Algebraic methods toward higher-order probability inequalities

Transcription

1 Algebraic methods toward higher-orderprobability inequalities p. 1/3 Algebraic methods toward higher-order probability inequalities Donald Richards Penn State University and SAMSI

2 Algebraic methods toward higher-orderprobability inequalities p. 2/3 Background L: A finite distributive lattice with partial order Least upper bound: Greatest lower bound: L is order-isomorphic to the lattice of subsets of a finite set There exists a finite set A such that L = 2 A, the set of all subsets of A 2 A is a finite distributive lattice Partial order: Least upper bound: Greatest lower bound:

3 Algebraic methods toward higher-orderprobability inequalities p. 3/3 µ: A probability measure on 2 A µ is multivariate totally positive of order 2 (MTP 2 ) if µ(a b)µ(a b) µ(a)µ(b) for all a,b A Other terminology: log-supermodular FKG measure

4 Algebraic methods toward higher-orderprobability inequalities p. 4/3 A function f : 2 A R is increasing if f(a) f(b), a b The expected value of f with respect to µ: E(f) := a Aµ(a)f(a) The FKG inequality (Fortuin, Kasteleyn, Ginibre, 1971): If f 1,f 2 : L R are increasing andµ is MTP 2 then Cov(f 1,f 2 ) := E(f 1 f 2 ) E(f 1 )E(f 2 ) 0

5 Algebraic methods toward higher-orderprobability inequalities p. 5/3 Applications Statistics: Multivariate statistical analysis, dependence properties of random variables, observational studies Probability theory: Diffusion equations, reliability theory, percolation Mathematical physics: Interacting particle systems, Ising models Total positivity, finite reflection groups, analysis on Lie groups Combinatorics: Monotonicity of partial orders, Sperner theory, graph theory, Ramsey theory

6 Algebraic methods toward higher-orderprobability inequalities p. 6/3 The case of R n x = (x 1,...,x n ) and y = (y 1,...,y n ) in R n The partial order: x y if x j y j, j = 1,...,n The lattice operations x y = (max(x 1,y 1 ),...,max(x n,y n )) x y = (min(x 1,y 1 ),...,min(x n,y n )) f : R n R is increasing if f(x) f(y) whenever x y

7 Algebraic methods toward higher-orderprobability inequalities p. 7/3 A probability density function K : R n R is MTP 2 if K(x y)k(x y) K(x)K(y), x,y R n A remarkable result: For suitably smooth K, the MTP 2 condition is equivalent to 2 x i x j lnk(x) 0 for all i j The FKG inequality: If f 1,f 2 : R n R are increasing and K is MTP 2 then ( ) ( ) f 1 f 2 K dx f 1 K dx f 2 K dx 0 R n R n R n

8 Algebraic methods toward higher-orderprobability inequalities p. 8/3 The normal distribution (X 1,...,X n ) N n (0,Σ) K(x) = (2π) n/2 Σ 1/2 exp ( 1 2 x Σ 1 x) K is MTP 2 if (Σ 1 ) ij 0 for all i j For a = (a 1,...,a n ) R n, construct the indicator function I a (x) = I a is increasing { 1, x 1 a 1,...,x n a n 0, otherwise

9 Algebraic methods toward higher-orderprobability inequalities p. 9/3 If (X 1,...,X n ) is a MTP 2 normal random vector then EI a (X 1,...,X n )I b (X 1,...,X n ) EI a (X 1,...,X n ) EI b (X 1,...,X n ) Equivalently, P(X 1 a 1 b 1,...,X n a n b n ) P(X 1 a 1,...,X n a n ) P(X 1 b 1,...,X n b n ) There are many applications of this result in statistical inference

10 Algebraic methods toward higher-orderprobability inequalities p. 10/3 Cumulants X R n : A random vector f 1,...,f m : R n R Cumulants: The coefficients in the Taylor-Maclaurin expansion of the cumulant-generating function log E exp(t 1 f t n f m ) κ 2 = E(f 1 f 2 ) E(f 1 )E(f 2 ) is the simplest cumulant The next three cumulants are κ 3 := E(f 1 f 2 f 3 ) [E(f 1 f 2 )E(f 3 ) + E(f 1 f 3 )E(f 2 ) + E(f 1 )E(f 2 f 3 )] + 2E(f 1 )E(f 2 )E(f 3 )

