Lecture 33. The Malliavin matrix and existence of densities

More generally, by using the same methods as in the previous Lecture, we can introduce iterated derivatives. If F \in \mathcal{S}, we set
\mathbf{D}^k_{t_1,...,t_k} F = \mathbf{D}_{t_1} ...\mathbf{D}_{t_k} F.
We may then consider \mathbf{D}^k F as a square integrable random process indexed by [0,1]^{k} and valued in \mathbb{R}^n. By using the integration by parts formula, it is possible to prove, as we did it in the previous Lecture, that for any p \geq 1, the operator \mathbf{D}^k is closable on \mathcal{S}. We denote by \mathbb{D}^{k,p} the domain of \mathbf{D}^k in L^p, it is the closure of the class of cylindric random variables with respect to the norm
\left\| F\right\| _{k,p}=\left( \mathbb{E}\left( F^{p}\right) +\sum_{j=1}^k \mathbb{E}\left( \left\| \mathbf{D}^j F\right\|_{\mathbf{L}^2 ([0,1]^j, \mathbb{R}^n)}^{p}\right) \right)^{\frac{1}{p}},
and
\mathbb{D}^{\infty}=\bigcap_{p \geq 1} \bigcap_{k \geq 1} \mathbb{D}^{k,p}.
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 F=(F_1,...,F_m) be a \mathcal{F}_1 measurable random vector such that:

  • For every i=1,...,m, F_i \in \mathbb{D}^{\infty};
  • The matrix
    \Gamma= \left( \int_0^1 \langle \mathbf{D}_s F^i , \mathbf{D}_s F^j \rangle_{\mathbb{R}^n} ds \right)_{1 \leq i,j \leq m}
  • is invertible.

Then F has a density with respect to the Lebesgue measure. If moreover, for every p > 1,
\mathbb{E} \left( \frac{1}{\mid \det \Gamma \mid ^p} \right) < \infty,
then this density is C^\infty.

The matrix \Gamma is often called the Malliavin matrix of the random vector F.

This theorem relies on the following lemma of Fourier analysis for which we shall use the following notation: If \phi: \mathbb{R}^n \rightarrow \mathbb{R} is a smooth function then for \alpha =(i_1,...,i_k) \in  \{1,...,n\}^k, we denote
\partial_\alpha \phi =\frac{\partial^k}{\partial x_{i_1} \cdots \partial x_{i_k} } \phi.
Lemma. Let \mu be a probability measure on \mathbb{R}^n such that for every smooth and compactly supported function \phi :\mathbb{R}^n \rightarrow \mathbb{R},
\left| \int_{\mathbb{R}^n} \partial_\alpha \phi d\mu \right| \le C_\alpha \| \phi \|_\infty,
where \alpha  \in  \{1,...,n\}^k, k \ge 1, C_\alpha > 0. Then \mu is absolutely continuous with respect to the Lebesgue measure with a smooth density.

Proof. The idea is to show that we may assume that \mu is compactly supported and then use Fourier transforms techniques. Let x_0 \in \mathbb{R}^n, R > 0 and R' > R. Let \Psi be a smooth function on \mathbb{R}^n such that \Psi =1 on the ball \mathbf{B} (x_0,R) and \Psi=0 outside the ball \mathbf{B} (x_0,R'). Let \nu be the measure on \mathbb{R}^n that has a density \Psi with respect to \mu. It is easily seen, by induction and integrating by parts that for every smooth and compactly supported function \phi :\mathbb{R}^n \rightarrow \mathbb{R},
\left| \int_{\mathbb{R}^n} \partial_\alpha \phi d\nu \right| \le C'_\alpha \| \phi \|_\infty,
where \alpha  \in  \{1,...,n\}^k, k \ge 1, C'_\alpha > 0. Now, if we can prove that under the above assumption \nu has a smooth density, then we will able to conclude that \phi has a smooth density because x_0 \in \mathbb{R}^n and R,R' are arbitrary. Let
\hat{\nu}(y) =\int_{\mathbb{R}^n} e^{i \langle y,x \rangle} \nu (dx)
be the Fourier transform of the measure \mu. The assumption implies that \hat{\nu} is rapidly decreasing (apply the inequality with \phi(x)=e^{i \langle y,x \rangle}). We conclude that \nu has a smooth density with respect to the Lebesgue measure and that this density f is given by the inverse Fourier transform formula:
f(x)=\frac{1}{(2\pi)^n} \int_{\mathbb{R}^n} e^{-i \langle y,x \rangle} \hat{\nu} (y) dy \square

We may now turn to the proof of the Theorem.

The proof relies on the integration by parts formula for the Malliavin derivative. Let \phi be a smooth and compactly supported function on \mathbb{R}^n. Since F_i \in \mathbb{D}^\infty, we easily deduce that \phi(F) \in \mathbb{D}^\infty and that
\mathbf{D} \phi (F) =\sum_{i=1}^n \partial_i \phi (F) \mathbf{D} F_i.
Therefore we obtain
\int_0^1 \langle  \mathbf{D}_t \phi (F),\mathbf{D}_t F_j \rangle dt= \sum_{i=1}^n \partial_i \phi (F)  \int_0^1 \langle \mathbf{D}_t F_i, \mathbf{D}_t F_j \rangle dt.
We conclude that
\partial_i \phi (F)=\sum_{j=1}^n (\Gamma^{-1})_{i,j}  \int_0^1 \langle  \mathbf{D}_t \phi (F),\mathbf{D}_t F_j \rangle dt .
As a consequence, we obtain
\mathbb{E} \left(\partial_i \phi (F) \right)  = \mathbb{E} \left(\sum_{j=1}^n (\Gamma^{-1})_{i,j}  \int_0^1 \langle  \mathbf{D}_t \phi (F),\mathbf{D}_t F_j \rangle dt \right)
=\sum_{j=1}^n \mathbb{E} \left(  \int_0^1 \langle  \mathbf{D}_t \phi (F), (\Gamma^{-1})_{i,j}\mathbf{D}_t F_j \rangle dt   \right)
=\sum_{j=1}^n \mathbb{E} \left( \phi (F) \delta ( (\Gamma^{-1})_{i,j}\mathbf{D} F_j ) \right)
= \mathbb{E} \left( \phi (F) \delta \left( \sum_{j=1}^n (\Gamma^{-1})_{i,j}\mathbf{D} F_j \right) \right)
By using inductively this integration by parts formula, it is seen that for every \alpha  \in  \{1,...,n\}^k, k \ge 1, there exists an integrable random variable Z_\alpha such that,
\mathbb{E} \left( \partial_\alpha \phi (F)\right)=\mathbb{E} \left( \phi (F) Z_\alpha \right).

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