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

ISSN:1083-589X in PROBABILITY

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

Davide Giraudo

Dalibor Volný

Abstract

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 theo- rem.

Keywords: Central limit theorem ; Hilbert space ; mixing conditions ; strictly stationary pro- cess.

AMS MSC 2010:60F05 ; 60G10.

Submitted to ECP on January 9, 2014, final version accepted on August 24, 2014.

1 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(Xn)_n∈ZfromΩtoSisstrictly stationaryif for all in- tegerdand all integerk, thed- uple(X1, . . . , Xd)has the same law as(Xk+1, . . . , Xk+d).

Rosenblatt introduced in [18] the measure of dependence between two sub-σ-algebrasAandB:

α(A,B) := sup{|µ(A∩B)−µ(A)µ(B)|, A∈ A, B∈ B}. Another one isβ-mixing, which is defined by

β(A,B) := 1 2sup

I

X

i=1 J

X

j=1

|µ(A_i∩B_j)−µ(A_i)µ(B_j)|,

where the supremum is taken over the finite partitions{A1, . . . , AI}and{B1, . . . , BJ}of Ω, which consist respectively of elements ofAandB. It was introduced by Volkonskii and Rozanov in [21].

In order to measure dependence of a sequence of random variables, say X :=

(X_j)_j∈

Z(assumed strictly stationary for simplicity), we defineF_mⁿ as theσ-algebra gen- erated by theX_jform6j6n, where−∞6m6n6+∞.

Then mixing coefficients are defined by

αX(n) :=α F_−∞⁰ ,F_n^+∞

(1.1)

∗Université de Rouen, France. E-mail:[email protected]

†Université de Rouen, France.

E-mail:[email protected]

β_X(n) :=β F_−∞⁰ ,F_n^+∞

, (1.2)

which will be simply writenα(n)(respectivelyβ(n)) when there is no ambiguity.

We say that the strictly stationary sequence(Xj)_jisα-mixing(respectivelyβ-mixing) iflim_n→∞α(n) = 0(respectivelylim_n→∞β(n) = 0). Sequences which areα-mixing are also called strong-mixing. Notice that the inequality 2α(A,B) 6 β(A,B)for any two sub-σ-algebrasAandBimplies that eachβ-mixing sequence is strong mixing. We refer the reader to Bradley’s book [4] for further information about mixing conditions.

Let (V,k·k) be a separable normed space. We can represent a strictly stationary sequence(X_j)_jbyX_j =f◦T^j, whereT: Ω→Ωis measurable and measure preserving, that is,µ(T⁻¹(S)) =µ(S)for allS∈ F (see [8], p.456, second paragraph).

Given an integerN, we defineSN(f) :=

N−1

X

j=0

f◦T^j and(σN(f))²:=Eh

kSN(f)k²i . When V = R^d, d ∈ N^∗ it is well-known that if f ◦T^j

j>0 satisfies the following assumptions:

1. limN→+∞σN(f) = +∞; 2. R

f dµ= 0:

3. lim_n→+∞α(n) = 0; 4. the familyn_kS

N(f)k²

(σ_N(f))², N>1o

is uniformly integrable, then

1

σ_N(f)SN(f)

N>1converges in distribution to a Gaussian law. It was established ford= 1by Denker [7], Mori and Yoshihara [14] using a blocking argument. Volný [22]

gave a proof fordarbitrary based on approximation by an array of independent random variables.

A natural question would be: what if we replaceR^dby another normed space?

First, we restrict ourselves to separable normed spaces in order to avoid measurabil- ity issues of sums of random variables. Corollary 10.9. in [11] asserts that a separable Banach space B with norm k·k_B is isomorphic to a Hilbert space if and only if for all random variablesX with values inB, the conditionsE[X] = 0andEh

kXk²_Bi

<∞are necessary and sufficient forX to satisfy the central limit theorem. By "Xsatisfies the CLT", we mean that if(X_j)_j

>1is a sequence of independent random variables, with the same law asX, the sequence

n^−1/2Pn j=1Xj

n>1

weakly converges inB. Hence we cannot expect a generalization in a class larger than separable Hilbert spaces. Such a space is necessarily isomorphic toH:=`²(R), the space of square sumable sequences (xn)_n>1 endowed with the inner producthx, yi_H :=P+∞

n=1xnyn. We shall denote byen

the sequence whose all terms are0, except then-th which is1. Bold letters denote both randoms variables taking their values inHand elements of this space.

