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# hw2sol - Probablistic Graphical Models Spring 2007 Homework...

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Probablistic Graphical Models, Spring 2007 Homework 2 solutions December 6, 2007 1 Markov Networks Solution due to Yongjin Park Strong Union: X Y | Z = X Y | Z, W (1) Transitivity: ¬ ( X A | Z ) ∧¬ ( A Y | Z ) = ⇒ ¬ ( X Y | Z ) (2) 1. See Figure 1. Figure 1: BN G of which I ( G ) does not satisfy these two properties. 2. The global Markov assumption defines X Y | Z if and only if sep H ( X ; Y | Z ), or there is no other path between X and Y that does not pass through Z . Due to the monotonicity of separation, sep H ( X ; Y | Z ) = se H ( X ; Y | Z ) , (3) where Z Z and { X, Y } / Z . Then it follows that any set of nodes Z s.t. Z Z , X Y | Z = X Y | Z . Since { Z }⊂{ Z, W } , the strong union property holds in I ( H ). As for the transitivity, we can prove by showing the contrapositive proposition is correct: A, X Y | Z = X A | Z A Y | Z (4) There can be three cases of A without violating X Y | Z , A belongs to the only path between X and Y where sep H ( X ; Y | Z ). If A is between X and Z then ¬ sep H ( X ; A | Z ) but sep H ( A ; Y | Z ); thus A Y | Z . The mirror case is also true. A is connected to either X or Y , but not simultaneously. If a path between X and A , but not between Y and A , ¬ sep H ( A ; X | Z ) but sep H ( A ; Y ). The mirror case is also true. Thus, either X A | Z or A Y | Z . A is not connected with either X or Y at all. Trivially, X A Y A . 1

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3. Show I l ( H ) = I p ( H ). Equivalently we can prove by showing that if X Y | N H ( X ), then X Y |X− { X, Y } , for all disjoint sets X, Y . By the definition of I-map, X Y | N H ( X ) ⇐⇒ sep H ( X ; Y | N H ( X )) . (5) Again exploiting the monotonicity of d separation (Eq. 3), if N H ( X ) Z , Z d-separates X and Y , i.e. sep H ( X ; Y | Z ). Inductively, adding each node A [ X−{ X, Y }− N H ( X )] to Z , i.e. Z Z ∪{ A } , where Z is initially N H ( X ), sep H ( X ; Y | N H ( X )) = ⇒ ··· = sep H ( X ; Y | Z ) = sep H ( X ; Y | Z ) = ⇒··· = sep H ( X ; Y |X−{ X, Y } ) . (6) In sum, X Y | N H ( X ) = X Y |X−{ X, Y } , and I l ( H ) = I p ( H ). 2 Constructing Junction Tree Solution due to Steve Gardiner . I will proceed by induction on the length of a path in the tree. All paths mentioned here are assumed to be in a junction tree.
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hw2sol - Probablistic Graphical Models Spring 2007 Homework...

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