Reimer s Inequality on a Finite Distributive Lattice

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1 Reimer s Inequality on a Finite Distributive Lattice Clifford Smyth Mathematics and Statistics Department University of North Carolina Greensboro Greensboro, NC USA cdsmyth@uncg.edu May 1, 2013 Abstract We generalize Reimer s Inequality [6] (a.k.a the BKR Inequality or the van den Berg Kesten Conjecture [1]) to the setting of finite distributive lattices. (MSC 60C05) 1 Introduction Let n be a positive integer. Let Ω 1,..., Ω n be finite sets and let Ω = n i=1 Ω i. For b Ω and S [n] (where [n] := {1,..., n}) we define the cylinder Ω(b, S) = {x Ω : i S x i = b i }. Note that b Ω(b, S), Ω(b, ) = Ω, and Ω(b, [n]) = {b}. Let b Ω and let A, B Ω. We say b A and b B hold disjointly if there exist S, T [n] with S T = such that Ω(b, S) A and Ω(b, T ) B. It does not change the definition to additionally require that S and T be a partition of [n], for if S and T satisfy the conditions of the definition, then so do S 0 = [n] \ T and T 0 = T since S S 0 implies Ω(b, S 0 ) Ω(b, S). The box product of A and B is defined (in [1]) to be A B = {b Ω : b A and b B hold disjointly}. (1) 1

2 Note A B A B. We say a probability measure µ on Ω = n i=1 Ω i is a product probability measure if µ = n i=1 µ i where µ i is a probability measure on Ω i for each i with 1 i n. We have the following theorem. Theorem 1.1 (Reimer s Inequality [6]) If Ω = n i=1 Ω i is a finite product of finite sets and µ is a product probability measure on Ω then µ(a B) µ(a)µ(b), for all A, B Ω. (2) Inequality (2) was originally conjectured by van den Berg and Kesten [1] who proved it for increasing sets on products of chains. Theorem 1.1 is sometimes known as the BKR Inequality. We generalize this theorem to finite distributive lattices, see Theorem 1.2 below. Let L be a finite bounded lattice. Let 0 L = min L and 1 L = max L. If x, y L, let [x, y] L = {z L : x z y}. Let b L and let A, B L. We say b A and b B hold disjointly if there exist x, y, z, w L with x y, z w, x z = 0 L, y w = 1 L, and b = x z = y w, such that b [x, y] L A and b [z, w] L B. Note that [x, y] L [z, w] L = {b}. (In general, if x y and z w, S = [x, y] L [z, w] L if and only if x z y w, in which case S = [x z, y w] L.) The L-box product of A and B is A L B = {b L : b A and b B hold disjointly}. (3) Note that A L B A B. If L is a lattice, a function µ : L [0, + ) is log-supermodular if and is log-modular if µ(a)µ(b) µ(a b)µ(a b), for all a, b L, (4) µ(a)µ(b) = µ(a b)µ(a b), for all a, b L. (5) Our main theorem is the following. Theorem 1.2 Let L be a finite distributive lattice and let µ be a log-modular probability measure on L. Then µ(a L B) µ(a)µ(b), for all A, B L. (6) 2

