Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In our lecture on Brownian motion & stochastic calculus we proved:

If $ X $ is a canonical RCLL process having the strong Markov property and $ \tau $ is a stopping time with $ \tau < + \infty, P-a.s $, then the process $ \sideset{^\tau}{}X $ (started at $ \tau $ ) has the weak Markov property.

After the proof there is a remark:

In particular, if $ X $ is Feller, then so is $ \sideset{^\tau}{}X $, and therefore $ \sideset{^\tau}{}X $ is again a strong Markov process.

Now, my question is:

If $ X $ is a strong Markov process, isn't $ \sideset{^\tau}{}X $ a strong Markov process in any case? Or, to put it differently, is there a mistake in my proof (see below)?

Let $ \sigma $ be a stopping time, let $ f \geq 0 $ be a measurable function on the state space, and let $ \mathcal{G}_\sigma := \mathcal{F}_{\tau + \sigma} $. Then,

$ E[f( \sideset{^\tau}{_{\sigma+h}}X) | \mathcal{G}_{\sigma} ] = E[f(X_{\tau + \sigma + h}) | \mathcal{F}_{\tau + \sigma}] = E[f(X_{\tau + \sigma + h}) | \sigma (X_{\tau + \sigma})] = E[f(\sideset{^\tau}{_{\sigma + h}}X) | \sigma (\sideset{^\tau}{_\sigma}X)], $

$P-a.s.$ on $ \{ \tau + \sigma < + \infty \} $ (where the second equality follows from the strong markov property applied to $ \tau + \sigma $). And since $ \tau < + \infty, P-a.s$, the above holds even $P-a.s.$ on $ \{ \sigma < + \infty \} $, which is exactly the strong Markov property for the process $ \sideset{^\tau}{}X $.

Thanks for you help!

Regards, Si

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.