11 Algebraic methods toward higher-orderprobability inequalities p. 11/3 κ 4 := E(f 1 f 2 f 3 f 4 ) [E(f 1 f 2 f 3 )E(f 4 ) + ] [E(f 1 f 2 )E(f 3 f 4 ) + ] + 2[E(f 1 f 2 )E(f 3 )E(f 4 ) + ] 6E(f 1 )E(f 2 )E(f 3 )E(f 4 ) κ 5 := E(f 1 f 2 f 3 f 4 f 5 ) [E(f 1 f 2 f 3 f 4 )E(f 5 ) + ] [E(f 1 f 2 f 3 )E(f 4 f 5 ) + ] + 2[E(f 1 f 2 f 3 )E(f 4 )E(f 5 ) + ] + 2[E(f 1 f 2 )E(f 3 f 4 )E(f 5 ) + ] 6[E(f 1 f 2 )E(f 3 )E(f 4 )E(f 5 ) + ] + 24E(f 1 )E(f 2 )E(f 3 )E(f 4 ) Question: Are there FKG inequalities for the cumulants?

12 Algebraic methods toward higher-orderprobability inequalities p. 12/3 Answer: Probably not For the normal distribution, cumulants of order 3 or more are identically zero It is not difficult to find MTP 2 random vectors with negative cumulants Percus (1975), Sylvester (1975): Interesting correlation inequalities for cumulants Evidence of cumulant-type inequalities were obtained with indicator functions

13 Algebraic methods toward higher-orderprobability inequalities p. 13/3 Recall that κ 3 := E(f 1 f 2 f 3 ) [E(f 1 f 2 )E(f 3 ) + E(f 1 f 3 )E(f 2 ) + E(f 1 )E(f 2 f 3 )] + 2E(f 1 )E(f 2 )E(f 3 ) Define the conjugate cumulant κ 3 := 2E(f 1 f 2 f 3 ) [E(f 1 f 2 )E(f 3 ) + E(f 1 f 3 )E(f 2 ) + E(f 1 )E(f 2 f 3 )] + E(f 1 )E(f 2 )E(f 3 ) Conjugate : Reverse the order of the absolute value of the coefficients in the cumulant Analogy with the conjugate of a partition

14 Algebraic methods toward higher-orderprobability inequalities p. 14/3 L: finite distributive lattice µ: An MTP 2 probability measure on L f 1, f 2 and f 3 : Nonnegative increasing functions on L After years of hard work... Theorem: κ 3 (f 1,f 2,f 3 ) 0 My gratitude to the Institute for Advanced Study

15 Algebraic methods toward higher-orderprobability inequalities p. 15/3 Corollary: The FKG inequality Proof: Set f 3 1 The theorem cannot be deduced from the FKG inequality κ 3 is an alternating sum κ 3 = Cov(f 1 f 2,f 3 ) E(f 1 )Cov(f 2,f 3 ) + Cov(f 1 f 3,f 2 ), κ 4 and κ 5 : Conjugating the coefficients still works f 1,...,f 5 : Nonnegative increasing functions on L

16 Algebraic methods toward higher-orderprobability inequalities p. 16/3 Define the fourth- and fifth-order conjugate cumulants κ 4 := 6E(f 1 f 2 f 3 f 4 ) 2[E(f 1 f 2 f 3 )E(f 4 ) + ] [E(f 1 f 2 )E(f 3 f 4 ) + ] + [E(f 1 f 2 )E(f 3 )E(f 4 ) + ] E(f 1 )E(f 2 )E(f 3 )E(f 4 ) κ 5 := 24E(f 1 f 2 f 3 f 4 f 5 ) 6[E(f 1 f 2 f 3 f 4 )E(f 5 ) + ] 2[E(f 1 f 2 f 3 )E(f 4 f 5 ) + ] + 2[E(f 1 f 2 f 3 )E(f 4 )E(f 5 ) + ] + [E(f 1 f 2 )E(f 3 f 4 )E(f 5 ) + ] [E(f 1 f 2 )E(f 3 )E(f 4 )E(f 5 ) + ] + E(f 1 )E(f 2 )E(f 3 )E(f 4 ) Theorem: Under the same hypotheses as before, κ 4,κ 5 0

17 Algebraic methods toward higher-orderprobability inequalities p. 17/3 The proof that κ 3 0 We adapt a proof of the FKG inequality due to den Hollander and Keane (1986) The proof is by induction on the length of maximal chains in L L = 2 A for some finite set A If A = then the result is trivial, so assume that A WLOG, assume that µ(a) > 0 for all a A Why? Ef depends only on the support of µ Choose and fix B, an arbitrarily chosen subset of A

