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I am trying to prove that $\mathbb{P}(\sup_{t\in[0,1]}|B_t|<\epsilon)>0$ where $(B_t)_{t\geq 0}$ is the standard Brownian motion, and $\epsilon>0$, I came up with the following proof but I feel that something is wrong with it.

Suppose that instead $\mathbb{P}(\sup_{t\in[0,1]}|B_t|<\epsilon)=0$, then there exists $t_0\in [0,1]$ such that $B_{t_0}=\infty$ a.s.

Take $t_1\in[0,1]$ such that $t_1 \neq t_0$ and such that $B_{t_1}$ is bounded (here we can take $t_1=0$), then we know that $$E((B_{t_1}-B_{t_0})^2)=t_1-t_0 $$ But $$E((B_{t_1}-B_{t_0})^2)=E(\infty)=\infty $$

which leads to a contradiction, i.e. $\mathbb{P}(\sup_{t\in[0,1]}|B_t|<\epsilon)>0$

I've already seen the answers to this question here as no one thought of it, which made me doubt my own :/ what do you think?

Houda
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  • You want to show that the first claim is true $\forall \epsilon>0$ when you work with absurd you have the hypothesis that $\exists \epsilon>0 $ such that the opposite claim is true so you can't say that there is $t_0$ such that $B_{t_0}= \infty$ a.s. – chak Jan 01 '18 at 00:05
  • oh right ! omg that was stupid... thank you – Houda Jan 01 '18 at 00:07
  • @chak, houda even if you had that for all $\epsilon$ I don't see how you can show there exists $t_0$ such that $B_{t_0}=\infty$ a.s. – clark Jan 01 '18 at 00:10
  • @clark yes, indeed being $B_t$ continuous that would not be possible. – chak Jan 01 '18 at 00:14
  • @chak, clark under which condition this can be possible? – Houda Jan 01 '18 at 00:38
  • The standard Brownian motion is a.s. continuous and starts from finite value (0) so it's impossible to have $B_{t_0}= \infty$ a.s. for finite $t_0$, if you take a drifted brownian motion with positive drift then i think that taking the limit $t \rightarrow \infty$ you get that it is $\infty$ a.s. – chak Jan 01 '18 at 00:46
  • yes, that makes sense... Thanks @chak ! – Houda Jan 01 '18 at 00:59

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