Research and Lecture notes

MA694 Rough paths theory

Posted on December 13, 2012 by Fabrice Baudoin

During the Spring 2013 semester I will teach a class on rough paths theory and post the lectures on this blog.

The rough paths theory was discovered by Terry Lyons in the 1990’s. The theory allows to solve differential equations driven by rough signals. The theory is deterministic but perfectly applies to the study of differential equations driven by rough random signals as the Brownian motion or even potentially rougher signals such as the fractional Brownian motion. The main reference for the course will be the book:

P. Friz, N. Victoir: Multidimensional stochastic processes as rough paths. Theory and Applications. Cambridge Studies in Advanced Mathematics (CUP, 2009)

A rough table of contents is as follows:

Ordinary differential equations
Young’s integration theory
Estimating iterated integrals
p-rough paths
Rough linear differential equations
Carnot groups
Geometric rough paths
Rough differential equations
Stochastic processes as rough paths

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Lecture 35. Weak differentiability for solutions of stochastic differential equations and the existence of a smooth density

Posted on November 8, 2012 by Fabrice Baudoin

As usual, we consider a filtered probability space $\left( \Omega , (\mathcal{F}_t)_{t \geq 0} , \mathcal{F},\mathbb{P} \right)$ which satisfies the usual conditions and on which is defined a $n$ -dimensional Brownian motion $(B_t)_{t \ge 0}$ . Our purpose here, is to prove that solutions of stochastic differential equations are differentiable in the sense of Malliavin.
The following lemma is easy to prove by using the Wiener chaos expansion.

Lemma. Let $(u_s)_{0 \le s \le 1}$ be a progressively measurable process such that for every $0 \le s \le 1$ , $u^i_s \in \mathbb{D}^{1,2}$ and
$\mathbb{E} \left(\int_0^1 \|u_s\|^2 ds \right)<+\infty, \quad \mathbb{E} \left(\int_0^1 \int_0^1 \|\mathbf{D}_s u_t\|^2 ds dt\right)<+\infty.$
Then $\int_0^1 u_s dB_s \in \mathbb{D}^{1,2}$ and
$\mathbf{D}_t\left( \int_0^1 u_s dB_s\right)=u_t+\sum_{i=1}^n \int_t^1 (\mathbf{D}_t u^i_s) dB^i_s.$

Proof. We make the proof when $n=1$ and use the notations introduced in the Wiener chaos expansion Lecture. For $f \in L^2([0,1])$ , we have
$\mathbf{D}_t I_n( f^{\otimes n } )= f(t) I_{n-1} (f^{\otimes {(n-1)}}).$
But we can write,
$I_n( f^{\otimes n } )=\int_0^1f(t) \left( \int_{\Delta_{n-1}[0,t]} f^{\otimes {(n-1)}} dB_{t_1} \cdots dB_{t_{n-1}}\right) dB_t,$
and thus
$I_n( f^{\otimes n } )=\int_0^1 u_s dB_s,$
with $u_t= f(t)\int_{\Delta_{n-1}[0,t]} f^{\otimes {(n-1)}} dB_{t_1} \cdots dB_{t_{n-1}}$ . Since
$f(t) I_{n-1} (f^{\otimes {(n-1)}})= f(t) \left( \int_{\Delta_{n-1}[0,t]} f^{\otimes {(n-1)}} dB_{t_1} \cdots dB_{t_{n-1}}\right)$
$+ f(t) \int_t^1 f(s)\left( \int_{\Delta_{n-2}[0,s]} f^{\otimes {(n-1)}} dB_{t_1} \cdots dB_{t_{n-2}}\right)dB_s$ ,
we get the result when $\int_0^1 u_s dB_s$ can be written as $I_n( f^{\otimes n } )$ . By continuity of the Malliavin derivative on the space of chaos of order $n$ , we conclude that the formula is true if $\int_0^1 u_s dB_s$ is a chaos of order $n$ . The result finally holds in all generality by using the Wiener chaos expansion $\square$

We consider two functions $b : \mathbb{R}^n \to \mathbb{R}^n$ and $\sigma:\mathbb{R}^n \to \mathbb{R}^{n \times n}$ and we assume that $b$ and $\sigma$ are $C^\infty$ with derivatives at any order (more than 1) bounded.

As we know, there exists a bicontinuous process $(X_t^{x})_{t\ge 0, x \in \mathbb{R}^d}$ such that for $t \ge 0$ ,
$X_t^{x} =x +\int_0^t b(X_s^{x}) ds +\sum_{k=1}^n \int_0^t \sigma_k(X_s^{x}) dB^k_s.$
Moreover, for every $p \ge 1$ , and $T \ge 0$
$\mathbb{E} \left( \sup_{0 \le t \le T} \| X^x_t \|^p\right) < +\infty.$

Theorem. For every $i=1,...,n$ , $0 \le t \le 1$ , $X_t^{x,i} \in \mathbb{D}^{\infty}$ and for $r \le t$ ,
$\mathbf{D}^j_r X_t^{x,i}= \sigma_{i,j}(X_r^{x}) +\sum_{l=1}^n \int_r^t \partial_l b_i(X_s^{x})\mathbf{D}^j_r X_s^{x,l} ds +\sum_{k,l=1}^n \int_r^t \partial_l\sigma_{i,k}(X_s^{x})\mathbf{D}^j_r X_s^{x,l} dB^k_s,$
where $\mathbf{D}^j_r X^i_t$ is the $j$ -th component of $\mathbf{D}_r X^i_t$ . If $r > t$ , then $\mathbf{D}^j_r X_t^{x,i}=0$ .

