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Let $(X_{s},\mathcal{F}_{s})$ be a stochastic process adapted to a given filtration. I was told that, in order to prove that $X$ is Markov, it suffice to prove that for any nonnegative, Borel-measurable function $f$, $E[f(X_{t}) \vert \mathcal{F}_{s}]=g(X_{s}) \quad a.s.$ for $\textbf{some}$ Borel function $g$. I get confused here about the arbitrariness of the function g, shouldn't we expect that $g=P_{s,t} \circ f$, where $P_{s,t}$ is a transition function?

So I'm wondering if the above characterization for Markov processes is correct. I know this may be a stupid question. Any help or comment is greatly appreciated.

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    @Sasha: Yes, that definitely won't hold in general. (It holds iff $f$ is harmonic. In particular, for a finite-state irreducible Markov chain it will fail for every non-constant $f$.)2012-08-18

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Suppose that for $t\geq s$, we have some function $g$ so that $E[f(X_{t}) \,\vert\, \mathcal{F}_{s}]=g(X_{s}) \quad \mbox{a.s.}\tag1$ Conditioning on $X_s$ in (1) gives $E[f(X_{t}) \,\vert\, X_s]=g(X_{s}) \quad \mbox{a.s.}\tag2$ From this we deduce $E[f(X_{t}) \,\vert\, \mathcal{F}_{s}]=E[f(X_{t}) \,\vert\, X_s] \quad \mbox{a.s.}\tag3 $ Equation (3) is one of the equivalent ways of expressing the Markov property. Notice that the arbitrary function $g$ has dropped out of the picture.