Random vector $X$ follows a pairwise Markov property on graph $G=(V,E)$ if for any $(i,j) \notin E$, $X_i$ and $X_j$ are conditionally independent given $X_{V \setminus \{i,j\}}$.

My question is, why do we not impose the converse? That is, why do we also not impose that if $X_i$ and $X_j$ are conditionally independent given $X_{V \setminus \{i,j\}}$, then $(i,j) \notin E$?

For example, a completely connected graph would always satisfy pairwise Markov property, right?

I'm working with $X$ with positive and continuous density, so pairwise and global Markov properties are equivalent.


1 Answer 1


Who says we don't impose the converse? Pasting a block quote from these CMU 10-702 course notes as taught by Singh and Wasserman. The notes also comment on the complete graph.

Pairwise Markov Property - An edge between two nodes $X_i$ and $X_j$ is absent in the graph if and only if $X_i$ and $X_j$ are conditionally independent given the other variables, i.e. $X_i \bot X_j | X_{\setminus i \setminus j}$.

Notice that the complete graph encodes no conditional independence assumptions. It is the absence of edges that makes a graphical representation useful for describing the distribution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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