# Why can't Bayesian and Markov networks represent all conditional independencies in a joint distribution?

From here:

An I-map is said to be perfect if $$I(G)=I(p)$$. Given a distribution $$p$$, it is not always possible to find a DAG $$G$$ such that $$I(G)=I(p)$$. Consider a joint distribution over four random variables such that $$X_A \perp X_C|X_B,X_D$$ and $$X_B \perp X_D|X_A,X_C$$ are the only conditional independence relations. One can verify that there is no four vertex DAG that implies only these conditional independence assumptions.

I am also aware that not even Markov networks are perfect maps for all probability distributions and so can't represent all conditional independencies. Why is this? I thought the whole point of probabilistic graphical models was to represent a joint distribution as efficiently as possible? If that is not possible, then how this problem overcome?

You can work out the probability of $$X_A$$, and then get the probabilities $$P(X_B | X_A)$$ and $$P(X_D | X_A)$$, and then $$P(X_C | X_B, X_D)$$. This gives you a directed graph that contains a cycle. There is no way to represent this accurately without a cycle.