# On Complete Convergence of Moving Average Process for AANA Sequence

Authors: Wenzhi Yang; Xuejun Wang; Nengxiang Ling; Shuhe Hu
Journal: Discrete Dynamics in Nature and Society
 (2012)

---

## Abstract

We investigate the moving average process such thatXn=∑i=1∞aiYi+n, n≥1, where ∑i=1∞|ai|<∞ and {Yi,1≤i<∞} is a sequence of asymptotically almost negatively associated (AANA) random variables. The complete convergence, complete moment convergence, and the existence of the moment of supermum of normed partial sums are presented for this moving average process.


---

## Body

## 1. Introduction

We assume that{Yi,-∞<i<∞} is a doubly infinite sequence of identically distributed random variables with E|Y1|<∞. Let {ai,-∞<i<∞} be an absolutely summable sequence of real numbers and
(1.1)Xn=∑i=-∞∞aiYi+n,n≥1
be the moving average process based on the sequence {Yi,-∞<i<∞}. As usual, Sn=∑k=1nXk,n≥1 denotes the sequence of partial sums.Under the assumption that{Yi,-∞<i<∞} is a sequence of independent identically distributed random variables, various results of the moving average process {Xn,n≥1} have been obtained. For example, Ibragimov [1] established the central limit theorem, Burton and Dehling [2] obtained a large deviation principle, and Li et al. [3] gave the complete convergence result for {Xn,n≥1}.Many authors extended the complete convergence of moving average process to the case of dependent sequences, for example, Zhang [4] for φ-mixing sequence, Li and Zhang [5] for NA sequence. The following Theorems A and B are due to Zhang [4] and Kim et al. [6], respectively.Theorem A.
Suppose that{Yi,-∞<i<∞} is a sequence of identically distributed φ-mixing random variables with ∑m=1∞φ1/2(m)<∞ and {Xn,n≥1} is as in (1.1). Let h(x)>0(x>0) be a slowly varying function and 1≤p<2, r≥1. If Y1=0 and E|Y1|rph(|Y1|p)<∞, then
(1.2)∑n=1∞nr-2h(n)P(|Sn|≥ɛn1/p)<∞,∀ɛ>0.Theorem B.
Suppose that{Yi,-∞<i<∞} is a sequence of identically distributed φ-mixing random variables with EY1=0,EY12<∞ and ∑m=1∞φ1/2(m)<∞ and {Xn,n≥1} is as in (1.1). Let h(x)>0(x>0) be a slowly varying function and 1≤p<2,r>1. If E|Y1|rph(|Y1|p)<∞, then
(1.3)∑n=1∞nr-2-1/ph(n)E(|Sn|-ɛn1/p)+<∞,∀ɛ>0,
where x+=max{x,0}.Chen and Gan [7] investigated the moments of maximum of normed partial sums of ρ-mixing random variables and gave the following result.Theorem C.
Let0<r<2 and p>0. Assume that {Xn,n≥1} is a mean zero sequence of identically distributed ρ-mixing random variables with the maximal correlation coefficient rate ∑n=1∞ρ2/s(2n)<∞, where s=2 if p<2 and s>p if p≥2. Denote Sn=∑i=1nXi,n≥1. Then
(1.4){E|X1|r<∞,ifp<r,E[|X1|rlog⁡(1+|X1|)]<∞,ifp=r,E|X1|p<∞,ifp>r,E(sup⁡n≥1|Xnn1/r|p)<∞,E(sup⁡n≥1|Snn1/r|p)<∞
are all equivalent.Chen et al. [8] and Zhou [9] also studied limit behavior of moving average process under φ-mixing assumption. For more related details of complete convergence, one can refer to Hsu and Robbins [10], Chow [11], Shao [12], Li et al. [3], Zhang [4], Li and Zhang [5], Chen and Gan [7], Kim et al. [6], Sung [13–15], Chen and Li [16], Zhou and Lin [17], and so forth.Inspired by Zhang [4], Kim et al. [6], Chen and Gan [7], Sung [13–15], and other papers above, we investigate the limit behavior of moving average process under AANA sequence, which is weaker than NA and obtain some similar results of Theorems A, B, and C. The main results can be seen in Section 2 and their proofs are given in Section 3.Recall that the sequence{Xn,n≥1} is stochastically dominated by a nonnegative random variable X if
(1.5)sup⁡n≥1P(|Xn|>t)≤CP(X>t)forsomepositiveconstantCand∀t≥0.Definition 1.1.