General considerations about probability measures and central limit theorem in Ba- nach spaces are contained in Araujo and Giné’s book [2].

Notation 1. If(an)_n>1,(bn)_n>1 are sequences of non- negative real numbers,an . bn

means thata_n 6Cb_n, whereC does not depend onn. In an analogous way, we define a_n&b_n. Whena_n.b_n.a_n, we simply writea_nb_n.

Our main results are

Theorem A. There exists a probability space(Ω,F, µ)such that given0 < q <1, we can construct a strictly stationary sequenceX= (f◦T^k) = (X_k)_k∈Ndefined onΩ, taking its values inH, such that:

a) E[f] = 0,E[kfk^p_H]is finite for eachp; b) the limitlim_N→∞σ_N(f)is infinite;

c) the process(Xk)_k∈N isβ-mixing, more precisely,βX(j) =O

1 j^q

; d) the familyn_kS

N(f)k²_H

σ²_N(f) , N>1o

is uniformly integrable;

e) ifI⊂Nis infinite, the familyn_S

N(f)

σ_N(f), N∈Io

is not tight inH; furthermore, given a sequence(cN)_N>1of real numbers going to infinity, we have either

• lim_N→+∞^σN_c^(f)

N = 0, hence_S

N(f) c_N

N>1converges to0_Hin distribution, or

• lim sup_N→+∞^σN_c^(f)

N >0, and in this case the collectionn_S

N(f)

c_N , N >1o is not tight.

Theorem A’. Let (bN)N>1 and (hN)N>1 be sequences of positive real numbers, with lim_N→∞bN = 0andlimN→∞hN =∞. Then there exists a strictly stationary sequence X:= (f◦T^k)k∈N = (Xk)k∈N of random variables with values inHsuch that A, A, A of Theorem A and the following two properties hold:

b’) we haveσ_N²(f).N·h_N and ^σ2N_N^(f) → ∞;

c’) the process(X_k)_k∈N is β-mixing, and there is an increasing sequence(n_k)_k>1 of integers such that for eachk,β_X(n_k)6b_nk.

Remark 2. Theorem A shows that Denker’s result does not remain valid in its full gen- erality in the context of Hilbert space valued random variables.

Furthermore, a careful analysis of the proof of Proposition 6 shows that for the construction given in Theorem A, we have σ_N²(f) = N ·h(N)with hslowly varying in the strong sense. Theorem 1 of [12] does not remain valid in the Hilbert space setting.

Indeed, the arguments given in pages 654-655 show that the conditions of Denker’s theorem together with the assumption thatσ_N² =N·h(N)withhslowly varying in the strong sense imply those of Theorem 1. These arguments are still true in the Hilbert space setting.

Remark 3. Theorem A’ gives a control of the mixing coefficients on a subsequence.

WhenbN :=N⁻²for example, the construction gives a better estimation for the consid- ered subsequence than what we get by Theorem A.

Tone has established in [20] a central limit theorem for strictly stationary random fields with values in Hunder ρ⁰-mixing conditions. For sequences, these coefficients are defined by

ρ⁰(n) := sup

(|E[hf,gi_H]− hE[f],E[g]i_H| kfk_L2(H)kgk_L2(H)

) ,

where the supremum is taken over all the non-zero functionsf andgsuch thatf andg are respectivelyσ(X_j, j∈S₁)andσ(X_j, j ∈S₂)-measurable, whereS₁andS₂are such thatmin_{s∈S1,t∈S2}|s−t|>n, whileL²(H)denote the collection of equivalence classes of random variablesX: Ω→ Hsuch thatkXk²_His integrable.

So "interlaced index sets" can be considered, which is not the case for α and β- mixing coefficient. Takingf andg as characteristic functions of elements ofF_−∞⁰ and

F_n^+∞ respectively, one can see that α(n) 6 ρ⁰(n), hence ρ⁰-mixing condition is more restrictive thanα-mixing condition.

A partial generalization of the finite dimensional result was proved by Politis and Ro- mano [15], namely, the conditionsEkX1k^2+δ_H finite for some positiveδandP

jαX(j)^2+δδ guarantees the convergence ofn^−1/2Pn

j=1X_jto a Gaussian random variableN, whose covariance operatorSsatisfies

E

hN, hi²

=hSh, hiH= Var(hX1, hi) + 2

+∞

X

i=1

Cov (hX1, hi,hX1+i, hi). Similar results were obtained by Dehling [6].

