Lecture 29. Stochastic differential equations as rough differential equations

Based on the results of the previous Lecture, it should come as no surprise that differential equations driven by the Brownian rough path should correspond to Stratonovitch differential equations. In this Lecture, we prove that it is indeed the case. Let us first remind to the reader the following basic result about existence and uniqueness for solutions of stochastic differential equations.

Let $(B_t)_{t \geq 0}=(B^1_t,...,B^d_t)_{t \geq 0}$ be a $d$ -dimensional Brownian motion defined on some filtered probability space $\left( \Omega , (\mathcal{F}_t)_{t \geq 0} , \mathbb{P} \right)$ that satisfies the usual conditions.

Theorem: Assume that $V_1,\cdots,V_d$ are $C^2$ 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 $(X_t)_{t \geq 0}$ such that for $t \geq 0$ ,
$X_t=x_0 + \sum_{i=1}^d \int_0^t V_i (X_s) \circ dB^i_s.$

Thanks to Ito’s formula the corresponding Ito’s formulation is
$X_t=x_0 + \frac{1}{2} \sum_{i=1}^d \int_0^t \nabla_{V_i} V_i (X_s) ds +\sum_{i=1}^d \int_0^t V_i (X_s ) dB^i_s,$
where for $1 \leq i \leq d$ , $\nabla_{V_i} V_i$ is the vector field given by
$\nabla_{V_i} V_i (x)=V_i^2 \mathbf{I} (x)= \sum_{j=1}^n \left( \sum_{k=1}^n v_i^k (x) \frac{\partial v^j_i}{\partial x_k}(x)\right)\frac{\partial}{\partial x_j}, \text{ }x \in \mathbb{R}^n.$

The main result of the Lecture is the following:

Theorem: Let $\gamma > 2$ and let $V_1,\cdots,V_d$ be $\gamma$ -Lipschitz vector fields on $\mathbb{R}^n$ . Let $x_0 \in \mathbb{R}^n$ . The solution of the rough differential equation
$X_t=x_0 + \sum_{i=1}^d \int_0^t V_i (X_s) \ dB^i_s,$
is the solution of the Stratonovitch differential equation:
$X_t=x_0 + \sum_{i=1}^d \int_0^t V_i (X_s) \circ dB^i_s.$