A finite collection of random variablesX1,X2,…,Xn is said to be negatively associated (NA) if for every pair of disjoint subsets A1,A2 of {1,2,…,n},
(1.6)Cov{f(Xi:i∈A1),g(Xj:j∈A2)}≤0,
whenever f and g are coordinatewise nondecreasing such that this covariance exists.An infinite sequence{Xn,n≥1} is NA if every finite subcollection is NA.Definition 1.2.
A sequence{Xn,n≥1} of random variables is called asymptotically almost negatively associated (AANA) if there exists a nonnegative sequence q(n)→0 as n→∞ such that
(1.7)Cov(f(Xn),g(Xn+1,Xn+2,…,Xn+k))≤q(n)[Var⁡(f(Xn))Var⁡(g(Xn+1,Xn+2,…,Xn+k))]1/2,
for all n,k≥1 and for all coordinate-wise nondecreasing continuous functions f and g whenever the variances exist.The concept of NA sequence was introduced by Joag-Dev and Proschan [18]. For the basic properties and inequalities of NA sequence, one can refer to Joag-Dev and Proschan [18] and Matula [19]. The family of AANA sequence contains NA (in particular, independent) sequence (with q(n)=0,n≥1) and some more sequences of random variables which are not much deviated from being negatively associated. An example of an AANA which is not NA was constructed by Chandra and Ghosal [20, 21]. For various results and applications of AANA random variables can be found in Chandra Ghosal [21], Wang et al. [22], Ko et al. [23], Yuan and An [24], and Wang et al. [25, 26] among others.For simplicity, in this paper we consider the moving average process:(1.8)Xn=∑i=1∞aiYi+n,n≥1,
where ∑i=1∞|ai|<∞ and {Yi,1≤i<∞} is a mean zero sequence of AANA random variables.The following lemmas are our basic techniques to prove our results.Lemma 1.3.
Let{Xn,n≥1} be a sequence of AANA random variables with mixing coefficients {q(n),n≥1}. If f1,f2,… are all nondecreasing (or nonincreasing) continuous functions, then {fn(Xn),n≥1} is still a sequence of AANA random variables with mixing coefficients {q(n),n≥1}.Remark 1.4.
Lemma1.3 comes from Lemma 2.1 of Yuan and An [24], but the functions of f1,f2,… in Lemma 2.1 of Yuan and An [24] are written to be all nondecreasing (or nonincreasing) functions. According to the definition of AANA, f1,f2,… should be all nondecreasing (or nonincreasing) continuous functions.Lemma 1.5 (cf. Wang et al. [25, Lemma 1.4]).
Let1<p≤2 and {Xn,n≥1} be a mean zero sequence of AANA random variables with mixing coefficients {q(n),n≥1}. If ∑n=1∞q2(n)<∞, then there exists a positive constant Cp depending only on p such that
(1.9)E(max⁡1≤k≤n|∑i=1kXi|p)≤Cp∑i=1nE|Xi|p,
for all n≥1, where Cp=2p[22-p+(6p)p(∑n=1∞q2(n))p-1].Lemma 1.6 (cf. Wu [27, Lemma 4.1.6]).
Let{Xn,n≥1} be a sequence of random variables, which is stochastically dominated by a nonnegative random variable X. For any α>0 and b>0, the following two statements hold:
(1.10)E[|Xn|αI(|Xn|≤b)]≤C1{[EXαI(X≤b)]+bαP(X>b)},E[|Xn|αI(|Xn|>b)]≤C2E[XαI(X>b)],
where C1 and C2 are positive constants.Throughout the paper,I(A) is the indicator function of set A, x+=max{x,0} and C,C1,C2,… denote some positive constants not depending on n, which may be different in various places.


## 2. The Main Results

Theorem 2.1.
Letr>1,  1≤p<2 and  rp<2. Assume that  {Xn,n≥1} is a moving average process defined in (1.8), where  {Yi,1≤i<∞} is a mean zero sequence of AANA random variables with  ∑n=1∞q2(n)<∞ and stochastically dominated by a nonnegative random variable  Y. If  EYrp<∞, then for every  ɛ>0,
(2.1)∑n=1∞nr-2P(max⁡1≤k≤n|Sk|>ɛn1/p)<∞,(2.2)∑n=1∞nr-2P(sup⁡k≥n|Skk1/p|>ɛ)<∞.Theorem 2.2.
