Is there an example of progressively measurable process that is not predictable?

This question is motivated by Revuz-Yor, Continuous Martingales and Brownian Motion http://www.springer.com/gb/book/9783540643258.

They define the space of integrands with respect to a continuous $L^2$-bounded martingale $M$ as the progressively measurable processes $\phi$ such that:

$$\int_0^\infty \phi^2 dM<\infty$$

So they ask $\phi$ to be progressively measurable instead of the very usual, stronger, hypothesis of $\phi$ being predictable.

${}{}$ I wonder whether this space is actually bigger than the one we get with the predictability imposition, or if this yields the same result in this case.

  • $\begingroup$ I think so, although I don't know any examples off the top of my head. They might coincide under some mild assumptions though. To be honest I really don't know, but it is a great question. $\endgroup$ – Chill2Macht Jul 15 '16 at 20:41

This depends, of course, on the underlying probability space, most notably the filtration. In general, there are plenty of progressive processes which are not predictable, and this should be fairly standard material. If one works with the raw (un-augmented) filtration on the canonical path space of continuous functions, then the notions of predictability and progressive measurability coincide. I don't have the reference accessible at the moment, but I would bet any amount of money that it's in the first volume of Dellacherie-Meyer.

On the other hand, if you augment your filtration to make it complete and right-continuous, as one so often does, there are plenty of easy examples of progressive processes which are not predictable. (Actually, right-continuity is all that matters in the following.) For example, define $(X_t)_{t \ge 0}$ by $X_t=0$ for $t < 1$ and $X_t=\pm 1$ with some nontrivial (say, equal) probabilities. Consider the augmented filtration generated by $X$,

$\mathcal{F}_t = \cap_{s > t} \sigma(X_s) \vee \mathcal{N}$,

where $\mathcal{N}$ is the set of null sets of the ambient probability space. Then the right-limit process $X_+=(X_{t+})_{t \ge 0}$ is progressive but not predictable, with respect to this filtration. To see this, recall that the predictable processes are (by definition) those generated by the left-continuous adapted processes. Because $X$ is constant strictly before time $1$, every adapted process must be a.s. constant and deterministic on $[0,1)$. A left-continuous process is then also a.s. constant and deterministic on $[0,1]$, and we conclude that the same is true of predictable processes. As $X_{1+}$ is random, it follows that $X_+$ is not predictable. To see that $X_+$ is progressive, just note that it is right-continuous and adapted, because $X_{1+}$ is $\mathcal{F}_1$-measurable due to the right-continuity imposed on the filtration.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.