3 Distributivity is a necessary condition in Theorem 1.2. In the nondistributive lattice N 5 = ({, {1}, {2}, {2, 3}, {1, 2, 3}}, ), the sets A = {{2}, {2, 3}, {1, 2, 3}} and B = {, {2}, {2, 3}} furnish a counterexample when µ is uniform: µ(a L B) = µ(a B) = 2 > ( )2 = µ(a)µ(b). Essentially the same counterexample can be found in the non-distributive lattice M 3 = ({, {1}, {2}, {3}, {1, 2, 3}}, ). If P is a partially ordered set and X P, then X is increasing if for all x, y P, x X and x y imply y X. X is decreasing if for all x, y P, x X and x y imply y X. If f : P [0, + ), then f is increasing if for all x, y P, x y implies f(x) f(y). f is decreasing if for all x, y P, x y implies f(x) f(y). We state the following well-known theorem. Theorem 1.3 (The FKG Inequality [2]) If L is a finite distributive lattice and µ is a log-supermodular probability measure on L then g(x)µ(x), (7) x L f(x)g(x)µ(x) x L f(x)µ(x) x L for all increasing f, g : L [0, + ). Equivalently, µ(a B) µ(a)µ(b), for all increasing A, B L. (8) Let Ω = n i=1 Ω i be a finite product of finite sets. Let { n } L(Ω) = Ω i : Ω i is a total ordering of Ω i for 1 i n i=1 be the set of lattices obtained from Ω by viewing each factor Ω i as a chain. Theorem 1.4 (The Harris Kleitman Theorem [4, 5]) If Ω is a finite product of finite sets, µ is a product probability measure on Ω, and L L(Ω) then µ(a B) µ(a)µ(b), for all increasing A, B L. (9) The proofs in [4, 5] only handled the case where Ω = {0, 1} n. As a product measure is log-modular (see Proposition 1.6, statement 1), Theorem 1.3 implies Theorem 1.4. We make the following observations on our definitions, proved in Section 4. 3

4 Proposition 1.5 Let Ω = n i=1 Ω i be a finite product of finite sets and let L L(Ω). i. For all A, B Ω, A B A L B A B. ii. If A L is increasing and B L is decreasing then A L B = A B. iii. If Ω i 2 for all 1 i n then for all A, B Ω, A B = A L B. iv. If Ω i 3 for some i with 1 i n, then there are examples of A, B Ω for which = A B A L B = A B. Proposition 1.6 i. Suppose Ω is a finite product of finite sets, µ is a probability measure on Ω, and L L(Ω). Then the condition that µ is a product probability measure on Ω is equivalent to the condition that µ is a log-modular probability measure on L. ii. Suppose L and K are finite distributive lattices and L is a sublattice of K. If µ : L (0, + ) is log-modular, then there exists a log-modular ν : K (0, + ) such that µ = ν L. iii. If L is a finite distributive lattice and µ is a probability measure on L, then there is a product of chains, K, that contains L as a sublattice and a product probability measure ν on K such that µ( ) = ν( L). Given Propositions 1.5 and 1.6 it is easy to prove the following observations on how Theorem 1.2 relates to Theorems 1.1, 1.3, and 1.4. Proposition 1.7 We have the following implications. i. Theorem 1.2 implies Reimer s Inequality, Theorem 1.1, and is a stronger statement. ii. Theorem 1.2 implies the FKG Theorem, Theorem 1.3, in the case that µ is log-modular. iii. Theorem 1.2 implies the Harris Kleitman theorem, Theorem 1.4. We prove Theorem 1.2 in Section 3 after developing the preliminaries we need in Section 2. We ll prove Propositions 1.5, 1.6 and 1.7 in Section 4. This last section only requires the material from the introduction and the first paragraph of Section 2. 4