Let the conditions of Theorem2.1 hold. Then for every ɛ>0,
(2.3)∑n=1∞nr-2-1/pE(max⁡1≤k≤n|Sk|-ɛn1/p)+<∞,(2.4)∑n=1∞nr-2E(sup⁡k≥n|Skk1/p|-ɛ)+<∞.Theorem 2.3.
Let0<r<2 and 0<p<2. Assume that {Xn,n≥1} is a moving average process defined in (1.8), where {Yi,1≤i<∞} is a mean zero sequence of AANA random variables with ∑n=1∞q2(n)<∞ and stochastically dominated by a nonnegative random variable Y with EY<∞. Suppose that
(2.5){forp<r,{E[Ylog⁡(1+Y)]<∞,ifr=1,EYr<∞,ifr>1,forp=r,{E[Ylog⁡(1+Y)]<∞,if0<r<1,E[Ylog⁡2(1+Y)]<∞,ifr=1,E[Yrlog⁡(1+Y)]<∞,ifr>1,forp>r,{E[Ylog⁡(1+Y)]<∞,ifp=1,EYp<∞,ifp>1.
Then
(2.6)E(sup⁡n≥1|Snn1/r|p)<∞.


## 3. The Proofs of Main Results

Proof of Theorem2.1.
Firstly, we show that the moving average process (1.8) converges almost surely under the conditions of Theorem 2.1. Since rp>1, it has EY<∞, following from the condition EYrp<∞. On the other hand, by Lemma 1.6 with α=1 and b=1, one has
(3.1)E|Yi|≤1+C2E[YI(Y>1)]≤1+C2EY<∞,1≤i<∞.
Consequently, by the condition ∑i=1∞|ai|<∞, we have that
(3.2)∑i=1∞E|aiYi+n|≤C3∑i=1∞|ai|<∞,
which implies ∑i=1∞aiYi+n converges almost surely.
Note that(3.3)∑k=1nXk=∑k=1n∑i=1∞aiYi+k=∑i=1∞ai∑k=i+1i+nYk,n≥1.
Since r>1 and EYrp<∞, one has EYp<∞. Combining EYp<∞ with EYj=0, ∑i=1∞|ai|<∞ and Lemma 1.6, we can find that
(3.4)n-1/pmax⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kE[YjI(|Yj|≤n1/p)]|=n-1/pmax⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kE[YjI(|Yj|>n1/p)]|≤n-1/p∑i=1∞|ai|∑j=i+1i+nE[|Yj|I(|Yj|>n1/p)]≤CE[(n1/p)p-1YI(Y>n1/p)]≤CE[YpI(Y>n1/p)]⟶0asn⟶∞.
Meanwhile,
(3.5)n-1/pmax⁡1≤k≤n|∑i=1∞ai∑j=i+1i+k(-n1/p)E[I(Yj<-n1/p)]|≤n-1/p∑i=1∞|ai|∑j=i+1i+nE[|Yj|I(|Yj|>n1/p)]≤CE[YpI(Y>n1/p)]⟶0asn⟶∞,n-1/pmax⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kn1/pE[I(Yj>n1/p)]|≤n-1/p∑i=1∞|ai|∑j=i+1i+nE[|Yj|I(|Yj|>n1/p)]≤CE[YpI(Y>n1/p)]⟶0asn⟶∞.
Let
(3.6)Ynj=-n1/pI(Yj<-n1/p)+YjI(|Yj|≤n1/p)+n1/pI(Yj>n1/p),j≥1,Ỹnj=Ynj-EYnj,j≥1.
Hence, for for all ɛ>0, there exists an n0 such that
(3.7)n-1/pmax⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kEYnj|<ɛ4,n≥n0.
Denote
(3.8)Ynj*=n1/pI(Yj<-n1/p)-n1/pI(Yj>n1/p)+YjI(|Yj|>n1/p),j≥1.
Noting that Yj=Ynj*+Ynj,j≥1, we can find
(3.9)∑n=1∞nr-2P(max⁡1≤k≤n|Sk|>ɛn1/p)≤∑n=1∞nr-2P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kYnj*|>ɛn1/p2)+∑n=1∞nr-2P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kYnj|>ɛn1/p2)≤∑n=1∞nr-2P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kYnj*|>ɛn1/p2)+∑n=1∞nr-2P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kỸnj|>ɛn1/p4)+C+∑n=n0∞nr-2P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kEYnj|>ɛn1/p4)≤C+∑n=1∞nr-2P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kYnj*|>ɛn1/p2)+∑n=1∞nr-2P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kỸnj|>ɛn1/p4)=:C+I+J.
