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Let $(B_t)$ be a standard $\mathbb R$-valued Brownian motion starting from 0. Let $$T = \inf\{t \in [0, 1] : B_t = \sup_{s \in [0, 1]} B_s\}. $$ I wish to show that $T < 1$ almost surely. Intuitively this should be true, with how jagged Brownian motion is I expect that almost surely for all $\epsilon > 0$ there exists $1 - \epsilon < t < \epsilon$ such that $B_t > B_1$. I'm unsure as to how to make this rigorous though.

Feng
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bitesizebo
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    Shift your focus to $t=0$ and ask if $B_t\ge B_0=0$ for all $t\in[0,1]$, with prob. 1, and use time inversion (as in https://math.stackexchange.com/questions/182107/prove-the-time-inversion-formula-is-brownian-motion) and something like$B_t\ge0$ for all $t\ge 1$, which is ruled out by LIL (say). – kimchi lover Jun 02 '19 at 13:25
  • @kimchilover Could you please elaborate? – bitesizebo Jun 02 '19 at 13:43
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    Let $C(t)=B(1)-B(1-t)$ for $t\in[0,1]$; this gives you another BM. You are asking if $C$ is positive on $(0,1]$ is possible. Etc. – kimchi lover Jun 02 '19 at 13:54

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