René Blacher

Keywords: Central Limit Theorem, moments, strongly mixing sequence, weak dependence, martingale, dependence density, higher order correlation coefficients

The higher order correlation coefficients are able to detect any dependence. So, in a previous paper, we obtained conditions about these coefficients equivalent to the convergence of moments. We have deduced a central limit theorem with minimal assumptions. However, it was assumed that all random variables have the same distribution. In this report, we remove this condition. This allows us to reduce the assumptions necessary for the convergence of moments for martingales and even to replace this assumption by a weaker hypothesis. On the other hand, we shall prove that these assumptions can be simplified when the random variables are bounded.

On the other hand, we will compare the different assumptions of asymptotic independence between them, in particular, strong mixing condition, weak dependence and condition HmI which we introduced in a previous paper We understand that it is this condition HmI which is closest to the minimum conditions to ensure asymptotic normality. Finally, we see that, if one has a process whose moments converge, moments converge also for almost all processes which has only the same multilinear correlation coefficients that the first process.

Résumé : Les coefficients de corrélation d'ordre supérieur sont capables de détecter toute dépendance. Aussi, dans un article précédent, on a obtenu des conditions sur ces coefficients équivalentes à la convergence des moments. On en a déduit un théorème de la limite centrale avec des hypothèses minimales. On supposait cependant que tous les variables aléatoires aient la même loi. Dans ce rapport, nous supprimons cette condition. Cela nous permet de diminuer les hypothèses nécessaires à la convergence des moments pour les martingales et même de remplacer cette hypothèse par une hypothèse plus faible. D'autre part, nous montrons que l'on peut simplifier ces thèoremes lorsque on utilise des variables alèatoires bornèes. D'autre part, nous allons comparer les différentes hypothèses d'indépendance asymptotique entre elles, en particulier, la condition fortement mélangeante, la faible dépendance et la condition HmI que nous avons introduite dans un précédent article. On verra que c'est cette condition HmI qui est la plus proche des conditions minimales permettant d'assurer la normalité asymptotique. Enfin, on verra aussi que si on a un processus dont les moments convergent, les moments convergent aussi pour presque tous les processus ayant seulement les mêmes coefficients de corrélation multilinéaire que ce premier processus.

Chapter 1

Higher Order Correlation Coefficients and MCLT

We first introduce the notations which we use throughout this report.

Notations 1.0.1 Let X n be a sequence of real random variables defined on a probability space (Ω, A, P ). We suppose E{X s } = 0 for all s ∈ N * and we set σ(n) 2 = E{(X 1 + X 2 + ..... + X n ) 2 } where E{.} is the expectation. We suppose E{|X s | p } < ∞ for all s ∈ N * and for all p ∈ N.

Hypothesis 1.0.1 We assume that X s has the law m s for each s ∈ N * . Then, we denote by {P s j } j∈N the family of orthonormal polynomials associated to m s . We suppose that there exists. If all the moments E{Z q n } converges to the real E{Z q }, one writes Z n M → Z. Moreover, by misuse of our notations, one writes Z n M → N (0, M 2 ) if Z has the normal distribution N (0, M 2 ).

Higher Order Correlation Coefficients

At first, we recall the definition of polynomial coefficients of correlation ρ j1,j2,....,jn , (j 1 , ...., j n ) ∈ N n . Notations 1.1.1 For all n ∈ N * , for all (j 1 , j 2 , j 3 , ....., j n ) ∈ N n , we set ρ j1,j2,j3,...........,jn = E{P 1 j1 (X 1 )P 2 j2 (X 2 ).......P n jn (X n )} and α j1,j2,........,jn = E{ P 1 j1 (X 1 ) P 2 j2 (X 2 )......... P n jn (X n )} where P s j = σ s,j P s j when σ s,j = E{(X s ) j P s j (X s ) }. If m s = m for all s ∈ N * , we set P s j = P j and σ s,j = σ j .

These dependence coefficients have been defined by Lancaster [START_REF] Lancaster | Orthogonal models for contingency tables[END_REF]. Each one measures a particular type of dependence between X 1 , .....,X n . For example, ρ j1,j2,....,jn = 0 if one of the X j 's is independent of the others. Moreover, if n=2, ρ j1,j2 is the polynomial correlation coefficient of order (j 1 , j 2 ) between X 1 and X 2 . In particular, α 1,1 is the covariance and ρ 1,1 is the classical correlation coefficient : ρ 1,1 measures the linear dependence.

More generally the α j1,j2,...,jn 's such that j s ≤ 1 measure multilinear dependence. Indeed, if j s = 0 or 1, there existe t 1 , ..., t p such that α j1,j2,...,jn = E{X t1 X t2 ....X tp }.

Moreover, if {P s j }, j ∈ N, is a basis of L 2 (R, m s ) for each s, dependence is completely determined by these coefficients. For better understanding the part of the ρ j1,j2,....,jn in dependence, we generalize the definitions of dependence density (cf [START_REF] Blacher R | Higher Order Correlation Coefficients[END_REF]), i. e. the density with respect to m ⊗ = m 1 ⊗ m 2 ⊗ ......... ⊗ m n : Definition 1.1.2 Assume that, for all s, {P s j }, j ∈ N, is a basis of L 2 (R, m s ). Then, we call dependence density of (X 1 , ...., X n ) the formal series: f (x 1 , ...., x n ) = 1 + (j1,...,jn)∈N n , at least 2 js =0 ρ j1,j2,....,jn P 1 j1 (x 1 )....P n jn (x n ) .

Indeed, one can generalize the results of [START_REF] Blacher R | Higher Order Correlation Coefficients[END_REF] by the following way.

Proposition 1.1.1 Let F Xs and F X be the distribution functions of X s and (X 1 , ...., X n ). Then, for all x = (x 1 , ...., x n ) ∈ R n , In particular, if (X 1 , ...., X n ) has a density f * with respect to the product measure m ⊗ = m 1 ⊗ m 2 ⊗ ........ ⊗ m n , f * ∈ L 2 (R n , m ⊗ ), j1≤k1,...,jn≤kn ρ j1,j2,....,jn P 1 j1 (x 1 )....P n jn (x n ) converges in L 2 (R n , m ⊗ ) to f * . Then, one can identify f and f * .

F X (x) = * u≤x f (u)m ⊗ (du) ,
On the other hand, X 1 , X 2 , ....., X n are independent if f ≡ 1, that is ρ j1,j2,....,jn = 0 for all (j 1 , j 2 , ...., j n ) = (0, 0, ...., 0). The use of dependence density allows a better understanding of the contribution of the ρ j1,...,jn 's in dependence. Moreover, it simplifies the notations. Of course, f can be not a density because * is not inevitablty a Stieljes Riemann integral.

The interest of this definition is that the ρ j1,j2,...,jn 's are indeed dependence coefficients. As a matter of fact, the ρ j1,j2,...,jn 's measure polynomial dependence. For example ρ 1,2 , ρ 2,1 and ρ 2,2 measure quadratic dependence, ρ 1,3 , ρ 3,1 , etc, measure cubic dependence. Moreover, ∞ j=1 ρ 2 j,1 ≤ 1 and X 2 = g(X 1 ), g ∈ L 2 (R, m 1 ), if and only if ∞ j=1 ρ 2 j,1 = 1. As a matter of fact, by using the ρ j1,j2,...,jn 's we can have a complete study of dependence. The most interesting property of these coefficients is that they can detect the most of the functional dependence.

The ρ j1,j2,...,jn 's have many applications and enable a better understanding of certain processes. For example, it is easy to express the fact that a process is a martingale because it is an orthogonal projection (cf appendix A. (where q ′ ≤ q) ≈ e -t 2 /2 + Q q=0

(it) q √ n q j1+....+jn=q σ j1 ....σ jn ρ j1,j2,....,jn j 1 !....j n ! + o( |t| q ) e -t 2 /2 + o( |t| Q ) .

Thanks to this result, a necessary and sufficient condition of convergence of moments was deduced in th 1-5 of [START_REF] Blacher R | Central limit theorem by polynomial dependence coefficients[END_REF] (cf also theorem 2 ). This is not surprising: orthogonal polynomials have interesting applications in probability. Thus, we have obtained in [START_REF] Blacher R | Multivariate quadratic forms of random vectors[END_REF] the exact distributions of quadratic forms by using the Hermite polynomials H j and Laguerre polynomials L j which have properties even stronger : R e itx Hj (x).e -x 2 /2 dx (2π) 1/2 = σj j! (it) j . This has provided a simple formula to calculate the distributions of quadratic forms of Gaussian vectors without assuming the independence and whatever the dimension.

1.2 Central Limit Theorem 1.2.1 Case of random variables with the same distribution

One has proved in [START_REF] Blacher R | Central limit theorem by polynomial dependence coefficients[END_REF] the following theorem.

Theorem 2 One assumes that, for all s ∈ N * , m s = m. Then, all the moments M n q = E (X1+X2+....+Xn) q √ n q converges to M q ∈ R if and only if, for all q ∈ N, there existe S q ∈ R such that q! √ n q j1+j2+...+jn=q; js≤2 α j1,j2,...,jn → S q .

Moreover, for all q ∈ N, M q is the moment of order q of N (0, M 2 ) if and only if, for all q ∈ N, S q is the moment of order q of N (0, S 2 ). In this case, M 2 = S 2 + σ 2 0 where σ 2 0 = E{X 2 1 }.

The interest of this theorem is that the ρ j1,j2,...,jn 's are indeed dependence coefficients. Now, theorem 2 gives only an equivalence to the convergence of the moments. In other words, we only turn this convergence into a condition on the dependence coefficients ρ j1,j2,...,jn . Then, in these theorems there is no asymptotical independence assumption. Besides, we can easily build up some sequences {X n } whose the moments converge without that the X j are asymptotically independent. For example, let us take X n = e n Y when Y has a distribution N(0,1) and e n = ±1 is correctly chosen : M n q → M q for all q. Though, in this case, the X n 's has the most strong dependence, the linear dependence with a linear correlation coefficient ρ 1,1 = ±1.

Then, in order to have asymptotical independence condition it is enough to choose assumptions a little stronger on the ρ j1,j2,...,jn 's. By this method, we can obtain minimal conditions for the central limit theorem. For example the following theorem holds (cf [START_REF] Blacher R | Central limit theorem by polynomial dependence coefficients[END_REF]).

Theorem 3 One assumes that, for all s ∈ N * , m s = m. We suppose that

n -2 E n s=1 X 2 s -E{X 2 s } 2 → 0 .
We suppose also that, for all q ∈ N * , q! √ n q n t1=1 n t2=t1+1 .....

n tq=tq-1+1

E{X t1 X t2 .....X tq } converges to the moment of order q of N (0, S 2 ) and that 1 √ n q j1+j2+...+jn=q; js≤2, only one js=2 ρ j1,j2,...,jn is bounded. Then, (X 1 + X 2 + ....

+ X n )/ √ n M → N (0, M 2 ) with M 2 = S 2 + σ 2 1 .
We recall that (X 1 + X 2 + ....

+ X n )/ √ n d → N (0, M 2 ) if (X 1 + X 2 + .... + X n )/ √ n M → N (0, M 2 ).
Remark that S 2 < 0 is possible because M 2 = σ 2 0 + S 2 . In this case, the moment of order q of N (0, S 2 ) is the moment of i|S 2 |Y G where Y g ∼ N (0, 1).

Note that the ρ j1,....,jn 's or the α j1,....,jn 's appear well in each of these conditions. Indeed, there exists (j 1 , ..., j n ) where j s ≤ 1 such that E{X t1 X t2 .....X tq } = α j1,....,jn . Moreover, by proposition Now, it seems natural to choose this condition in a CLT. Then the theorem 3 seems a theorem with minimum conditions of asymptotic independence for the MCLT. We can therefore assume that this is the case. In fact, we shall see in section 2.3.5 that this condition is maybe too weak because it does not require asymptotic normality.

A.2.1, n -2 E n s=1 X 2 s -E{X 2 s } 2 → 0 is equivalent to n -2 j1+j2+

Generalization

Theorems 2 and 3 are given under the assumption that the X j 's have the same law m. It is a too restrictive condition which prevents application of these theorems for martingales, for example. So we will study the case where the laws of the X j 's are different.

At first, we will need a sequence of normalization Ψ(n) which can often be replaced by σ(n).

Notations 1.2.1 Let Ψ(n) > 0. One supposes that c Ψ √ n ≤ Ψ(n) where c Ψ > 0. Let M n p = E (X1+....+Xn) p Ψ(n) p . Let h ∈ N. We set B n h = max{1, |M n h |}.

Note that we could impose a weaker hypothesis than c Ψ √ n ≤ Ψ(n) : in this case, we get conditions more complicated in the MCLT. Now, because we study the case where the laws of the X j 's are different, we have to impose minimal assumptions in order to avoid, for example that E{X 2 n } → ∞. Also we will impose the following assumptions. Hypothesis 1.2.1 One supposes that, for all p ∈ N * , for all j ≥ 2

E n t=1 (X t ) j Ψ(n) j p ≤ C n (j, p) ≤ C(j, p)
where C(j,p) depends only on j and p and where C n (j, p) = ǫ n (j, p) → 0 as n → ∞ if j ≥ 3.

Let β s = E{X 2 s }. One assumes that nm s=1 βs

Ψ(n) 2 → σ 2 0 ∈ R + .
Of course these conditions are checked if the m r 's have the same law. More generally, the first condition is checked if, for all p ∈ N * , there exists C 1 (p) > 0 such that |E{X p n }| ≤ C 1 (p). Remark that condition "for all p ∈ N * , for all j ≥ 2, E P n t=1 (Xt) j Ψ(n) j p ≤ C n (j, p) " is equivalent to condition "for all p ∈ N * , for all j ≥ 2, E P n t=1 (Xt) j Ψ(n) j p ≤ C n (j, p) ". It suffices to consider p even and Holder's inequality.

Then, with these conditions, one can generalize theorem 3 by the following way.

Theorem 4 We suppose that

Ψ(n) -4 E n s=1 X 2 s -E{X 2 s } 2 → 0 .
All the moments M n q = E (X1+X2+....+Xn) q Ψ(n) q converges to a real M q if and only if, for all q ∈ N, there existe S q ∈ R and Sb r q ∈ R , r=2,3, such that s1 =s2 =..... =sq E{X s1 X s2 ......X sq } Ψ(n) q → S q s1 =s2 =..... =sq-1

E{ P s1 2 (X s1 )X s2 ......X sq-1 } Ψ(n) q ≤ Sb 2 q , s1 =s2 =...... =sq γ s1 E{X s1 X s2 ......X sq } Ψ(n) q+1 ≤ Sb 3 q where P s1 2 (x) = x 2 -γ s1 x -β s1 with γ s1 = E{X 3 s1 }/E{X 2 s1 }.

Moreover, M q , is the moment of order q of N (0, M 2 ) if and only if, for all q ∈ N * , S q = ν q , the moment of order q of N (0, S 2 ). In this case M 2 = σ 2 0 + S 2 .

This theorem is proved in chapter 3.

Remark that if all the laws m j 's are the same, the third condition can be removed. Now, when the X j 's are bounded, we shall prove a simpler theorem.

Theorem 5 We suppose that there exists F > 0 such that |X s | ≤ F for all s ∈ N * . We suppose that

Ψ(n) -4 E n s=1 X 2 s -E{X 2 s } 2 → 0 .
All the moments M n q = E (X1+X2+....+Xn) q Ψ(n) q converges to a real M q if and only if, for all q ∈ N, there existe S q ∈ R such that s1 =s2 =..... =sq E{X s1 X s2 ......X sq } Ψ(n) q → S q .

Moreover, M q is the moment of order q of N (0, M 2 ) if and only if, for all q ∈ N * , S q = ν q .

Remark that if |X m,t | ≤ F , the hypothesis 1.2.1 holds.

Chapter 2

Applications

Process with the same first coefficients of correlation

Theorem 4 allows to better understand if an asymptotical independence condition is useful or not. For example, for fixed n, asymptotic normality depends only on a finite number of correlation coefficients : that is a countable number of those are useless.

We have a simple application of this result: if a sequence X n satisfies the MCLT, an infinity of other sequences which have the same first correlation coefficients will also check MCLT.

Proposition 2.1.1 Assume that, for all s ∈ N * , m s = m. Let {Y n } be a process such that, for all s ∈ N * , Y s has the same distribution m as X s . Let ρ j1,j2,....,jn and ρ ′ j1,j2,....,jn be the higher order correlation coefficients associated to {X n } and {Y n }, respectively.

Assume that, for all s ∈ N * , {P s j } j∈N is a basis of L 2 (R, m s ). Assume that, for all n ∈ N * , the dependence density of the process {Y n } satisfies : for all n, Then,

f Y (x 1 , ...., x n ) = 1 + (j1,...,
X 1 + ..... + X n Ψ(n) M → N (0, M 2 ) if and only if Y 1 + ..... + Y n Ψ(n) M → N (0, M 2 ) .
Thus we obtain a set of processes which satisfy the MCLT as soon as one of them satisfies it.

For example by using proposition A.2.1, we have the following propeties.

Example 2.1.1 Let X n be a bounded strictly stationary φ-mixing process such that σ(n) 2 ≥ c 2 Ψ n. Then, the MCLT holds (cf [START_REF] Mairoboda | The Central limit Theorem for empirical moment generating functions[END_REF] and [START_REF] Yokohama R | Moment bounds for stationary mixing sequences[END_REF]).

Then, Y1+.....+Yn σ(n) M → N (0, 1) for all process {Y n } such that {Y n }, for all n, the distribution of Y n is m and has the dependence density

f Y (x 1 , ...., x n ) = 1+ (j1,...,

Martingale theory

We have understood in proposition 1.1.3 that, if (X 1 +...+X n , F n ) is a martingale, ρ j1,j2,....,jn,1 = 0 for all n ∈ N * . Then, E{X s1 X s2 ......X sq } = 0 for all s 1 < ... < s q . The condition of theorem 4

s1 =s2 =..... =sq E{Xs 1 Xs 2 ......Xs q } Ψ(n) q
→ S q is automatically checked.

It is therefore not surprising that we obtain quite simple CLT for martingale. This result clearly shows that the ρ j1,j2,....,jn 's which define all dependence, allow to better understand the importance of classical assumptions in the CLT and what they really mean. Now we can also consider inovation processes :

X n+1 = Z n+1 -E{Z n+1 |Z n , Z n-1 , ....}
where Z n is any stochastic process : ρ j1,j2,....,jn,1 = 0 for all n .

But in order that the MCLT holds, one can simplify this condition : in theorem 4 it is enough to assume ρ j1,....,jn = 0 if j s ≤ 1 in order to obtain E{X s1 X s2 ......X sq } = 0. Then, one use the following notation. Notations 2.2.1 Let Z n be a stochastic process. We denote by P{Z n+1 |Z n , Z n-1 , ....} the orthogonal projection of Z n+1 onto the subspace generated by linear combination of random variables Z t1 Z t2 ....Z tp , t 1 < t 2 < .... < t p ≤ n where p ∈ N * . Indeed, one can use process much simpler than innovation process in order to apply theorem 4 with ρ j1,....,jn = 0 if j s ≤ 1 : one uses X n+1 = Z n+1 -P{Z n+1 |Z n , Z n-1 , ....}. This condition is less strong than the martingale assumption. Indeed, one can write

E{Z n+1 |Z n , Z n-1 , ....} = P{Z n+1 |Z n , Z n-1 , ....} + R{Z n+1 |Z n , Z n-1 , ....} where R{Z n+1 |Z n , Z n-1 , ....} is orthogonal to P{Z n+1 |Z n , Z n-1 , ....}.
More generally, one can use process much simpler than martingales in order to apply theorem 4. Instead of assuming E{X n+1 |F n } = 0, one can suppose P{X n+1 |X n , X n-1 , ....} = 0 for all n ∈ N * , i.e. ρ j1,....,jn = 0 if j s ≤ 1.

Example 2.2.1 Let X t = ∞ i=0 C i (Θ t+i )f i+1 (Ψ t+i ) where C n (x) = √ 2.cos(4 n x), where {Θ i } is IID with uniform distribution on [0, 2π],
where {Ψ i } is a strictly stationary process independent of {Θ i } and where |f i+1 (y)| ≤ 1 (i+1) 1/2+a with a > 0 . Then, we shall prove in appendix B that P{X n+1 |X n , X n-1 , ....} = 0 for all n ∈ N * .

Then, in order to apply theorem 4 with E{X s1 X s2 ......X sq } = 0, it is not necessary that X n is a martingale. So we can state the following theorem.

Theorem 6 Assume that the hypotheses 3.1.2 hold with ψ(n) = n. Assume that P{X n+1 |X n , X n-1 , ........} = 0 for all n ∈ N * . We suppose that

n -2 E n s=1 X 2 s -E{X 2 s } 2 → 0 .
All the moments M n q = E (X1+X2+....+Xn) q √ n q converges to a moment of order q of N (0, σ 2 0 ) if and only if, for all q ∈ N, there existe Sb 2 q ∈ R , such that

s1 =s2 =..... =sq-1 E{X 2 s1 X s2 ......X sq-1 } √ n q ≤ Sb 2 q ,
Theorem 7 Assume that there exists F > 0 such that |X s | ≤ F for all s. We suppose that n t=1

E{X 2 s } n → σ 2 0 . Assume that P{X n+1 |X n , X n-1 , ....} = 0 for all n ∈ N * . We suppose that n -2 E n s=1 X 2 s -E{X 2 s } 2 → 0 .
Then, all the moments M n q = E (X1+X2+....+Xn) q √ n q converges to a moment of order q of N (0, σ 2 0 ).

Compare these results to classical theorems about martingales (cf [START_REF] Heyde | Martingale limit theory and its applications[END_REF] pages 58 and 71).

Theorem 8 Let {Σ ni , F ni , 1 ≤ i ≤ k n , n ≥ 1}
) be a zero mean square integrable martingale array with difference X ni and let η 2 be an a.s. finite random variables. Assume that the σ-fields are nested :

F n,i ⊂ F n+1,i for 1 ≤ i ≤ k n , n ≥ 1. Assume that A) max i |X ni | P → 0 . B) E max i (X 2 ni ) is bounded in n. C) U 2 n,kn = i X 2 ni P → η 2 .
Then,

Σ n,kn = i X ni d → Z ,
where the random variable Z has the characteristic function E{exp(-η 2 t 2 /2)}.

For example, we can choose

k n = n, F n,i = F i , Σ n,kn = (X 1 + ... + X n )/ √ n, X ni = X i / √ n and η 2 = σ 2 0 .
Then, in [START_REF] Heyde | Martingale limit theory and its applications[END_REF], we have also the following result about the convergence of moments.

Theorem 9 Let p > 1. Let µ p be the moment of order p of N(0,1). Assume that (X 1 +...+X n , F n ) is a martingale. Assume that the following conditions hold.

A)

1 n max i∈{2,3,....,n} E{X 2 i |F i-1 } P → 0 . B) E 1 n n n=1 E{X 2 i |F i-1 } -σ 2 0 p → 0 . C) E 1 n n n=1 [X 2 i -σ 2 0 ] p → 0 .
Then,

E X 1 + .... + X n √ n 2p → µ 2p σ 2p 0 .
Let us compare this theorem and theorem 6. At first, P{X n+1 |X n , X n-1 , ....} = 0 holds if (X 1 + ... + X n , F n ) is a martingale. It is a condition much weaker than the martingale assumption.

Moreover, by lemma A.2.1 , condition C) with p=2, involves that

n -2 E n s=1 X 2 s -E{X 2 s } 2 → 0 . Now consider condition A) : (1/n) max i∈{2,3,....,n} E{X 2 i |F i-1 } P → 0. We know that E{X 2 i |F i-1
} can be written with the ρ j1,....,jn : e.g. E{X 2 2

|F 1 } = E{X 2 2 } + σ 1,2 j ρ j,2 P 1 j (X 1 )
. That is the ρ j1,....,jn are implicitly in this theorem. But many are useless for the MCLT. The aim of theorem 4 is to suppress these useless parameters e.g. the ρ j,2 such that j > 2.

Moreover, in theorem 7 we do not need of use the maximum as in theorem 9. On the other hand, we do not need condition B) for all p :

E 1 n n n=1 E{X 2 i |F i-1 } -σ 2 0 p → 0.
Then, clearly theorem 7 obtained by using the the ρ j1,....,jn 's gives conditions much simpler than theorem 9.

Example 2.2.2 Consider the sequences

X t = ∞ i=0 C i (Θ t+i )f i+1 (Ψ t+i ) defined in example 2.2.1. One chooses f i+1 (Ψ t ) = Ψt (1+i) 1/2+a
. One supposes Ψ t stricly stationary and bounded. We know that P{X n+1 |X n , X n-1 , ....} = 0 for all n ∈ N * . Then, (X 1 + .....

+ X n )/ √ n M → N 0, E{(X 1 ) 2 } if E{Ψ t ) 2 (Ψ t ′ ) 2 } -E{(Ψ t ) 2 }E{(Ψ t ′ ) 2 } ≤ ǫ(t -t ′ ) ,
where 1 ≥ ǫ(t) > 0 and where ǫ(t) is decreasing and converges to 0. This condition of asymptotic independence about Ψ t is therefore very weak, especially compared to the strong mixing condition or to the condition of weak dependence.

This shows clearly that the use of the ρ j1,....,jn 's simplifies the CLT for martingales and allows also to better understand why the classical CLT conditions are relatively simple in the case of martingales.

Comparison of the conditions of asymptotic independence 2.3.1 Classical conditions

We first recall the definition of the strong mixing condition.

Definition 2.3.1 : Assume that {X n } n∈N * is a sequence of random variables. Then, {X n } is strongly mixing with coefficient α if sup A∈M n 1 , B∈M ∞ n+h P (A ∩ B) -P (A)P (B) = α(h) → 0 as h → ∞, where for a ≤ b, M b a is the σ-field generated by X a , X a+1 , ....., X b .
For example, suppose now that the X j 's have the same law m and that X n is strong mixing with coefficient α. Suppose that all the orthonormal polynomials P j exist. We know that we can express the ρ i1,i2,........in 's in the form : ρ j1,j2,........jn = E P j1 (X t1 )P j2 (X t2 )......P je (X te )P je+1 (X te+1 )P je+2 (X te+2 )......P jq (X tq ) where t s ∈ N * , s=1,2,...,q, and t 1 < t 2 < .... < t q .

