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.)


1 Answer 1


The issue is zero probability states. If the distribution is positive everywhere, then it can be written as the product of local clique potentials (this is known as the Gibbs' potential construction). Zero-probability states ruin this because conditional independence is trivially satisfied for conditions with zero probability. A simple example is provided by Pearl in exercise 3.3 of his book "Probabilistic Reasoning in Intelligent Systems" (1988). Section 3.2.3 of that book provides more discussion of the Gibbs' potential construction for Markov fields.


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