The set  of all independence statements defined by (3

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Presentation transcript:

The set  of all independence statements defined by (3 The set  of all independence statements defined by (3.11) is called the pairwise basis of G. These are the independence statements that define the graph.

Edge minimal and unique.

The set  of all independence statements defined by (3 The set  of all independence statements defined by (3.12) is called the neighboring basis of G.

Testing I-mapness Proof: (2) (1)(3) (2). (2) (1): Holds because G is an I-map of G0 is an I map of P. (1)(3): True due to I-mapness of G (by definition). (3) (2):

Insufficiency of Local tests for non strictly positive probability distributions Consider the case X=Y=Z=W. What is a Markov network for it ? Is it unique ? The Intersection property is critical !

Markov Networks that represents probability distributions (rather than just independence) 1. Define for each (maximal) clique Ci a non-negative function g(Ci) called the compatibility function. 2. Take the product i g(Ci) over all cliques. 3. Define P(X1,…,Xn) = K· i g(Ci) where K is a normalizing factor (inverse sum of the product).

The two males and females example

P(, BG(), U-  -BG()) = f1(,BG()) f2 (U-) (*) Theorem 6 [Hammersley and Clifford 1971]: If a probability function P is formed by a normalized product of non negative functions on the cliques of G, then G is an I-map of P. Proof: It suffices to show (Theorem 5) that the neighborhood basis of G holds in P. Namely, show that I(,BG(), U-  -BG() hold in P, or just that: P(, BG(), U-  -BG()) = f1(,BG()) f2 (U-) (*) Let J stand for the set of indices marking all cliques in G that include . = f1(,BG()) f2 (U-) The first product contains only variables adjacent to  because Cj is a clique. The second product does not contain . Hence (*) holds.

Note: The theorem and converse hold also for extreme probabilities but the presented proof does not apply due to the use of Intersection in Theorem 5. Theorem X: Every undirected graph G has a distribution P such that G is a perfect map of P. (In light of previous notes, it must have the form of a product over cliques).

Drawback: Interpreting the Links is not simple Another drawback is the difficulty with extreme probabilities. There is no local test for I-mapness. Both drawbacks disappear in the class of decomposable models, which are a special case of Bayesian networks