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It's well known that 3 random variables may be pairwise statistically independent but not mutually independent, for an illustration see: example pairwise vs. mutual relations.

Can mutual statistical independence be modeled with Bayesian Networks aka Graphical Models? These are nonparametric structured stochastic models encoded by Directed Acyclic Graphs.

"Each vertex represents a random variable (or group of random variables), and the links express probabilistic relationships between these variables. The graph then captures the way in which the joint distribution over all of the random variables can be decomposed into a product of factors each depending only on a subset of the variables." -- CM Bishop Pattern Recognition and Machine Learning, Ch. 8 p. 630.

Here's a basic example from Wikipedia:

enter image description here

It would appear that hypergraphs are needed to represent the higher order (in)dependence. Is there some trick based on "d-separation", "Markov Blankets" or maybe grouping variables that would enable such a representation?

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reference? The mutual independence is strictly stronger than the pairwise one in a sense that the former implies the latter, but the latter does not always imply the former. – Ilya Jan 21 '13 at 17:26
As @Ilya (+1) alludes to, it might help you to review the formal definition of independence. From that it will be clear that your conjecture is false. – cardinal Jan 21 '13 at 17:31
@Ilya, I could have sworn I read in a recently published text (~2010) that the conditions are independent. But it's the first time I had seen that hence I was looking for references. I will edit my Q until I find that book again. It does not affect the underlying representation issue. – alancalvitti Jan 21 '13 at 17:46
I am not aware of the most recent results on the independence, but would be surprised if they make old ones being wrong :) though I'm not aware of the statistical independence either – Ilya Jan 21 '13 at 18:11
@cardinal, I understand the formal definition as factorization conditions, stated in the link provided. (examples pairwise vs. mutual relations). Are you saying that P(ABC)=P(A)P(B)P(C) implies: (P(AB)=P(A)P(B) and P(AC)=P(A)P(C) and P(BC)=P(B)P(C))? – alancalvitti Jan 21 '13 at 19:09

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