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Consider a random walk on $\mathbb{Z}$ with step distribution $$\mathbb{P}(X_1 = n) = \frac{1}{2}\left ( \frac{1}{|n|^{\alpha}} - \frac{1}{(|n| + 1)^{\alpha}} \right ), \ \ \ \ n \neq 0.$$ I am trying to find the values $\alpha >0$ that makes the random walk recurrent. I know that, if the random walk starts at $1$, I need to show that $$\sum_{n=1}^{\infty}\mathbb{P}(S_n = 1) = \infty.$$ However, I cannot find a closed formula for the return probability. For example, if the random walk starts at $1$, then $\mathbb{P}(S_2 = 1)$ is easy to find since that means the random walk went up $n$ steps and then went down $n$ steps. But for $n>2$ it becomes much more complicated.

Can anyone help? Thanks!

  • Not that $\sum_{n=1}^{\infty} \mathbb{P}(S_n=1) = \lim_{N\to\infty}\sum_{n=1}^{N} \mathbb{P}(S_n=1)$. Try to write up the whole formula, not just the part after the limit, and expand the sum into its terms. Now depending on the value of $\alpha$, most terms may just cancel out in the limit. – Daniel P Jul 21 '22 at 16:17
  • @DanielP Thanks. But how does looking at the “finite sum” help? –  Jul 21 '22 at 16:20
  • You could look at Lemma 2.21 of these notes to see if it helps. – Greg Martin Jul 23 '22 at 16:53

2 Answers2

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I'd reason like this:

we know from the general theory that

$$\mathbb{E}[N] = \lim_{t \to 1}\int_{[-\pi,\pi]}\frac{dk}{\sqrt{2\pi}(1-t \varphi(k))}$$ where $\varphi(k)$ is the characteristic function of $X_1$ and $N = \sum_{n \ge 0}\mathbb{I}_{S_n = S_0}$.

First we notice that

$$ \varphi(k) = \sum_{n \ne 0}\frac{e^{ikn}}{2} \bigg[ \frac{1}{|n|^{\alpha}} -\frac{1}{(|n| + 1)^{\alpha}} \bigg] = \sum_{n = 1}^{\infty} \cos(nk) \bigg[ \frac{1}{|n|^{\alpha}} -\frac{1}{(|n| + 1)^{\alpha}} \bigg]$$

Then

$$ 1 - \varphi(k) = \sum_{n = 1}^{\infty} (1-\cos(nk)) \bigg[ \frac{1}{|n|^{\alpha}} -\frac{1}{(|n| + 1)^{\alpha}} \bigg] \le \sum_{n \ge 1} \frac{1-\cos(nk)}{n^{\alpha}}$$

Remembering the inequality $\cos(x) \ge 1 - x^2 / 2$, we finally obtain $$ 1 -\varphi(k) \le \sum_{n \ge 1} \frac{k^2}{2 n^{\alpha -2}} = C k^2$$

when it is convergent, hence when $\alpha > 3$. By Fatou's Lemma we can say that $$ \liminf_{t \to 1}\int_{[-\pi,\pi]}\frac{dk}{\sqrt{2\pi}(1-t \varphi(k))} \ge \int_{[-\pi,\pi]}\frac{dk}{\sqrt{2\pi}(1-\varphi(k))} = \infty$$

Therefore it is recurrent whenever $\alpha > 3$.

finch
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  • It seems to me the $\sqrt{2\pi}$ should be $2\pi$ in your first formula – puppyHuazai Jul 26 '22 at 07:39
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    Also, instead of dropping the term $\frac{1}{(|n|+1)^{\alpha}}$, the difference is of order $\frac{1}{|n|^{\alpha+1}}$, so your conclusion can be strengthened to $\alpha>2$ – puppyHuazai Jul 26 '22 at 07:48
  • You are absolutely right in both cases. I completely forgot the additional $1/n$ that comes from the limit! Thanks for the remark! – finch Jul 26 '22 at 08:01
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This is a complement to finch's answer. We show the walk is recurrent when $\alpha\gt1$. By finch's method, if suffices to show $(1-\phi(k))\leq c*|k|$ for $c>0$. Note that for $\alpha>1$, $E[|X_{1}|] <\infty$ (see the remarks after finch's answer). Then by the mean value theorem,

$(1-\phi(k)) = E[1-\cos(kX_{1})]\leq E[|k||X_{1}|] = c|k|$.

I also suspect that $\alpha=1$ is the marginal case, that is for $\alpha\lt 1$, the walk is transient.

Added: For $\alpha =1 $, the walk is also recurrent. since in this case, $(1-\phi(k))\leq a\sum_{n=1}^{n=\infty}\frac{1-cos(nk)}{n^2}$ for some $a>0$ and the infinite sum can be computed according to Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$ as $-\frac{k^2}{4}+\frac{\pi k}{2}$ for $0<k<2\pi$. Hence the integral is again divergent.

puppyHuazai
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