# How can I show that the first exit time by a planar Brownian motion is a.s. finite, i.e. $\mathbb{P}_z(\tau_D<\infty)=1$?

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.

• @saz can you maybe help me here? Apr 2 at 21:01

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.

• 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 $\{z\in D: |f(z)|<1/2\}$ Apr 3 at 15:52
• 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 $1/2$ Apr 3 at 16:00
• 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 $1/2$ Apr 3 at 16:11
• @user123234 what is the reference for this proof? It does seem strange. I want to read it from there. Apr 3 at 16:15
• 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? Apr 3 at 18:04