For I, by Markov's inequality, ∑i=1∞|ai|<∞,|Ynj*|≤|Yj|I(|Yj|>n1/p), Lemma 1.6 and EYrp<∞, it has
(3.10)I≤2ɛ∑n=1∞nr-2n-1/pE(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kYnj*|)≤C1∑n=1∞nr-2n-1/p∑i=1∞|ai|E(max⁡1≤k≤n∑j=i+1i+k|Yj|I(|Yj|>n1/p))≤C2∑n=1∞nr-1-1/pE[YI(Y>n1/p)]=C2∑n=1∞nr-1-1/p∑m=n∞E[YI(m<Yp≤m+1)]=C2∑m=1∞E[YI(m<Yp≤m+1)]∑n=1mnr-1-1/p≤C3∑m=1∞mr-1/pE[YI(m<Yp≤m+1)]≤CEYrp<∞.
Since fj(x)=-n1/pI(x<-n1/p)+xI(|x|≤n1/p)+n1/pI(x>n1/p) is a nondecreasing continuous function of x, we can find by using Lemma 1.3 that {Ỹnj,1≤j<∞} is a mean zero AANA sequence and EỸnj2≤EYnj2, Ynj2=Yj2I(|Yj|≤n1/p)+n2/pI(|Yj|>n1/p), j≥1. Consequently, by the property of AANA, Markov's inequality, Hölder's inequality, Lemma 1.5, Cr inequality, and Lemma 1.6, we can check that
(3.11)J≤(4ɛ)2∑n=1∞nr-2n-2/pE{max⁡1≤k≤n(∑i=1∞ai∑j=i+1i+kỸnj)2}≤(4ɛ)2∑n=1∞nr-2n-2/pE{∑i=1∞(|ai|1/2)(|ai|1/2max⁡1≤k≤n|∑j=i+1i+kỸnj|)}2≤(4ɛ)2∑n=1∞nr-2n-2/p(∑i=1∞|ai|)2sup⁡i≥1E{max⁡1≤k≤n(∑j=i+1i+kỸnj)2}≤C1∑n=1∞nr-2n-2/psup⁡i≥1∑j=i+1i+nEỸnj2≤C2∑n=1∞nr-2n-2/psup⁡i≥1∑j=i+1i+n{E[Yj2I(|Yj|≤n1/p)]+n2/pE[I(|Yj|>n1/p)]}≤C3∑n=1∞nr-1-2/pE[Y2I(Y≤n1/p)]+C4∑n=1∞nr-1P(Y>n1/p)≤C3∑n=1∞nr-1-2/pE[Y2I(Y≤n1/p)]+C4∑n=1∞nr-1-1/pE[YI(Y>n1/p)]=:C3J1+C4J2.
Since rp<2, it can be seen by EYrp<∞ that
(3.12)J1=∑n=1∞nr-1-2/p∑i=1nE[Y2I((i-1)1/p<Y≤i1/p)]=∑i=1∞E[Y2I((i-1)1/p<Y≤i1/p)]∑n=i∞nr-1-2/p≤C1∑i=1∞E[YrpY2-rpI((i-1)1/p<Y≤i1/p)]ir-2/p≤C2EYrp<∞.
By the proof of (3.10), we have J2≤CEYrp<∞. Therefore, (2.1) follows from (3.9), (3.10), (3.11), and (3.12).
Inspired by the proof of Theorem 12.1 of Gut [28], it can be checked that
(3.13)∑n=1∞nr-2P(sup⁡k≥n|Skk1/p|>22/pɛ)=∑m=1∞∑n=2m-12m-1nr-2P(sup⁡k≥n|Skk1/p|>22/pɛ)≤22-r∑m=1∞P(sup⁡k≥2m-1|Skk1/p|>22/pɛ)∑n=2m-12m-12m(r-2)≤22-r∑m=1∞2m(r-1)P(sup⁡k≥2m-1|Skk1/p|>22/pɛ)=22-r∑m=1∞2m(r-1)P(sup⁡l≥mmax⁡2l-1≤k<2l|Skk1/p|>22/pɛ)≤22-r∑m=1∞2m(r-1)∑l=m∞P(max⁡1≤k≤2l|Sk|>ɛ2(l+1)/p)=22-r∑l=1∞P(max⁡1≤k≤2l|Sk|>ɛ2(l+1)/p)∑m=1l2m(r-1)≤C1∑l=1∞2l(r-1)P(max⁡1≤k≤2l|Sk|>ɛ2(l+1)/p)=22-rC1∑l=1∞∑n=2l2l+1-12(l+1)(r-2)P(max⁡1≤k≤2l|Sk|>ɛ2(l+1)/p)≤22-rC1∑l=1∞∑n=2l2l+1-1nr-2P(max⁡1≤k≤n|Sk|>ɛn1/p)(sincer<2)≤22-rC1∑n=1∞nr-2P(max⁡1≤k≤n|Sk|>ɛn1/p).
