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

This entry was posted in Rough paths theory. 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