# Markov and strong Markov properties

In my study of strong Markov property of an RCLL canonical Markov process I encounter the following definition:

Suppose $Y_t:\omega\rightarrow \omega(t)$ is canonical Markov process with respect to its raw filtration $\mathbb{F}^0$, with transition probability kernel $(K_t)_t$, where $\omega$ is RCLL function of $t$ taking values in polish space S, then $Y$ has strong Markov property wrt to the RC filtration $\mathbb{F}$ if for all $\mathbb{F}$-stopping time $\tau$,

$$E[f(Y_{\cdot+\tau})\mid\mathcal{F}_\tau]=E_{Y_\tau}[f(Y)]$$ on $\{\tau<\infty\}$, where $E_\mu$ refer to taking expectation conditional on initial value $\mu$ and the same transition probability, and $f(\omega)$ bounded measurable real positive RV of the sample path.

Then there is this claim that the above condition is equivalent to for all $\mathbb{F}$-stopping time $\tau$,

$$E[f(Y_{\cdot+\tau}){\bf 1}_{\tau<\infty}]=E[E_{Y_\tau}[f(Y)]{\bf 1}_{\tau<\infty}],$$

i.e., the first equation only need to hold in integrated form. I tried to prove this claim but was not able to make any progresses. Any suggestions? Thank you.

• Is it not just taking the expectation of both sides and applying iterated conditioning? Oh wait, that holds in one direction. Are you sure about the equivalency part? Aug 5, 2015 at 12:18
• I think so. In the next step this equivalence was used to prove RCLL Feller processes have strong MP, using countable approximations to stopping times and the regular MP. Aug 5, 2015 at 13:15
• @user138668 By finite stopping time, do you mean it only takes finitely many values? Or do you mean a bounded stopping time? There is a big difference and I think bounded is what you want. Could you tell us the name of your textbook?
– user940
Aug 5, 2015 at 13:51
• @ByronSchmuland, Yes I refer to bounded stopping times. This is from our lecture notes rather than a textbook, many lemmas are left as exercises. Aug 5, 2015 at 14:08

Let $\tau$ be a bounded $\mathbb{F}$-stopping time and $F \in \mathcal{F}_{\tau}$. We define a new stopping time by setting

$$\varrho(\omega) := \begin{cases} \tau(\omega), & \omega \in F, \\ \infty, &\text{otherwise} \end{cases}.$$

Then, by assumption,

$$\mathbb{E}(f(Y_{\cdot+\varrho}) 1_{\{\varrho<\infty\}}) = \mathbb{E}(\mathbb{E}_{Y_{\varrho}}(f(Y)) 1_{\{\varrho<\infty\}}),$$

i.e.

$$\mathbb{E}(f(Y_{\cdot+\tau}) 1_F) = \mathbb{E}(\mathbb{E}_{Y_{\tau}}(f(Y)) 1_F).$$

Since this holds for any $F \in \mathcal{F}_{\tau}$, we conclude

$$\mathbb{E}(f(Y_{\cdot+\tau}) \mid \mathcal{F}_{\tau}) = \mathbb{E}_{Y_{\tau}}(f(Y)).$$

Remark: A very similar result holds true for martingales. In fact, an adapted integrable process $(M_t,\mathcal{F}_t)_{t \geq 0}$ is a martingale if and only if

$$\mathbb{E}(M_{\varrho}) = \mathbb{E}(M_{\tau})$$

for all bounded ($\mathcal{F}_t)$-stopping times $\varrho$ and $\tau$.

• Your $\varrho$ is not a stopping time. Usually we set $\varrho$ to be equal to $\tau$ on $F$, and $\infty$ otherwise, not zero otherwise. With $\infty$, this new stopping time is not bounded, or finite (whichever of these the OP wants)
– user940
Aug 5, 2015 at 14:04
• @ByronSchmuland, sorry, I made a bad mistake in the statement of the problem, causing your confusion. Now it has been fixed. Aug 5, 2015 at 14:58
• @saz, following your approach with Byron Schmuland's comments in mind, I was able to fix the problem statement and make the proof. Thank you guys very much! Aug 5, 2015 at 15:29
• @ByronSchmuland Argh, you are right of course; thanks for pointing this out.
– saz
Aug 5, 2015 at 16:40
• @saz +1 Glad we got it all straightened out!
– user940
Aug 5, 2015 at 17:04