18 Algebraic methods toward higher-orderprobability inequalities p. 18/3 For a B, define µ B (a) := b A\B µ(a b) (1) µ B is the marginal probability measure on the lattice 2 B Easy fact (well-known to statisticians!): µ B is MTP 2

19 Algebraic methods toward higher-orderprobability inequalities p. 19/3 For f : 2 A R, define f B (a) := 1 µ B (a) b A\B µ(a b)f(a b) (2) f B (a) is the conditional expectation of f, given a B Easy fact (well-known to statisticians!): If f is increasing then so is f B

20 Algebraic methods toward higher-orderprobability inequalities p. 20/3 For g : 2 B R, define E B (g) := a B µ B (a)g(a) The Double-Expectation Theorem (well-known to undergrad. statistics & probability students!): For B A, E(f) = a A µ(a)f(a) = a B µ B (a)f B (a) = E B (f B ) In words: The overall average value of f equals the average value of its conditional averages

21 Algebraic methods toward higher-orderprobability inequalities p. 21/3 Suppose B = A\{z} where z A is chosen arbitrarily Shorthand notation: f ib denotes (f i ) B, i = 1,2,3 The Claim: ( ) 2E B (f1 f 2 f 3 ) B [ ( ) ( ) ( )] E B (f1 f 2 ) B f 3B + EB (f1 f 3 ) B f 2B + EB f1b (f 2 f 3 ) B + E B (f 1B f 2B f 3B ) 0 (3)

22 Algebraic methods toward higher-orderprobability inequalities p. 22/3 For a B, it follows from (1) that µ B (a) = µ(a) + µ(a {z}) (4) and, from (2), we get f B (a) = 1 µ B (a) ( µ(a)f(a) + µ(a {z})f(a {z}) ) (5) Conclude: µ B (a) 3 (f 1 f 2 f 3 ) B (a) = µ B (a) 2[ µ(a)f 1 (a)f 2 (a)f 3 (a) ] + µ(a {z})f 1 (a {z})f 2 (a {z})f 3 (a {z}) (6)

23 Algebraic methods toward higher-orderprobability inequalities p. 23/3 For {i,j,k} = {1,2,3}, µ B (a) 3 (f i f j ) B (a)f kb (a) = µ B (a) [ µ(a)f i (a)f j (a) + µ(a {z})f i (a {z})f j (a {z}) ] [ µ(a)f k (a) + µ(a {z})f k (a {z}) ] (7) µ B (a) 3 f 1B (a)f 2B (a)f 3B (a) = 3 [ µ(a)fi (a) + µ(a {z})f i (a {z}) ] (8) i=1 Collect together (6) - (8), simplify the algebraic expressions using (4) and (5):

24 Algebraic methods toward higher-orderprobability inequalities p. 24/3 µ B (a) 3{ 2(f 1 f 2 f 3 ) B (a) [ (f 1 f 2 ) B (a)f 3B (a) + (f 1 f 3 ) B (a)f 2B (a) + f 1B (a)(f 2 f 3 ) B (a) ] } + f 1B (a)f 2B (a)f 3B (a) = µ(a)µ(a {z}) [ (f1 (a {z}) f 1 (a) ) ] Φ 1B (a) + f 1 (a {z})φ 2B (a) (9) where

25 Algebraic methods toward higher-orderprobability inequalities p. 25/3 Φ 1B (a) = µ(a) ( f 2 (a {z})f 3 (a {z}) f 2 (a)f 3 (a) ) +µ(a {z})f 3 (a) ( f 2 (a {z}) f 2 (a) ) +µ(a {z})f 2 (a) ( f 3 (a {z}) f 3 (a) ) Φ 2B (a) = ( µ(a {z}) + µ(a) ) ( f 2 (a {z}) f 2 (a) ) ( f 3 (a {z}) f 3 (a) ) Each f i is nonnegative and increasing Therefore Φ 1B,Φ 2B are sums of products of nonnegative terms Conclude: (9) is nonnegative

26 Algebraic methods toward higher-orderprobability inequalities p. 26/3 Divide both sides of (9) by µ B (a) 2 and sum over all a B Double-Expectation Theorem gives: a B µ B (a)(f 1 f 2 f 3 ) B (a) = E B ( (f1 f 2 f 3 ) B ) = E(f1 f 2 f 3 ), For {i,j,k} = {1,2,3}, ( ) µ B (a)(f i f j ) B (a)f kb (a) = E B (fi f j ) B f kb a B Also, a B µ B (a)f 1B (a)f 2B (a)f 3B (a) = E B (f 1B f 2B f 3B )

