Let's take a simple random walk on $\mathbb{Z}$, $(S_n)_{n\geq0}$, started at zero. If $\tau^+_0 = \inf\{n \geq 1: S_n = 0\}$ is the first time the walk returns on zero, we know that $\mathbb{E}[\tau^+_0] = +\infty$, since the walk is recurrent null. Now I need to know the tail, $\mathbb{P}[\tau^+_0> K]$, when $K\to+\infty$. I've understood that it's $\tfrac{C}{\sqrt{K}}$, but does someone have a proof or a reference ? Thanks.
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3$\begingroup$ Probably more suitable at math.stackexchange.com than here, but here are some hints for one approach. The probability that the first return is at time $2n$ is $2^{-2n}$ times the number of Dyck paths of length $2n$. The number of such paths is the $n$th Catalan number $C_n=\frac{1}{n+1}\binom{2n}{n}$, which behaves as $C_n\sim\frac{2^{2n}}{n^{3/2}\sqrt{\pi}}$ as $n\to\infty$. This already gives you $\mathbb{P}(\tau_0^+=2n)\sim Cn^{-3/2}$. For a start on Dyck paths, Catalan numbers, and many related things, you could look at en.wikipedia.org/wiki/Catalan_number . $\endgroup$– James MartinCommented Jul 19, 2022 at 23:35
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$\begingroup$ Oh sorry I didn't know there was a difference between stackexchange avec overflow. I guess overflow is more professional in a way ? Thanks anyway ! $\endgroup$– RafaëlCommented Jul 20, 2022 at 13:41
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This can be done by the reflection principle. Also, one can use Theorem 0.6, which implies
$$P(\tau_0^+>k)=\tfrac1k\,E|S_k|.$$
By the central limit theorem and uniform integrability, for $k\to\infty$,
$$E|S_k|\sim\sqrt k\,E|Z|=\sqrt k\,\sqrt{\frac2\pi},$$
where $Z$ is a standard normal random variable.
So,
$$P(\tau_0^+>k)\sim\sqrt{\frac2\pi}\frac1{\sqrt k}.$$
answered Jul 19, 2022 at 23:40