Proof. We first prove that $X_1^{x,i} \in \mathbb{D}^{1,p}$ for every $p \ge 1$ . We consider the Picard approximations given by $X_0(t)=x$ and
$X_{n+1}(t) =x +\int_0^t b(X_n(s)) ds +\sum_{k=1}^n \int_0^t \sigma_k(X_n(s)) dB^k_s.$
By induction, it is easy to see that $X_n(t) \in \mathbb{D}^{1,p}$ and that for every $p \ge 1$ , we have
$\Psi_n(t)=\sup_{0 \le r \le t} \mathbb{E} \left( \sup_{s \in [r,t]} \| \mathbf{D}_r X_n(s) \|^p \right)< +\infty,$
and
$\Psi_{n+1}(t)\le \alpha +\beta\int_0^t \Psi_n(s)ds.$
Then, we observe that $X_n(t)$ converges to $X_t^x$ in $L^p$ and that the sequence $\| X_n(t) \|_{1,p}$ is bounded. As a consequence $X_1^{x,i} \in \mathbb{D}^{1,p}$ for every $p \ge 1$ . The equation for the Malliavin derivative is obtained by differentiating the equation satisfied by $X_t^x$ . Higher order derivatives may be treated in a similar way with a few additional work $\square$

Combining this theorem with the uniqueness property for solutions of linear stochastic differential equations, we obtain the following representation for the Malliavin derivative of a solution of a stochastic differential equation:

Corollary:
$\mathbf{D}^j_r X^x_t=\mathbf{J}_{0 \rightarrow t}(x) \mathbf{J}_{0 \rightarrow r}^{-1}(x) \sigma_j (X^x_r),~~j=1,...,n, ~~ 0\leq r \leq t,$

where $(\mathbf{J}_{0 \rightarrow t}(x))_{ t \geq 0}$ is the first variation process defined by
$\mathbf{J}_{0 \rightarrow t}(x)=\frac{\partial X^x_t}{\partial x}(x).$

We now fix $x \in \mathbb{R}^n$ as the initial condition for our equation and denote by $\Gamma_t=\left( \sum_{j=1}^n \int_0^1 \mathbf{D}_r^j X_t^{i,x}\mathbf{D}_r^j X^{i',x}_t dr\right)_{1 \le i,i' \le n}$ the Malliavin matrix of $X^x_t$ . From the previous corollary, we deduce that
$\Gamma_t(x)=\mathbf{J}_{0 \rightarrow t}(x) \int_0^t \mathbf{J}_{0 \rightarrow r}^{-1}(x) \sigma (X^x_r) \sigma (X^x_r)^* \mathbf{J}_{0 \rightarrow r}^{-1}(x)^* dr \mathbf{J}_{0 \rightarrow t}(x)^*.$

We are now finally in position to state the main theorem of the section:

Theorem. Assume that there exists $\lambda > 0$ such that for every $x \in \mathbb{R}^n$ ,
$\| \sigma (x) \|^2 \ge \lambda \| x \|^2,$
then for every $t > 0$ and $x \in \mathbb{R}^n$ , the random variable $X_t^x$ has a smooth density with respect to the Lebesgue measure.

Proof:
We want to prove that $\Gamma_t(x)$ is invertible with inverse in $L^p$ for $p \ge 1$ . Since $\mathbf{J}_{0 \rightarrow t}(x)$ is invertible and that its inverse solves a linear equation, we deduce that for every $p \ge 1$ ,
$\mathbb{E}\left( \| \mathbf{J}_{0 \rightarrow t}^{-1}(x) \|^p \right) < +\infty.$

We conclude that it is enough to prove that $C_t(x)$ is invertible with inverse in $L^p$ where
$C_t(x)= \int_0^t \mathbf{J}_{0 \rightarrow r}^{-1}(x) \sigma (X^x_r) \sigma (X^x_r)^* \mathbf{J}_{0 \rightarrow r}^{-1}(x)^* dr.$
By the uniform ellipticity assumption, we have
$C_t(x) \ge \lambda \int_0^t \mathbf{J}_{0 \rightarrow r}^{-1}(x) \mathbf{J}_{0 \rightarrow r}^{-1}(x)^* dr,$
where the inequality is understood in the sense that the difference of the two symmetric matrices is non negative. This implies that $C_t(x)$ is invertible. Moreover, it is an easy exercise to prove that if $M_t$ is a continuous map taking its values in the set of positive definite matrices, then we have
$\left(\int_0^ t M_s ds\right)^{-1} \le \frac{1}{t^2} \left(\int_0^ t M^{-1}_s ds\right).$
As a consequence, we obtain
$C^{-1}_t(x) \le \frac{1}{t^2 \lambda} \int_0^t \mathbf{J}_{0 \rightarrow r}(x)^* \mathbf{J}_{0 \rightarrow r}(x) dr.$
Since $\mathbf{J}_{0 \rightarrow r}(x)$ has moments in $L^p$ for all $p \ge 1$ , we conclude that $C_t(x)$ is invertible with inverse in $L^p$ $\square$

To conclude, we note that this approach to prove existence and smoothness of the density for solutions of stochastic differential equations can also be extended to stochastic differential equations driven by other processes than Brownian motion. For instance, this approach applies to stochastic differential driven by fractional Brownian motions.

