# Markov random fields and Gibbs measures

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1 Chapter Markov random fields and Gibbs measures 1. Conditional independence Suppose X i is a random element of (X i, B i ), for i = 1, 2, 3, with all X i defined on the same probability space (.F, P). The random elements X 1 and X 3 are said to be conditionally independent given X 2 if, at least for all bounded, B i -measuable real functions f i, P ( ) ( ) ( ) f 1 (X 1 ) f 3 (X 3 ) X 2 = P f1 (X 1 ) X 2 P f3 (X 3 ) X 2 almost surely. Equivalently, with F i (X 2 ) = P ( ) f i (X i ) X 2 for i = 1, 3, <1> P f 1 (X 1 ) f 2 (X 2 ) f 3 (X 3 ) = PF 1 (X 2 ) f 2 (X 2 )F 3 (X 2 ) for all f i. Notice that equality <1> involves the random elements only through their joint distribution. We could, with no loss of generality, assume = X 1 X 2 X 3 equipped with its product sigma-field B = B 1 B 2 B 3, with P the joint distribution of (X 1, X 2, X 3 ) and X i (ω) = ω i, the ith coordinate map. Things are greatly simplified if P has a density p(ω) with respect to a sigma finite product measure λ = λ 1 λ 2 λ 3 on B. As in the case of densities with respect to Lebesgue measure, the marginal distributions have densities obtained by integrating out some coordinates. For example, the marginal distribution of ω 2 has density p 2 (ω 2 ) = λ 1 λ 3 p(ω 1,ω 2,ω 3 ) with respect to λ 2. Similarly, the various conditional distributions are given by conditional densities (cf. Pollard 2001, Section 5.4). For example, the conditional distribution of (ω 1,ω 3 ) given ω 2 has density p 13 2 (ω 1,ω 3 ω 2 ) = p(ω 1,ω 2,ω 3 ) {p 2 (ω 2 )>0} with respect to λ 1 λ 3. p 2 (ω 2 ) Remark. It is traditional to use less formal notation, for example, writing p(ω 1,ω 3 ω 2 ) instead of p 13 2 (ω 1,ω 3 ω 2 ). As long as the arguments are specified symbolically there is no ambiguity. But an expression like p(a, b c) could refer to several different conditional densities evaluated at values (a, b, c). The conditional expectations are given by integrals involving conditional densities. For example, P ( ) f 1 (ω 1 ) ω 2 = λ1 f 1 (ω 1 )p 1 2 (ω 1 ω 2 ) almost surely. 6 February 2006 Stat 606, version: 6feb06 c David Pollard 1

