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

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

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