Markov Conditional Expectations and the Feynman–Kac Connection

# markov property for stochastic differential equation


# markov property for stochastic differential equation












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 by Gordon (score 5, accepted)

https://quant.stackexchange.com/a/33994

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.