1
$\begingroup$

Let $(\mathbb{P}_x)_{x \in \mathbb{R}^d}$ be a continuous Markov process. So the $\mathbb{P}_x$ are the laws on the space of continuous function from the positive real half line to $\mathbb{R}^d$, subsequently denoted by $\Omega$. In the book by Stroock\Varadhan on multidimensional diffusions, they define the strong Markov property as follows:

For every $x \in \mathbb{R}^d$ and every finite stopping time $\tau$, the family $(\delta_{\omega} \otimes_{\tau(\omega)}\mathbb{P_{\pi_{\tau(\omega)}(\omega)}})_{\omega \in \Omega}$ is a regular conditional probability distribution of $\mathbb{P}_x$ w.r.t. $\mathcal{F}_{\tau}$, where the latter one denotes the usual $\sigma$-field of the $\tau$-past. $\pi$ denotes the canonical projection from the path space to the value at time $t$.

For some $\omega \in \Omega$ (i.e. a continuous path), a time $t \geq 0$ and a probability measure $\mu$ on $\Omega$, the measure $\delta_{\omega}\otimes_{t}\mu$ is defined as follows: On sets $A \in \mathcal{F}_t$, it is defined as $\delta_{\omega}(A)$ and on $B \in \sigma(\pi_r, r \geq t)$, it is given by $\mu(B)$. Intuitively, this means that the measure gives mass to paths only if they look like the reference path $\omega$ until time $t$ and after time $t$, the measurement of paths is inherited by the action of $\mu$.

If you spell this out, you can conclude $$\mathbb{E}_x[\phi|\mathcal{F}_{\tau}] = \mathbb{E}_{\delta_{\omega}\otimes_{\tau(\omega)}\mathbb{P}_{\pi_{\tau(\omega)}(\omega)}}[\phi] \,\,\,a.s.$$ for all measurable and bd. $\phi:\Omega \to \mathbb{R}$. (How) Is this equivalent to the "usual" definition of the strong Markov property, i.e. $$\mathbb{E}_x[\phi\circ \mathcal{v}_{\tau}|\mathcal{F}_{\tau}] = \mathbb{E}_{\pi_{\tau}}[\phi]\,\, a.s.?$$ Intuitively it seems close, but I am not able to make the rigorous connection. I would apprectiate any help!

$\endgroup$
  • $\begingroup$ I'd be grateful for any input on this! $\endgroup$ – Marco Mar 1 at 9:53

Your Answer

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

Browse other questions tagged or ask your own question.