A counter example to central limit theorem in Hilbert spaces under a strong mixing condition

We show that in a separable infinite dimensional Hilbert space, uniform integrability of the square of the norm of normalized partial sums of a strictly stationary sequence, together with a strong mixing condition, does not guarantee the central limit theorem.


Introduction and notations
Let (Ω, F, µ) be a probability space and (S, d) a separable metric space.We say that the sequence of random variables (X n ) n∈Z from Ω to S is strictly stationary if for all integer d and all integer k, the d-uple (X 1 , . . ., X d ) has the same law as (X k+1 , . . ., X k+d ).
Rosenblatt introduced in [Ros56] the measure of dependence between two subσ-algebras A and B: An other one is β-mixing, which is defined by where the supremum is taken over the finite partitions {A 1 , . . ., A I } and {B 1 , . . ., B J } of Ω, which consist respectively of elements of A and B. It was introduced by Volkonskii and Rozanov in [VR59].
In order to measure dependence of a sequence of random variables, say X := (X j ) j∈Z (assumed strictly stationary for simple), we define F n m as the σ-algebra generated by the X j for m j n, where −∞ m n +∞.
Then mixing coefficients are defined by (1) which will be simply writen α(n) (respectively β(n)) when there is no ambiguity.We say that the strictly stationary sequence (X j ) j is αmixing (respectively β-mixing) if lim n→∞ α(n) = 0 (respectively lim n→∞ β(n) = 0).Sequences which are α-mixing are also called strong-mixing.Notice that the inequality α(A, B) 2β(A, B) for any two sub-σ-algebras A and B implies that each β-mixing sequence is Date: January 10, 2014.
strong mixing.We refer the reader to Bradley's book [Bra07] for further information about mixing conditions.Let (V, • ) be a separable normed space.We can represent a strictly stationary sequence (X j ) j by X j = f • T j , where T : Ω → Ω is measurable and measure preserving, that is, µ(T −1 (S)) = µ(S) for all S ∈ F.
Given an integer N , we define S N (f ) := satisfies the following assumptions: (1) lim N →+∞ σ N (f ) = +∞; (2) f dµ = 0: (3) lim n→+∞ α(n) = 0; (4) the family converges in distribution to a Gaussian law.First, we restrict ourselves to separable normed spaces in order to avoid measurability issues of sums of random variables.Corollary 10.9. in [LT91] asserts that a separable Banach space B with norm • B is isomorphic to a Hilbert space if and only if for all random variable X with values in B, the conditions E [X] = 0 and E X 2 B < ∞ are necessary and sufficient for X to satisfy the central limit theorem.By "X satisfies the CLT", we mean that if (X j ) j 1 is a sequence of independent random variables, with the same law as X, the sequence n −1/2 n j=1 X j n 1 weakly converges in B. Hence we cannot expect a generalization in a class larger than separable Hilbert spaces.Such a space is necessarily isomorphic to H := ℓ 2 (R), the space of square sumable sequences (x n ) n 1 endowed with the inner product x, y H := +∞ n=1 x n y n .We shall denote by e n the sequence whose all terms are 0, except the n-th which is 1.Bold letters denote both randoms variables taking their values in H and elements of this space.
General considerations about probability measures and central limit theorem in Banach spaces are contained in Araujo and Giné's book [AG80].
are sequences of non-negative real numbers, a n b n means that a n Cb n , where C doesn't depend on n.In an analogous way, we define a n b n .When a n b n a n , we simply write a n ≍ b n .

Our main results are
Theorem A. There exists a probability space (Ω, F, µ) such that given 0 < q < 1, we can construct a strictly stationary sequence X = (f σ N (f ) , N ∈ I is not tight in H; furthermore, given a sequence (c N ) N 1 of real numbers going to infinity, we have either Then there exists a strictly stationary sequence

random variables with values in H such that a), d), e) of Theorem A and the following two properties hold
Remark 2. Theorem A' gives a control of the mixing coefficients on a subsequence.When b N := N −2 for example, the construction gives a better estimation for the considered subsequence than what we get by Theorem A.
