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I am following Karatzas and Shreve- Brownian Motion and Stochastic Calculus. In the context of Stochastic integration one defines the martingale transform for simple functions:

enter image description here

Now we would like to extend the this definition for more general processes, therefore we need to approximate those processes by simple functions.

Then the following Lemma is an important step in this program

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This Lemma is proved in tree steps,

1) consider X continuous 2) consider X progressively measurable 3) Consider X measurable and adapted.

I can't see why step 2 is needed. It seems to me that the arguments made in step 2 could be done for any measurable and adapted process.

see

enter image description here

Where do we use progressive measurability (in a sense that goes beyond measurable + adapted) of X in step 2?

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    $\begingroup$ Also thank for you asking this question -- I was looking for a proof somewhere that progressively measurable functions could be approximated by simple predictable processes in $L^2$. $\endgroup$ May 23, 2016 at 1:13

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There are at least two reasons. I am certain of the second, only 60% of the first one, so please keep this in mind.

1. Look at equation (2.7); $F_t$ is a stopped version of the integral of the progressively measurable process.

It should follow that (at least restricted to locally bounded/integrable progressively measurable processes) that the integral itself is progressively measurable.

To the best of my understanding, an arbitrary stopped version of a measurable and adapted process does not necessarily have to be either measurable or adapted (or at least can fail one of the other two).

On the other hand, any progressively measurable function stopped at a stopping time is again progressively measurable. To the best of my understanding, the definition of progressively measurable also implies that progressively measurable processes are exactly the processes which are still adapted after stopping at deterministic times (and even progressively measurable still since deterministic times are stopping times).

2. I am 100% certain of this one. Look at the set $A$. As a result of its definition and the definition of progressive measurability, measurable and adapted is in general insufficient for $A$ to be measurable with respect to the product sigma-algebra; we need progressive measurability for this to be true. If $A$ is not measurable with respect to the product sigma-algebra, then Fubini is not applicable and we cannot say anything about the cross-sections of $A$, and therefore we cannot prove that there exists a sequence of simple predictable processes converging to $X_t$ in $L^2$.


Also, as a final note, even if adapted and measurable were sufficient for this proof (although I believe they are not), we still would only want to focus on progressively measurable processes anyway.

That is because the semimartingale stochastic integral only accepts locally bounded progressively measurable processes as integrands, so even if we could approximate in $L^2$ a wider class of functions by simple processes, we still would not be able to use that approximation in defining/generalizing stochastic integrals, so there would be no point in considering it.

Again though, I am fairly certain that progressive measurability is necessary for the proof to go through for either one of the two reasons mentioned above.

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    $\begingroup$ I am going through your answer again. Let's think about the proof in 2) (the one you are 100% sure). You say that you need progressive measurability of X obtain the measurability of $A$ right? Why? Could you explain a bit where do you use the progressive measurability? Now, conside the sets $A_\omega$. If it follows that every one is a nul set of $[0,t]$ doesn't that imply that $A$ is a null measure set on the product measure? In which case that would imply that $A$ is measurable. Hear from you! $\endgroup$ Jul 28, 2017 at 16:07
  • $\begingroup$ @ConradoCosta Thank you for your interest in my answer. I probably could have explained it to you a year ago, but since I have not been studying or using stochastic processes at all since then (i.e. for around a year), I am extremely rusty on the topic and looking at this answer it looks like it was written by another person. Since the question is old enough, it might make sense for you to post a new question on Math.SE, whose content is essentially the same as your comment. Other users will possibly be willing and able to explain why my answer was or wasn't wrong. I am only willing, not able. $\endgroup$ Jul 30, 2017 at 18:43
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I think you're right here, step two isn't technically needed and I believe it is added purely for structural reasons in the book.

I think this is evident from the next part of the proof, c), where they show that if $X_t(\omega)$ is measurable, bounded and adapted then $F_t(\omega):=\int_0^{t\wedge T}X_s(\omega)\>ds$ is still progressively measurable and we can therefore use the arguments in b).

Part c) is really only used to show that part b) holds when the progressive measurability condition on $X_t(\omega)$ is relaxed to measurable and adapted. I think part b) is used in that way because they had shown in Problem 1.2.19 that $F_t(\omega)$ is progressively measurable if $X_t(\omega)$ is also progressively measurable. This simply makes the proof flow better, using previous results, than just going straight to c) and incorporating the logic of b).

I hope that makes sense?

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