Given the following setting:

Let $\{W_t:t\geq 0\}$ be a Brownian motion. for arbitrary $a>0$, define the exit time of the interval $[-a,a]$ as $$\tau=\inf\{t\geq 0:|W_t|>a\}$$

The question is to show that this is a stopping time, i.e. that for a filtration $\mathscr{F}_t=\sigma(W_s,0\leq s\leq t)$, we have that $(\tau\leq t)\in\mathscr{F}_t$.

This is an exercise that came after proving that a hitting time, i.e. $\tau=\inf\{t\geq 0:W_t=a\}$, is a stopping time. I know how to prove that this is a stopping time, but the method I used there cannot be applied here because we do not have that $(\tau\leq t)\in\mathscr{F}_t$, but instead we have that $(\tau\leq t)\in\mathscr{F}_{t^+}$.

I'm not even sure where to start with this exercise, but the hint given was that we can ''make'' $\mathscr{F}_{t^+}=\mathscr{F}_{t}$ by adding things of measure $0$ to the filtration. However, this hint does not bring me any closer to knowing where to start. What is meant by adding things of measure $0$ to the filtration? Why does this help? How do I proceed with solving this exercise?

Any help is appreciated.


1 Answer 1


By considering right-continuous $(\mathcal{F}_t)$ it suffices to show that $\{\tau<t\}\in\mathcal{F}_t$. Then since $t\mapsto W_t$ is continuous,

$$ \{\tau<t\}=\bigcup_{q<t,q\in\mathbb{Q}}\{|W_q|>a\}. $$

The same applies to any open set $A$ and $\tau:=\inf\{t\ge 0:W_t\in A\}$.

The completed natural filtration of a Brownian motion, $(\mathcal{F}_t\bigvee \mathcal{N})$ ($\mathcal{N}$ are the $\mathsf{P}$-null sets of $\mathcal{F}$), is right-continuous (e.g. Theorem 8.2.2 on page 309 here).

  • $\begingroup$ Thanks for the answer. Any idea on what could be meant with the hint ''adding things of measure zero to the filtration''? $\endgroup$ Jan 25, 2019 at 18:57
  • $\begingroup$ This shows only that $\tau$ is a.s. equal to an $(\mathscr F_t)$ stopping time. $\endgroup$ Jan 26, 2019 at 21:07

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