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Lecture 34. The Wiener chaos expansion

Posted on November 4, 2012 by Fabrice Baudoin

As in the previous Lectures, we consider a filtered probability space $(\Omega, (\mathcal{F}_t)_{0 \le t \le 1}, \mathbb{P})$ on which is defined a Brownian motion $(B_t)_{0 \le t \le 1}$ , and we assume that $(\mathcal{F}_t)_{0 \le t \le 1}$ is the usual completion of the natural filtration of $(B_t)_{0 \le t \le 1}$ . Our goal is here to write an orthogonal decomposition of the space $L^2(\mathcal{F}_1)$ that is particularly suited to the study of the space $\mathbb{D}^{1,2}$ . For simplicity of the exposition, we restrict ourselves to the case where the Brownian motion $(B_t)_{0 \le t \le 1}$ is one-dimensional.

In the sequel, for $n \ge 1$ , we denote by $\Delta_n$ the simplex $\Delta_n =\{ 0\le t_1 \le \cdots \le t_n \le 1\}$ and if $f_n \in L^2( \Delta_n)$ ,
$I_n (f_n) =\int_0^1 \int_0^{t_n} \cdots \int_0^{t_2} f_n(t_1,\cdots,t_n) dB_{t_1}...dB_{t_n}$
$=\int_{\Delta_n} f_n(t_1,\cdots,t_n) dB_{t_1}...dB_{t_n}.$

The set
$\mathbf{K}_n=\left\{\int_{\Delta_n} f_n(t_1,\cdots,t_n) dB_{t_1}...dB_{t_n}, f_n \in L^2( \Delta_n) \right\}$
is called the space of Wiener chaos of order $n$ . By convention the set of constant random variables shall be denoted by $\mathbf{K}_0$ .

By using the Itō’s isometry, we readily compute that
$\mathbb{E} \left(I_n (f_n)I_p (f_p) \right)= \begin{cases} 0 & \text{if }p \neq n \\ \| f_n \|^2_{L^2(\Delta_n)} & \text{if }p=n. \end{cases}$
As a consequence, the spaces $\mathbf{K}_n$ are orthogonal in $L^2$ . It is easily seen that $\mathbf{K}_n$ is the closure of the linear span of the family
$\left\{ I_n (f^{\otimes n}), f \in L^2([0,1]) \right\},$
where for $f \in L^2([0,1])$ , we denoted by $f^{\otimes n}$ the map $\Delta_n \to \mathbb{R}$ such that $f^{\otimes n}(t_1,\cdots,t_n)=f(t_1)\cdots f(t_n)$ . It turns out that $I_n (f^{\otimes n})$ can be computed by using Hermite polynomials. The Hermite polynomial of order $n$ is defined as
$H_n (x)=(-1)^n \frac{1}{n!} e^{\frac{x^2}{2}} \frac{d^n}{dx^n} e^{-\frac{x^2}{2}}.$
By the very definition of $H_n$ , we see that for every $t, x \in \mathbb{R}$ ,
$\exp \left( t x -\frac{t^2}{2}\right)=\sum_{k=0}^{+\infty} t^k H_k(x).$

Lemma. If $f \in L^2([0,1])$ then $I_n (f^{\otimes n})=\| f \|^n_{L^2([0,1])} H_n \left(\frac{ \int_0^1 f(s) dB_s}{\| f \|_{L^2([0,1])} } \right) .$

Proof. On one hand, we have for $\lambda \in \mathbb{R}$ ,
$\exp \left( \lambda \int_0^1 f(s) dB_s-\frac{\lambda^2}{2} \int_0^1 f(s)^2 ds \right)=\sum_{n=0}^{+\infty} \lambda^n \| f \|^n_{L^2([0,1])} H_n \left(\frac{ \int_0^1 f(s) dB_s}{\| f \|_{L^2([0,1])} } \right) .$

On the other hand, for $0 \le t \le 1$ , let us consider
$M_t(\lambda)=\exp \left( \lambda \int_0^t f(s) dB_s-\frac{\lambda^2}{2} \int_0^t f(s)^2 ds \right).$
From Itō’s formula, we have
$M_t(\lambda)=1+\lambda \int_0^t M_s f(s) dB_s.$
By iterating the previous linear relation, we easily obtain that for every $n \ge 1$ ,
$M_1(\lambda)=1+\sum_{k=1}^n \lambda^k I_k( f^{\otimes k})+\lambda^{n+1} \int_0^1 M_tf(t)\left(\int_{\Delta_n([0,t])} f(t_1)\cdots f(t_n) dB_{t_1}...dB_{t_n}\right) dB_t.$
We conclude,
$I_n( f^{\otimes n})=\frac{1}{n!} \frac{ d^k M_1}{d \lambda^n}(0)=\| f \|^n_{L^2([0,1])} H_n \left(\frac{ \int_0^1 f(s) dB_s}{\| f \|_{L^2([0,1])} } \right) \square$

As we pointed it out, for $p \neq n$ , the spaces $\mathbf{K}_n$ and $\mathbf{K}_p$ are othogonal. We have the following orthogonal decomposition of $L^2$ :

