Brownian Zeros Accumulate at the Origin

From a Fixed Interval to Arbitrarily Small Times

Brownian motion starts at zero, but its path does not remain on just one side of zero immediately afterward. In fact, with probability one, it takes both positive and negative values in every interval \((0,\varepsilon)\), no matter how small \(\varepsilon>0\) is. Continuity then forces the path to cross zero again between such positive and negative values. The key probability calculation is the reflection principle for the running maximum.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((B_t)_{t\ge0}\) be standard Brownian motion: \(B_0=0\), its paths are continuous almost surely, and its increments are independent centered normal random variables with \(\operatorname{Var}(B_t-B_s)=t-s\) for \(0\le s<t\). For \(h>0\), define the running maximum \(M_h=\sup_{0\le t\le h}B_t\) and running minimum \(m_h=\inf_{0\le t\le h}B_t\). The reflection principle gives, for \(a>0\),

where \(\Phi(x)=\mathbb{P}(Z\le x)\) for a standard normal random variable \(Z\). The principle pairs paths that reach level \(a\) and end below it with reflected paths that reach \(a\) and end above it. The formula is stated here in the form needed; its reflection argument is the standard reflection-principle result from the earlier material on Brownian motion. Since \(-B\) is also standard Brownian motion, symmetry gives \(\mathbb{P}(m_h\le-a)=2\mathbb{P}(B_h\ge a)\).

Proof. Because \(B_0=0\), the event that \(B_t\le0\) for every \(t\in[0,h]\) is \(\{M_h=0\}\). For each positive integer \(r\), let \(A_r=\{M_h\ge1/r\}\). These events increase as \(r\) increases, and their union is \(\{M_h>0\}\): a positive value of \(M_h\) exceeds \(1/r\) for some sufficiently large \(r\). By the reflection formula, \(\mathbb{P}(A_r)=2\mathbb{P}(B_h\ge1/r)\). The normal distribution of \(B_h\) is continuous, so as \(r\to\infty\), \(\mathbb{P}(B_h\ge1/r)\to\mathbb{P}(B_h\ge0)=1/2\). Continuity of probability for increasing events now gives \(\mathbb{P}(M_h>0)=1\). Thus a positive value occurs at some time in \((0,h]\), since \(B_0=0\). Applying the same argument to \(-B\), which has the same Brownian distribution, gives \(\mathbb{P}(m_h<0)=1\), and hence a negative value occurs in \((0,h]\) almost surely. \(\square\)

The proposition is for one fixed \(h\). To conclude that the path has both signs arbitrarily close to zero, we need a single probability-one event that works for every interval length. It is enough to establish the property for the countable sequence \(h=1/n\), because every positive interval contains one of these intervals.

Proof. For each positive integer \(n\), let \(E_n^+\) be the event that \(B_t>0\) for some \(t\in(0,1/n]\), and let \(E_n^-\) be the event that \(B_t<0\) for some \(t\in(0,1/n]\). The proposition gives \(\mathbb{P}(E_n^+)=\mathbb{P}(E_n^-)=1\) for each \(n\). Therefore the countable intersection \(E=\bigcap_{n\ge1}(E_n^+\cap E_n^-)\) has probability one: its complement is a countable union of probability-zero events. On \(E\), fix any \(\varepsilon>0\) and choose a positive integer \(n\) with \(1/n<\varepsilon\). The events \(E_n^+\) and \(E_n^-\) supply times \(s,t\in(0,1/n]\), so \(s,t<\varepsilon\), with \(B_s>0\) and \(B_t<0\). This proves the assertion simultaneously for every \(\varepsilon>0\). \(\square\)

Why the Sign Changes Force Zeros

The zero set of a Brownian path is the set of times \(t\ge0\) at which \(B_t=0\). Saying that zeros accumulate at the origin means more than the fact that \(B_0=0\): it means there are positive zero times arbitrarily close to zero. The sign result gives those times by the intermediate value theorem, using path continuity.

Proof. Take a sample path in the probability-one event of the preceding theorem and in the probability-one event on which the path is continuous. For each positive integer \(n\), choose \(s_n,t_n\in(0,1/n)\) such that \(B_{s_n}>0\) and \(B_{t_n}<0\). The times are distinct, since one path value cannot be both strictly positive and strictly negative. Let \(u_n=\min(s_n,t_n)\) and \(v_n=\max(s_n,t_n)\). The values \(B_{u_n}\) and \(B_{v_n}\) have opposite signs. By continuity on \([u_n,v_n]\), the intermediate value theorem supplies \(z_n\in(u_n,v_n)\) with \(B_{z_n}=0\). Since \(0<u_n<z_n<v_n<1/n\), these are positive zero times and \(z_n\to0\). More explicitly, given \(\varepsilon>0\), choose \(n\) with \(1/n<\varepsilon\); then \(z_n\in(0,\varepsilon)\). Thus the zeros accumulate at the origin. \(\square\)

A Common Reflection-Principle Pitfall

For \(h=1\) and \(a=1\), an incorrect calculation is \(\mathbb{P}(M_1\ge1)=\mathbb{P}(B_1\ge1)=1-\Phi(1)\). It counts paths that end at or above the barrier but omits paths that hit \(1\) and then end below \(1\). Reflection pairs this latter group with paths ending above the barrier, which is why the correct calculation is

This factor of two applies to the event of reaching a positive barrier by a fixed time, with a positive barrier and standard Brownian motion starting at zero. It should not be confused with the probability of being positive at the terminal time, which is \(\mathbb{P}(B_1>0)=1/2\). Nor can one infer a path event from the terminal-time probability alone: the reflection calculation concerns the maximum over the whole interval.

The countability step is another important distinction. The proposition establishes probability one for each fixed interval length. The theorem uses a countable intersection to obtain one event on which the conclusion holds for every interval length. Taking an intersection over an uncountable collection of probability-one events would not, by itself, justify the same conclusion.