Let $(B_t)_{t \ge 0}$ be a Brownian motion starting from $0$. Then, do we have that, almost surely, for all $s \ge 0$, there exist $t$, $u \ge s$ with $B_t < 0 < B_u$?
2025-01-13 02:16:45.1736734605
Almost surely, for all $s \ge 0$, there exist $t$, $u \ge s$ with $B_t < 0 < B_u$?
49 Views Asked by user265817 https://math.techqa.club/user/user265817/detail AtRelated Questions in REAL-ANALYSIS
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