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I've seen definitions of Markov Property for a process $X$ indexed with the positive integers with values in $S$, that are supposedly equavalent. I consider the canonical filtration $\mathcal F=(\mathcal F_n)_{n\geq 0}$ generated by the process.

$1)\ \ \mathbb P(X_{n+1}=s\ \ |\mathcal F_n)=\mathbb P(X_{n+1}=s|X_n)$

$2)\ \ \mathbb E[f(X_{n+1})|\mathcal F_n]=\mathbb E[f(X_{n+1})|X_n]$ for all bounded functions $f:S\rightarrow \mathbb R$.

$2\rightarrow 1$ holds trivially on finite $S$. But what is the idea in the general case?

Do I have to go through the usual thing of first proving it for indicator functions, simple functions and limits of simple functions and dominated convergence? Or is there a simpler way?

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    $\begingroup$ Where did you read that (1) is an equivalent definition of the Markov property of a process? $\endgroup$
    – Snoop
    Commented Nov 19, 2022 at 17:38
  • $\begingroup$ I'm sorry, i was thinking of indicator functions, but didn't write it... I corrected it above. $\endgroup$
    – Peter Strouvelle
    Commented Nov 19, 2022 at 18:21

1 Answer 1

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For any bounded map $f:S\to\mathbb R$ and $\sigma$-algebra $\mathcal G$, $$ \mathbb E[f(X_{n+1})\vert\mathcal G]=\sum_{s\in S}f(s)\mathbb P(X_{n+1}=s\vert\mathcal G), $$ from which you can easily deduce the equivalence.

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