Theorem.[Wiener chaos expansion]
$L^2 =\bigoplus_{n \ge 0} \mathbf{K}_n.$

Proof. As a by-product of the previous proof, we easily obtain that for $f \in L^2([0,1])$ ,
$\exp \left( \lambda \int_0^1 f(s) dB_s-\frac{\lambda^2}{2} \int_0^1 f(s)^2 ds \right)=\sum_{n=1}^{+\infty} I_n( f^{\otimes n}),$
where the convergence of the series is almost sure but also in $L^2$ . Therefore, if $F \in L^2$ is orthogonal to $\bigoplus_{n \ge 1} \mathbf{K}_n$ , then $F$ is orthogonal to every $\exp \left( \lambda \int_0^1 f(s) dB_s-\frac{\lambda^2}{2} \int_0^1 f(s)^2 ds \right)$ , $f \in L^2([0,1])$ . This implies that $F=0$ $\square$

As we are going to see, the space $\mathbb{D}^{1,2}$ or more generally $\mathbb{D}^{k,2}$ is easy to describe by using the Wiener chaos expansion. The keypoint is the following proposition:

Proposition. Let $F=I_n(f_n) \in \mathbf{K}_n$ , then $F \in \mathbb{D}^{1,2}$ and $\mathbf{D}_t F=I_{n-1} ( \tilde{f}_n (\cdot, t)),$ where for $0\le t_1 \le \cdots \le t_{n-1} \le 1$ ,
$\tilde{f}_n (t_1,\cdots, t_{n-1},t) =f_n (t_1,\cdots, t_k, t, t_{k+1}, \cdots, t_{n-1}) \quad \text{if } t_{k} \le t \le t_{k+1}.$

Proof. Let $f \in L^2([0,1])$ . We have
$I_n (f^{\otimes n})=\| f \|^n_{L^2([0,1])} H_n \left(\frac{ \int_0^1 f(s) dB_s}{\| f \|_{L^2([0,1])} } \right) .$
Thus $F=I_n (f^{\otimes n})$ is a smooth cylindric functional and
$\mathbf{D}_t F =\| f \|^{n-1}_{L^2([0,1])} f(t) H'_n \left(\frac{ \int_0^1 f(s) dB_s}{\| f \|_{L^2([0,1])} } \right).$
It is easy to see that $H_n'=H_{n-1}$ , therefore we have
$\mathbf{D}_t F =\| f \|^{n-1}_{L^2([0,1])} f(t) H_{n-1} \left(\frac{ \int_0^1 f(s) dB_s}{\| f \|_{L^2([0,1])} } \right) =f(t) I_{n-1} (f^{\otimes {(n-1)}}).$
As a consequence, we compute that $\mathbb{E} \left(\int_0^1 (\mathbf{D}_t F)^2 dt \right)=n \mathbb{E} (F^2).$ We now observe that $\mathbf{K}_n$ is the closure in $L^2$ of the linear span of the family
$\left\{ I_n (f^{\otimes n}), f \in L^2([0,1]) \right\}$
to conclude the proof of the proposition $\square$

We can finally turn to the description of $\mathbb{D}^{1,2}$ using the chaos decomposition:

Theorem. Let $F \in L^2$ and let
$F=\mathbb{E}(F) +\sum_{m \ge 1} I_m (f_m),$
be the chaotic decomposition of $F$ . Then $F \in \mathbb{D}^{1,2}$ if and only if
$\sum_{m \ge 1} m \mathbb{E}\left( I_m (f_m)^2\right) < +\infty,$
and in that case,
$\mathbf{D}_t F= \mathbb{E}(\mathbf{D}_tF) + \sum_{m \ge 2} I_{m-1} ( \tilde{f}_m (\cdot, t)).$

Proof. It is a consequence of the fact that for $F \in \mathbf{K}_n$ , $\mathbb{E} \left(\int_0^1 (\mathbf{D}_t F)^2 dt \right)=n \mathbb{E} (F^2)$ $\square$ .

An immediate but useful corollary of the previous theorem is the following result:

Corollary. Let $(F_n)_{n \ge 0}$ be a sequence in $\mathbb{D}^{1,2}$ that converges to $F$ in $L^2$ and such that
$\sup_{ n \ge 0} \mathbb{E} \left(\int_0^1 (\mathbf{D}_t F_n)^2 dt \right) < +\infty.$
Then, $F \in \mathbb{D}^{1,2}$ .

Exercise. Let $F \in L^2$ and let
$F=\mathbb{E}(F) +\sum_{m \ge 1} I_m (f_m),$
be the chaotic decomposition of $F$ . Show that that $F \in \mathbb{D}^{k,2}$ , $k \ge 1$ if and only if
$\sum_{m \ge 1} m^k \mathbb{E}\left( I_m (f_m)^2\right) < +\infty.$

Exercise. Let $L=\delta \mathbf{D}$ . Show that for $F \in \mathbf{K}_n$ , $LF=nF$ .

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Some advices to young mathematicians

Posted on November 3, 2012 by Fabrice Baudoin

Being a mathematician constantly ranks among the best jobs. The research in mathematics certainly offers an exciting intellectual adventure.

For the young mathematicians and graduate students who wish to pursue a career in this domain, this article by Gian-Carlo Rota contains thoughts that, I think, can specially be useful.

