More generally, by using the same methods as in the previous Lecture, we can introduce iterated derivatives. If , we set
We may then consider as a square integrable random process indexed by and valued in . By using the integration by parts formula, it is possible to prove, as we did it in the previous Lecture, that for any , the operator is closable on . We denote by the domain of in , it is the closure of the class of cylindric random variables with respect to the norm
We have the following key result which makes Malliavin calculus so useful when one wants to study the existence of densities for random variables.
Theorem.(P. Malliavin) Let be a measurable random vector such that:
- For every , ;
- The matrix
Then has a density with respect to the Lebesgue measure. If moreover, for every ,
then this density is .
The matrix is often called the Malliavin matrix of the random vector .
This theorem relies on the following lemma of Fourier analysis for which we shall use the following notation: If is a smooth function then for , we denote
Lemma. Let be a probability measure on such that for every smooth and compactly supported function ,
where , , . Then is absolutely continuous with respect to the Lebesgue measure with a smooth density.
Proof. The idea is to show that we may assume that is compactly supported and then use Fourier transforms techniques. Let , and . Let be a smooth function on such that on the ball and outside the ball . Let be the measure on that has a density with respect to . It is easily seen, by induction and integrating by parts that for every smooth and compactly supported function ,
where , , . Now, if we can prove that under the above assumption has a smooth density, then we will able to conclude that has a smooth density because and are arbitrary. Let
be the Fourier transform of the measure . The assumption implies that is rapidly decreasing (apply the inequality with ). We conclude that has a smooth density with respect to the Lebesgue measure and that this density is given by the inverse Fourier transform formula:
We may now turn to the proof of the Theorem.
The proof relies on the integration by parts formula for the Malliavin derivative. Let be a smooth and compactly supported function on . Since , we easily deduce that and that
Therefore we obtain
We conclude that
As a consequence, we obtain
By using inductively this integration by parts formula, it is seen that for every , , there exists an integrable random variable such that,