point was outside that particular open ball, continuity forces an earlier crossing of its sphere. This proves Brownian hitting of lattice spheres:A uniform chance of hitting some member of the periodic family replaces recurrence to any one prescribed sphere; there is no contradiction with (b).