Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

3 Answers 3

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.

share|improve this answer
    
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:

http://books.google.co.uk/books?id=0o0r21x1Ns8C&pg=PA317&dq=moderate+markov+proces&hl=da&sa=X&ei=pnv9T6zCNMOp0QXRk-yPCQ&ved=0CDUQ6AEwAQ#v=onepage&q&f=false

Hope it's to some help.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.