Let $G=(V,E)$ be a finite simple graph, and let $\{X\_i\}\_{i \in V}$ be a collection of random variables associated with the vertices of $G$. The joint distributions of these r.v.s is a *Markov Random Field* if, for any three subsets $U,W,C \subset V$ such that $C$ separates $U$ from $W$ (i.e., any path from $U$ to $W$ passes through $C$), it holds that conditioned on the r.v.s in $C$, the r.v.s in $U$ are independent from those in $W$. This is called the *global Markov property*. An implication is the *pairwise Markov property*, which states that for all $(i,j) \not \in E$, $X\_i$ is independent of $X\_j$, conditioned on the rest of the r.v.s. 


A finite distribution over $\{X\_i\}$ is called *positive* if every combination of assignments has positive probability (here "every" means those whose marginals are positive). I am looking for a reference for the fact that for positive distributions, the pairwise Markov property implies the global Markov property. Thanks!



Local Markov implies global Markov
src: <http://web.engr.illinois.edu/~swoh/courses/IE598/handout/markov.pdf>


copied below


(3) being if $x\_A ⊥ x\_B|(x\_C, x\_D)$ and $x\_A ⊥ x\_C|x\_B ∪ x\_D,$ then $x\_A ⊥ (x\_B, x\_C)|x\_D$


which always holds for positive P.


Proof of (P)$⇒$(G) when (3) holds: [Pearl,Paz 1987]
by induction over s $\triangleq$ |B|
when s = n − 2, (P)$⇔$(G)
assume (G) for any B with |B| ≥ s and prove it for |B| = s − 1


by induction assumption
$$x\_C ⊥ x\_A|(x\_B, x\_i) \\
x\_C ⊥ x\_i|(x\_B, x\_A) by (3),\\
x\_C ⊥ (x\_A, x\_i)|x\_B$$
by induction, we see that (G)
holds for all sizes of B.