Rio’s inequality [16] asserts that given two real valued random variables X andY with finite two order moments,

|E[XY]−E[X]E[Y]|62

Z α(σ(X),σ(Y)) 0

QX(u)QY(u)du.

It was extented by Merlevède et al. [13], namely, ifXandYare two random variables with values inH, with respective quantile functionQ_kXk

H andQ_kYk

H, then

|E[hX,YiH]− hE[X],E[Y]iH|618 Z α

0

Q_kXkHQ_kYkHdu, whereα:=α(σ(X), σ(Y)).

From this inequality, they deduce a central limit theorem for a stationary sequence (X_j)_j∈ZofH-valued zero-mean random variables satisfying

Z 1 0

α⁻¹(u)Q²_kX0k

H(u)du <∞, (1.3)

whereα⁻¹is the inverse function ofx7→αX(bxc).

Discussion after Corollary 1.2 in [17] proves that the later result implies Politis’ one.

Relative optimality of condition (1.3) (cf. [9]) 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.

Defining α2,X(n) := sup_i>j>nα(F_−∞⁰ , σ(Xi,Xj)) and QX₀ the right-continuous in- verse of the functiont7→µ{kX0k_H> t}(that is,

Q_X0(u) := inf{t∈R, µ{kX0k_H> t}6u}), Dedecker and Merlevède have shown in [5] that under the assumption

+∞

X

k=1

Z α_2,X(k) 0

Q²_X0(u)du <∞,

we can find a sequence(Zi)_i∈Nof Gaussian random variables with values inHsuch that almost surely,

Sn−

n

X

i=1

Zi

H

=op

nlog logn .

2 The proof

2.1 Construction off

In order to construct a counter-example, we shall need the following lemma, which will be proved later.

We will denoteU the Koopman operator associated toT, which acts on measurable functions byU(f)(x) :=f(T(x)).

Lemma 4. Let(u_k)_k>1⊂(0,1)be a sequence of numbers. Then there exists a dynami- cal system(Ω,F, µ, T)and a sequence of random variables(ξ_k)_k>1such that

1. for eachk>1,µ(ξk= 1) =µ(ξk=−1) = ^u₂^k andµ(ξk = 0) = 1−uk; 2. the random variables(Uⁱξ_k, k>1, i∈Z)are mutually independent.

Recall thatekis thek-th element of the canonical orthonormal system ofH=`²(R). We define

f_k:=

nk−1

X

i=0

U⁻ⁱξ_kandf :=

+∞

X

k=1

f_ke_k, (2.1)

where theξi’s are constructed using to Lemma 4 takinguk :=n⁻²_k . Conditions on the increasing sequence of integers(nk)_k>1will be specified latter.

ThenXk :=f ◦T^k is a strictly stationary sequence. Note thatkfk²_H is an integrable random variable whenever P

k 1

n_k is convergent. In the sequel, the choice ofnk will guarantee this condition.

2.2 Preliminary results

We express SN(fk) as a linear combination of independent random variables. By direct computations, we get

fk=nkξk+ (I−U)

−1

X

i=1−nk

(nk+i)Uⁱξk, (2.2) hence

S_N(f_k) =n_k

N−1

X

j=0

U^jξ_k+

−1

X

i=1−n_k

(n_k+i)Uⁱξ_k−

N−1

X

i=N−nk+1

(n_k+i−N)Uⁱξ_k.

This formula can be simplified if we distinguish the casesN >nk andnk< N(we break the third sum at the indexi= 0if necessary). This gives

SN(fk) =

N−1

X

j=0

(N−j)U^jξk+

N−nk

X

j=1−nk

(nk+j)U^jξk

+N

−1

X

j=1+N−nk

U^jξk, ifN < nk, (2.3)

SN(fk) =nk N−nk

X

j=0

U^jξk+

N−1

X

j=N−nk+1

(N−j)U^jξk

+

−1

X

j=1−nk

(nk+j)U^jξk, ifN >nk. (2.4) The computation of the expectation of the square of partial sums gives

σ²_N(fk) =



















1 n²_k



2

N

X

j=1

j²+ (nk−N−1)N²



 ifN < nk,

1 n²_k



n²_k(N−n_k+ 1) + 2

nk−1

X

j=1

j²



 ifN >n_k.