Tone has established in [Ton11] a central limit theorem for strictly stationary random fields with values in H under ρ ′ -mixing conditions.For sequences, these coefficients are defined by where the supremum is taken over all the non-zero functions f and g such that f and g are respectively σ(X j , j ∈ S 1 ) and σ(X j , j ∈ S 2 )-measurable, where S 1 and S 2 are such that min s∈S 1 ,t∈S 2 |s − t| n, while L 2 (H) denote the collection of equivalence classes of random variables X : Ω → H such that X 2 H is integrable.So "interlaced index sets" can be considered, which is not the case for α and β-mixing coefficient.Taking f and g as characteristic functions of elements of Dedecker and Merlevède have shown in [DM10] that under the assumption we can find a sequence (Z i ) i∈N of Gaussian random variables with values in H such that almost surely, A partial generalization of the finite dimensional result was proved by Politis and Romano [PR94], namely, the conditions E X 1 2+δ H finite for some positive δ and j α X (j) δ 2+δ guarantees the convergence of n −1/2 n j=1 X j to a Gaussian random variable N , whose covariance operator S satisfies Similar results were obtained by Dehling [Deh83].Rio's inequality [Rio93] asserts that given two real valued random variables X and Y with finite two order moments, It was extented by Merlevède et al. [MPU97], namely, if X and Y are two random variables with values in H, with respective quantile function where α := α(σ(X), σ(Y)).
From this inequality, they deduce a central limit theorem for a stationary sequence (X j ) j∈Z of H-valued zero-mean random variables satisfying (3) where α −1 is the inverse function of x → α X (⌊x⌋).
Discussion after Corollary 1.2 in [Rio00] proves that the later result implies Politis' one.
Relative optimality of condition (3) (cf.[DMR94]) can give a finite-dimensional counter-example to the central limit theorem when this condition is not satisfied.Here, the condition of uniform integrability prevents such counter-examples.
2. The proof 2.1.Construction of f .In order to construct a counter-example, we shall need the following lemma, which will be proved latter.
We will denote U the Koopman operator associated to T , which acts on measurable functions by U (f )(x) := f (T (x)).
Lemma 3. Let (u k ) k 1 ⊂ (0, 1) be a sequence of numbers.Then there exists a dynamical system (Ω, F, µ, T ) and a sequence of random variables Recall that e k is the k-th element of the canonical orthonormal system of H = ℓ 2 (R).We define (4) where the ξ i 's are constructed using to Lemma 3 taking u k := n −2 k .Conditions on the increasing sequence of integers (n k ) k 1 will be specified latter.
Then X k := f • T k is a strictly stationary sequence.Note that f 2 H is an integrable random variable whenever k 1 n k is convergent.In the sequel, the choice of n k will guarantee this condition.
2.2.Preliminary results.We express S N (f k ) as a linear combination of independent random variables.By direct computations (cf.[Gor69] also e.g.[Vol93]), we get (5) This formula can be simplified if we distinguish the cases N n k and n k < N (we break the third sum at the index i = 0 if necessary).This gives The computation of the expectation of the square of partial sums gives Notation 4. If N is a positive integer and (n k ) k 1 is an increasing sequence of integers, denote by i(N ) the unique integer for which n i(N ) N < n i(N )+1 .
Proposition 5. Assume that (n k ) k 1 satisfies the condition Proof.Using (8), the fact that M 3 ≍ M j=1 j 2 and σ 2 From (8) in the case n k N , we deduce (10) Since n i(N )+1 N and the series k 1 n 1−p k is convergent, we obtain (11) Combining ( 9) and ( 11), we get Proposition 6. Assume that k n −a k is convergent for any positive real number a. Then for each integer p, f H has a finite moment of order p.
