In the previous two posts of the stochastic calculus notes, I began by introducing the basic concepts of a stochastic process and filtrations. As we often observe stochastic processes at a random time, a further definition is required. A *stopping time* is a random time which is adapted to the underlying filtration. As discussed in the previous post, we are working with respect to a filtered probability space .

Definition 1A stopping time is a map such that for each .

This definition is equivalent to stating that the process is adapted. Equivalently, at any time , the event that the stopping time has already occurred is observable.

One common way in which stopping times appear is as the first time at which an adapted stochastic process hits some value. The *debut theorem* states that this does indeed give a stopping time.

Theorem 2 (Debut theorem)Let be an adapted right-continuous stochastic process defined on a complete filtered probability space. If is any real number then defined by

(1)

is a stopping time.

If is defined by equation (1) then it does seem intuitively obvious that it will be a stopping time. Clearly, will be less than or equal to precisely when for some ,

(2) |

As is an adapted process, each of the sets inside the union on the right hand side is -measurable, and it seems reasonable to conclude that should also be -measurable, so that is a stopping time. However, the right hand side of (2) is an *uncountable* union, and sigma-algebras are only closed under countable unions and intersections in general. This result demonstrates the added difficulties in looking at continuous-time processes versus the discrete-time case. In discrete-time, the union in (2) is only over finitely many times and in that case the debut theorem follows easily.

For continuous adapted processes (e.g. Brownian motion), the debut theorem is relatively easy to prove. Continuous processes always achieve their supremum value on any compact interval, and it is enough to look at the maximum process . By continuity, this supremum can be restricted to the countable set of rational numbers. Equation (2) reduces to the following,

which expresses in terms of countable intersections of countable unions of and hence is in .

For right-continuous processes it is still true that is fully determined by its values at rational times, so it might seem that the debut theorem can be proved in a similar way as for continuous processes. However, this is not the case, and it is not possible to express using countable unions and intersections of sets in . In fact, need not be measurable in general, and the completeness of the filtered probability space is required. Still, it is not difficult to prove using only elementary techniques, and I give a proof of this below.

The debut theorem for right-continuous processes is only a special case of a more general result for arbitrary progressively measurable processes. However, the more general case relies on properties of analytic sets, which is a subject going well outside of these notes (I added a proof of the general case to PlanetMath), and right-continuous processes are more than general enough for these notes.

The value of a jointly measurable stochastic process at a random time is a measurable random variable, as mentioned in the previous post. As well as simply observing the value at this time, as the name suggests, stopping times are often used to stop the process. A process stopped at the random time is denoted by ,

It is important that stopping an adapted process at a stopping time preserves the basic measurability properties.

Lemma 3Let be a stopping time. If the stochastic process satisfies any of the following properties then so does the stopped process .

- left-continuous and adapted.
- right-continuous and adapted.
- predictable.
- optional.
- progressively measurable.

*Proof:* First, recall that if is jointly measurable and is any random time then is measurable (see here). It follows from the decomposition

that is also jointly measurable. Now suppose that is progressive and is any fixed time. By definition, is -measurable and, if is a stopping time, then is -measurable. Then, by what we have just shown above, the stopped process

is -measurable. This shows that is progressive.

Now let be left (resp. right) continuous and adapted. Then it is progressively measurable and, as has just been shown, is progressive. So, is adapted and it is clearly also left (resp. right) continuous.

Finally, we note that the collection of all processes such that is predictable (resp. optional) includes the left (resp. right) continuous adapted processes and is closed under the limit of a sequence of processes. So, by the functional monotone class theorem it follows that is predictable (resp. optional) whenever is predictable (resp. optional). ⬜

Other than the proof of the debut theorem given below, this covers the main results on stopping times for this post. All that remains are some very useful lemmas which are almost trivial to prove. First, it is often useful to replace the inequality in the definition of a stopping time by a strict inequality. This can be done as long as the filtration is right-continuous.

Lemma 4A map is a stopping time with respect to the right-continuous filtration if and only if for each .

*Proof:* For a stopping time using the fact that for each gives

Conversely, if for each time, then for any ,

As this is true for all it shows that . ⬜

Finally, the class of stopping times is closed under basic operations such as taking the maximum or minimum of two times or, for right-continuous filtrations, taking the limit of a sequence of times.