Combining (2.1) with the inequality above, we obtain (2.2) immediately.Proof of Theorem2.2.
For allɛ>0, it has
(3.14)∑n=1∞nr-2-1/pE(max⁡1≤k≤n|Sk|-ɛn1/p)+=∑n=1∞nr-2-1/p∫0∞P(max⁡1≤k≤n|Sk|-ɛn1/p>t)dt=∑n=1∞nr-2-1/p∫0n1/pP(max⁡1≤k≤n|Sk|-ɛn1/p>t)dt+∑n=1∞nr-2-1/p∫n1/p∞P(max⁡1≤k≤n|Sk|-ɛn1/p>t)dt≤∑n=1∞nr-2P(max⁡1≤k≤n|Sk|>ɛn1/p)+∑n=1∞nr-2-1/p∫n1/p∞P(max⁡1≤k≤n|Sk|>t)dt.
By Theorem 2.1, in order to prove (2.3), we only have to prove that
(3.15)∑n=1∞nr-2-1/p∫n1/p∞P(max⁡1≤k≤n|Sk|>t)dt<∞.
For t>0, let
(3.16)Ytj=-tI(Yj<-t)+YjI(|Yj|≤t)+tI(Yj>t),j≥1,Ỹtj=Ytj-EYtj,j≥1,Ytj*=tI(Yj<-t)-tI(Yj>t)+YjI(|Yj|>t),j≥1.
Since Yj=Ytj*+Ytj, j≥1, it is easy to see that
(3.17)∑n=1∞nr-2-1/p∫n1/p∞P(max⁡1≤k≤n|Sk|>t)dt≤∑n=1∞nr-2-1/p∫n1/p∞P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kYtj*|>t2)dt+∑n=1∞nr-2-1/p∫n1/p∞P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kỸtj|>t4)dt+∑n=1∞nr-2-1/p∫n1/p∞P(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kEYtj|>t4)dt=:I1+I2+I3.
For I1, by Markov's inequality, |Ytj*|≤|Yj|I(|Yj|>t), Lemma 1.6 and EYrp<∞, we get that
(3.18)I1≤2∑n=1∞nr-2-1/p∫n1/p∞t-1E(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kYtj*|)dt≤2∑n=1∞nr-2-1/p∫n1/p∞t-1∑i=1∞|ai|E(max⁡1≤k≤n∑j=i+1i+k|Yj|I(|Yj|>t))dt≤C1∑n=1∞nr-1-1/p∫n1/p∞t-1E[YI(Y>t)]dt=C1∑n=1∞nr-1-1/p∑m=n∞∫m1/p(m+1)1/pt-1E[YI(Y>m1/p)]dt≤C2∑n=1∞nr-1-1/p∑m=n∞m1/p-1-1/pE[YI(Y>m1/p)]=C2∑m=1∞m-1E[YI(Y>m1/p)]∑n=1mnr-1-1/p≤C3∑m=1∞mr-1-1/pE[YI(Y>m1/p)]≤C4EYrp<∞.
From the fact that {Ỹtj,1≤j<∞} is a mean zero AANA sequence and EỸtj2≤EYtj, Ytj2=Yj2I(|Yj|≤t)+t2I(|Yj|>t),j≥1, similar to the proof of (3.11), we have
(3.19)I2≤42∑n=1∞nr-2-1/p∫n1/p∞t-2E(max⁡1≤k≤n(∑i=1∞ai∑j=i+1i+kỸtj)2)dt≤42∑n=1∞nr-2-1/p∫n1/p∞t-2(∑i=1∞|ai|)2sup⁡i≥1E{max⁡1≤k≤n(∑j=i+1i+kỸtj)2}dt≤C1∑n=1∞nr-2-1/p∫n1/p∞t-2sup⁡i≥1∑j=i+1i+nEỸtj2dt≤C2∑n=1∞nr-2-1/p∫n1/p∞t-2sup⁡i≥1∑j=i+1i+n{E[Yj2I(|Yj|≤t)]+t2E[I(|Yj|>t)]}dt≤C3∑n=1∞nr-1-1/p∫n1/p∞t-2E[Y2I(Y≤t)]dt+C4∑n=1∞nr-1-1/p∫n1/p∞P(Y>t)dt=:C3I21+C4I22.