(2.5)

Notation5. IfN is a positive integer and(n_k)_k>1is an increasing sequence of integers, denote byi(N)the unique integer for whichn_i(N)6N < n_i(N)+1.

Proposition 6. Assume that(nk)k>1satisfies the condition

there isp >1such that for eachk, nk+1>n^p_k. (C) Thenσ_N²(f)N·i(N).

Proof. Using (2.5), the fact thatM³PM

j=1j²andσ²_N(f) =P

k>1σ²_N(fk), we have

σ²_N(f)>

i(N)

X

k=1

σ²_N(f_k)N

i(N)

X

j=1

1 =N·i(N). (2.6)

From (2.5) in the casenk >N, we deduce X

k>i(N)+1

σ²_N(f_k). X

k>i(N)+1

N²

n_k 6 N²

n_i(N)+1 + X

k>i(N)+1

N² n_k

1

n^p−1_k . (2.7) Sincen_i(N)+1>Nand the seriesP

k>1n^1−p_k is convergent (by the ratio test), we obtain X

k>i(N)+1

σ_N²(f_k).N+N X

k>i(N)+1

1

n^p−1_k .N. (2.8)

Combining (2.6) and (2.8), we get

N·i(N).σ_N²(f).

i(N)

X

k=1

σ_N²(fk) + X

k>i(N)+1

σ_N²(fk).N·i(N) +N .N·i(N). (2.9)

Proposition 7. Assume thatP

kn^−a_k is convergent for any positive real numbera. Then for each integerp,kfk_Hhas a finite moment of orderp.

Proof. We shall use Rosenthal’s inequality (Theorem 3, [19]): there exists a constantC depending only onqsuch that ifM is an integer,Y1, . . . , YM are independent real valued zero mean random variables for whichE|Yi|^q <∞for eachi, then

E

M

X

j=1

Yj

q

6C







M

X

j=1

E|Yj|^q+





M

X

j=1

E Y_j²





q/2



. (2.10)

Ifq= 2pis given then we have

E|fk|^2p .n⁻¹_k +n^−p_k .n⁻¹_k . (2.11)

We provide a sufficient condition for the uniform integrability of the family S :=

n_kS

N(f)k²_H

σ²_N(f) , N>1o .

Proposition 8. If(n_k)_k>1satisfies (C), thenSis uniformly integrable.

Proof. ForN>1, we have:

kSN(f)k²_H σ²_N(f) =

i(N)−1

X

j=1

|SN(fj)|² σ²_N(f) +

SN(f_i(N))

2

σ²_N(f) +

SN(f_i(N)+1)

2

σ_N²(f) + X

j>i(N)+2

|SN(fj)|² σ_N²(f) ,

hence it is enough to prove that the families

S1:=







i(N)−1

X

k=1

|SN(fk)|² σ²_N(f) , N >1





 ,

S2:=

( SN(fi(N))

2

σ_N²(f) , N >1 )

=:{uN, N >1},

S₃:=

( SN(f_i(N)+1)

2

σ²_N(f) , N >1 )

=:{v_N, N >1}, and

S4:=





 X

k>i(N)+2

|SN(f_k)|² σ²_N(f) , N>1







are uniformly integrable. ForS1andS4, we shall show that these families are bounded inL^pforp∈(1,2]as in (C).

• forS1: using the expression in (2.4) and (2.10) withq:= 2p >2, we have

Eh

|SN(fk)|^2pi 6C



2

n_k

X

j=1

j^2p

n²_k +n^2p_k (N−nk) n²_k



+C



2

n_k

X

j=1

j²

n²_k +(N−nk)n²_k n²_k





p

. 1

n²_k

n^2p+1_k + (N−nk)n^2p_k + 1

n^2p_k n³_k+ (N−nk)n²_k^p

=N n^2p_k

n²_k +N^pn^2p_k n^2p_k

=N n^2(p−1)_k +N^p hence

SN(fk)²

_p.N^1/pn²

p−1 p

k +N,

which gives

i(N)−1

X

k=1

|SN(fk)|² σ_N²(f)

_p

.

Pi(N)−1

k=1 (N^1/pn²

p−1 p

k +N) σ²_N(f)

.

i(N)n²

p−1 p

i(N)−1+N i(N) σ²_N(f) .

From (2.6), we get

i(N)−1

X

k=1

|SN(fk)|² σ²_N(f)

_p

. n²

p−1 p

i(N)

n_i(N) + 1 =n

p−2 p

i(N)+ 1.

