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 $\kappa$?

  • $\begingroup$ I wonder if there is a typo in the first question, as $(X_t)$ need not be adapted to $(\mathcal G_t)$. $\endgroup$ Jun 22, 2021 at 18:25

1 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.

  • $\begingroup$ 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. $\endgroup$
    – Joe_Affine
    Jun 22, 2021 at 12:12
  • $\begingroup$ 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. $\endgroup$ Jun 22, 2021 at 12:15
  • $\begingroup$ Do you know were I can find a reference to this later fact? $\endgroup$
    – Joe_Affine
    Jun 22, 2021 at 12:26
  • 1
    $\begingroup$ @Joe_Affine : It probably should be an exercise/remark in some book on Markov processes/chains, but I don't know such a reference. $\endgroup$ Jun 22, 2021 at 12:36

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