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Published byValentin Cartier Modified over 6 years ago
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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.
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Edge minimal and unique.
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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.
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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):
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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 !
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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).
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The two males and females example
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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.
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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).
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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
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