Preservation of the Markov Property under Conditioning

Question

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 $\kappa$?

Answer 1

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.