The concept of a stopping times was introduced a couple of posts back. Roughly speaking, these are times for which it is possible to observe when they occur. Often, however, it is useful to distinguish between different types of stopping times. A random time for which it is possible to predict when it is about to occur is called a predictable stopping time. As always, we work with respect to a filtered probability space .
Definition 1 A map is a predictable stopping time if there exists a sequence of stopping times satisfying whenever .
Predictable stopping times are alternatively referred to as previsible. The sequence of times in this definition are said to announce . Note that, in this definition, the random time was not explicitly required to be a stopping time. However, this is automatically the case, as the following equation shows.
One way in which predictable stopping times occur is as hitting times of a continuous adapted process. It is easy to predict when such a process is about to hit any level, because it must continuously approach that value.
Theorem 2 Let be a continuous adapted process and be a real number. Then
is a predictable stopping time.
Proof: Let be the first time at which which, by the debut theorem, is a stopping time. This gives an increasing sequence bounded above by . Also, whenever and, by leftcontinuity, setting gives whenever . So, , showing that the sequence increases to . If then, by continuity, . So, whenever and the sequence announces .
In fact, predictable stopping times are always hitting times of continuous processes, as stated by the following result. Furthermore, by the second condition below, it is enough to prove the much weaker condition that a random time can be announced `in probability’ to conclude that it is a predictable stopping time.
Lemma 3 Suppose that the filtration is complete and is a random time. The following are equivalent.
 is a predictable stopping time.
 For any there is a stopping time satisfying
(1)
 for some continuous adapted process .
Proof: If is predictable then there is a sequence of stopping times announcing . This means that or for large . By bounded convergence, (1) is satisfied with and large .
Now suppose that the second statement holds. In order to construct the process satisfying the third condition, it is enough to find an adapted continuous process taking values in , and which is finite for and infinite for . Then, can be obtained by . Furthermore, can be replaced by on any zero probability set and, by completeness of the filtration, it will still be adapted. So, only has to have the above properties with probability one.
By equation (1), there is a sequence of stopping times satisfying

(2) 
The BorelCantelli lemma shows that, with probability one, or for all large . In particular, equals up to a zero probability set, so is measurable. Choosing a sequence of positive reals , set on the set and
otherwise. As , this reduces to a finite sum when . It only remains to show that for large enough , the left limit at
will be almost surely infinite whenever . By monotone convergence, equation (2) gives for large . In this case, another application of the BorelCantelli lemma gives, with probability one, for all large when , so is indeed infinite.
Finally, using Theorem 2, the first statement follows from the second.
As with general stopping times, the class of predictable stopping times is closed under basic operations such as taking the maximum and minimum of two times.
Proof: If announce and then it is clear that and announce and respectively.
Furthermore, the predictable stopping times are closed under taking limits of sequences, under some conditions. For example, increasing sequences or eventually constant sequences of predictable stopping times are predictable.
Lemma 5 Let be a sequence of predictable stopping times. Then,
 is a predictable stopping time.
 if there is a limit satisfying eventually (for all ), and the filtration is complete, then is a predictable stopping time.
Proof: Let announce . For the first statement, the sequence announces .
For the second statement, choose any and large enough such that . By Lemma 3, there is a stopping time satisfying
It follows that equation (1) is satisfied and, by Lemma 3, is predictable.
As shown in the previous post, given a stopping time and a set , the time as defined below will also be a stopping time.

(3) 
If, however, is a predictable stopping time then need not be predictable. For this to be true, we need to restrict the set to be in .
Lemma 6 Suppose that the filtration is complete. If is a predictable stopping time and , then is a predictable stopping time.
Proof: Let be stopping times announcing and set . As proven in the previous post, . The monotone class theorem shows that for any there is a set with .
Then, for large so, as shown in the previous post, will be a stopping time. For any ,
which, by bounded convergence, is greater than for large . Lemma 3 completes the proof.
Predictable stopping times are sometimes used to define the predictable sigmaalgebra. In these notes, it was defined as the sigmaalgebra generated by the adapted and leftcontinuous (or, just continuous) processes, which is a bit more direct. However, an alternative definition is that it is the sigmaalgebra generated by the stochastic intervals for predictable stopping times . A stochastic interval is a `random interval’ of the real numbers, whose endpoints are random variables. This expresses a subset of ,
Similarly, the optional sigmaalgebra can also be generated by such intervals, where is an arbitrary stopping time, not necessarily predictable. However, optional processes are not used in these notes beyond the definition in the first few posts, so we do not prove that here.
