# 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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