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Let $X$ be a Markov chain with state space $\mathcal{S}$ and denote $\mathbb{N} := \{0,1, \cdots\}$. I need to show that for any stopping time $\tau < \infty$ and any bounded measurable function $\phi : \mathcal{S} \longrightarrow \mathbb{R}$, there exists a function $\psi: \mathbb{N}\times \mathcal{S} \longrightarrow \mathbb{R}$ such that:

$\mathbb{E}[\phi(X_{\tau +1}) | \mathcal{F}_{\tau}] = \psi(\tau, X_{\tau})$

Here is the way I tried to show this, using iterated property of expectation:

\begin{align} \mathbb{E}[\phi(X_{\tau +1}) | \mathcal{F}_{\tau}] &= \mathbb{E}[ \; \mathbb{E}[\phi(X_{\tau +1}) \; | \; \tau, \; \mathcal{F}_{\tau}] \; | \; \mathcal{F}_{\tau}] \\ & = \mathbb{E}[ \; \mathbb{E}[\mathbb{1}_{\{\tau = n\}} \;\phi(X_{\tau +1}) \; | \; \tau, \;\mathcal{F}_{\tau}] \; | \; \mathcal{F}_{\tau}] \\ &= \mathbb{E}[ \; \mathbb{1}_{\{\tau = n\}} \; \mathbb{E}[\phi(X_{n+1}) \; | \; \mathcal{F}_n] \; | \; \mathcal{F}_{\tau}] \\ &= \mathbb{E}[\; \mathbb{1}_{\{\tau = n\}} \;h(X_n) \; | \; \mathcal{F}_{\tau}]\\ &= \mathbb{1}_{\{\tau = n\}} \; h(X_n) \; , \; \forall n \\ &= \psi(X_{\tau}, \tau) \end{align}

I am not sure whether the way I tried to employ smoothing property of expectation is sound and correct. I wonder if you could provide your insights about my approach. Thanks.

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  • $\begingroup$ The step $\mathbb{E}[\phi(X_{\tau +1}) \; | \; \tau, \; \mathcal{F}_{\tau}]=\mathbb{E}[\mathbb{1}_{\{\tau = n\}} \;\phi(X_{\tau +1}) \; | \; \tau, \;\mathcal{F}_{\tau}]$ is obviously incorrect (what is $n$?). $\endgroup$
    – Did
    May 5, 2016 at 21:05
  • $\begingroup$ So I am writing the expression for an alternative $\tau = n$ to be able to apply Markov property and drop the unnecessary information in the filtration. How exactly that is wrong? $\endgroup$ May 6, 2016 at 18:04

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