What is (not) the Markov property

Let (\Omega, \mathcal{F}, \{\mathcal{F}_k\}_k, \mathrm{P}) be a filtered probability space with a discrete filtration (although the results we are going to discuss hold for continuous random processes as well). We say that a random process \{X_k\}_k on (\Omega, \mathcal{F}, \{\mathcal{F}_k\}_k, \mathrm{P}) possesses the Markov property if

\mathrm{P}[X_{k+1} \in A\mid \mathcal{F}_{s}] = \mathrm{P}[X_{k+1} \in A\mid \mathcal{F}_k]

for all s\leq k. We see this often in the following form

\mathrm{P}[X_{k+1} \in A\mid X_s, X_{s-1},\ldots, X_0] = \mathrm{P}[X_{k+1} \in A\mid X_s]

or simply with with s=k. This is the Markov property.

The following is just wrong: For a sequence of sets \{B_k\}_{k} where B_k\in\mathcal{F}_k

\mathrm{P}[X_{k+1} \in A\mid X_s\in B_{s}, X_{s-1}\in B_{s-1},\ldots, X_0\in B_0] = \mathrm{P}[X_{k+1} \in A\mid X_s\in B_{s}]

A very easy and straightforward way to verify that this is false is to set B_{k}=\Omega, B_{k-2}=\Omega, \ldots, B_{0}=\Omega, that is, provide no information about X_{k}, X_{k-2}, \ldots, X_{0} and provide the information X_{k-1}=x, i.e., B_{k-1}=\{x\} which actually offers some information. Then, according to the wrong statement above, it would be

\mathrm{P}[X_{k+1} \in A\mid X_s\in \Omega, X_{k-1}=x, X_{k-2}\in \Omega, \ldots] = \mathrm{P}[X_{k+1} \in A].

Although, it should naturally be \mathrm{P}[X_{k+1} \in A\mid X_{k-2}=x]. That is, the information that X_{k-2}=x is completely expunged.

Reference: K.L. Chung, Green, Brown and Probability, World Scientific, 1995 [Chap. 5].


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