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

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).