Motivation: If I recall correctly, the simple symmetric random walk from i.i.d binary steps converges in distribution to the Wiener measure (if scaled with $a_n = \sqrt{n}$). What I am wondering is if something like that can be true for the simplest opposite example I can think of: walks constructed from irrational rotation.

Let $\alpha$ be an irrational number and $T_\alpha: S^1 \to S^1$ be the rotation by $\alpha$, i.e., the transformation $T_\alpha$ is defined by $$z = e^{2\pi i \cdot \theta} \implies T_\alpha(z) = e^{2\pi i \cdot (\theta + \alpha)}$$.

Let $f: S^1 \to \{1,-1\}$ be the measurable function defined by assigning +1 to the first half $\{\exp(2\pi i \cdot \theta): \theta \in [0, \frac12]\}$ and -1 to the remaining half. We consider $\alpha$ to be a fixed parameter. The triple $(S^1, T_\alpha, f)$ defines a binary-valued stationary process:

$$z \in S^1 \mapsto (f(z), f(T(z)), f(T^2(z)) \dots) \in \{+1,-1\}^{\mathbb N}$$

where the ambient probability space for the process in our construction is $S^1$ equipped with the uniform probability distribution on it (its Haar measure).

Consider the random walk constructed from partial sums of this stationary process, that is, we define $$S_n(z) := f(z) + f(Tz) + \dots + f(T^{n-1}z)$$ and consider the new process: $$z \in S^1 \mapsto (0, S_1(z),S_2(z),\dots) \in {\mathbb Z}^{\mathbb N}$$

For the purpose of the following question, we extend the discrete-time random walk $\{S_n\}_{n \in \mathbb N}$ to the continuous-time random walk $\{S_t\}_{t \in [0,\infty)}$ by extending in the zig zag way.

Q1. As $n \to \infty$, does this random walk approach some distribution (with appropriate scaling)? In other words, is there a sequence $0 < a_n \nearrow \infty$ (maybe depending on $\alpha$) such that the path-valued random variable $B_N: S^1 \to C[0,\infty)$ defined by $$B_N(z)(t) = \frac{S_{tN}(z)}{a_N}$$ converges in distribution to some nontrivial probability measure on $C[0,\infty)$ (which may depend on $\alpha$) as $N\to\infty$?

Q2. If the answer to Q1 is no for some $\alpha$, what are some sufficient conditions on the parameter $\alpha$ to ensure existence of a limiting distribution? .

  • $\begingroup$ Just to make sure I understand: this walk is not random at all, but is a deterministic process depending only on $\alpha$ - is that correct? $\endgroup$ Commented Dec 25, 2014 at 18:49
  • $\begingroup$ @GregMartin Depending on what is meant by a deterministic process, if it means a process such that the present state determines future states, then the process in question is not a deterministic process, but if it means a process of zero entropy rate, then the process in question is a deterministic process. $\endgroup$
    – Jisang Yoo
    Commented Dec 25, 2014 at 19:29
  • 3
    $\begingroup$ @GregMartin: There is some lack of determinism: the initial point is a random variable. The sequence of $\pm 1$ depends in a fairly subtle way on the initial point. For what it's worth, you can similarly encode the entire randomness of a Brownian motion into a single real number. So the question would be: if you take larger and larger $N$'s and then look at the distribution of $B_N(z)(\cdot)$ (a step function taking values for positive values of the argument) as $z$ runs over the circle, then does this distribution converge to a limiting distribution as $N\to\infty$? $\endgroup$ Commented Dec 27, 2014 at 21:07
  • $\begingroup$ I think there is a related theorem by Sarig. $\endgroup$
    – Asaf
    Commented Dec 28, 2014 at 22:24

2 Answers 2


The paper mentioned by Asaf can be found here http://www.wisdom.weizmann.ac.il/~sarigo/CylinderMap6.pdf In particular, the introduction gives a very nice account of existing results.


One can start by looking at the behaviour of the $S_n$ indexed by integers. If you don't have a CLT there, you won't have either on the real line.

In the probabilitic axiomatization, $z$ is usually written down $\omega $, and we consider the uniformly distributed variable $V=\theta$ on $[0,2\pi]$. Let us call $U_{n}:=f(T^{n}(z))=f(V+2\pi n\alpha )$ with an abuse of notation where $f$ is $2\pi$ periodical on the real line.

Those variables are not independent, but they are centred \begin{align*} \mathbf{E} f(T^{n}(z))=\mathbf{P}(T^{n}(z)\in [0,\pi])-\mathbf{P}(T^{n}(z)\in [\pi,2\pi]) \end{align*}since $z$ is uniformly distributed, it is $0$.

They are unlikely independent but they form a stationnary sequence: ($U_{n}$ has the same law as $U_{n+m}$) and they have a short memory \begin{align*} \mathbf{E} U_{0}U_{m}&=\mathbf{P}(z\in [0,\pi ], T^m(z)\in [0,\pi ])+\mathbf{P}(z\in [\pi ,2\pi ], T_{m}(z)\in [\pi ,2\pi ])\\ &-\mathbf{P}(z\in [0,\pi ], T_{ m}(z)\in [\pi ,2\pi ])-\mathbf{P}(z\in [\pi ,2\pi ], T_{ m}(z)\in [0,\pi ])\\ &= 2| \{z:z \in [0,\pi ],z+m\alpha \in [0,\pi ]\} | -| \{z:z+m\alpha \in [0,\pi ],z+m\alpha \in [ \pi,2\pi ]\} | \end{align*} The first expression is \begin{align*} [0,\pi /m]\cup [2\pi /m,3\pi /m]\cup \dots \cup [2k\pi /m,\pi ] \end{align*}where $2k/m<1\leq (2k+1)/m$, which gives $1/2+O(1/m)$, and idem for the other terms, therefore the whole is in $1/m$, it is not summable.

In this case the variance of $S_n$ is in $n*log(n)$ (and not $n$) and a CLT is unlikely, you should try to look convergence to a stable limit instead.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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