On the other hand, E [P j1 (X v1 )......

P j k (X v k )] 4 k ≤ E P j1 (X v1 ) 4k .......E P j k (X v k ) 4k
which is equal to a constant C j1,...,.j k . Then, by theorem 17-2-2 of [START_REF] Ibragimov | Independent and stationary sequences of random variables[END_REF], we know that the strong mixing condition involves E P j1 (X t1 )....P jq (X tq ) -E P j1 (X t1 )...P je (X te ) E P je+1 (X te+1 )...P jq (X tq ) ≤ K a α(t e+1 -t e ) 1-a where K a is a constant and a > 0 arbitrarily small. Of course, this relationship is written with the ρ j1,j2,........jn such that j s = 0 if r = t e < s < t e+1 = r + h as : |ρ j1,j2,........jn -ρ j1,j2,........jr,0,0,....0 ρ 0,0,....,0,j r+h ,j r+h+1 ,........jn | ≤ K a α(h) 1-a .

On the other hand, Doukhan and Louhichi [START_REF] Doukhan | A new weak dependence condition and application to moments inequalities[END_REF] have introduced the (θ, L, Ψ) weak-dependence.

Definition 2.3.2 : Let L = ∪ ∞ p=1 L p where L p = {f : R p → R} . Let Ψ : L ⊗ L ⊗ (N * ) 2 → R + and (θ r ) r∈N ց 0. The sequence {X n } n∈Z is (θ, L, Ψ) weakly dependent if ∀r ∈ N, ∀u, v ∈ N * , ∀(h, k) ∈ L u ⊗ L v , ∀ i 1 < i 2 < ..... < i u < i u + r ≤ j 1 < ..... < j v , Cov h(X i1 , ...., X iu ), k(X j1 , ...., X jv ) ≤ θ r Ψ(h, k, u, v).
Clearly, under this assumption of weak dependence, we find the same kind of relationship as when the strong mixing condition holds : E P j1 (X t1 )......P jq (X tq ) -E P j1 (X t1 ).....P je (X te ) E P je+1 (X te+1 ).....P jq (X tq ) ≤ C 1 θ te+1-te , where C 1 depends on e, q-e, j 1 , j 2 , ...., j q . Now, remark that

E (X t ) 2 (X t+h ) 2 -E (X t ) 2 E (X t+h ) 2 → 0 .
In fact, it's as true for the strong mixing condition as for the weak dependence, That means by theorem 4, that, if all moments converge, then in addition to one or other of these conditions, it will be required inter alia that

s1 =s2 =..... =sq E{X s1 X s2 ......X sq } σ(n) q → ν q .

Condition H mI

Conditions of asymptotic independence H mI and of asymptotic stationarity H mS were introduced in [START_REF] Blacher R | Theoreme de la limite centrale par les moments[END_REF] and [START_REF] Blacher R | Central limit theorem by moments[END_REF].

Notations 2.3.3 We denote by κ(n) ∈ N , an increasing sequence such that κ(1) = 0, κ(n) ≤ n and κ(n)/n → 0 as n → ∞ . We define the sequences u(n) and τ (n) by : u

(1)=1, u(n) = max m ∈ N * 2m + κ(m) ≤ n and τ (1) = 0, τ (n) = n -2u(n) if n ≥ 2.
Moreover, we simplify u(n) and τ (n) in u n = u and τ n = τ .

Let σ(u) 2 be the variance of

X 1 + X 2 + ... + X u . One sets Σ u = X1+X2+.........+Xu σ(u) , ξ u = Xu+1+Xu+2+.........+Xu+τ σ(u)
and Σ ′ u = Xu+τ+1+Xu+τ+2+.........+Xu+τ+u σ(u)

.

In [START_REF] Blacher R | Theoreme de la limite centrale par les moments[END_REF] , one has proved that n/u → 2 and τ /u → 0 as n → ∞. Moreover, one chooses E{(ξ u ) 2 } → 0 . Notations 2.3.4 : We define conditions H mS and H mI by the following way :

H mS : ∀p ∈ N , E (Σ u ) p -E (Σ ′ u ) p → 0 as n → ∞. H mI : ∀(p, q) ∈ (N * ) 2 , E (Σ u ) p (Σ ′ u ) q -E (Σ u ) p E (Σ ′ u ) q → 0 as n → ∞.
In fact, in [START_REF] Blacher R | Central limit theorem by moments[END_REF], we define conditions a little less strong because we consider the asymptotic independence of moments between Σ u + v u and Σ ′ u + v ′ u where {v u } and {v ′ u } are two sequences of random variables such that E{|v u | p } + E{|v ′ u | p } → 0 for all p ∈ N. Then, in [START_REF] Blacher R | Central limit theorem by moments[END_REF] one has proved the following result.

Theorem 10 : Assume that E{|ξ u | k } → 0 as n → ∞ for all k ∈ N . Assume that H mS and H mI hold. Then, Σ n M → N (0, 1).
In fact, H mS and H mI implies also the convergence in dimension 2. Proof By theorem 10, E{(Σ u ) k } → µ k , the moments of order k of N(0,1). By

H mS , E{(Σ ′ u ) k } → µ k . Then, E{(Σ u ) q }E{(Σ ′ u ) p )} → µ q µ p . By H mI , E{(Σ u ) q (Σ ′ u ) p )} -E{(Σ u ) q }E{(Σ ′ u ) p )} → 0. Then, E{(Σ u ) q (Σ ′ u ) p )} → µ q µ p .
Note that the convergence of moments involves the convergence in distribution

Corollary 2.3.2 Assume that E{|ξ u | k } → 0 as n → ∞ for all k ∈ N . Assume that H mS and H mI hold. Then, Σ n d → N (0, 1). Proof By our assumptions, E{(Σ u ) k } → µ k . By H mS , E{(Σ ′ u ) k } → µ k . By H mI , for all (a, b) ∈ R 2 , E{(aΣ u + bΣ ′ u ) k } converges to the moments of order k of N (0, a 2 + b 2 ). One deduces that Σ n d → N (0, 1). Example 2.3.1 Let ζ t = ∞ i=0 b i+1 h i (Θ t+i ), where |h i (Θ 1 )| ≤ 1 , |b| ≤ 1/2
i ∈ N . Assume that ∞ s=0 ∞ i=s k i+1 (Θ 1 , ζ 1 ) < C < ∞. Let X t = ∞ i=0 k i+1 (Θ t+i , ζ t+i ). Then H mI hold and S n M → N (0, 1) (cf [25])

Condition H mI and correlation coefficients of higher order

We will compare these results about H mI with the results about the correlation coefficients of higher order. We will see that we obtain almost minimal conditions more similar to classical conditions. This is not surprising. We introduced the conditions H mI by trying to find conditions slightly stronger than those about correlation coefficients of higher order (Partie B-I of [START_REF] Blacher R | Loi de la somme de N variables aléatoires et théorème de la limite centrale par les coefficients de dépendance polynômiale et les moments[END_REF]) and closer to the classical conditions : cf Partie B-II of [START_REF] Blacher R | Loi de la somme de N variables aléatoires et théorème de la limite centrale par les coefficients de dépendance polynômiale et les moments[END_REF].

So we obtain the following theorem.

Theorem 11 Suppose that |X n | ≤ F where F > 0. One assumes that there exists

c Ψ > 0 such that σ(n) ≥ c Ψ √ n. One supposes that 1 σ(un) 2 un s=1 E{X 2 s } → σ 2 0 and 1 σ(un) 2 n s=1+un+τn E{X 2 s } → σ 2 0 . One assumes that E un t=1 [(X t ) 2 -E{X 2 t }] σ(u n ) 2 2 + E un t=1 [(X un+τn+t ) 2 -E{X 2 un+τn+t }] σ(u n ) 2 2 → 0 . One assumes that, for all k ∈ N, E X u+1 + X u+2 + ......... + X u+τn σ(u n ) k → 0 .
Then, H mI and H mS hold if and only if for all q ∈ N, for all p ∈ N, s1 =s2 =.... =sq, sr≤un t1 =t2 =.... =tp, un+τn<tr≤n

E{X s1 X s2 ....X sq X un+τn+t1 X un+τn+t2 ....X un+τn+tp } σ(u n ) p+q
converges to ν ′ q ν ′ p , where ν ′ q is the moment of order p de la la loi N (0, 1 -σ 2 0 ).

Proof We apply corollaries 3.10.1 and 4.5.1 with Ψ(n) = σ(n), n m = u n , X m,t = X t for t = 1, .., u n , Y m,t = X un+τn+t for t = 1, .., u n .

Indeed, if H mI and H mS hold, by corollary 2.3.1, all the moments

M n q,p = E (X 1 + .... + X un ) q (X un+τn+1 + .... + X un+τn+u ) p σ(u n ) p+q
converges to µ q µ p . Then, all the conditions of corollary 4.5.1 are checked. That proves the necessary condition.

Conversely, let us prove the sufficient condition. Suppose that the conditions of this theorem are checked. By corollaries 3.10.1 and 4.5.1 , M n q,p → µ q µ p , M n q = E (X1+....+Xu) q σ(u) q → µ q M ′n p = E (Xu+τ+1+....+Xu+τ+u) p σ(u) p+q → µ p . Therefore, M n q,p -M n q M ′n p → 0. Then, H mI and H mS hold.

If X n is not bounded, one can use corollary 4.4.1 : conditions are more complicated. But this is hardly important: this theorem 11 suffices to show how H mI results in terms of correlation coefficients of higher order. In particular, the main condition about the coefficients of multilinear correlation implies the following condition (when the sequences are bounded or not).

Corollary 2.3.3 One assumes that there exists c Ψ > 0 such that σ(n

) ≥ c Ψ √ n. One supposes that (1/σ(u n ) 2 ) un s=1 E{X 2 s } → σ 2 0 and (1/σ(u n ) 2 ) n s=1+un+τn E{X 2 s } → σ 2 0 . One assumes that E un t=1 [(X t ) 2 -E{X 2 t }] σ(u n ) 2 2 + E un t=1 [(X un+τn+t ) 2 -E{X 2 un+τn+t }] σ(u n ) 2 2 → 0 .
One assumes that, for all k ∈ N,

E X u+1 + X u+2 + ......... + X u+τn σ(u n ) k → 0 .
Then, if H mI and H mS hold

s1 =s2 =..... =sq t1 =t2 =..... =tp E{X s1 X s2 ......X sq X un+τn+t1 X un+τn+t2 ......X un+τn+tp } σ(u) p+q - s1 =s2 =..... =sq E{X s1 X s2 ......X sq } σ(u) q t1 =t2 =..... =tp E{X un+τn+t1 X un+τn+t2 ......X un+τn+tp } σ(u) p
converges to 0

Finally we see that the condition H mI leads to a condition about the ρ j1,....,jn 's which is hardly stronger than that of Theorem 4.

Comparison of conditions

We will therefore compare the strength of the different conditions of asymptotic independence in spite of the fact that all are not directly comparable.

At first, it is not necessary that H mI holds in the case of weak dependence a priori, at least. So we can not say that the condition H mI is weaker than the condition of weak dependence (a priori, at least) and one can not directly compare these two conditions. It will be the same between the Martingale hypothesis and the assumptions of Theorem 4.

On the other hand, let us remark that if strong mixing condition holds and if the MCLT holds, it is necessary that H mI holds. Now, the ρ j1,....,jn 's determine all dependence. So we must be able to formulate the various conditions of asymptotic independence as conditions about the ρ j1,....,jn 's. However, it may be difficult to give an equivalence. So we shall just give some consequences that these conditions of asymptotic independence lead about the ρ j1,....,jn 's and the MCLT1 .

Conditions of theorem 4

It is easy to understand that if the conditions of theorem 4 are checked, Q conditions about the ρ j1,....,jn 's have to be checked in order that the first q moments converge where Q is approximately equal to 3q.

In this case the conditions which we obtain are conditions being about sums of ρ j1,....,jn 's.

Conditions of theorem 4 and H mS

Clearly, if we impose moreover that H mS holds Q' conditions about the ρ j1,....,jn 's have to be checked in order that the first q moments converge where Q' is approximately equal to 6q (cf theorem 12 ).

In this case also the conditions which we obtain are conditions being about sums of ρ j1,....,jn 's.

Conditions of theoreme 4 and H mI

If we impose moreover that H mI holds Q" conditions about the ρ j1,....,jn 's have to be checked in order that the first q moments M n q ′ ,p ′ , q ′ + p ′ ≤ q, converge where Q" is approximately equal to q 2 /2 (cf theorem [START_REF] Yokohama R | Moment bounds for stationary mixing sequences[END_REF].

In this case again the conditions which we obtain are conditions being about sums of ρ j1,....,jn 's.

Strong mixing condition Now suppose that the X j 's have the same law m and that {X j } is strong mixing with coefficient α . Therefore, for all h, for all t 1 < t 2 < .... < t h , E P j1 (X t1 )....P j h (X t h ) -E P j1 (X t1 )...P je (X te ) E P je+1 (X te+1 )...P j h (X t h ) ≤ K a α(t e+1 -t e ) 1-a .

So there is an countable number of relations about the ρ j1,....,jn 's. Since only the first ρ j1,....,jn 's are useful for the MCLT by theorem 2 (j s ≤ 2), so there is a countable number of unnecessary relationships which are also checked.

Remark also that E

(X t ) 2 (X t+h ) 2 -E (X t ) 2 E (X t+h ) 2 → 0.
This means that the conditions of MCLT for strong mixing processes will be stronger than those of Theorem 4.

Moreover, it is easy to see that if all moments M n q of a strictly stationary strong mixing process are bounded, H mI holds. Then the MCLT holds also. Now, the conditions are relationship between groups of 3 ρ j1,....,jn 's : |ρ i1,i2,........in -ρ i1,i2,........ir,0,0,....0 ρ 0,0,....,0,i r+h ,i r+h+1 ,........in | ≤ K a α(h) 1-a . Furthermore it is the supremum which converges to 0. Then, these relations are stronger than those of sums of ρ j1,....,jn 's.

Weak dependence Now, we suppose that the sequence {X n } n∈Z is (θ, L, Ψ) weakly dependent. Then, for all t 1 < t 2 < .... < t h , E P j1 (X t1 )......P j h (X t h ) -E P j1 (X t1 )......P je (X te ) E P je+1 (X te+1 )......P j h (X t h )

≤ C 1 θ te+1-te .
By the same way,

E X t ) 2 (X t+h ) 2 -E X t ) 2 E X t+h ) 2 → 0.
We obtain the same conclusions as for strong mixing processes. However, it is not sure a priori that, if all moments M n q converges, H mI holds.

In the case of weak dependence, conditions that we have about the ρ j1,....,jn 's are always relations between 3 groups of ρ j1,....,jn 's.

Martingale if {X j } is a martingale, ρ j1,j2,....,jn,1 = 0, i.e. E P t1 j1 (X t1 ).....P

t h-1 j h-1 (X t h-1
)X t h = 0 for all h, for all t 1 < t 2 < .... < t h . Clearly, in this case also, there is a countable number of relations useless for the MCLT.

Moreover, these conditions are equalities on some ρ j1,....,jn : ρ j1,j2,....,jn,1 = 0. These relations are much stronger than the convergence of sums of ρ j1,....,jn .

Processus such that P{X n+1 |X n , X n-1 , ....} = 0. In this case, E X t1 X t2 ......X t h } = 0 for all h, for all t 1 < t 2 < .... < t h . It is a weaker condition than the martingale condition, but stronger than those of theorem 4 and furthermore, there are a countable number of relations necessary for the convergence of the moment of order q when we consider that the relationship must be true for all n.

In this case the conditions that we have are still equalities on some ρ j1,....,jn . But there are less than for martingales.

Conclusion

All these conditions are not always directly comparable. For example, a martingale does not necessarily satisfy the MCLT.

But there is a way to get an idea of the strength of each of the hypotheses: by using the correlation coefficients of higher order.

For example, one could say that a condition is stronger than another if it requires more relations about the ρ j1,....,jn 's. Now we can also consider what type of relationship it is. In this case we understand that the conditions of theorem 4 are conditions which we can consider as a minimum. But we also understand that the condition H mI is almost minimal.

In fact we shall even wonder if the true minimum condition is not H mI .

Condition H mI and minimal condition

Minimal condition It arises indeed a question: Is that the conditions of theorem 4 are actually conditions which can be considered as conditions of asymptotic independence? These conditions which are directly related to the correlation coefficients of higher order are actually stronger than the conditions of theorem 2 (which give an equivalence to the convergence of moments and which involve no dependence a priori). But they have a default: the conditions of theorem 4 does not necessarily mean that asymptotic distributions are normal unless we impose the S q are normal moments.

But to impose that the S q 's are normal moments is more like a chance because it does not change much about the condition of asymptotic independence itself. Indeed, to say that s1 =s2 =..... =sq E{Xs 1 Xs 2 ......Xs q } Ψ(n) q → S q where S q is arbitrary and s1 =s2 =..... =sq E{Xs 1 Xs 2 ......Xs q } Ψ(n) q → ν q are two conditions which require the same type of asymptotic independence.

If one wants this condition of asymptotic normality, we wonder if we should not therefore impose conditions slightly stronger. One wonders if, as a matter of fact, condition H mI is the minimum condition for the asymptotic independence in order that MCLT holds with asymptotic normality.

Study of conditions H mI and H mS At first, note that if the conditions of Theorem 4 are checked, it is necessary that other aditional condition about the correlation coefficients of higher order holds in order that H mS holds. However, if the conditions of Theorem 4 are checked and if H mS does not hold, we will have a very strange case where we have Σ n M → N (0, 1), Σ u M → N (0, 1) and Σ ′ u M → N (0, 1). Intuitively, we feel that it is logical to assume moreover that H mS holds if we want a minimal regularity in the asymptotic convergence. Now, if the conditions of Theorem 4 are checked and if H mI does not hold, there will be also conditions on the correlation coefficients of higher order which seems rather strange.

For example suppose that σ(n) 2 = n, that X n is strictly stationary, and that it is the moment of order 3 which does not check H mI , but that conditions of theorem 4 holds. Then,

r<s<t≤n E{X r X s X t } n 3/2 ∼ r<s<t≤u E{X r X s X t } u 3/2 ∼ u+τ <r<s<t E{X r X s X t } u 3/2 → 0 where we set x n ∼ y n if x n -y n → 0 as n → ∞ for all real sequences x n et y n . Therefore, r<s<t≤n E{X r X s X t } n 3/2 ∼ r<s≤u<u+τ <t≤n E{X r X s X t } n 3/2 + r≤u<u+τ <s<t E{X r X s X t } n 3/2 → 0 .
Then, under H mI , by corollary 2.3.3,

r<s≤u<u+τ <t≤n E{X r X s X t } n 3/2
→ 0 , and

r≤u<u+τ <s<t E{X r X s X t } n 3/2 → 0 .
On the other hand, under the assumption of theorem 4 ,

r<s≤u<u+τ <t≤n E{X r X s X t } n 3/2 + r≤u<u+τ <s<t E{X r X s X t } n 3/2 → 0 .
Therefore, if H mI does not hold for the moment of order 3 and if the hypotheses of theorem 4 are checked

r<s≤u<u+τ <t≤n E{X r X s X t } n 3/2 + r≤u<u+τ <s<t E{X r X s X t } n 3/2 → 0 r<s≤u<u+τ <t≤n E{X r X s X t } n 3/2 → 0 and r≤u<u+τ <s<t E{X r X s X t } n 3/2 → 0 .
This is a case which seems strange when we admit that there is some asymptotic independence.

Indeed, is what one can speak of asymptotic independence if r<s≤u<u+τ <t≤n

E{XrXsXt} n → 0 and r≤u<u+τ <s<t

E{XrXsXt} n → 0? Thus, if X n is strictly stationary, it will be difficult to find examples where H mI does not hold for the moment of order 3 if the assumptions of theorem 4 are checked.

In order to find more easily a such example, we must give up some of our assumptions.

Example 2.3.2 We suppose that X n+1 = √ n + 1Z n+1 - √ nZ n , X 1 = Z 1 where Z n is IID and Z 1 has the distribution N(0,1).
Study Clearly X n is 2-dependent, and then there exists a simple condition about the ρ j1,....,jn 's : E{P t1 j1 (X t1 ).....P tp jp (X tp )} = E{P t1 j1 (X t1 ).....P ti ji (X ti )}E{P ti+1 ji+1 (X ti+1 ).....P tp jp (X tp )} if there exists i such that t i+1 -t i > 2.

Moreover,

X 1 +X 2 = Z 1 + √ 2Z 2 - √ 1Z 1 = √ 2Z 2 , X 1 +X 2 +X 3 = √ 2Z 2 + √ 3Z 3 - √ 2Z 2 = √ 3Z 3 . Therefore, X 1 + X 2 + X 3 + .... + X n = √ nZ n and one can choose Ψ(n) = σ(n) = √ n. Moreover, (Σ u , Σ ′ u ) = ( √ u n Z un )/ √ u n , √ nZ n - √ u n + τ n Z un+τn )/ √ u n converge to the same distribution as Z un , ( √ 2Z n -Z un+τn ) which does not converge to N (0, 1) ⊗ N (0, 1). Remark that, in this case, (1/n) β s does not converge because n s=0 E{X 2 s+1 } = n s=1 E{( √ s + 1Z s+1 - √ sZ s ) 2 } + E{Z 2 1 } = n s=1 [(s + 1) + s] + 1 .

Conclusion If r<s≤u<u+τ <t≤n

E{XrXsXt} n 3/2
→ 0 and r≤u<u+τ <s<t

E{XrXsXt} n 3/2 → 0, is that the condition r<s≤u<u+τ <t≤n E{XrXsXt} n 3/2 + r≤u<u+τ <s<t E{XrXsXt} n 3/2
→ 0 is sufficient to say that there is asymptotic independence? If it is not the case, we must choose conditions a little stronger. Precisely, a condition a little stronger is the condition H mI .

On the other hand, intuitively, the condition

(Σ u , Σ ′ u ) M → N 2 (0, I 2
) is a condition which appears minimum as a condition of asymptotic independence.

Yet in terms of correlation coefficients of higher order for the MCLT, the condition H mI is not minimal : it is only almost minimal. But we understood that, if we impose only assumptions of theorem 4, there will be quite strange conditions. Moreover, there is no asymptotic normality. That shows indeed that the condition H mI is very close to conditions of asymptotic independence which can be required for convergence to the normal law.

Conclude by saying that the convergence of (Σ u , Σ ′ u ) to N (0, I 2 ) is a requirement nearly asymptotic independence which can perhaps be considered as minimal for the asymptotic normality.

Chapter 3 MCLT in dimension 1

In this chapter, we prove theorems 4 and 5 .

Then, we will study the case where the laws of X j 's are possibly different. Unfortunately, in this case, we can not apply the same technique as in [START_REF] Blacher R | Central limit theorem by polynomial dependence coefficients[END_REF]. It is then easier to prove theorems without using the correlation coefficients of higher order ρ j1,....,jn and orthogonal polynomials. Unfortunately the proof is much longer.

Then, we shall prove several MCLT before deducting the MCLT with the correlation coefficients of higher order.

On the other hand, in order to prove Theorem 11, the easiest way is to prove a MCLT in dimensions 2 for sequences with double array X m,s , Y m,s . Then, it is easier to prove these results under larger assumptions which we will introduce now.

Notations and assumptions

Notations 3.1.1 Let x n and y n be two real sequences. We set

x n ∼ y n if x n -y n → 0 as n → ∞. In particular, x n ∼ x if x n → x as n → ∞.
Let Z n and T n be two sequences of random variables defined on (Ω, A, P ). We set Z n ∼ T n if Z n and T n have asymptotically the same distribution.

By misuse of our notations, we set also S n ∼ N (0, σ 2 ) if S n has asymptotically the distribution N (0, σ 2 ).

In chapter 4, we shall generalize by natural way these notations to double triangular array of random variables (X m,s , Y m,s ).

Notations 3.1.2 Let X m,s , s = 1, 2, ...., n m , m=1,2,.
. be a triangular array of random variables defined on a probability space (Ω, A, P ).

We suppose E{X m,s } = 0 and |E{(X m,s ) p }| < ∞ for all p ∈ N.

Hypothesis 3.1.1 Let Ψ(n) > 0. We suppose that √ c Ψ n ≤ Ψ(n) where c Ψ > 0. We set B nm h = max{1, |M nm h |}.
Hypothesis 3.1.2 We suppose that, for all p ∈ N * , for all j ≥ 2

E nm t=1 (X m,t ) j Ψ(n m ) j p ≤ C nm (j, p) ≤ C(j, p) ,
where C(j,p) depends only on j and p and where

C nm (j, p) = ǫ m (j, p) → 0 as m → ∞ if j ≥ 3. Hypothesis 3.1.3 Let β m,s = E{X 2 m,t }. One assumes that nm t=1 βm,t Ψ(nm) 2 → σ 2 0 ∈ R + . Indeed, if E (Xm,1+Xm,2+Xm,3+....+Xm,n) 2 Ψ(nm) 2
converges and

P s =t Xm,sXm,t Ψ(nm) 2 converges also, then, (1/Ψ(n m ) 2 ) s E(X 2
m,s } converges also. Now, in all the MCLT of this report, we impose that

P s =t Xm,sXm,t Ψ(nm) 2
converges.

General lemma

Lemma about sets

At first, we need the following notations.

Notations 3.2.1 Let k and r be two integers such that 1 ≤ r ≤ k. We set

t 1 = t 2 = .... = t r , t r+1 , ..., t k = (t 1 , t 2 , ...., t r , t r+1 , ..., t k ) ∈ {1, 2, ...., n m } k t s = t s ′ if s < s ′ ≤ r , t 1 = t 2 = .... = t r-1 , t r+1 , ..., t k = (t 1 , t 2 , ...., t r-1 , t r+1 , ..., t k ) ∈ {1, 2, ...., n m } k-1 t s = t s ′ if s < s ′ ≤ r -1 . In particular, t 1 , t 2 , ....., t k = 1, 2, ..., n m k . Moreover, if r=2, t 1 = t 2 = .... = t r-1 , t r+1 , ..., t k = t 1 , t 3 , t 4 , ....., t k and if r=1, t 1 = t 2 = .... = t r , t r+1 , ..., t k = t 1 , t 2 , ....., t k . Lemma 3.2.1 Let r ≥ 3. Then, t 1 = t 2 = .... = t r-1 , t r , ..., t k = t 1 = t 2 = .... = t r , t r+1 , ..., t k ∪ t 1 = t r = t 2 = ... = t r-1 , t r+1 , .., t k ∪ t 1 = t 2 = t r = ... = t r-1 , t r+1 ,
∪ t 1 = t 2 = ... = t r-1 = t r , t r+1 , .., t k .
For example, the following lemma holds.

