# Recurrence and Transience of BM

In this post, we consider Brownian motion as a Markov process, and consider the recurrence and transience properties in several dimensions. As motivation, observe from Question 5 of this exam paper that it is a very much non-trivial operation to show that Brownian motion in two-dimensions almost surely has zero Lebesgue measure. We would expect this to be true by default, as we visualise BM as a curve. So it is interesting to see how much we can deduce without significant technical analysis. We will make use of Ito’s formula. This material comes from the Part III course Advanced Probability, which doesn’t explicitly mention Ito’s result, and instead proves the result required separately, making use of the nature of solutions to the diffusion equation. In this context we assume that for $f\in C_b^{1,2}$:

$M_t:=f(t,B_t)-f(0,B_0)-\int_0^t(\frac{\partial}{\partial t}+\frac12\Delta)f(s,B_s)ds$

is a martingale. Of course, precisely, from Ito’s formula, this can be expressed as the stochastic integral of a bounded function with respect to Brownian motion, which is therefore a (continuous local, but bounded) martingale.

d=1: In one dimension, BM is point-recurrent. This means that almost surely, BM returns to zero infinitely many times. This is easiest shown by using the time-reversal equivalence to deduce that $\lim\sup B_t=-\lim\inf B_t=\infty$.

d=2: BM is two dimensions is point-transient. That means that the probability of returning to a given point is <1. In fact it is 0, as one might suspect from the fact that BM is space-invariant and, intuitively at least, has measure 0. However, it is neighbourhood-recurrent, meaning that it almost surely returns to a ball around a given point infinitely often. We discuss small balls around 0, but obviously the conclusions apply equally well elsewhere.

The aim is to choose a function so that the expression in Ito’s formula as above is as simple as possible. Taking f a function of space alone and harmonic causes the integral term to vanish. In this case, $f(y)=\log|y|$ will suffice. Obviously we have to restrict attention to $\epsilon\leq |y|\leq R$. We stop M at $T_\epsilon\wedge T_R$, that is the first time that the BM hits the boundary of the annulus on which f is defined, and apply OST, since $\log|B_t|$ is bounded here and the stopping time is a.s. finite. We obviously have to assume the BM starts in this annulus, but then we obtain:

$\mathbb{E}_x\log|B_{T_\epsilon\wedge T_R}|=\log|x|$

and so we can consider the two possibilities for $B_{T_\epsilon\wedge T_R}$ to deduce:

$\mathbb{P}_x(T_\epsilon

Now let $\epsilon\downarrow 0$ to see that $\mathbb{P}_x(B_t=0,\text{ some }t>0)=0$. Now apply the (weak) Markov property at a small fixed time a, to deduce, with a mild abuse of notation:

$\mathbb{P}_0(B_t=0,\text{ some }t>a)=\int \mathbb{P}_x(B_t=0,t>0)\mathbb{P}_0(B_a=dx)=0$

as the first term in the integral we have shown to be 0 $B_a$-a.e. Then let $a\downarrow 0$ to obtain the result about point-transience.

For neighbourhood recurrence, instead let $R\uparrow\infty$, so $\mathbb{P}_x(T_\epsilon<\infty)=1$. As before, can integrate over law of $B_n$ to obtain

$\mathbb{P}_0(|B_t|\leq \epsilon,\text{ some }t\geq n)=1$

which is precisely what we require for transience.

d=>3: BM is transient. That is, $|B_t|\rightarrow\infty$ a.s. Note that for d>3, the first three components have the same distribution as BM in three dimensions, and so it suffices to consider the case d=3.

Here, the correct choice of harmonic function is $f(y)=\frac{1}{|y|}$, so conclude as before that

$\mathbb{P}_x(T_\epsilon

From this, we can take a limit to see that

$\mathbb{P}_x(T_\epsilon<\infty)\leq \frac{\epsilon}{|x|}$

We deploy a neat trick to lift this result to a global statement about transience. Define the events that the modulus never returns to $n$ after hitting $n^3$

$A_n:=\{|B_t|>n\quad \forall t\geq T_{n^3}\}$

Calculate

$\mathbb{P}_0(A_n^c)\stackrel{\text{SMP}}{=}\mathbb{E}_0[\mathbb{P}_{B_{T_{n^3}}}(T_n<\infty)]=\mathbb{E}_0[\frac{1}{n^2}]=\frac{1}{n^2}$

Applying Borel-Cantelli 1, $A_n$ eventually holds almost surely, which certainly implies the desired result.