On the Dovbysh-Sudakov representation result

We present a detailed proof of the Dovbysh-Sudakov representation for symmetric positive definite weakly exchangeable infinite random arrays, called Gram-de Finetti matrices, which is based on the representation result of Aldous and Hoover for arbitrary (not necessarily positive definite) symmetric weakly exchangeable arrays.

We consider an infinite random matrix R = (R l,l ′ ) l,l ′ ≥1 which is symmetric, nonnegative definite in a sense that (R l,l ′ ) 1≤l,l ′ ≤n is nonnegative definite for any n ≥ 1, and weakly exchangeable, which means that for any n ≥ 1 and for any permutation ρ of {1, . . ., n} the matrix (R ρ(l),ρ(l ′ ) ) 1≤l,l ′ ≤n has the same distribution as (R l,l ′ ) 1≤l,l ′ ≤n .Following [6], we will call a matrix with the above properties a Gram-de Finetti matrix.Since all its propertiessymmetric, positive definite and weakly exchangeable -are expressed in terms of its finite dimensional distributions, we can think of R as a random element in the product space M = 1≤l,l ′ R with the pointwise convergence topology and the Borel σ-algebra M. Let P denote the set of all probability measures on M. Suppose that P ∈ P is such that for all A ∈ M, where Q : Ω × M → [0, 1] is a probability kernel from some probability space (Ω, F , Pr) to M such that (a) Q(u, •) ∈ P for all u ∈ Ω and (b) Q(•, A) is measurable on F for all A ∈ M.
In this case we will say that P is a mixture of laws Q(u, •).We will say that a law Q ∈ P of a Gram-de Finetti matrix is generated by an i.i.d.sample if there exists a probability measure η on ℓ 2 × R + such that Q is the law of where (h l , a l ) is an i.i.d.sequence from η and h • h ′ denotes the scalar product on ℓ 2 .For simplicity, we will often say that a matrix (rather than its law on M) is generated by an i.i.d.sample from measure η.The result of L.N. Dovbysh and V.N.Sudakov in [6] states the following.
Proposition 1 A law P ∈ P of any Gram-de Finetti matrix is a mixture (1.1) of laws in P such that for all u ∈ Ω, Q(u, •) is generated by an i.i.d.sample.
Proposition 1 has recently found important applications in spin glasses; for example, it played a significant role in the proof of the main results in [3] and [11], where a problem of ultrametricity of an infinite matrix (R l,l ′ ) l,l ′ ≥1 was considered under various hypotheses on its distribution.For this reason, it seems worthwhile to have an accessible proof of this result which was, in fact, the main motivation for writing this paper.Currently, there are two known proofs of Proposition 1.The proof in the original paper [6] contains all the main ideas that will appear, maybe in a somewhat different form, in the present paper but the proof is too condensed and does not provide enough details necessary to penetrate these ideas.Another available proof in [8] is much more detailed but, unfortunately, it is applicable not to all Gram-de Finetti matrices even though it works in certain cases.
In the present paper we will give a detailed proof of Proposition 1 which starts with exactly the same idea as [8].Namely, we will deduce Proposition 1 from the representation result for arbitrary weakly exchangeable arrays that are not necessarily positive definite, due to D. Aldous ([1], [2]) and D.N. Hoover ([9], [10]), which states that for any weakly exchangeable matrix there exist two measurable functions f : [0, 1] 4 → R and g : [0, 1] 2 → R such that the distribution of the matrix coincides with the distribution of R l,l = g(u, u l ) and R l,l ′ = f (u, u l , u l ′ , u l,l ′ ) for l = l ′ , ( where random variables u, (u l ), (u l,l ′ ) are i.i.d.uniform on [0, 1] and function f is symmetric in the middle two coordinates, u l and u l ′ .It is customary to define the diagonal elements as a function of three variables R l,l = g(u, u l , v l ) where (v l ) is another i.i.d.sequence with uniform distribution on [0, 1]; however, one can always express a pair (u l , v l ) as a function of one uniform random variable u ′ l in order to obtain the representation (1.3).We will consider a weakly exchangeable matrix defined by (1.3) and, under an additional assumption that it is positive definite with probability one, we will prove that its distribution is a mixture of distributions generated by an i.i.d.sample in the sense of (1.2).First, in Section 2 we will consider a uniformly bounded case, |f |, |g| ≤ 1, and then in Section 3 we will show how the unbounded case follows by a truncation argument introduced in [6].In the general case of Section 3 we do not require any integrability conditions on g rather than g < +∞.Finally, to a reader interested in the proof of (1.3) we recommend a comprehensive recent survey [4] of the representation results for exchangeable arrays.
Acknowledgment.The author would like to thank Gilles Pisier and Joel Zinn for several helpful conversations.
We will start with the case when the matrix elements |R l,l ′ | ≤ 1 for all l, l ′ ≥ 1 with probability one, so we can assume that both functions |f |, |g| ≤ 1.Of course, the representation of the law of R as the mixture (1.1) will be simply the disintegration of the law of (1.3) on M over the first coordinate u.The main problem is now to show that for a fixed u the (law of) matrix R can be represented as (1.2).In other words, if we make the dependence of f and g on u implicit, then assuming that a weakly exchangeable matrix given by is positive definite with probability one, we need to show that its law can be represented as (1.2).Our first step is to show that f does not depend on the last coordinate, which is exactly the same as Lemma 3 in [8].
) is positive definite with probability one then for Proof.We will give a sketch of the proof for completeness.Since (R l,l ′ ) is positive definite, for any sequence of bounded measurable functions (h l ) on [0, 1], almost surely, where E ′ denotes the expectation in (u l ).Let us take n = 4m and given two measurable sets A 1 , A 2 ⊂ [0, 1], let h l (x) be equal to With this choice of (h l ), the sum over the diagonal terms Off-diagonal elements in the sum in (2.2) will all be of the type and for each of the three combination terms of each type will be of order n 2 , while the difference between the number of terms with opposite signs of each type will be at most n/2.Therefore, by the central limit theorem, the distribution of the left hand side of (2.2) converges weakly to some normal distribution and (2.2) can hold only if the variance of the terms in (2.3) is zero, i.e. these terms are almost surely constant.In particular, with probability one.The same holds for some countable collection of sets A 1 × A 2 that generate the product σ-algebra on [0, 1] 2 and this proves that for almost all z on [0, 1], f (x, y, z) = f(x, y) for almost all (x, y) on [0, 1] 2 .
For simplicity of notations we will keep writing f instead of f so that now is positive definite with probability one and |f |, |g| ≤ 1.
Lemma 2 If R in (2.4) is positive definite with probability one then there exists a measurable map φ : [0, 1] → B where B is the unit ball of ℓ 2 such that almost surely on [0, 1] 2 .
Remark.It is an important feature of the proof (similar to the argument in [6]) that the representation (2.5) of the off-diagonal elements is determined independently of the function g that defines the diagonal elements.The diagonal elements play an auxiliary role in the proof of (2.5) simply through the fact that for some function g the matrix R in (2.4) is positive definite.Once the representation (2.5) is determined, the representation (1.2) will immediately follow.
Proof.Let us begin the proof with a simple observation that the fact that the matrix (2.4) is positive definite implies that f (x, y) is a symmetric positive definite kernel on [0, 1] 2 , f (x, y)h(x)h(y) dx dy ≥ 0 (2.6) then lim inf n→+∞ S n ≥ 0 a.s. and (2.6) follows by the law of large numbers for U-statistics (Theorem 4.1.4in [5]), the proof of which we will recall for completeness.Namely, if we consider σ-algebra F n = σ(u (1) , . . ., u (n) , (u l ) l>n ) where u (1) , . . ., u (n) are the order statistics of u 1 , . . ., u n then (S n , F n ) is a reversed martingale and n≥1 F n is trivial by the Hewitt-Savage zero-one law since it is in the tail σ-algebra of i.i.d.(u l ) l≥1 .Therefore, a.s.
which proves (2.6).Since f (x, y) is symmetric and in L 2 ([0, 1] 2 ), there exists an orthonormal sequence (ϕ l ) in L 2 ([0, 1]) such that (Theorem 4.2 in [12]) where the series converges in L 2 ([0, 1] 2 ).By (2.6), all λ l ≥ 0 and it is clear that now we would like to define φ in (2.5) by However, we still need to prove that the series in (2.7) converges a.s. on [0, 1] 2 and that Since the series in (2.7) converges in L 2 ([0, 1] 2 ), we can choose a subsequence (n j ) such that the (2.9) and for x ∈ [i2 −m , (i + 1)2 −m ) we can define Therefore, for x and y in the same dyadic interval [i2 −m , (i + 1)2 −m ), and since |E(f |D m )| ≤ 1, (2.9) implies that lim The fact that all λ l ≥ 0 implies that for any n ≥ 1 By the martingale convergence theorem, E(ϕ l |D m ) → ϕ l a.s. as m → +∞ and, therefore, The first term goes to zero a.s.by the Borel-Cantelli lemma since f n j − f 2 ≤ j −2 and the second term can be bounded by and, by (2.10), also goes to zero a.s. as n → +∞.Finally, the map φ is measurable since for any open ball B ε (h) in ℓ 2 of radius ε centered at h, φ −1 (B ε (h)) can be written as and the left hand side is obviously a measurable function.This finishes the proof.
Lemma 2 proves that if (h l , t l ) is an i.i.d.sequence from distribution η = λ • (φ, g) −1 on ℓ 2 × R + then the law of R in (2.4) coincides with the the law of To prove (1.2) it remains to show that h l 2 ≤ t l a.s. and define a l = t l − h l 2 .