Lemma 3.2.2 We simplify X m,t in X t . Let k ∈ N, k ≥ 3 and h ≥ k.
For all s ∈ {1, 2, ..., k}, we denote by R t s , t = 1, 2, ..., n m , s=1,...,k, a sequence of polynomials of degree j s . Let r ≥ 3. Then, for all k ≤ h,

E t1 =t2 =.... =tr-1,tr,...,t k R t1 1 (X t1 )R t2 2 (X t2 ).....R t k k (X t k ) Ψ(n) h = E  P t 1 =t 2 =.... =tr ,t r+1 ,...,t k R t 1 1 (Xt 1 )R t 2 2 (Xt 2 ).....R t k k (Xt k ) Ψ(n) h ff +E  P t 1 =t 2 =... =t r-1 ,t r+1 ,..,t k [R t 1 1 (X t 1 )R t 1 r (X t 1 )]R t 2 2 (Xt 2 )...R t r-1 r-1 (Xt r-1 )R t r+1 r+1 (Xt r+1 )...R t k k (Xt k ) Ψ(n) h ff +E  P t 1 =t 2 =... =t r-1 ,t r+1 ,..,t k R t 1 1 (Xt 1 )[R t 2 2 (X t 2 )R t 2 r (X t 2 )]...R t r-1 r-1 (Xt r-1 )R t r+1 r+1 (Xt r+1 )...R t k k (Xt k ) Ψ(n) h

+E

 P t 1 =t 2 =... =t r-1 ,t r+1 ,..,t k R t 1 1 (Xt 1 )R t 2 2 (Xt 2 )...[R t r-1 r-1 (X t r-1 )R t r-1 r (X t r-1 )]R t r+1 r+1 (Xt r+1 )...R t k k (Xt k ) Ψ(n) h ff .
Example Suppose r=3. Then,

E t1 =t2,t3,...,t k X t1 X t2 .........X t k √ n h = E t1 =t2 =t3,t4,...,t k X t1 X t2 X t3 X t4 .........X t k √ n h +E t1 =t2,t4,...,t k X 2 t1 X t2 X t4 .........X t k √ n h +E t1 =t2,t4,...,t k X t1 X 2 t2 X t4 .........X t k √ n h .
If r=2, the following lemma hold.

Lemma 3.2.3 We suppose r=2. Then,

t 1 , t 2 , ..., t k = t 1 = t 2 , t 3 , ..., t k ∪ t 1 = t 2 , t 3 , .., t k . Lemma 3.2.4
The following equalities holds.

E t1,t2,...,t k R t1 1 (X t1 )R t2 2 (X t2 ).....R t k k (X t k ) Ψ(n) h = E t1 =t2,t3,...,t k R m1 1 (X t1 )R t2 2 (X t2 ).....R t k k (X t k ) Ψ(n) h +E t1,t3,..,t k [R t1 1 (X t1 )R t1 2 (X t1 )]R t3 3 (X t3 )...R t k k (X t k ) Ψ(n) h .

Number of coefficients of moments

We simplify n m in n. Because

E (X m,1 + .... + X m,nm ) q Ψ(n m ) q = j1+.....+jn m =q q! j 1 !......j nm ! E X j1 m,1 .....X jn m m,nm Ψ(n m ) q ,
one wants to study the sets {j 1 + ..... + j nm = q, j s ≤ 1} in order to know the sums

j1+.....+jn m =q,js≤1 E X j1 m,1 .....X jn m m,nm Ψ(n m ) 2q .
First study We study the set

{j 1 + ..... + j n = q, j s ≤ 1} = {(j 1 , ....., j n ) ∈ N n |j 1 + ..... + j n = q, j s ≤ 1}
when q=2, n=6. We have {j 1 + ..... + j n = q, j s ≤ 1} = {(1, 1, 0, 0, 0, 0), (1, 0, 1, 0, 0, 0), ....., (1, 0, 0, 0, 0, 1)} ∪{(0, 1, 1, 0, 0, 0), (0, 1, 0, 1, 0, 0), ....., (0, 1, 0, 0, 0, 1)} ∪{(0, 0, 1, 1, 0, 0), (0, 0, 1, 0, 1, 0), (0, 0, 1, 0, 0, 1)} ∪{(0, 0, 0, 1, 1, 0), (0, 0, 0, 1, 0, 1)} ∪{(0, 0, 0, 0, 1, 1)} .

Its cardinal is (n -1) + (n -2) + (n -3) + ...... + 2 + 1 = (n-1)n 2 = C 2 n .
Indeed, there is C 2 n ways to select two terms among n. Now, we choose q=3, n=8. We have {j 1 + ..... + j n = q, j s ≤ 1} = {(1, 1, 1, 0, 0, 0, 0, 0), (1, 1, 0, 1, 0, 0, 0, 0), ....., (1, 1, 0, 0, 0, 0, 0, 1)} : n -2 events ∪{(1, 0, 1, 1, 0, 0, 0, 0), (1, 0, 1, 0, 1, 0, 0, 0), ....., (1, 0, 1, 0, 0, 0, 0, 1)} : n -3 events ∪{(1, 0, 0, 1, 1, 0, 0, 0), (1, 0, 0, 1, 0, 1, 0, 0), .....(1, 0, 0, 1, 0, 0, 0, 1)} : n -4 events ................................ ∪{(1, 0, 0, 0, 0, 0, 1, 1)} : n -2 -(n -2 -1) events ∪{(0, 1, 1, 1, 0, 0, 0, 0), (0, 1, 1, 0, 1, 0, 0, 0), ....., (0, 1, 1, 0, 0, 0, 0, 1)} : n -3 events ∪{(0, 1, 0, 1, 1, 0, 0, 0), (0, 1, 0, 1, 0, 1, 0, , 0), ....., (0, 1, 0, 1, 0, 0, 0, 1)} : n -4 events ∪{(0, 1, 0, 0, 1, 1, 0, 0), (0, 1, 0, 0, 1, 0, 1, 0), .....(0, 1, 0, 0, 1, 0, 0, 1)} : n -5

The number of possible combinations is

Card = (n -2)(n -1) 2 + (n -3)(n -2) 2 +......+ (n -(n -2))(n -(n -2) + 1) 2 + (n -(n -1))(n -(n -1) + 1) 2 = (n -2)(n -2) 2 + (n -3)(n -3) 2 +......+ (n -(n -2))(n -(n -2)) 2 + (n -(n -1))(n -(n -1)) 2 + (n -2) 2 + (n -3) 2 + ...... + (n -(n -2)) 2 + (n -(n -1)) 2 .

Now,

n i=1 i 2 = (2n+1)(n+1)n 6 . Donc n-2 i=1 i 2 = (2(n-2)+1)((n-2)+1)(n-2) 6 = (2n-3)(n-1)(n-2) 6 .
Therefore,

Card = (1/2) (2n -3)(n -1)(n -2) 6 + (n -2)(n -1) 2 = (1/2) (2n -3)(n -1)(n -2) 6 + 3(n -2)(n -1) 6 = (1/2) (2n -3 + 3)(n -1)(n -2) 6 = n(n -1)(n -2) 6 = n! 3!(n -3)! = C 3 n .
This is normal : it is C 3 n ways to select three terms among n.

Finally card({j 1 + ..... + j n = q, j s ≤ 1}) = C q n . Moreover, for all random variables X 1 , ...., X n ,

X j1 1 ....X jn n j 1 + ..... + j n = q, j s ≤ 1 = X s1 ....X sq (s 1 , s 2 , ...., s q ) ∈ {1, 2, ...., n} q , 1 ≤ s 1 < s 2 < ..... < s q ≤ n . Lemma 3.2.5 Let p ∈ N * . We simplify X m,s in X s , n m in n. Then, j1+.....+jn=pq, js=0 ou p E X j1 1 .....X jn n Ψ(n) pq = 1 q! t1 =t2 =.... =tq E X p t1 .....X p tq Ψ(n) pq Proof One can suppose p=1. When p > 1, it is enough to set X t = Y p t . Then, X j1 1 ....X jn n j 1 + ..... + j n = q, j s ≤ 1 = X s1 ....X sq 1 ≤ s 1 < s 2 < ..... < s q ≤ n .
Therefore,

E{(X j1 1 ....X jn n )} j 1 + ..... + j n = q, j s ≤ 1 = E{(X s1 ....X sq )} 1 ≤ s 1 < s 2 < ..... < s q ≤ n .
Now, let P q the set of permutations of q terms. Then, (s p(1) , .....s p(q) ) ∈ {1, ...., n} q s 1 < s 2 < ..... < s q , p ∈ P q = (s 1 , ...., s q ) ∈ {1, ...., n} q s 1 = s 2 = .... = s q .

Moreover, if p ∈ P q , then, E{(X s p(1) ....X s p(q) )} = E{(X s1 ....X sq )}. Therefore, because there is q! permutations which belongs to P q , j1+.....+jn=q, js=0 ou 1

E X j1 1 .....X jn n Ψ(n) q = t1<t2<....<tq E X t1 .....X tq Ψ(n) q = 1 q! t1 =t2 =.... =tq E X t1 .....X tq Ψ(n) q .
Second study We study the sets {j 1 + ..... + j n = q, j s ≤ 2, one j s = 2}. At first, choose q=4. We have {j 1 + ..... + j n = q, j s ≤ 2, un j s = 2} = {(2, 1, 1, 0, 0, 0, ...., 0), (2, 1, 0, 1, 0, 0, ...., 0), ....., (2, 1, 0, 0, ...., 0, 0, 1)} ∪{(2, 0, 1, 1, 0, 0, ...., 0), (2, 0, 1, 0, 1, 0, ...., 0), ....., (2, 0, 1, 0, ...., 0, 0, ∪{(2, 0, 0, ...., 0, 1, 1, 0), (2, 0, 0, ...., 0, 1, 0, 1)} ∪{(2, 0, 0, 0, ...., 0, 1, 1)} ∪{(1, 2, 1, 0, 0, 0, ...., 0), (1, 2, 0, 1, 0, 0, ...., 0), ....., (1, 2, 0, 0, ...., 0, 0, 1)} ∪{(0, 2, 1, 1, 0, 0, ...., 0), (0, 2, 1, 0, 1, 0, ...., 0), ....., (0, 2, 1, 0, ...., 0, 0, ∪{(0, 2, 0, ...., 0, 1, 1, 0), (0, 2, 0, ...., 0, 1, 0, 1)} ∪{(0, 2, 0, 0, ...., 0, 1, 1)} ∪{(1, 1, 2, 0, 0, 0, ...., 0), (1, 0, 2, 1, 0, 0, ...., 0), ....., (1, 0, 2, 0, ...., 0, 0, 1)} ∪{(0, 1, 2, 1, 0, 0, ...., 0), (0, 1, 2, 0, 1, 0, 0, ...., 0), ....., (0, 1, 2, 0, 0, ...., 0, 0, 1)} ∪{(0, 0, 2, 1, 1, 0, 0, ...., 0), (0, 0, 2, 1, 0, 1, 0, ...., 0), ....., (0, 0, 2, 1, 0, ...., 0, 0, 1) ∪{(0, 0, 2, 0, 1, 1, 0, 0, ...., 0), (0, 0, 2, 0, 1, 0, 1, 0, ...., 0), ....., (0, 0, 2, 0, 1, 0, ...., 0, 0, 1 ∪{(0, 0, 2, 0, ...., 0, 1, 1, 0), (0, 0, 2, 0, ...., 0, 1, 0, 1)} ∪{(0, 0, 2, 0, 0, ...., 0, 1, 1)} ∪{(1, 1, 0, 0, ....0, 2, 0), (1, 0, 1, 0, ....0, 2, 0), ...., (1, 0, 0, ...., 1, 2, 0), (1, 0, 0, ...., 0, 2, 1)} ∪{(0, 1, 1, 0, ....0, 2, 0), (0, 1, 0, 1, 0, ....0, 2, 0), ...., (0, 1, 0, ...., 1, 2, 0), (0, 1, 0, ...., 0, 2, 1)} ∪{(0, 0, 0, ...., 0, 1, 1, 2, 0), (0, 0, 0, ...., 1, 0, 2, 1)} ∪{(0, 0, 0, 0, ...., 0, 1, 2, 1)} ∪{(1, 1, 0, 0, ....0, 2), (1, 0, 1, 0, ....0, 2), ...., (1, 0, 0, ...., 1, 2)} ∪{(0, 1, 1, 0, ....0, 2), (0, 1, 0, 1, 0, ....0, 2), ...., (0, 1, 0, ...., ∪{(0, 0, 0, ...., 0, 1, 1, 0, 2), (0, 0, 0, ...., 1, 0, 1, 2)} ∪{(0, 0, 0, 0, ...., 0, 1, 1, 2)} .

Therefore, {X j1 1 .....X jn n | j 1 + ..... + j n = q, j s ≤ 2, one j s = 2} = {X s1 X s2 X 2 u1 | (s 1 , s 2 , u 1 ) ∈ {1, 2, ...., n} q | s 1 < s 2 , s i = u 1 f or i = 1, 2} .
Then, it is clear that to know {j 1 + ..... + j n = q, j s ≤ 2, "k" j s = 1, "h" j s = 2}, it is the same thing as to know all the k-tuple S = (s 1 , ...., s k ) ∈ {1, 2, ...., n} k and all the h-tuple U S = (u 1 , ...., u h ) ∈ {{1, 2, ...., n}\S} h in the n-k remaining elements. It is clear that the order within the h-tuples and k-tuples has no interest. Therefore, we have to consider the s 1 < .... < s k and the u 1 < ... < u h . Now, the following lemma holds.

Lemma 3.2.6 Let S * = {S = (s 1 , ...., s k ) ∈ {1, 2, ...., n} k |s 1 < s 2 < .... < s k } and U * S = {U S = (u 1 , ...., u h ) ∈ {{1, 2, ...., n}\S} h | u 1 < u 2 < .... < u h }. Then, {j 1 + ..... + j n = q, j s ≤ 2, "k" j s = 1, "h" j s = 2} = ∪ S∈S * ∪ U S ∈U * S {(j 1 , ....., j n ) | j si = 1 if s i ∈ S, j ui = 2 if u i ∈ U S , j i = 0 if not} . Lemma 3.2.7 Let S = {S = (s 1 , ...., s k ) ∈ {1, 2, ...., n} k |s 1 = s 2 = .... = s k } and U S = {U S = (u 1 , ...., u h ) ∈ {{1, 2, ...., n}\S} h | u 1 = u 2 = .... = u h }.
Then, we have

{j 1 + ..... + j n = q, j s ≤ 2, "k" j s = 1, "h" j s = 2} = ∪ S∈S * ∪ U S ∈U * S {(j 1 , ....., j n ) | j si = 1 if s i ∈ S, j ui = 2 if u i ∈ U S , j i = 0 if not} = ∪ S∈S ∪ U S ∈U S {(j 1 , ....., j n ) | j si = 1 if s i ∈ S, j ui = 2 if u i ∈ U S , j i = 0 if not} . Lemma 3.2.8 We simplify X m,s in X s . Then, j1+.....+jn=q, js=0,1,2, "k" js=1, "h" js=2 E X j1 1 .....X jn n Ψ(n) q = 1 h!k! s1 =s2 .... =s k =u1 =u2 .... =u h E X s1 .....X s k X 2 u1 .....X 2 u h Ψ(n) q .
Proof We have the following equalities

{X j1 1 .....X jn n | j 1 + ..... + j n = q, j s = 0, 1, 2, "k" j s = 1, "h" j s = 2} = {X s1 .....X s k X 2 u1 .....X 2 u h | (s 1 , s 2 , ...., s k , u 1 , ..., u h ) ∈ {1, 2, ...., n} k+h ∩ B} where B = {(s 1 , s 2 , ...., s k , u 1 , ..., u h ) | s 1 < s 2 < ..... < s k , u 1 < u 2 < ..... < u h , s i = u j } .
Let P k be the set of the permutations of k elements. Then, (s 1 , s 2 , ...., s k , u 1 , ..., u h ) ∈ {1, ..., n} k+h s 1 = s 2 ....

= s k = u 1 = u 2 .... = u h . = (s p(1) , .....s p(k) , u p ′ (1) , .....u p ′ (h) ) s 1 < ..... < s k , u 1 < ..... < u h , s i = u j , p ∈ P k , p ′ ∈ P h . Then, if p ∈ P k , p ′ ∈ P h , we have E{X s p(1) ....X s p(k) X 2 u p ′ (1) .....X 2 u p ′ (h) } = E{X s1 .....X s k X 2 u1 .....X 2 u h }.
Then, because there are q! permutations belonging to P q , j1+.....+jn=q, js=0,1,2, r js=1, t js=2

E X j1 1 .....X jn n Ψ(n) q = s1<s2<.....<s k , u1<u2<.....<u h , si =uj E X s1 .....X s k X 2 u1 .....X 2 u h Ψ(n) q = 1 h!k! s1 =s2 .... =s k =u1 =u2 .... =u h E X s1 .....X s k X 2 u1 .....X 2 u h Ψ(n) q .
One can generalize easily this lemma Lemma 3.2.9 Let p ∈ N * . Then, j1+.....+jn=q, ji≤p, "ht" ji=t, t=1,2,...,p

E X j1 1 .....X jn n Ψ(n) q = 1 h 1 !h 2 !....h p ! s t 1 =s t 2 .... =s t h t , s t i =s t ′ j if t =t ′ E p t=1 (X t s t 1 .....X t s t h t ) Ψ(n) q .
3.3 First equivalence to bounded moments

Lemma of recurence

In all these lemma, we shall use the following notations.

Notations 3.3.1 For all s ∈ {1, 2, ..., k}, for all m ∈ N * , we denote by R m,t s , t = 1, 2, ..., n m , a sequence of polynomials of degree j s . We set

k 0 = card{j s = 0}, k 1 = card{j s = 1}, k 2 = card{j s = 2}, and k 3 = card{j s > 2}. Let k ′ t = card{j s = t} and h = 2k ′ 0 + t tk ′ t .
Then, we have the following lemma.

Lemma 3.3.1 We assume k 0 = 0. Let H 1 ≥ k. Then, E t1,t2,...,t k k s=1 R m,ts s (X m,ts ) Ψ(n m ) h ≤ js=1 E nm t=1 R m,t s (X m,t ) Ψ(n m ) H1 1 H 1 js =1 E nm t=1 R m,t s (X m,t ) Ψ(n m ) js H1 1 H 1 . Proof Of course, h ≥ k. Then, we can write E t1,t2,...,t k k s=1 R m,ts s (X ts ) Ψ(n m ) h = E k s=1 nm t=1 R m,t s (X m,t ) Ψ(n m ) js = E js=1 nm t=1 R m,t s (X m,t ) Ψ(n m ) js js =1 nm t=1 R m,t s (X m,t ) Ψ(n m ) js ≤ js=1 E n t=1 R m,t s (X m,t ) Ψ(n m ) H1 1 H 1 js =1 E n t=1 R m,t s (X m,t ) Ψ(n m ) js H1 1 H 1
, by Holder's inequality.

Then, by using hypothesis 3.1.2, we have the following corollary. Lemma 3.3.2 For all s ∈ {1, 2, ..., k}, we suppose that R m,t s (x) = x js , t = 1, 2, ..., n m . We assume k 0 = 0. Then,

E t1,t2,...,t k k s=1 R m,ts s (X m,ts ) Ψ(n m ) h ≤ js>1 C nm (j s , H 1 ) 1 H 1 js=1 E nm t=1 (X m,t ) Ψ(n m ) H1 1 H 1 .
Then, by using lemma 3.2.2 and 3.2.4 , we deduce :

Lemma 3.3.3 For all s ∈ {1, 2, ..., k}, we suppose that R m,t s (x) = x js , t = 1, 2, ..., n m . We assume k 0 = 0. Then, for all r,

E t1 =t2 =.... =tr,tr+1,...,t k k s=1 R m,ts s (X m,ts ) Ψ(n m ) h ≤ e(n)K 1 M ax 1 , js=1 E nm t=1 (X m,t ) Ψ(n m ) H1 1 H 1 ,
where K 1 > 0 and where e(n) → 0 if there exists s such that j s ≥ 3.

Proof By lemma 3.3.2 , the lemma holds for r=1.

In order to prove the lemma if r ≥ 1, one uses lemma 3.

k 3 > 0 or k 2 ≥ 2 if k 3 = 0.
We define H * by H * = h -2. Then, we define H by H = H * if H * is even and H = H * + 1 if not. Then, there exists K 3 > 0 and K 4 > 0 which do not depend on m, such that

E t1,t2,...,t k k s=1 R m,ts s (X m,ts ) Ψ(n m ) h ≤ e(n m )K 3 B nm H ,
where e(n) ≤ K 4 and e(n m ) → 0 as m → ∞ if k 3 > 0.

Proof Indeed, H is even. Moreover, by our assumptions,

k ≤ h -2. Then, H ≥ k ≥ k 1 . Then, one can choose H 1 = H in lemma 3.3.2. Then, k 1 /H ≤ 1. Moreover, E nm t=1 X m,t Ψ(n m ) H = E nm t=1 X m,t Ψ(n m ) H . If E P nm t=1 Xm,t Ψ(nm) H ≤ 1, E P nm t=1 Xm,t Ψ(nm) H k 1 H ≤ 1.
If not, E

P nm t=1 Xm,t Ψ(n) H k 1 H ≤ E P nm t=1 Xm,t Ψ(n) H .
Then, E

P nm t=1 Xm,t Ψ(nm) H k 1 H ≤ B nm H .
Then, it is enough to apply lemma 3.3.2.

Lemma 3.3.5 Under the assumptions of lemma 3.3.4 , there exists K ′ 3 > 0 and K ′ 4 > 0 which do not depend on m, such that, for all r ∈ N,

1 ≤ r ≤ k, E t1 =t2 =.... =tr,tr+1,...,t k k s=1 R m,ts s (X m,ts Ψ(n m ) h ≤ e ′ (n m )K ′ 3 .B nm H , where e ′ (n m ) ≤ K ′ 4 and e ′ (n m ) → 0 as m → ∞ if k 3 > 0.
Proof Indeed by lemma 3.

First proposition about bounded moments

By using the previous lemma, we can prove the following proposition.

Proposition 3.3.1 All the moments M nm q = E (Xm,1+Xm,2+....+Xm,n m ) q Ψ(nm) q
are bounded by a real B q > 0 if and only if, for all q ∈ N, there exists Sb 1 q ∈ R + and Sb 2 q ∈ R + such that

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n m ) q ≤ Sb 1 q , s1 =s2 =..... =sq-1 E{X 2 m,s1 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q .
Proof of the sufficiency condition of prop 3.3.1 We prove the theorem by recurrence on q. If q=0,1 and 2, it is obvious.

So, we suppose that it holds for all q ′ ≤ q -1. Let S * q , be the substitution of (j 1 , j 2 , ..., j n ) ∈ N n j 1 +....+j n = q defined by S * q (j 1 , j 2 ...,

j n ) = {u 1 , u 2 , ...., u n } where u 1 ≥ u 2 ≥ .... ≥ u n .
We suppose n > q because n m → ∞. Then, u q+1 = u q+2 = .... = u n = 0 and we define S q by S q (j 1 , j 2 ..., j n ) = {u 1 , u 2 , ...., u q }. Let P q = S q (j 1 , j 2 , ...., j n ) j 1 + j 2 + .... + j n = q . Then, by lemma 3.2.9 ,

M nm q = E (X m,1 + X m,2 + .... + X m,nm ) q Ψ(n m ) q = j1+....+jn m =q q! j 1 !...j nm ! E X j1 m,1 ....X jn m m,nm Ψ(n m ) q = Oq∈Pq (j1,....,jn m ): S * q (j1,....,jn m )=Oq q! j 1 !...j nm ! E X j1 m,1 ....X jn m m,nm Ψ(n m ) q = Oq=(u1,...,uq)∈Pq N ′ Oq s1 =.... =sq q! u 1 !...u nm ! E X u1 m,s1 ....X uq m,sq Ψ(n m ) q = Oq=(u1,...,uq)∈Pq N Oq s1 =.... =sq E X u1 m,s1 ....X uq m,sq Ψ(n m ) q ,
where N ′ Oq ∈ R and N Oq ∈ R.

Let {u 1 , u 2 , ..., u k } where u 1 ≥ 3 or u 2 ≥ 2. By Lemma 3.3.5

E t1 =t2 =.... =t k X u1 m,t1 ....X u k m,t k Ψ(n m ) q
is bounded.

By our assumption,

s1 =s2 =..... =sq E{X s1 X m,s2 ......X m,sq } Ψ(n m ) q ≤ Sb 1 q , s1 =s2 =..... =sq-1 E{X 2 m,s1 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q .
Therefore Oq=(u1,...,uq): u1+....+uq=q

N Oq s1 =.... =sq E X u1 m,s1 ....X uq m,sq Ψ(n m ) q is bounded.
Proof of the necessity condition of propostion 3.3.1 Now we suppose that all the moments are bounded. Then, by lemma 3.3.3, for all O h = (u 1 , ...., u h ),

E s1 =.... =s h X u1 m,s1 ....X u h m,s h Ψ(n m ) q is bounded In particular s1 =s2 =..... =s h E{X m,s1 X m,s2 ......X m,s h } Ψ(n m ) h s1 =s2 =..... =s h-1 E{X 2 m,s1 X m,s2 ......X m,s h-1 } Ψ(n m ) h are bounded.
3.4 Second equivalence to bounded moments

3.4.1 Lemma Lemma 3.4.1 Let h = 2k ′ 0 + k-1 t=1 tk ′ t . We assume R m,t 1 = β m,t = E{X 2 m,t }
for all s ∈ {1, 2, ..., k}, and, for s ≥ 2, R m,t s (x) = x js where j s ≥ 1 . Then, there exists K 0 > 0 and

H 2 ∈ N, H 2 ≤ h -2 such that E t1,t2,...,t k-1 k-1 s=1 R m,ts s (X m,ts ) Ψ(n m ) h ≤ n t=1 β m,t Ψ(n m ) 2 B nm H2 K 0 . Proof Let H = card{j s |j s ≥ 1}. Then, H ≤ h -2. If H = h -2, the result is obvious. Indeed, E t1,t2,...,t k-1 k-1 s=1 R m,ts s (X m,ts ) Ψ(n m ) h = E nm t=1 β m,t Ψ(n m ) 2 k-1 s=2 nm t=1 X m,t Ψ(n m ) . If H < h -2, we set H 2 = H if H is even and H 2 = H + 1 if H is odd. Then we can write E t1,t2,...,t k-1 k-1 s=1 R m,ts s (X m,ts ) Ψ(n m ) h = nm t=1 β m,s Ψ(n m ) 2 E js=1 nm t=1 R m,t m,s (X m,t ) Ψ(n m ) js js>1 nm t=1 R m,t m,s (X m,t ) Ψ(n m ) js ≤ nm t=1 β m,s Ψ(n m ) 2 js=1 E nm t=1 X m,t Ψ(n m ) H2 1/H2 js>1 E nm t=1 R m,t m,s (X m,t ) Ψ(n m ) js H2 1/H2
.

