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Unformatted text preview: 10708 Graphical Models: Homework 4 Solutions 1 Markov Network Representations 1. We can convert the triangle graph on ( A,B,C ) with potential ( A,B,C ) into a pair wise Markov Random Field by introducing a variable X and connecting the nodes as shown in figure 1. X will take on 8 possible values, with each value corresponding to a joint setting of ( A,B,C ). The edge potentials are given in the following tables. (The leftmost column of each table gives the 8 possible values of X and each table belongs to the edge potential between X and one of A , B or C ) A 1 000 1 001 1 010 1 011 1 100 1 101 1 110 1 111 1 B 1 000 1 001 1 010 1 011 1 100 1 101 1 110 1 111 1 C 1 000 1 001 1 010 1 011 1 100 1 101 1 110 1 111 1 The node potential of X is just set to mirror : ( X ) 000 15 001 5 010 1 011 7 100 12 101 14 110 22 111 8 2. The original graph ( A,B,C,D ) is chordal, but not after converting to a pairwise Markov Random Field. 1 A X B C Figure 1: Pairwise Markov Random Field for part (a) A X B Y C D Figure 2: Pairwise Markov Random Field for part (b) 2 HammersleyClifford Assume for the sake of contradiction that P factors over the graph H : P = AB BC CD DA Now observe that P (0010) = 0, which implies that AB (0 , 0) BC (0 , 1) CD (1 , 0) DA (0 , 0) = 0 This means that at least one of the factors in this product, AB (0 , 0), BC (0 , 1), CD (1 , 0), or DA (0 , 0) is zero. But on the other hand, we have that P (0000) > 0 which implies that AB (0 , 0) is nonzero, and P (0011) > 0 which implies that BC (0 , 1) is nonzero, and P (1110) > 0 which implies that CD (1 , 0) is nonzero, and P (0000) > 0 which implies that DA (0 , 0) is also nonzero. This gives us the desired contradiction, and so P cannot factor over H . 2 3 Importance Sampling Since the question changed after its initial release, we were lenient in grading....
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 Fall '07
 CarlosGustin

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