Consider the simple symmetric random walk on $\mathbb{Z}$. That is, let $X_1, X_2, \dots$ be i.i.d. random variables with $$ P(X_i=1)=P(X_i=-1)=1/2, $$ and define $S_n=X_1+\dots+X_n$ with $S_0=0$. As is well known, the sum $S_n$ is (null) recurrent and satisfies the law of the iterated logarithm $$ P\left(\limsup_{n\to \infty} \frac{S_n}{\sqrt{2n \log \log n}}=1\right)=1, $$

Assume now that we restrict ourselves to the subset of realisations of $S_n$ where each realisation satisfies

  1. $\frac{1}{n} S_n \to 0$.

  2. For any integer $m$, there are infinitely many values of $n$ such that $S_n=m$ or $S_n=-m$.

Does the law of the iterated logarithm take a stronger form in this case, in the sense:

Q: If $S_n$ is satisfies the criteria above, does this imply $$ \limsup_{n\to \infty} \frac{|S_n|}{\sqrt{2n \log \log n}}\leq 1? $$

Or are there even in this case subsets of realisations for which this is not true?


closed as off-topic by Anthony Quas, Alexey Ustinov, Franz Lemmermeyer, Wolfgang, Jan-Christoph Schlage-Puchta Aug 25 '16 at 17:25

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  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Anthony Quas, Alexey Ustinov, Franz Lemmermeyer, Wolfgang, Jan-Christoph Schlage-Puchta
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  • $\begingroup$ Isn't the realization $(1,0,1,0,1,0,\dots)$ in your subset? I think if you write out your statements carefully, being explicit about the null sets, you'll see they aren't plausible. $\endgroup$ – Nate Eldredge Aug 25 '16 at 8:39
  • $\begingroup$ I changed the equality to an inequality. That should account for all cases when $S_n$ is trivially bounded. I'm interested in whether there are any counterexamples where $S_n$ is surely recurrent, but infinitely many times has values larger than $\sqrt{2n \log \log n}$. This should be a null set, but it would be good to see a constructed example. $\endgroup$ – user45947 Aug 25 '16 at 9:03
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    $\begingroup$ Still no. Consider (1,0,1,1,0,0,1,1,1,1,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,....) (that is $2^n$ 1's followed by $2^n$ 0's followed by $2^{n+1}$ 1's then $2^{n+1}$ 0's etc. For this one, the growth is linear in $n$ (not $\sqrt{n\log\log n}$). $\endgroup$ – Anthony Quas Aug 25 '16 at 9:14
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    $\begingroup$ This is still hopeless. The mention of probability is a distraction; it plays no role in your question. If $f_n$ is any function that grows slower than linearly, you can find an up-down path $s_n$ that grows slower than linearly but satisfies $s_n=f_n$ for infinitely many values of $n$. To take an arbitrary example, $(+)^1(-)^3(+)^5(-)^7(+)^9....$ gives you $|s_{k^2}|=k$ for every $k$. $\endgroup$ – James Martin Aug 25 '16 at 14:00
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    $\begingroup$ (but also, parenthetically, if something is "not a good question" I had the impression that explaining the reason why in the comments, possibly along with a down-vote, is the recommended thing to do?) $\endgroup$ – James Martin Aug 25 '16 at 17:20

Consider a path satisfying $S_{k^3} = (-1)^k k^2$ for all sufficiently large $k$. Since $$(k+1)^3 -k^3 \approx 3k^2 \gg 4k \approx |S_{(k+1)^3} - S_{k^3}|$$ this is clearly possible. Moreover, we can ensure that $|S_n| \le |S_{k^3}|$ for all $1 \le n \le k^3$ simply by replacing all excursions outside that value by a zigzag path of alternating increments $\pm 1$. To see that (1) is satisfied, note that for $k^3 \le n \le (k+1)^3$ we have by assumption $|S_n| \le |S_{(k+1)^3}| = (k+1)^2$, so that $$\frac{|S_n|}{n} \le \frac{(k+1)^2}{k^3} \to 0.$$ (2) is also clearly satisfied. And since $|S_n| = n^{2/3}$ when $n = k^3$ the limsup in the LIL is infinite.

  • $\begingroup$ Nice construction that in addition to the comments above helped improving my understanding of random walks. Thanks! $\endgroup$ – user45947 Aug 25 '16 at 22:31
  • $\begingroup$ Variants of this can give you walks whose asymptotic growth rate is anything smaller than $n$. Here I just chose $n^{2/3}$ but if you want $n/\log n$ or $n/\log\log\log\log n$ that would work just as well. $\endgroup$ – Nate Eldredge Aug 26 '16 at 1:27
  • $\begingroup$ "Nice construction that in addition to the comments above helped improving my understanding of random walks." The construction is allright but that it may help to advance the understanding of random walks is dubious since it is based on behaviours with zero probability. As already mentioned in the comments on main, this question is not related to probability (rather, to combinatorics). $\endgroup$ – Did Aug 27 '16 at 12:36
  • $\begingroup$ I guess maybe the relevance to probability is it illustrates that we cannot expect to obtain the LIL as an immediate consequence of SLLN and recurrence. $\endgroup$ – Nate Eldredge Aug 27 '16 at 14:38

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