Let $(B_t)$ be a standard Brownian motion and $\mathcal F_t$ the associated canonical filtration. It's a standard result that the hitting time for a closed set is a stopping time for $\mathcal F_t$ and the hitting time for an open set is a stopping time for $\mathcal F_{t+}$.

Is there an elementary way to see that the hitting time for an open set is not in general a stopping time for $\mathcal F_t$? Say the hitting time for an open interval $(a,b)$?

I'm interested in this question because it would show the filtration generated by a right-continuous process need not be right-continuous. There are other counterexamples for this on M.SE but they're all somewhat artificial.

My apologies if this is obvious. I just started learning about such things.


No, it's not at all obvious.

If we interpret $\mathcal{F}_t$ as "information up to time $t$", it's not surprising that the hitting time of an open interval $(a,b)$ is not an $\mathcal{F}_t$-stopping time. For instance if $B_t(\omega)=a$ for some $t>0$, then the information about the past, i.e. $(B_s(\omega))_{s \leq t}$, is not enough to decide whether $\tau(\omega)=t$; we need a small glimpse into the future. enter image description here

Making this intuition rigorous is, however, not easy. One possibility is to apply Galmarino's test which states that a mapping $\tau: \Omega \to [0,\infty]$ is a stopping time (with respect to $(\mathcal{F}_t)_{t \geq 0}$) if and only if $$\tau(\omega)=t, B_s(\omega) = B_s(\omega') \, \, \text{for all $s \leq t$} \implies \tau(\omega')=t, \tag{1}$$ see this question. If $\tau$ is the hitting time of an open interval $(1)$ is not satisfied; the easiest way to see this is to consider the canonical Brownian motion, i.e. consider $(B_t)_{t \geq 0}$ as a process on the space of continuous mappings.

Let me finally remark that there are other ways to prove that the filtration generated by a Brownian motion is not right-continuous, see, for instance, this answer.

  • $\begingroup$ Thank you! I've benefited greatly from your answers on this site, including this one. $\endgroup$ – Potato Feb 11 '17 at 18:57
  • $\begingroup$ @Potato Thanks; you are welcome. $\endgroup$ – saz Feb 11 '17 at 19:37

Consider the open interval $(1,2)$ and two (extremely stylized) sample paths: $$ \omega_1(t) =\cases{t,&$0\le t\le 1$,\cr 2-t,&$t>1$,\cr} $$ and $$ \omega_2(t)=t,\quad t\ge 0. $$ These two paths agree up to time $t=1$. Letting $T$ denote the hitting time of $(1,2)$, you have $T(\omega_1) =1$ but $T(\omega_2)=+\infty$. If $T$ were a stopping time for $(\mathcal F_t)$ you would have $\{T\le 1\}\in\mathcal F_1$, which would in turn imply that if $T(\omega_1)\le 1$ and $\omega_1|_{[0,1]}=\omega_2|_{[0,1]}$ then also $T(\omega_2)\le 1$. As this fails to hold, $T$ cannot be a stopping time for $(\mathcal F_t)$.

  • $\begingroup$ This seems like the right idea, but I don't think it works as written. The paths you give are not sample paths of Brownian motion, because they're too nice (differentiable almost everywhere, for example). $\endgroup$ – Potato Feb 10 '17 at 23:07
  • $\begingroup$ Hence "extremely stylized". Draw two more Brownian-like paths, one with a local max value of 1 occuring at $t=1$ and the other hitting (and surpassing) level 1 at time $t=1$. $\endgroup$ – John Dawkins Feb 10 '17 at 23:21
  • $\begingroup$ It's not entirely clear to me such paths exist. It's also not clear to me that exhibiting one bad path is enough to tank the whole thing, given that we usually only care about properties up to a.e. equivalence. (Not sure if that's the case here...) $\endgroup$ – Potato Feb 10 '17 at 23:38
  • $\begingroup$ If you "complete" the natural filtration the problem goes away, for then the filtration is right continuous. $\endgroup$ – John Dawkins Feb 11 '17 at 0:12

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