Lemma 7 The predictable sigma algebra is
Proof: If is a sequence of stopping times announcing , then are adapted leftcontinuous and hence predictable processes. So,
So, the collection of sets of the form for predictable stopping times is contained in the predictable sigmaalgebra. Conversely, is generated by the following sets
 for . Letting be the identically zero stopping time, this is .
 for . Letting be the stopping time identically equal to shows that is in the sigmaalgebra generated by .
So, is contained in, and hence equals, .
Finally, it is useful to realize that the set of predictable stopping times is almost the same, regardless of whether it is defined with respect to the filtration or the rightcontinuous filtration . Many approaches to stochastic calculus assume that the filtration satisfies `usual conditions’ and, in particular, is rightcontinuous. However, the results can often be directly carried across to the more general situation simply by applying them to the rightcontinuous filtration .
Lemma 8 A random time is a predictable stopping time if and only if it is a predictable stopping time with respect to the rightcontinuous filtration and .
Proof: Any predictable stopping time is still clearly a predictable stopping time under the larger filtration , so only the converse needs to be proven. However, in this case, Lemma 3 shows that is the first time at which some continuous and adapted process hits zero. Multiplying by if necessary, this process can be assumed to satisfy , which is measurable. Furthermore, by leftcontinuity, will in fact be measurable at each positive time, so it is adapted. The result now follows from Theorem 2.
Accessible and inaccessible times
The flip side of the coin to predictable stopping times are totally inaccessible stopping times. These are times which cannot be predicted with any positive probability. Simple examples include the jump times of a Poisson process.
Definition 9 A stopping time is
 totally inaccessible if for all predictable stopping times .
 accessible if there exists a sequence of predictable stopping times such that
for all .
The graph of a random time is the set , so a stopping time is accessible if its graph is contained in the graphs of a countable set of predictable stopping times. Accessible and totally inaccessible stopping times are always disjoint in the following sense. If is accessible and is totally accessible then . This follows directly from the definition above.
Arbitrary stopping times have a unique decomposition into their accessible and totally inaccessible parts. Actually, the proof below doesn’t make any use if the properties of predictable stopping times, and will still work if the term `predictable stopping time’ is replaced by any other type of stopping time.
Theorem 10 If is a stopping time, then there exists an such that (defined by equation 3) is an accessible stopping time and is a totally inaccessible stopping time.
Furthermore, is uniquely defined up to a set of zero probability.
Proof: First, as is a stopping time, , so it is necessarily true that . Set
If is any sequence and , then the graph of is contained in the union of the graphs of which, in turn, are each contained in the graphs of countable sets of predictable stopping times. As countable unions of countable sets are themselves countable, it follows that is accessible. So, is closed under countable unions. In particular, choosing such that gives . Therefore, there is an with .
For , the next step is to show that is totally inaccessible if and only if . First, if and is a predictable stopping time then is in and . From the definition of , this gives and is totally inaccessible. Conversely, let be totally inaccessible. For any , and, therefore, and, consequently, . So, .
This shows that there is an satisfying , which then satisfies the conclusion of the lemma. Finally, if there is a also satisfying the required properties, then so do and . This gives and differ by a zero probability set.
Thin sets and jump times of a process
Stopping times are often used to control what happens when a process jumps. As we shall see, the set of jump times of a càdlàg adapted process can be covered by a sequence of stopping times and, furthermore, can be decomposed into predictable and totally inaccessible times. To state this in slightly more generality, we start with the definition of thin sets, the standard example being the set of jump times of an adapted process.
Definition 11 A set is said to be a thin set if for a sequence of stopping times .
A process X is said to be thin if it is optional and
is a thin set.
So, countable unions of thin sets are thin. Also, every progressively measurable subset and, in particular, every optional subset of a thin set is itself thin.
Lemma 12 If A is a thin set then any progressively measurable subset of A is thin.
Proof: Suppose that for stopping times and is progressive. Set . As is the value of the progressively measurable process at time , we have . So, are stopping times and
is thin.
For example, if X and Y are thin processes then this shows that the sum is also thin. Being the sum of optional processes, will be optional and, is thin.