27 Algebraic methods toward higher-orderprobability inequalities p. 27/3 Collect together these identities, apply the nonnegativity of (9); we obtain The Claim: 2E B ( (f1 f 2 f 3 ) B ) [ E B ( (f1 f 2 ) B f 3B ) + EB ( (f1 f 3 ) B f 2B ) + EB ( f1b (f 2 f 3 ) B )] Finally, set B = ; observe that E (f i ) E(f i ); E ((f i f j ) f k ) E(f i f j )E(f k ); E (f 1 f 2 f 3 ) E(f 1 )E(f 2 )E(f 3 ) + E B (f 1B f 2B f 3B ) 0 The Claim reduces to: κ 3 0 Q.E.D.

28 Algebraic methods toward higher-orderprobability inequalities p. 28/3 The proof of κ 4,κ 5 0: Similar We used MAPLE to carry out the algebraic computations What about κ 6? We had some evidence that κ 6 0 Siddhartha Sahi (Rutgers Univ.) has found stronger evidence recently Nevertheless, there is a conjecture...

29 Algebraic methods toward higher-orderprobability inequalities p. 29/3 A partition λ = (λ 1,λ 2,...) is a sequence of nonnegative integers with λ 1 λ 2 The parts of λ are the non-zero λ i The weight of λ is λ := λ 1 + λ 2 + The length of λ, denoted by l(λ), is the number of parts of λ For i = 1,2,..., let λ i denote the cardinality of the set {j : λ j i} Clearly, λ 1 λ 2 The partition λ = (λ 1,λ 2,...) is the partition conjugate to λ Easy homework: Verify that (λ ) = λ and that λ 1 = l(λ)

30 Algebraic methods toward higher-orderprobability inequalities p. 30/3 S m : The symmetric group on m symbols The standard action of S m on R m : For τ S m and u = (u 1,...,u m ) R m, set τ u := (u τ(1),...,u τ(m) ) For each partition λ = (λ 1,...,λ m ) of weight m, define P λ (f 1,...,f m ) := l(λ) j=1 ( λ j E k=1 f λ1 + +λ j 1 +k Denote by D(λ) the set of all τ S m which give rise to distinct permutations P λ (τ (f 1,...,f m )) of P λ (f 1,...,f m ) )

31 Algebraic methods toward higher-orderprobability inequalities p. 31/3 Conjecture: Let µ be an MTP 2 probability measure on L, and f 1,...,f m be nonnegative increasing functions on L. Then (i) There exist non-zero constants {c λ Z : λ = m} such that P m (f 1,...,f m ) := P λ (τ (f 1,...,f m )) is nonnegative λ =m c λ τ D(λ) (ii) For m 3, there exists a constant d m such that P m (f 1,...,f m 1,1) d m P m 1 (f 1,...,f m 1 ) (iii) If f j 1, j = 1,...,m then P m (1,...,1) = 0; equivalently, card(d(λ))c λ = 0 λ =m

32 Algebraic methods toward higher-orderprobability inequalities p. 32/3 For m = 2,3,4,5, the conjecture is valid Choose for P m the conjugate cumulant κ m For λ = (λ 1,λ 2,...), all coefficients c λ = ( 1) l(λ) 1 (λ 1 1)! in the expansion of κ m(f 1,...,f m ) are non-zero and satisfy (iii) For m = 3,4,5 each κ m satisfies (ii) with d m = m 2 I found evidence that κ m(1,...,1) > 0, m 6

33 Algebraic methods toward higher-orderprobability inequalities p. 33/3 Sahi has constructed a class of functionals τ m such that τ m κ m, m = 2,3,4,5, and τ m κ m, m 6 For special MTP 2 measures µ, Sahi has proved that τ m 0 for all m Sahi explains that there is a crucial difference between the class of partitions of weight 5 and those of weight > 5 Sahi conjectures that, for λ = (λ 1,λ 2,...), we can take c λ = ( 1) l(λ) 1 (λ 1 1)!(λ 2 1)! This differs from my coefficients only if λ 2 3, in which case λ 6

34 Algebraic methods toward higher-orderprobability inequalities p. 34/3 Sahi also has obtained FKG inequalities for functions taking values in very general partially ordered algebras, e.g., the algebra of square matrices, symmetric matrices, or polynomials His proofs are, in a sense, simpler than the original FKG An amazing coincidence? Connections with k-characters, group determinants Ken Johnson s paper Johnson believes that there is a connection between FKG-type inequalities and Chern classes