5 2 Butterflies Let L be a bounded lattice. An L-butterfly is a 6-tuple β = (x, y, z, w, b, L) where x, y, z, w, b L, x y, z w, x z = 0 L, y w = 1 L and b = x z = y w. R(β) = [x, y] L is the red wing of β, Y(β) = [z, w] L is the yellow wing of β, and b(β) = b is the body of β. If A, B L we say β is an (A, B)-butterfly if R(β) A and Y(β) B. Note that A L B = {b Ω : (A, B)-butterfly β with b(β) = b}. Butterflies have been defined in [6], but only for Ω = {0, 1} n. We ll soon see that the two definitions are equivalent on {0, 1} n. Lemma 2.1 Let L be a finite distributive lattice, let c, d L with c d. Consider the map f = f c,d : L [c, d] L defined by f(x) = (x c) d = (x d) c. Then f is a lattice homomorphism and hence order preserving. Thus if K L is a sublattice (respectively convex) then f 1 (K) is a sublattice (respectively convex). Note f(x) = x for all x [c, d] L, f 1 (c) [0 L, c] L and f 1 (d) [d, 1 L ] L. Proof: (Sketch.) Let x 1, x 2 L. f(x 1 ) f(x 2 ) = (x 1 c) (x 2 c) d = ((x 1 x 2 ) ((c x 1 x 2 ) c)) d = ((x 1 x 2 ) c) d = f(x 1 x 2 ). The proof that f preserves joins is similar. If K is a sublattice of L and f(x 1 ), f(x 2 ) K then f(x 1 x 2 ) = f(x 1 ) f(x 2 ) K. Similarly, f 1 (K) is also closed under joins and hence is a lattice. The other statements are just as straightforward. Butterflies on intervals restrict to subintervals in a nice way. Lemma 2.2 Let L be a finite distributive lattice. Let c, d, C, D L with C c d D. Let f(x) = f c,d (x). Let β = (x, y, z, w, b, [C, D]) be a [C, D]- butterfly. Then f(β) = (f(x), f(y), f(z), f(w), f(b), [c, d]) is a [c, d]-butterfly. If b [c, d] then b(β) = b = f(b) = b(f(β)), R(f(β)) = R(β) [c, d] and Y(f(β)) = Y(β) [c, d]. Proof: (Sketch.) We ll prove that if b [c, d] then R(f(β)) = R(β) [c, d]. The proof that Y(f(β)) = Y(β) [c, d] is similar and all the other statements 5

6 follow straightforwardly from Lemma 2.1: f(y) f(w) = f(y w) = f(d) = d, x b y implies f(x) f(b) = b f(y), etc. Since all the sets we now consider are intervals of L we ll drop the subscripts, R(f(β)) = [f(x), f(y)], R(β) = [x, y], etc. Since b [x, y] [c, d], [x, y] [c, d] = [x c, y d]. Since x c b d, f(x) = (x c) d = x c. Similarly, c b y d and f(y) = y d. Thus R(f(β)) = [f(x), f(y)] = [x, y] [c, d] = R(β) [c, d]. Let L be a bounded lattice. We say x, y L are L-complementary if x y = 0 L and x y = 1 L. In this case, we say x and y are complemented in L and that the complement of x in L is y, which is also denoted as (x) L. If L is distributive and x L is complemented, then there is a unique y L such that (x) L = y ( [3], p.62). If every element in L has a complement, we say L is complemented. Let C(L) = {x L : x is complemented in L}. Note that 0 L, 1 L C(L). By DeMorgan s identities, C(L) is a lattice ( [3], p.63). Since C(L) is distributive and complemented, it is by definition a Boolean lattice ( [3], p.63). By [3] (Corollary 21, p.85), C(L) is thus isomorphic to the lattice of subsets of some finite set, or, equivalently, to {0, 1} n for some n 1. In [6] Reimer defined a butterfly B b,a on {0, 1} n to be an ordered pair (b, a) with b, a {0, 1} n and also defined R(B b,a ) = [a, b], Y(B b,a ) = [a, b], and b(b b,a ) = b. Here [a, b] := {x {0, 1} n : x i {a i, b i }, for all 1 i n} and a is the complement of a in {0, 1} n. As can be easily checked, B b,a corresponds to the butterfly β = (a b, a b, a b, a b, b, {0, 1} n ) in the sense that R(B b,a ) = R(β), Y(B b,a ) = Y(β) and b(b b,a ) = b = b(β). In this same sense, the butterfly β = (x, y, z, w, b, {0, 1} n ) corresponds to the butterfly B b,a where a is the complement of b in [x, y]. Lemma 2.3 Let L be a finite distributive lattice and let c, d L with c d. Let β = (x, y, z, w, b, [c, d]) be an [c, d]-butterfly. If b C(Ω) then β 0 = (x, y, z, w, b, C([c, d])) is a C([c, d])-butterfly. Furthermore, x = b a, y = b a, z = b a, and w = b a where a = f x,y (b ), a = f z,w (b ), and b = (b) [c,d]. R(β 0) = R(β) C([c, d]) and Y(β 0 ) = Y(β) C([c, d]). Proof: First we show that a = (b) [x,y] and a = (b) [z,w]. This will mean that the formulas given for x, y, z, w hold. We then show that a = (a) [c,d]. This will mean a, a, b, b are all members of the lattice C([c, d]) and thus so 6

