There's a relation between two-dimensional Brownian motion and conformal maps, see e.g. Thurston's answer to [this question](https://mathoverflow.net/questions/51863/does-riemann-map-depend-continuously-on-the-domain). Given two non-empty simply-connected domains $U$ and $V$ in the complex plane which are not equal to the complex plane, the Riemann mapping theorem guarantees the existence of a conformal bijection $U\to V$. Is there a way to use Brownian motion to construct such a map?



Constructing Riemann maps using Brownian motion? 
Let $U$ be a Jordan domain and $b\_0,b\_1,b\_2$ be points appearing in order on $\partial U$. Let $f : U \to \mathbb{H}$ be the Riemann map normalized so that $f(b\_0)=0$, $f(b\_1)=1$ and $f(b\_2)=\infty$. Chris Bishop taught me a poor man's way to construct $f$ using Brownian motion.


Let $A\_k$ be the arc of $\partial U$ between $b\_k$ and $b\_{k+1}$, where the index is taken modulo 3. Similarly, let $B\_0$ denote the interval $(0,1)$, $B\_1$ the interval $(1,+\infty)$ and $B\_2$ the interval $(-\infty,0)$, so that $f$ maps $A\_k$ to $B\_k$ for each $k$. For $z\in U$, let $p\_k$ denote the probability that Brownian motion started at $z$ first hits $\partial U$ along $A\_k$. The probability $p\_k=\omega(z,U)(A\_k)$ is known as the harmonic measure of $A\_k$ with respect to $(z,U)$ and is invariant under conformal maps. One can show there is a unique point $w \in \mathbb{H}$ such that $\omega(w,\mathbb{H})(B\_k)=p\_k$ for each $k$. By conformal invariance, we have $w=f(z)$.


Here is how to find $w$ given $(p\_0,p\_1,p\_2)$. If $C\_0,C\_1,C\_2$ are consecutive arcs on the unit circle such that $C\_k$ measures $2\pi p\_k$ radians, then the harmonic measure $\omega(0,\mathbb{D})(C\_k)$ is obviously $p\_k$. Now, let $c\_k$ be the point at the junction between $C\_{k-1}$ and $C\_k$ and let $M$ be the Moebius transformation sending $c\_0,c\_1,c\_2$ to $0,1,\infty$ respectively. You can write $M$ explicitly as
$$
M(\zeta) = \frac{(c\_1-c\_2)(\zeta-c\_0)}{(c\_1-c\_0)(\zeta-c\_2)}.
$$ 
It is easy to see that $M$ maps the unit disk to the upper half-plane. By conformal invariance, we have $$\omega(M(0),\mathbb{H})(B\_k)=\omega(0,\mathbb{D})(C\_k)=p\_k$$
for each $k$ and hence $M(0)=w=f(z)$.


I implemented this in Matlab some years ago. What you do is for each $z$ where you want to evaluate the Riemann map, throw a bunch of Brownian motions to estimate the probabilities $p\_0$,$p\_1$,$p\_2$. It was not very efficient, but was quite easy to program.