Proof.We shall use Rosenthal's inequality (Theorem 3, [Ros70]): if M is an integer, Y 1 , . . ., Y M are independent real valued zero mean random variables for which E |Y i | q < ∞, q > 2, for each i, then there exists a constant C depending only on q for which ( 13) If q = 2p is given then we have We provide a sufficient condition for the uniform integrability of the family S := Proof.For N 1, we have: , hence it is enough to prove that the families , N 1 =: {v N , N 1} , and , N 1    are uniformly integrable.For S 1 and S 4 , we shall show that these families are bounded in L p for p > 1 as in C.
From (9), we get Since p − 2 0, we obtain that S 1 is bounded in L p hence uniformly integrable.
• for S 2 : using (7) in the case n k N and Proposition 5, we get .
Since u N 1 → 0 and u N ∈ L 1 for each N , the family S 2 is uniformly integrable.• for S 3 : using (6) in the case n k > N and Proposition 5, we get .
Since v N 1 → 0 and v N ∈ L 1 for each N , the family S 3 is uniformly integrable.• for S 4 : as for S 1 , we shall show that this family is bounded in L p .We have, using ( 6) and ( 13) Also, using (8), we have The condition n k+1 n p k gives boundedness in L p of S 4 .This concludes the proof of d).
Proposition 8. Assume that (n k ) k 1 is such that S is uniformly integrable and k n −1 k is convergent.Then for each I ⊂ N infinite, the collection S N (f ) σ N (f ) , N ∈ I is not tight in H. Its finite-dimensional distributions converge to 0 in probability.
Furthermore, if (c N ) N 0 is a sequence of positive numbers going to infinity, we have either > 0, and in this case the sequence S N (f ) c N , N 1 is not tight.
Proof.We first prove that the finite dimensional distributions of S N (f ) σ N (f ) converge weakly to 0.
For each d ∈ N, we have S N (f ),e d H σ N (f ) → 0 in distribution.Indeed, we have by ( 5) to 0 in probability as N goes to infinity, using Proposition 5 and the estimate This can be extended replacing e d by any v ∈ H by an application of Theorem 4.2. in [Bil68].By Proposition 4.15 in [AG80], the only possible limit is the Dirac measure at 0 H .
Assume that the sequence S N (f ) σ N (f ) , N 1 is tight.The sequence is a uniformly integrable sequence of random variables of mean 1.A weakly convergent subsequence would go to 0 H .According to Theorem 5.4 in [Bil68], we should have that the limit random variable has expectation 1.This contradiction gives the result when I = N \ {0}.Applying this reasonning to subsequences, one can see that for any infinite subset I of N \ {0}, the family S N (f ) σ N (f ) , N ∈ I is not tight.Let (c N ) N 1 be a sequence of positive real numbers such that lim N →+∞ c N = +∞.
• first case: σ N (f ) c N converges to 0. In this case, the sequence S N (f ) 2 c 2 N N 1 converges to 0 in L 1 , hence the sequence S N (f ) Hence there is some r > 0 and a sequence of integers l i ↑ ∞ such that for each i, Assume that the family , i 1 is tight.This means that given a positive ε, one can find a compact set K = K(ε) such that for each i, µ We can assume that this compact set is convex and contains 0 (we consider the closed convex hull of K ∪ {0}, which is compact by Theorem 5.35 in [AB06]).Then we have and we would deduce tightness of σ l i (f ) , i 1 , which cannot happen.
Remark 9.In the second case, it may happen that the finite dimensional distributions doesn't converge to degenerate ones, for example with c N := N .
2.3.Proof of Theorem A. Notice that if n k+1 n p k for some p > 1 and n 1 = 2, then n k 2 p k , hence the condition of Proposition 6 is fulfilled.We get a) since each f k has expectation 0.
We denote ⌊x⌋ := sup {k ∈ Z, k x} the integer part of the real number x. Proof.We define β k (n) as the n-th β-mixing coefficient of the sequence (f k • T i ) i 0 .