set of zeros on brownian motion is unbounded

Question

Let $W_t$ be a brownian motion. Show that the set $S=\{t: W_t=0 \}$ is unbounded with probability one.

My attempt: Suppose that there is a set $A$ with positive measure such that $S$ is bounded. We know that brownian motion is a strong markov process so I'd assume we try to make some sort of stopping time to help us. For example, $\sup\{t: W_t=0\}$.

Does this answer your question? Proof that a standard Brownian motion visits zero infinitely often at the beginning — Snoop, May 09 '21 at 23:09

score 1 · Accepted Answer · answered May 11 '21 at 13:22

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Suppose that's not true. We will have there exists $s>0$, such that with positive probability, $W_t$ won't hit zero after time $t=s$, i.e., $$\mathbb{P}^0(W_t\neq 0 \text{ for any }t\ge s)>0.$$ Note that using simple Markov property, the left hand side equals $$\mathbb{E}^{0}(\mathbb{P}^{W_s}(W_t\neq 0 \text{ for any }t\ge 0)).$$ Now for any $x$, (for simplicity we assume $x>0$) $$\mathbb{P}^{x}(W_t\neq 0 \text{ for any }t\ge 0)=\lim_{M\uparrow \infty}\mathbb{P}^{x}(\text{hit }M\text{ before hit }0)=\lim_{M\uparrow \infty}\frac{x}{M}=0.$$ Hence $$\mathbb{P}^0(W_t\neq 0 \text{ for any }t\ge s)=\mathbb{E}^{0}(\mathbb{P}^{W_s}(W_t\neq 0 \text{ for any }t\ge 0))=0,$$ which leads to a contradiction.

answered May 11 '21 at 13:22

Yushu Zheng

51

What is $P^{W_s}$ and $P^{x}$? – Literally an Orange May 11 '21 at 20:53
$\mathbb{P}^x$ is the law of Brownian motion starting at $x$. $\mathbb{P}^{W_s}$ is a composite function. We can write this as $\mathbb{P}^x|_{x=W_s}$ . – Yushu Zheng May 12 '21 at 02:01
alright, thanks – Literally an Orange May 12 '21 at 02:29

set of zeros on brownian motion is unbounded

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