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Take a simple random walk $\gamma$ in the complex plane conditioned to start at point $a$ and end at point $b$. For this random walk, we can define the winding number $W_\gamma(a,b)$ around $b$ in the usual way for complex curves.

If instead we have a 2D Brownian motion $Z=X+iY$, then this definition becomes more complicated. For example, if we have a Brownian motion starting at the origin, we can talk about the winding number $\theta_t$ at time $t$ around the origin by solving the stochastic differential equation

$d\theta_t=\frac{X_tdY_t-Y_tdX_t}{|Z_t|^2}$

with initial condition $\theta_0=0$.

The issue is that for Brownian motion, we cannot condition on the path $Z$ to hit a particular point, because this has probability zero. Moreover, by considering annuli around $b$ and the fact that planar Brownian motion moves between concentric annuli with positive probability, it seems to me the situation becomes rather singular.

Question: Is there a sensible generalization for the winding number of a Brownian motion conditioned to hit a single point?

For example, can we look at the limit of the winding number around an annulus about point $b$ whose radius shrinks to zero? I would imagine we would require the Brownian motion to be conditioned to hit some region of positive area just outside the shrinking annulus.

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What's the usual way of defining $W_\gamma(a,b)$? It seems to me that if the curve passes through $b$ there isn't any reasonable winding number around $b$. Also, shouldn't the curve be closed in order to have a well defined winding number? –  Pablo Lessa Oct 13 '10 at 22:36
    
@Pablo: Re your 2nd question: One can define a winding number for an open curve, but it will not in general be an integer, just the total angular turn from end to end divided by $2 \pi$ –  Joseph O'Rourke Oct 14 '10 at 0:06
    
@Pablo: so to be clear, b is fixed to be the endpoint of the random walk. As in you are given the walk starts at a and ends at b. –  Alex R. Oct 14 '10 at 0:13
    
Why do you need to hit a point? I see that you need to start from a point, which is no problem. Could you clarify, in which place you need hitting a point? –  zhoraster Oct 14 '10 at 6:40
    
@Joseph: Ok, thanks! I guess the problem with the endpoint being $b$ is that the winding number might be infinite. But I see now how one could define it as you say. –  Pablo Lessa Oct 14 '10 at 11:25

2 Answers 2

Actually, it is quite possible to condition Brownian motion to hit a given point at a given time. The process is a called a Brownian bridge and the distribution of the winding number of the Brownian bridge is even known explicitly. It has been computed by Marc Yor in the paper

Marc Yor: Loi de l'indice du lacet brownien, et distribution de Hartman-Watson. Z. Wahrsch. Verw. Gebiete 53 (1980), no. 1, 71–95.

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I think you need to take the log of the Brownian motion and look at the imaginary coordinate.

$$R + i\theta = \log(X+iY)$$

Since $z \mapsto \log z$ is conformal, there exists a time change taking one Brownian motion to the other.

Also, notice that $R_t$ is related to the Cauchy distribution.

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