It follows from rp<2 and EYrp<∞ that
(3.20)I21=∑n=1∞nr-1-1/p∑m=n∞∫m1/p(m+1)1/pt-2E[Y2I(Y≤t)]dt≤C1∑n=1∞nr-1-1/p∑m=n∞m1/p-1-2/pE[Y2I(Y≤(m+1)1/p)]=C1∑m=1∞m-1/p-1E[Y2I(Y≤(m+1)1/p)]∑n=1mnr-1-1/p≤C2∑m=1∞mr-1-2/pE[Y2I(Y≤(m+1)1/p)]=C2∑m=1∞mr-1-2/p∑i=1m+1E[Y2I((i-1)1/p<Y≤i1/p)]=C2∑m=1∞mr-1-2/pE[Y2I(m1/p<Y≤(m+1)1/p)]+C2∑m=1∞mr-1-2/p∑i=1mE[Y2I((i-1)1/p<Y≤i1/p)]=C2∑m=1∞mr-1-2/pE[YrpY2-rpI(m1/p<Y≤(m+1)1/p)]+C2∑i=1∞E[Y2I((i-1)1/p<Y≤i1/p)]∑m=i∞mr-1-2/p≤2(2-rp)/pC2∑m=1∞m-1E[YrpI(m1/p<Y≤(m+1)1/p)]+C3∑i=1∞E[YrpY2-rpI((i-1)1/p<Y≤i1/p)]ir-2/p≤C4EYrp<∞.
By the proof of (3.18), one has
(3.21)I22≤∑n=1∞nr-1-1/p∫n1/p∞t-1E[YI(Y>t)]dt≤CEYrp<∞.
On the other hand, by the property EYj=0, we have
(3.22)max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kEYtj|=max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+k{E[YjI(|Yj|≤t)]-tE[I(Yj<-t)]+tE[I(Yj>t)]}|=max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+k{E[YjI(|Yj|>t)]+tE[I(Yj<-t)]-tE[I(Yj>t)]}|≤2∑i=1∞|ai|∑j=i+1i+nE[|Yj|I(|Yj|>t)].
Thus, by Lemma 1.6 and the proof of (3.18), it can be seen that
(3.23)I3≤4∑n=1∞nr-2-1/p∫n1/p∞t-1(max⁡1≤k≤n|∑i=1∞ai∑j=i+1i+kEYtj|)dt≤8∑n=1∞nr-2-1/p∫n1/p∞t-1∑i=1∞|ai|∑j=i+1i+nE[|Yj|I(|Yj|>t)]dt≤C1∑n=1∞nr-1-1/p∫n1/p∞t-1E[YI(Y>t)]dt≤C2EYrp<∞.
Consequently, by (3.14), (3.17), (3.18), (3.19), (3.20), (3.21), and (3.23) and Theorem 2.1, (2.3) holds true.
Next, we prove (2.4). It is easy to see that
(3.24)∑n=1∞nr-2E(sup⁡k≥n|Skk1/p|-ɛ22/p)+=∑n=1∞nr-2∫0∞P(sup⁡k≥n|Skk1/p|>ɛ22/p+t)dt=∑m=1∞∑n=2m-12m-1nr-2∫0∞P(sup⁡k≥n|Skk1/p|>ɛ22/p+t)dt≤22-r∑m=1∞∫0∞P(sup⁡k≥2m-1|Skk1/p|>ɛ22/p+t)dt∑n=2m-12m-12m(r-2)≤22-r∑m=1∞2m(r-1)∫0∞P(sup⁡k≥2m-1|Skk1/p|>ɛ22/p+t)dt=22-r∑m=1∞2m(r-1)∫0∞P(sup⁡l≥mmax⁡2l-1≤k<2l|Skk1/p|>ɛ22/p+t)dt≤22-r∑m=1∞2m(r-1)∑l=m∞∫0∞P(max⁡1≤k≤2l|Sk|>(ɛ22/p+t)2(l-1)/p)dt=22-r∑l=1∞∫0∞P(max⁡1≤k≤2l|Sk|>(ɛ22/p+t)2(l-1)/p)dt∑m=1l2m(r-1)≤22-r∑l=1∞2l(r-1)∫0∞P(max⁡1≤k≤2l|Sk|>(ɛ22/p+t)2(l-1)/p)dt(lets=2(l-1)/pt)≤C1∑l=1∞2l(r-1-1/p)∫0∞P(max⁡1≤k≤2l|Sk|>ɛ2(l+1)/p+s)ds=2(2+1/p-r)C1∑l=1∞∑n=2l2l+1-12(l+1)(r-2-1/p)∫0∞P(max⁡1≤k≤2l|Sk|>ɛ2(l+1)/p+s)ds≤2(2+1/p-r)C1∑l=1∞∑n=2l2l+1-1nr-2-1/p∫0∞P(max⁡1≤k≤n|Sk|>ɛn1/p+s)ds(sincer<2)≤2(2+1/p-r)C1∑n=1∞nr-2-1/pE(max⁡1≤k≤n|Sk|-ɛn1/p)+<∞.
