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Hi,

Consider $\theta_n = (\theta_0 + n \theta) \mod 1$, $\theta$ being an irrational number, and $\theta_0$ an uniform random variable in $(0,1)$. Is there any estimates for the time it will take this process to hit $(0,\alpha)$ ? From the ergodic theorem I know that, if I denote $N(n)$ the number of times $\theta_n \in (0,\alpha)$, then $N(n)/n \to \alpha$. What I want to know is how much time it will take for this limit to be attained.

Another way of framing this question is : is there any "central limit theorem" (or weakening thereof ; I'm mainly interested in guaranteed bounds for $P(N\geq 1)$) for ergodic processes? From what I've read, there is no general answer to this for a generic ergodic process and function f. There are some results that depend on $f$ being smooth, which it isn't here.

The same question was asked on Quantitative versions of ergodic theorem, but I haven't found anything there that relates to my question.

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  • $\begingroup$ This is a tricky one. In Chapter 2, Section 3 of Kuipers and Niederreiter, Uniform Distribution of Sequences there are some results that might help you. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 22, 2012 at 13:25
  • $\begingroup$ Thanks! This is very interesting. If I understand correctly, the strategy is to use the Koksma–Hlawka inequality. This fails in my case because $f$ is an indicator function, which is not BV. $\endgroup$
    – Antoine Levitt
    Commented Oct 22, 2012 at 13:47
  • $\begingroup$ An indicator function is BV with total variation $2$. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 22, 2012 at 14:39
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    $\begingroup$ Oh, of course, sorry, how stupid of me. I'm a bit lost in this maze of theorems, but that does imply an upper bound on the hitting time, independent on $\theta_0$. The downside is that this bound depends on the diophantine approximation of $\alpha$. Even in what seems to be the most favorable case of $D_N = O(log N / N)$, I get lower bounds which are solutions of $\alpha n = log n$, and so grow (a bit) faster than $1/\alpha$. I was hoping for hitting times on the order of $1/\alpha$, but hey, that's life. Maybe other methods can do better though. Thanks! $\endgroup$
    – Antoine Levitt
    Commented Oct 22, 2012 at 15:08
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    $\begingroup$ The following is only anecdotal: The lowest natural power of $2$ whose decimal representation begins with a '$9$' is $2^{53}$. $\endgroup$
    – Christian Blatter
    Commented Oct 22, 2012 at 19:31

2 Answers 2

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There is a theorem of Kesten, which roughly says, that if you take $(\theta, \theta_0)$ random, and the number of times you hit $(0, \alpha)$ in the first $N$ iterations, subtract the expected $N \times \alpha$, and normalize by $\rho \times ln(n)$, the result will converge to Cauchy distribution. This can be viewed as an analogue of CLT in this case.

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  • $\begingroup$ That's amazing. I'm accepting that as an answer, thanks! The reason I'm interested in this is that I've got some numbers from a simulation that I'm trying to explain. This is exactly what I was looking for. I'll try and fit this Cauchy distribution, and see where that takes me. $\endgroup$
    – Antoine Levitt
    Commented Oct 30, 2012 at 8:01
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This is a very nice question! A lot of results (and references) are given in Zaq Coelho's "The loss of tightness of time distributions for homeomorphisms of the circle".

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  • $\begingroup$ Thanks a lot! The question seems harder than I thought! The limit $\alpha \to 0$ is precisely the case I'm interested in. If I understand correctly the paper, it says that when $\varepsilon \to 0$, one can find subsets of nonzero measure whose hitting time arbitrarily exceeds the expected return time $1/\varepsilon_n$. That's unsettling, but fair enough. What about the case $\varepsilon$ fixed? I did not know about Kac's Lemma, is it invalid outside $J_\varepsilon$? Ie is $\int_{S^1} \tau(\omega) d\omega \neq 1/\varepsilon? If so, what's it equal to? This would be an answer to my question. $\endgroup$
    – Antoine Levitt
    Commented Oct 22, 2012 at 14:20

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