## Lecture 22. Recurrence and transience of the multidimensional Brownian motion

In the next few Lectures we will illustrate through several examples of application the power of the stochastic integration theory.

We start with a study of the multidimensional Brownian motion. As already pointed out, a multidimensional stochastic process $(B_t)_{t \ge 0}= \left( B^1_t,\cdots, B^n_t\right)_{t \ge 0}$, is called a Brownian motion if the processes $(B^1_t)_{t \ge 0}$ , $\cdots$, $(B^n_t)_{t \ge 0}$ are independent Brownian motions. In the sequel we denote by $\Delta$ the Laplace operator on $\mathbb{R}^n$, that is

$\Delta =\sum_{i=1}^n \frac{\partial^2}{\partial x_i^2}.$

The following result is an easy consequence of the Itō’s formula.

Proposition. Let $f: \mathbb{R}_{\ge 0} \times \mathbb{R}^n\rightarrow \mathbb{R}$ be a function that is once continuously differentiable with respect to its first variable and twice continuously differentiable with respect to its second variable and let $(B_t)_{t \ge 0}=(B^1_t,...,B^n_t)_{t \ge 0}$ be a $n$-dimensional Brownian motion. The process
$X_t=f(t,B_t)-\left( \int_0^t \frac{1}{2} \Delta f (s,B_s)+\frac{\partial f}{\partial t}(s,B_s) ds \right)$
is a local martingale. If moreover $f$ is such that
$\sum_{i=1}^n \left( \frac{\partial f}{\partial x_i}(t,x) \right)^2 \le \phi (t) e^{K \| x \|},$
for some continuous function $\phi$ and some constant $K \in \mathbb{R}$, then $(X_t)_{t \ge 0}$ is a martingale.

In particular, if $f$ is a harmonic function, i.e. $\Delta f=0$, and if $(B_t)_{t \ge 0}$ is a multidimensional Brownian motion, then the process $(f(B_t))_{t \ge 0}$ is a local martingale. As we will see it later, this nice fact has many consequences. A first nice application is the study of recurrence or transience of the multidimensional Brownian motion paths. As we have seen before, the Brownian motion recurrent: It reaches any value with probability 1. In higher dimensions, the situation is more subtle.

Let $(B_t)_{t \ge 0}=(B^1_t,...,B^n_t)_{t \ge 0}$ be a $n$-dimensional Brownian motion with $n \ge 2$. For $a > 0$ and $x \in \mathbb{R}^n$, we consider the stopping time
$T_a^x=\inf \{ t \ge 0, \| B_t +x \| =a \}.$
Proposition. For $a < \| x \| < b$,
$\mathbb{P}\left( T_a^x < T_b^x \right)= \begin{cases} \frac{\ln b -\ln \|x \|}{\ln b-\ln a}, & n=2 \\ \frac{\| x\|^{2-n}-b^{2-n}}{a^{2-n}-b^{2-n}}, & n \ge 3. \end{cases}$

Proof. For $a < \| x \| < b$, we consider the function
$f(x)=\Psi(\|x \|)= \begin{cases} \ln \| x\| , & n=2 \\ \| x \|^{2-n}, & n \ge 3. \end{cases}$
A straightforward computation shows that $\Delta f=0$. The process $(f(B_{t \wedge T_a^x \wedge T_b^x}))_{t \ge 0}$ is therefore a martingale, which implies $\mathbb{E}\left( f(B_{ T_a^x \wedge T_b^x})\right)=f(x)$. This yields
$\Psi(a) \mathbb{P}\left( T_a^x < T_b^x \right)+\Psi(b) \mathbb{P}\left( T_b^x < T_a^x \right)=f(x).$
Since
$\left( T_a^x < T_b^x \right)+\mathbb{P}\left( T_b^x < T_a^x \right)=1,$
we deduce that
$\mathbb{P}\left( T_a^x < T_b^x \right)= \begin{cases} \frac{\ln b -\ln \|x \|}{\ln b-\ln a}, & n=2 \\ \frac{\| x\|^{2-n}-b^{2-n}}{a^{2-n}-b^{2-n}}, & n \ge 3. \end{cases}$ $\square$

By letting $b \to \infty$, we get
Corolllary. For $0 < a < \| x \|$,
$\mathbb{P}\left( T_a^x < +\infty \right)= \begin{cases} 1, & n=2 \\ \frac{\| x\|^{2-n}}{a^{2-n}}, & n \ge 3. \end{cases}$
As a consequence, for $n=2$ the Brownian motion is recurrent, that is, for every non empty set $\mathcal{O} \subset \mathbb{R}^2$,
$\mathbb{P}\left( \exists t \ge 0, B_t \in \mathcal{O}\right)=1.$

Though the two-dimensional Brownian motion is recurrent, points are always polar.

Proposition.
For every $x \in \mathbb{R}^n$, $\mathbb{P}( \exists t \ge 0, B_t=x)=0.$

Proof. It suffices to prove that for every $x \in \mathbb{R}^n$, $x \neq 0$, $\mathbb{P}\left( T_0^x < +\infty \right)=0$. We have
$\{ T_0^x < +\infty \} =\cup_{n \ge 0} \cap_{m \ge \frac{1}{\| x \|}} \{ T_{1/m}^x \le T_n^x\}.$
Since $\mathbb{P} \left( \cap_{m \ge \frac{1}{\| x \|}} \{ T_{1/m}^x \le T_n^x\} \right)=\lim_{m \to \infty} \mathbb{P} \left( T_{1/m}^x \le T_n^x \right)=0$, we get
$\mathbb{P}\left( T_0^x < +\infty \right)=0$ $\square$

As we have just seen, the two-dimensional Brownian motion will hit every non empty open set with probability one. The situation is different in dimension higher than 3: Brownian motion paths will eventually leave any bounded set with probability one.

Proposition. Let $(B_t)_{t \ge 0}=(B^1_t,...,B^n_t)_{t \ge 0}$ be a $n$-dimensional Brownian motion. If $n \ge 3$ then almost surely
$\lim_{t \to \infty} \|B_t \| =+\infty.$

Proof. Let us assume $n \ge 3$. Let $\Phi(x)=\frac{1}{\| x+a \|^{n-2}}$ where $a \in \mathbb{R}^n$. Since $(B_t)_{t \ge 0}$ will never hit the point $-a$, we can consider the process $(\Phi(B_t))_{t \ge 0}$ which is seen to be a positive local martingale from Itō’s formula. A positive local martingale is always a supermartingale. Therefore from the Doob’s convergence theorem, the process $(\Phi(B_t))_{t \ge 0}$ converges almost surely when $t \to \infty$ to an integrable and non negative random variable $Z$. From Fatou’s lemma, we have $\mathbb{E}(Z) \le \lim \inf_{t \to +\infty} \mathbb{E} ( \Phi(B_t))$. By the scaling property of the Brownian motion, it is clear that $\lim \inf_{t \to +\infty} \mathbb{E} ( \Phi(B_t))=0$. We conclude $Z=0$ $\square$

Exercise (Probabilistic proof of Liouville theorem) By using martingale methods, prove that if $f:\mathbb{R}^n \to \mathbb{R}$ is a bounded harmonic function, then $f$ is constant.

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