# 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). Commented Feb 21, 2018 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. Commented Feb 21, 2018 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. Commented Feb 21, 2018 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. Commented Feb 21, 2018 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. Commented Feb 21, 2018 at 21:45

(I'll only do the case in which $$S:\Omega\to[0,\infty)$$. A truncation argument reduces the general case to this special case.)
Consider a bounded $$\mathcal F\otimes \mathcal B[0,\infty)$$ measurable function $$F$$. Then for $$\mathcal F_\tau$$ measurable $$S:\Omega\to[0,\infty)$$, $$\Bbb E[F(\cdot,S) \mid\mathcal F_\tau](\omega) = \Bbb E[F(\cdot,u) \mid\mathcal F_\tau](\omega)\Big|_{u=S(\omega)},$$ for $$\Bbb P$$-a.e. $$\omega$$ in $$\{\tau<\infty\}$$. This is an evident consequence of the "usual" SMP as displayed in the question provided $$F$$ is of the form $$F(\omega,u)=G(\omega)H(u)$$. The assertion then follows for general $$F$$ by the functional form of the monotone class theorem. To finish, take $$F(\omega,u):=1_A(X_{\tau(\omega)+u}(\omega))$$, which has the required measurability because $$X$$ is progressive, by hypothesis.
• Thanks. You said that your display equation is a consequence of SMP for $F$ of product form; but SMP is a property of the stochastic process $(X_t)$, which does not appear in your display equation. But more importantly, the RHS of your display makes no sense: for each $u$, the conditional expectation is only defined up to $\mathbb{P}$-almost sure equality; and so once the $u$-input becomes $\omega$-dependent, null-set modifications of this $u$-dependent conditional expectation can make the overall RHS equal to literally any $\mathcal{F}_\tau$-measurable random variable that you like. Commented Aug 1, 2023 at 20:20
• Good points. 1. The displayed equation is general and doesn't depend on SMP. SMP only comes in when applying it to $F$ of the form $1_A(X_{\tau(\omega)+u}(\omega))$. Commented Aug 1, 2023 at 23:01
• 2 Also, you'll need to use a regular conditional distribution of $\Bbb P$, given $\mathcal F_\tau$, call it $\Bbb P_\tau(\omega,\cdot)$. Then use $$\int_\Omega F(\omega',u) \Bbb P_\tau(\omega,d\omega')$$ as a version of $\Bbb E[F(\cdot,u)\mid \mathcal F_\tau](\omega)$. Some mild regularity of $(\Omega,\mathcal F,\Bbb P)$ is needed for this. For example, if $\Omega$ is a suitable path space, as is the custom in Markov process theory. Commented Aug 1, 2023 at 23:01