Lecture 32. The Malliavin derivative

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 (\Omega, (\mathcal{F}_t)_{0 \le t \le 1}, \mathbb{P}) on which is defined a Brownian motion (B_t)_{0 \le t \le 1}. We assume that (\mathcal{F}_t)_{0 \le t \le 1} is the usual completion of the natural filtration of (B_t)_{0 \le t \le 1}.

A \mathcal{F}_{1} measurable real valued random variable F is said to be cylindric if it can be written
F=f \left( \int_0^{1} h^1(s) dB_s,...,\int_0^{1} h^m(s) dB_s \right)
where h^i \in \mathbf{L}^2 ([0,1], \mathbb{R}^n) and f:\mathbb{R}^m \rightarrow \mathbb{R} is a C^{\infty} function such that f and all its partial derivatives have polynomial growth. The set of cylindric random variables is denoted by \mathcal{S}. It is easy to see that \mathcal{S} is dense in L^p for every p \ge 1.

The Malliavin derivative of F \in \mathcal{S} is the \mathbb{R}^n valued stochastic process (\mathbf{D}_t F )_{0 \leq t \leq 1} given by
\mathbf{D}_t F=\sum_{i=1}^{m} h^i (t) \frac{\partial f}{\partial x_i} \left( \int_0^{1} h^1(s) dB_s,...,\int_0^{1} h^m(s)dB_s \right).
We can see \mathbf{D} as an (unbounded) operator from the space \mathcal{S} \subset L^p into the Banach space
\mathcal{L}^p=\left\{ (X_t)_{0 \le t \le 1},\mathbb{E}\left( \left(  \int_0^1 \| X_t \|^2 dt\right)^p \right) < +\infty \right\}.
Our first task will be to prove that \mathbf{D} 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 F \in \mathcal{S} and (h(s))_{0 \le s \le 1} be a progressively measurable such that \mathbb{E}\left( \int_0^1 \| h(s)\|^2 ds \right) < +\infty. We have
\mathbb{E} \left( \int_0^1( \mathbf{D}_s  F)h(s) ds \right)=\mathbb{E}\left( F   \int_0^{1} h(s)dB_s\right).

Proof.
Let
F=f \left( \int_0^{1} h^1(s) dB_s,...,\int_0^{1} h^m(s) dB_s \right) \in \mathcal{S}.
Let us now fix \varepsilon  \ge 0 and denote
F_\varepsilon =f \left( \int_0^{1} h^1(s) d\left( B_s +\varepsilon  \int_0^{s} h(u)du \right),...,\int_0^{1} h^m(s) d\left( B_s +\varepsilon  \int_0^{s} h(u)du \right) \right).
From Girsanov’s theorem, we have
\mathbb{E} ( F_\varepsilon)=\mathbb{E} \left(\exp \left(\varepsilon \int_0^{1} h(u)dB_u  -\frac{\varepsilon^2}{2}\int_0^{1} \|h(u)\|^2du   \right) F \right).
Now, on one hand we compute
\lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \left( \mathbb{E} ( F_\varepsilon)-\mathbb{E} (F) \right) =\mathbb{E} \left( \int_0^1\sum_{i=1}^{m}  \frac{\partial f}{\partial x_i} \left( \int_0^{1} h^1(s)dB_s,...,\int_0^{1} h^m(s) dB_s \right) h^i(s)h(s) dt \right)
=\mathbb{E} \left( \int_0^1( \mathbf{D}_s  F)h(s) dt \right),
and on the other hand, we obtain
\lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \left( \mathbb{E} ( F_\varepsilon)-\mathbb{E} (F) \right)=\mathbb{E}\left( F   \int_0^{1} h(s)dB_s\right)
\square

Proposition. Let p \ge 1. As a densely defined operator from L^p into \mathcal{L}^p, \mathbf{D} is closable.

