# limiting distribution of the random walk from irrational rotation

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? .

• 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? Dec 25, 2014 at 18:49
• @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. Dec 25, 2014 at 19:29
• @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 $N\to\infty$? Dec 27, 2014 at 21:07
• I think there is a related theorem by Sarig.
– Asaf
Dec 28, 2014 at 22:24

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.