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Nov 26 at 8:33 comment added user8965 In general that should follow from the strong ratio limit theorem, see Section 4.3 in arxiv.org/abs/1511.01721 for a recent reference.
Nov 25 at 22:32 history edited Viktor B CC BY-SA 4.0
Forgot limit n to ininity
Oct 9, 2022 at 0:06 comment added Iosif Pinelis The real challenge here is to show that this holds for all zero-mean $\mu$ (or maybe to construct a counterexample).
Oct 6, 2022 at 21:35 comment added Viktor B @IosifPinelis Indeed, $\mathbb{P}\{|S_n| \leq a \} > 0$ is required at least for large $n$. If $S_n$ is non-lattice, then $\mathbb{P}\{|S_n| \leq a \} > 0$ follows from the local limit theorem, whereas if $S_n$ is lattice, it has to be aperiodic (or (1) has to be adapted). The local limit theorem holds for a distribution in the domain of attraction of a stable distribution, so it seems to me that at least for those distributions the proof (1) can follow pretty much along the same lines as in the post.
Oct 6, 2022 at 20:36 comment added Iosif Pinelis You also have to assume that $P(|S_n|\le a)\ne0$. Then, I think, this should be true for all zero-mean $\mu$ -- but the proof is probably quite nontrivial.
Oct 6, 2022 at 20:21 history edited Viktor B CC BY-SA 4.0
Added proof idea based on Thomas Kojar's comment
Oct 6, 2022 at 20:01 comment added Viktor B I think you are correct and for a wide class of $\mu$ (1) is going to follow a local limit theorem. I modify the post.
Oct 6, 2022 at 19:28 comment added Thomas Kojar How about doing it by hand by writing $$S_{n+1}=X_1+R_n$$ and then applying strong law to show that the effect of X1 is negligible in the second event in $$P(X_1>t, -a/n<X_1/n+R_n/n<a/n)\approx P(X_1>t) P( -a/n-c_n<R_n/n<a/n+c_n)+o(g_n)$$ for some $g_n,c_n\to 0$ at least when if we assume that X1 is upper/lower bounded?
Oct 6, 2022 at 18:35 history asked Viktor B CC BY-SA 4.0