Therefore, (2.4) holds true following from (2.3).Proof of Theorem2.3.
Similar to the proof of Theorem2.1, by ∑i=1∞|ai|<∞ and EY<∞, ∑i=1∞aiYi+n converges almost surely. It can be seen that
(3.25)E(sup⁡n≥1|Snn1/r|p)=∫0∞P(sup⁡n≥1|Snn1/r|>t1/p)dt≤2p/r+∫2p/r∞P(sup⁡n≥1|Snn1/r|>t1/p)dt≤2p/r+∫2p/r∞P(sup⁡k≥1max⁡2k-1≤n<2k|Snn1/r|>t1/p)dt≤2p/r+∫2p/r∞∑k=1∞P(max⁡2k-1≤n<2k|Snn1/r|>t1/p)dt≤2p/r+∫2p/r∞∑k=1∞P(max⁡1≤n≤2k|Sn|>2(k-1)/rt1/p)dt(lets=2(k-1)p/rt)=2p/r+2p/r∑k=1∞2-kp/r∫2kp/r∞P(max⁡1≤n≤2k|Sn|>s1/p)ds.
For s1/p>0, let
(3.26)Ysj=-s1/pI(Yj<-s1/p)+YjI(|Yj|≤s1/p)+s1/pI(Yj>s1/p),j≥1,Ỹsj=Ysj-EYsj,j≥1,Ysj*=s1/pI(Yj<-s1/p)-s1/pI(Yj>s1/p)+YjI(|Yj|>s1/p),j≥1.
Since Yj=Ysj*+Ysj,j≥1, then
(3.27)∑k=1∞2-kp/r∫2kp/r∞P(max⁡1≤n≤2k|Sn|>s1/p)ds≤∑k=1∞2-kp/r∫2kp/r∞P(max⁡1≤n≤2k|∑i=1∞ai∑j=i+1i+nYsj*|>s1/p2)ds+∑k=1∞2-kp/r∫2kp/r∞P(max⁡1≤n≤2k|∑i=1∞ai∑j=i+1i+nỸsj|>s1/p4)ds+∑k=1∞2-kp/r∫2kp/r∞P(max⁡1≤n≤2k|∑i=1∞ai∑j=i+1i+nEYsj|>s1/p4)ds=:H1+H2+H3.
For H1, similar to (3.18), by Markov's inequality and Lemma 1.6, one has
(3.28)H1≤2∑k=1∞2-kp/r∫2kp/r∞s-1/pE(max⁡1≤n≤2k|∑i=1∞ai∑j=i+1i+nYsj*|)ds≤2∑k=1∞2-kp/r∫2kp/r∞s-1/p∑i=1∞|ai|E(max⁡1≤n≤2k∑j=i+1i+n|Yj|I(|Yj|>s1/p))ds≤C1∑k=1∞2k-kp/r∫2kp/r∞s-1/pE[YI(Y>s1/p)]ds=C1∑k=1∞2k-kp/r∑m=k∞∫2mp/r2(m+1)p/rs-1/pE[YI(Y>s1/p)]ds≤C2∑k=1∞2k-kp/r∑m=k∞2mp/r-m/rE[YI(Y>2m/r)]=C2∑m=1∞2mp/r-m/rE[YI(Y>2m/r)]∑k=1m2k-kp/r≤{C3∑m=1∞2m-m/rE[YI(Y>2m/r)],ifp<r,C4∑m=1∞m2m-m/rE[YI(Y>2m/r)],ifp=r,C5∑m=1∞2mp/r-m/rE[YI(Y>2m/r)],ifp>r.
For the case p<r, if 0<r<1, then
(3.29)∑m=1∞2m-m/rE[YI(Y>2m/r)]=∑m=1∞2m-m/r∑k=m∞E[YI(2k/r<Y≤2(k+1)/r)]=∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]∑m=1k2m(1-1/r)≤C1∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]≤C1EY.
