It turns out that stochastic integrals may be defined for other stochastic processes than Brownian motions. The key properties that were used in the above approach were the martingale property and the square integrability of the Brownian motion.
As above, we consider a filtered probability space that satisfies the usual conditions. A martingale defined on this space is said to be square integrable if for every , .
For instance, if is a Brownian motion on and if is a process which is progressively measurable with respect to the filtration such that for every , then, the process is a square integrable martingale.
The most important theorem concerning continuous square integrable martingales is that they admit a quadratic variation. Before proving this theorem, we state a preliminary lemma.
Lemma. Let be a continuous martingale such that
Then is constant.
We may assume . For , let us consider the stopping time
The stopped process is a martingale and therefore for ,
Consider now a sequence of subdivisions whose mesh tends to 0. By summing up the above inequality on the subdivision, we obtain
By letting , we get . This implies . Letting now , we conclude .
Theorem. Let be a martingale on which is continuous and square integrable and such that .There is a unique continuous and increasing process denoted that satisfies the following properties:
- The process is a martingale.
Actually for every and for every sequence of subdivisions such that
the following convergence takes place in probability:
The process is called the quadratic variation process of .
We first assume that the martingale is bounded and prove that if is a sequence of subdivisions of the interval such that
then the limit
exists in and thus in probability.
Toward this goal, we introduce some notations. If is a subdivision of the time interval and if is a stochastic process, then we denote
where is such that ,
An easy computation on conditional expectations shows that if is a martingale, then the process is also a martingale. Also, if and are two subdivisions of the time interval , we will denote by the subdivision obtained by putting together the points and the points of . Let now be a sequence of subdivisions of such that
Let us show that the sequence is a Cauchy sequence in . Since the process is a martingale (as a difference of two martingales), we deduce that
Let us denote by ‘s the points of the subdivision and for fixed , we denote by the point of which is the closest to and such that . We have
Therefore, from Cauchy-Schwarz inequality,
Since the martingale is assumed to be continuous, when ,
Thus, in order to conclude, it suffices to prove that is bounded. This fact is an easy consequence of the fact that is assumed to be bounded. Therefore, in the sense the following convergence holds
The process is seen to be a martingale because for every and , the process is a martingale. Let us now show that the obtained process is a continuous process. From Doob’s inequality, for and ,
From Borel-Cantelli lemma, there exists therefore a sequence such that the sequence of continuous stochastic processes almost surely uniformly converges to the process . This proves the existence of a continuous version for . Finally, to prove that is increasing, it is enough to consider a an increasing sequence of subdivisions whose mesh tends to . Let us now prove that is the unique process such that is a martingale. Let and be two continuous and increasing stochastic processes such that and such that and are martingales. The process is then seen to be a martingale that has a bounded variation. From the previous lemma, this implies that is constant and therefore equal to 0 due to its initial condition.
We now turn to the case where is not necessarily bounded. Let us introduce the sequence of stopping times:
According to the previous arguments, for every , there is an increasing process such that is a martingale. By uniqueness of this process, it is clear that , therefore we can define a process by requiring that provided that . By using convergence theorems, it is then checked that is a martingale.
Finally, let be a sequence of subdivisions whose mesh tends to . We have for every ,
This easily implies the announced convergence in probability of the quadratic variations to
Exercise. Let be a square integrable martingale on a filtered probability space . Assume that . If is a subdivision of the time interval and if is a stochastic process, we denote
where is such that . Let be a sequence of subdivisions of such that
Show that the following convergence holds in probability,
Thus, in the previous theorem, the convergence is actually uniform on compact intervals.
We have already pointed out that stochastic integrals with respect to Brownian motion provide an example of square integrable martingale, they therefore have a quadratic variation. The next proposition explicitly computes this variation.
Proposition. Let be a Brownian motion on a filtered probability space that satisfies the usual conditions. Let be a progressively measurable process such that for every , . For :
Since the process is continuous, increasing and equals when , we just need to prove that
is a martingale.
If , is a simple process, it is easily seen that for :
We may then conclude by using the density of in
As a straightforward corollary of the existence of a quadratic variation for the square integrable martingales, we immediately obtain:
Corollary. Let and be two continuous square integrable martingales on such that . There is a unique continuous process with bounded variation that satisfies:
- The process is a martingale.
Moreover, for and for every sequence such that , the following convergence holds in probability:
The process is called the quadratic covariation process of and .
We may actually just use the formula
as a definition of the covariation and then check that the above properties are indeed satisfied
Exercise. Let and be two independent Brownian motions. Show that .