Sincep−260, we obtain thatS1is bounded inL^phence uniformly integrable.

• forS2: using (2.4) in the casen_k 6N and Proposition 6, we get kuNk₁. N

σ_N²(f) . 1

i(N). (2.12)

SincekuNk₁→0andu_N ∈L^{1for each}N, the familyS2is uniformly integrable.

• forS3: using (2.3) in the casen_k > N and Proposition 6, we get kvNk₁. N²

n_i(N)+1σ_N²(f) . N

N·i(N). (2.13)

Sincekv_Nk₁→0andv_N ∈L^{1for each}N, the familyS₃is uniformly integrable.

• forS₄: as forS₁, we shall show that this family is bounded inL^pwithp∈(1,2]. We have, using (2.3) and (2.10)

Eh

|SN(fk)|^2pi

. 1

n²_k(N^2p+1+N^2p(nk−N)) + 1

n^2p_k (N³+ (nk−N)N²)^p

=N^2p n_k +N^2p

n^p_k .N^2p

n_k asN 6nk. We thus get that

X

k>i(N)+2

|S_N(f_k)|² _p

.N² X

k>i(N)+2

1 n^1/p_k

.

Also, using (2.5), we have

σ_N²(f)&N² X

k>i(N)+1

1 nk

.

The conditionnk+1>n^p_k gives boundedness inL^pofS4. This concludes the proof of A.

Proposition 9. Assume that(n_k)_k>1is such thatS is uniformly integrable andP

kn⁻¹_k is convergent. Then for eachI⊂Ninfinite, the collectionn_S

N(f)

σN(f), N ∈Io

is not tight in H. Its finite-dimensional distributions converge to0in probability.

Furthermore, if(cN)N>0is a sequence of positive numbers going to infinity, we have either

• limN→+∞σN(f)

c_N = 0, hence_S

N(f) c_N

N>1converges to0Hin distribution, or

• lim sup_N→+∞^σN_c^(f)

N >0, and in this case the sequencen_S

N(f)

c_N , N>1o

is not tight.

Proof. We first prove that the finite dimensional distributions of ^S_σ^N(f)

N(f) converge weakly to0.

For eachd ∈N^{, we have hSN}σ^(f),e_N(f)^diH → 0in distribution. Indeed, we have by (2.2) thathSN(f),edi_H =ndPN−1

i=0 Uⁱξd+ (I−U^N)P−1

i=1−nd(nd+i)Uⁱξd. We conclude noticing thatσN(f)⁻¹(I−U^N)P−1

i=1−nd(nd+i)Uⁱξdgoes to0in probability asN goes to infinity, using Proposition 6 and the estimate

E nd N−1

X

i=0

Uⁱξd

!²

=N .σ_N²(f) i(N)

This can be extended replacing e_d by anyv∈ Hby an application of Theorem 4.2.

in [3]. By Proposition 4.15 in [2], the only possible limit is the Dirac measure at0_H. Assume that the sequencen_S

N(f)

σ_N(f), N >1o

is tight. The sequence_kS

N(f)k²_H σ²_N(f)

N>1

is a uniformly integrable sequence of random variables of mean1. A weakly convergent subsequence would go to 0_H. According to Theorem 5.4 in [3], we should have that the limit random variable has expectation1. This contradiction gives the result when I=N\ {0}. Applying this reasonning to subsequences, one can see that for any infinite subsetIofN\ {0}, the familyn_S

N(f)

σ_N(f), N ∈Io

is not tight.

Let(c_N)_N>1be a sequence of positive real numbers such thatlim_N→+∞c_N = +∞.

• first case: ^σN_c^(f)

N converges to 0. In this case, the sequence_kS

N(f)k² c²_N

N>1

con- verges to 0 in L¹, hence the sequence _S

N(f) cN

N>1

converges in distribution to 0_H.

• second case:lim sup_N→∞^σN_c^(f)

N >0. Hence there is somer >0and a sequence of integersli↑ ∞such that for eachi, ^σli_c^(f)

li > ¹r, that is,cl_i 6rσl_i(f). Assume that the familyn_S

li(f)

c_li , i>1o

is tight. This means that given a positiveε, one can find a compact setK =K(ε)such that for eachi,µn_S

li(f) c_li ∈Ko

>1−ε. We can assume that this compact set is convex and contains 0 (we consider the closed convex hull ofK∪ {0}, which is compact by Theorem 5.35 in [1]). Then we have

S_li(f) cl_i

∈K

=

S_li(f) σl_i(f) ∈ c_li

σl_i(f)K

⊂

S_li(f) σl_i(f) ∈rK

,

and we would deduce tightness ofn_S

li(f)

σ_li(f), i>1o

, which cannot happen.

