I am interested in a proof of the following statement which seems intuitive, but is somehow really tricky:

Let $X$ be a stochastic process and let $(\mathcal{F}(t) : t \geq 0)$ be the filtration that it generates (unaugmented). Let $T$ be a bounded stopping time. Then we have $\mathcal{F}(T) = \sigma(X(T \wedge t) : t \geq 0)$

I have a proof at hand (Bain and Crisan, Fundamentals of Stochastic Filtering, page 309), but in my opinion there is a major gap. I will try to explain the idea of proof.

Let $V$ be the space of functions $[0,\infty) \rightarrow \mathbb{R}$ equipped with the sigma algebra generated by the cylinder sets. Consider the canonical map $X^T:\Omega \rightarrow V$ which maps $\omega$ to the trajectory $t \mapsto X(t \wedge T(\omega),\omega)$. Then we have $\sigma(X(T \wedge t) : t \geq 0) = \sigma(X^T)$.

The difficult part is $\subseteq$. Let $A \in \mathcal{F}(T)$. We want to find a measurable map $g:V \rightarrow \mathbb{R}$ such that $1_A = g \circ X^T$, then we're done. It is now straightforward to show that $1_A$ is constant on sets where the sample paths of $X^T$ are constant. (To be more precise: for $\rho \in \Omega$ consider the set $\mathcal{M}(\rho) = \lbrace \omega : X(\omega,t) = X(\rho,t), 0 \leq t \leq T(\rho) \rbrace$. Then $T$ and $1_A$ are constant on every set of this form).

The problem is: this is not sufficient! It suffices to construct a map $g$ such that $1_A = g \circ X^T$, but how we can we know that $g$ is measurable? This is where the proof of Bain and Crisan comes up short IMO.

I can show this result only under the assumption that the map $X:\Omega \rightarrow V$ be surjective: Since $A \in \mathcal{F}(\infty)$, we have a measurable map $g$ such that $1_A = g \circ X$. Let $x \in V$. Then $T$ and $1_A$ are constant on the preimage of $x$ under $X$. Therefore, $g(x)$ does not depend on the values of $x$ after time $T$ (which is constant on the preimage of $x$). Since $X$ is surjective, we have $g(x) = g(K^Tx)$, where $K$ is the killing functional $K^tx(s) = x(t \wedge s)$. Hence, $g \circ X = g \circ X^T$, and we are done.

I think that this result could be a little bit deeper. I have seen two proofs of this for the special case that $X$ is the coordinate process on $C[0,\infty)$, one is given in the book of Karatzas & Shreve, Lemma 5.4.18. The fact that Karatzas proves this late in the book only in this special case somehow makes me think that the general case is not so easy.

I would really appreciate any comment or other reference for this result.


1 Answer 1


I've been worried about the same thing. Here's what I came up with:

Assume that $X$ is progressively measurable (e.g. cadlag), then the inclusion $\sigma (X_s^T : s \le t)\subset\mathcal{X}_T^0$ trivial. Without this condition, this inclusion won't hold in general, I believe. For the other direction:

$A \in \mathcal{X}_T^0 \Leftrightarrow \mathbb{1}_A$ is $\mathcal{X}_T^0$ measurable $\Leftrightarrow \mathbb{1}_A = Y_T$ for some $\mathcal{X}_t^0$-optional process $Y_t$.

Thus it suffices to show that for all optional processes $Y_t$, $Y_T$ is $\sigma (X_s^T: s \ge 0)$ measurable.

Now, the optional processes are generated by stochastic intervals of the form $[\sigma,\infty)$, so using a functional class argument it is enough to show that $\{\sigma \le T\} \in \sigma (X_s^T : s \ge 0)$ for all $\mathcal{X}_t^0$-stopping times $\sigma$, $T$. Do this via discretisation of $\sigma$, $T$ and taking limits. (Similarly we can prove that a $\mathcal{X}_t^0$-stopping time $T$ is a $\sigma (X_s^T : s \ge 0)$ stopping time, I believe.)

EDIT: I spoke too soon, this discretisation argument doesn't work unless the filtration is right continuous.... I have no idea how to proceed.

EDIT2: It appears that the proof is IV.100 of Probabilities and Potential (Dellacherie and Meyer) though I do not currently have this to hand.

  • 2
    $\begingroup$ Dellacherie and Meyer as well as all other sources only proof this for the canonical process on the space of samplepaths. That's the problem. It might not be true on general probability spaces. $\endgroup$
    – hoirkman
    Apr 17, 2013 at 14:18
  • $\begingroup$ Other authors also prove this using additional requirements. See math.stackexchange.com/questions/1618328/… $\endgroup$
    – Mr_3_7
    Nov 8, 2017 at 10:00

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