# Does the strong Markov property imply the “really strong Markov” property?

Let $\mathbf{\Omega}=(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \in [0,\infty)},\mathbb{P})$ be a filtered probability space satisfying the Usual Conditions.

Let $P \colon [0,\infty) \times \mathbb{R} \times \mathcal{B}(\mathbb{R}) \to [0,1]$ be a family of Markov transition probabilities on $\mathbb{R}$ (that is: $P(t,x,\cdot)$ is a probability measure, $P(t,x,A)$ is measurable in $(t,x)$, and the Chapman-Kolmogorov relations hold).

Let $(X_t)_{t \geq 0}$ be a real-valued progressively measurable stochastic process over the filtered probability space $\mathbf{\Omega}$. Suppose that $X$ has the strong Markov property with transition probabilities $P$; that is, for every $(\mathcal{F}_t)$-stopping time $\tau \colon \Omega \to [0,\infty]$, every $t \geq 0$ and every $A \in \mathcal{B}(\mathbb{R})$, $$\mathbb{P}(X_{\tau+t} \in A | \mathcal{F}_\tau) \ = \ P(t,X_\tau,A) \hspace{4mm} \mathbb{P}\textrm{-a.e. on }\{\tau < \infty\}.$$

Is it necessarily the case that for every $(\mathcal{F}_t)$-stopping time $\tau \colon \Omega \to [0,\infty]$, for every $\mathcal{F}_\tau$-measurable function $s \colon \Omega \to [0,\infty]$ and every $A \in \mathcal{B}(\mathbb{R})$, $$\mathbb{P}(X_{\tau+s} \in A | \mathcal{F}_\tau) \ = \ P(s,X_\tau,A) \hspace{4mm} \mathbb{P}\textrm{-a.e. on }\{\tau,s < \infty\}\,?$$

(I once saw a paper that used the "strong Markov property" at a certain point, but at this point it actually seemed to be using the above "really strong Markov" property. In the context, it was actually fairly easy to prove the necessary claim just by restricting to rational times, since everything was continuous and the sets in question were open/closed; but still, it would be interesting to know if the logic holds in general.)

• I believe that the answer is yes, under reasonable conditions. For Feller processes this is given as Theorem 3 in Section 2.3 of Chung–Walsh. (This is by far my favourite book on Markov processes. For example, the authors point out that this "really strong Markov property" is precisely what is needed for the reflection principle). – Mateusz Kwaśnicki Feb 21 '18 at 20:20
• Certainly I expect it to be true of Feller processes (assuming right-continuity of sample paths), since basically everything is as continuous as can be, and so one can use approximations by rationals. But I don't really want "reasonable conditions" of a topological nature, as the concepts of Markov / strong Markov / really strong Markov make no reference whatsoever to topological notions. – Julian Newman Feb 21 '18 at 21:04
• The point is that we often make a big deal of the strong Markov property as though it truly captures the idea of the future probabilities only depending on the present where the mere Markov property fails to do so; but if strong Markov doesn't imply "really strong Markov", then perhaps we should make less big a deal of the "strong Markov" property as it is currently defined, and use the stronger definition as our main definition of the "stronger version of the Markov property". And as you say, we can still prove that cadlag Feller processes have this stronger property. – Julian Newman Feb 21 '18 at 21:13
• But all of this would be pointless if the strong Markov property as currently defined really does imply the "really strong Markov" property, which is why I ask the question. – Julian Newman Feb 21 '18 at 21:14
• While I do not know the answer to your question, I do not think people that work with Markov processes ever use this property without further (semi)-topological conditions, like those in definitions of Feller, standard Markov, Hunt or Ray processes. Also, note that while your statement of the strong Markov property appears to be purely measure-theoretic, it implies, for example, the Blumenthal 0–1 law and measurability of $X_\tau$ w.r.t. $\mathcal{F}_\tau$ (or perhaps $\mathcal{F}_{\tau+}$?), which are rather problematic when the topology of the state space is completely ignored. – Mateusz Kwaśnicki Feb 21 '18 at 21:45