Proof: Let us work on a fixed interval $[0,T]$ and consider a sequence $D_n$ of subdivisions of $[0,T]$ such that $D_{n+1} \subset D_n$ and whose mesh goes to 0 when $n \to +\infty$ . As in the previous lectures, we denote by $B^n$ the piecewise linear process which is obtained from $B$ by interpolation along the subdivision $D_n$ , that is for $t_i^n \le t \le t_{i+1}^n$ ,
$B^n_t= \frac{t_{i+1}^n -t}{ t_{i+1}^n-t_i^n} B_{t^n_i} + \frac{t-t_i^n}{ t_{i+1}^n-t_i^n} B_{t^n_{i+1}}.$
Let us then consider the process $X_n$ that solves the equation
$X^n_t=x_0 + \sum_{i=1}^d \int_0^t V_i (X^n_s) \ dB^{i,n}_s,$
and the process $\tilde{X}^n$ , which is piecewise linear and such that
$\tilde{X}^n_{t^n_{k+1}}=\tilde{X}^n_{t^n_{k}}+\sum_{i=i}^d V_i ( X^n_{t_k^n}) ( B^i_{t^n_{k+1}}- B^i_{t^n_{k}})+\frac{1}{2} \sum_{i=1}^d V_i^2 \mathbf{I} (X^n_{t_k^n})(t_{k+1}^n -t_k^n).$
We can write
$X_{t_{k+1}^n} - \tilde{X}_{t_{k+1}^n} = \sum_{\nu=0}^k ( X_{t_{\nu+1}^n} - X_{t_{\nu}^n} )- (\tilde{X}_{t_{\nu+1}^n} - \tilde{X}_{t_{\nu}^n} ).$
Now,
$( X_{t_{\nu+1}^n} - X_{t_{\nu}^n} )- ( \tilde{X}_{t_{\nu+1}^n} - \tilde{X}_{t_{\nu}^n} )$
$= ( X_{t_{\nu+1}^n} - X_{t_{\nu}^n} )- \sum_{i=i}^d V_i ( X^n_{t_\nu^n}) ( B^i_{t^n_{\nu+1}}- B^i_{t^n_{\nu}}) -\frac{1}{2} \sum_{i=1}^d V_i^2 \mathbf{I} (X^n_{t_\nu^n})(t_{\nu+1}^n -t_\nu^n).$
From Davie’s estimate, we have, with $2 < p < \gamma$ ,
$\left\|( X_{t_{\nu+1}^n} - X_{t_{\nu}^n} )- \sum_{i=i}^d V_i ( X^n_{t_\nu^n}) ( B^i_{t^n_{\nu+1}}- B^i_{t^n_{\nu}}) -\sum_{i,j=1}^d (V_i V_j \mathbf{I}) (X^n_{t_\nu^n}) \int_{t_\nu^n}^{t_{\nu+1}^n} (B^{n,i}_u-B^{n,i}_{t_\nu^n})dB^{n,j}_u \right\|$
$\le C \| V \|_{Lip^{\gamma-1}} \| S_2(B^n) \|^\gamma_{p-var, [t^n_{\nu}, t^n_{\nu+1}]}$
$\le C \| V \|_{Lip^{\gamma-1}} \| B^n \|^\gamma_{p-var, [t^n_{\nu}, t^n_{\nu+1}]}$
$\le C' \| V \|_{Lip^{\gamma-1}} \| B \|^\gamma_{p-var, [t^n_{\nu}, t^n_{\nu+1}]}.$
We deduce that, almost surely when $n \to \infty$ ,
$\sum_{\nu=0}^k \left\|( X_{t_{\nu+1}^n} - X_{t_{\nu}^n} )- \sum_{i=i}^d V_i ( X^n_{t_\nu^n}) ( B^i_{t^n_{\nu+1}}- B^i_{t^n_{\nu}}) -\sum_{i,j=1}^d (V_i V_j \mathbf{I}) (X^n_{t_\nu^n}) \int_{t_\nu^n}^{t_{\nu+1}} (B^{n,i}_u-B^{n,i}_{t_\nu^n})dB^{n,j}_u \right\| \to 0.$
On the other hand,
$\left\| \sum_{i,j=1}^d (V_i V_j \mathbf{I}) (X^n_{t_\nu^n}) \int_{t_\nu^n}^{t_{\nu+1}^n} (B^{n,i}_u-B^{n,i}_{t_\nu^n})dB^{n,j}_u -\frac{1}{2} \sum_{i=1}^d V_i^2 \mathbf{I} (X^n_{t_\nu^n})(t_{\nu+1}^n -t_\nu^n)\right\|$
$\le\| V \|_{Lip^{\gamma}} \sum_{i,j=1}^d \left| \int_{t_\nu^n}^{t_{\nu+1}^n} (B^{n,i}_u-B^{n,i}_{t_\nu^n})dB^{n,j}_u - \frac{1}{2} \delta_{ij} (t_{\nu+1}^n -t_\nu^n) \right|$
$\le \frac{1}{2} \| V \|_{Lip^{\gamma}} \sum_{i,j=1}^d \left| (B^{n,i}_{t_{\nu+1}^n}-B^{n,i}_{t_\nu^n}) (B^{n,j}_{t_{\nu+1}^n}-B^{n,j}_{t_\nu^n})- \delta_{ij} (t_{\nu+1}^n -t_\nu^n) \right|$
We deduce that in probability,
$\sum_{\nu=0}^k \left\| \sum_{i,j=1}^d (V_i V_j \mathbf{I}) (X^n_{t_\nu^n}) \int_{t_\nu^n}^{t_{\nu+1}^n} (B^{n,i}_u-B^{n,i}_{t_\nu^n})dB^{n,j}_u -\frac{1}{2} \sum_{i=1}^d V_i^2 \mathbf{I} (X^n_{t_\nu^n})(t_{\nu+1}^n -t_\nu^n)\right\| \to 0.$
We conclude that in probability,
$X_{t_{k+1}^n} - \tilde{X}_{t_{k+1}^n} \to 0.$
Up to an extraction of subsequence, we can assume that almost surely
$X_{t_{k+1}^n} - \tilde{X}_{t_{k+1}^n} \to 0.$
We now know that from the Lyons’ continuity theorem, almost surely $X_t^n \to X_t$ where $(X_t)_{t \in [0,T]}$ is the solution of the rough differential equation
$X_t=x_0 + \sum_{i=1}^d \int_0^t V_i (X_s) \ dB^i_s.$
Thus almost surely, we have that $\tilde{X}_t^n \to X_t$ . On the othe hand, by definition, we have
$\tilde{X}^n_{t^n_{k+1}}=\tilde{X}^n_{t^n_{k}}+\sum_{i=i}^d V_i ( X^n_{t_k^n}) ( B^i_{t^n_{k+1}}- B^i_{t^n_{k}})+\frac{1}{2} \sum_{i=1}^d V_i^2 \mathbf{I} (X^n_{t_k^n})(t_{k+1}^n -t_k^n),$
which easily implies that $\tilde{X}^n$ converges in probability to $x_0+\sum_{i=i}^d \int_0^t V_i (X_s)\circ dB^i_s$ . This proves that
$X_t=x_0 + \sum_{i=1}^d \int_0^t V_i (X_s) \circ dB^i_s$
$\square$