The previous post introduced the notion of a stopping time . A stochastic process can be sampled at such random times and, if the process is jointly measurable, will be a measurable random variable. It is usual to study adapted processes, where is measurable with respect to the sigma-algebra at that time. Then, it is natural to extend the notion of adapted processes to random times and ask the following. What is the sigma-algebra of observable events at the random time , and is measurable with respect to this? The idea is that if a set is observable at time then for any time , its restriction to the set should be in . As always, we work with respect to a filtered probability space . The sigma-algebra at the stopping time is then,

The restriction to sets in is to take account of the possibility that the stopping time can be infinite, and it ensures that . From this definition, a random variable us -measurable if and only if is -measurable for all times .

Similarly, we can ask what is the set of events observable strictly before the stopping time. For any time , then this sigma-algebra should include restricted to the event . This suggests the following definition,

The notation denotes the sigma-algebra generated by a collection of sets, and in this definition the collection of elements of are included in the sigma-algebra so that we are consistent with the convention used in these notes.

With these definitions, the question of whether or not a process is -measurable at a stopping time can be answered. There is one minor issue here though; stopping times can be infinite whereas stochastic processes in these notes are defined on the time index set . We could just restrict to the set , but it is handy to allow the processes to take values at infinity. So, for the moment we consider a processes where the time index runs over , and say that is a predictable, optional or progressive process if it satisfies the respective property restricted to times in and is -measurable.

Lemma 1Letbe a stochastic process andbe a stopping time.

Ifis progressively measurable thenis-measurable.Ifis predictable thenis-measurable.

*Proof:* If is progressive then, as proven in the previous post, the stopped process is also progressive and, hence, is adapted. It follows that is -measurable which, from the definition above, implies that is -measurable.

Furthermore, is -measurable and is zero when restricted to the set for all , so is also -measurable.

Now, consider a predictable process . Write for the predictable sigma-algebra on . That is, the subsets of which are predictable when restricted to and such that is -measurable. Then, is -measurable. By the functional monotone class theorem, it is enough to prove the result for processes of the form for some pi-system of sets generating .

The predictable sigma algebra is generated by the sets of the following forms,

- for times and . If then which, by definition, is -measurable.
- for . If then which is -measurable, and so is also -measurable.

⬜

So, the `adaptedness’ of measurable processes extends to stopping times. In fact, it is possible to go further and use this as an alternative definition of these sigma-algebras.

Lemma 2Letbe a random variable andbe a stopping time. Then,

is-measurable if and only iffor some progressively measurable (or, optional) process.is-measurable if and only iffor some predictable process.

*Proof:* If is -measurable, then the process is adapted and right-continuous. Therefore, it is optional (and hence, progressive) and clearly .

For the second statement, consider the set of random variables which can be expressed as for a predictable process . The functional monotone class theorem can be used to show that contains all -measurable random variables. First, is clearly closed under taking linear combinations. Second, if is increasing to the limit then there exists predictable processes with . Then, is also in .

Finally, it just needs to be shown that for all in a pi-system generating . By definition, the following sets generate .

- . In this case, with .
- for and . Then, with .

In both these cases, is left-continuous and adapted and, hence, is predictable. ⬜

This result gives the main motivation for the definitions of and . For the remainder of this post, I state and prove several simple results which are useful for general applications of stopping times.

Lemma 3Any stopping timeis bothand-measurable.

*Proof:* The deterministic process is trivially adapted and both left and right-continuous, so it is predictable and optional. Consequently, by the previous lemma, is and -measurable. ⬜

Next, the sigma-algebras are increasing in the sense that we would hope.

Lemma 4For any stopping time,

Ifis any other stopping time then,

If, furthermore,wheneverthen.

*Proof:* This proof makes use of Lemma 2. First, by the lemma, every -measurable set can be written in the form for a predictable process . However, as predictable processes are progressive, will also be in .

Now suppose that is (resp. ) measurable. Then, there is a progressive (resp. predictable) process satisfying . As the stopped process is also progressive (resp. predictable) it follows that is (resp. )-measurable.