Proof. Let (F_n)_{n \in \mathbb{N}} be a sequence in \mathcal{S} that converges in L^p(\mathcal{B}_{1}) to 0 and such that \mathbf{D}F_n converges in \mathcal{L}^p to X. We want to prove that X=0. Let (h(s))_{0 \le s \le 1} be a function in L^2([0,1]). Let us first assume p > 1. We have
\lim_{n \to \infty} \mathbb{E} \left( \int_0^1( \mathbf{D}_sF  _n)h(s) ds \right)=\mathbb{E} \left( \int_0^1 X_s h(s) ds \right),
and
\lim_{n \to \infty}\mathbb{E}\left( F_n   \int_0^{1} h(s)dB_s\right)=0.
As a consequence, we obtain
\mathbb{E} \left( \int_0^1 X_s h(s) ds \right)=0.
Since h is arbitrary, we conclude X=0. Let us now assume p=1. Let \eta be a smooth and compactly supported function and let \Theta=\eta \left( \int_0^{1} h(s)dB_s \right) \in \mathcal{S}. We have
\mathbf{D} (F_n \Theta)=F_n (\mathbf{D} \Theta )+( \mathbf{D}F_n) \Theta.
As a consequence, we get
\mathbb{E}\left(\int_0^1  \mathbf{D}_s (F_n \Theta) h(s)ds \right) =\mathbb{E}\left(F_n \int_0^1 (\mathbf{D}_s \Theta ) h(s)ds \right) +\mathbb{E}\left( \Theta \int_0^1 ( \mathbf{D}_sF_n) h(s)ds \right),
and thus
\lim_{n \to \infty} \mathbb{E} \left( \int_0^1 \mathbf{D}_s(F  _n \Theta) h(s) ds \right)=\mathbb{E} \left( \Theta \int_0^1 X_s h(s) ds \right).
On the other hand, we have
\mathbb{E} \left( \int_0^1 \mathbf{D}_s(F  _n \Theta) h(s) ds \right)=\mathbb{E} \left( F  _n \Theta \int_0^1 h(s) dB_s\right) \to_{n \to \infty} 0.
We conclude
\mathbb{E} \left( \Theta \int_0^1 X_s h(s) ds \right)=0 \square

The closure of \mathbf{D} in L^p shall still be denoted by \mathbf{D}. Its domain \mathbb{D}^{1,p} is the closure of \mathcal{S} with respect to the norm
\left\| F\right\| _{1,p}=\left( \mathbb{E}\left( F^{p}\right) + \mathbb{E}\left( \left\| \mathbf{D} F\right\|_{\mathbf{L}^2 ([0,1], \mathbb{R}^n)}^{p}\right) \right)^{\frac{1}{p}},
For p > 1, we can consider the adjoint operator \delta of \mathbf{D}. This is a densely defined operator \mathcal{L}^q \to L^q(\mathcal{B}_{1}) with 1/p+1/q=1 which is characterized by the duality formula
\mathbb{E} (F \delta u)=\mathbb{E} \left(\int_0^1 (\mathbf{D}_s F) u_s ds \right) , \quad F \in \mathbb{D}^{1,p}.
From the integration by parts formula and Burkholder-Davis-Gundy inequalities, it is clear that the domain of \delta in \mathcal{L}^q contains the set of progressively measurable processes (u_t)_{0 \le t \le 1} such that \mathbb{E} \left(\left( \int_0^1 \| u_s \|^2ds\right)^{q/2} \right) < + \infty and in that case, \delta u =\int_0^1 u_s dB_s. The operator \delta can thus be thought as an extension of the Itō’s integral. It is often called the Skohorod integral.

Exercise.(Clark-Ocone formula)
Show that for F \in \mathbb{D}^{1,2},
F=\mathbb{E}(F)+\int_0^1 \mathbb{E} \left( \mathbf{D}_1F \mid \mathcal{F}_t \right)dB_t.

This entry was posted in Stochastic Calculus lectures. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s