Let $(X\_t,Z\_t)\_t$ be an $\mathbb{R}^{n}\times \mathbb{R}^m$-valued time-homogeneous Markov process on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}\_t)\_t,\mathbb{P})$ with transition kernel $\kappa$ and where $\mathcal{F}\_t$ is the right-continuous filtration generated by this process. Let $\mathcal{G}\_t:=\sigma(\{Z\_s\}\_{0\leq s<t})$, for each $t\geq 0$. Then, is $X\_t$ still Markovian under the smaller filteration $(\mathcal{G}\_t)\_{t}$?  

Let $f\in C(\mathbb{R}^n,\mathbb{R})$.


Is the process
$
(\mathbb{E}\left[f(X\_t)|\mathcal{G}\_t\right])\_{t\geq 0}
$ Markovian on the reduced space $(\Omega,(\mathcal{G}\_t)\_t,\mathbb{P})$?


If so, how is the Markov kernel of this process related to $\kappa$?



Preservation of the Markov Property under Conditioning
No. E.g., let $n=m=1$ and $X\_t=Z\_t=B\_t$, where $B$ is the standard Brownian motion. Take the natural filtrations, so that
$E(f(X\_t)|\mathcal G\_t)=f(B\_t)$. Let $f(x)$ to be something like $\max(0,x)$. It should be easy to show that the process $(f(B\_t))$ is not Markov.


Indeed, to simplify calculations, let $f(x):=1(x>0)$. Then
$$P(f(B\_3)=1|f(B\_2)=0)=\frac{1}{2}-\frac{\tan ^{-1}\left(\sqrt{2}\right)}{\pi }=0.19591\ldots
\ne\frac16=P(f(B\_3)=1|f(B\_2)=0,f(B\_1)=0).$$
If one insists that $f$ be continuous, this may be achieved by approximation.


