I would like to restate the question.
I'm reading Revuz/Yor's definition of Markov process (P81), they started from transition function, and define the $P_t f(x)$ as usual (let's only consider the time homogenous case here).
A process $X_t$ is called Markov w.r.t $\mathcal F_t$ if
$E(f(X_{t+s})|\mathcal F_t) = P_sf(X_t)\tag{1}$
My previous impression had been that, once a transition function is defined, the induced process $X_t$ should be naturally Markov (w.r.t the natural filtration $\mathcal F^X_t$, of course). It is not clear to me, what is the intuition of using the more general filtration $\mathcal F_t$ in the definition.
More specifically, consider
$\mathcal F_t = \mathcal F_t^X \vee \mathcal F_t^Y $
where $Y_t$ is another process. If $X_t$ is Markov under $\mathcal F^X_t$, what kind $Y_t$ will make equation(1) hold?
Since equation (1) implies that
$E(f(X_{t+s})|\mathcal F_t) = E(f(X_{t+s})|\mathcal F_t^X)$
which means, knowing the additional information about $Y$ (up to time $t$) will not change how $X$ evolves after $t$. This seems to suggest that $X$ is somehow independent of $Y$, or $Y$ does not provide additional information on how $X$ will evolve. Is there more accurate (and hopefully non-trivial) characterization of such $Y$?
[Below is the old post, will delete later]
Suppose $X_t$ is a Markov process with respect to its natural filtration. Let $Y_t$ be another process. Let $\mathcal F_t=\mathcal F_t^{X,Y}$ be the filtration generated by $X_t,Y_t$ jointly.
Question: under what condition, is $X_t$ Markovian with respect to $\mathcal F_t$?
Intuitively, as long as $\mathcal F_t^Y$ does not have any "future" information about $X_t$, it should be OK. So I guess something like
$E(f(X_{t+s})|\mathcal{F}_t^Y)\in \mathcal F_t^X$
for any positive or bounded measurable $f$.
Thanks
[EDIT] Some interesting special cases: (1) $X$ and $Y$ are independent. (2) $X$ and $Y$ are jointly Markovian.
I'm not quite sure about (2). To be more specific, if $X$ is Markovian with respect to the natural filtration, and $(X,Y)$ is also Markovian with respect to the natural filtration, prove:
$X_t$ is Markovian with respect to $\mathcal F_t^{X,Y}$
[EDIT] Sorry, keep changing my mind. I guess the above is not true, but need a counter example.
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