The next Lectures will be devoted to the study of the problem of the existence of a density for solutions of stochastic differential equations. The basic tool to study such questions is the so-called Malliavin calculus.
Let us consider a filtered probability space on which is defined a Brownian motion . We assume that is the usual completion of the natural filtration of .
A measurable real valued random variable is said to be cylindric if it can be written
where and is a function such that and all its partial derivatives have polynomial growth. The set of cylindric random variables is denoted by . It is easy to see that is dense in for every .
The Malliavin derivative of is the valued stochastic process given by
We can see as an (unbounded) operator from the space into the Banach space
Our first task will be to prove that is closable. This will be a consequence of the following fundamental integration by parts formula which is interesting in itself.
Proposition. (Integration by parts formula) Let and be a progressively measurable such that . We have
Let us now fix and denote
From Girsanov’s theorem, we have
Now, on one hand we compute
and on the other hand, we obtain
Proposition. Let . As a densely defined operator from into , is closable.
Proof. Let be a sequence in that converges in to and such that converges in to . We want to prove that . Let be a function in . Let us first assume . We have
As a consequence, we obtain
Since is arbitrary, we conclude . Let us now assume . Let be a smooth and compactly supported function and let . We have
As a consequence, we get
On the other hand, we have
The closure of in shall still be denoted by . Its domain is the closure of with respect to the norm
For , we can consider the adjoint operator of . This is a densely defined operator with which is characterized by the duality formula
From the integration by parts formula and Burkholder-Davis-Gundy inequalities, it is clear that the domain of in contains the set of progressively measurable processes such that and in that case, The operator can thus be thought as an extension of the Itō’s integral. It is often called the Skohorod integral.
Show that for ,