# How to show that $P( \sup_{0 \leq s \leq 1} |B_s| \leq \epsilon) > 0$ for any $\epsilon > 0$?

Let $(B_s)_{s \geq 0}$ be a standard Brownian motion.

In order to solve an interesting exercise, I need to show that $P( \sup_{0 \leq s \leq 1} |B_s| \leq \epsilon) > 0$ for any $\epsilon$.

I've tried the following general methods, and failed at them:

1. Doob's Maximal inequality. This only works for sufficiently large $\epsilon$.
2. Reflection principle / using what I know about $B_t^*$. Again, this only works for sufficiently large $\epsilon$. (Essentially what I did boiled down to this result, which is not strong enough: Show that $\mathbb{P}(\sup_{s\leq t} |B_s|\geq x)\leq 2\mathbb{P}(|B_t|\geq x)$ )
3. Various fiddling with time change symmetries of Brownian motion.
4. Using the distributions of the stopping time for one sided barriers. (Similar to 2.)

Perhaps one of these approaches works, but I'm totally stuck. I would appreciate a hint.

• Using the cameron-martin theorem, this is almost trivial. Without that, you can also show this just using the standard construction of brownian motion using haar functions, and borel-cantelli. May 5, 2018 at 11:24
• @Shalop At this point I know a solution to this problem, but I would love it if you could expand on your comment in an answer. May 5, 2018 at 20:04

I learned how to do this from David Freedman's book "Brownian motion and Diffusion" (Lemma 37)

Let $T = \inf \{ t : B(t) = \pm \epsilon /2 \}$.

There is a $\delta > 0$ so that $P (T \geq \delta) > 0$. (For example, one can argue using Doob's maximal inequality: $PP ( T \geq \delta) = P( \sup_{0 \leq t \leq \delta} |B_t| < \epsilon/2 ) \geq 1 - \frac{ 2 E |B_{\delta}|}{\epsilon} = 1 - \frac{ 2 \delta E |B_1| }{\epsilon}$.)

$P ( T \geq \delta , B(T) = \epsilon /2) = 1/2 P( T \geq \delta$ by symmetry. (That is, because $\{ T \geq \delta , B(T) = \epsilon /2 \} \cup \{ T \geq \delta , B(T) = -\epsilon /2 \} = \{ T \geq \delta \}$.)

Choose (even) $n$ large enough so that $n \delta > 1$.

Let $T_0 = 0$ and $T_{j+1} = \inf \{ t > T_j : |B(t) - B(T_j)| = \epsilon /2 \}$.

(So $T = T_1$.)

(Now this is the crucial idea) consider this event:

$E = \{ T_1 \geq \delta, T_2 - T_1 \geq \delta,\ldots, T_n - T_{n-1} \geq \delta, B(T_1) = \epsilon /2, B(T_2) = 0, B(T_3) = \epsilon/2, \ldots, B(T_n) = 0\}$.

In this event, $B_t$ goes up to $\epsilon/2$ before $-\epsilon/2$ in in time $\geq \delta$, then back $0$ before getting to $\epsilon$ (taking time $\geq \delta$). Then it repeats this $n/2$ times. Hence, $T_n \geq n \delta > 1$ on $E$. Because of our description, on the event $E$, $B_s$ always stays within $(- \epsilon/2, \epsilon)$ up to time $T_n > 1$. So in particular:

$E \subset \{ \sup_{0 \leq s \leq t} |B(s)| < \epsilon \}$.

Moreover, by the strong Markov property, we have $P(E) = (1/2 P( T \geq \delta))^n > 0$.

This proves the claim.

• Wonderful answer. Let me just point out that to prove that $\mathbb{P}(T\geq \delta)>0$, you can just use that $B$ is continuous and hence $\mathbb{P}(\sup_{[0,1/n]}|B_t|<\epsilon/2)\to 1$ as $n\to \infty$. Oct 27, 2022 at 21:38