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

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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