# Relation between the strong Markov property of a process and the strong Markov property of the associated canonical process on the path space

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space;
• $$(E,\mathcal E)$$ be a measurable space;
• $$\pi_I$$ denote the projection from $$E^{[0,\:\infty)}$$ onto $$I\subseteq[0,\infty)$$ and $$\pi_t:=\pi_{\{t\}}$$ for $$t\ge0$$;
• $$(X_t)_{t\ge0}$$ be an $$(E,\mathcal E)$$-valued Markov process on $$(\Omega,\mathcal A,\operatorname P)$$ with transition family $$(\kappa_{s,\:t}:0\le s\le t)$$.

We know that there is a Markov kernel $$\kappa$$ with source $$(E,\mathcal E)$$ and target $$(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)})$$ such that $$\kappa(x,\;\cdot\;)\circ\pi_{t_0,\:\ldots\:,\:t_n}^{-1}=\delta_x\otimes\bigotimes_{i=1}^n\kappa_{t_{n-1},\:t_n}\tag1,$$ where $$\delta_x$$ is the Dirac measure at $$x$$, for all $$x\in E$$, $$n\in\mathbb N_0$$ and $$0=t_0<\cdots. Moreover, if $$\mu$$ is a probability measure on $$(E,\mathcal E)$$ and $$\operatorname P_\mu:=\mu\kappa$$, then $$(\pi_t)_{t\ge0}$$ is a Markov process on $$(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)},\operatorname P_\mu)$$ with initial distribution $$\mu$$ and transition family $$(\kappa_{s,\:t}:0\le s\le t)$$.

(So, choosing $$\mu=\mathcal L(X_0)$$, the process $$(\pi_t)_{t\ge0}$$ is somehow the canonical realization of $$(X_t)_{t\ge0}$$.)

Let $$\mathcal F^Y$$ denote the filtration generated by any process $$Y$$. Say that $$(X_t)_{t\ge0}$$ is strongly Markov on $$(\Omega,\mathcal A,\operatorname P)$$ if $$\operatorname E[f((X_{\tau+t})_{t\ge0})\mid\mathcal F^X_\tau]=(\kappa f)(X_\tau)\tag2$$ for all bounded $$\mathcal E^{\otimes[0,\:\infty)}$$-measurable $$f:E^{[0,\:\infty)}\to\mathbb R$$ and every finite $$(\mathcal F^X_t)_{t\ge0}$$-stopping time $$\tau$$ on $$(\Omega,\mathcal A)$$.

Are we able to show that $$(X_t)_{t\ge0}$$ is strongly Markov on $$(\Omega,\mathcal A,\operatorname P)$$ if and only if $$(\pi_t)_{t\ge0}$$ is strongly Markov on $$(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)},\operatorname P_\mu)$$ for every probability measure $$\mu$$ on $$(E,\mathcal E)$$?

Note that, by $$(1)$$, the distribution $$\mathcal L(X)$$ of the process $$(X_t)_{t\ge0}$$ is equal to $$\mathcal L(X)=\operatorname P_{\mathcal L(X_0)}\tag3.$$ Moreover, if $$\theta_s:E^{[0,\:\infty)}\to E^{[0,\:\infty)}\;,\;\;\;x\mapsto(x_{s+t})_{t\ge0}$$ for $$s\ge0$$, then the left-hand side of $$(2)$$ is equal to $$\operatorname E[f\circ\theta_\tau\circ X\mid\mathcal F^X_\tau]\tag4$$ and $$(\theta_\tau\circ X)^{-1}(B)=X^{-1}(\theta_\tau^{-1}(B))\;\;\;\text{for all }B\in\mathcal E^{\otimes[0,\:\infty)}\tag5.$$ Moreover, if $$\mu$$ is a probability measure on $$(E,\mathcal E)$$, then $$(\pi_t)_{t\ge0}$$ is stronlgy Markov at $$\tau$$ on $$(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)},\operatorname P_\mu)$$ iff $$\operatorname E_\mu\left[f\circ\theta_\sigma\mid\mathcal F^\pi_\sigma\right]=(\kappa f)(\pi_\sigma)=\operatorname E_{\pi_\sigma}[f]\tag6$$ for all bounded $$\mathcal E^{\otimes[0,\:\infty)}$$-measurable $$f:E^{[0,\:\infty)}\to\mathbb R$$ and every finite $$(\mathcal F^\pi)_{t\ge0}$$-stopping time $$\sigma$$ on $$(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)}$$, where $$\operatorname E_\mu$$ is the expectation associated to $$\operatorname P_\mu$$.

And it might be useful to note that, by definition, $$\mathcal E^{\otimes[0,\:\infty)}=\sigma(\pi_t,t\ge0)\tag7$$ and if $$\mathcal R_n:=\left\{B_0\times\cdots\times B_n:B_0,\ldots,B_n\in\mathcal E\right\},$$ then $$\sigma(\pi_{t_0},\ldots,\pi_{t_n})=\sigma(\pi^{-1}_{\{t_0,\:\ldots\:,\:t_n\}}(\mathcal R_n))\tag8$$ for all $$n\in\mathbb N_0$$.

1. How do we know that $$\kappa$$ exists?
2. You'll have no trouble chasing down the definitions to show that the strong Markov property on $$(\Omega,\mathcal A,\operatorname P)$$ implies the strong Markov property on the canonical space, for $$\mu=\mathcal L(X_0)$$ (under $$\operatorname P$$). The converse is dicier, the problem being that given a stopping time $$T$$ of $$(X_t)$$, how do you construct a stopping time $$\tau$$ on $$E^{[0,\infty)}$$ such that $$T=\tau\circ X$$?
• I'm sorry for my very late response. For some reason, I think I wasn't notified. (1.) We need to assume that $E$ is a Polish space to ensure that $\kappa$ exists. (2.) In order to avoid this problem, we might fix a stopping time $\tau$ on $E^{[0,\:\infty)}$ beforehand and reformulate the statement to: "$X$ is strongly Markov at $\tau\circ X$ iff $(\pi_t)_{t\ge0}$ is strongly Markov at $\tau$ with respect to $\operatorname P_\mu$ for all $\mu$. Do you think the equivalence can then be shown? (Please see also my related question: math.stackexchange.com/q/4288294/47771.) Oct 27, 2021 at 9:16