By
Ψ(n m ) 2j = nm t=1 E{X 2 m,t } j Ψ(n m ) 2j ≤ nm t=1 E{(X m,t ) 2j } Ψ(n m ) 2j = E nm t=1 (X m,t ) 2j Ψ(n m ) 2j = E nm t=1 (X m,t ) 2j Ψ(n m ) 2j 1 ≤ ǫ m (2j, 1) → 0 as n → ∞ . Lemma 3.4.3 Let Q, u 1 ,v 1 ∈ N where Q > 0. If u 1 > 0 and v 1 > 0, E s1 β u1 m,s1 X v1 m,s1 Ψ(n m ) 2u1+v1 Q ≤ C nm (2v 1 , 2Q) 1/4 s1 β 2u1 m,s1 Ψ(n m ) 4u1 Q/2 → 0 . If u 1 = 0 and v 1 > 1 or if u 1 > 0 and v 1 = 0, E P s 1 β u 1 m,s 1 X v 1 m,s 1 Ψ(nm) 2u 1 +v 1 Q is bounded. If u 1 = 0 and v 1 = 1, E P s 1 β u 1 m,s 1 X v 1 m,s 1 Ψ(nm) 2u 1 +v 1 Q
is bounded if all the moments M nm q are bounded.

Proof We suppose u 1 > 0 and v 1 > 0. We have the following relations

E s1 β u1 m,s1 X v1 m,s1 Ψ(n m ) 2u1+v1 Q ≤ E s1 β 2u1 m,s1 Ψ(n m ) 4u1 1/2 s1 X 2v1 m,s1 Ψ(n m ) 2v1 1/2 Q ≤ s1 β 2u1 m,s1 Ψ(n m ) 4u1 Q/2 E s1 X 2v1 m,s1 Ψ(n m ) 2v1 Q/2 ≤ s1 β 2u1 m,s1 Ψ(n m ) 4u1 Q/2 E s1 X 2v1 m,s1 Ψ(n m ) 2v1 2Q 1/4 ≤ C nm (2v 1 , 2Q) 1/4 s1 β 2u1 m,s1 Ψ(n m ) 4u1 Q/2
Ψ(nm) 4u 1 → 0 if u 1 ≥ 1. Therefore, C nm (2v 1 , 2Q) 1/4 s1 β 2u 1 m,s 1 Ψ(nm) 4u 1 Q/2 converges always to 0 if u 1 > 0 and v 1 > 0.
Moreover, if u 1 = 0 and v 1 > 0 or if u 1 > 0 and v 1 = 0, the lemma is obvious. Lemma 3.4.4 Let 0 ≤ q" ≤ q ′ and h = q" s=1 (2u s + v s ) + (q ′ -q"). One assumes that all the moments M n q are bounded. Then, for all

P ∈ N * , E s1,....s q ′ β u1 m,s1 X v1 m,s1 β u2 m,s2 X v2 m,s2 ....β u q" m,s q" X v q"
m,s q" X m,s q"+1 ......X m,s q ′ Ψ(n m ) h P is bounded and converges to 0 if u 1 ≥ 1 and v 1 ≥ 1.

Proof Suppose P=1. We have

s1,....s q ′ E β u1 m,s1 X v1 m,s1 β u2 m,s2 X v2 m,s2 ....β u q" m,s q" X v q" m,s q" X m,s q"+1 ......X m,s q ′ Ψ(n m ) h = E s1 β u1 m,s1 X v1 m,s1
Ψ(n m ) 2u1+v1 ........

s q" β u q" m,s q" X v q" m,s q"
Ψ(n m ) 2u q" +v q"

s q"+1 X m,s q"+1 Ψ(n m ) .......

s q ′ X m,s q ′ Ψ(n m ) = E s1 β u1 m,s1 X v1 m,s1
Ψ(n m ) 2u1+v1 ........

s q" β u q"
m,s q" X v q" m,s q" Ψ(n m ) 2u q" +v q" s q"+1 X m,s q"+1 Ψ(n m )

q ′ -q" ≤ q" r=1 E sr β ur m,sr X vr m,sr Ψ(n m ) 2ur+vr q"+1 1
q"+1 E s q"+1 X m,s q"+1 Ψ(n m ) (q ′ -q")(q"+1)

1 q"+1 ≤ |M nm 2(q ′ -q")(q"+1) | 1 2(q"+1) E s1 β u1 m,s1 X v1 m,s1 Ψ(n m ) 2u1+v1 q"+1 1 q"+1 ......E s q" β u q"
m,s q" X v q" m,s q" Ψ(n m ) 2u q" +v q" q"+1 1 q"+1 .

It is enough to apply lemma 3.4.3 in order to conclude. In order to prove the lemma when P > 1, it is enough to use the same technique.

Lemma 3.4.5 For all s ∈ {1, 2, ..., k -1}, we assume R m,t s (x) = x for s ≥ 2 and R m,t 1 = β m,t for t = 1, 2, 3, ..., n m . We set H=k-2 if k is even and H=k-1, if not.

Then, for all r,

E t1 =t2 =.... =tr,tr+1,...,t k-1 k-1 s=1 R m,ts s (X m,ts ) Ψ(n m ) k ≤ K 5 B nm H .
where K 5 is a constants K 5 > 0.

Proof By hypothesis 3.1.3,

P nm t=1 βm,t Ψ(nm) 2
is bounded. Then, by lemma 3.4.1 this lemma is proved if r=1. Now, we prove this lemma by recurence on r. For example, for r=2, and for

E t1,t2,t3,...,t k-1 β m,t1 X m,t2 .........X m,t k-1 Ψ(n m ) k = E t1 =t2,t3,t4,...,t k-1 β m,t1 X m,t2 X m,t3 X m,t4 .........X m,t k-1 Ψ(n m ) k +E t1,t3,t4,...,t k-1 β m,t1 X m,t1 X m,t3 X m,t4 .........X m,t k-1 Ψ(n m ) k ,
where

E t1,t3,t4,...,t k-1 β m,t1 X m,t1 X m,t3 X m,t4 .........X m,t k-1 Ψ(n m ) k = E t1 β m,t1 X m,t1 Ψ(n m ) 3/2 t3 X m,t3 Ψ(n m ) .............. t k-1 X m,t k-1 Ψ(n m ) = E t1 β m,t1 X m,t1 Ψ(n m ) 3/2 t X m,t Ψ(n m ) k-3 ≤ E t1 β m,t1 X m,t1 Ψ(n m ) 3/2 k-2 1/(k-2) E t X m,t Ψ(n m ) k-2 (k-3)/(k-2) ≤ E t1 β m,t1 X m,t1 Ψ(n m ) 3/2 k-2 1/(k-2)
B nm H ,

where

E P t 1 βm,t 1 Xm,t 1 Ψ(nm) 3/2 k-2 1/(k-2)
→ 0 by lemma 3.4.3.

Or, in order to prove the recurrence for r=2, by lemma 3.4.1 , one proves the result for sums of the type

E t1,t2,t3,...,t k-2 β m,t1 X 2 m,t2 .........X m,t k-2 Ψ(n m ) k .

Second proposition about bounded moments

Now, one can prove the following proposition.

Proposition 3.4.1 All the moments M nm q = E (Xm,1+Xm,2+....+Xm,n m ) q Ψ(nm) q are bounded by a real B q > 0 if and only if, for all q ∈ N, there existe Sb r q ∈ R , r=1,2, such that

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n m ) q ≤ Sb 1 q s1 =s2 =..... =sq-1 E{[(X m,s1 ) 2 -β m,s1 ]X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q .
Proof of the sufficiency condition of proposition 3.4.1 We prove the result by recurrence on q. When q=0, 1 and 2, it is obvious.

One supposes that all the moment of order q' are bounded if q ′ ≤ q -1. Then, by lemma 3.4.5, and by recurence assumption,

s1 =s2 =...... =sq E{ β m,s1 X m,s2 ......X m,sq-1 } Ψ(n m ) q
is bounded. Therefore

s1 =s2 =...... =sq-1 E{ X 2 m,s1 X m,s2 ......X m,sq-1 } Ψ(n m ) q
is also bounded. Therefore, it is enough to refer tot the proof of the recurrence in proposition 3.3.1 which proves that M nm q is bounded. That proves the recurence.

Proof of the necessity condition of proposition 3.4.1 Now we suppose that all the moments are bounded. Then, by proposition 3.3.1 ,

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n m ) q s1 =s2 =..... =sq-1 E{X 2 m,s1 X m,s2 ......X m,sq-1 } Ψ(n m ) q are bounded
Moreover, by lemma 3.4.5,

s1 =s2 =...... =sq E{ β m,s1 X m,s2 ......X m,sq-1 } Ψ(n m ) q is bounded. Therefore, s1 =s2 =..... =sq-1 E{[(X m,s1 ) 2 -β m,s1 ]X m,s2 ......X m,sq-1 } Ψ(n m ) q
is bounded.

Third equivalence to bounded moments

We now give the condition expressed in terms of ρ j1,...,jn 's.

Lemma 3.5.1 Let P m,s1

2 (x) = x 2 -γ m,s1 x -β m,s1 where γ m,s1 = E{X 3 m,s1 }/E{X 2 m,s1 }. We suppose that nm s=1 γ 2 m,s Ψ(n) 2 is bounded. Then, for all h ∈ N * , E nm s=1 γm,s(Xm,s) h Ψ(n) h+1 2 is bounded. Proof For all h ∈ N * , E nm s=1 γ m,s (X m,s ) h Ψ(n) h+1 2 ≤ E nm s=1 γ 2 m,s Ψ(n) 2 nm s=1 (X m,s ) 2h Ψ(n) 2h ≤ nm s=1 γ 2 m,s Ψ(n) 2 C(2h, 1)
which is bounded by our assumptions.

Then, we have the following proposition.

Proposition 3.5.1 We suppose that

nm s=1 γ 2 m,s Ψ(n) 2 is bounded. Then, all the moments M nm q = E (Xm,1+Xm,2+....+Xm,n m ) q Ψ(n) q
are bounded by a real B q > 0 if and only if, for all q ∈ N, there existe Sb r q ∈ R , r=1,2,3, such that

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n) q ≤ Sb 1 q s1 =s2 =..... =sq-1 E{ P m,s1 2
(X m,s1 )X m,s2 ......X m,sq-1 } Ψ(n) q ≤ Sb 2 q , s1 =s2 =...... =sq γ m,s1 E{X m,s1 X m,s2 ......X m,sq } Ψ(n) q+1 ≤ Sb 3 q .
These conditions are indeed conditions about the ρ j1,...,jn 's. Indeed, we know that P m,s 2 is an orthogonal polynomial with degree 2 associated to X m,s . Therefore, there exists (j 1 , ..., j n ) with j s ≤ 2 and only one j t = 2 such that E{ P m,s1 2 (X m,s1 )X m,s2 ......X m,sq-1 } = α j1,...,jn .

Proof of the sufficiency of proposition 3.5.1 By our assumption,

s1 =s2 =...... =sq E{ γ m,s1 X m,s1 X m,s2 ......X m,sq-1 } Ψ(n) q is bounded. Therefore, s1 =s2 =...... =sq E{ [X 2 m,s1 -β m,s1 ]X m,s2 ......X m,sq-1 } Ψ(n) q
is bounded. Then, all the conditions of proposition 3.4.1 are checked. Then the sufficient condition is proved.

Proof of the necessity of proposition 3.5.1 Now we suppose that all the moments are bounded. Then, by proposition 3.4.1 ,

s1 =s2 =..... =s h E{X m,s1 X m,s2 ......X m,s h } Ψ(n) h s1 =s2 =..... =s h-1 E{[X 2 m,s1 -β m,s1 ]X m,s2 ......X m,s h-1 } Ψ(n) h are bounded Moreover, s1,s2,....,sq-1 E{ γ m,s1 X m,s1 X m,s2 ......X m,sq-1 } Ψ(n) q = E s1 γ m,s1 X m,s1 Ψ(n) 2 s2 X m,s2 Ψ(n) ........ sq-1 X m,sq-1 Ψ(n) = E s1 γ m,s1 X m,s1 Ψ(n) 2 s2 X m,s2 Ψ(n) q-2 ≤ E s1 γ m,s1 X m,s1 Ψ(n) 2 2 1/2 E s2 X m,s2 Ψ(n) 2(q-2) 1/2
which is bounded by lemma 3.5.1. Then, by lemma 3.2.2 and 3.2.4 , we shall understand, by using the same technique with the lemma 3.5.1 and by using lemma 3.3.2 that it is true also for

s1 =s2 =...... =sq E{ γ m,s1 X m,s1 X m,s2 ......X m,sq } Ψ(n) q .
That proves necessary condition.

Fourth equivalence to bounded moments

In this section, we suppose that there exists F > 0 such that |X m,t | ≤ F .

Lemma

Lemma 3.6.1 Let j 1 ≥ 2 and j s ≥ 1 for s=2,3,....,k. We define H * by H * = k -1. Then, we define H by H = H * if H * is even and H = H * + 1 if not. Then, there exists

K 6 > 0 such that E t1,t2,...,t k k s=1 (X m,ts ) js Ψ(n m ) h ≤ K 6 B nm H .
Proof We have

E t1,t2,...,t k k s=1 (X m,ts ) js Ψ(n m ) h ≤ E nm t=1 (X m,t ) j1 Ψ(n m ) j1 s≥2 nm t=1 (X m,t ) js Ψ(n m ) js ≤ n m F j1 c j1 Ψ (n m ) j1/2 E s≥2 nm t=1 (X m,t ) js Ψ(n m ) js ≤ F j1 c j1 Ψ E s≥2 nm t=1 (X m,t ) js Ψ(n m ) js ≤ F j1 c j1 Ψ E s≥2, js≥2 nm t=1 (X m,t ) js Ψ(n m ) js s≥2, js=1 nm t=1 (X m,t ) js Ψ(n m ) js
(therefore for the same reasons as previously, there exists K 5 > 0 such that )

≤ K 6 E s≥2, js=1 nm t=1 (X m,t ) js Ψ(n m ) js ≤ K 6 E nm t=1 X m,t Ψ(n m ) k1 (then, because k 1 ≤ H * ≤ H where H is even) ≤ K 6 E nm t=1 X m,t Ψ(n m ) H k 1 H ≤ K 6 B nm H . Lemma 3.6.2 Let k ∈ N * , k ≥ 2. We define H * by H * = k -1. Then, we define H by H = H * if H * is even and H = H * + 1 if not. There exists K 7 > 0 such that E t1 =t2 =.... =tr,tr+1,...,t k X 2 m,t1 X m,t2 X m,t3 ....X m,t k Ψ(n m ) k+1 ≤ K 7 .B nm H .
Proof By lemma 3.6.1, this result holds for r=1. Now suppose that it holds for all r ′ ≤ r -1.

Then, it is enough to apply lemma 3.2.2, 3.3.4, 3.2.4 and 3.6.1.

Fourth proposition about bounded moments

Remark that if |X m,t | ≤ F , the hypothesis 3.1.2 holds.

Proposition 3.6.1 One supposes that there exists

F > 0 such that |X m,t | ≤ F . All the moments M nm q = E (Xm,1+Xm,2+....+Xm,n m ) q Ψ(n) q
are bounded by a real B q > 0 if and only if, for all q ∈ N, there existe Sb 1 q ∈ R such that

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n) q ≤ Sb 1 q
Proof of the sufficiency condition of proposition 3.6.1 We prove this proposition by recurence. It holds for q=0,1,2. Suppose that it holds for all q ′ ≤ q -1. Then, by lemma 3.6.2,

s1 =s2 =...... =sq E{ X 2 m,s1 X m,s2 ......X m,sq-1 } Ψ(n) q is bounded.
By proof of the sufficient condition of proposition 3.3.1,

M nm q =
Oq=(u1,...,uq)∈Pq

N Oq s1 =.... =sq E X u1 m,s1 ....X uq m,sq Ψ(n m ) q ,
where N Oq ∈ R.

Let {u 1 , u 2 , ..., u k } where u 1 ≥ 3 or u 2 ≥ 2. By lemma 3.3.3

E t1 =t2 =.... =t k X u1 m,t1 ....X u k m,t k Ψ(n m ) q
is bounded. By our assumption,

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n m ) q ≤ Sb 1 q .
Then, M nm q is bounded. The sufficient condition is proved.

Proof of the necessity condition of proposition 3.6.1 Now we suppose that all the moments are bounded. Then, it is enough to use proposition 3.3.1.

First Theorem of Convergence

Lemma

We shall need the following lemma.

Lemma 3.7.1 We suppose that all the moments M nm q are bounded. We suppose that R m,t

1 (x) = Q m,t 2 (x) = x 2 -β m,t and that R m,t
s is a polynomial of degree j s . If j s = 0, one assume that

(1/Ψ(n m ) 2 ) nm t=1 R m,t s (X m,t ) is bounded. If j s ≥ 1 , one assume that R m,t s (x) = x js . Moreover let h = k s=1 j ′ s where j ′ s = j s if j s ≥ 1 and j ′ 0 = 2 if j s = 0. We assume 1 Ψ(nm) 4 E nm t=1 Q m,t 2 (X m,t ) 2 → 0. Then, for all r, 1 ≤ r ≤ k, 1 Ψ(n m ) h .E t1 =t2 =.... =tr,tr+1,...,t k k s=1 R m,ts s (X m,ts ) converges to 0.
Proof By the Schwartz inequality and by the Holder Inequality

1 Ψ(n m ) 2h E t1 t2,...,t k k s=1 R m,ts s (X m,ts ) 2 = 1 Ψ(n m ) 2h E k s=1 nm ts=1 R m,ts s (X m,ts ) 2 ≤ E 1 Ψ(n m ) 4 nm t=1 R m,t 1 (X m,t ) 2 E 1 Ψ(n m ) 2h-4 k s=2 nm ts=1 R m,ts s (X m,ts ) 2 ≤ E 1 Ψ(n m ) 4 nm t=1 R m,t 1 (X m,t ) 2 k s=2 E nm ts=1 R m,ts s (X m,ts ) Ψ(n m ) js 2(k-1) 1 k-1 .
By assumption and by hypothesis 3.1.2 , the term on right-hand side is bounded. Moreover,

1 Ψ(n m ) 4 E nm t=1 R m,t 1 (X m,t ) 2 → 0 .
We deduce the lemma if r=1. When r > 1, we prove the result by recurrence by using the lemma 3. Ψ(n m ) q → 0 .

Proof For example,

s1 =.... =sq-2 E X 2 m,s1 X 2 m,s2 X m,s3 ....X m,sq-2 Ψ(n m ) q = s1 =.... =sq-2 E (X 2 m,s1 -β m,s1 )X 2 m,s2 X m,s3 ....X m,sq-2 Ψ(n m ) q + s1 =.... =sq-2 E β m,s1 X 2 m,s2 X m,s3 ....X m,sq-2 Ψ(n m ) q = s1 =.... =sq-2 E (X 2 m,s1 -β m,s1 )X 2 m,s2 X m,s3 ....X m,sq-2 Ψ(n m ) q + s1 =.... =sq-2 E β m,s1 (X 2 m,s2 -β m,s2 )X m,s3 ....X m,sq-2 Ψ(n m ) q + s1 =.... =sq-2 E β m,s1 β m,s2 X m,s3 ....X m,sq-2 Ψ(n m ) q .
Then for example by lemma 3.7.1

s1 =.... =sq-2 E β m,s1 (X 2 m,s2 -β m,s2 )X m,s3 ....X m,sq-2 Ψ(n m ) q → 0 .
Lemma 3.7.3 Let q ∈ N and e ∈ N such that 1 ≤ e ≤ q -e. We suppose that all the moments M nm h are bounded. We suppose that, for all q ′ ≤ q -1,

s1 =s2 =..... =s q ′ E{X m,s1 X m,s2 ......X m,s q ′ } Ψ(n m ) q ′ → S q ′ .
We set S 0 = 1. Then, s1 =.... =sq-e E β m,s1 β m,s2 .....β m,se X m,se+1 ....X m,sq-e Ψ(n m ) q → σ 2e 0 S q-2e .

Proof We set {s 1 = .... = s e , s e+1 = .... = ...s q-e } = (s 1 , s 2 , ..., s e , s e+1 , .., s q-e ) ∈ {1, 2, ..., n m } q-e s r = s r ′ if r < r ′ ≤ e or e+1 ≤ r < r ′ ≤ q-e .

Then, we study at first , we shall have to add or to subtract sums such that there exists i, j satisfying s i = s e+j . For example, we shall have to subtract terms of the form s2 =.... =se,se+1 =.... =...sq-e

E

β m,s2 ....β m,se [β m,se+1 X m,se+1 ]X m,se+2 ....X m,sq-e Ψ(n m ) q .

One will prove that this sum converges to 0 by recurence. With this aim, we study first as a sum of products. Or by using lemma 3.4.4, we always find in these products, terms which are bounded by products of bounded expectations (cf lemma 3.3.1 and hypothesis 3.1.2) in the form

E P s 1 β u m,s X v m,s Ψ(nm) 2u+v 2w
, u, v, w ∈ N : among these terms there is at least one term of the form

se+1 β u m,s e+1 X v m,s e+1 Ψ(nm) 2u+v
which converges to 0. That proves the recurence, and then, that s2 =.... =se,se+1 =.... =...sq-e E β m,s2 ....β m,se β m,se+1 X m,se+1 X m,se+2 ....X m,sq-e Ψ(n m ) q → 0 .

We deduce that s1 =....... =sq-e E β m,s1 β m,s2 ....β m,se X m,se+1 X m,se+2 ....X m,sq-e Ψ(n m ) q converges to the same limit as s1 =.... =se,se+1 =.... =...sq-e E β m,s1 β m,s2 ....β m,se X m,se+1 X m,se+2 ....X m,sq-e Ψ(n m ) q , i.e. as

s1 β m,s1 Ψ(n m ) 2 ....... se β m,se Ψ(n m ) 2 E se+1 =.... =sq-e
X m,se+1 ....X m,sq-e Ψ(n m ) q-2e → σ 2e 0 S q-2e .

First theorem of convergence.

Theorem 12 One assumes that

E nm t=1 (X m,t ) 2 -E{(X m,t ) 2 } Ψ(n m ) 2 2 → 0 .
All the moments M nm q = E (Xm,1+Xm,2+....+Xm,n) q Ψ(nm) q converges to a real M q if and only if, for all q ∈ N, there exists S q ∈ R and Sb 2 q ∈ R + :

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n m ) q → S q s1 =s2 =..... =sq-1 E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q .
Proof of sufficient Condition of theorem 12 By proposition 3.3.1, all the moments are bounded.

Thanks to lemma 3.7.3 and 3.7.2, we deduce that, for all q and all e,

s1 =.... =sq-e E X 2 m,s1 X 2 m,s2 ....X 2 m,se X m,se+1 ....X m,sq-e Ψ(n m ) q → σ 2e 0 S q-2e
.

By lemma 3.3.5 , s1 =.... =s k E X u 1 m,s 1 ....X u k m,s k Ψ(nm) h
→ 0 if there exist t such that u t ≥ 3. Therefore

s1 =.... =s k E X u 1 m,s 1 ....X u k m,s k Ψ(nm) h
converges for all (u 1 , ..., u k ). Therefore, because

E (X m,1 + .... + X m,n ) q ψ(n m ) q = j1+.....+jn m =q q! j 1 !.......j nm ! E X j1 m,1 .....X jn m m,nm ψ(n m ) q ,
all the moments converge

Proof of Necessary Condition of theorem 12

We suppose that all the moments converge. Then, the second relation is a consequence of proposition 3.3.1 .

Then, we prove the first relation by recurrence. If q= 1 or 2, it is obvious. Then, we suppose that it holds for all q ′ ≤ q -1.

We deduce from lemma 3.7.3 and 3.7.2 that, for all e ≥ 1,

s1 =.... =sq-e E X 2 m,s1 X 2 m,s2 ....X 2 m,se X m,se+1 ....X m,sq-e Ψ(n m ) q → σ 2e 0 S q-2e . By lemma 3.3.5 , s1 =.... =s k E X u 1 m,s 1 ....X u k m,s k Ψ(nm) h → 0 if there exist t such that u t ≥ 3. Therefore s1 =.... =s k E X u 1 m,s 1 ....X u k m,s k Ψ(nm) h
converges for all (u 1 , ..., u k ) such that there exists u t ≥ 2. Therefore, because

E (X m,1 + .... + X m,n ) q ψ(n m ) q = j1+.....+jn m =q q! j 1 !.......j nm ! E X j1 m,1 .....X jn m m,nm ψ(n m ) q ,
then, s1 =.... =sq E Xm,s 1 ....Xm,s q Ψ(nm) q converges.

Convergence to the normal distribution

One can specify theorem 12 in the case of convergence to the normal distribution.

Proposition 3.8.1 One assumes that

E nm t=1 ( (X m,t ) 2 -E{(X m,t ) 2 } Ψ(n m ) 2 2 → 0 .
The moments of order q, M nm q converges to the moment of order q of N (0, σ 2 0 + S 2 ) where S 2 ∈ R, if and only if, for all q ∈ N * ,

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n m ) q → S q ,
where S q = ν q , the moment of order q of N (0, S 2 ) for q ≥ 1 and if there exists Sb 2 q > 0 such that, for all q ∈ N * ,

s1 =s2 =..... =sq-1 E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q .

