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see this question

claiming that $\mathbb{E}[B_{\tau}] = E[B_0]$ means that either $\tau$ is bounded or $B_\infty = 0 $ since the brownian motion isn't uniformly integrable,

$\varphi_{B_t} (s) = \exp(-\frac12ts^2) \to 0 \neq 1$

so the brownian motion doesn't even converge to $0$ in distribution, we can forget about $B_\infty = 0 $ and this leaves us with $\tau$ being bounded but I don't understand why, can't $B_t$ just swirl around a neighboorhood of $0$ forever ?

Also, I remember I read somewhere that $\limsup B_t = +\infty$ and $\liminf B_t = -\infty$, could this be the reason why $\tau$ is bounded ?

the_firehawk
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    There certainly isn't a real constant $C$ such that $\tau < C$ with probability 1, but I think it is $L^p$ bounded for all $p < \infty$. – user217285 Jan 04 '20 at 18:19
  • @user217285 you mean the stopping time is $L^p$ bounded, not the Brownian motion right, because the Brownian motion isn't even uniformly continuous ? – the_firehawk Jan 04 '20 at 18:21
  • Yeah $\tau$, but for any fixed $t$, $B_t$ is a normally distributed random variable so it's in $L^p$ for all $p < \infty$ as well. – user217285 Jan 04 '20 at 18:24

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