Lemma 3
The measure η is concentrated on the set {(h, t) : h 2 ≤ t}.
Using Lemma 4, we will now construct a "consistent" sequence of laws (η N ) recursively as follows.Suppose that the measure η N that generates Ψ N (R) as in (2.11) has already been defined.Suppose now that Ψ N +1 (R) is generated by an i.i.d.sample (h l , t l ) from some measure η N +1 .Since Ψ N (Ψ N +1 (R)) = Ψ N (R), (3.6) observation (b) above implies that Ψ N (R) can also be generated by ψ N (h l , t l ) from measure η N +1 • ψ −1 N .Lemma 4 then implies that there exists a unitary operator q on ℓ 2 such that since ψ N and (q, id) obviously commute.We now redefine η N +1 to be equal to η N +1 •(q, id) −1 .
Clearly, Ψ N +1 (R) is still generated by an i.i.d.sequence from this new measure η N +1 and in addition we have Let A N := ℓ 2 × [0, N).Since ψ N (h, t) ∈ A N if and only if (h, t) ∈ A N and ψ N (h, t) = (h, t) on A N , the consistency condition (3.7) implies that the restrictions of measures η N and η N +1 to A N are equal.Therefore, η N converges in total variation to η = η N ⇂ A N \A N−1 .Since N and since ψ N is continuous, letting N ′ → +∞ gives η N = η • ψ −1 N .Finally, letting N → +∞ proves that R is generated by an i.i.d.sample from η which proves representation (2.11) in the unbounded case, and Lemma 3 again implies (1.2).
.10)so the map φ in (2.8), indeed, maps [0, 1] into the unit ball of ℓ 2 .Let us now show that (2.5) holds, i.e. the series in (2.7) converges a.s.Given n ≥ 1, let us take n j ≥ n and write