# Twisted random walks

Suppose the points of two random walks in $\mathbb{R}^2$ are given the step number (or time) as a third coordinate, so that they become paths in $\mathbb{R}^3$. Here are several pairs of walks of $n=100$ steps, both starting at the origin, with steps normally distributed with $\sigma=1$:

I would like to know the expected number of times that one path winds about the other, as a function of $n$, the number of steps. I believe for the three pairs illustrated, there is zero winding by $n=100$. Experiments indicate winding becomes less likely as $n$ grows. This is a bit counterintuitive to me.

The winding number of path $b(t)$ about path $a(t)$ up to $t=T$ could be defined by counting the number of times (the normalization of) the vector $b(t)-a(t)$ turns around the origin for $t\in[0,T]$.

• What do you mean by "turns around the origin"? -- For random walks $a(t)$ and $b(t)$, the difference $a(t)-b(t)$ will "go hence and forth" in some sense, I think -- I don't see how you get your winding number from this. Commented Aug 6, 2013 at 12:38
• Sorry for not being clear. If one normalizes $b(t)-a(t)$, it can be viewed as a point on the unit circle centered on the origin. (Assume $a(t)=b(t)$ occurs with zero probability.) This point wanders around that unit circle. Commented Aug 6, 2013 at 12:49
• The normalization is clear -- but the issue is that the point won't wander around the unit circle in one direction. It will rather move randomly hence and forth, in both directions -- how do you count this? -- Or have I misunderstood something? Commented Aug 6, 2013 at 13:06
• Can't you equivalently consider a single random walk's winding number about the origin? I think you must put some barrier at the origin, because (in the plane) I think it will hit the origin with non-zero probability. Once you figured out the right set-up, you can convert the discrete random walk into a diffusion equation on the Riemann surface for $\log z$, and if you can solve the diffusion equation you will have your answer. Commented Aug 6, 2013 at 15:31
• By symmetry the expectation would be zero winding, but the interesting question would be how the variance grows with time. Here is the surface on which I was suggesting to solve the diffusion equation: en.wikipedia.org/wiki/File:Riemann_surface_log.jpg Commented Aug 6, 2013 at 15:43

The upshot is that the root mean square winding number grows logarithmically with the number of steps N for $N \to \infty$.