2 Equality <1> then becomes λ f 1 (ω 1 ) f 2 (ω 2 ) f 3 (ω 3 )p(ω 1,ω 2,ω 3 ) = λ 2 f 2 (ω 2 )p 2 (ω 2 ) ( λ 1 f 1 (ω 1 )p 1 2 (ω 1 ω 2 ) )( λ 1 f 3 (ω 3 )p 3 2 (ω 3 ω 2 ) ) = λf 1 (ω 1 ) f 2 (ω 2 ) f 3 (ω 3 )p 2 (ω 2 )p 1 2 (ω 1 ω 2 )p 3 2 (ω 3 ω 2 ) A simple generating-class argument then shows that the conditional independence is equivalent to the factorization <2> p(ω 1,ω 2,ω 3 ) = p 2 (ω 2 )p 1 2 (ω 1 ω 2 )p 3 2 (ω 3 ω 2 ) a.e. [λ] Note that the right-hand side of the last display has the form φ(ω 1,ω 2 )ψ(ω 2,ω 3 ). Actually, any factorization <3> p(ω 1,ω 2,ω 3 ) = φ(ω 1,ω 2 )ψ(ω 2,ω 3 ) for nonnegative φ and ψ implies the conditional independence. For, if we define (ω 2 ) = λ 1 φ(ω 1,ω 2 ) and (ω 2 ) = λ 3 ψ(ω 2,ω 3 ), we then have p 2 (ω 2 ) = λ 1 λ 3 p = (ω 2 ) (ω 2 ) p 1,2 (ω 1,ω 2 ) = λ 3 p = φ(ω 1,ω 2 ) (ω 2 ) p 2,3 (ω 2,ω 3 ) = λ 1 p = (ω 2 )ψ(ω 2,ω 3 ) p 1 2 (ω 1 ω 2 ) = φ(ω 1,ω 2 )/ (ω 2 ){p 2 > 0} p 3 2 (ω 3 ω 2 ) = ψ(ω 2,ω 3 )/ (ω 2 ){p 2 > 0} from which <2> follows. Similar arguments apply when P is the joint distribution for more than three random elements. Notation. Let S be a finite index set. (Later the points of S will be called sites at which random variables are defined.) Consider probability densities p(ω i : i S) with respect to a product measure λ = i S λ i on the product sigma-field B = i S B i for a product space = X i S X i.for each A S write ω A for (ω i : i A). Better: The rhs of <4> is a version of p. If A, B, and C are disjoint subsets with union S, then it follows by an argumnet similar to the one leading from <3> that ω A and ω B are conditionally independent given ω C if and only if the density factorizes as <4> p(ω) = φ(ω A,ω C )ψ(ω B,ω C ) In fact, this condition follows directly from the earlier argument if we take X 1 = ω A, X 3 = ω B, and X 2 = ω C. If A, B, and C are disjoint subsets whose union is not the whole of S, the extra variables can complicate the checking of conditional independence. However a sufficient condition for conditional independence of ω A and ω B given ω C is existence of a factorization <5> p(ω) = φ(ω A,ω C,ω D )ψ(ω B,ω C,ω E ) where D and E are disjoint subsets of S\ ( A B C ). The integration over ω D and ω E, which is needed to find the density for ω A B C, preserves the factorization needed for conditional independence. 2. Gibbs distributions There is a particularly easy way to create probability measures with recognizable conditional independence properties. Let F be a collection of subsets of S. For 2 6 February 2006 Stat 606, version: 6feb06 c David Pollard

3 each a in F, let a (ω) = a (ω a ) be a nonnegative function that depends on ω only through ω a. Provided the number Z := λ a F a is neither zero nor infinite, we can define a Gibbs measure by means of the density <6> p(ω) = 1 Z a F a(ω a ) with respect to λ. The conditional independence properties are easily seen from the corresponding factor graph, which has vertex set V = S F with edges drawn only between those i S and a F for which i a. For example, if S ={1, 2, 3, 4, 5, 6, 7} and F ={{1, 2, 3}, {2, 3, 4, 5}, {4, 5, 6, 7}}: The same connectivities could be represented, less economically, without the F vertices by joining all sites i, j for which there is some a F with {i, j} a, as shown in the graph on the right of the display. If i and j share a common neighbor a in the factor graph (equivalently, if i and j are neighbors in the representation without F), write i j. Write N i ={j S\{i} : i j}. <7> Theorem. Suppose A, B, andc are disjoint subsets of S with the separation property: each path joining a site in A to a site in B must pass through some site in C. Then ω A and ω B are conditionally independent given ω C under the Gibbs measure given by the density <6>. Proof. Define V A as the set of vertices from V that can be connected to some site in A by a path that passes through no sites from C. By the separation assumption, A V A and B V c A. Sites in C might belong to either V A or V c A. Note that p(ω) = φ(ω)ψ(ω) where φ = a V A a and ψ = a / V A a. The function φ depends on ω only through the sites in V A. Conservatively, we could write it as φ(ω A,ω C,ω D ), where D denotes the set of all sites in V A \ ( A C ). Similarly, ψ(ω) = ψ(ω B,ω C,ω E ) where E S\ ( B C ) consists of all those sites i connected to some a V c A. There can be no path joining i to a point k in A without passing through C, for otherwise there would be a path from k to a via i that would force a to belong to V A.In other words, D and E are disjoint subsets of S\ ( A B C ). That is, we have a factorization of p as in <5>, which implies the asserted conditional independence. <8> Exercise. For each site i, show that the conditional distribution of ω i given ω S\i depends only on (ω j ; j N i ). Specifically, show that p i S\i (ω i ω S\i ) = This fact is called the Markov property for the Gibbs distribution. 6 February 2006 Stat 606, version: 6feb06 c David Pollard 3

