Converse of pairwise Markov property

Question

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.

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.