Let $(X_t)$ and $(Y_t)$ be sample continuous stochastic processes on $[0,1]$ such that $Z_t (\sigma):= X_t + \sigma Y_t$, where $\sigma >0$, is Markov with regard to the filtration generated by $Z_t (\sigma)$, denoted by $(\mathcal F^{Z(\sigma)}_t)$, for any choice of $\sigma >0$, i.e for any $s<t$, and bounded mesurable function $f$,
$$E[f(Z_t (\sigma) )| \mathcal F^{Z(\sigma)}_s ]=E[f(Z_t (\sigma)) | Z_s (\sigma) ].$$
Equivalently,
$$E[f(Z_t (\sigma) )| \mathcal F^{Z(\sigma)}_s ]=E[f(Z_t (\sigma)) | X_s + \sigma Y_s ].$$
Does that mean that forcibly, $(X_t)$ is Markov with regard to its own filtration? This is the limit case when $\sigma \rightarrow 0$.
