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Statement

We say $\Phi$ has the strong Markov Property if for any initial distribution $\mu$, any real-valued bounded measurable function $h$ on $\Omega$, and any stopping time $\zeta$, $$ \mathsf{E}_{\mu}[\theta^\zeta H \mid \mathcal{F}_\zeta^\Phi] = \mathsf{E}_{\Phi_\zeta}[H] \hspace{10mm} \mathsf{P}_\mu \text{-a.s.} $$ on the set $\{\zeta < \infty\}$.

Proof

\begin{align} \mathsf{E}_{\mu}[\theta^\zeta H \mid \mathcal{F}_\zeta^\Phi] &= \mathsf{E}_{\mu}[\theta^\zeta H \sum_{k \ge 1} \mathbb{I}(\zeta = k) \mid \mathcal{F}_\zeta^\Phi] \tag{mult by $1$}\\ &= \sum_{k \ge 1} \mathsf{E}_{\mu}[\theta^\zeta H \mathbb{I}(\zeta = k) \mid \mathcal{F}_\zeta^\Phi] \tag{linearity} \\ &= \sum_{k \ge 1} \mathsf{E}_{\mu}[\theta^k H \mathbb{I}(\zeta = k) \mid \mathcal{F}_\zeta^\Phi] \tag{$\theta^\zeta = \theta^k$ on $\{\zeta = k\}$ by defn.} \\ &= \sum_{k \ge 1} \mathbb{I}(\zeta = k) \mathsf{E}_{\mu}[\theta^k H \mid \mathcal{F}_\zeta^\Phi] \tag{indicator is $\mathcal{F}_\zeta^\Phi$-measurable} \\ &= \sum_{k \ge 1} \mathbb{I}(\zeta = k) \mathsf{E}_{\mu}[\theta^k H \mid \mathcal{F}_k^\Phi] \tag{?} \\ &= \sum_{k \ge 1} \mathbb{I}(\zeta = k) \mathsf{E}_{\Phi_k}[ H] \tag{(the other) Markov property} \\ &= \sum_{k \ge 1} \mathbb{I}(\zeta = k) \mathsf{E}_{\Phi_\zeta}[ H] \tag{?} \\ &= \mathsf{E}_{\Phi_\zeta}[ \sum_{k \ge 1} \mathbb{I}(\zeta = k) H] \\ &= \mathsf{E}_{\Phi_\zeta}[H] \end{align}

Definitions

  • $\Phi = \{\Phi_0, \Phi_1, \ldots\}$ is the chain defined on $\Omega = \mathsf{X}^{\infty}$
  • $\mu$ is the chain's initial distribution on $(\mathsf{X}, \mathcal{X})$
  • $\mathsf{E}_\mu$ and $\mathsf{P}_\mu$ are the expectation and probability measure defined on the entire product measurable space $(\Omega, \mathcal{F})$
  • $\zeta$ is the stopping time
  • $\mathcal{F}_{\zeta}^\Phi$ is the sigma-field that describes all the events that happen up to time $\zeta$
  • $\theta^{\zeta} : \{x_0, x_1, \ldots\} \mapsto \{x_{\zeta}, x_{\zeta+1}, \ldots\}$ is the (random) shift operator
  • $H = h(\Phi_0, \Phi_1, \ldots)$ is a random variable made from the bounded measurable function $h$
  • $\theta^\zeta H = H \circ \theta^\zeta(\omega)$ is the function $h$ applied to randomly-delayed chain

Questions And Comments

I can do $$ \mathbb{I}(\zeta = k) \mathsf{E}_{\mu}[\theta^k H \mid \mathcal{F}_\zeta^\Phi] = \mathbb{I}(\zeta = k) \mathsf{E}_{\mu}[\theta^k H \mid \mathcal{F}_k^\Phi] $$ because

$$ \int_A \mathbb{I}(\zeta = k) \mathsf{E}_{\mu}[\theta^k H \mid \mathcal{F}_\zeta^\Phi] \mathsf{P}_\mu(d\omega) = \int_A \mathbb{I}(\zeta = k) \mathsf{E}_{\mu}[\theta^k H \mid \mathcal{F}_k^\Phi] \mathsf{P}_\mu(d\omega) \tag{*} $$ for any $A \in \mathcal{F}_\zeta^\Phi$, right? By definition of $\mathcal{F}_\zeta^\Phi$ $$ A \cap \{\zeta = k\} \in \mathcal{F}_k $$ so $\mathbb{I}(\omega \in A)\mathbb{I}(\zeta = k)$ is $\mathcal{F}_k$-measurable. This gives us that both sides of $(*)$ equal $\mathsf{E}_{\mu}[\mathbb{I}(\omega \in A)\theta^k H ]$.

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  • $\begingroup$ You write $$ \mathsf{E}_{\mu}[\theta^\zeta H \sum_{k \ge 1} \mathbb{I}(\zeta = k) \mid \mathcal{F}_\zeta^\Phi] $$ but what is $\zeta$ to which $k$ can be compared in the indicator function? $\endgroup$ Commented Nov 17, 2020 at 13:09
  • $\begingroup$ @SextusEmpiricus it's the stopping time. I'm doing the thing where I partition up $\Omega$ and multiply by $\mathbb{I}(A) + \mathbb{I}(A^c)$ $\endgroup$
    – Taylor
    Commented Nov 17, 2020 at 14:49
  • $\begingroup$ But isn't the stopping time multivalued? For instance, if you have a Brownian motion and consider the stopping time the time when the particle hits some boundary or passes some level, what is then $\mathbb{I}(\zeta = k)$? Which value do you put in? $\endgroup$ Commented Nov 17, 2020 at 15:08
  • $\begingroup$ @SextusEmpiricus this is a discrete-time Markov chain, and so this is a countable sum. Random variables that take into account stopping times are defined piecewise on individual sets. On the set $\{\zeta = k\} \subseteq \Omega$, $\zeta$ evaluates to $k$ $\endgroup$
    – Taylor
    Commented Nov 17, 2020 at 15:48
  • $\begingroup$ So you are looking for the property $\sum_{k} \delta_{k,\zeta} \mathsf{E}_{\mu}[\theta^\zeta H \mid \mathcal{F}_\zeta^\Phi] = \mathsf{E}_{\mu}[\theta^k H \mid \mathcal{F}_k^\Phi]$ with $\delta_{k,\zeta}$ the Kronecker delta. $\endgroup$ Commented Nov 17, 2020 at 15:55

1 Answer 1

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The two question marked steps are indeed correct. This proof also follows the proof given for Proposition 3.4.6 here. Note that the assumption of $P(\tau = \infty) = 0$ comes into play when we sum over $1,2,\ldots$.

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