# Markov process that is not Markov wrt the right-continuous filtration

Let $(\Omega, \mathcal{A}, P)$ be a probability space, $(E, \mathcal{E})$ a measurable space and $X_t : \Omega \to E$ a continuous-time Markov process meaning that $P(X_{t+s} \mid \mathcal{F}_t) = P(X_{t+s} \mid X_t)$ a.s., where $\mathcal{F}_t = \sigma(X_u \mid u \leq t)$ is the canonical filtration of $X_t$. In general, the filtration $\mathcal{F}_t$ need not be right-continuous. If $X_t$ is a Feller process on a locally compact second-countable Hausdorff space $E$ (with $\mathcal{E}$ the Borel $\sigma$-algebra), then it is know that $\mathcal{F}_t$ is right-continuous.

Is there an example of a Markov process $X_t$ (relative to its own filtration $\mathcal{F}_t$) such that it is not Markov relative to its right-continuous filtration $\mathcal{F}_{t+}$?

Note: The filtration $(\mathcal F_t)$ is right continuous (when $(X_t)$ is Feller) only if augmented by the addition of the appropriate null sets. This is true even for Brownian motion.
Example: The state space is the subset of $\Bbb R^2$ comprising the rays $\{(0,x): x<0\}$, $\{(x,x): x\ge 0\}$, and $\{(x,-x): x\ge 0\}$. Suppose $X_0=-1$ (for example). The process moves to the right at rate 1 until reaching $(0,0)$. At that point a fair coin is tossed; if Heads the process continues along the upper ray $\{(x,x): x\ge 0\}$ at unit rate, and if Tails then likewise along the lower ray. This process is (simple) Markov with respect to $(\mathcal F_t)$ but not with respect to $(\mathcal F_{t+})$. This because $\mathcal F_{1+}$ contains information about which ray is followed at the split, but $\mathcal F_1$ does not.
• Thank you for the example. I think, I got the idea. One can also have a Markov chain on 3 states, starting in $1$ at time $0$ and immediately jumping at time $0+$ to state $2$ or $3$ (and stay there forever), i.e. $P(t) = ((0,1/2,1/2),(0,1,0),(0,0,1))$ for $t > 0$. You are right about the fact that the completed canonical filtration of a Feller process is right-continuous, but not necessarily the raw canonical filtration. – yadaddy Jul 14 '18 at 18:47