Let $X$ be a continuous Markov process on $\mathbb{R}^d$ that is also a semimartingale. Let $V=(V_1,...,V_d)$ be a collection of suitably nice vector fields on $\mathbb{R}^d$ such that there exists a process $Y$ that solves

$ dY_t=V(Y_t)dX_t=\sum_{i=1}^d V_i(Y_t)dX_t^i, Y_0=y_0\in\mathbb{R}^d. $

Now I have read that 1) $Y$ will not possess the Markov property unless $X=B$, the Brownian motion. 2) Protter did some work on Markovian drivers of SDEs and the Markov property of solutions, however, all I get out of his book is that $Y$ would possess the Markov property if all components of $X$ were Levy processes (i.e., semimartingales with independent increments). 3) Here ([When is a stochastic process defined via a SDE Markovian?), somebody makes quite a general statement.

Now, intuitively, I would think that for $t>s:$

$ \mathbb{E}[Y_t\vert \mathcal{F}_s]=\mathbb{E}[Y_0+\int_0^t V(Y_r)dX_r\vert \mathcal{F}_s]=\mathbb{E}[Y_0+\int_0^s V(Y_r)dX_r+\int_s^t V(Y_r)dX_r\vert \mathcal{F}_s ]=Y_s + \mathbb{E}[\int_s^t V(Y_r)dX_r\vert \mathcal{F}_s] $

Now why should the $\int_s^t V(Y_r)dX_r$ be not independent of $\mathcal{F}_s$, $V$ is a vector field and as such should not somehow dependent on the history of $Y$ itself, or could it (as a vector field, it is simply a map from $\mathbb{R}^d$ to itself)? Maybe I am getting it all wrong and there is a simple counterexample of a Markov process for which the solution to the above SDE is not Markovian... Any hints and ideas would be very welcome.


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