Proof

We know that

E (X m,1 + .... + X m,nm ) q Ψ(n m ) q = j1+.....+jn m =q q! j 1 !.......j nm ! E X j1 m,1 .....X jn m m,nm Ψ(n m ) q .
Now, because, by theorem 12 , all the moments converges, then, by lemma 3.3.5, if there exists s such that j s ≥ 3, j1+.....+jn=q, at least one js≥3

q! j 1 !.......j nm ! E X j1 m,1 .....X jn m m,nm Ψ(n m ) q → 0 .
Then, it is enough to study the sums where j s ≤ 2.

Then, we have the following lemma Lemma 3.8.1 We set S 0 = 1. Let q" = ⌊q/2⌋, the integer part of q/2. Then,

E (X m,1 + .... + X m,nm ) q Ψ(n m ) q ∼ q! 1 σ 0 0 0! S q (q)! + q! 2 σ 2 0 1! S q-2 (q -2)! + q! 2 2 σ 4 0 2!
S q-4 (q -4)! + ............. + q! 2 q" σ 4 0 q"! S q-2q" (q -2q")! .

Proof

We have

E (X m,1 + .... + X m,nm ) q Ψ(n m ) q = j1+.....+jn m =q q! j 1 !.......j nm ! E X j1 m,1 .....X jn m m,nm Ψ(n m ) q ∼ j1+.....+jn m =q, js≤1 q! j 1 !.......j nm ! E X j1 m,
q! j 1 !.......j nm ! E X j1 m,1 .....X jn m m,nm Ψ(n m ) q ∼ q! 1 j1+.....+jn m =q, js≤1 E X j1 m,1 .....X jn m m,nm Ψ(n m ) q + q! 2 j1+.....+jn m =q, js≤2, one js=2 E X j1 m,1 .....X jn m m,nm Ψ(n m ) q + q! 2 2
Ψ(n m ) q ∼ q! 1 σ 0 0 0! S q (q)! + q! 2 σ 2 0 1! S q-2 (q -2)! + q! 2 2 σ 4 0 2!
S q-4 (q -4)! + ............. + q! 2 q" σ 2q" 0 q"! S q-2q" (q -2q")! , by lemma 3.7.2, 3.7.3 and 3.2.9.

Proof of sufficient condition If q is odd, q-2e is odd and S q-2e = 0. Then, by lemma 3.8.1 M nm q → 0. Now, we study the case where q is even. Then, by lemma 3.8.1 and because ν 2q =

ν q 2 (2q)! 2 q q! , E (X m,1 + .... + X m,nm ) 2q Ψ(n m ) 2q ∼ 2q! 1 σ 0 0 0! S 2q (2q)! + 2q! 2 σ 2 0 1! S 2q-2 (2q -2)! + 2q! 2 2 σ 4 0 2! S 2q-4 (2q -4)! + ............. + 2q! 2 q σ 2q 0 q! S 0 0! ∼ 2q! 1 σ 0 0 0! S q 2 (2q)! (2q)!2 q q! + 2q! 2 σ 2 0 1! S q-1 2 (2q -2)! (2q -2)!2 q-1 (q -1)! + 2q! 2 2 σ 4 0 2! S q-2 2 (2q -4)! (2q -4)!2 q-2 (q -2)! +................+ 2q! 2 q σ 2q 0 q! S q-q 2 (2q -2q)! 0!2 q-q (q -q)! ∼ 2q! 1 σ 0 0 0! S q 2 2 q q! + 2q! 2 σ 2 0 1! S q-1 2 2 q-1 (q -1)! + 2q! 2 2 σ 4 0 2! S q-2 2 2 q-2 (q -2)! + ................ + 2q! 2 q σ 2q 0 q! S q-q 2 2 q-q (q -q)! ∼ 2q! 2 q S q 2 0!q! + σ 2 0 S q-1 2 1!(q -1)! + σ 4 0 S q-2 2 2!(q -2)! + ................ + σ 2q 0 S 0 2 q!0! ∼ 2q! q!2 q (σ 2 0 + S 2 ) q
which is the moment of order 2q of N (0, σ 2 0 + S 2 ).

Proof of necessary condition

We suppose that the moments M nm q converge to M q which is the moment of order q of N (0, M 2 ). Then, we prove by recurence that S q is the moment of order q of N (0, S 2 ). Indeed, it is true if q=1. If q=2,

E (X m,1 + .... + X m,nm ) 2 Ψ(n m ) 2 = s E X 2 m,s Ψ(n m ) 2 + s =t E X m,s X m,t Ψ(n m ) 2 → σ 2 0 + S 2 = M 2 .
Then, we suppose that S q ′ is the moment of order q' of N (0, S 2 ) if q ′ ≤ 2q -2. By lemma 3.8.1, if q is even,

M 2q = M q 2 (2q!) 2 q q! = (S 2 + σ 2 0 ) q (2q!) 2 q q! ∼ E (X m,1 + .... + X m,nm ) 2q Ψ(n m ) 2q ∼ 2q! 1 σ 0 0 0! S 2q (2q)! + 2q! 2 σ 2 0 1! S 2q-2 (2q -2)! + 2q! 2 2 σ 4 0 2! S 2q-4 (2q -4)! + ........ + 2q! 2 q σ 2q 0 q! S 0 0! ∼ 2q! 1 σ 0 0 0! S 2q (2q)! + 2q! 2 σ 2 0 1! S q-1 2 2 q-1 (q -1)! + 2q! 2 2 σ 4 0 2! S q-2 2 2 q-2 (q -2)! + ................ + 2q! 2 q σ 2q 0 q! S q-q 2 2 q-q (q -q)! ∼ S 2q + 2q! 2 q σ 2 0 S q-1 2 1!(q -1)! + σ 4 0 S q-2 2 2!(q -2)! + ................ + σ 2q 0 S 0 2 q!0! . Therefore (S 2 + σ 2 0 ) q = S 2q 2 q q! (2q!) + q! σ 2 0 S q-1 2 1!(q -1)! + σ 4 0 S q-2 2 2!(q -2)! + ................ + σ 2q 0 S 0 2 q!0! . Therefore, (S 2 + σ 2 0 ) q = S 2q 2 q q! (2q!) + q! σ 0 0 S q 2 q! + σ 2 0 S q-1 2 1!(q -1)! + σ 4 0 S q-2 2 2!(q -2)! + ................ + σ 2q 0 S 0 2 q!0! -S q 2 .
Therefore,

S2q2 q q! (2q!) = S q 2 and S 2q = S q 2 S q 2 (2q!) 2 q q! .
If q is odd, this result is easier to prove because S 2q+1 = 0 and M 2q+1 = 0.

Second theorem of convergence

In this section, we assume that that

nm s=1 γ 2 m,s Ψ(n) 2 is bounded. Moreover, we suppose that E nm s=1 γm,sXm,s Ψ(n) 2
2 → 0. For example, it is an hypothesis which holds as soon as γ m,s is bounded, as E X 2 m,s is bounded and as |E X m,s X m,t | ≤ α(s -t) where α(h) → 0 when h → ∞ 3.9.1 Lemma Lemma 3.9.1 We suppose that E nm s=1 γm,sXm,s

Ψ(n) 2
2 is bounded. Then

E nm t=1 P m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 . if and only if E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 . Proof We know that P m,s1 2 (x) = x 2 -γ m,s1 x -β m,s1 where γ m,s1 = E{X 3 m,s1 }/E{X 2 m,s1 }. Then, E nm t=1 P m,t 2 (X m,t ) Ψ(n m ) 2 2 = E nm s=1 (X 2 m,s -γ m,s X m,s -β m,s ) Ψ(n m ) 2 nm t=1 (X 2 m,t -γ m,t X m,t -β m,t ) Ψ(n m ) 2 = E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 -2 nm s=1 γ m,s X m,s Ψ(n m ) 2 nm t=1 (X 2 m,t -β m,t ) Ψ(n m ) 2 + nm s=1 γ m,s X m,s Ψ(n m ) 2 2
.

By our assumption

E nm s=1 γ m,s X m,s Ψ(n m ) 2 2 → 0 .
Therefore, by Schwartz inequality,

E nm s=1 γ m,s X m,s Ψ(n m ) 2 nm t=1 (X 2 m,t -β m,t ) Ψ(n m ) 2 2 ≤ E nm s=1 γ m,s X m,s Ψ(n m ) 2 2 E nm t=1 (X 2 m,t -β m,t ) Ψ(n m ) 2 2 where E nm t=1 (X 2 m,t -β m,t ) Ψ(n m ) 2 2 = E nm t=1 (X 2 m,t -β m,t ) Ψ(n m ) 2 nm t=1 (X 2 m,t -β m,t ) Ψ(n m ) 2 = E nm t=1 X 2 m,t Ψ(n m ) 2 2 -2E nm t=1 β m,t Ψ(n m ) 2 nm t=1 X 2 m,t Ψ(n m ) 2 + E nm t=1 β m,t Ψ(n m ) 2 2 = E nm t=1 X 2 m,t Ψ(n m ) 2 2 -2 nm t=1 β m,t Ψ(n m ) 2 E nm t=1 X 2 m,t Ψ(n m ) 2 + nm t=1 β m,t Ψ(n m ) 2 2 ≤ C(2, 2) + 2C(2, 1) nm t=1 β m,t Ψ(n m ) 2 + nm t=1 β m,t Ψ(n m ) 2 2
which is bounded by hypothesis 3.1.3.

Therefore

E nm s=1 γ m,s X m,s Ψ(n m ) 2 nm t=1 (X 2 m,t -β m,t ) Ψ(n m ) 2 2 → 0. We deduce that E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 is equivalent to E nm t=1 P m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 .

Statement and proof of the theorem

This second theorem is expressed in terms of ρ j1,...,jn .

Theorem 13

We assume that that

nm s=1 γ 2 m,s Ψ(n) 2 is bounded and that E nm s=1 γm,sXm,s Ψ(n) 2 2 → 0 .
One assumes that

E nm t=1 P m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 . All the moments M nm q = E (Xm,1+Xm,2+....+Xm,n m ) q Ψ(nm) q
converge to a real M q if and only if, for all q ∈ N, there exists S q ∈ R and Sb r q ∈ R , r=2,3, such that

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n) q → S q s1 =s2 =..... =sq-1 E{ P m,s1 2
(X m,s1 )X m,s2 ......X m,sq-1 } Ψ(n) q ≤ Sb 2 q , s1 =s2 =...... =sq γ m,s1 E{X s1 X s2 ......X sq } Ψ(n) q+1 ≤ Sb 3 q .
Proof of the sufficiency condition of theorem 13 By lemma 3.9.1,

E nm t=1 P m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 . is equivalent to E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 .
By proposition 3.5.1 , all the moments are bounded. Then, by proposition 3.3.1

s1 =s2 =..... =sq-1 E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q .
Then, by theorem 12, all the moments converge.

Proof of the necessity condition of theorem 13 Now we suppose that all the moments converge. Then, by proposition 3.5.1 ,

s1 =s2 =..... =sq-1 E{ P m,s1 2
(X m,s1 )X m,s2 ......X m,sq-1 } Ψ(n) q ≤ Sb 2 q , s1 =s2 =...... =sq γ m,s1 E{X s1 X s2 ......X sq } Ψ(n) q+1 ≤ Sb 3 q .
Moreover, by theorem 12,

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n) q → S q .
By using proposition 3.8.1, it is easy to deduce the following theorem.

Corollary 3.9.1 One assumes that

E nm t=1 ( P m,t 2 (X m,t )) Ψ(n m ) 2 2 → 0 .
All the moments of order q M nm q converge to M q , the moment of order q of N (0, σ 2 0 + S 2 ), S 2 ∈ R, if and only if, for all q ∈ N,

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n m ) q → S q
where S q = ν q the moment of order q of N (0, S 2 ),

s1 =s2 =..... =sq-1 E{ P m,s1 2
(X m,s1 )X m,s2 ......X m,sq-1 } Ψ(n) q ≤ Sb 2 q , s1 =s2 =...... =sq γ m,s1 E{X s1 X s2 ......X sq } Ψ(n) q+1 ≤ Sb 3 q .

Third theorem of Convergence

Theorem 14 One supposes that there exists F > 0 such that |X m,t | ≤ F . One assumes that

E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 .
All the moments M nm q = E (Xm,1+Xm,2+....+Xm,n m ) q Ψ(nm) q converges to a real M q if and only if, for all q ∈ N, there existe S q ∈ R such that

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n) q → S q .
Proof of the sufficient condition of theorem 14 By proposition 3.6.1, all the moments M nm q are bounded. By proposition 3.3.1,

s1 =s2 =..... =sq-1 E{X 2 m,s1 X m,s2 ......X m,sq-1 } Ψ(n) q
is bounded. By theorem 12, all the moments M nm q converges.

Proof of the Necessary condition of theorem 14 By theorem 12, s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n) q → S q .

By using proposition 3.8.1, it is easy to deduce the following theorem.

Corollary 3.10.1 We suppose that there exists F > 0 such that |X m,t | ≤ F . One assumes that

E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 . All the moments M nm q = E (Xm,1+Xm,2+....+Xm,n m ) q Ψ(nm) q
converge to the moment of order q of N (0, σ 2 0 + S 2 ), S 2 ∈ R, if and only if, for all q ∈ N * ,

s1 =s2 =..... =sq E{X m,s1 X m,s2 ......X m,sq } Ψ(n m ) q → S q ,
where S q = ν q , the moment of order q of N (0, S 2 ).

Chapter 4

MCLT in dimension 2

Notations and assumptions

We use again notation introduced in the previous chapter. Moreover, they are completed by the following way.

E nm t=1 (X m,t ) j Ψ(n m ) j p ≤ C nm (j, p) ≤ C(j, p) . E nm t=1 (Y m,t ) j Ψ(n m ) j p ≤ C nm (j, p) ≤ C(j, p) .
Hypothesis 4.1.2 One supposes that there exists σ 2 0 and σ 2 1 such that

(1/Ψ(n) 2 ) s E{X 2 m,s } → σ 2 0 and (1/Ψ(n) 2 ) s E{Y 2 m,s } → σ 2 1 . We set β ′ m,s = E{Y 2 m,s }.
In the case of random vectors, higher order correllation coefficients are defined by the same way.

Notations 4.1.2 For all n, n ′ ∈ N * , for all (j 1 , j 2 , ..., j n ) ∈ N n , for all (i 1 , i 2 , ..., i n ′ ) ∈ N n ′ , we set ρ j1,j2,...,jn,i1,i2,...,i n ′ = E{P m,1 j1 (X m,1 )P m,2 j2 (X m,2 )....P m,n jn (X m,n )P ′m,1 i1 (Y m,1 )P ′m,2 i2 (Y m,2 )....P ′m,n ′ i n ′ (Y m,n ′ )} ,
where {P ′m,t j }, j ∈ N, is the family of orthonormal polynomials associated to Y m,t .

First theorem of Convergence

lemmas about sets

One generalizes easily the lemmas to the case with two dimensions by using natural notations.

Then, the following lemma holds.

Lemma 4.2.1 Let r ≥ 3. Then, for all k ≤ h and all r ′ ≤ r -1,

t 1 = t 2 = .... = t r-1 , t r , ..., t k , s 1 = s 2 = .... = s r ′ , s r ′ +1 , ..., s k ′ = t 1 = t 2 = .... = t r , t r+1 , ..., t k , s 1 = s 2 = .... = s r ′ , s r ′ +1 , ..., s k ′ ∪ t 1 = t r = t 2 = ... = t r-1 , t r+1 , .., t k , s 1 = s 2 = .... = s r ′ , s r ′ +1 , ..., s k ′ ∪ t 1 = t 2 = t r = ... = t r
∪ t 1 = t 2 = ... = t r-1 = t r , t r+1 , .., t k , s 1 = s 2 = .... = s r ′ , s r ′ +1 , ..., s k ′ .
It is enough to sum under the sign E and to use the proof of lemma 3.2.9 in order to obtain the following propositions.

Lemma 4.2.2 We simplify X m,t in X t and Y m,t in Y t . Let p, q ∈ N * . Then, j1+...+jn=q, js≤2, "k" js=1, "h" js=2 i1+...+in=p, is≤2, "a" is=1, "b" is=2 E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) k+2h+a+2b = 1 h!k!a!b! s1 =.... =s k =u1 =.... =u h t1 =.... =ta =w1 =.... =w b E X s1 .....X s k X 2 u1 .....X 2 u h Y t1 .....Y ta Y 2 w1 .....Y 2 w b Ψ(n) k+2h+a+2b .
Lemma 4.2.3 Let p, q ∈ N * . Then, j1+...+jn=q, "ht" js=t, t=0,1,..,r i1+...+in=p, "at" is=t, t=0,1,..,o

E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) p+q = 1 h 1 !...h r !a 1 !...a o ! s t 1 =....
= r ′ 1 + .... + r ′ k ′
We suppose that all the moments M nm q,p are bounded. We suppose that there exists

d s ≥ 3 or d ′ s ≥ 3. Then, u1 =u2 =.... =ur,ur+1,...,u k v1 =v2 =.... =v r ′ ,v r ′ +1 ,...,v k ′ E X d1 m,u1 ....X d k m,u k Y d ′ 1 m,v1 ....Y d ′ k ′ m,v k ′ Ψ(n) h+h ′ → 0 .
Proof This lemma holds if r=r'=1. For example, if d 1 ≥ 3, by Holder Inequality

u1 =u2 =.... =ur,ur+1,...,u k v1 =v2 =.... =v r ′ ,v r ′ +1 ,...,v k ′ E X d1 m,u1 ....X d k m,u k Y d ′ 1 m,v1 ....Y d ′ k ′ m,v k ′ Ψ(n) h+h ′ = E s X d1 m,s Ψ(n) d1 ..... s X d k m,s Ψ(n) d k s Y d ′ 1 m,s Ψ(n) d ′ 1 ..... s Y d ′ k ′ m,s Ψ(n) d ′ k ′ ≤ E ˛P s X d 1 m,s Ψ(n) d 1 ˛k+k ′ ff 1 k+k ′ .....E ˛P s X d k m,s Ψ(n) d k ˛k+k ′ ff 1 k+k ′ E ˛P s Y d ′ 1 m,s Ψ(n) d ′ 1 ˛k+k ′ ff 1 k+k ′ ......E ˛P s Y d ′ k ′ m,s Ψ(n) d ′ k ′ ˛k+k ′ ff 1 k+k ′ ≤ 2 4 E " P s X d 1 m,s Ψ(n) d 1 « 2k+2k ′ ff ....E " P s X d k m,s Ψ(n) d k « 2k+2k ′ ff E " P s Y d ′ 1 m,s Ψ(n) d ′ 1 « 2k+2k ′ ff .....E " P s Y d ′ k ′ m,s Ψ(n) d ′ k ′ « 2k+2k ′ ff 3 5 1 2k+2k ′ ≤ ǫ(n) 2 4 E " P s X d 2 m,s Ψ(n) d 2 « 2k+2k ′ ff ....E " P s X d k m,s Ψ(n) d k « 2k+2k ′ ff E " P s Y d ′ 1 m,s Ψ(n) d ′ 1 « 2k+2k ′ ff ....E " P s Y d ′ k ′ m,s Ψ(n) d ′ k ′ « 2k+2k ′ ff 3 5 e
where ǫ(n) → 0 and e = 1 2k+2k ′ . Moreover, by our assumption and by hypothesis 3.1.2, all the other terms are bounded. Then, the lemma is proved when r=r'=1. For r ≥ 1 or r ′ ≥ 1, it is enough to apply lemma 3.2.3 or 4.2.1 for example. Lemma 4.2.5 We suppose that all the moments M nm q,p are bounded. We suppose R m,t

1 (x) = Q m,t 2 (x) = x 2 -β m,t and R m,t s (resp S m,t s ) is a polynomial of degree j s ≥ 0 (resp, i s ≥ 0). If j s = 0 (resp i s = 0), one assume that (1/Ψ(n m ) 2 ) nm t=1 R m,t s (X m,t ) (resp (1/Ψ(n m ) 2 ) nm t=1 S m,t s (X m,t ) ) is bounded. If j s ≥ 1 (resp i s ≥ 1) , one assume that R m,t s (x) = x js (resp S m,t s (x) = x is ). Moreover let h = k s=1 j ′ s where j ′ s = j s if j s ≥ 1 and j ′ s = 2 if j s = 0 and h ′ = k ′ s=1 i ′ s where i ′ s = i s if i s ≥ 1 and i ′ s = 2 if i s = 0. We assume 1 Ψ(nm) 4 E nm t=1 Q m,t 2 (X m,t ) 2 → 0.
Then, for all r, r' ,

1 ≤ r ≤ k, 1 ≤ r ′ ≤ k ′ , E t1 =t2 =.... =tr,tr+1,...,t k o1 =o2 =.... =o r ′ ,o r ′ +1 ,...,o k ′ k s=1 R m,ts s (X m,ts ) k ′ s ′ =1 S m,o s ′ s ′ (Y m,o s ′ ) Ψ(n m ) h+h ′ converges to 0.
Proof By the Schwartz inequality and the Holder Inequality,

1 Ψ(n m ) 2h+2h ′ .E t1 t2,...,t k o1,o2...,o k ′ k s=1 R m,ts s (X m,ts ) k ′ s ′ =1 S m,o s ′ s ′ (Y m,o s ′ ) 2 ≤ E 1 Ψ(n m ) h+h ′ k s=1 nm ts=1 R m,ts s (X m,ts ) k ′ s=1 nm os=1 S m,os s (Y m,os ) 2 ≤ E 1 Ψ(n m ) 4 nm t=1 R m,t 1 (X m,t ) 2 E 1 Ψ(n m ) 2h+2h ′ -4 k s=2 nm ts=1 R m,ts s (X m,ts ) 2 k ′ s=1 nm os=1 S m,os s (Y m,os ) 2 ≤ E 1 Ψ(n m ) 4 nm t=1 R m,t 1 (X m,t ) 2 k s=2 E nm ts=1 R m,ts s (X m,ts ) Ψ(n m ) js 2(k+k ′ -1) 1 k+k ′ -1 . k ′ s ′ =1 E nm o s ′ =1 S m,o s ′ s ′ (Y m,o s ′ ) Ψ(n m ) i s ′ 2(k+k ′ -1) 1 k+k ′ -1
.

By our assumption and by hypothesis 3.1.2 , the term on right-hand side is bounded. Moreover,

1 Ψ(n m ) 4 E nm t=1 R m,t 1 (X m,t ) 2 → 0 .
We deduce the lemma if r=1 and r'=1. When r > 1 or r ′ > 1, we prove the result by recurrence by using the lemma 3.2.2, 3.2.4 , 4.2.1 and 3.4.3.

Lemma 4.2.6 Let 0 ≤ q" ≤ q ′ , 0 ≤ p" ≤ p ′ . Let h = q" s=1 (2u s + v s ) + (q ′ -q") and let h ′ = p" s=1 (2c s + d s ) + (p ′ -p").
One assumes that M nm q,p is bounded pour tout q,p. Let

S nm H = t1,....t p ′ ∈H β c1 m,t1 Y d1 m,t1 β c2 m,t2 Y d2 m,t2 ....β c p" m,t p" Y d p" m,t p" Y m,t p"+1 ......X m,t p ′ . Then, s1,....s q ′ E β u1 m,s1 X v1 m,s1 β u2 m,s2 X v2 m,s2 ....β u q" m,s q" X v q"
m,s q" X m, q"+1 ......X m,s q ′ S nm

H Ψ(n m ) h+h ′ → 0 if u 1 ≥ 1 and v 1 ≥ 1.
Proof By lemma 3.4.4 E{(S nm H /Ψ(n m ) h ′ ) P } is bounded for all P. Then, we have

s1,....s q ′ E β u1 m,s1 X v1 m,s1 β u2 m,s2 X v2 m,s2 ....β u q" m,s q" X v q" m,s q" X m,s q"+1 ......X m,s q ′ S nm H Ψ(n m ) h+h ′ = ˛E" P s 1 β u 1 s 1 X v 1 m,s 1 Ψ(nm) 2u 1 +v 1 « ........ " P s q" β u q"
m,s q" X v q" m,s q" Ψ(nm) 2u q" +v q" «" P s q"+1 Xm,s q"+1 Ψ(nm) « .......

" P s q ′ Xm,s q ′ Ψ(nm) «" S nm H Ψ(nm) h ′ «ff= E s1 β u1 m,s1 X v1 m,s1
Ψ(n m ) 2u1+v1 ........

s q" β u q" m,s q" X v q" m,s q" Ψ(n m ) 2u q" +v q" s q"+1 X m,s q"+1 Ψ(n m ) q ′ -q" S nm H Ψ(n m ) h ′ ≤ q" r=1 E sr β ur m,sr X vr m,sr Ψ(n m ) 2ur+vr q"+2 1 q"+2 E s q"+1 X m,s q"+1 Ψ(n m ) (q ′ -q")(q"+2) 1 q"+2 E S nm H Ψ(n m ) h ′ q"+2 1 q"+2 ≤ |M nm 2(q ′ -q")(q"+2) | 1 2(q"+2) q" r=1 E sr β ur m,sr X vr m,sr Ψ(n m ) 2ur+vr q"+2 1 q"+2 E S nm H Ψ(n m ) h ′ 2q"+4 1 2q"+4 .
It is enough to apply lemma 3.4.3 in order to conclude.