As mentioned above, thin sets are often used to represent the set of jump times of a process. For the concept of the jumps of a process X to make sense, it is necessary to apply some constraints on its sample paths. We suppose that is rightcontinuous with left limits everywhere. Such processes are called càdlàg from the French “continu à droite, limites à gauche”. Alternatively, they are sometimes known as rcll processes (rightcontinuous with left limits). The left limit at any time is denoted by , and we set . Then, is a leftcontinuous process. The jumps of X are denoted by .
Proof: First, as are adapted and respectively right and leftcontinuous, they are progressive. So, is progressive. For each define the stopping time
Consider any sample path of X and time t such that . For any positive rational , the existence of left limits along the sample path implies the existence of a positive rational such that for all u in the interval . It can be seen that . So,
is thin.
We shall refer to a thin set A as accessible if every stopping time satisfying is accessible. Similarly, we shall refer to A as totally inaccessible if every such stopping time is totally inaccessible. Then the collection of accessible (respectively, totally inaccessible) thin sets is closed under taking countable unions and progressively measurable subsets. Note that if A is both accessible and totally inaccessible then every stopping time is both accessible and totally inaccessible, so . Therefore, A is evanescent. Similarly, if A is accessible and B is totally inaccessible then is evanescent.
Theorem 10 above generalises to give a decomposition of thin sets into their accessible and totally inaccessible parts.
Theorem 14 Let A be a thin set. Then, up to evanescence, there is a unique decomposition into thin sets , where B is accessible and C is totally inaccessible.
Proof: Write for a sequence of stopping times . By Theorem 10 there are sets such that are accessible stopping times and are totally inaccessible. The required decomposition is given by
Suppose that is any other such decomposition. Then,
However, is evanescent. So, up to evanescence. Combining this with the reverse inclusion gives and up to evanescence.
We finally show that every thin set is contained in the disjoint graphs of a sequence of stopping times, each of which is either predictable or totally inaccessible. This can be useful when looking at the jumps of a process to break the problem down into the separate cases of predictable jump times and totally inaccessible jump times.
Theorem 15 Suppose that the filtration is complete and that is a thin set. Then for a sequence of stopping times satisfying,
 For each n, is either predictable or totally inaccessible.
 For each we have whenever .
Proof: First, by definition, there is a sequence of stopping times such that . For each n define the set

(4) 
As is just the value of the optional process at time we have . So, are stopping times with
Also, by construction, we have whenever and, hence, for each as required. It only remains to be shown that the stopping times can each be chosen to be either predictable or totally inaccessible.
Consider the case where A is accessible. By definition, writing for stopping times , we have that are accessible. So, there is a doublyindexed sequence of predictable times with . Reindexing the sequence gives a sequence of predictable stopping times such that the thin set contains A. Now apply the argument above to . The set defined by (4) is in , since is the value of the predictable process at time . By Lemma 6 (which requires completeness of the filtration), the times are predictable. This proves the result in the case where A is accessible.
Finally, consider any thin set A. By Theorem 14 we can decompose for B accessible and C totally inaccessible. As shown above, the result can be applied to the accessible set B to get predictable stopping times such that and whenever and . Then apply the argument above to the thin set to obtain stopping times with and whenever and .
We have constructed predictable stopping times and stopping times . The graphs () are all disjoint and have union containing A. Furthermore, as are all contained in C, are totally inaccessible times. The result now follows by merging the sequences and together into a single sequence.
Update: I added an extra section to this post. This introduces the notion of thin sets, and shows that the collection of jump times of a cadlag process can be covered by a sequence of stopping times, each of which is predictable or totally inaccessible. These are very simple ideas, but very useful, especially in some of the `general theory’ which I have been writing recently.
Comment by George Lowther — 3 November 11 @ 12:53 AM 
Hi. I like your blog and learnt a lot. I have a question related to this topic. Since the predictable sigma field is contained in the optional sigma field generated by rcll processes, a predictable process is an optional process. My question is: when a rcll process is a predictable process?
Comment by Anonymous — 5 September 12 @ 10:40 PM 
Hi. Actually I wrote out a lemma in a later post answering precisely this question (link).
Comment by George Lowther — 6 September 12 @ 2:36 AM 
Hi, I’m a little confused how you got the second inequality in Lemma 6. It seems to me that the event in the second line implies the event in the third (giving an inequality in the opposite direction), but hopefully I’m mistaken here.
Comment by Dominic — 10 December 12 @ 3:39 AM 