We know that for an asymmetric random walk or Brownian motion with drift, then it will across zero finitely often. I want to know that if I am looking at the following random variable

$$ M =\mbox{Number of times the process across 0 }$$

where "the process" refers to asymmetric random walk or Brownian motion with drift with time $t\in[0,+\infty)$, then what is the support of $M$? Is it finite? And what is the distribution of $M$?

Intuitively, what does it mean that the process across zero finitely often? Does it mean, this process can cross the zero $1,2,3,...n,...$ times, but the corresponding probability goes to 0 as $n$ goes to infinity? But then, this means the support of this random variable $M$ is still infinite?


1 Answer 1


The fact that the random walk crosses zero finitely often, simply means that $\mathbb P[M=\infty]=0$. But it does not mean that the support of $M$ is finite. In fact, $M$ has a geometric distribution, as shown below.

Let $0=S_0,S_1,S_2,\ldots$ be an asymmetric random walk such that $\mathbb P[S_1=1]=p$ and $\mathbb P[S_1=-1]=1-p$. Assume $p>\frac 1 2$. Exercise 1.13 of Durret's book shows that $\mathbb P[M=0] = p(2p-1)=:q$. Now, the fact that $M$ has a geometric distribution follows from noting that $M=k$ if and only if the random walk returns to zero $k$ times and does not return to zero afterwards, hence, $\mathbb P[M=k]=(1-q)^k q$. This is true for any transient Markov chain (see Theorem 3.1 of Durret's book).

For the Brownian motion, the number of returns to zero is necessarily infinite. But you might be interested in the local time at zero. By the invariance principle, I guess that the local time has an exponential distribution.

  • $\begingroup$ +1. You can also say that if the probability of starting at $0$ and never again crossing $0$ is $q$ then $\mathbb P(M\ge k) = (1-q)\mathbb P(M\ge k-1) =(1-q)^k$ $\endgroup$
    – Henry
    Commented Nov 9, 2022 at 14:34

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