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Let $\{D_i\}_{i=0,1,2,\dots }$ be independent $\exp(1)$ random variables. We use the collection $\{D_i\}$ to define a random walk on $\mathbb Z$ by $S_0 = 0$ and $S_n = \sum_1^n X_i$ with $X_i \in \{-1,0,1\}$ given by $$X_i = \mathbf 1 \{ D_{i} > 2 D_{i+1} \} - \mathbf 1 \{ D_{i+1} > 2 D_{i } \}.$$ I am interested in showing $S_n$ visits 0 infinitely often a.s. The pairwise dependence between $X_i$ and $X_{i+1}$ is making this (frustatingly) difficult.

The assumption that $D_i \sim \exp(1)$ is just for convenience and should not be important.

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  • $\begingroup$ The expected value of $X_i$ is not zero. Birkhoff's ergodic theorem implies that $S_n$ goes off to infinity almost surely. $\endgroup$ Jul 25, 2014 at 23:00
  • $\begingroup$ Mixed up my inequalities. The expected value of X_i should be 0 and I think my latest edit makes it so. $\endgroup$
    – mathjunge
    Jul 25, 2014 at 23:07
  • $\begingroup$ Ok. Now $X_i$ can be $0$ you should fix that too. $\endgroup$ Jul 25, 2014 at 23:12
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    $\begingroup$ Your summands have the property of 1-dependence: if $|i-j|>1$, then $X_i$ and $X_j$ are independent. There are CLT for 1-dependent random variables. The events that $\limsup S_n=\infty$ and $\liminf S_n=-\infty$ are tail events, and therefore have probability 0 or 1 by Kolmogorov's 0-1 law. It's not hard to see the probability is not 0. $\endgroup$ Jul 26, 2014 at 1:18
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    $\begingroup$ In this case you can just apply the Kolmogorov 0-1 law for the sequence $D_i$, since the events are also tail events for that sequence. $\endgroup$ Jul 27, 2014 at 23:46

2 Answers 2

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If I remove a summand $X_n$ at regular intervals, then the increments become independent. More concretely, let's introduce $Y_1=X_1+X_2$, $Y_2=X_4+X_5$ etc. and write $$ S_{3n}=\sum_{j=1}^n Y_j + \sum_{j=1}^n X_{3j} . $$ Both sums now have iid summands, so the law of the iterated logarithm applies to both: On a probability $1$ set, we have that $\sum Y_j \approx \pm \sigma_Y \sqrt{2n\ln\ln n}$ for infinitely many $n$. Since $\sigma_X<\sigma_Y$ and the law of the iterated logarithm applies to $\sum X_{3j}$ as well, this second sum can't change the sign of $S_{3n}$. Thus for either sign $\pm$, there are arbitrarily large $n$ for which $S_{3n}$ takes this sign.

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  • $\begingroup$ Am I mistaken to believe it isn't obvious that $\sigma_X < \sigma_Y$? The variables $X_1, X_2$ are negatively correlated. I suppose the same argument works when $\sigma_X > \sigma_Y$, but fails if the two are equal. $\endgroup$
    – mathjunge
    Jul 26, 2014 at 16:45
  • $\begingroup$ @mathjunge: No, it's not obvious (I was lazy), but I think it's strongly expected. Also, the argument is very robust; I could also make the blocks defining $X$ or $Y$ longer. Anyway, I did the calculation now: $\sigma^2_X=2/3$ and $P(Y=2)=1/15$, which contributes $4/15$ to $\sigma^2_Y=EY^2$, as does the event $Y=-2$, so to verify that $\sigma_Y>\sigma_X$, I now need that $P(Y=1)>1/15$, which is easy to check. $\endgroup$ Jul 26, 2014 at 17:28
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An old theorem of Atkinson [Recurrence of co-cycles and random walks, J. London Math. Soc. (2) 13 (1976), 486–488. MR0419727] states that for sums of stationary $\mathbb Z$-valued random variables with a finite first moment recurrence is equivalent to vanishing of the expectation.

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