# Preservation of the Markov Property under Conditioning

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 $$\mathcal{G}_t:=\sigma(\{Z_s\}_{0\leq s, for each $$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 $$\kappa$$?

• I wonder if there is a typo in the first question, as $(X_t)$ need not be adapted to $(\mathcal G_t)$. Jun 22 '21 at 18:25

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.
• Amazing example, but if I left out $f$ and instead considered $(\mathbb{E}(X_t)|mathcal{G}_t])_t$ would this always be Markovian (since it seems the trucation of $\max\{0,\cdot\}$ causes the issue. Jun 22 '21 at 12:12
• That's right, here basically we use the fact that, if $(X_t)$ is Markov, then $(f(X_t))$ does not have to be Markov. Jun 22 '21 at 12:15