The blog of Terence Tao is also an excellent source for advices on writing mathematics and academic careers.

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Lecture 33. The Malliavin matrix and existence of densities

Posted on October 29, 2012 by Fabrice Baudoin

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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Lecture 32. The Malliavin derivative

Posted on October 25, 2012 by Fabrice Baudoin

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 $p > 1$ . 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 bounded progressively measurable process. 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 $X=0$ . $\square$

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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Lecture 31. Then, a miracle occurs

Posted on October 25, 2012 by Fabrice Baudoin

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Lecture 30. Stratonovitch stochastic differential equations

Posted on October 24, 2012 by Fabrice Baudoin

As usual, let $\left( \Omega , (\mathcal{F}_t)_{t \geq 0} , \mathbb{P} \right)$ be a filtered probability space which satisfies the usual conditions. It is often useful to use the language of Stratonovitch ‘s integration to study stochastic differential equations because the Itō’s formula takes a much nicer form. If $(N_t)_{0 \leq t \leq T}$ , $T > 0$ , is an $\mathcal{F}$ -adapted real valued local martingale and if $(\Theta_t)_{0 \leq t \leq T}$ is an $\mathcal{F}$ -adapted continuous semimartingale satisfying $\mathbb{P} \left( \int_0^T \Theta_t^2 d \langle N \rangle_t < +\infty \right)=1$ , then by definition the Stratonovitch integral of $(\Theta_t)_{0 \leq t \leq T}$ with respect to $(N_t)_{t \ge 0}$ is defined as
$\int_0^T \Theta_t \circ d N_t =\int_0^T \Theta_t d N_t+\frac{1}{2} \langle \Theta, N \rangle_T,$
where:

$\int_0^T \Theta_t d N_t$ is the Itō integral of $(\Theta_t)_{0 \leq t \leq T}$ against $(N_t)_{0 \leq t \leq T}$ ;
$\langle \Theta, N \rangle_T$ is the quadratic covariation at time $T$ between $(\Theta_t)_{0 \leq t \leq T}$ and $(N_t)_{0 \leq t \leq T}$ .

By using Stratonovitch integral instead of Itō’s, the Itō formula reduces to the classical change of variable formula.

Theorem. Let $(X_t)_{t \geq 0}=\left( X^1_t , \cdots , X^n_t \right)_{t \geq 0}$ be a $n$ – dimensional continuous semimartingale. Let now $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a $C^2$ function. We have
$f(X_t) =f(X_0)+\sum_{i=1}^n \int_0^t \frac{\partial f}{\partial x_i} (X_s) \circ dX^i_s, \quad t \ge 0.$

Let $\mathcal{O} \subset \mathbb{R}^n$ be a non empty open set. A smooth vector field $V$ on $\mathcal{O}$ is simply a smooth map
$\begin{array}{llll} V: & \mathcal{O} & \rightarrow & \mathbb{R}^{n} \\ & x & \rightarrow & (v_{1}(x),...,v_{n}(x)). \end{array}$
The vector field $V$ defines a differential operator acting on smooth functions $f: \mathcal{O} \rightarrow \mathbb{R}$ as follows:
$Vf(x)=\sum_{i=1}^n v_i (x) \frac{\partial f}{\partial x_i}.$
We note that $V$ is a derivation, that is a map on $\mathcal{C}^{\infty} (\mathcal{O} , \mathbb{R} )$ , linear over $\mathbb{R}$ , satisfying for $f,g \in \mathcal{C}^{\infty} (\mathcal{O} , \mathbb{R} )$ , $V(fg)=(Vf)g +f (Vg).$
An interesting result is that, conversely, any derivation on $\mathcal{C}^{\infty} (\mathcal{O} , \mathbb{R} )$ is a vector field.

Let now $(B_t)_{t \geq 0}=(B^1_t,...,B^d_t)_{t \geq 0}$ be a $d$ -dimensional Brownian motion and consider $d+1$ $C^1$ vector fields $V_i : \mathbb{R}^n \rightarrow \mathbb{R}^n$ , $n \geq 1$ , $i=0,...,d$ . By using the language of vector fields and Stratonovitch integrals, the fundamental theorem for the existence and the uniqueness of solutions for stochastic differential equations is the following:

Theorem. Assume that $V_0,V_1,\cdots,V_d$ are bounded vector fields with bounded derivatives up to order 2. Let $x_0 \in \mathbb{R}^n$ . On $\left( \Omega , (\mathcal{F}_t)_{t \geq 0} , \mathbb{P} \right)$ , there exists a unique continuous and adapted process $(X_t^{x_0})_{t \geq 0}$ such that for $t \geq 0$ ,
$X_t^{x_0}=x_0 + \sum_{i=0}^d \int_0^t V_i (X_s^{x_0}) \circ dB^i_s,$
with the convention that $B^0_t=t$ .

