Is every killed Markov process still a Markov process?

Suppose we've got $X=(X(t))_{t\geq 0}$. $X$ is a strong Markov process with respect to filtration $\mathcal{F}_t$, taking values in some subset of $\mathbb{R}$. We take $\tau$ - a stopping time w.r.t $\mathcal{F}_t$ and we kill $X$ in $\tau$ obtaining new process $Y$ which is given by

$$Y(t)=\left\{ \begin{array}{ll} X(t), & \textrm{ for } t<\tau\\ \Delta, & \textrm{ for } t\geq \tau \end{array} \right..$$ where $\Delta$ is an isolated point.

We can then ask if $Y$ is still strong Markov process. There are a lot examples when it is (e.g. when $\tau$ is terminal time) however I am interested in an example when it isn't. I've got simple examples when killing destroys some other properties (e.g time-homogenity) but Markov property seems not so easy to eliminate.

I would be very thankful for any ideas.

• Try a stopping time that depends on the path of $X$ and not just its current state (like the 2nd hitting time of $0$ for a random walk). – pgassiat May 11 '12 at 12:49
• I think that it is always true for a continuous time Markov chain (assuming that paths are right continuous). – angry_pacifist Jun 14 '12 at 16:17
• @angry_pacifist: Obviously not, by the previous comment. – Did Aug 6 '12 at 11:34