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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 $0=t_0<\cdots<t_n$. 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 $n\in\mathbb N_0$.

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  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 $T=\tau\circ X$?

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  • $\begingroup$ 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.) $\endgroup$
    – 0xbadf00d
    Oct 27, 2021 at 9:16

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