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Divide the unit circle into three arcs, and let $z$ be a point in the open unit disk. Is there a simple formula for the probability that Brownian motion started at $z$ will hit one particular arc rather than the other two when it first hits the unit circle? It would also be nice to have a way of representing $z$ in terms of the three probabilities (assuming that the mapping from points to probability triples is invertible); indeed, this would give a natural way to assign "conformal coordinates" to points in the interior of any simply-connected domain (given a choice of three points on the boundary). Is there literature on this?

In a related vein, given a triangle, one might ask for a formula that Brownian motion started at some point $z$ inside the triangle will hit one particular side of the triangle rather than the other two when it first hits the boundary of the triangle. This is equivalent to the preceding question (via the conformal map between the triangle and the disk that sends the vertices to the three marked points) but one picture might lead to nicer formulas than the other.

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In view of some of the answers, which seem to assume you know less probability theory than I suspect you do, perhaps you could edit your question to emphasize that you are looking for nice and suggestive formulas, rather than "integrate the density of the exit distribution/Poisson kernel over each arc"? (Sorry if I've misunderstood your intent.) –  Yemon Choi Oct 29 '10 at 18:23
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4 Answers 4

up vote 11 down vote accepted

I think zhoraster's answer is on the right track, but it can be said in a simple geometric way:

You can think of the interior of the disk as the hyperbolic plane, using the Poincaré disk model. Brownian paths are the same in any conformally equivalent metric. It follows (by symmetry) that the hitting probability for paths starting at a point $x$ is proportional to the angle subtended by the three arcs in the hyperbolic plane.

For the same problem in the upper half plane, there's an even simpler description: The hyperbolic angle subtended by an arc on the real line from a point in upper half space is exactly twice the Euclidean angle. Roughly speaking: If you're in a big stuffy room with an open door in the middle of one wall, the amount of fresh air you get is proportional to the visual width of the door.

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It is a real pleasure to read your geometric insights :) –  Mariano Suárez-Alvarez Oct 29 '10 at 13:36
    
Cute interpretation! However, I don't understand why the probabilities should be proportional to the angles. –  zhoraster Oct 29 '10 at 17:27
    
@zhoraster: I understood now. Beautiful idea, thanks for sharing it. –  zhoraster Oct 29 '10 at 17:42
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For those who might still not understand, note that the proportionality between probabilities and angles holds when $x$ is at the center of the disk; so its holds more generally (since hyperbolic angles are invariant under Mobius transformations of the unit disk). This is what Bill meant by "(by symmetry)". –  James Propp Oct 29 '10 at 19:33
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Let $B$ be the two-dimensional Brownian motion, $D$ be the unit disk, $\tau$ be the moment of hitting the circle (equivalently, of escaping the interior of the disk), $A$ be one of the arcs.

You're speaking about $u(z) = E^z[\mathbf{1}_{A}(B(\tau))]$, where $E^z$ is the expectation given $B(0)=z$. By Kakutani, this is exactly the solution to the following Dirichlet problem for Laplace's equation: \begin{equation}\tag{LE} \begin{cases} \Delta u = 0 & \quad\text{inside }D, \\\\ u=\mathbf{1}_A & \quad\text{on }\partial D. \end{cases} \end{equation}

For example, it can be expressed via the Poisson integral formula.


Update: a solution of (LE)

(This must be known and written somewhere, though I don't know any reference.)

Let $O$ be the disk center, $A$ be the arc from $e_1$ to $e_2$ (counterclockwise) and $|A|$ its length.

Start by the following simple observation. Let us be given a point $e$ on the circle. Define the function $$u_e(z) = \angle(zeO),$$ which to a point $z$ assigns an angle in $(-\pi/2,\pi/2)$ by which one should rotate the ray $ez$ counterclockwize around $e$ to get $eO$. Then $u_e$ is harmonic inside $D$.

Now on the circle, $u_{e_1} -u_{e_2} = \pi -|A|/2$ inside the arc $A$ and $u_{e_1}-u_{e_2} = -|A|/2$ outside. Thus we can immediately write the solution of (LE): $$ u(z) = \frac{1}{\pi}(u_{e_1}(z)-u_{e_2}(z)+|A|/2). $$

Consequently, the map you are interested in is invertible.

Hope I didn't mess with those angles. Made it community wiki so one can correct if needed.

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I think that an extremely explicit expression for the solution may be written. Will write in an hour or two. –  zhoraster Oct 29 '10 at 10:28
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Use a conformal mapping to transform the circle to the upper half plane. Since Brownian motion is conformally invariant you have reduced the problem to studying the probability that a Brownian motion started in the upper half plane hits hits the x-axis in a given interval. However, the distribution of the hitting location of the x-axis is a Cauchy distribution with centering and scaling depending on the starting location in the upper half-plane (see the related Math Overflow question).

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