4 3. Hammersley-Clifford When P is defined by a strictly positive density p(ω) with respect to λ, there is a representation for P as a Gibbs measure with a factor graph that represents any Markov properties that P might have. The representation will follow from an application of the following Lemma to log p(ω). <9> Lemma. Let g be any real-valued function on = X i S X i. Let ω be any arbitrarily chosen, but fixed, point of. There exists a representation g(ω) = a S a(ω) with the following properties. (i) a (ω) = a (ω a ) depends on ω only through the coordinates ω a.in particular, is a constant. (ii) If a and if ω i = ω i for any i in a then a (ω) = 0. Proof. and For each subset a S define g a (ω a ) = g a (ω) = g(ω a, ω S\a ) a (ω) = b a ( 1)#(a\b) g b (ω b ). Clearly depends on ω only through ω a. To prove (ii), for simplicity suppose 1 a and ω 1 = ω 1. Divide the subsets of a into a set of pairs: those b for which 1 / b and the corresponding b = b {1}. The subsets b and b together contribute ( 1) #(a\b) ( g(ω b, ω 1, ω S\b ) g(ω b,ω 1, ω S\b ) ) = 0 because ω 1 = ω 1. Finally, to establish the representation for g, consider the coefficient of g b in a S a(ω) = {b a a,b S}( 1)#(a\b) g b (ω b ) When b = a = S, we recover ( 1) 0 g S (ω) = g(ω). When b is a proper subset of S, the coefficient of g b equals {b a a S}( 1)#(a\b) = D S\b ( 1)#D, which equals zero because half of the subsets D of S\b have #D even and the other half have #D odd. Applying the Lemma with g(ω) = log p(ω) gives p(ω) = exp a S a(ω a ), which is a trivial sort Gibbs density. We could, of course, immediately discard any terms for which a (ω) = 0 for all ω. The factorizations that follow from any conditional independnce properties of P lead in this way to more interesting representations. The following argument definitely works when is finite and λ is counting measure. I am not so confident for the general case because I haven t thought carefully enough about the role of negligible sets and versions. <10> Lemma. Suppose ω i and ω j are conditionally independent given ω C, where C = S\{i, j}. Then a 0 for every a with {i, j} a S. Proof. Invoke equality <4> with A ={i} and B ={j} to get log p(ω) = f (ω i,ω C ) + h(ω j,ω C ) where f = log φ and h = log ψ. 4 6 February 2006 Stat 606, version: 6feb06 c David Pollard

5 Notice that the representation for a from Lemma <9> is linear in g. Thus a = a f + a h, the sum of the terms by applying the same construction to f then to h. For example, a f (ω a) = b a ( 1)#a\b f (ω b, ω S\b ) If j a, the right-hand side is unaffected if we replace ω j by ω j, because f does not depend on ω j. However, the left-hand side is zero when ω j = ω j. It follows that f a (ω a ) 0 when j a. A similar argument shows that h a (ω a) 0 when i a. References Griffeath, D. (1976), Introduction to Markov Random Fields, Springer. Chapter 12 of Denumerable Markov Chains by Kemeny, Knapp, and Snell (2nd edition). Pollard, D. (2001), A User s Guide to Measure Theoretic Probability, Cambridge University Press. 6 February 2006 Stat 606, version: 6feb06 c David Pollard 5

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