7 are x, y, z, w. The conditions that remain to be proved about x, y, z, w all immediately follow from the fact that β is a [c, d]-butterfly: y w = d = 1 C([c,d]), etc. Since b [x, y], b a = f x,y (b) f x,y (b ) = f x,y (b b ) = f x,y (c) = x. Similarly, b a = y. Thus a = (b) [x,y]. Similarly, a = (b) [z,w]. a a = ((b x) y) ((b z) w) = (b x) (b z) (y w) = (b x) (b z) b = (b b) (b z b) (x b b) (x z b) = c c c c = c. Similarly a a = d and thus a = (a) [c,d]. Clearly, R(β 0 ) = [x, y] C([c,d]) = [x, y] [c,d] C([c, d]) = R(β) C([c, d]). Similarly Y(β 0 ) = Y(β) C([c, d]). Let L be a bounded lattice and let B be a family of butterflies on L. Let R(B) = β B R(β) and Y(B) = β B Y(β). We say B has distinct bodies if for all β 1, β 2 B with β 1 β 2, b(β 1 ) b(β 2 ). If X L = {0, 1} n, let X = {(x) L : x X}. Reimer proved Lemma 2.4, below, in [6], where it was the crux of his proof of Theorem 1.1. We will use it in the next section to prove Theorem 1.2. Lemma 2.4 (Reimer s Butterfly Lemma [6]) If L = {0, 1} n is a finite Boolean lattice and B is a family of L-butterflies with distinct bodies, then B R(B) (Y(B)). 3 Proof of Theorem 1.2 Proof of Theorem 1.2. We write µ(a) = b L 1 A(b)µ(b) and µ(b) = b L 1 A(b )µ(b ), where 7

8 1 A (x) := 1 if x A, 0 otherwise. Thus µ(a)µ(b) = 1 A (b)1 B (b )µ(b)µ(b ) (b,b ) L 2 = 1 A (b)1 B (b )µ(b b )µ(b b ) (b,b ) L 2 = 1 A (b)1 B (b ) µ(c)µ(d) = (c,d) L 2 :c d (c,d) L 2 :c d (b,b ) L 2 :b b =c,b b =d (A C([c, d]) (B C([c, d])) µ(c)µ(d). The second equality follows from the log-modularity of µ. Similarly we multiply µ(a L B) = b L 1 A B(b)µ(b) and 1 = µ(l) = b L µ(b ) to get L µ(a L B)µ(L) = 1 A L B(b) µ(c)µ(d) or µ(a L B) = (c,d) L 2 :c d (c,d) L 2 :c d (b,b ) L 2 :b b =c,b b =d (A L B) C([c, d]) µ(c)µ(d). To complete the proof, we will show that (A L B) C([c, d]) (A C([c, d]) (B C([c, d])) for all c, d L with c d. Given b (A L B) C([c, d]), let β b = (x, y, z, w, b, L) be a butterfly witnessing the fact that b A L B, i.e. b(β b ) = b, R(β b ) A, and Y(β b ) B. Let f = f c,d. By Lemma 2.2, γ b = (f(x), f(y), f(z), f(w), b, [c, d]) is a [c, d]-butterfly with R(γ b ) = R(β b ) [c, d] A and Y(γ b ) = Y(β b ) [c, d] B. By Lemma 2.3, δ b = (f(x), f(y), f(z), f(w), b, C([c, d])) is a C([c, d])-butterfly with R(δ b ) = R(γ b ) C([c, d]) A C([c, d]). Similarly, Y(δ b ) B C([c, d]). By construction, B = {δ b : b (A L B) C([c, d])} is a family of butterflies of C([c, d]) with distinct bodies. Since C([c, d]) is isomorphic to {0, 1} n for some n 1, we apply Lemma 2.4 to finish the proof: (A L B) C([c, d]) = B R(B) (Y(B)) (A C([c, d])) (B C([c, d])). Note that the last inequality in the chain holds because of the inclusions R(B) A C([c, d]) and Y(B) B C([c, d]). Indeed, R(δ b ) A C([c, d]) for all δ b B, so R(B) = b R(δ b) A C([c, d]). 8

