Stochastic Processes – Problem 2: Prove that with probability 1 there exist sequences with , such that and

Let $B_t$ be standard Brownian motion and let $\tau = \sup\{0\le t\le1:B_t=0\}$. Prove that with probability 1 there exist sequences $t_n < s_n$ with $t_n \uparrow \tau$, such that $B_{t_n} < 0$ and $B_{s_n} > 0$.

Step 1. Let $\tau = \sup\{t\le1:B_t=0\}$. By continuity of Brownian paths, $B_\tau = 0$ almost surely.

Step 2. Brownian motion is symmetric: $B_t$ and $-B_t$ have the same law. So it is enough to rule out one-sided behavior near $\tau$.

Step 3. Suppose, for contradiction, that with positive probability there exists $\delta > 0$ such that $B_t \ge 0$ for all $t \in (\tau-\delta, \tau)$. Then the path has no sign changes in that left neighborhood of $\tau$.

Step 4. This contradicts the local oscillation property of Brownian motion: in every nontrivial interval, almost surely the path attains both positive and negative values infinitely often. Therefore in every left neighborhood of $\tau$, there are times with negative and positive values.

Step 5. Hence we can choose sequences $t_n < s_n < \tau$ with $t_n \uparrow \tau$, $B_{t_n} < 0$, and $B_{s_n} > 0$ almost surely.

QED.