Consider a (continuous time) simple symmetric random walk on $\mathbb Z$, starting from the origin. Let us denote this random walk by $\{X_t: t\geq 0\} $. It is well known that this random walk is recurrent. My question is, before the first time the walker returns to the origin, how far does it travel? More precisely, define the first returning time to the origin $$\tau:=\inf\{t>0: \text{there exists} \,\,s<t \,\,\text{such that}\,\, X_s\neq 0\,\, \text{and}\,\,X_t=0 \}.$$ What can we say about $$\mathbb P[\max_{t\leq \tau}|X_t|\geq a]?$$ I feel like this question might be elementary, but I can`t find any reference for it. Any help would be appreciated.
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$\begingroup$ This question is not for this forum. $\endgroup$– Iosif PinelisCommented Nov 23, 2023 at 15:51
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$\begingroup$ Yes, you are right. After reading Carlo's answer, I realize that I should have posted it on stackexchange. $\endgroup$– TiagoCommented Nov 23, 2023 at 20:17
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$\begingroup$ @Tiago --- no harm is done, you did write yourself that "I feel like this question might be elementary, but I can`t find any reference for it." This applies to many questions on MO, we frown on silly questions, but your question is OK. Incidentally, since the $1/a$ answer is so simple, is there a way to arrive at it without having to solve a recurrence relation? $\endgroup$– Carlo BeenakkerCommented Nov 24, 2023 at 7:30
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2$\begingroup$ @CarloBeenakker: WLOG suppose the first step is to the right. Then the random walk starting at 1, and stopped when it reaches $0$ or $a$, is a martingale $M_n$. By recurrence it does eventually reach $0$ or $a$, so $E[M_\infty] = a P(M_n = a)$. On the other hand, $M_n$ is bounded and so uniformly integrable, hence $E[M_\infty] = E[M_0] = 1$. $\endgroup$– Nate EldredgeCommented Nov 24, 2023 at 20:17
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$\begingroup$ @NateEldredge --- neat argument, worthy of an answer, I think. $\endgroup$– Carlo BeenakkerCommented Nov 24, 2023 at 20:52
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1 Answer
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The probability $\mathbb P[\max_{t\leq \tau}|X_t|\geq a]$ is the probability to reach a point at a distance $a>0$ from the origin before returning to the origin, which is just $1/a$, see https://math.stackexchange.com/q/4076273/87355 .
answered Nov 23, 2023 at 13:18