Can we find a function being equal to a nonmartingale random variable?

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Consider a prob. space $(\Omega, \mathcal{F}, \mathbb{P})$ and $\mathbb{F} = (\mathcal{F}_t)_{0\leq t\leq T}$ being a filtration for a scalar Brownian motion $W = (W_t)_{0\leq t\leq T}$.

Consider $X=(X_t)_{0\leq t\leq T}$, which is a unique solution of $$dX_t = \mu(X_t)dt + \sigma(X_t)dW_t.$$

Consider a process $v = (v_t)_{0\leq t \leq T}$ $$v_t = \mathbb{E}(h(X_t,X_T)\mid\mathcal{F}_t).$$


I have checked that $v$ is NOT a martingale.

My question is that can we find a function $u:[0,T]\times \mathbb{R} \rightarrow \mathbb{R}$, such that $u(t,X_t) = v_t$?

My answer is that we cannot by Feynman-Kac formula. Since $v$ is not martingale, so we cannot form a PDE with solution $v$. Ant it does not have a stochastic representation.

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The solution $X$ is a time-homogeneous Markov process. Let $P_t(x,\cdot)$ be the distribution of $X_t$ under the condition that $X_0=x$. Then $v_t = u(t,X_t)$ (a.s.) where $u$ is given by $$ u(t,x)=\int_{\Bbb R} P_{T-t}(x,dy) h(x,y). $$ (I'm assuming that $h$ is bounded, for simplicity.)