In his texts on causality, Judea Pearl always refers to the simplest graphs he uses, i.e. the acyclic graphs with independent confounders, as Markovian. I don't see why these graphs contain anything like a Markov-property.
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
He is referring to the Parental Markov Condition (see theorems 1.2.7 and 1.4.1 of Causality). Given a graph $G$, we say a distribution $P$ is Markov relative to $G$ if every variable is independent of all its non descendants conditional on its parents. An acyclic causal model $M$ with jointly independent error terms induce a probability distribution over the observed variables which is Markovian relative to $G(M)$.
answered Apr 16, 2019 at 21:15
$\begingroup$
$\endgroup$
These graphs do satisfy the Markov property - once you condition on the parent node, from which the causal arrow comes, the variable is independent of earlier ancestors that causally affect that parent (unless there is a separate arrow directly from the ancestor node to the present node).
