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Let $B_t, t\ge0$ be a standard Brownian motion that starts at 0. Prove that, for any $\epsilon > 0$, $$P(\#\{t\in(0,\epsilon]\mid B_t = 0\}=\infty)=1.$$

This is a well-known fact, but surprisingly I cannot find a single proof on this site. The closest I get is this question, which proves the zero set of a standard Brownian motion restricted to $[0,1]$ is homeomorphic to the Cantor set. However, this proof seems to have utilised unfamiliar properties such as the ternary-expansion construction of the zero set.

Does there exist an elementary proof that only uses the familiar properties of the Brownian motion (normality, independent increments etc.)?

Did
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Vim
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    A shortcut is to note that $W_t=tB_{1/t}$ defines another Brownian motion $(W_t)$ such that $B_t=0$ for some given $t>0$ if and only if $W_{1/t}=0$, hence $$P(#{t\in(0,\epsilon]\mid B_t = 0}=\infty)=P(#{t\geqslant1/\epsilon\mid B_t = 0}=\infty)$$ a statement which you might know a proof of. – Did Feb 03 '19 at 16:40
  • @Did yes I know the proof that W is Brownian motion. Thanks for this elegant one-liner. – Vim Feb 03 '19 at 16:52
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    not just at the beginning, but at every root. The set of roots has no isolated points. – Albert Chen Mar 03 '22 at 16:15
  • @Did Does the property you show fulfills only for $B_t=0$ or it is also fulfilled for every $B_t = a$ such $a\in\mathbb{R}$? I have made a related question here now with bounty, where I have the hipothesis that it would still work still for $B_t = a\ /\ |a|\leq \sqrt{2t\sqrt{\pi^2+\log(\log(1+t))^2}}$ (the absolute value of the Modulus of Continuity of the Wiener process), but I don't even know how to start a proof. Hope you could check it. – Joako Nov 16 '23 at 16:15
  • @Joako $B_t$ seems to be translation invariant so I'm pretty sure it works for $C_t:=B_t-a$ as well which is still a Brownian motion – Vim Nov 16 '24 at 17:31
  • @Vim It is not the only force involved: concentration inequalities tells you also it don't going to be too far from some initial values if some mild assumptions hold – Joako Nov 17 '24 at 01:12
  • I was misunderstanding your question then. I thought you wanted to investigate whether the solution set ${t\mid B_t = a}$ behaves similarly (no isolated points) across all $a\in\Bbb R$, the answer to which I'm pretty sure is yes because $B_t$ is Markovian. – Vim Nov 17 '24 at 09:41

2 Answers2

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Here's another way. Define $T_a:=\inf\{t>0: B_t=a\}$. It's well known that $E[\exp(-\lambda T_a)]=\exp(-|a|\sqrt{2\lambda})$ for $\lambda>0$ and $a\not=0$. As $a$ decreases to $0$, $T_a$ decreases to $T_{0+}:=\inf\{t>0:B_t>0\}$, and by the Laplace transform cited above, $P[T_{0+}=0]=1$. Similarly, with $T_{0-}:=\inf\{t>0:B_t<0\}$ you have $P[T_{0-}=0]=1$. By path conitnuity and the Intermediate Value Theorem, the zero set $\{t>0: B_t=0\}\cap(0,\epsilon)$ is non-empty for all $\epsilon>0$, with probability 1.

John Dawkins
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Per @Did's comment, the key is to note that $tB_{1/t}$ also defines a standard BM. And thus $$P(\#\{t\in(0,\epsilon]\mid B_t = 0\}=\infty)=P(\#\{t\geqslant1/\epsilon\mid B_t = 0\}=\infty)$$ and what's left is to show $B_t$ visits $0$ i.o. in $t\ge M$ for any $M>0$.

Here's how I proved it: consider the "first return time" $$\tau:=\inf\{t\color{red}{>} 0\mid B_t=0\}$$ Then it suffices to show $\tau<\infty$ a.s.

Now set any $h>0$. Note that among all the possibilities that $B_t$ returns to 0, one special possibility is that $B_t$ first hits $+h$ and then moves $-h$ to hit $0$. We investigate this special possibility. Define $\tau_1 = \{t>0\mid B_t = h\}$ and $\tau_2$ to be the time it takes $B_t$ to hit 0 starting from $+h$. Clearly $\tau_{1,2}$ are independent and $\tau_2$ equals $\tau_3:=\{t>0\mid B_t=-h\}$ in distribution.

Since "first hitting $+h$ then hitting $0$" is just one among many ways to return to $0$, we have $\tau\le\tau_1+\tau_2$. Therefore $$P(\tau<\infty)\ge P(\tau_1+\tau_2 <\infty)=P(\tau_1<\infty)P(\tau_2<\infty)=P(\tau_1<\infty)P(\tau_3<\infty)=1.$$

Vim
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