Draw the moralized graph and the fill in edges

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, draw the moralized graph and the fill-in edges created in the process of elimination. X 7 X 6 X 1 X 10 X 9 X 7 X 4 X 8 X 2 X 3 X 5 X 10 X 6 X 1 X 10 X 9 X 4 X 8 X 2 X 3 X 5 X 4 X 6 X 1 X 9 X 4 X 8 X 2 X 3 X 5 X 9 X 6 X 1 X 9 X 8 X 2 X 3 X 5 X 5 X 6 X 1 X 8 X 2 X 3 X 5 X 8 X 6 X 1 X 8 X 2 X 3 X 1 X 6 X 1 X 2 X 3 X 2 X 6 X 2 X 3 X 3 X 6 X 3 4
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B A M E J T N B A M E J T N A M J T N B E A M J T N B E a. G b. G B A c. G B A d. G ′′ B A Figure 1: G with arc reversals applied 3. For the elimination order X 6 , X 9 , X 4 , X 8 , X 7 , X 10 , X 1 , X 2 , X 5 , draw the resulting junction tree. X 1 ,X 4 ,X 7 ,X 8 ,X 10 X 1 ,X 7 ,X 8 ,X 10 X 1 ,X 7 ,X 10 X 1 ,X 2 ,X 3 ,X 5 X 1 ,X 10 X 6 ,X 9 X 3 ,X 5 X 5 X 1 ,X 7 ,X 8 ,X 10 X 1 ,X 7 ,X 10 X 2 ,X 3 ,X 5 X 1 ,X 10 X 3 ,X 5 X 1 X 5 X 6 X 2 ,X 3 ,X 5 ,X 6 X 2 ,X 3 ,X 5 X 2 ,X 3 ,X 5 5 Arc Reversal 1. Let G = ( V, E ) be the original directed graph. Let G B A = ( V, E - { B A } ∪ { A B } ) be G with the edge simply reversed (see Figure 1b). Note that G B A is not an I-map of the distribution since the independency B E | A is entailed by G B A but not by G . Let G B A be the graph with the edge reversed and a moralizing edge added for each parent of A . Specifically, in this case, G B A will have an additional edge E B (see Figure 1c). Note that the direction of the moralizing edge is forced by the acyclicity constraint on the graph; specifically, if we add B E then we would have a cycle A B E A . 2. Given G = ( V, E ) the original BN with X Y the only directed path from X to Y, let G X Y = ( V, E - { ( X Y ) } ∪ { ( Y X ) } ∪ { ( Z X ) | ( Z Y ) E } ). In words, this is G with the edge from X Y reversed, plus the addition of moralizing edges Z X for any parents of Y . Note that no parent of Y may be a child of X by supposition; if there were such a node Z , we would have an additional directed path from X to Y, namely X Z Y . 5
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Proof. M Y U X Y U M X Y U M X a. H b. H X Y c. H ′′ X Y Figure 2: H with arc reversals applied Proposition 4. G X Y contains no cycles Proof. Suppose otherwise; suppose there are cycles in G X Y . Then there is a shortest such cycle N 1 N 2 . . . N n N 1 . At least one of the edges in the cycle must be a “new” edge, otherwise the cycle would have existed in G . Without loss of generality, assume the new edge is the first edge in the cycle.
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  • Fall '07
  • CarlosGustin
  • Graph Theory, probability density function, graphical model

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