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 1Let be a stochastic process and be a stopping time.

- If is progressively measurable then is -measurable.
- If is predictable then is -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 2Let be a random variable and be a stopping time. Then,

- is -measurable if and only if for some progressively measurable (resp. optional) process .
- is -measurable if and only if for 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.

*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 ,If is any other stopping time then,

If, furthermore, whenever then .

*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 5Let be stopping times. Then

- If the filtration is right-continuous and for each then
- 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 6If is equal to the constant value then,

*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.

Lemma 7If are stopping times then

*Proof:* If is measurable then is in 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 8Let be stopping times such that for all .

A set is in if and only if for 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 9Let be 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

Comment by kenneth — 30 April 10 @ 4:02 PM |

Yes! I’ll fix it. Thanks

Comment by George Lowther — 30 April 10 @ 5:33 PM |

Dear Georges,

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

Comment by josh — 3 October 12 @ 9:40 AM |

You’re right. I fixed it, thanks.

Comment by George Lowther — 3 October 12 @ 11:32 AM |

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

Comment by josh — 8 October 12 @ 9:40 AM |

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!

Comment by Zhenyu (Rocky) Cui — 1 November 12 @ 7:55 PM |