4
$\begingroup$

I have found the following interesting proposition on planar Brownian motions, but I don't really understand the proof.

Let $D$ be a proper simply connected domain and $z\in D$. Let $B$ be a complex valued Brownian motion starting from $z$. Then $\mathbb{P}_z(\tau_D<\infty)=1$ where $\tau_D:=\inf\{t\geq 0: B_t\notin D\}$

They have given the following proof:

By the Riemann mapping theorem there exists a conformal isomorphism $f:D\rightarrow \mathbb{D}$. By the conformal invariance theorem there exists a time change $\sigma(t)$ such that $W_t:=f(B_{\sigma(t)})$ is a complex Brownian motion for $t<\tau_{f(D)}=\inf\{t\geq 0: W_t\notin f(D)\}=\inf\{t\geq 0: B_{\sigma(t)}\notin D\}$. Hence if $t\rightarrow \tau_{f(D)}$ then eventually $|W_t|\geq 1/2$. But $B$ is neighbourhood recurrent by a previous corollary, so it visits the open set $\{z\in D: |z|<1/2\}=f^{-1}(\{|z|<1/2\})$ at an unbounded set of times a.s. Hence $\tau_D<\infty$ a.s.

My first question is, where do i need that $W_t$ is a complex Brownian motion, I mean why can't I only work with $B_t$ instead of $W$?

Futhermore how does they conclude that $\tau_D<\infty$? I thouth about if I can show that $\tau_{f(D)}<\infty$ then $\tau_D$ needs to be finite a.s. since there is only a time change in the game.

Can someone explain me this proof or help me how to proof it similarly. I know the conformal invariance theorem and that a complex Brownian motion is neighbourhood recurrent.

$\endgroup$
1
  • $\begingroup$ @saz can you maybe help me here? $\endgroup$
    – user123234
    Apr 2 at 21:01

1 Answer 1

2
$\begingroup$

My first question is, where do i need that $W_t$ is a complex Brownian motion, I mean why can't I only work with $B_t$ instead of $W$?

You could do it with just using neighborhood-recurrence for 2d-Brownian motion. Namely since the domain is a strict subset $D\subset \mathbb C$ and also simply connected, there exists a ball $B_{r}(z_{0})$ in the complement $D^{c}$.

Futhermore how does they conclude that $\tau_D<\infty$? I thouth about if I can show that $\tau_{f(D)}<\infty$ then $\tau_D$ needs to be finite a.s. since there is only a time change in the game.

Here they just use conformal invariance, namely $W_{t}$ is equal in law to standard 2d-BM $\tilde{B}$

$$P[\tau_{B,D}<\infty]=P[\tau_{\tilde{B},f(D)}<\infty],$$

namely the $"\tau_{\tilde{B},f(D)}<\infty"$ is the probability that standard 2d-BM $\tilde{B}$ escapes domain $f(D)$ in finite time.

Hence if $t\rightarrow \tau_{f(D)}$ then eventually $|W_t|\geq 1/2$.

I think this part is clear, since $f(D)=\mathbb D$, we have that due to path-continuity of $W_{t}$, as we near the exit time, it will have to first exit the 1/2-ball $B_{1/2}(0)$.

But $B$ is neighbourhood recurrent by a previous corollary, so it visits the open set $\{z\in D: |z|<1/2\}=f^{-1}(\{|z|<1/2\})$ at an unbounded set of times a.s.

The neighbourhood-recurrence is the statement that for every neighbourhood $U$ around a fixed point $z_{0}$, there is a infinite increasaing sequence of $t_{k}$ s.t. $B_{t_{k}}\in U$.

However, I think they made a slight typo here because it could be that $D$ and the set $|z|<1/2$ are disjoint. I think they meant to use the starting point $z_{0}\in D$ of BM and then study the 1/2-ball centered at it

$$B_{1/2}(z_{0})\cap D:=\{z\in D: |z-z_{0}|\leq \frac{1}{2}\}.$$

Hence $\tau_D<\infty$ a.s.

So their argument up to that point prove the nice statement that Brownian is neighbourhood-recurrent around any arbitrary point $z_0$ rather than just the origin. But this of course implies the statement as mentioned above because it means BM will exit to go hit a ball $B_{r}(x)$ that is fully in the complement $B_{r}(x)\subset D^{c}$.

More proofs

1) The neighbourhood recurrence is also shown in Brownian Motion by Peter Mörters and Yuval Peres.

2) General proof for all dimensions and all closed/open sets with finite positive measure. Let $D\subset \mathbb{R}^{d}$ have measure $m(D)\in (0,\infty)$ and $x\in D$. Then

$$P_{x}(\tau_{D}>t)\leq P_{x}(B_{t}\in D)=\int_{D}p_{t}(x,y)dy=\frac{1}{2\pi t^{d/2}}\int e^{-|x-y|^{2}/2t}dy\leq \frac{m(D)}{2\pi t^{d/2}}.$$

So as $t\to +\infty$, we get $P_{x}(\tau_{D}=\infty)=0$.

In the case of planar-Browmian motion, we can use conformal-invariance to map to the bounded domain $\mathbb{D}$ and thus finite measure.

3) In the notes TOPICS IN COMPLEX ANALYSIS CONFORMAL FRACTALS, PART II: BROWNIAN MOTION by C.Bishop, he has a nice proof for nice-enough domains by studying the expected exit time $E[\tau_{D}]$.

We start with conformal $f:\mathbb{D}\to D$. We may assume that $f(0) = 0$ since translating the domain and starting point does not change the expected exit time. We use the Wald-identity

$$2E[\tau_{D}]=E_{f(0)}[|B_{\tau_{D}}|^{2}].$$

Then the expectation on the right side above is

$$RHS=\int_{\partial D}|z|^{2}d\omega_{f(0)}(z),$$

where $\omega_{f(0)}$ is harmonic measure on $\partial D$ with respect to $f(0)$, i.e., the hitting distribution of Brownian motion started at $f(0)$. By the conformal invariance of Brownian motion, we get

$$2E[\tau_{D}]=E_{f(0)}[|B_{\tau_{D}}|^{2}]=\int_{\partial \mathbb D}|f(z)|^{2}(\theta)d\theta.$$

So here we need a nice enough domain so that the RHS is finite eg. continuous extension of the conformal map Carathéodory-Torhorst theorem.

$\endgroup$
9
  • $\begingroup$ Thanks a lot for your answer. Could you explain again why $P[\tau_{W,f(D)}<\infty]=1$? I don't see where the argument of $|f(B_t)|\geq 1/2$ comes into play and why one uses $\{z\in D: |f(z)|<1/2\}$ $\endgroup$
    – user123234
    Apr 3 at 15:52
  • $\begingroup$ I think I don't get the argument compleatly since I don't see why the unit disk is useful and the disk with radius $1/2$ $\endgroup$
    – user123234
    Apr 3 at 16:00
  • $\begingroup$ My thought was that since a complex BM is neighbourhood recurrent, it should intersect a small neighbourhood of for example $50+i50$, i.e. $W$ should exit $f(D)$ in finite time but I don't think this works since they did it with the disk of radius $1/2$ $\endgroup$
    – user123234
    Apr 3 at 16:11
  • $\begingroup$ @user123234 what is the reference for this proof? It does seem strange. I want to read it from there. $\endgroup$ Apr 3 at 16:15
  • $\begingroup$ Thanks for your extension of the answer. But I already know that a complex Brownian motion is neighbourhood recurrent around any point, how can I then shorten the argument? $\endgroup$
    – user123234
    Apr 3 at 18:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .