## 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 $F \in \mathbb{D}^{1,2}$,
$F=\mathbb{E}(F)+\int_0^1 \mathbb{E} \left( \mathbf{D}_1F \mid \mathcal{F}_t \right)dB_t.$

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