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C 1
b
v 1
1
C 1
l 2 (0.2)
a 1
b
v
v
v 3
2
3
2
l 4 (0.4)
v 4
C 2
a 2
b 3
C 2
v 5
v
l 6 (0.2)
6
a 3
b 4
v 5
v 7
C 3
b 5
C 3
(b) PME-graph.
(a) A set of linkages
L
.
(c) Clique graph.
Fig. 7.2 A set of linkages and the corresponding PME-graph and clique graph .
2. v i and v j are mutually exclusive if there is an edge e
E.
3. v i and v j are conditionally independent given another vertex v if there is a path
between v i and v j passing v.
=(
v i ,
v j )
Theorem 7.1. A PME-graph G L , A , B is a Markov random field.
Proof. We need to show that G L , A , B satisfies the Markov property, which states that,
in a set of random variables, the probability of any given random variable X being
in a state x only depends on a subset of random variables in the system [134]. In a
PME-graph G L , A , B , a vertex v is mutually exclusive with its adjacent vertices N v .
For any other vertex v
N v , v and v are independent conditional on N v .
V
−{
v
}−
The Markov property is satisfied.
A PME-graph has two interesting and useful properties.
Lemma 7.1 (Property of PME-graphs). For a PME-graph G corresponding to
linkages
between tuple sets A and B:
1. For a tuple t
L
B, the edges corresponding to the linkages in the mutual
exclusion rule R t form a maximal clique in G.
2. Any two cliques in G can share at most one common vertex.
Aort
Proof. Since a maximal clique in a PME-graph G captures a mutual exclusion rule
in the linkages
L A , B , the first item holds.
To prove the second item, since a vertex v in G L , A , B is a linkage and can involve
only one tuple in A and another tuple in B , v can participate in at most 2 maximal
cliques in the PME-graph.
For example, two maximal cliques C 1
= {
v 1
,
v 2
,
v 3
}
and C 2
= {
v 3
,
v 4
,
v 5
}
in Fig-
ure 7.2(b) only share one common vertex v 3 .
A PME-graph captures two types of dependencies among linkages: first, two
linkages associated with the same tuple are mutually exclusive; second, two link-
ages are conditionally independent given the linkages connecting the two linkages.
 
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