$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ > [!idea] Informal > Markov Properties take the form, "if after some (stopping) time $T$, we are at $X_T$, then $X_{T+s}$ forever looks like $X_s$." This is a statement about distributions on the space of (cadlag) processes. Hence, we will recall definitions, and then phrase our results in terms of expectations of test functions (measurable $\Phi: D(E)\to \RR_+$). Let $(Q_t)_{t\geq 0}$ be a possibly non-Feller [[Definition and Existence of Markov Processes|transition semigroup]] on some metric space $E$. For each $x\in E$, suppose we can construct path-cadlag Markov processes $(X^x_t)_{t\geq 0}$; this follows immediately if $Q$ is Feller. Now, the [[cadlag|space of all cadlag paths]] is called $D(E)$, and we can get pushforward laws $\PP_x$ for each $X^x$. (Note that even if multiple $X^x$ exist corresponding to some $Q$, the finite-dimensional marginals are completely determined by $Q$, so these laws $\PP_x$ are really a property of $Q$, not of $X^x$). >[!claim] >$\EE_x$ denotes expectation with respect to $\PP_x$. For any measurable $\Phi: D(E)\to \RR_+$, $x\mapsto \EE_x[\Phi]$ is measurable. >[!proof]- Easy cylinder bootstrap > It suffices to consider $\Phi = \id_A$ where $A\in D(E)$ is a cylinder set. In this case,$\PP_x\left[\id_{X^x_{t_1}\in A_1}\dots \id_{X^x_{t_p}\in A_p}\right]$has the un-enlightening explicit formula$\int_E \id_{x_1\in A_1}Q_{t_1}(x,dx_1)\dots \int_E \id_{x_t\in A_t}Q_{t_p - t_{p-1}}(x_{p-1},dx_p)$which is clearly $E$-measurable (by definition). The cylinder sets generate $D(E)$ and the indicators generate measurable $D(E)\to \RR_+$ so we conclude. >[!theorem] Simple Markov Property >Let $s\geq 0$ and $\Phi: D(E)\to \RR_+$ be measurable. Let $Y$ be a specific $Q$ $(\FFF_t)$ path-cadlag Markov process. Then,$\EE\left[\Phi\left(\left(Y_{s+t}\right)_{t\geq 0}\right) | \FFF_s\right] = \EE_{Y_s}[\Phi].$ Type-checking, both sides are $\FFF_s$-measurable. Actually, the full proof is also an easy cylinder bootstrap. >[!proof]- Another easy cylinder boostrap > For any $\phi_1,\dots, \phi_p\in B(E)$ (bounded measurable functions with $\infty$ norm), obviously$\EE\left[\phi_1(Y_{s + t_1})\dots | \FFF_s\right] = \int Q_{t_1}(Y_s, dx_1) \phi_q(x_1)\dots \int Q_{t_p - t_{p-1}}(x_{p-1}, dx_p) \phi_p(x_p).$So we can specialize to cylinders $\Phi = \id_A$ and win. # Strong Markov Property This time, make sure $(Q_t)_{t\geq 0}$ is a Feller subgroup. > [!theorem] Strong Markov Property > Let $T$ be a stopping time of $(\FFF_{t+})$ and $\Phi: D(E)\to \RR_+$ be measurable. Then,$\EE\left[\id_{T < \infty} \Phi\left(\left(Y_{T+t}\right)_{t\geq 0}\right) | \FFF_T\right] = \id_{T < \infty} \EE_{Y_T}[\Phi].$ Note that $T$ is a stopping time of $\FFF_{t+}$, which is extra sweet. Note that $\omega\mapsto Y_T(\omega)$ is $\FFF_T$-measurable because we are [[Index Progressive Process at Stopping Time|indexing a progressive process at a stopping time]]. Proof boring, details omitted, use dyadics to employ cadlag.