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 left-continuity, 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 Borel-Cantelli 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 Borel-Cantelli 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.
Lemma 4 If
are predictable stopping times, then so are
.
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 sigma-algebra. In these notes, it was defined as the sigma-algebra generated by the adapted and left-continuous (or, just continuous) processes, which is a bit more direct. However, an alternative definition is that it is the sigma-algebra 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 sigma-algebra 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 left-continuous and hence predictable processes. So,
So, the collection of sets of the form
for predictable stopping times
is contained in the predictable sigma-algebra. 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 sigma-algebra 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 right-continuous filtration
. Many approaches to stochastic calculus assume that the filtration satisfies `usual conditions’ and, in particular, is right-continuous. However, the results can often be directly carried across to the more general situation simply by applying them to the right-continuous 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 right-continuous 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 left-continuity,
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 inaccessible 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 right-continuous 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 (right-continuous with left limits). The left limit at any time
is denoted by
, and we set
. Then,
is a left-continuous process. The jumps of X are denoted by
.
Lemma 13 If X is a càdlàg adapted process then
is thin.
Proof: First, as are adapted and respectively right and left-continuous, 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 doubly-indexed 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.
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?
Hi. Actually I wrote out a lemma in a later post answering precisely this question (link).
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.
Thanks for the blog. Very helpful. Typo: In “If
is accessible and
is totally accessible then
, the 2nd "accessible" should be "inaccessible".
Fixed. Thanks