Let $\Omega$ be the set of càdlàg functions $f : [0,\infty) \to \mathbb R^d$ equipped with the Skorokhod topology for any $d \geq 1$, and let $X_t(\omega) = \omega(t)$ for any $\omega \in \Omega, t \geq 0$. Then, $\{X_t(\omega)\}_{t \geq 0}$ is a stochastic process on $\mathbb R^d$ with associated natural filtration $\{\mathcal F_{t}\}_{t \geq 0}$ (and we let $\mathcal F_{\infty} = \sigma(\cup_{n \geq 0} \mathcal F_n)$). Let $\tau$ be an a.s. finite stopping time, and recall the stopped filtration $\mathcal F_{\tau}$, which contains all $A \in \mathcal F_{\infty}$ such that $A \cap \{\tau \leq t\} \in \mathcal F_t$ for all $t \geq 0$.

My question is concerned with a very "logical" but, to me, intractable characterization of $\mathcal F_{\tau}$ for a stopping time $\tau$ in this particular case. Let $\tau_1 \leq \tau_2$ be two stopping times. For any stopping time $\sigma$ we can define the "shift" operator $\theta_{\sigma}:\Omega \to \Omega$ by $(\theta_{\sigma}(\omega))(t) = \omega(t + \sigma(\omega))$ (i.e. just shifting time ahead by $\sigma$).

Motivation and question

Informally, what is the filtration $\mathcal F_{\tau_2}$? It consists of all events $A$ whose occurrence or lack of is completely determined by "all coordinates before $\tau_2$" (or determined by observing the stochastic process till time $\tau_2$). Now, break any such observation into two parts :

  • A part based on the observations purely from $\tau_1$.

  • A part based on the observations between $\tau_1$ and $\tau_2$.

Now, we know that events determined purely by the first category of observations are $\mathcal F_{\tau_1}$. On the other hand, events determined purely by the second category of observations are captured by sets of the form $\{\omega : \theta_{\tau_1}(\omega) \in B\}$ where $B \subset \mathcal F_{\tau_{2} - \tau_1}$.

Hence, the following question arises.

For $A \in \mathcal F_{\tau_1}$, $B\in \mathcal F_{\tau_2 - \tau_1}$, let $$ E_{A,B} = A \cap \{\omega : \theta_{\tau_1}(\omega) \in B\}. $$ Is it true that $$\mathcal F_{\tau_2} = \sigma(\{E_{A,B} : A \in \mathcal F_{\tau_1}, B\in \mathcal F_{\tau_2 - \tau_1}\})?$$

Ideas and known results

This question on Mathematics Stack Exchange is a corollary of my question above, so that's why I'd like an answer.

Here are some known results.

  • $\mathcal F_{\tau} = \sigma\{X_{\tau \wedge t} : t\geq 0\}$ for any a.s. finite stopping time $\tau$.

  • A function $f : \Omega \to \mathbb R^d$ is $\mathcal F_{\tau}$ measurable if and only if, for any $\omega_1,\omega_2$ such that $\tau_1(\omega) = \tau_2(\omega)$, we have $f(\omega_1) = f(\omega_2)$. In particular, if $f$ is $\mathcal F_{\tau}$ measurable, then $f(\omega) = f(\omega^{\tau})$ for all $\omega \in \Omega$, where $\omega^{\tau}(t) = \omega(t \wedge \tau(\omega))$.

More ideas may be found in the MSE question linked above. I'm surprised an answer isn't available in a standard text.


1 Answer 1


$\newcommand\F{\mathcal F}$This question does not seem to make sense, because the symbol $\F_{\tau_2-\tau_1}$ does not make sense in general, because $\tau_2-\tau_1$ will not be a stopping time in general even if $\tau_1$ and $\tau_2$ are stopping times such that $\tau_1\le\tau_2$.

For instance, suppose that $X_t=1+B_t$, where $B_\cdot$ is a standard Brownian motion, $$\tau_2:=\inf\{t\ge0\colon X_t=2\},\quad \tau_0:=\inf\{t\ge0\colon X_t=0\},$$ $$\tau_1:=\tau_0\wedge\tau_2=\inf\{t\ge0\colon X_t\in\{0,2\}\}\le\tau_2.$$ Then the event $$\{\tau_2-\tau_1\le1\} =\big\{\exists s\ge0\ X_s=2\ \&\ \forall u\in[0,s-1)\ X_u\notin\{0,2\}\big\}$$ is not determined by $(X_t)_{t\in[0,1]}$.

  • $\begingroup$ Ah right, that is an issue. In any case I have probably solved what I needed to, but I cannot interfere with the question at present so I'll just accept this. $\endgroup$ Commented Apr 24 at 14:55

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