This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 10-708 Probabilistic Graphical Models Homework 2 Solutions Adapted from solutions of Anton Chechetka 1 By the decomposition property, we can assume X ∪ Y = V . Also, from the Running Intersection Property (RIP), we have that X ∩ Y ⊆ S ij This is because any u ∈ X ∩ Y lies in all cliques in the tree-path between the cliques containing X and Y , and consequently, in both C i and C j ; and thus in S ij = C i ∩ C j . Now, by the Family Preservation Property (FPP), the scope of every factor of the BN lies in at least one of the cliques of the Clique Tree. Thus, we can distribute all the factors of the BN to the cliques of the tree. By RIP, any factor which involves variables in both X and Y has to be within the scope of S ij . If there is a factor which involves a variable u ∈ X \ S ij and v ∈ Y \ S ij , then there has to be a clique containing both u and v . WLOG, let this clique lies on the X side of the tree; then as v ∈ Y it also lies in a clique in the Y side of the tree and by RIP v ∈ S ij ; which contradicts v ∈ Y \ S ij . Thus, we can separate the factors of the BN into those having scope completely either in the X side or the Y side. That is, P ( X ) = f ( X,S ij ) g ( Y,S ij ) which was proved in homework 1 to be equivalent to ( X ⊥ Y | S ij )....
View Full Document
This note was uploaded on 05/25/2008 for the course MACHINE LE 10708 taught by Professor Carlosgustin during the Fall '07 term at Carnegie Mellon.
- Fall '07