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.

  • 2
    $\begingroup$ As I understand it, behaviour of this type of random walk is not particularly sensitive to the underlying lattice, so you should expect the walk to look like Brownian motion. $\endgroup$
    – Ben Barber
    Commented Feb 25, 2014 at 16:53
  • $\begingroup$ @Ben: Regarding your comment that it is not sensitive to the lattice; I had one question: The usual Brownian motion (obtained from the square lattice) is a conformal invariant, where we are looking at the usual conformal structure on $\mathbb{R}^2$. Assuming this Random Walk on the Hexagon converges, I would expect it to be "conformally" invariant. But what is this conformal structure? Is it the usual conformal structure or a different one? In fact I would guess it is the conformal structure induced by the lattice. So certain things do depend on the Lattice. Don't you agree? $\endgroup$
    – Ritwik
    Commented Feb 25, 2014 at 17:22
  • $\begingroup$ I'm afraid I'm not really a probabilist, so can't offer any more details. $\endgroup$
    – Ben Barber
    Commented Feb 25, 2014 at 17:44
  • $\begingroup$ @Ritwik: you can see that the conformal structure is the standard one from the fact that the hexagonal lattice has lots of symmetries. But you do raise an interesting (at least for me) question: how can one determine the conformal structure corresponding to Brownian motion on a more general lattice? $\endgroup$ Commented Feb 26, 2014 at 4:53
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    $\begingroup$ @Ritwik: put another way, your choice of $\pi/2$ breaks the symmetry of the lattice and hence cannot be the correct answer since any symmetry of the lattice gives a symmetry of the limit brownian motion. $\endgroup$ Commented Feb 26, 2014 at 7:43

2 Answers 2


If you want to derive the Functional CLT for this process from Donsker's theorem, you have to add a couple of small ingredients.

Donsker's applies to random walks with i.i.d. steps. Yours is not quite of that kind. Two consecutive steps are not i.i.d: (i)one imposes a direction restriction on another; (ii) also if you color the vertices of the lattice in two colors appropriately, even steps will go from black to white and odd steps will go from white to black.

There are many ways to overcome this. One is to notice that if you observe your process only at even times, then the resulting process has independent steps, so you can apply Donsker's to it. It remains to estimate the difference between the new process and the original one, that is easy.

A more general way is to use the martingale problems approach. See the chapter of Ethier & Kurtz on diffusion approximation.

  • $\begingroup$ @ Yuri: Two questions to be sure I understand your comments: If I look at the sequence of measures $\tilde{\mu}_n := \mu_{2n}$, then you are saying Donsker's theorem immediately implies $\tilde{\mu}_n$ converges weakly? Secondly, I assume this requirement of i.i.d steps is the reason that one can't apply the theorem to Self Avoiding Walk? I just wanted to understand why the machinery used for Random Walks don't go through for Self Avoiding Walks (it is an open question whether the SAW converges to anything). $\endgroup$
    – Ritwik
    Commented Feb 26, 2014 at 4:12
  • $\begingroup$ 1. Sort of, but not quite. If you are using the version of Donsker's theorem where you interpolate linearly between consecutive positions, then this interpolation applied to only odd (or only even) times will not coincide with your original process: you walk along the edges of the hexagonal lattice, but the new process walks along the edges of a certain triangular lattice. The difference between those processes is small and easy to estimate. 2. SAW is a very different beast. It is not a Markov process and, moreover, SAW of length $n$ is not a continuation of SAW of length $m$ for $n>m$. $\endgroup$ Commented Feb 26, 2014 at 4:36
  • $\begingroup$ The argument about looking at pairs of steps makes this straightforward. $\endgroup$
    – jwg
    Commented Feb 26, 2014 at 9:36

Yes, it converges to Brownian motion by Donsker's theorem. This is an example of what's known as universality in statistical mechanics: the large scale dynamics of a system should be independent of its microscopic geometry.

  • $\begingroup$ Can you explain how exactly does it follow from Donsker's theorem? I looked up Donsker's theorem on wikipedia en.wikipedia.org/wiki/Donsker's_theorem I am not seeing immediately how it porves the Random walk converges (on the Hexagon). $\endgroup$
    – Ritwik
    Commented Feb 25, 2014 at 17:23
  • $\begingroup$ I assume user looked it up (or knows it): The only question would be the relative rate at which the steps contract and time speeds up. If they are not properly balanced, you get something degenerate. $\endgroup$ Commented Feb 25, 2014 at 19:03

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