For every positive integer $n$, consider a regular hexagon $\mathrm{H}_n$ such that the distance of each vertex from the center is $\frac{1}{\sqrt{n}}$. That in turn induces a tiling of $\mathbb{R}^2$. Let us call that tiling $\mathcal{T}_n$ (assume that one of the vertices is at the origin).

For each such tiling consider the following random walk: Start from the origin at $t=0$. At $t= \frac{1}{n}$ there is a $\frac{1}{3}$ probability to move to any of the neighboring vertex; continue this $n$ times till $t=1$. Join these $n$ points by a straight line to get a continuous map from $[0,1]$ to $\mathbb{R}^2$. There are obviously $3^n$ different continuous paths this way.

Notice that each $n$ gives us a probability measure $\mu_n$ on the space $$ \Omega:= \{ f:[0,1] \rightarrow \mathbb{R}^2: f(0) =0, ~~~~f ~~\text{is continuous}\}.$$

(choose each of those walks with probability $\frac{1}{3^n}$, any other path with probability zero).

My question is the following: Do these probability measures converge in a weak sense to some measure $\mu$ on $\Omega$?

By weak convergence I mean that for any bounded continuous function $\Phi: \Omega \rightarrow \mathbb{R}$ $$ \int \Phi(f) d \mu_n \rightarrow \int \Phi(f) d\mu$$

Here $\Omega$ is to be thought of as a metric space with supremum norm.

Note that, if the tiling was a square tiling with length $\frac{1}{\sqrt{n}}$, then these measures would converge. I believe this is one way to construct the standard Brownian Motion.

shouldexpect the walk to look like Brownian motion. $\endgroup$ – Ben Barber Feb 25 '14 at 16:53