9 4 Proofs of Propositions 1.5, 1.6 and 1.7 Proof of Proposition 1.5. (i.) Suppose A, B Ω. A L B A B by definition. We now show A B A L B. Suppose b A B. Then there exist S and T, a partition of [n] such that Ω(b, S) A and Ω(b, T ) B. For U [n] we define b U, b U L as (b U ) i = { bi, if i U 0 Ω i, otherwise and (b U ) i = { bi, if i U 1 Ω i, otherwise. (10) Let x = b S, y = b S, z = b T, w = b T. It is easy to check that x z = b = y w, x z = 0 L, y w = 1 L, [x, y] L = Ω(b, S) A, and [z, w] L = Ω(b, T ) B. Thus b A L B as desired. (ii.) Suppose A L is increasing and B L is decreasing. By (i.), we only need to show that A B A L B. Suppose b A B. Then β = (b, 1 L, 0 L, b, b, L) is an (A, B)-butterfly with b(β) = b. Hence b A L B. (Note R(β) = [b, 1 L ] L A, since b A and A is increasing. A similar argument gives Y(β) B.) (iii.) We may assume Ω = L = {0, 1} n for some n 1 where x L y iff x i y i for all i. By (i.), we need only prove A L B A B. Suppose b A L B. Let β = (x, y, z, w, b, Ω ) be an (A, B) butterfly. To prove b A B, we exhibit a partition S,T of [n] so that Ω(b, S) = [x, y] L A and Ω(b, T ) = [z, w] L B. Let S = {i [n] : x i = y i } and T = {i [n] : z i = w i }. Since 1 L = y w, b = y w = x z, x z = 0 L, we have 1 = max(y i, w i ), b i = min(y i, w i ) = max(x i, z i ), and 0 = min(x i, z i ) for all i [n]. In particular, if b i = 1, y i = w i = 1 and {x i, z i } = {0, 1} while if b i = 0, x i = z i = 0 and {y i, w i } = {0, 1}. This means that S and T restricted to {i : b i = 1} form a partition. Similarly, S and T restricted to {i : b i = 0} form a partition. Thus S and T are a partition of [n]. It is not hard to prove that Ω(b, S) = [x, y] L. a [x, y] L is equivalent to x i a i y i for all i, which by definition of S is equivalent to a i = x i = y i for all i S. Since x i b i y i for all i, this last statement is equivalent to a i = b i for all i S or a Ω(b, S). (iv.) Suppose k 1,..., k n, i 0 are positive integers with 1 i 0 n and k i0 3. Let Ω = L = n i=1 [k i] where x L y iff x i y i for all i. Let A = {x Ω : 9