Finally, suppose that whenever and that . Then

is a left-continuous and adapted at finite times, and is -measurable. Hence, it is predictable process and is -measurable. ⬜

The sigma-algebras satisfy the expected left and right-limits. In the following lemma, the first statement says that right-continuity of a filtration extends to arbitrary stopping times. The second says that can indeed be interpreted as a left-limit. However, this statement does not say anything much for arbitrary stopping times, because it is not in general possible to strictly approximate them from the left in this way. If such a sequence does indeed exist then the stopping time is called *predictable*.

Lemma 5Letbe stopping times. Then

If the filtrationis right-continuous andfor eachthen

If, with a strict inequality whenever, then

*Proof:* Starting with the first statement, we know that . So, it just needs to be shown that any is in . Any such set satisfies

Then, by right-continuity of the filtration, for any

as required.

For the second statement, we know that , so it is only necessary to prove that there is a generating set for lying in . As it is enough to consider sets of the form for . However

as required. ⬜

As should be the case, the definition of the sigma-algebras at a constant stopping time is consistent with the filtration.

Lemma 6Ifis equal to the constant valuethen,

*Proof:* If then for all times ,

showing that . Conversely, if then as required.

This shows that . The equality follows by taking left limits and applying the previous lemma. ⬜

Given two stopping times it follows from the definitions that and . So, is in and, by symmetry, is also in . Furthermore, these two sigma-algebras coincide when restricted to this set. For a sigma-algebra on , and a subset , we use to denote the sigma-algebra on *S* consisting of sets for .

Lemma 7Ifare stopping times then

*Proof:* If is measurable then is in and . The reverse inclusion follows by exchanging and .

Next, is generated by sets in , and by sets of the form for which, restricted to , coincide with . So, , and the reverse inclusion follows by exchanging and . ⬜

Given a stopping time taking values in a countable set of times, the following result is often useful to show that a set is in the sigma algebra by checking it at each of the fixed times.

Lemma 8Letbe stopping times such thatfor all.

A setis inif and only iffor each.

*Proof:* By the previous lemma, if then . Conversely,

as required. ⬜

Finally, the following result is used to construct new stopping times out of old ones. If we wait until a time occurs, and then decide to either use that time or not based on an -measurable event, the result is again a stopping time.

Lemma 9Letbe a stopping time and. Then,

is also a stopping time.

*Proof:* This follows from the following

for all . ⬜

Dear Almost sure,

In lemma 7, do you want to show

Thanks

Yes! I’ll fix it. Thanks

Dear Georges,

Excuse me if I am wrong, but in the proof of Lemma 2, last but one line, shouldn’t it be ?

You’re right. I fixed it, thanks.

Thanks for your answer. Did you also by any chance see my question on the Stochastic Integral ?

Hello George,

I recently meet the problem on consistency of probability measures. Given a sequence of probability measures Q_n, how is it possible to check that they are consistent? The definition seems impractical. What I want is a method that we can actually apply facing concrete examples. Also, where shall I start if I want to find counter-examples?

Thank you very much!

Hi, I am struggling with the definition of the “stopping-time sigma-algebra” (aka sigma algebra at stopping time). You state that, “The idea is that if a set is observable at time then for any time , its restriction to the set should be in .” Is it possible to get a concrete example please? Thanks.

Hi, Thanks for the excellent posts. One question: Why is the sigma algebra at a stopping time “tau” not defined as the sigma algebra generated by tau, in the first place? How is this notion related to the (standard) definition that you have given above? I vaguely recall some mention (not sure in which reference) that the former definition is not operationally useful to perform the related analysis–is that so or are there other reasons?

Consider the case where is constant, equal to a deterministic time . Then . With the definition you suggest, would be trivial. More generally, the sigma algebra generated by would miss out lots of events observable at the stopping time. We want to be -measurable for any reasonably regular adapted process X.

I have been trying to prove if and are stopping times and , then . Any suggestions on how to proceed?

You can try using . From the definitions of the two sigma-algebras, this is in . Take the union over a countable dense set for t. Alternatively use the fact that is left-continuous and adapted, so is measurable.

Nitpicking: “resp.” in Lemma 2 is quite confusing. Probably you meant just “or”?

[GL: Fixed, thanks!]