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We consider on a filtered probability space $(\Omega,\mathcal{F},\mathcal{F}_{r \in \mathbb{R}_+},P)$ with the usual conditions. Let $B$ be a $(\mathcal{F}_r)_{r \geq 0}$-Brownian motion. Let $\theta:=\inf\{r \geq 0,r\sup_{u \in [0,r]}|B_u|>1\}.$

Can we claim that $\theta$ is a stopping time?

Attempt: it's sufficient consider for $v >0,\{\theta \geq v\} \in \mathcal{F}_v.$

We have $\{\theta \geq v\}=\bigcap_{r \in [0,v[}\{r \sup_{u \in [0,r]}|B_u| \leq 1\}$ where we obtained an uncountable intersection.

What do you think?

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Since $r \sup_{u \in [0,r]} |B_u|$ is a continuous function of $r$, $$\bigcap_{r \in [0,v]} \{r \sup_{u \in [0,r]} |B_u| \le 1\} = \bigcap_{r \in [0,v]\cap \mathbb{Q}} \{r \sup_{u \in [0,r]} |B_u| \le 1\}$$ is a countable intersection of $\mathcal F_v$-measurable events and is hence $\mathcal F_v$-measurable. Therefore, $\theta$ is a stopping time.


Given a continuous function $f:[0,\infty) \rightarrow \mathbb{R}$, the function $g(t) := \sup_{s \in [0,t]} f(s)$ is continuous. This can be seen by considering that, for $g$ to be discontinuous at a point $t$, $f$ must also be discontinuous at that point. For a formal proof, we'll show left-continuity; the proof of right-continuity is similar. Let $(t_n) \uparrow t$. Then $\lim g(t_n) = \sup_{s \in [0,t)} f(s)$, but by continuity of $f$, $\sup_{s \in [0,t)} f(s) = \sup_{s \in [0,t]} f(s) = g(t)$.

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  • $\begingroup$ How do you know that $\sup_{u\in[0,r]}|B_u|$ is continuous? $\endgroup$
    – mathex
    Feb 1 at 21:13
  • $\begingroup$ @mathex Because Brownian motion is continuous $\endgroup$ Feb 1 at 21:17
  • $\begingroup$ By continuity of Brownian motion we can say that $r\to \sup_{u\in [0,r]}|B_u|$ is measurable, can you provide a proof for the continuity of the supremum? $\endgroup$
    – mathex
    Feb 1 at 21:20
  • $\begingroup$ @mathex I added a proof that the running supremum of any continuous function is left-continuous. You can apply this to $|B_t|$ to get the result. $\endgroup$ Feb 1 at 21:35
  • $\begingroup$ You forget $\leq 1$ $\endgroup$
    – mathex
    Feb 1 at 21:41

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