Remark 10. In the second case, it may happen that the finite dimensional distributions does not converge to degenerate ones, for example withcN :=N.

2.3 Proof of Theorem A

Notice that if nk+1 > n^p_k for some p > 1 and n1 = 2, then nk > 2^pk, hence the condition of Proposition 7 is fulfilled. We get A since eachf_k has expectation0.

We denotebxc:= sup{k∈Z, k6x}the integer part of the real numberx.

Proposition 11. Letp > 1. With n_k := b2^pkc(which satisfies (C)), we have for each positive integerl,

β_X(l). 1 l^p1 .

Proof. We defineβk(n)as then-thβ-mixing coefficient of the sequence(fk◦Tⁱ)i>0. By Lemma 5 of [10], we have the estimate β_k(0) 6 4n⁻¹_k for each k. Using then Proposition 4 of this paper (cf. [4] for a proof), we get thatβ_X(n_k).P

j>k 1

nj for each integerk. Sincepⁱ>iforilarge enough,

X

j>k

1 nj

=

+∞

X

i=0

1 2^pi+k =

+∞

X

i=0

1 2^pipk .

+∞

X

i=0

1 2ⁱ

1 2^pk = 2

2^pk,

we get

βX(N)6βX(n_i(N)). 1

n_i(N) = 1

n^1/p_i(N)+1 6 1 N^1/p.

This proves A. For any p, the choicenk := b2^pkcsatisfies the condition of Proposi- tion 8, which proves A. We conclude the proof by Proposition 9.

Remark 12. For each of these choices, σ²_N(f)behaves asymptotically likeNlog logN. Theorem A’ shows that we can construct a process which satisfies the same asymptotic behavior of partial sums and has a variance close to a linear one.

A question would be: can we construct a strictly stationary sequence with all the properties of Theorem A, except A which is replaced by an assumption of linear vari- ance?

2.4 Proof of Theorem A’

Let(hN)_N>1be the sequence involved in Theorem A’. We define for an integeruthe quantityh⁻¹(u) := inf{j∈N, hj>u}.

If(bk)k>1 is the given sequence (that can be assumed decreasing), we define induc- tively

n_k+1:= max

n²_k,b2^k bn_k

c, h⁻¹(k)

. (2.14)

Let N be an integer. We assume without loss of generality that the growth of the sequence(h_N)_N>1 is slow enough in order to guarantee that there exists ksuch that N =h⁻¹(k). We then havei(N)6k+ 16h_N+ 1, hence using Proposition 6, we get b’).

We have nk > 2^2k hence by a similar argument as in the proof of Theorem A, A is satisfied.

By a similar argument as in [10], we getβX(nk)6bn_k, hence c’) holds.

Remark 13. By (1.3), we cannot expect the relationship βX(·) 6 b_· for the whole se- quence.

Since for eachk,n_k+1 >n²_k, Proposition 8 and 9 apply. This concludes the proof of Theorem A’.

Proof of Lemma 4. Let Ω := [0,1]^N∗×Z, where [0,1] is endowed with Borel σ- algebra and Lebesgue measure, andΩwith the product structure.

For(k, j)∈N^∗×Z^andS⊂[0,1], letPk,j(S) :=Q

(i₁,i₂)∈N^∗×ZSi₁,i₂, whereSi₁,i₂ =S if(i1, i2) = (k, j)and[0,1]otherwise. Then we define

A⁺_k,j :=Pk,j([0,2⁻¹(uk)⁻¹]),

A⁻_k,j :=Pk,j([2⁻¹(uk)⁻¹,(uk)⁻¹]),

A⁽⁰⁾_k,j:=P_k,j([(u_k)⁻¹,1]), the mapT byT

(xk,j)_(k,j)∈N∗×Z

:= (xk,j+1)_(k,j)∈N∗×Z, and ξk:=χ_A+

k,0−χ_A⁻

k,0.

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Acknowledgments. The authors would like to thank both referees for helpful com- ments, and for suggesting Remark 2.