How can a probability distribution P not factorize over a graph H when P satisfies the all the global independencies implied by H?
Here's an example: Let $X_1, \dots X_4$ be 4 random variables that can take on 0 or 1. The graph $H$, is a circle: $X_1 \rightarrow X_2 \rightarrow X_3 \rightarrow X_4 \rightarrow X_1$.
You can give a (valid) CPT that specifies the joint distribution of $X_1,X_2,X_3,X_4$ s.t. it satisfies $(X_1\bot X_3) \mid (X_2,X_4)$ and $(X_2 \bot X_4) \mid (X_1,X_3)$.
However, you can specify an assignment to $X_1 \dots X_4$ s.t. such that the joint distribution does not equal the product of the clique factors.
I'm trying to get some intuition for this, and I don't have any right now.
(Note: I've been told that allowing zero probabilities in Markov models plays a role in understanding this.)