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Let $\omega$ be an irrational number, and $X$ a random variable taking values $1,-1,\omega,-\omega$ each with probability $1/4$. Let then $X_i$ be iid variables with the same law as $X$ and $S_n=\sum_{i=1}^n X_i,n\in \mathbb N$ be the corresponding random walk.

Is it possible to have a precise asymptotics for $P(|S_n|<\epsilon)$ for $\epsilon>0$? Ultimately I would like to know the behaviour of $$\sum_{n=1}^\infty n^{-3/2} P(|S_n|<\epsilon)$$ as $\epsilon\to 0$.

I feel like the diophantine properties of $\omega$ are relevant for this asymptotics.

How would you proceed to get such an estimate? Ideally I would like to consider $X$ with any discrete law, with eventually infinitely many atoms.

EDIT: to be clear, I think there are ad-hoc methods to solve this kind of problems, as Mateusz shows below. I want to make sure not to miss any kind of general theory that solves this kind of problems in the theory of random walks.

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  • $\begingroup$ Ok, thanks! But the sum is at least bounded by the sum of the $n^{-3/2}$, so it should be finite. I expect it at least to go to $0$ with $\epsilon$, as your estimate suggests. Or maybe I misunderstood your answer (and btw $\epsilon$ is not necessarily irrational, but $\omega$ is) $\endgroup$ – kaleidoscop Feb 12 '20 at 10:32
  • $\begingroup$ One thing that may or may not be relevant is the following. If we think of this as a random walk on the lattice $\mathbb{Z} + \sqrt{-1} \omega \mathbb{Z} $, then this is a two dimensional random walk, as noted by Mateusz Kwaśnicki. Random walks in two dimensions are recurrent, so we expect $S_n$ to be zero infinitely often. As such, even for $\epsilon=0$, we expect there to be a lower bound on the sum you are studying. Of course, when we consider this as a 1-dimensional random walk, it will be close to the origin more frequently, but that seems much more complicated. $\endgroup$ – Gabe K Feb 12 '20 at 21:27
  • $\begingroup$ As a follow-up, one further thing to notice is that getting estimates also depends on the size of $\omega$. In particular, if $\omega \ll \epsilon \ll 1$, then the behavior will be very similar to a $1$-dimensional random walk (because you need many steps in the $\omega$ direction to get something $O(\epsilon)$. On the other hand, if $\omega \gg 1 \gg \epsilon$, the behavior will be more like a two dimensional random walk, where the steps in the $\omega$ direction have to cancel out to get back to the origin. $\endgroup$ – Gabe K Feb 12 '20 at 21:36
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    $\begingroup$ Ok thanks, but here $\omega$ is fixed and $\epsilon$ goes to $0$ $\endgroup$ – kaleidoscop Feb 13 '20 at 8:23
  • $\begingroup$ Right, the point is only that the size of $\omega$ will affect the behavior and that when $\epsilon$ is sufficiently small, it's almost equivalent to $\epsilon =0$. $\endgroup$ – Gabe K Feb 13 '20 at 16:12
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Just an extended comment. Let $X_n$ be the simple random walk in $\mathbb{Z}^2$. Then $$\mu(\{x\}) = \sum_{n = 1}^\infty n^{-3/2} P(X_n = x)$$ is comparable with $$\sum_{n = 1}^\infty n^{-3/2} \times n^{-1} \exp(-|x|^2 / (2 n)) \approx (1 + |x|)^{-3}.$$ So your question boils down to estimating $$\sum_{k \in \mathbb{Z}} \frac{1}{(1 + |k|)^3} \, \mathbb{1}_{(-\epsilon, \epsilon)}(k \omega - \lfloor k \omega\rfloor) . $$ This indeed seems closely related to how well one can approximate $\omega$ with rationals.

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  • $\begingroup$ Ok thanks for this nice approach! So you transfer the problem to a random walk on higher dimensions. I was actually interested by general results from the theory of random walks: is it possible to relate the distribution of the increments to how often the chain will pass close to some state? This way that might directly lead to a condition on the law of the $X_i$ (instead of some estimates on $\omega$) $\endgroup$ – kaleidoscop Feb 12 '20 at 10:42
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    $\begingroup$ @kaleidoscop: Sorry, I do not quite understand your comment. I do not think there is a "soft" way to answer this kind of questions, so I suppose that one cannot simply link the rate of convergence of the sum to zero with distributional properties of the increment $X_i$. $\endgroup$ – Mateusz Kwaśnicki Feb 12 '20 at 12:04
  • $\begingroup$ Well, you actually understood my comment, because that is exactly what I was trying to get :) $\endgroup$ – kaleidoscop Feb 12 '20 at 12:10

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