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


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.

  • $\begingroup$ 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? $\endgroup$ Oct 13 '10 at 22:36
  • $\begingroup$ @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 $2 \pi$ $\endgroup$ Oct 14 '10 at 0:06
  • $\begingroup$ @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. $\endgroup$
    – Alex R.
    Oct 14 '10 at 0:13
  • $\begingroup$ 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? $\endgroup$
    – zhoraster
    Oct 14 '10 at 6:40
  • $\begingroup$ @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. $\endgroup$ Oct 14 '10 at 11:25

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.