4
$\begingroup$

A Markov process $X_t$ on $E$ is a Feller-Dynkin (or sometimes just Feller) process if its semigroup is a strongly continuous, sub-Markov semigroup $\{P_t:t\geq 0\}$ of linear operators on $C_0(E)$ (my definition comes from the text of Rogers and Williams or see Wikipedia).

I am wondering what is known about when a time change of a Feller-Dynkin process is still Feller-Dynkin. I believe that I have read that this is not true in general (and can be a tricky thing to determine). I would be curious to see an example where this fails and also to know if the Feller-Dynkin property is preserved if the time change is nice enough.

For what it's worth, the time changes I am interested in are actually $C^{\infty}$. I do not need the Feller property to be preserved, but thinking about it made me realize I do not have a very good feeling for these sort of things.

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ By the so-called Dambis, Dubins-Schwarz theorem every continous local martingale is a time-changed Wiener process. Now it is enough to take your favourite martingale which is not a Markov process. $\endgroup$
    – Piotr Miłoś
    Jan 29, 2012 at 10:57
  • $\begingroup$ Piotr Miłoś: as you gave the answer in your comment, I would recommend that you add an answer to this question so that Shawn can accept it. Best Regards $\endgroup$
    – The Bridge
    Jan 30, 2012 at 13:15
  • $\begingroup$ Yes, this sort of idea seems to be a useful source of counterexamples. $\endgroup$
    – ShawnD
    Feb 14, 2012 at 1:16

1 Answer 1

Reset to default
1
$\begingroup$

If you time change $X_t$ using a "clock" of the form $B_t = \int_0^t b(X_s) ds$, with $b$ a bounded positive continuous function on the state space of $X_t$ that is also bounded below away from $0$, then the time-changed process will be a Feller process. This follows directly from the Hille-Yosida theorem.

Improve this answer
$\endgroup$
1
  • $\begingroup$ This is the type of thing I was looking for--now I'll try to convince myself I understand details. $\endgroup$
    – ShawnD
    Feb 1, 2012 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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