# Surely recurrent random walks and the law of the iterated logarithm [closed]

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-PuchtaAug 25 '16 at 17:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "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
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – Nate Eldredge Aug 25 '16 at 8:39
• 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. – user45947 Aug 25 '16 at 9:03
• 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 $\sqrt{n\log\log n}$). – Anthony Quas Aug 25 '16 at 9:14
• 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 $k$. – James Martin Aug 25 '16 at 14:00
• (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?) – 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.
• 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. – Nate Eldredge Aug 26 '16 at 1:27