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

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.

• Would you care to explain your reasoning? Mar 12 '20 at 23:10
• @NateEldredge I roughly edit it. Mar 12 '20 at 23:15

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.)