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Dear mathoverflow.

This is a question to a proof in a graduate text. I have asked two professors at my university without help, so I hope it suffices in difficulty for this forum otherwise I appologize.

I can't add an image but the theorem is at scribd page 336.

The book

My problem is with the $S_m-T_m$ is a random walk part. Since the waiting times are iid after the first I know he means $S_m-T_m$ without the first step will be a random walk. What he want to use it for is to conclude it is at some point zero so $S_m=T_m$, but if we disregard the first term would we not have to have to consider the event $S_m-T_m = S_0-T_0$ instead? I know the random walk will be recurrent from what he calls Chung-Fuchs theorem.

I really hope you are willing to help, since I really need to understand this step and I don't know where else to go.

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up vote 0 down vote accepted

You already know that the random walk $S_m-T_m$ is recurrent. It sounds like the piece that you are missing is that recurrent random walks (especially on $\mathbb{R}$) not only visit their initial value infinitely often but any other finite value as well as long as it has a positive probability of reaching this other value in finite time. A similar example to consider is a simple random walk on $\mathbb{Z}$ started at $10$ and it's visits to $0$. This is short Borel-Cantelli argument.

Edit: Suppose that $S_0-T_0 = \beta >0$ for the moment. Then you need the process $Y$ to hit it's initial value $\beta$ times before $X$ does. Since the times of these events are all i.i.d. then it is a question of the probability of $\beta$ returns for $Y$ before 1 return for $X$. Which can be crudely lower bounded by the probability that the return times for $Y$ are less than 1 and the return time for $X$ is greater than $10$ which have a positive, albeit small, probability.

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Thank you for your reply. I actually already realized that, but maybe I'm missing something even more elementary, but how do I know the probability is positive? – Henrik Oct 28 '12 at 17:07

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