2
$\begingroup$

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 $\kappa$?

$\endgroup$
1
  • $\begingroup$ 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 at 18:25
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 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 at 12:15
  • $\begingroup$ Do you know were I can find a reference to this later fact? $\endgroup$
    – Joe_Affine
    Jun 22 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 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.