Thanks to Itō’s formula the corresponding Itō’s formulation is
$X_t^{x_0} =x_0 + \frac{1}{2} \sum_{i=1}^d \int_0^t \nabla_{V_i} V_i (X_s^{x_0}) ds +\sum_{i=0}^d \int_0^t V_i (X_s^{x_0}) dB^i_s,$
where for $1 \leq i \leq d$ , $\nabla_{V_i} V_i$ is the vector field given by
$\nabla_{V_i} V_i (x)=\sum_{j=1}^n \left( \sum_{k=1}^n v_i^k (x) \frac{\partial v^j_i}{\partial x_k}(x)\right)\frac{\partial}{\partial x_j}, \text{ }x \in \mathbb{R}^n.$
If $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is a $C^2$ function, from Itō’s formula, we have for $t \geq 0$ ,
$f(X_t^{x_0})=f(x_0) + \sum_{i=0}^d \int_0^t (V_i f) (X_s^{x_0}) \circ dB^i_s,$
and the process
$\left( f(X_t^{x_0})-\int_0^t (Lf)(X_s^{x_0})ds \right)_{t \geq 0}$
is a local martingale where $L$ is the second order differential operator
$L = V_0+\frac{1}{2} \sum_{i=1}^d V_i^2.$

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Lecture 29. The strong Markov property for solutions of stochastic differential equations

Posted on October 22, 2012 by Fabrice Baudoin

In the previous section, we have seen that if $(X_t^x)_{t \ge 0}$ is the solution of a stochastic differential equation
$X_t^{x} =x +\int_0^t b(X_s^{x}) ds + \int_0^t \sigma(X_s^{x}) dB_s,$
then $(X_t^x)_{t \ge 0}$ is a Markov process, that is for every $t,T \ge 0$ ,
$\mathbb{E}(f(X_{t+T}^x) \mid \mathcal{F}_T)=(P_{t}f )(X_T^x),$
where $P_tf(x)=\mathbb{E}( f(X_t^x))$ . It is remarkable that this property still holds when $T$ is now any finite stopping time. This property is called the strong Markov property.
The key lemma is the following:

Lemma. Let $(B_t)_{t\ge 0}$ be a standard Brownian motion and let $T$ be a finite stopping time. The process, $(B_{T+t}-B_T)_{t\ge 0}$ is a standard Brownian motion independent from $\mathcal{F}_T$ .

Proof. Let $T$ be a finite stopping time of the filtration $(\mathcal{F}_t)_{t \ge 0}$ . We first assume $T$ bounded. Let us consider the process $\tilde{B}_t=B_{T+t}-B_T, \quad t \ge 0.$ Let $\lambda \in \mathbb{R}$ , $0\le s \le t$ . Applying Doob’s stopping theorem to the martingale $\left( e^{i\lambda B_t +\frac{\lambda^2}{2} t}\right)_{t \ge 0},$ with the stopping times $t+T$ and $s+T$ , yields:
$\mathbb{E} \left( e^{i\lambda B_{T+t} +\frac{\lambda^2}{2} (T+t)}\mid \mathcal{F}_{T +s} \right)=e^{i\lambda B_{T+s} +\frac{\lambda^2}{2} (T+s)}.$
Therefore
$\mathbb{E} \left( e^{i\lambda (B_{T+t} -B_{T+s})}\mid \mathcal{F}_{T+s} \right)=e^{-\frac{\lambda^2}{2} (t-s) }.$
The increments of $(\tilde{B}_t)_{t \ge 0}$ are therefore independent and stationary. The conclusion then easily follows. If $T$ is not bounded almost surely, then we can consider the stopping time $T \wedge N$ and from the previous result the finite dimensional distributions $(B_{ t_1 +T\wedge N}-B_{T \wedge N}, \cdots , B_{ t_n +T\wedge N}-B_{T \wedge N})$ do not depend on $N$ and are the same as a Brownian motion. We can then let $N \to + \infty$ to conclude $\square$

Theorem. For every $x \in \mathbb{R}^n$ , $(X_t^{x})_{t\ge 0, x \in \mathbb{R}^d}$ is a strong Markov process with semigroup $(P_t)_{t \ge 0}$ : For every Borel function $f:\mathbb{R}^n \to \mathbb{R}$ with polynomial growth, every $t \ge 0$ , and every finite stopping time $T$ ,
$\mathbb{E}(f(X_{t+T}^x) \mid \mathcal{F}_T)=(P_{t}f )(X_T^x).$

Proof. The proof is identical to the proof of the usual Markov property with the additional ingredient given by the previous proposition $\square$

The strong Markov property for solutions of stochastic differential equations is useful to solve boundary value problems in partial differential equations theory. Let $K$ be a bounded closed set in $\mathbb{R}^n$ . For $x \in \Omega$ , we denote $T_x= \inf \{ t \ge 0, X_t \in \partial K \}$ . If $f$ is bounded Borel function such that $f_{\partial K}=0$ , we define
$P^K_t f(x)=\mathbb{E}\left( f(X^x_t) \mathbf{1}_{t \le T_x } \right)$ .
The proof of the following theorem is let to the reader.