10 x i0 2} and let B = {x Ω : x i0 k i0 1}. Note that for any x Ω, Ω(x, S) is not contained in A or in B if i 0 S, thus A B =. Since A is increasing and B is decreasing, (ii.) implies A L B = A B = {x Ω : 1 x i0 k i0 1} A B. A similar example is given by Ω = L = {0, 1, 2} n, A = {x : x i 1, i}, B = {x : x i 1, i}. A L B = A B = {(1,..., 1)} while A B =. Proof of Proposition 1.6. (i.) If µ = µ i is a product measure on Ω = n i=1 Ω i then µ is a log-modular measure on the product of chains L = n i=1 Ω i: µ(a)µ(b) = n i=1 µ i(a i )µ i (b i ) = n i=1 µ i(a i b i )µ i (a i b i ) = µ(a b)µ(a b). The second equality follows because in the chain Ω i, {a i, b i } = {a i b i, a i b i }. Suppose now that µ is log-modular on L, then for any x 1, x 2,..., x k L, µ(x 1 )µ(x 2 ) µ(x k ) = µ(y 1 )µ(y 2 ) µ(y k ) where y 1,..., y k L have the property that for each i [n], ((y 1 ) i,... (y k ) i ) is a permutation of ((x 1 ) i,..., (x k ) i ). Indeed, both products are equal to µ(z 1 ) µ(z k ) where ((z 1 ) i,..., (z k ) i ) is a permutation of ((x 1 ) i,..., (x k ) i ) that is weakly increasing in Ω i, for all i. If i < j consider the products µ(a 1 ) µ(a k ) and µ(b 1 ) µ(b k ) where b i = a i a j, b j = a i a j, and b l = a l for l i, j. By log-modularity, the products are equal, but also for each l, (b i ) l = (a i a j ) l (a i a j ) l = (b j ) l. One can thus bring the product µ(x 1 ) µ(x k ) to the form µ(z 1 ) µ(z k ) by repeatedly using such two-factor replacements. If µ is additionally a probability measure, we claim that µ(x) = n i=1 µ i(x i ) is a product probability measure where µ i (x i ) = µ((z i,1,..., z i,i 1, x i, z i,i+1,..., z i,n )). z i,j Ω j,j i Clearly µ i is a probability distribution on Ω i, as x i Ω i µ i (x i ) = z L µ(z) = 1. Consider n µ i (x i ) = i=1 z i,j Ω,i j µ((x 1, z 1,2,..., z 1,n )) µ((z n,1,..., z n,n 1, x n )). 10

11 Let π S n such that π(i) is the unique number in [n] equivalent to i + 1 mod n. Since one can arbitrarily permute the ith coordinates of the products occurring in the summation, n µ i (x i ) = i=1 z i,j Ω,i j k 1 µ((x 1,..., x n )) µ((z π t (1),1,..., z π t (n),n)) t=1 = µ(x)(µ(l)) k 1 = µ(x). (ii.) It suffices to prove the result in the case that K = {0, 1} n. In the general case, one may use the fundamental theorem on finite distributive lattices ( [3], p.82) to embed K (and hence L) as a sublattice of some {0, 1} n. One can find ν log-modular on {0, 1} n restricting to µ on L and then restrict ν to K to get ν. It is completely straightforward to verify the following facts. If L is not trivial, pick some y L such that y < 1 L. Then L 1 = {x L : x y = 1 L } is a sublattice of L and furthermore L 1 is an interval L 1 = [a, 1 L ] L (where a L). L y = {x L : x y = y} is a sublattice of L and L 1 is isomorphic to the sublattice [a y, y] L y. The maps f : L 1 [a y, y] and g : [a y, a] L 1 given by f(x) = x y and g(x) = x a are inverse lattice isomorphisms. We assume now that K = {0, 1} n and proceed by induction on L. If L = 1 there is nothing to prove. If L = 2, let ν(x) = ar {i:x i=1} where a, r > 0. Note that ν is log-modular. Let w 0 = {i : (0 L ) i = 1} and w 1 = {i : (1 L ) i = 1}. Note w 0 < w 1. Pick a, r > 0 such that ν(0 l ) = ar w 0 = µ(0 L ) and ν(1 L ) = ar w 1 = µ(1 L ). Suppose now that L > 2. By induction, we get ν 1 log-modular on [0 K, y] K restricting to µ on L y. Let ν 2 be log-modular on [y, 1 K ] K restricting to the measure µ on {y, 1 L } given by µ (y) = 1 and µ(1 L ) = µ(1 L )/µ(y). We claim that ν(x) = ν 1 (x y)ν 2 (x y) is log-modular on K and restricts to µ on L. Indeed, since ν 1 is log-modular, ν 1 (x 1 y)ν 1 (x 2 y) = ν 1 ((x 1 y) (x 2 y))ν 1 ((x 1 y) (x 2 y)) = ν 1 ((x 1 x 2 ) y)ν 1 ((x 1 x 2 ) y). Thus the function ν 1 ( y) : K (0, + ) is log-modular. One can prove the same for ν 2 ( y) : K (0, + ). This means their product, ν, is also log-modular. If x L y then ν(x) = ν 1 (x)ν 2 (y) = µ(x) 1. If x L 1, ν(x) = ν 1 (x y)ν(1 L ) = µ(x y)µ(1 L )/µ(y) = µ(x). (This holds by the log-modularity of µ and the fact that x y = 1 L.) 11

