As was mentioned in the initial post of these stochastic calculus notes, it is important to choose good versions of stochastic processes. In some cases, such as with Brownian motion, it is possible to explicitly construct the process to be continuous. However, in many more cases, it is necessary to appeal to more general results to assure the existence of such modifications.
The theorem below guarantees that many of the processes studied in stochastic calculus have a right-continuous version and, furthermore, these versions necessarily have left limits everywhere. Such processes are known as càdlàg from the French for “continu à droite, limites à gauche” (I often drop the accents, as seems common). Alternative terms used to refer to a cadlag process are rcll (right-continuous with left limits), R-process and right process. For a cadlag process , the left limit at any time is denoted by (and ). The jump at time is denoted by .
We work with respect to a complete filtered probability space .
Theorem 1 below provides us with cadlag versions under the condition that elementary integrals of the processes cannot, in a sense, get too large. Recall that elementary predictable processes are of the form
for times , -measurable random variable and -measurable random variables . Its integral with respect to a stochastic process is
An elementary predictable set is a subset of which is a finite union of sets of the form for and for nonnegative reals and . Then, a process is an indicator function of some elementary predictable set if and only if it is elementary predictable and takes values in .
The following theorem guarantees the existence of cadlag versions for many types of processes. The first statement applies in particular to martingales, submartingales and supermartingales, whereas the second statement is important for the study of general semimartingales.
- X is integrable and, for every ,
- For every the set
is bounded in probability.
This existence of cadlag versions stated by this theorem is a special case of the following slightly more general result, which drops the requirement for the process to be right-continuous in probability. In the following, the modification has a right limit at every point. If the process is right-continuous in probability, then at each time, with probability one. By countable additivity, this remains true simultaneously at all times in any given countable set and therefore is a cadlag version.
Theorem 2 Let X be an adapted stochastic process, and suppose that either of the two conditions of Theorem 1 holds. Then, it has a version Y which has left and right limits everywhere and such that there is a countable subset for which is right-continuous at every .
A proof of this theorem is given below, using the ideas on upcrossings of a process as discussed in the previous post.
The main result for existence of cadlag martingales is as follows.
Proof: By applying the statement to , it suffices to prove the result for submartingales. However, in this case, for any elementary predictable set ,
The first condition of Theorem 1 is satisfied, showing that cadlag versions exist.
Even if the condition of right-continuity in probability is dropped, then it is still possible to pass to well-behaved modifications. Although a martingale can fail to have a cadlag version, it is always the case that there exists a version which is cadlag everywhere outside of a fixed countable set of times.
Theorem 4 Let X be a martingale, submartingale or supermartingale. Then, it has a version Y which has left and right limits everywhere and such that there is a countable subset for which is right-continuous at every .
Often, the underlying filtrations used are assumed to satisfy the usual conditions. That is, they are required to be right-continuous as well as being complete. In this case the existence of cadlag versions is particularly general. Every martingale has a cadlag version.
More generally, a submartingale or supermartingale X has a cadlag version if and only if is right-continuous.
Proof: By applying the result to , we may suppose that the process is a submartingale. Furthermore, for martingales the function , being constant, is trivially right-continuous. So, it suffices to prove the second more general statement. Theorem 4 gives a version Y which has left and right limits everywhere and is cadlag outside of some countable set . It only remains to be shown that , almost surely, for each .
Choose a sequence strictly decreasing to . Then , by the submartingale property. The idea is to commute the limit as with the conditional expectation to obtain
This can be done under the condition that the sequence is uniformly integrable. For a martingale, these are all conditional expectations , so uniform integrability is guaranteed. In fact, Lemma 6 states that this sequence is uniformly integrable whenever is a submartingale, so inequality (2) holds.
Similarly, using uniform integrability together with the right-continuity of gives
so that is a nonnegative random variable with zero expectation. This shows that almost surely. Finally, the right-continuity of the filtration, , is applied. As is necessarily -measurable, .
The proof above made use of the following simple, but useful, statement regarding uniform integrability of submartingales.
Then, is uniformly integrable for any decreasing sequence bounded below in .
Proof: The idea is to apply a `Doob-style decomposition’ of the submartingale into a martingale and an increasing process at the sequence of times. Let be a lower bound for and set
which, by the submartingale property, is a sum of nonnegative terms. Furthermore, by monotone convergence
so are integrable random variables, and is decreasing in . Furthermore, from the definition, the martingale property
is satisfied, so is a uniform integrable sequence. Finally, are uniformly integrable, and the result follows for .
Proof of Cadlag Versions
I now give a proof of Theorem 2. This will make use of the ideas from previous post on upcrossings and downcrossings to show that, when restricted to a countable set of times, has left and right limits everywhere. The result will follow from this.
For any , let be a finite subset of . The number of upcrossings of an interval for times in satisfies the bound
for some elementary set . Furthermore, letting be the first time in at which , the stochastic interval is elementary and,
Applying the same idea to shows that there are elementary predictable sets such that
Letting increase to a countably infinite subset of and applying monotone convergence, these inequalities generalize to all countable subsets . In particular, the number of upcrossings of and the supremum of are almost surely bounded on .
Again, letting increase to a countably infinite subset of and applying monotone convergence extends these inequalities to all countable subsets . Letting go to infinity then shows that the number of upcrossings of and the supremum of are bounded on .
In either case, applying countable additivity, the above shows that for time restricted to a countable subset the process , with probability one, is bounded and has finitely many upcrossings of for all rational on bounded time intervals. Replacing by the identically zero process outside this set of probability one, it can be assumed that this holds everywhere.
By the results of the previous post, this implies that for time restricted to such a countable set , almost surely has left and right limits everywhere. Assuming that is also dense in (e.g., ), this defines a cadlag process
for all , and where is restricted to in this limit.
Note that enlarging by any countable set will not change the process (up to a set of probability zero), by the arbitrariness of sequences in (5).
It remains to be shown that outside of some countable set of times. In fact, for any positive integer , the set of times at which is finite. If not, there would be an increasing or decreasing sequence of such times and, enlarging to include this sequence, would converge to zero, giving the contradiction as . Consequently, letting , there are only countably many times at which . Without loss of generality, we may suppose that includes all such times.
Finally, the process is defined by
As with time restricted to the set has left and right limits everywhere, it follows that also has left and right limits everywhere. Furthermore, is cadlag outside of the countable set .