Theorem. Let $f:K \to \mathbb{R}$ be a bounded Borel function and assume that the function $u(t,x)=(P^K_tf)(x)$ is $C^{1,2}$ . Then $u$ is the unique solution of the Dirirchlet boundary value problem
$\frac{\partial u}{\partial t} (t,x)=Lu(t,x)$
in $[0,+\infty) \times \mathbb{R}^n$ , with the initial condition
$u(0,x)=f(x),$
and the boundary condition
$u(t,x)=0, \quad x \in \partial K.$

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Lecture 28. The Feynman-Kac formula

Posted on October 18, 2012 by Fabrice Baudoin

It is now time to give some applications of the theory of stochastic differential equations to parabolic second order partial differential equations. In particular we are going to prove that solutions of such equations can represented by using solutions of stochastic differential equations. This representation formula is called the Feynman–Kac formula.
As usual, we consider a filtered probability space $\left( \Omega , (\mathcal{F}_t)_{t \geq 0} , \mathcal{F},\mathbb{P} \right)$ which satisfies the usual conditions and on which is defined a $n$ -dimensional Brownian motion $(B_t)_{t \ge 0}$ . Again, we consider two functions $b : \mathbb{R}^n \to \mathbb{R}^n$ and $\sigma: \mathbb{R}^{n \times n}$ and we assume that there exists $C > 0$ such that
$\| b(x)-b(y) \| + \| \sigma (x) - \sigma (y) \| \le C \| x-y \|, x,y \in \mathbb{R}^n.$
Let $L$ be the second order differential operator
$L=\sum_{i=1}^n b_i(x) \frac{\partial}{\partial x_i} +\frac{1}{2}\sum_{i,j=1}^n a_{ij}(x)\frac{\partial^2}{\partial x_i \partial x_j} ,$
where $a_{ij}(x)=(\sigma(x)\sigma^*(x))_{ij}$ .

As we know, there exists a bicontinuous process $(X_t^{x})_{t\ge 0, x \in \mathbb{R}^d}$ such that for $t \ge 0$ ,
$X_t^{x} =x +\int_0^t b(X_s^{x}) ds + \int_0^t \sigma(X_s^{x}) dB_s.$
Moreover, as it has been stressed before, for every $p \ge 1$ , and $T \ge 0$
$\mathbb{E} \left( \sup_{0 \le t \le T} \| X^x_t \|^p\right) < +\infty.$
As a consequence, if $f:\mathbb{R}^n \to \mathbb{R}$ is a Borel function with polynomial growth, we can consider the function
$P_t f (x) =\mathbb{E}(f(X_t^x)).$
Theorem. For every $x \in \mathbb{R}^n$ , $(X_t^{x})_{t\ge 0, x \in \mathbb{R}^d}$ is a Markov process with semigroup $(P_t)_{t \ge 0}$ . More precisely, for every Borel function $f:\mathbb{R}^n \to \mathbb{R}$ with polynomial growth and every $t \ge s$ ,
$\mathbb{E}(f(X_t^x) \mid \mathcal{F}_s)=(P_{t-s}f )(X_s^x).$

Proof. The key point, here, is to observe that solutions are actually adapted to the natural filtration of the Brownian motion $(B_t)_{t \ge 0}$ . More precisely, there exists on the space of continuous functions $[0,+\infty) \rightarrow \mathbb{R}^n$ a predictable functional such that for $t \ge 0$ :
$X_t^{x_0}=F(x_0, (B_u)_{0 \le u \le t}).$
Indeed, let us first work on $[0,T]$ where $T$ is small enough. In that case, as seen previously, the process $(X^{x_0}_t)_{0 \le t \le T}$ is the unique fixed point of the application $\Phi$ defined by
$\Phi(X)_t =x_0+\int_0^t b(X_s^{x_0}) ds + \int_0^t \sigma(X_s^{x_0}) dB_s.$

Alternatively, one can interpret this by observing that $(X^{x_0}_t)_{0 \le t \le T}$ is the limit of the sequence of processes $(X^{n}_t)_{0 \le t \le T}$ inductively defined by
$X^{n+1}=\Phi (X^n), \quad X^0=x_0.$
It is easily checked that for each $X^n$ there is a predictable functional $F_n$ such that
$X_t^{n}=F_n(x_0, (B_u)_{0 \le u \le t}),$
which proves the above claim when $T$ is small enough. To get the existence of $F$ for any $T$ , we can proceed

With this hands, we can now prove the Markov property. Let $s \ge 0$ . For $t \ge 0$ , we have
$X_{s+t}^{x_0} =X_s +\int_s^{s+t} b(X_u^{x_0}) du + \int_s^{s+t} \sigma(X_u^{x_0}) dB_u$
$=X_s +\int_0^{t} b(X_{u+s}^{x_0}) du + \int_0^{t} \sigma(X_s^{x_0}) d(B_{u+s}-B_s).$
Consequently, from uniqueness of solutions,
$X_{s+t}^{x_0}=F(X^{x_0}_s, (B_{u+s}-B_s)_{0 \le u \le t}).$
We deduce that for a Borel function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ with polynomial growth,
$\mathbb{E} \left( f(X_{s+t}^{x_0}) \mid \mathcal{F}_s \right)=\mathbb{E} \left( f(F(X^{x_0}_s, (B_{u+s}-B_s)_{0 \le u \le t}))\mid \mathcal{F}_s \right)=P_t f (X^{x_0}_s),$
because $(B_{u+s}-B_s)_{0 \le u \le t}$ is a Brownian motion independent of $\mathcal{F}_s$ $\square$

Theorem Let $f:\mathbb{R}^n \to \mathbb{R}$ be a Borel function with polynomial growth and assume that the function
$u(t,x)=(P_tf)(x)$
is $C^{1,2}$ , that is once differentiable with respect to $t$ and twice differentiable with respect to $x$ . Then $u$ solves the Cauchy problem
$\frac{\partial u}{\partial t} (t,x)=Lu(t,x)$
in $[0,+\infty) \times \mathbb{R}^n$ , with the initial condition $u(0,x)=f(x)$ .

