markov property for stochastic differential equation

Question

Suppose the stochastic equation： \begin{equation*} d X(u)=\beta(u,X(u))d u+\gamma(u,X(u))d W(u). \end{equation*} Suppose $X(T)$ is the solution of above stochastic differential equation with initial condition $X(t)=x$ and $h(x)$ is a Borel-measurable function. Denote by $$g(t,x)=E^{t,x}h(X(T))$$ We assume $E^{t,x}|h(X(T))|<\infty$

Let $X(u)$ is the solution of above stochastic differential equation with initial condition given at time $0.$

Use the markov property of $X(t),$ we have existing $g(t,x)$ s.t $$E[h(X(T))|\mathcal{F}(t)]=g(t,X(t))$$

My question is are those two $g(t,x)$ same? Since we want to use Feynman-Kac equation, but I am not sure whether it is true for first $$g(t,x)=E^{t,x}h(X(T)).$$ since the proof of Feynman-Kac equation needs the martingale of $g(t,X(t)),$ but I don't think here $g(t,X(t))$ is martingale?

Answer 1

Here, we assume that \begin{align*} g(t, x) = \mathbb{E}\left(h(X_T) \mid X_t = x \right). \end{align*} Note that, by Shiryaev, $g(t, x)$ is a Borel measurable function such that, for any Borel measurable set $A$, \begin{align*} \int_{\{X_t \in A\}} h(X_T) d\mathbb{P} &= \int_A g(t, x) \mathbb{P}_{X_t}(dx), \end{align*} where $\mathbb{P}_{X_t}(dx)$ is the Lebesgue-Stieltjes measure generated by the distribution function of $X_t$, that is, for any Borel measurable set $B$, \begin{align*} \mathbb{P}_{X_t}(B) = \mathbb{P}(X_t \in B). \end{align*} It can also be shown that (see Page 196 of Shiryaev, starting with indicator and simple functions, then, by monotone convergence theorem, to all positive functions, and, by decomposition, to all integrable measurable functions), \begin{align*} \int_A g(t, x) \mathbb{P}_{X_t}(dx) = \int_{\{X_t \in A\}} g(t, X_t) d\mathbb{P}. \end{align*} That is, \begin{align*} \int_{\{X_t \in A\}} h(X_T) d\mathbb{P} = \int_{\{X_t \in A\}} g(t, X_t) d\mathbb{P}. \end{align*} In other words, \begin{align*} g(t, X_t) = \mathbb{E}(h(X_T) \mid X_t) = \mathbb{E}(h(X_T) \mid \mathcal{F}_t), \end{align*} by the Markov property. Moreover, $\{g(t, X_t), \, 0\le t \le T \}$ is obviously a martingale.