Lemma 5

- If are stopping times then so are and .
- Let be a sequence of stopping times converging to a limit and suppose that for each , for large enough . Then is a stopping time. Note, in particular, that this includes the case where is increasing to the limit .
- If is a sequence of stopping times then is a stopping time.
- If is a sequence of stopping times and the filtration is right-continuous, then and are stopping times.

*Proof:* If are stopping times then

So, and are stopping times.

Now let be a sequence of stopping times such that, for each , for large . Then,

so that is a stopping time.

If is any sequence of stopping times then

so is a stopping time.

Finally, suppose that the filtration is right-continuous and that is a sequence of stopping times. Note that whenever infinitely often for some , enabling us to write

If the filtration is right-continuous, the lemma above shows that is a stopping time. Similarly, whenever for all large and some giving

so is also a stopping time. ⬜

Lemma 6LetXbe a cadlag adapted process. Then, there exists a sequence of stopping times such that whenever and , and

*Proof:* For any positive real numbers define the random time

It can be seen that the union of the graphs over positive rationals is equal to the set of times at which so, to complete the proof of the lemma, it is enough to show that is a stopping time. That is, the set is -measurable. For we have , so it is trivially measurable. So, we can suppose that . In that case, letting be any countable dense subset of with , set

As this is the supremum of a countable set of -measurable random variables, it is -measurable. Also, using the cadlag property of ,

so this is -measurable. Then, iff , which is -measurable. ⬜

#### Proof of the debut theorem

I now give a proof of the debut theorem for a right-continuous adapted process . For a fixed real number , let be the first time at which . We need to show that this is a stopping time.

Given any stopping time , it is possible to define the larger time

Clearly, and it is easily seen that this is a stopping time. Indeed, is less than or equal to a positive time precisely when, for each , there is a time in the range satisfying . By right-continuity, it is enough to restrict to rational multiples of giving,

Also, using right continuity, will be strictly greater than whenever .

The idea is to start with any stopping time bounded above by , for example will do. Then, by iteratively replacing by , approach from below by successively closer approximations. Unfortunately, right-continuous processes can be badly behaved enough that this can fail, even after infinitely many steps, and transfinite induction would be required. A quicker approach, which I use here, is to make use of the idea of the essential supremum of a set of random variables.

Let consist of the set of all stopping times satisfying . By properties of the essential supremum, there exists a sequence such that is an essential supremum of . As shown above, this is a stopping time and, therefore, . The stopping time defined above satisfies and is therefore in . From the definition of the essential supremum, this implies that and, therefore, with probability one. However, as mentioned above, whenever , which therefore has zero probability.

We have shown that the stopping time satisfies almost surely. Finally, completeness of the filtered probability space implies that is a stopping time.

With reference to your note on stopping times, I understand that if we consider a countable sequence of stopping times, then both the supremum and the infimum of such a sequence are stopping times. Now if we consider an uncountable sequence of stopping times. Is the supremum and infimum of this sequence a stopping time? If not, whats a counter-example? ( I am aware that the sigma algebra properties require countable unions and intersections )

Comment by Joseph — 23 September 10 @ 1:15 PM |

Hi. Counterexamples are,

– Take your probability space to be the unit interval with the standard Lebesgue measure, A ⊂ [0,1] be a non-measurable set. Let τ

_{x}(ω) be 1 if ω = x and 0 otherwise. This is a stopping time, but sup_{x∈A}τ_{x}is 1 precisely on the non-measurable set A so is not a stopping time.– A more subtle example is that, on the space of cadlag processes (with the natural filtration), the first time τ at which which the coordinate process hits a level K is not a stopping time unless you complete the filtration (but it is the supremum of all stopping times T ≤ τ). This is harder to see though, but is true because the set {τ ≤ t} can be any analytic set, and need not be measurable (but is universally measurable).

[Also, moved your comment to the relevant post]

Comment by George Lowther — 25 September 10 @ 3:39 AM |

According to the definition of stopping time, the most simple example of it would be T = c, c a real number, right?

Comment by Jhonny — 9 January 11 @ 8:27 PM |

Right.

Comment by George Lowther — 10 January 11 @ 9:41 PM |

Dear George, could you tell what does the notation means in the statement of Lemma 5? Also, in the last formula for should it be ? I also have the following question: it seems that without an assumption on the completeness of the filtration we can at least state that there is a stopping time such that the set is a null set (not necessarily measurable). Is it right?

Comment by Ilya — 19 February 12 @ 4:58 PM |

– Yes, I seem to have used the notation without explaining what it means, which is a bit sloppy. It is standard notation though — it is a stochastic interval,

You can think of as being a `random set’ , which depends on .