Proof. Let $T > 0$ and consider the function $v(t,x)=u(T-t,x)$ . According the previous theorem, we have
$\mathbb{E}(f(X_T^x) \mid \mathcal{F}_t)=v(t,X_t^x).$
As a consequence, the process $v(t,X_t^x)$ is a martingale. But from Ito’s formula the bounded variation part of $v(t,X_t^x)$ is $\int_0^t \left( \frac{\partial v}{\partial t}(s, X_s^x) + L v(s, X_s^x) \right)ds$ which is therefore 0. We conclude
$\frac{\partial v}{\partial t}(0, x) + L v(0, x)=\lim_{t \to 0} \frac{1}{t} \int_0^t \left( \frac{\partial v}{\partial t}(s, X_s^x) + L v(s, X_s^x) \right)ds=0$ $\square$

Exercise Show that if $f$ is a $C^2$ function such that $\nabla f$ and $\nabla^2 f$ have polynomial growth, then the function $P_tf(x)$ is $C^{1,2}$ . Here, we denote by $\nabla^2 f$ the Hessian matrix of $f$ .

Theorem. Let $f:\mathbb{R}^n \to \mathbb{R}$ be a Borel function with polynomial growth. Let $u:[0,+\infty)\times \mathbb{R}^n \to \mathbb{R}$ be a solution of the Cauchy problem
$\frac{\partial u}{\partial t} (t,x)=Lu(t,x)$
with the initial condition $u(0,x)=f(x)$ .
If there exists a locally integrable function $C$ and $p \ge 0$ , such that for every $t \ge 0$ and $x \in \mathbb{R}^n$ ,
$\| \nabla u (t,x) \| \le C(t) (1+\|x \|^p),$
then $u(t,x)= P_t f(x)$ .

Proof. Let $T > 0$ and, as before, consider the function $v(t,x)=u(T-t,x)$ . As a consequence of Ito’s formula, we have
$v(t,X_t^x)=u(T,x)+M_t,$
where $M_t$ is a local martingale with quadratic variation $\sum_{i,j}^n \int_0^t a_{ij}(X_s^x) \frac{\partial u}{\partial x_i}(X_s^x) \frac{\partial u}{\partial x_j} (X_s^x) ds$ . The conditions on $\sigma$ and $u$ imply that this quadratic variation is integrable. As a consequence, $v(t,X_t^x)$ is a martingale and thus $\mathbb{E} (v(T,X_t^x))=u(T,x)$ $\square$

The previous results may be extended to study parabolic equations with potential as well. More precisely, let $V:\mathbb{R}^n \to \mathbb{R}$ be a bounded function. If $f:\mathbb{R}^n \to \mathbb{R}$ is a Borel function with polynomial growth, we define
$P^V_t f (x) =\mathbb{E}\left(e^{\int_0^t V(X_s^x) ds }f(X_t^x)\right)$ .
The same proofs as before will give the following theorems.

Theorem. For every $x \in \mathbb{R}^n$ and every Borel function $f:\mathbb{R}^n \to \mathbb{R}$ with polynomial growth and every $t \ge s$ ,
$\mathbb{E}\left( e^{\int_0^t V(X_u^x) du } f(X_t^x) \mid \mathcal{F}_s\right)=e^{\int_0^s V(X_u^x) du }(P^V_{t-s}f )(X_s^x).$

Theorem. Let $f:\mathbb{R}^n \to \mathbb{R}$ be a Borel function with polynomial growth and assume that the function
$u(t,x)=(P^V_tf)(x)$
is $C^{1,2}$ , that is once differentiable with respect to $t$ and twice differentiable with respect to $x$ . Then $u$ solves the Cauchy problem
$\frac{\partial u}{\partial t} (t,x)=Lu(t,x)+V(x)u(t,x)$
in $[0,+\infty) \times \mathbb{R}^n$ , with the initial condition
$u(0,x)=f(x)$ .

Theorem. Let $f:\mathbb{R}^n \to \mathbb{R}$ be a Borel function with polynomial growth. Let $u:[0,+\infty)\times \mathbb{R}^n \to \mathbb{R}$ be a solution of the Cauchy problem
$\frac{\partial u}{\partial t} (t,x)=Lu(t,x)+V(x)u(t,x)$
with the initial condition $u(0,x)=f(x)$ . If there exists a locally integrable function $C$ and $p \ge 0$ , such that for every $t \ge 0$ and $x \in \mathbb{R}^n$ ,
$\| \nabla u (t,x) \| \le C(t) (1+\|x \|^p)$ ,
then $u(t,x)= P^V_t f(x)$ .

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Research and Lecture notes

MA694 Rough paths theory

Lecture 35. Weak differentiability for solutions of stochastic differential equations and the existence of a smooth density

Lecture 34. The Wiener chaos expansion

Some advices to young mathematicians

Lecture 33. The Malliavin matrix and existence of densities

Lecture 32. The Malliavin derivative

Lecture 31. Then, a miracle occurs

Lecture 30. Stratonovitch stochastic differential equations

Lecture 29. The strong Markov property for solutions of stochastic differential equations

Lecture 28. The Feynman-Kac formula

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