Lemma 4.2.7 Let β ′ m,s = E{Y 2 m,s }. We suppose that all the moments M nm q,p are bounded. We assume that

1 Ψ(nm) 4 E nm t=1 Q m,t 2 (X m,t ) 2 → 0. Then, s1 =.... =sq-e v1 =.... =v q ′ -f E X 2 m,s1 .......X 2 m,se X m,se+1 ....X m,sq-e Y 2 m,v1 ........Y 2 m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ - s1 =.... =sq-e v1 =.... =v q ′ -f E β m,s1 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 .....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ converges to 0 Proof For example, s1 =.... =sq-2 v1 =v2 =.... =v q ′ -2 E X 2 m,s1 X 2 m,s2 X m,s3 ....X m,sq-2 Y 2 m,v1 Y 2 m,v2 Y m,v3 .....Y m,v q ′ -2 Ψ(n m ) q+q ′ = X s 1 =.... =s q-2 X v 1 =v 2 =.... =v q ′ -2 E  (X 2 m,s 1 -βm,s 1 )X 2 m,s 2 Xm,s 3 ....Xm,s q-2 Y 2 m,v 1 Y 2 m,v 2 Ym,v 3 .....Ym,v q ′ -2 Ψ(nm) q+q ′ ff + X s 1 =.... =s q-2 X v 1 =v 2 =.... =v q ′ -2 E  βm,s 1 X 2 m,s 2 Xm,s 3 ....Xm,s q-2 Y 2 m,v 1 Y 2 m,v 2 Ym,v 3 .....Ym,v q ′ -2 Ψ(nm) q+q ′ ff = X s 1 =.... =s q-2 X v 1 =v 2 =.... =v q ′ -2 E  (X 2 m,s 1 -βm,s 1 )X 2 m,s 2 Xm,s 3 ....Xm,s q-2 Y 2 m,v 1 Y 2 m,v 2 Ym,v 3 .....Ym,v q ′ -2 Ψ(nm) q+q ′ ff + X s 1 =.... =s q-2 X v 1 =v 2 =.... =v q ′ -2 E  βm,s 1 (X 2 m,s 2 -βm,s 2 )Xm,s 3 ....Xm,s q-2 Y 2 m,v 1 Y 2 m,v 2 Ym,v 3 .....Ym,v q ′ -2 Ψ(nm) q+q ′ ff + X s 1 =.... =s q-2 X v 1 =v 2 =.... =v q ′ -2 E  βm,s 1 βm,s 2 Xm,s 3 ....Xm,s q-2 (Y 2 m,v 1 -β ′ m,v 1 )Y 2 m,v 2 Ym,v 3 .....Ym,v q ′ -2 Ψ(nm) q+q ′ ff + X s 1 =.... =s q-2 X v 1 =v 2 =.... =v q ′ -2 E  βm,s 1 βm, 2 Xm,s 3 ....Xm,s q-2 β ′ m,v 1 (Y 2 m,v 2 -β ′ m,v 2 )Ym,v 3 .....Ym,v q ′ -2 Ψ(nm) q+q ′ ff + X s 1 =.... =s q-2 X v 1 =v 2 =.... =v q ′ -2 E  βm,s 1 βm,s 2 Xm,s 3 ....Xm,s q-2 β ′ v 1 β ′ m,v 2 Ym,v 3 .....Ym,v q ′ -2 Ψ(nm) q+q ′ ff .
Then for example by lemma 4.2.5

X s 1 =.... =s q-2 X v 1 =v 2 =.... =v q ′ -2 E  βm,s 1 βm, 2 Xm,s 3 ....Xm,s q-2 β ′ m,v 1 (Y 2 m,v 2 -β ′ m,v 2 )Y 2 m,v 2 Ym,v 3 .....Ym,v q ′ -2 Ψ(nm) q+q ′ ff converges to 0. Lemma 4.2.8
We suppose that all the moments M nm h,k are bounded. One assumes that, for all, p, q, s1 =..... =sq t1 =..... =tp

E{X m,s1 X m,s2 ......X m,sq Y m,t1 Y m,t2 ......Y m,tp } Ψ(n m ) p+q → S q,p .
We set S 0,0 = 1. Then,

s1 =.... =sq-e v1 =.... =v p-f E β m,s1 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 .....β ′ m,v f Y m,v f +1 .....Y m,v p-f Ψ(n m ) q+p converges to σ 2e 0 σ 2e 1 S q-2e,p-2f
Proof At first, we study

s1 =.... =se,se+1 =.... =sq-e v1 =.... =v f ,v f +1 =.... =v q ′ -f E β m,s1 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ = s1 =.... =se se+1 =.... =sq-e v1 =.... =v f v f +1 =.... =v q ′ -f β m,s1 ....β m,se Ψ(n m ) 2e β ′ m,v1 ....β ′ m,v f Ψ(n m ) 2f E X m,se+1 ....X m,sq-e Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q-2e Ψ(n m ) q ′ -2f .
It is easy to understand that s1 =.... =se

β m,s1 β m,s2 ....β m,se Ψ(n m ) 2e → σ 2e 0 . v1 =.... =v f β ′ m,v1 ....β ′ m,v f Ψ(n m ) 2f → σ 2f 1 .
Indeed, s1,...,se

β m,s1 β m,s2 ....β m,se Ψ(n m ) 2e = m,s1 β m,s1 Ψ(n m ) 2 ....... se β m,se Ψ(n m ) 2 → σ 2e 0 ,
and, for example, by lemma 3.4.2,

s1 β 2 m,s1 Ψ(n m ) 4 → 0 .
Therefore, s1 =.... =se,se+1 =.... =sq-e v1 =...

. =v f ,v f +1 =.... =v q ′ -f E β m,s1 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ converges to σ 2e 0 σ 2f 1 S q-2e,q ′ -2f .
Now, in order to obtain s1 =..... =sq-e v1 =.... =v q ′ -f E βm,s 1 .....βm,s e Xm,s e+1 ....Xm,s q-e β ′ m,v 1

....β ′ m,v f Ym,v f +1 .....Ym,v q ′ -f Ψ(nm) q+q ′
, we shall have to add or to subtract sums such that there exists i, j satisfying s i = s e+j or i',j' such that v i ′ = v f +j ′ . For example, , we shall have to subtract the sum s2 =.... =se,se+1 =.... =sq-e v1 =...

. =v f ,v f +1 =.... =v q ′ -f E β m,s2 ....β m,se β m,se+1 X m,se+1 X m,se+2 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ .
One will prove that this sum converges to 0 by recurence. With this aim, we study first

X s 2 ........,s q-e X v 1 ,.......,v q ′ -f E ( βm,s 2 ....βm,s e βm,s e+1 Xm,s e+1 Xm,s e+2 ....Xm,s q-e β ′ m,v 1 ....β ′ m,v f Ym,v f +1 .....Ym,v q ′ -f Ψ(nm) q+q ′ )
which can be written as

e i=2 si β m,si Ψ(n m ) 2 f j=1 sj β ′ m,sj Ψ(n m ) 2 E se+1 β m,se+1 X m,se+1 Ψ(n m ) 3 se+2,....,sq-e X m,se+2 ....X m,sq-e Ψ(n m ) q-e-1 v1,...,v q ′ -f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q ′ -f . Now, P s 2 βm,s 2 Ψ(nm) 2 is bounded and E se+2,....,sq-e X m,se+2 ....X m,sq-e Ψ(n m ) q-2e-1 2 = E s X m,s Ψ(n m ) q-2e-1 2 ≤ E s X m,s Ψ(n m ) 2(q-2e-1)
is bounded by our assumption. At last, E

P s 1 βm,s e+1 Xm,s e+1 Ψ(nm) 3
→ 0 by lemma 3.4.3. It is enough to apply Holder Inequality to conclude.

It is general for all the steps of recurrence : thanks to lemma 3.2.2 and 3.2.4, one can always write the terms s1 =.... =sr,sr+1 ...,sq-e v1 =...

. =v r ′ ,v r ′ +1 ,....,v q ′ -f E β m,s2 ....β m,se β m,se+1 X m,se+1 X m,se+2 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′
as a sum of products. Or by using proof of lemma 4.2.6, we always find in these products, terms which are bounded by products of bounded expectations (cf lemma 3.3.1 and hypothesis 3.1.2) for example, in the form

E P s β u m,s X v m,s Ψ(nm) 2u+v 2w
, u, v, w ∈ N : among these terms there is at least one term of the form s1

β u m,s 1 X v m,s 1
Ψ(nm) 2u+v which converges to 0. That proves the recurence, and then, that s2 =.... =se,se+1 =.... =sq-e v1 =...

. =v f ,v f +1 =.... =v q ′ -f E β m,s2 .....β m,se β m,se+1 X m,se+1 X m,se+2 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ converges to 0.
One can reason by the same way about other terms, for example terms such

X s 3 =.... =se,s e+1 =.... =s q-e X v 1 =.... =v f ,v f +1 =.... =v q ′ -f E  βm,s 3 .....βm,s e βm,s e+1 Xm,s e+1 βm,s e+2 Xm,s e+2 Xm,s e+3 ....Xm,s q-e β ′ m,v 1 ....β ′ m,v f Ym,v f +1 .....Ym,v q ′ -f Ψ(nm) q+q ′ ff or X s 2 .... =se,s e+1 =.... =s q-e X v 2 =.... =v f ,v f +1 =.... =v q ′ -f E  βm,s 2 .....βm,s e βm,s e+1 Xm,s e+1 Xm,s e+2 ....Xm,s q-e β ′ m, 2 ....β ′ m,v f β ′ m,v f +1 Ym,v f +1 Ym,v f +2 .....Ym,v q ′ -f Ψ(nm) q+q ′ ff .
We deduce that s1 =.... =sq-e v1 =... =v q ′ -f

E β m,s1 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ converges to the same limit as s2 =.... =se,se+1 =.... =sq-e v1 =.... =v f ,v f +1 =.... =v q ′ -f E β m,s2 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ ,
i.e. converges to σ 2e 0 σ 2f 1 S q-2e,q ′ -2f .

Lemma 4.2.9 We suppose that all the moments M nm h,k converge. Then, for all e, for all f, for all p, for all q, s1 =.... =sq-e v1 =.... =v p-f

E β m,s1 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 .....β ′ m,v f Y m,v f +1 .....Y m,v p-f Ψ(n m ) q+p converges.
Proof We study at first s1,.....,sq-e v1,....,v q ′ -f

E β m,s1 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ = s β m,s Ψ(n m ) 2 e s β ′ m,s Ψ(n m ) 2 f E s X m,s Ψ(n m ) q-2e s Y m,s Ψ(n m ) p-2f which converges to σ 2e 0 σ 2f 1 M q-2e,q ′ -2f .
Now, in order to obtain

s1 =.... =sq-e v1 =.... =v p-f E β m,s1 .....β m,se X m,se+1 ....X m,sq-e β ′ m,v1 .....β ′ m,v f Y m,v f +1 .....Y m,v p-f Ψ(n m ) q+p ,
we shall have to add or to subtract sums such that there exists i, j satisfying

s i = s e+j or i',j' such that v i ′ = v j ′ .
For example, we shall have to subtract terms of the form

X s 1 ,....,se,s e+2 ,...,s q-e X v 1 ,...,v q ′ -f E  βm,s 1 .....βm,s e X 2 m,s e+2 Xm,s e+3 ....Xm,s q-e β ′ m,v 1 ....β ′ m,v f Ym,v f +1 .....Ym,v q ′ -f Ψ(nm) q+q ′
ff which, by using the same technique as in lemma 4.2.7, converges to the same limit as X s 1 ,....,se,s e+2 ,...,s q-e X v 1 ,...,v q ′ -f E  βm,s 1 .....βm,s e βm,s e+2 Xm,s e+3 ....Xm,s q-e β ′ m,v 1 ....

β ′ m,v f Ym,v f +1 .....Ym,v q ′ -f Ψ(nm) q+q ′ ff .
Then, we prove the result by using a recurence on e. Now, we shall have also to substract the sum

X s 2 ,....,s q-e X v 1 ,...,v q ′ -f E  βm,s 2 .....βm, e [βm,s e+1 Xm,s e+1 ]Xm,s e+2 ....Xm,s q-e β ′ m, 1 ....β ′ m,v f Ym,v f +1 .....Ym,v q ′ -f Ψ(nm) q+q ′ ff which is written as e Y i=2 " P s i βm,s i Ψ(nm) 2 « f Y j=1 " P s j β ′ m,s j Ψ(nm) 2 « E " P s 1 βm,s e+1 Xm,s e+1 Ψ(nm) 3 «" X s e+2 ,....,s q-e
Xm,s e+2 ....Xm,s q-e Ψ(nm) q-e-1

«" X

v 1 ,...,v q ′ -f Ym,v f +1 .....Ym,v q ′ -f Ψ(nm) q ′ -f «ff . Now, P s 2 βm,s 2 Ψ(nm) 2 is bounded and E se+2,....,sq-e X m,se+2 ....X m,sq-e Ψ(n m ) q-2e-1 2 = E s X m,s Ψ(n m ) q-2e-1 2 ≤ E s X m,s Ψ(n m ) 2(q-2e-1)
is bounded by our assumption. At last, E More generally, we shall have to substract tems of the form s1+c,....,se,s e+d+1 ,...,sq-e v1,...,v q ′ -f

E β s1+c ....β m,se [β c m,se+1 X d m,se+1 ]X m,s e+d+1 ...X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 ....Y m,v q ′ -f Ψ(n m ) q+q ′
which converges to 0 by the same reasoning, but by applying lemma 3.4.4 .

We shall have also to substract tems of the form s2,....,sq-e v1,...,v q ′ -f

E β 2 m,s2 β m,s3 .....β m,se X m,se+1 X m,se+2 ....X m,sq-e β ′ m,v1 ....β ′ m,v f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′
which converges to 0 because by lemma 3.4.2,

P s 2 β 2 m,s 2 Ψ(nm) 4 → 0 .
We shall deduce that s1 =..... =sq-e v1 =.... =v q ′ -f

E β m,s1 .....β m,se X m,se+1 ....X m,q-e β ′ m,1 ....β ′ m, f Y m,v f +1 .....Y m,v q ′ -f Ψ(n m ) q+q ′ converges.

Statement and proof of first theorem

Theorem 15 Let Q m,t 2 (x) = x 2 -β m,t , Q ′m,t 2 (y) = y 2 -β ′ m,t where β ′ m,t = E{Y 2 m,t }. One assumes that E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 , E nm t=1 Q ′m,t 2 (Y m,t ) Ψ(n m ) 2 2 → 0 .
All the moments M nm q,p = E (Xm,1+....+Xm,n m ) q (Ym,1+....+Ym,n m ) p Ψ(nm) p+q converges to a real M q,p if, for all q ∈ N, for all p ∈ N, there exists S q,p ∈ R and Sb 2 q such that

s1 =s2 =..... =sq t1 =t2 =..... =tp E{X m,s1 X m,s2 ......X m,sq Y m,t1 Y m,t2 ......Y m,tp } Ψ(n m ) p+q → S q,p s1 =s2 =..... =sq-1 E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q , t1 =t2 =..... =tp-1 E{(Y m,t1 ) 2 Y m,t2 ......Y m,tp-1 } Ψ(n m ) p ≤ Sb 2 p .
Proof of sufficient condition of theorem 15 By proposition 3.

u1 =u2 =..... =u k v1 =v2 =.... =v k ′ E X r1 m,u1 ....X r k m,u k Y r ′ 1 m,v1 ....Y r ′ k ′ m,v k ′ Ψ(n m ) h+h ′ → 0 . Therefore u1 =u2 =..... =u k v1 =v2 =.... =v k ′ E X r1 m,u1 ....X r k m,u k Y r ′ 1 m,v1 ....Y r ′ k ′ m,v k ′ Ψ(n m ) h+h ′ converges for all (u 1 , ..., u k , v 1 , ..., v k ′ ). Then, because E (X m,1 + .... + X m,nm ) q (Y m,1 + .... + Y m,
u1 =u2 =..... =u k v1 =v2 =.... =v k ′ E X r1 m,u1 ....X r k m,u k Y r ′ 1 m,v1 ....Y r ′ k ′ m,v k ′ Ψ(n m ) p+q → 0 .
Therefore,

u1 =u2 =..... =u k v1 =v2 =.... =v k ′ E X r1 m,u1 ....X r k m,u k Y r ′ 1 m,v1 ....Y r ′ k ′ m,v k ′ Ψ(

Convergence to the normal distribution

In case of convergence to the normal distribution, one specifies theorem 15 by the following way.

Proposition 4.3.1 One assume σ 2 0 = σ 2 1 . One assumes that

E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 , E nm t=1 Q ′m,t 2 (Y m,t ) Ψ(n m ) 2 2 → 0 .
All the moments M n q,p converges to a real M q,p = µ q µ p , where µ q is the moment of order q of N (0, 1) , if and only if , for all q ∈ N, for all p ∈ N , s1 =s2 =..... =sq t1 =t2 =..... =tp

E{X m,s1 X m,s2 ......X m,sq Y m,t1 Y m,t2 ......Y m,tp } Ψ(n m ) p+q → ν q ν p ,
where ν q is the moment of order q of N (0, ν 2 ) with 1 = σ 2 0 + ν 2 , and, if there exists Sb 2 q ∈ R s1 =s2 =..... =sq-1

E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q , t1 =t2 =..... =tq-1 E{(Y m,t1 ) 2 Y m,t2 ......Y m,tp-1 } Ψ(n m ) p ≤ Sb 2 p .

Proof of proposition 4.3.1

In this proof we simplify X m,t in X t and Y m,t in Y t and n m in n. Of course, we use theorem 15. At first, under necessary or sufficient assumption, we have the following lemma.

Lemma 4.3.1 Let q" = ⌊q/2⌋, p" = ⌊p/2⌋. Then,

E (X 1 + .... + X nm ) q (Y 1 + .... + Y nm ) p Ψ(n m ) q+p ∼ q!p! 2 0 2 0 σ 0 0 σ 0 1 0!0! S q,p q!p! + q!p! 2 1 2 0 σ 2 0 σ 0 1 1!0! S q-2,p (q -2)!p! + ............. + q!p! 2 q" 2 0 σ 2q" 0 σ 0 1 q"!0! S q-2q",p (q -2q")!p! + q!p! 2 0 2 σ 0 0 σ 2 1 0!1! S q,p-2 q!(p -2)! + q!p! 2 1 2 σ 2 0 σ 2 1 1!1! S q-2,p-2 (q -2)!(p -2)! + ................. + q!p! 2 q" 2 σ 2q" 0 σ 2 1 q"!1! S q-2q",p-2 (q -2q")(p -2)! + q!p! 2 0 2 2 σ 0 0 σ 4 1 0!2! S q,p-4 q!(p -4)! + q!p! 2 1 2 2 σ 2 0 σ 4 1 1!2!
S q-2,p-4 (q -2)!(p -4)! + ............ + q!p! 2 q" 2 2 σ 2q" 0 σ 4 1 q"!2! S q-2q",p-4 (q -2q")!(p -4)! . q!p! 2 q" 2 p" σ 2q" 0 σ 2p" 1 q"!p"! S q-2q",p-2p" (q -2q")!(p -2p")! .

+ q!p! 2 0 2 p" σ 0 0 σ 2p" 1 0!p"! S q,p-2p" q!(p -2p")! + q!p! 2 1 2 p" σ 2 0 σ 2p" 1 1!p"! S q-2,p-2p" (q -2)
Proof In order to prove necessary or sufficient condition, by theorem 15

q! j 1 !.......j nm ! p! i 1 !.......i nm ! E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p → 0 .
Then, it is enough to study the sums where j s ≤ 2 and i t ≤ 2.

Then,

E (X 1 + .... + X n ) q (Y 1 + .... + Y n ) p Ψ(n) q+p ∼ j1+.....
q! j 1 !.......j n ! p! i 1 !.......i n ! E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p = q!p! 2

+

q!p! 2 q" !2 p" j1+.....+jn=q, q" js=2 i1+.....+in=p, p" it=2

E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p ∼ q!p! 2 0 2 0 σ 0 0 σ 0 1 0!0! S q,p q!p! + q!p! 2 1 2 0 σ 2 0 σ 0 1 1!0!
S q-2,p (q -2)!p! + ............. + q!p! 2 q" 2 0 σ 2q" 0 σ 0 1 q"!0! S q-2q",p (q -2q")!p!

+ q!p! 2 0 2 σ 0 0 σ 2 1 0!1! S q,p-2 q!(p -2)! + q!p! 2 1 2 σ 2 0 σ 2 1 1!1! S q-2,p-2 (q -2)!(p -2)! + ................. + q!p! 2 q" 2 σ 2q" 0 σ 2 1 q"!1! S q-2q",p-2 (q -2q")(p -2)! + q!p! 2 0 2 σ 0 0 σ 4 1 0!2! S q,p-4 q!(p -4)! + q!p! 2 1 2 2 σ 2 0 σ 4 1 1!2!
S q-2,p-4 (q -2)!(p -4)! + ................ + q!p! 2 q" 2 2 σ 2q" 0 σ 4 1 q"!2! S q-2q",p-4 (q -2q")!(p -4)! q!p! 2 q" 2 p" σ 2q" 0 σ 2p" 1 q"!p"! S q-2q",p-2p" (q -2q")!(p -2p")! , by lemma 4.2.3, 4.2.8, 4.2.9 and 4.2.7.

+ q!p! 2 0 2 p" σ 0 0 σ 2p" 1 0!p"! S q,p-2p" q!(p -2p")! + q!p! 2 1 2 p" σ 2 0 σ 2p" 1 1!p"! S q-2,p-2p" (q -2)!(p - 2p
Proof of sufficient condition If q or p is odd, S p-2e,q-2f = ν q-2e ν ′ q-2f = 0. Therefore, by lemma 4.3.1, if p or q is odd, M q,p = 0. Now, suppose that q and p are even : in order to simplify, we study the moment of order (2q,2p). By lemma 4.3.1, We proceed by the same way for S Q,p-2p"+4 : S Q,p-2p"+4 = ν Q ν p-2p"+4 , and so on by recurrence untill S Q,p = ν Q ν p = S q+2,q = ν q+2 ν q . Therefore, S q+2,p ′ = ν q+2 ν p ′ for all p ′ ≤ p = q.

E (X 1 + .... + X n ) 2q (Y 1 + .... + Y n ) 2p Ψ(n) 2q+2p ∼ 2q!2p! 2 0 2 0 σ 0 0 σ 0 1 0!0! ν 2q (2q)! ν 2p (2p)! + 2q!2p! 2 1 2 0 σ 2 0 σ 0 1 1!0! ν 2q-2 (2q -2)! ν 2p (2p)! + 2q!2p! 2 2 2 0 σ 4 0 σ 0 1 2!0! ν 2q-4 (2q -4)! ν 2p (
2q!2p! 2 q 2 0 σ 2q 0 σ 0 1 q!0! ν 2q-2q (2q -2q)! ν 2p (2p)! + 2q!2p! 2 0 2 σ 0 0 σ 2 1 0!1! ν 2q (2q)! ν 2p-2 (2p -2)! + 2q!2p! 2 1 2 σ 2 0 σ 2 1 1!1! ν 2q-2 (2q -2)! ν 2p-2 (2p -2)! + 2q!2p! 2 2 2 σ 4 0 σ 2 1 2!1! ν 2q-4 (2q -4)!
+ 2q!2p! 2 q 2 σ 2q 0 σ 2 1 q!1! ν 2q-2q (2q -2q)! ν 2p-2 (2p -2)! + 2q!2p! 2 0 2 2 σ 0 0 σ 4 1 0!2! ν 2q (2q)! ν 2p-4 (2p -4)! + 2q!2p! 2 1 2 2 σ 2 0 σ 4 1 1!2! ν 2q-2 (2q -2)! ν 2p-4 (2p -4)! + 2q!2p! 2 2 2 2 σ 4 0 σ 4 1 2!2! ν 2q-4 (2q -4)!
+ 2q!2p! 2 q 2 2 σ 2q 0 σ 4 1 q!2! ν 2q-2q (2q -2q)!
+ 2q!2p! 2 0 2 p σ 0 0 σ 2p 1 0!p! ν 2q (2q)! ν 2p-2p (2p -2p)! + 2q!2p! 2 1 2 p σ 2 0 σ 2p 1 1!p! ν 2q-2 (2q -2)! ν 2p-2p (2p -2p)! + 2q!2p! 2 2 2 p σ 4 0 σ 2p 1 2!p! ν 2q-4 (2q -4)! ν 2p-2p (
+ 2q!2p! 2 q 2 p σ 2q 0 σ 2p 1 q!p! ν 2q-2q (2q -2q)! ν 2p-2p (2p -2p)! ∼ 2q!2p! 2 0 2 0 σ 0 0 σ 0 1 0!0! ν q 2 2 q q! ν p 2 2 p p! + 2q!2p! 2 1 2 0 σ 2 0 σ 0 1 1!0! ν q-1 2 2 q-1 (q -1)! ν p 2 2 p p! + 2q!2p! 2 2 2 0 σ 4 0 σ 0 1 2!0! ν q-2 2 2 q-2 (q -2)! ν p 2
+ 2q!2p! 2 q 2 0 σ 2q 0 σ 0 1 q!0! ν q-q 2 2 q-q (q -q)! ν p 2 2 p p! + 2q!2p! 2 0 2 σ 0 0 σ 2 1 0!1! ν q 2 2 q (q)! ν p-1 2 2 p-1 (p -1)! + 2q!2p! 2 1 2 σ 2 0 σ 2 1 1!1! ν q-1 2 2 q-1 (q -1)! ν p-1 2 2 p-1 (p -1)! + 2q!2p! 2 2 2 σ 4 0 σ 2 1 2!1! ν q-2 2 2 q-2 (q -2)! ν p-1 2 2 p-1 (p -
+ 2q!2p! 2 q 2 σ 2q 0 σ 2 1 q!1! ν q-q 2 2 q-q (q -q)! ν p-1 2 2 p-1 (p -1)! + 2q!2p! 2 0 2 2 σ 0 0 σ 4 1 0!2! ν q 2 2 q (q)! ν p-2 2 2 p-2 (p -2)! + 2q!2p! 2 1 2 2 σ 2 0 σ 4 1 1!2! ν q-1 2 2 q-1 (q -1)! ν p-2 2 2 p-2 (p -2)! + 2q!2p! 2 2 2 2 σ 4 0 σ 4 1 2!2! ν q-2 2 2 q-2 (q -2)! ν p-2 2 2 p-2 (p -
+ 2q!2p! 2 q 2 2 σ 2q 0 σ 4 1 q!2! ν q-q 2 2 q-q (q -q)! ν p-2 2 2 p-2 (p -
+ 2q!2p! 2 0 2 p σ 0 0 σ 2p 1 0!p! ν q 2 2 q (q)! ν p-p 2 2 p-p (p -p)! + 2q!2p! 2 1 2 p σ 2 0 σ 2p 1 1!p! ν q-1 2 2 q-1 (q -1)! ν p-p 2 2 p-p (p -p)! + 2q!2p! 2 2 2 p σ 4 0 σ 2p 1 2!p! ν q-2 2 2 q-2 (q -2)! ν p-p
+ 2q!2p! 2 q 2 p σ 2q 0 σ 2p 1 q!p! ν q-q 2 2 q-q (q -q)! ν p-p 2 2 p-p (p -p)! ∼ 2q!2p! 2 p 2 q σ 0 0 σ 0 1 0!0! ν q 2 q! ν p 2 p! + σ 2 0 σ 0 1 1!0! ν q-1 2 (q -1)! ν p 2 p! + σ 4 0 σ 0 1 2!0! ν q-2 2 (q -2)! ν p 2 p! + ...... + σ 2q 0 σ 0 1 q!0! ν q-q 2 (q -q)! ν p 2 p! + σ 0 0 σ 2 1 0!1! ν q 2 (q)! ν p-1 2 (p -1)! + σ 2 0 σ 2 1 1!1! ν q-1 2 (q -1)! ν p-1 2 (p -1)! + σ 4 0 σ 2 1 2!1! ν q-2 2 (q -2)! ν p-1 2 (p -
σ 2 1 q!1! ν q-q 2 (q -q)! ν p-1 2 (p -1)! + σ 0 0 σ 4 1 0!2! ν q 2 (q)! ν p-2 2 (p -2)! + σ 2 0 σ 4 1 1!2! ν q-1 2 (q -1)! ν p-2 2 (p -2)! + σ 4 0 σ 4 1 2!2! ν q-2 2 (q -2)! ν p-2 2 (p -
+ σ 0 0 σ 2p 1 0!p! ν q 2 (q)! ν p-p 2 (p -p)! + σ 2 0 σ 2p 1 1!p! ν q-1 2 (q -1)! ν p-p 2 (p -p)! + σ 4 0 σ 2p 1 2!p! ν q-2 2 (q -2)! ν p-
q!p! ν q-q 2 (q -q)! ν p-p 2 (p -p)! ∼ 2q!2p! 2 p 2 q ν p 2 p! σ 0 1 0! σ 0 0 0! ν q 2 q! + σ 2 0 1! ν q-1 2 (q -1)! + σ 4 0 2! ν q-2 2 (q -2)! + ...... + σ 2q 0 q! ν q-q 2 (q -q)! + ν p-1 2 (p -1)! σ 2 1 1! σ 0 0 0! ν q 2 q! + σ 2 0 1! ν q-1 2 (q -1)! + σ 4 0 2! ν q-2 2 (q -2)! + ...... + σ 2q 0 q! ν q-q 2 (q -q)! + ν p-2 2 (p -2)! σ 4 1 2! σ 0 0 0! ν q 2 q! + σ 2 0 1! ν q-1 2 (q -1)! + σ 4 0 2! ν q-2 2 (q -2)! + ...... + σ 2q 0 q! ν q-q 2 (q -q)! ....
We proceed by the same way for S q,p+2 by reversing p and q and we prove that S q ′ ,p+2 = ν q ′ ν p+2 for all q ′ ≤ q.