– I fixed the last formula. Thanks.

– For your final question, yes. There will always be a stopping time with being -null.

Comment by George Lowther — 22 February 12 @ 1:51 AM |

btw if you have two stopping times are their sum necessarily a stopping time?

Comment by ben — 24 October 12 @ 4:29 AM |

Yes. From an intuitive point of view, this seems obvious. Given that times σ, τ are observable when they occur, can you tell when σ + τ occurs? Yes, you clearly can, because σ and τ will both have already been observed by then.

More precisely, given any measurable function f:R

_{+}→ R_{+}with f(s,t) ≥ max(s,t), then f(σ,τ) will be a stopping time whenever σ and τ are. In particular this holds for f(s,t) = s + t. More generally, you just need to require that f(s,t)∧u = f(s∧u,t∧u) holds.Comment by George Lowther — 24 October 12 @ 9:09 PM |

Dear Almost Sure,

I thank you for your blog, it is very interesting and I’ve found here a lot of explanations. I’m reading the book “Stochastic calculus and financial applications” of Michael Steele and I have a doubt about the theorem “Doob’s Continuous-Time Stopping Theorem” on page 51. This is the statement: “Suppose {Mt} is a continuous martingale w.r.t. a filtration {Ft} that satisfies the usual conditions. If tau is a stopping time for {Ft}, the process Xt=M_min(t,tau) is also a continuous martingale w.r.t. {Ft}”

My doubt is about the condition that the filtration needs to satisfy the usual conditions. Your lemma 3 is a more general version of the theorem and you haven’t assumed the satisfation of the usual conditions. I don’t have very strong theoretical skills and so I’m worried that there is something that I don’t understand. In the proof of the theorem I am not able to find the place where the assumption is necessary. If you can give your opinion, I’ll appreciate it a lot. Thanks a lot

Comment by Ale — 4 July 14 @ 12:12 PM |

Sorry, I read badly and your lemma 3 is not a generalization of the theorem “Continuous-time stopping theorem”. The lemma doesn’t assure the martingale property of the stopped process.

Comment by Ale — 4 July 14 @ 1:29 PM |

No, my Lemma 3 here does not involve martingales. However, I do prove the stopping theorem in a later post, and usual conditions are not required at all.

Comment by George Lowther — 6 June 16 @ 4:06 AM |

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Pingback by Proof of the Début theorem | Question and Answer — 28 October 14 @ 2:50 AM |

Dear George, I am from engineering back ground. I see that when considering stopping times for continuous stochastic processes, you considered only over rational numbers. Also I see this in many texts when considering stopping time proof. I am quite confused about considring only rational numbers. Can you kindly clarify with some references also if possible.

Comment by Ram — 22 January 15 @ 11:42 AM |

It is because in probability and measure theory, you can only work easily with operations on countably many sets (or events) at once. This is because of countable additivity of measure. Unions of uncountably many sets can give non-measurable sets. As the set of reals is uncountable, you need to restrict to a countable subset in many probabilistic arguments.

Comment by George Lowther — 6 June 16 @ 3:49 AM |

In the proof of lemma 5, you define a stopping time. But how do you know that what you define actually is a stopping time?

Comment by Anonymous — 19 June 16 @ 11:56 PM |

You are right to ask – I did skip over the proof of that rather quickly. We need to show that is measurable. One way is is to set . Then, iff eventually.

Comment by George Lowther — 20 June 16 @ 12:37 AM |

Just one question regarding this, how are you allowed to use max, instead of sup? How do we know that the max exists?

Comment by Anonymous — 9 August 16 @ 4:52 PM |

Actually, I should have written sup rather than max. I’ll update the post. The argument should be unchanged though. Actually, the max of does exist over any bounded innterval, but that is not important to the result.

Comment by George Lowther — 9 August 16 @ 5:34 PM

I updated the proof to include this argument.

Comment by George Lowther — 20 June 16 @ 1:00 AM |

Thank you very much. I have been looking for arguments that jumps really are stopping times, and functions of jumps really are measurable etc.. Do you know if any book has these arguments? Or did you come up with it yourself? Anyway, great blog, really good work!

Comment by Anonymous — 21 June 16 @ 4:00 PM |

It is quite standard, although the precise arguments in these notes are my own. I can check my references for published statements of such facts.

Comment by George Lowther — 26 July 16 @ 9:26 PM