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?

  • $\begingroup$ Have a look at Shreve's Stochastic Calculus for Finance II Theorem 6.3.1. It states that $\mathbb{E}[h(X(T))|\mathcal{F}(t)] = \mathbb{E}^{t,x}[h(X(T))]$ (without proof) $\endgroup$
    – zer0hedge
    Apr 20, 2017 at 17:06
  • $\begingroup$ A conditional expectation of the form $Y_t = \Bbb{E} [ f(X_T) \vert \mathcal{F}_t ]$ (with no explicit dependence on $t$ inside $f$) will always be a martingale by the tower property, right? Just compute $\Bbb{E}[Y_t \vert \mathcal{F}_s ]$ to convince yourself that it is indeed equal to $Y_s$. Of course this is assuming the process is adapted to the filtration $\endgroup$
    – Quantuple
    Apr 20, 2017 at 17:07
  • $\begingroup$ @zer0hedge but in the proof of Kolmogorov backward equation we use $g(t,x)=E^{t,x}h(X(T))=\int^{\infty}_0h(y)p(t,T,x,y)d y,$ that is the first definition, and we will still use the martingale of $g(t,x).$ $\endgroup$
    – A.Oreo
    Apr 21, 2017 at 3:59
  • $\begingroup$ @Quantuple Yeah, I am sure $E[h(X(T))|\mathcal{F}(t)]=g(t,X(t))$ is martingale, my question is whether $g(t,x) = E^{t,x}h(X(T))$ is martingale, those two $g(t,x)$ are different, and we will use the martingale of later $g(t,x)$ to prove Kolmogorov backward equation. $\endgroup$
    – A.Oreo
    Apr 21, 2017 at 4:04
  • 1
    $\begingroup$ "Note that there is nothing random about $g(t,x)$; it is an ordinary (actually, Borel-measurable) function of the two dummy variables t and x" Shreve, p.266 $\endgroup$
    – zer0hedge
    Apr 21, 2017 at 6:01

1 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.


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