It remains to apply a last time lemma 4.3.1, and 4.3.2 in order to prove that S q+2,q+2 = ν q+2 ν p+2 . Indeed, let P=p+2. Then,

E (X 1 + .... + X n ) Q (Y 1 + .... + Y n ) P Ψ(n m ) Q+P ∼ Q!P ! 2 0 2 0 σ 0 0 σ 0 1 0!0! S Q,P Q!P ! + Q!P ! 2 1 2 0 σ 2 0 σ 0 1 1!0! ν Q-2 ν P (Q -2)!P ! + ............. + Q!P ! 2 Q" 2 0 σ 2Q" 0 σ 0 1 Q"!0! ν Q-2Q" ν P (Q -2Q")!P ! + Q!P ! 2 0 2 σ 0 0 σ 2 1 0!1! ν Q ν P -2 Q!(P -2)! + Q!P ! 2 1 2 σ 2 0 σ 2 1 1!1! ν Q-2 ν P -2 (Q -2)
!P ! 2 Q" 2 σ 2Q" 0 σ 2 1 Q"!1! ν Q-2Q" ν P -2 (Q -2Q")(P -2)! + Q!P ! 2 0 2 2 σ 0 0 σ 4 1 0!2! ν Q-4 ν P -4 Q!(P -4)! + Q!P ! 2 1 2 2 σ 2 0 σ 4 1 1!2! ν Q-2 ν P -4 (Q -2)
!P ! 2 Q" 2 2 σ 2Q" 0 σ 4 1 Q"!2! S Q-2Q",P -4 (Q -2Q")!(P -
+ Q!P ! 2 0 2 P " σ 0 0 σ 2P " 1 0!P "! ν Q ν P -2P " Q!(P -2P ")! + Q!P ! 2 1 2 P " σ 2 0 σ 2P " 1 1!P "! ν Q-2 ν P -2P " (Q -2)
!P ! 2 Q" 2 P " σ 2Q" 0 σ 2P " 1 Q"!P "! ν Q-2Q" ν P -2P " (Q -2Q")!(P -2P ")! .
This result proves the recurence, and then, the necessary condition.

Second theorem of convergence

Now we have a MCLT with conditions about the ρ j1,....,jn 's.

Theorem 16 Let P ′m,s

2 (y) = y 2 -γ ′ m,s y -β m,s where γ ′ m,s = E{Y 3 m,s }/E{Y 2 m,s }. We assume that nm s=1 γ 2 m,s Ψ(n) 2 and nm s=1 γ ′2 m,s Ψ(n) 2 are bounded. We assume that E nm s=1 γm,sXm,s Ψ(n) 2 2 → 0 and E nm s=1 γ ′ m,s Ym,s Ψ(n) 2 2 → 0. One assumes that E nm t=1 P m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 , E nm t=1 P ′m,t 2 (Y m,t ) Ψ(n m ) 2 2 → 0 .
All the moments M nm q,p = E (Xm,1+....+Xm,n m ) q (Ym,1+....+Ym,n m ) p Ψ(nm) p+q converges to a real M q,p if, for all q ∈ N, for all p ∈ N, there exists S q,p ∈ R and Sb r q ∈ R, r=2,3,4,5, such that

s1 =s2 =..... =sq t1 =t2 =..... =tp E{X m,s1 X m,s2 ......X m,sq Y m,t1 Y m,t2 ......Y m,tp } Ψ(n m ) p+q → S q,p , s1 =s2 =..... =sq-1 E{ P m,s1 2 (X m,s1 )X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q , s1 =s2 =..... =sq-1 E{ P ′m,s1 2
(Y m,s1 )Y m,s2 ......Y m,sq-1 } Ψ(n m ) q ≤ Sb 4 q , s1 =s2 =...... =sq γ m,s1 E{X s1 X s2 ......X sq } Ψ(n m ) q+1 ≤ Sb 3 q . s1 =s2 =...... =sq γ ′ m,s1 E{Y s1 Y s2 ......Y sq } Ψ(n m ) q+1 ≤ Sb 5 q .
Proof of sufficient Condition of theorem [START_REF] Yokohama R | The convergence of moments in the Central limit Theorem for stationary f-mixing processes[END_REF] The conditions of theorem are checked if p=0 or q=0. Therefore, all the conditions of theorem 13 are checked for the sequences X m,t and Y m,t . Therefore

s1 =s2 =..... =sq-1 E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q , t1 =t2 =..... =tp-1 E{(Y m,t1 ) 2 Y m,t2 ......Y m,tp-1 } Ψ(n m ) p ≤ Sb 2 p .
Then, it is enough to apply theorem 15 in order to prove the sufficient condition.

Proof of necessary condition of theorem 16

We suppose that all the moments converge. Therefore, the first condition is checked by theorem 15. Then, the moments of marginal distributions converge. Then, for the four other conditions, it is enough to use theorem 13 for sequences X m,t and Y m,t .

Corollary 4.4.1 One assumes that σ 2 0 = σ 2 1 . We assume that

nm s=1 γ 2 m,s Ψ(n) 2 and nm s=1 γ ′2 m,s Ψ(n) 2 are bounded. We assume that E nm s=1 γm,sXm,s Ψ(n) 2 2 → 0 and E nm s=1 γ ′ m,s Ym,s Ψ(n) 2 2 → 0. One assumes that E nm t=1 P m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 , E nm t=1 P ′m,t 2 (Y m,t ) Ψ(n m ) 2 2 → 0 .
All the moments M nm q,p = E (Xm,1+....+Xm,n m ) q (Ym,1+....+Ym,n m ) p Ψ(nm) p+q converges to µ q µ p if and only if , for all q ∈ N, for all p ∈ N, there exists Sb r q ∈ R, r=2,3,4,5, such that

s1 =s2 =..... =sq t1 =t2 =..... =tp E{X s1 X s2 ......X sq Y t1 Y t2 ......Y tp } Ψ(n m ) p+q → ν q ν p , s1 =s2 =..... =sq-1 E{ P m,s1 2
(X m,s1 )X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q , s1 =s2 =..... =sq-1 E{ P ′m,s1 2 (Y m,s1 )Y m,s2 ......Y m,sq-1 } Ψ(n m ) q ≤ Sb 4 q , s1 =s2 =...... =sq γ m,s1 E{X s1 X s2 ......X sq } Ψ(n m ) q+1 ≤ Sb 3 q . s1 =s2 =...... =sq γ ′ m,s1 E{Y s1 Y s2 ......Y sq } Ψ(n m ) q+1 ≤ Sb 5 q .
Proof of sufficient Condition of Corollary 4.4.1 If the conditions of corollary are checked, all the moments converge by theorem 16 . Therefore, all the conditions of theorem 15 are checked for the sequences X m,t and Y m,t . Therefore

s1 =s2 =..... =sq-1 E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q , t1 =t2 =..... =tp-1 E{(Y m,t1 ) 2 Y m,t2 ......Y m,tp-1 } Ψ(n m ) p ≤ Sb 2 p .
Then, it is enough to use proposition 4.3.1 in order to prove the sufficient condition.

Proof of necessary condition of Corollary 4.4.1 We suppose that all the moments converge to the moments of N (0, I 2 ). Then, by proposition 4.3.1, the first condition is checked. By theorem 16, the other conditions are also checked.

Third theorem of convergence

Now we suppose that X m,t and Y m,t are bounded. Remak that, in this case, hypothesis 4.1.1 holds.

Theorem 17 We suppose that there exists

F > 0 such that |X m,t | ≤ F and |Y m,t | ≤ F . One assumes that E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 , E nm t=1 Q ′m,t 2 (Y m,t ) Ψ(n m ) 2 2 → 0 .
All the moments M nm q,p = E (Xm,1+....+Xm,n m ) q (Ym,1+....+Ym,n m ) p Ψ(nm) p+q converges to a real M q,p if, for all q ∈ N, for all p ∈ N, there exists S q,p ∈ R :

s1 =s2 =..... =sq t1 =t2 =..... =tp E{X m,s1 X m,s2 ......X m,sq Y m,t1 Y m,t2 ......Y m,tp } Ψ(n m ) p+q → S q,p .
Proof of sufficient Condition of theorem 17 The conditions of theorem are checked when p=0 or q=0. Therefore, all the conditions of proposition 3.6.1 are checked by sequences X m,t and Y m,t . Then, all the moments are bounded. Then, by proposition, 3.3.1

s1 =s2 =..... =sq-1 E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q , t1 =t2 =..... =tp-1 E{(Y m,t1 ) 2 Y m,t2 ......Y m,tp-1 } Ψ(n m ) p ≤ Sb 2 p .
Then, it is enough to apply theorem 15 in order to prove the sufficient condition.

Proof of necessary condition of theorem 17

We suppose that all the moments converge.

Then, the condition is proved by theorem 15.

Corollary 4.5.1 One assumes σ 2 0 = σ 2 1 . We suppose that there exists

F > 0 such that |X m,t | ≤ F and |Y m,t | ≤ F . One assumes that E nm t=1 Q m,t 2 (X m,t ) Ψ(n m ) 2 2 → 0 , E nm t=1 Q ′m,t 2 (Y m,t ) Ψ(n m ) 2 2 → 0 .
All the moments M nm q,p = E (Xm,1+....+Xm,n m ) q (Ym,1+....+Ym,n m ) p Ψ(nm) p+q converges to µ q µ p if and only if , for all q ∈ N, for all p ∈ N,

s1 =s2 =..... =sq t1 =t2 =..... =tp E{X s1 X s2 ......X sq Y t1 Y t2 ......Y tp } Ψ(n m ) p+q → ν q ν p .
Proof of sufficient Condition of Corollary 4.5.1 If the conditions of the corollary are checked, all the moments converge by theorem 17 . Therefore, all the conditions of theorem 15 are checked by the sequences X m,t and Y m,t . Therefore, s1 =s2 =..... =sq-1 E{(X m,s1 ) 2 X m,s2 ......X m,sq-1 } Ψ(n m ) q ≤ Sb 2 q , t1 =t2 =...

= E » ∞ X i 1 =0 C i 1 (Θ t 1 +i 1 )f i 1 +1 (Ψ t 1 +i 1 ) -» ∞ X i 2 =0 C i 2 (Θ t 2 +i 2 )f i 2 +1 (Ψ t 2 +i 2 ) - ...... » ∞ X ip=0 C ip (Θ tp+ip )f ip+1 (Ψ tp+ip ) -ff = E  ∞ X i 1 =0 ∞ X i 2 =0 ..... ∞ X ip=0 C i 1 (Θ t 1 +i 1 )f i 1 +1 (Ψ t 1 +i 1 )C i 2 (Θ t 2 +i 2 )f i 2 +1 (Ψ t 2 +i 2 ).......C ip (Θ tp+ip )f ip+1 (Ψ tp+ip ) ff = ∞ X i 1 =0 ∞ X i 2 =0
....... If there exists s such that t s +i s = t r +i r when r = s, Θ ts+is is independent of the other Θ tr+ir . Therefore, in this case, this C is (Θ ts+is ) is independent of other ones and because E{C is (Θ ts+is )} = 0, E C i1 (Θ t1+i1 )C i2 (Θ t2+i2 ).......C ip (Θ tp+ip ) = 0.

Then, we study the E C is 1 (Θ ts 1 +is 1 )C is 2 (Θ ts 2 +is 2 ).......C is q (Θ ts q +is q ) where t s1 + i s1 = t s2 + i s2 = .... = t sq + i sq . Indeed, E C i1 (Θ t1+i1 )C i2 (Θ t2+i2 ).......C ip (Θ tp+ip ) is thus a product of terms of the form E C is 1 (Θ ts 1 +is 1 )C is 2 (Θ ts 2 +is 2 ).......C is q (Θ ts q +is q ) where t s1 + i s1 = t s2 + i s2 = .... = t sq + i sq . Suppose that t s1 < t s2 < .... < t sq and t s1 + i s1 = t s2 + i s2 = .... = t sq + i sq . Then, i s1 > i s2 > .... > i sq . Therefore, by lemma B.1.5, E C is 1 (Θ ts 1 +is 1 )C is 2 (Θ ts 2 +is 2 ).......C is q (Θ ts q +is q ) = 0 . If there exists s ≥ 3 such that t s + i s = t r + i r when r = s, C is (Θ ts+is ) is independent of other C ir (Θ tr+ir ) with E{C is (Θ ts+is )} = 0. Therefore, if E C i1 (Θ t1+i1 )C i2 (Θ t1+i2 )C i3 (Θ t3+i3 ).......C ip (Θ tp+ip ) = 0 , E C i1 (Θ t1+i1 )C i2 (Θ t1+i2 )C i3 (Θ t3+i3 ).....................C ip (Θ tp+ip ) is thus a product of terms of the form E C is 1 (Θ ts 1 +is 1 )C is 2 (Θ ts 2 +is 2 ).......C is q (Θ ts q +is q ) where t s1 +i s1 = t s2 +i s2 = .... = t sq +i sq .

Then, we have to study the products E C is 1 (Θ ts 1 +is 1 )C is 2 (Θ ts 2 +is 2 ).......C is q (Θ ts q +is q ) where t s1 + i s1 = t s2 + i s2 = .... = t sq + i sq .

In order that these terms are not zero, there is a priori only two possible cases 1) There exists h and a sequence s t such that t 1 + i 1 = t 1 + i 2 = t s3 + i s3 = .... = t s h + i s h .

2) There exists h and k and two sequences s ′ t and s" t such that t 1 +i 1 = t s ′ 3 +i s ′ 3 = .... = t s ′ h +i s ′ h and t 1 + i 2 = t s"3 + i s"3 = .... = t s" k + i s" k . Indeed, by the proof of previous proposition B.2.1, there is no other solutions, which can give a priori a nonzero expectation : if one find a sequence s" t = 1 such that t s"3 +i s"3 = .... = t s" k +i s" k , we shall have, for example, t s"3 < .... < t s" k and, therefore, i s"3 > .... > i s" k . We shall deduce, by lemma B.1.5, E C i3 (Θ t s" 3 +i s" 3 ).......C ip (Θ t s" k +i s" k ) = 0. Therefore, Therefore there are only two cases where we can find a priori a non-zero expectation 1) t 1 + i 1 = t 1 + i 2 = t s3 + i s3 = .... = t sp + i sp with p ≥ 3 because we are interested in E{X 2 t1 X t2 ......X tp } = 0 with p ≥ 2. 2) t 1 + i 1 = t s ′ 3 + i s ′ 3 = .... = t s ′ h + i s ′ h and t 1 + i 2 = t s"3 + i s"3 = .... = t s" k + i s" k .

E C i1 (Θ t1+i1 )C i2 (Θ t1+i2 )C i3 (Θ t3+i3
At first, suppose t 1 + i 1 = t 1 + i 2 = t s3 + i s3 = .... = t sp + i sp . Then, i 1 = i 2 . Then, C i1 (Θ t1+i1 )C i2 (Θ t1+i2 ) = C i1 (Θ t1+i1 )C i1 (Θ t1+i1 ) = Cos(2 * 4 i1 Θ t1+i1 ) + 1. Suppose, for example t s3 < t s4 < .... < t s h . Because t s3 + i s3 = .... = t s h + i s h , then, i s3 > i s4 > .... > i s h .

If t 1 + i 1 = t s3 + i s3 = T and t 1 > t s3 , i 1 < i s3 . Therefore, by lemma B.1. Therefore, E C i1 (Θ T )C i2 (Θ T )C is 3 (Θ T ).......C is h (Θ T ) = 0. where 1 ≥ ǫ(t) > 0 is decreasing and converges to 0 as t → ∞.

Suppose now that

We shall prove that |E{X 2 t X 2 t+h } -E{X 2 t }E{X 2 t+h }| ≤ α(h) where α(h) → 0 as h → ∞.

B.3.1 Elementary equalities

Because X t = ∞ i=0 C i (Θ t+i )f i+1 (Ψ t+i ), X 2 t = ∞ i=0 ∞ j=0 C i (Θ t+i )C j (Θ t+j )f i+1 (Ψ t+i )f j+1 (Ψ t+j ) , and
E{X 2 t } = ∞ i=0 E{C i (Θ t+i ) 2 }E{f i+1 (Ψ t+i ) 2 } ≤ ∞ i=0 1 (1 + i) 1+2a < ∞ . Moreover, X 2 t+h = ∞ r=0 ∞ s=0 C r (Θ
E{f i+1 (Ψ t+i ) 2 f r+1 (Ψ t+h+r ) 2 } -E{f i+1 (Ψ t+i ) 2 }E{f r+1 (Ψ t+h+r ) 2 } ≤ ∞ i=0 ∞ r=0 1 (1 + i) 1+2a 1 (1 + r) 1+2a ǫ(h + r -i) ≤ h/2 i=0 ∞ r=0 1 (1 + i) 1+2a 1 (1 + r) 1+2a ǫ(h + r -i) + ∞ i=h/2+1 ∞ r=0 1 (1 + i) 1+2a 1 (1 + r) 1+2a ǫ(h + r -i) .
Now, if i ≤ h/2, h + r -i ≥ h/2 and ǫ(h + r -i) ≤ ǫ(h/2). Moreover, ǫ(r) ≤ 1. Therefore, i,r,i =h+r

E{f i+1 (Ψ t+i ) 2 f r+1 (Ψ t+h+r ) 2 } -E{f i+1 (Ψ t+i ) 2 }E{f r+1 (Ψ t+h+r ) 2 } ≤ h/2 i=0 ∞ r=0 1 (1 + i) 1+2a 1 (1 + r) 1+2a ǫ(h/2) + ∞ i=h/2+1 ∞ r=0 1 (1 + i) 1+2a 1 (1 + r) 1+2a ≤ (K 0 ) 2 ǫ(h/2) + K h/

Notations 1 . 0 . 2

 102 Let Z n be a sequence of real random variables. If Z n converges in distribution to a random variable Z , one writes Z n d → Z. If Z n converges in probability to Z , one writes Z n P → Z.

  where* u≤x f (u)m ⊗ (du) = F X1 (x 1 ).....F Xn (x n ) dm n ..... .

Corollary 2 . 3 . 1 :M→ N 2

 2312 Assume that E{|ξ u | k } → 0 as n → ∞ for all k ∈ N . Assume that H mS and H mI hold. Then, (Σ u , Σ ′ u ) (0, I 2 ) = N (0, 1) ⊗ N (0, 1).

  .., t k .................................................................

  ff +...................................................................................................

  ................................ ∪{(0, 1, 0, 0, 0, 0, 1, 1)} : n -3 -(n -3 -1) events. ................................ ................................

  } ..................................................

  } ..................................................

  } .................................................. ..................................................

Notations 4 . 1 . 1

 411 Let (X m,s , Y m,s ) ∈ R 2 , s = 1, 2, ...., n m , m=1,2,.. be a triangular array of random vectors defined on a probability space (Ω, A, P ) such that n m → ∞.One supposes that E{X m,s } = E{Y m,s } = 0 and |E{(X m,s ) p }| < ∞ and |E{(Y m,s ) p }| < ∞ for all p ∈ N.

Hypothesis 4 . 1 . 1

 411 One keeps the notations of hypothesis 3.1.2. Then, we suppose that, for all p ∈ N * , for all j ≥ 2

P s 1

 1 βm,s e+1 Xm,s e+1 Ψ(nm) 3 → 0 by lemma 3.4.3.

  ............................................................................................

  !(p -2p")! +........................ ................................... +

  .............................................................................................

  ")! +................. ..................................... +

  2p)! +.................................................... +....................................................

ν 2p- 2 (

 2 2p -2)! .......................................... ..........................................

ν

  2p-4 (2p -4)! .......................................... ..........................................

ν 2p- 4 (

 4 2p -4)! ............................................................................................. ............................................................................................. ............................................................................................. .............................................................................................

  2p -2p)! .......................................... ..........................................

2

  p p! .......................................... ..........................................

  1)! .......................................... ..........................................

  2)! .......................................... ..........................................

2 2 p

 2 -p (p -p)! .......................................... ..........................................

  2)! +........ ........................................ .......................................................................................... .............................................................................................

  4)! .............................................................................................

  !(P -2P ")! +.............. .....................+ Q

  Then, one will use the relation cos(a)cos(b) = [cos(a+b)+cos(a-b)]/2 in order to prove the following proposition.

Proposition

  2 , C i3 (Θ T ) is orthogonal to Cos(2 * 4 i1 Θ T )C is 4 (Θ T )C is 5 (Θ T ).......C is h (Θ T ), and, therefore, orthogonal to C i1 (Θ T )C i2 (Θ T )C is 4 (Θ T )C is 5 (Θ T ).......C is h (Θ T ) = [Cos(2 * 4 i1 Θ T )+1]C is 4 (Θ T )C is 5 (Θ T ).................. C is h (Θ T ). Therefore, E C i1 (Θ T )C i2 (Θ T )C is 3 (Θ T )C is 4 (Θ T ).......C is h (Θ T ) = 0 Suppose always t s3 < t s4 < .... < t s h . If t 1 + i 1 = T and t 1 ≤ t s3 , i 1 ≥ i s3 . Thus, by lemmas B.1.1 and B.1.3, Cos(2 * 4 i1 Θ T ) is orthogonal to C is 3 (Θ T )C is 4 (Θ T )C is 5 (Θ T ).......C is h (Θ T ).

  1.2).

			ρ j1,j2,....,jn		e itx1/	√ n P = φ m (t/	√	n) n
	+	∞ q=0 j1+....+jn=q	ρ j1,j2,....,jn	σ j1 j 1 !	(it) j1 √ n j1	+ o( |t| j1 ) ......	σ jn j n !	(it) jn √ n

j1 (x 1 )m(dx 1 ) ...... e itxn/ √ n P jn (x n )m(dx n ) jn + o( |t| jn ) φ m (t/ √ n) n-q ′

  On the other hand, the conditions of Theorem 3 are actually stronger than those of Theorem 2 . Indeed, if n -2 E

	by lemma 4-1 of [23],	n s=1 [X 2 s -E{X 2 s }]	2 → 0 and if (X 1 + X 2 + .... + X n )/	√ n	M → N (0, M 2 ) ,
		1 √ n q	j1+j2+...+jn=q; js≤2, at least 1 js=2	ρ j1,j2,...,jn → 0 .
	Remark that the condition n -2 E sumptions. Indeed, by proposition A.2.2, it holds if |E{X 2 n s=1 X 2 s -E{X 2 s } s X 2 t } -E{X 2 2 → 0 is checked under weak as-s }E{X 2 t }| ≤ α(|t -s|) where α(h) → 0 as h → ∞.

...+jn=4; js=2 or 0 ρ j1,j2,...,jn → 0 .

  and where Θ t is an IID sequence independent of another IID sequence Θ t . Assume that ζ t is not strong mixing.

	Assume that k i (Θ t , ζ t ) = i -5/2 g i (Θ t ).sin(e(i)ζ t ) where E{g i (Θ 1 )} = 0 and |e(i)| ≤ 2π for all

)

  ..................................................

  .................................................. ..................................................

  2.2 and 3.2.4. Then, by lemma 3.3.2, We suppose that R m,t

	E	t1 =t2 =.... =tr,tr+1,...,t k Ψ(n m ) h	k s=1 R m,ts s	(X m,ts )
	is increased by sums of products of terms themselves bounded by some C nm (j, p)'s and by terms
	of the form E Now, let c ≤ d = H 1 . Suppose E P nm t=1 (Xm,t) Ψ(nm) c 1/c where c ≤ H 1 . P nm t=1 (Xm,t) Ψ(nm) E P nm t=1 (Xm,t) Ψ(nm) d 1/d by Holder Inequality.	d	≥ 1. Then, E	P nm t=1 (Xm,t) Ψ(nm)	c 1/c	≤
	Suppose now E It is enough to prove the lemma. P nm t=1 (Xm,t) Ψ(nm) d ≤ 1. Then, E	P nm t=1 (Xm,t) Ψ(nm)	c 1/c	≤ E	P nm t=1 (Xm,t) Ψ(nm)	d 1/d	≤ 1.
	Lemma 3.3.4

s (x) = x js for t = 1, 2, ..., n m where j s > 0. We assume

  hypothesis 3.1.2, if j s ≥ 1, E

				P nm t=1 R m,t m,s (Xm,t) Ψ(nm) js	H2	is bounded.
	Moreover, E	P nm t=1 Xm,t Ψ(nm)	H2 k1/H2	≤ B nm H2 .
			P nm t=1 β j m,s

Lemma 3.4.2 Pour tout j ≥ 2, Ψ(nm) 2j → 0. Proof By hypothesis 3.1.2 and by Holder Inequality, nm t=1 β j m,s

,

  by hypothesis 3.1.2. Now, C nm (2v 1 , 2Q) is bounded because v 1 > 0 and, moreover, converges to 0 if v 1 > 1. Moreover, by lemma 3.4.2, s1

	β 2u 1 m,s 1

  2.2, 3.2.4 and 3.4.3.