12 Note that we do need that µ to be positive. If L = {x 1, x 2, x 3 } with x 1 < x 2 < x 3 is a sublattice of K = {0, 1} n and µ(x 1 ) = µ(x 2 ) = 1 while µ(x 1 ) = 0, then µ does not extend to any log-modular function ν on K. If x 1 is the complement of x 1 in [x 0, x 2 ] K, then we would have to have 0 = ν(x 1 )ν(x 1) = ν(x 0 )ν(x 2 ) = 1, an impossibility. (iii.) Suppose µ is a log-modular probability measure on L. Let K be a product of chains containing L as a sublattice. Let ν : K (0, + ) be log-modular restricting to µ on L. ν(x) = ν (x)/ν (K) is a log-modular probability measure on K. Note ν(x L) = ν({x} L)/ν(L) = ν ({x} L)/ν (L) = ν ({x} L)/µ(L) = ν ({x} L). Thus ν(x L) = µ(x) if x L and ν(x L) = 0 otherwise. Proof of Proposition 1.7. (i.) Let Ω be a finite product of finite sets. Let µ be a product measure on Ω. Let L L(Ω). By Proposition 1.6 (i.), µ is log-modular on L. Thus we may apply Theorem 1.2 and Proposition 1.5 (i.), to get µ(a B) µ(a L B) µ(a)µ(b). Proposition 1.5 (iv.) shows that one can have 0 = µ(a B) < µ(a L B) µ(a)µ(b). (ii.) Let L be a finite distributive lattice and let µ be a log-modular probability measure on L. Let A, B L be increasing. Note B 0 = L \ B is decreasing so by Proposition 1.5 (ii.), A L B 0 = A B 0. By Theorem 1.2, µ(a B) = µ(a) µ(a B 0 ) µ(a) µ(a)µ(b 0 ) = µ(a)µ(b). (iii.) By Proposition 1.6 (i.), Theorem 1.4 is a special case of the log-modular case of Theorem 1.3. References [1] J. van den Berg and H. Kesten, Inequalities with applications to percolation and reliability, J. Appl. Probab. 22 (1985), no. 3, MR MR (87b:60027) [2] C. M. Fortuin, P. W. Kasteleyn, and J. Ginibre, Correlation inequalities on some partially ordered sets, Comm. Math. Phys. 22 (1971), MR (46 #8607) 12

13 [3] George Grätzer, General lattice theory, Birkhäuser Verlag, Basel, 2003, With appendices by B. A. Davey, R. Freese, B. Ganter, M. Greferath, P. Jipsen, H. A. Priestley, H. Rose, E. T. Schmidt, S. E. Schmidt, F. Wehrung and R. Wille, Reprint of the 1998 second edition [MR ]. MR [4] T. E. Harris, A lower bound for the critical probability in a certain percolation process, Proc. Cambridge Philos. Soc. 56 (1960), MR MR (22 #6023) [5] Daniel J. Kleitman, Families of non-disjoint subsets, J. Combinatorial Theory 1 (1966), MR MR (33 #1242) [6] David Reimer, Proof of the van den Berg-Kesten conjecture, Combin. Probab. Comput. 9 (2000), no. 1, MR MR (2001g:60017) 13