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Suppose $\{ X_t: t \in \mathbb{R} \}$ is a stochastic process on a probability space $(\Omega, \mathcal{F}, P)$, and it is adapted to a filtration $\{\mathcal{F}_t \}$ on the probability space.

  1. $\{ X_t\}$ is said to have Markov property with respect to the filtration $\{\mathcal{F}_t \}$, if $\forall t \in \mathbb{R}$ and $\forall A \in \mathcal{F}_{\geq t}$, $$P(A \mid \mathcal{F}_t) = P(A \mid X_t) \text{ a.s.}.$$
  2. $\{ X_t\}$ is said to have Markov property with respect to its natural filtration $\{\mathcal{F}_{\leq t} \}$, if $\forall t \in \mathbb{R}$, $\forall A_1 \in \mathcal{F}_{\geq t}$ and $\forall A_2 \in \mathcal{F}_{\leq t}$, $$P(A_1 \cap A_2 \mid \mathcal{F}_{=t}) = P(A_1 \mid \mathcal{F}_{=t}) \, P(A_2 \mid \mathcal{F}_{=t}) \text{ a.s.}.$$

    ADDED: $\mathcal{F}_{\leq t}:= \sigma(\{ X_s: s \leq t \})$, $\mathcal{F}_{\geq t}:= \sigma(\{ X_s: s \geq t \})$ and $\mathcal{F}_{= t}:= \sigma( X_t )$.

I was wondering if it is possible to formulate Markov property with respect to the general filtration $\{\mathcal{F}_t \}$, in a way similar to that with respect to the natural filtration $\{\mathcal{F}_{\leq t}\}$ defined in 2?

If yes, why is this new definition equivalent to the definition in 1?

Any references?

Thanks in advance!

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You should add to your post the definitions of $\mathcal{F}_{\le t}$ and $\mathcal{F}_{\ge t}$. – Did May 4 '11 at 6:37
@Didier: Thank you! I just added those. Note that I was asking about Markov property wrt the general filtration instead of the natural filtration. Will you have some idea? – Ethan May 4 '11 at 11:58

I've copied this from page 2 of General Theory of Markov Processes by Michael Sharpe, with some changes in notation.

Suppose $$P(A_1\cap A_2\,|\, {\cal F}_{=t} )= P(A_1 \,|\, {\cal F}_{=t} )P( A_2\,|\, {\cal F}_{=t} )$$ for all $A_1\in {\cal F}_{\leq t}$ and $A_2\in {\cal F}_{\geq t}$. Using well known properties of conditional expectations, \begin{eqnarray*} P(A_1\cap A_2) &=&P(P(A_1\cap A_2\ |\ {\cal F}_{=t}))\cr &=&P\left( P(A_1\ |\ {\cal F}_{=t})\ P(A_2 \ | \ {\cal F}_{=t}) \right)\cr &=&P(P(A_2\ |\ {\cal F}_{=t}) ; A_1). \end{eqnarray*}

As $A_1\in {\cal F}_{\leq t}$ was arbitrary, it follows that $$P(A_2\ |\ {\cal F}_{\leq t} ) =P(A_2\ | \ {\cal F}_{=t} )$$ for every $A_2\in {\cal F}_{\geq t}$.

That is, prediction of future behavior of $X$ based on the entire past is only as valuable as the predictor based on the present value $X_t$ alone.

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Thanks a lot! But my question is about how to generalize Markov property with respect to a general filtration, from Markov property with respect to the natural filtration. – Ethan May 4 '11 at 3:22
@Ethan Whoops! I misread your question. I will think about it. – Byron Schmuland May 4 '11 at 3:25

If I understand your question correctly the answer is yes and I guess this would suffice as source (page 2 definition 1) he even proves it equivalent to your other definition:

Hope it's to some help.

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Conditional independence is the relevant concept. Let $\cal F$ be the natural filtration of $(X_t)$ and $\cal G \supset \cal F$ a bigger filtration. Then $(X_t)$ is Markovian with respect to $\cal G$ means that $\cal F_\infty$ is conditionally independent of $\cal G_t$ given $\sigma(X_t)$. You can check that 1 and 2 (the case when $\cal G=\cal F$) are equivalent to this property by using the classical characterizations of conditional independence.

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