	Lemma 3.7.2 We assume M n q are bounded. Then,	1 Ψ(nm) 4 E	nm t=1 Q m,t 2 (X m,t )	2 → 0. We suppose that all moments
	s1 =.... =sq-e	E	X 2 m,s1 .......X 2 m,se X m,se+1 ....X m,sq-e Ψ(n m ) q	-

s1 =.... =sq-e E β m,s1 .....β m,se X m,se+1 ....X m,sq-e

  .......................................................... ..........................................................

	+	j1+.....+jn m =q, js≤2, one js=2	q! j 1 !.......j nm !	E	X j1 m,1 .....X m,nm jn m Ψ(n m ) q
	+	j1+.....+jn m =q, js≤2, two js=2	q! j 1 !.......j nm !	E	X j1 m,1 .....X m,nm jn m Ψ(n m ) q
	+
		j1+.....+jn m =q, js≤2, q" js=2
					jn m m,nm
					Ψ(n m ) q

1 .....X

  -1 , t r+1 , .., t k , s 1 = s 2 = .... = s r ′ , s r ′ +1 , ..., s k ′ .................................................................

  Let h = r 1 + .... + r k and h ′

	=s t h t	, s t i =s t ′ j	e t 1 =.... =e t a t , e t i =e t ′ j	E	r t=1 X t s t 1	.....X t s t h t Ψ(n) p+q	o t=1 Y t e t 1	.....Y t e t a t	.
	Lemma 4.2.4

  [START_REF] Dehling H. Denker M | Versik Processes and very weak Bernouilli processes with summable rates are in,dependent[END_REF].1, all the moments of marginal distributions are bounded. By using Holder inequality, we deduce that all the moments are bounded.

	By lemma 4.2.8 and 4.2.7 we deduce that
	s1 =.... =sq-e v1 =.... =v p-f	E	X 2 m,s1 ....X 2 m,se X m,se+1 ....X m,sq-e Y 2 m,v1 .....Y 2 m,v f Y m,v f +1 .....Y m,v p-f Ψ(n m ) q+p
	converges to σ 2e 0 σ 2e 1 S q-2e,p-2f . By lemma 4.2.4, if there exists r s ≥ 3 or r ′ s ≥ 3,

  nm ) p Ψ(n m ) p+q

	=	j1+.....+jn m =q i1+.....+in m =p	q! j 1 !.......j nm !	p! i 1 !.......i nm !	E	X j1 m,1 .....X m,nm Y i1 jn m m,1 .....Y m,nm in m Ψ(n m ) q+p	,
	all the moments converge.
	Proof of necessary condition of theorem 15 We suppose that all the moments converge.
	Then, the second and thirth relations are a consequence of proposition 3.3.1 .
	By lemma 4.2.9 and 4.2.7 , we deduce that, for all e, for all f,
	s1 =.... =sq-e v1 =.... =v p-f	E	X 2 m,s1 ....X 2 m,se X m,se+1 ....X m,sq-e Y 2 m,v1 .....Y 2 m,v f Y m,v f +1 .....Y m,v p-f Ψ(n m ) q+p
	converges.
	By lemma 4.2.4, if there exists r s ≥ 3 or r ′ s ≥ 3,

  n m ) p+q converges for all (u 1 , ..., u k , v 1 , ..., v k ′ ) if there exists u t ≥ 2 or v t ≥ 2. j1+.....+jn m =q i1+.....+in m =q ′ q! j 1 !.......j nm ! q ′ ! i 1 !.......i nm !

		E	X j1 m,1 .....X m,nm Y i1 jn m m,1 .....Y m,nm in m Ψ(n m ) q+q ′	,
	then,
	s1 =s2 =..... =sq t1 =t2 =..... =tp	E{X m,s1 X m,s2 ......X m,sq Y m,t1 Y m,t2 ......Y m,tp } Ψ(n m ) p+q
	converges.

Because

E (X m,1 + .... + X m,nm ) q (Y m,1 + .... + Y m,nm ) p Ψ(n m ) p+q =

  s1 =s2 =..... =sq t1 =t2 =..... =tp E{X s1 X s2 ......X sq Y t1 Y t2 ......Y tp } Ψ(n) p+q → S q,p .By lemma 4.2.4, if there exists s such that j s ≥ 3 or i s ≥ 3, for example, j1+.....+jn m =q, at least one js≥3 i1+.....+in m =p

  ......................................... .......................................... ......................................... .......................................... ............................................................................................ ............................................................................................. ............................................................................................. ............................................................................................. ......................................... ..........................................

	+jn=q, js≤1 i1+.....+in=p, it≤1 j1+.....+jn=q, 1 js=2 i1+.....+in=p, it≤1 j 1 !.......j n ! q! q! j 1 !.......j n ! i 1 !.......i nm ! p! p! i 1 !.......i n ! j1+.....+jn=q, q" js=2 i1+.....+in=p, it≤1 + + q! j 1 !.......j n ! p! i 1 !.......i n ! E E E + j1+.....+jn=q, js≤1 i1+.....+in=p, 1 it=2 q! j 1 !.......j n ! p! i 1 !.......i n ! E + j1+.....+jn=q, 1 js=2 i1+.....+in=p, 1 it=2 q! j 1 !.......j n ! p! i 1 !.......i n ! E j1+.....+jn=q, q" js=2 i1+.....+in=p, 1 it=2 q! j 1 !.......j n ! p! i 1 !.......i n ! E + j1+.....+jn=q, js≤1 i1+.....+in=p, 2 it=2 q! j 1 !.......j n ! p! i 1 !.......i n ! E + j1+.....+jn=q, 1 js=2 i1+.....+in=p, 2 it=2 q! j 1 !.......j n ! p! i 1 !.......i n ! E + j1+.....+jn=q, q" js=2 i1+.....+in=p, 2 it=2 q! j 1 !.......j n ! p! i 1 !.......i n ! E + j1+.....+jn=q, js≤1 i1+.....+in=p, p" it=2 q! j 1 !.......j n ! p! i 1 !.......i n ! E + j1+.....+jn=q, 1 js=2 i1+.....+in=p, p" it=2 q! j 1 !.......j n ! p! i 1 !.......i n ! E + Ψ(n) + X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p j1+.....+jn=q, q" js=2 i1+.....+in=p, p" it=2

.q+p .......................................... .......................................... ...

  q+p .......................................... .......................................... q+p .......................................... .......................................... ............................................................................................ ............................................................................................. ............................................................................................. ............................................................................................. q+p .......................................... ..........................................

	0 2 0 q!p! 2 1 2 0 j1+.....+jn=q, 1 js=2 i1+.....+in=p, it≤1 j1+.....+jn=q, js≤1 i1+.....+in=p, it≤1 E E q!p! + 2 q" 2 0 j1+.....+jn=q, q" js=2 i1+.....+in=p, it≤1 E + q!p! 2 0 2 j1+.....+jn=q, js≤1 i1+.....+in=p, 1 it=2 E + q!p! 2 1 2 j1+.....+jn=q, 1 js=2 i1+.....+in=p, 1 it=2 E Ψ(n) + X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p X j1 1 .....X jn n Y i1 1 .....Y in n q!p! 2 q" 2 j1+.....+jn=q, q" js=2 i1+.....+in=p, 1 it=2 E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p + q!p! 2 0 2 2 j1+.....+jn=q, js≤1 i1+.....+in=p, 2 it=2 E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p + q!p! 2 1 2 2 j1+.....+jn=q, 1 js=2 i1+.....+in=p, 2 it=2 E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) + q!p! 2 q" 2 2 j1+.....+jn=q, q" js=2 i1+.....+in=p, 2 it=2 E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p + q!p! 2 0 2 p" j1+.....+jn=q, js≤1 i1+.....+in=p, p" it=2 E X j1 1 .....X jn n Y i1 1 .....Y in n Ψ(n) q+p Ψ(n) + + q!p! 2 1 2 p" j1+.....+jn=q, 1 js=2 i1+.....+in=p, p" it=2 X j1 1 .....X jn n Y i1 1 .....Y in n E Ψ(n)

q+p .......................................... ..........................................

.

  2)! ............................................................................................. ............................................................................................. ............................................................................................. .............................................................................................

  !(P -2)! +...............

............................. + Q

  .. =tp-1 E{(Y m,t1 ) 2 Y m,t2 ......Y m,tp-1 } Ψ(n m ) p ≤ Sb 2 p .Then, it is enough to apply proposition 4.3.1 in order to prove sufficient condition. Proof of necessary condition of Corollary 4.5.1 We suppose that all the moments converge. Therefore, by proposition 4.3.1 , the condition is proved. cos([4 ip+1 -s]x) where 4 ip+1 -s > 0 and 4 ip+1 + s < 4 ip+1 + 4 ip+1 ≤ 2 * 4 ip+1 < 4 ip+1+1 . We deduce the following propositions. Lemma B.1.4 Let i 1 < i 2 < .... < i p , i s ∈ N. Then, 2π x=0 cos(4 i1 x)cos(4 i2 x)......cos(4 ip x)dx = 0. Lemma B.1.5 Let i s1 > i s2 > .... > i sq . Then, E C is 1 (Θ t )C is 2 (Θ t ).......C is q (Θ t ) = 0. B.2 Some properties Proposition B.2.1 Let t 1 < t 2 < ...... < t p . Then, E{X t1 X t2 ......X tp } = 0. Proof We have E{X t1 X t2 ......X tp }

				=		1 √ n = (1/2) E s=1 n s=1 X s √ n 4 ip+1 -1	n t=1 A s cos([4 ip+1 + s]x) + (1/2) X 2 t -E{X 2 t } n -n s=1 4 ip+1 -1 E{X 3 s } n 2 + s=1 A	n s=1	E{X s }E{X 2 n 2 s }
						=	1 √ n	E		n s=1	X s √ n	n t=1	X 2 t -E{X 2 t } n	-	E{X 3 1 } n	+	E{X 1 }E{X 2 1 } n	.
		Now, by Schwartz Inequality,
															1 √ n	E	n s=1	X s √ n	n t=1	X 2 t -E{X 2 n t }	2
											≤	1 √ n	E	n s=1	X s √ n	2	E	n t=1	X 2 t -E{X 2 n t }	2
												≤	M n 2 √ n	E	s,t	X 2 s -E{X 2 s } X 2 t -E{X 2 t } n 2
																≤	M n 2 √ n s,t	E{X 2 s X 2 t } -E{X 2 s }E{X 2 t } n 2
													≤	M n 2 √ n s,t	E{X 4 s } 1/2 E{X 4 t } 1/2 n 2	-E{X 2 1 } 2
																≤	M n 2 √ n	E{X 4 1 } -E{X 2 1 } 2 → 0 .
		Now, let γ = E{X 3 1 }/E{X 2 1 } and σ 2 2 = E X 2 1 -γX 1 -E{X 2 1 }	2 . Then,
						n -2	j1+j2+...+jn=4; js=2 or 0	ρ j1,j2,...,jn = E	n s=1	n t=s+1	P 2 (X s )P 2 (X t ) n 2
								= E	n s=1	n t=s+1	X 2 s -γX s -E{X 2 s } X 2 t -γX t -E{X 2 t } σ 2 2 n 2
								=	1 2σ 2 2	E		s =t	X 2 s -γX s -E{X 2 s } X 2 t -γX t -E{X 2 t } n 2
	=	1 2σ 2 2	E	s =t	X 2 s -E{X 2 s } X 2 t -E{X 2 t } n 2	-	γ σ 2 2	E	s =t	X s X 2 t -E{X 2 t } n 2	+	γ 2 2σ 2 2	E	s =t	X s X t n 2
				=	1 2σ 2 2	E	s,t	X 2 s -E{X 2 s } X 2 t -E{X 2 t } n 2	-	1 2σ 2 2	E	s	X 2 1 -E{X 2 1 } n 2	2
										-	γ σ 2 2	E		s =t	X s X 2 t -E{X 2 t } n 2	+	γ 2 2σ 2 2	E	s =t	X s X t n 2

s

  C i 1 (Θ t 1 +i 1 )C i 2 (Θ t 2 +i 2 ).......C ip (Θ tp+ip )

∞ X ip=0 E n o E n f i 1 +1 (Ψ t 1 +i 1 )......f ip+1 (Ψ tp+ip ) o .

  B.2.2 Let t 1 = t s if s ≥ 2 and t 2 < t 3 < ...... < t p . Then, E{X 2 t1 X t2 ......X tp } = 0.Proof In the proof of proposition B.2.1, we have provedE{X t1 X t2 ......X tp } i1 (Θ t1+i1 )C i2 (Θ t1+i2 )C i3 (Θ t3+i3 ).......C ip (Θ tp+ip ) E f i1+1 (Ψ t1+i1 )f i2+1 (Ψ t1+i2 )f i3+1 (Ψ t3+i3 )......f ip+1 (Ψ tp+ip ) .

	=	∞		∞	.......
		i1=0	i2=0
				∞	∞
		=	X	X
			i 1 =0	i 2 =0

....... ∞ X ip=0 E n C i 1 (Θ t 1 +i 1 )C i 2 (Θ t 2 +i 2 ).......C ip (Θ tp+ip ) o E n f i 1 +1 (Ψ t 1 +i 1 )......f ip+1 (Ψ tp+ip ) o .

Therefore,

E{X 2 t1 X t3 ......X tp } ∞ ip=0 E C

  ).......C ip (Θ tp+ip ) (Θ t 1 +i 1 )C i 2 (Θ t 1 +i 2 )C i s ′ (Θ t s" 3 +i s" 3 ).......C i s" k (Θ t s" k +i s" k )

	n							o	n
	= E	C i 1 3	(Θ t s ′ 3	+i s ′ 3	).......C i s ′ h	(Θ t s ′ h	+i s ′ h	)	E	C i s" 3

o = 0 .

  there exists h and k and two sequences s ′ t and s" t such thatt 1 + i 1 = t s ′ 3 + i s ′ 3 = .... = t s ′ h + i s ′ h and t 1 + i 2 = t s"3 + i s"3 = .... = t s" k + i s" k . For example, study E C i1 (Θ t1+i1 )C i s ′ ) where t 1 +i 1 = t s3 +i s3 = .... = t s ′ h + i s ′ h . Because, t 1 = t s3 = ..... = t s h , i 1 = i s3 = ..... = i s h . For example, i 1 > i s3 > ...... > i s h . Then, par lemmas B.1.1 and B.1.3, √ 2cos(4 i1 Θ T ) = C i1 (4 i1 Θ T ) is orthogonal to C is 3 (Θ ts 3 +is 3 ).......C is h (Θ ts h +is h ). Therefore, E C i1 (Θ t1+i1 )C is 3 (Θ ts 3 +is 3 ).......C is h (Θ ts h +is h ) = 0. B.3 Calculation of E{X 2 t X 2 t+h } -E{X 2 t }E{X 2 t+h } E{f i+1 (Ψ t+i ) 2 f r+1 (Ψ t+h+r ) 2 } -E{f i+1 (Ψ t+i ) 2 }E{f r+1 (Ψ t+h+r ) 2 }

			3	(Θ t s ′ 3	+i s ′ 3	).......C i s ′ h	(Θ t s ′ h	+i s ′ h
	We suppose
	≤	1 (1 + i) 1+2a	1 (1 + r)

1+2a ǫ(t + h + r -(t + i)) ,

  t+h+r )C s (Θ t+h+s )f r+1 (Ψ t+h+r )f s+1 (Ψ t+h+s ) . (Θ t+i ) 2 }E{f i+1 (Ψ t+i ) 2 }E{C j (Θ t+h+j ) 2 }E{f j+1 (Ψ t+h+j ) 2 } E{f i+1 (Ψ t+i ) 2 }E{f r+1 (Ψ t+h+r ) 2 } Cr(Θ t+h+r ) 2 }E{fi+1(Ψt+i) 2 fr+1(Ψ t+h+r ) 2 } -E{fi+1(Ψt+i) 2 }E{fr+1(Ψ t+h+r ) 2 } E{fi+1(Ψt+i) 2 fr+1(Ψ t+h+r ) 2 } -E{fi+1(Ψt+i) 2 }E{fr+1(Ψ t+h+r ) 2 } t+h+r ) 2 C r (Θ t+h+r ) 2 }E{f h+r+1 (Ψ t+h+r ) 2 f r+1 (Ψ t+h+r ) 2 } -E{f h+r+1 (Ψ t+h+r ) 2 }E{f r+1 (Ψ t+h+r ) 2 } E{f i+1 (Ψ t+i ) 2 f r+1 (Ψ t+h+r ) 2 } -E{f i+1 (Ψ t+i ) 2 }E{f r+1 (Ψ t+h+r ) 2 } .

	B.3.3 Study of E{X 2 t+h X 2 t } -E{X 2 t+h }E{X 2 t } We recall
				=	∞ i=0	∞ j=0	E{C i =	∞	∞	E{X 2 t }E{X 2 t+h }
								i=0	r=0
		Therefore,
								E{X 2 t+h X 2 t } -E{X 2 t+h }E{X 2 t }
	=	X	»	E{Ci(Θt+i) 2 -
		i,i=h+r
				+	X
		+
				i,r,i =h+r
		Let K t =	∞ i=t	1 (1+i) 1+2a . Then,
		Therefore,

i,r,i =h+r h i = r E{C r+h (Θ i,r,i =h+r

  2 K 0 whichi converges to 0 if h → ∞. E{C h+r (Θ t+h+r ) 2 C r (Θ t+h+r ) 2 }E{f h+r+1 (Ψ t+h+r ) 2 f r+1 (Ψ t+h+r ) 2 } -E{f h+r+1 (Ψ t+h+r ) 2 }E{f r+1 (Ψ t+h+r ) 2 } ≤ r 4E{f h+r+1 (Ψ t+h+r ) 2 f r+1 (Ψ t+h+r ) 2 } + E{f h+r+1 (Ψ t+h+r ) 2 }E{f r+1 (Ψ t+h+r ) 2 } At first, because {Θ t } is IID and {Ψ t } is strictly stationary, X t = ∞ i=0 C i (Θ t+i )f i+1 (Ψ t+i ) is also strictly stationary. Then, (1/n) E{X 2 s }E{X 2 t }| ≤ α(|t -s|) where α(h) → 0 as h → ∞. Then, by proposition, A.2.2 , n -2 E E{X 1 } = 0 and E{|X s | p } < ∞ for all p ∈ N.Moreover, by proposition, B.2.1 E{X t1 X t2 ......X tp } = 0. Therefore P{X n+1 |X n , X n-1 , ....} = 0. Then, the second condition of proposition 3.8.1 holds with ν 2 = 0.At last, by proposition B.2.2, E{X 2 t1 X t2 ......X tp } = 0. Then, the third condition of corollary 3.8.1 holds. One can also remark that X 1 is bounded and apply corollary 3.10.1.

	Moreover,
	≤	r	4	1 (1 + h + r) 1+2a	1 (1 + r) 1+2a +	1 (1 + h + r) 1+2a	1 (1 + r) 1+2a
				≤	1 (1 + h) 1+2a	r	4	1 (1 + r) 1+2a +	1 (1 + r) 1+2a
						≤	5K 0 (1 + h) 1+2a
	which converges to 0 if h → ∞.
	We deduce
	|E{X 2 t X 2 t+h } -E{X 2 t }E{X 2 t+h }| ≤ (K 0 ) 2 ǫ(h/2) + K h/2 K 0 +	5K 0 (1 + h) 1+2a .
	B.4 Conclusion
	n s=1 E{X 2 s } = E{X 2 1 }. s -E{X 2 s=1 X 2 t } -n s X 2 Then, |E{X 2 s } 2 → 0. Then, the first condition of proposition 3.8.1 holds.
	Moreover, clearly,

r We deduce that

E X 1 + X 2 + ...... + X n √ n M → N (0, E{X 2 1 }) .

Of course, in this case, one assumes Ψ(n) = σ(n) and σ(n)/ √ n ≥ c ψ .

which is the moment of order (2q,2p) of N (0, σ 2 0 + ν 2 ) ⊗ N (0, σ 2 1 + ν 2 ). Now we want to prove necessary condition. Then, we need the following lemma.

Lemma 4.3.2 We suppose that µ q µ p = q!p! 2 0 2 0 σ 0 0 σ 0 0 0!0! S q,p q!p! + q!p! 2 1 2 0 σ 2 0 σ 0 0 1!0! ν q-2 ν p (q -2)!p! + ............. + q!p! 2 q" 2 0 σ 2q" 0 σ 0 0 q"!0! ν q-2q" ν p (q -2q")!p! + q!p! 2 0 2 σ 0 0 σ 2 0 0!1! ν q ν p-2 q!(p -2)! + q!p! 2 1 2

ν q-2 ν p-2 (q -2)!(p -2)! + ................. + q!p! 2 q" 2 σ 2q" 0 σ 2 0 q"!1! ν q-2q" ν p-2 (q -2q")(p -2)!

ν q-2 ν p-4 (q -2)!(p -4)! +.................+ q!p! 2 q" 2 2 σ 2q" 0 σ 4 0 q"!2! S q-2q",p-4 (q -2q")!(p -

+

q!p! 2 0 2 p" σ 0 0 σ 2p" 0 0!p"! ν q ν p-2p" q!(p -2p")! + q!p! 2 1 2 p" σ 2 0 σ 2p" 0 1!p"! ν q-2 ν p-2p" (q -2)!(p -2p")! +............

Then, S q,p = ν q ν p .

Proof We know that 1 = M 2 = σ 2 0 + ν 2 . If q=2h and p=2k, one can write

We deduce that S q,p =

If q (or p) is odd, it is easier to prove because µ q = 0 = ν q-2e . .

Proof of necessary condition

We want to prove by recurence that S q,p = ν q ν p for all (q,p). It is easy to understand that the result is true if q = 0 or p = 0 : it is the result in dimension 1 . By definition, it is true also if p = q = 1.

At first, we will see by recurrence that it is true if p = 1 or q = 1. For example, if p=1, it is thus true for q=1. Then, we suppose that it holds for all q ′ ≤ q. Let Q=q+2. Let Q" = ⌊Q/2⌋. Then, by lemma 4.3.1,

Then, the results holds if p=1 or q=1.

Then, we suppose that the result holds for all (q',p') such that q ′ ≤ q and p ′ ≤ p = q. Let Q=q+2 and Q"=q"+1. Then, p-2p" = 0 or 1. Therefore, S Q,p-2p" = ν Q ν p-2p" . Let P= p-2p"+2. Then, by lemma 4.3.1,

Correlation coefficients of higher ordrer

First proposition

Now we prove proposition 1.1.1, that is that We need the following lemma.

Lemma A.1.1 Let M X be the probability of (X 1 , X 2 , ....., X n ). Let f ∈ L 4 (R n , M X ). Let g s ∈ L 4(n-1) (R, m s ) for s=1,2,...,n. For s=1,2,....,n, let g s k be a sequence of L 4(n-1) (R, m s ) such that

Proof By Holder's Inequality,

By the same method applied for k s , s=2,3,....,n, one proves the result. For example,

Then, for example, for n=3,

Then, for ǫ > 0, one chooses

Proof of proposition 1.1.1 Let B s , s=1,2,...,n, be n Borel sets of R. Assume that B s = [-∞, x s ]. Then, 1 Bs = ∞ j=0 γ s j P s j with γ s j = Bs P s j dm s . Then, one proves proposition 1.1.1 by using lemma A.1.1 with f=1 and g s = 1 Bs (x s ).

A.1.2 Second and third propositions

Now, we prove proposition 1.1.2 : (X 1 + ... + X n , F n ) is a martingale if and only if E{X n+1 |F n } = 0 for all n ≥ 1.

Proof of proposition 1.1.2 By definition, (X

Now, we prove proposition 1.1.3, i.e. :

If (X 1 + ... + X n , F n ) is a martingale, ρ j1,j2,....,jn,1 = 0 for all n ≥ 1. Conversely, if {P j s } is a basis of L 2 (R, m s ) and if ρ j1,j2,....,jn,1 = 0 for all n ∈ N * and for all (j 1 , ...., j n ), then

Then, X n+1 is orthogonal to the space L 2 generated by X 1 ,.....,X n . In particular X n+1 is orthogonal to the function P 1 j1 (X 1 )......P n jn (X n ). Then, ρ j1,j2,....,jn,1 = 0 for all n ∈ N * .

Reciprocally, assume that, for all s, {P s j } is a basis of L 2 (R, m s ), and that ρ j1,j2,....,jn,1 = 0 for all (j 1 , ...., j n ) and for all n ∈ N * .

Then one uses lemma A.1.1 with f (x 1 , ...., x n , x n+1 ) = x n+1 . One uses the same notations as in the proof of proposition 1.1.1 : Let B s , s=1,2,...,n, be n Borel sets of R. Then, 1 Bt = ∞ j=0 γ s j P s j with γ s j = Bs P s j dm s . One sets g s = 1 Bs and g s ks = ks j=0 γ s j P s j . By by lemma A.1.1 , for all Borel sets B 1 ,....,B n ,

A.2 Equivalences

In this section, we suppose that m s = m and X m,t = X t . Then, we have the following proposition.

Proof We know that

Proposition A.2.2 We assume that m s = m for all s. We suppose that

Then, it is enough to choose n such that

and therefore,

With these relations, we deduce the lemma.

Appendix B

Study of example 2.2.1

We recall that we suppose

where

where Θ i is IID with uniform distribution on [0, 2π], where {Ψ i } is a strictly stationary process, independent from {Θ i } and where |f i+1 (y)| ≤ 1 (i+1) 1/2+a with a > 0 .

B.1 Study of trigonometric functions

We recall

We have