If r=1, then
(3.30)∑m=1∞2m-m/rE[YI(Y>2m/r)]=∑m=1∞E[YI(Y>2m)]=∑m=1∞∑k=m∞E[YI(2k<Y≤2k+1)]=∑k=1∞E[YI(2k<Y≤2k+1)]∑m=1k1=∑k=1∞kE[YI(2k<Y≤2k+1)]≤C1∑k=1∞E[Ylog⁡(1+Y)I(2k<Y≤2k+1)]≤C1E[Ylog⁡(1+Y)].
Otherwise for r>1, it has
(3.31)∑m=1∞2m-m/rE[YI(Y>2m/r)]=∑m=1∞2m-m/r∑k=m∞E[YI(2k/r<Y≤2(k+1)/r)]=∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]∑m=1k2m-m/r≤C1∑k=1∞2k-k/rE[YI(2k/r<Y≤2(k+1)/r)]≤C1∑k=1∞E[YrI(2k/r<Y≤2(k+1)/r)]≤C1EYr.
Similarly, for the case p=r, if 0<r<1, then
(3.32)∑m=1∞m2m-m/rE[YI(Y>2m/r)]=∑m=1∞m2m-m/r∑k=m∞E[YI(2k/r<Y≤2(k+1)/r)]=∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]∑m=1km2m(1-1/r)≤∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]k∑m=1k2m(1-1/r)≤C1∑k=1∞kE[YI(2k/r<Y≤2(k+1)/r)]≤C2E[Ylog⁡(1+Y)].
If r=1, then
(3.33)∑m=1∞m2m-m/rE[YI(Y>2m/r)]=∑m=1∞m∑k=m∞E[YI(2k<Y≤2k+1)]=∑k=1∞E[YI(2k<Y≤2k+1)]∑m=1km≤C2∑k=1∞k2E[YI(2k<Y≤2k+1)]≤C2∑k=1∞E[Ylog⁡2(1+Y)I(2k<Y≤2k+1)]≤C2E[Ylog⁡2(1+Y)].
Otherwise, for r>1, it follows
(3.34)∑m=1∞m2m-m/rE[YI(Y>2m/r)]=∑m=1∞m2m-m/r∑k=m∞E[YI(2k/r<Y≤2(k+1)/r)]=∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]∑m=1km2m-m/r≤∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]k∑m=1k2m-m/r≤C1∑k=1∞k2k-k/rE[YI(2k/r<Y≤2(k+1)/r)]≤C2E[Yrlog⁡(1+Y)].
On the other hand, for the case p>r, if 0<p<1, then
(3.35)∑m=1∞2mp/r-m/rE[YI(Y>2m/r)]=∑m=1∞2m(p-1)/r∑k=m∞E[YI(2k/r<Y≤2(k+1)/r)]=∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]∑m=1k2m(p-1)/r≤C1∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]≤C1EY.
If p=1, then
(3.36)∑m=1∞2mp/r-m/rE[YI(Y>2m/r)]=∑m=1∞∑k=m∞E[YI(2k/r<Y≤2(k+1)/r)]=∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]∑m=1k1=∑k=1∞kE[YI(2k/r<Y≤2(k+1)/r)]≤C1EYlog⁡(1+Y).
For p>1, it has
(3.37)∑m=1∞2mp/r-m/rE[YI(Y>2m/r)]=∑m=1∞2m(p-1)/r∑k=m∞E[YI(2k/r<Y≤2(k+1)/r)]=∑k=1∞E[YI(2k/r<Y≤2(k+1)/r)]∑m=1k2m(p-1)/r≤C1∑k=1∞2k(p-1)/rE[YI(2k/r<Y≤2(k+1)/r)]≤C1∑k=1∞E[YpI(2k/r<Y≤2(k+1)/r)]≤C1EYp.
Consequently, by (3.28), the conditions of Theorem 2.3 and inequalities above, we obtain that
(3.38)H1≤{C1∑m=1∞2m-m/rE[YI(Y>2m/r)],ifp<r,C2∑m=1∞m2m-m/rE[YI(Y>2m/r)],ifp=r,C3∑m=1∞2mp/r-m/rE[YI(Y>2m/r)],ifp>r≤{forp<r,{C4EY<∞,if0<r<1,C5E[Ylog⁡(1+Y)]<∞,ifr=1,C6EYr<∞,ifr>1,forp=r,{C7E[Ylog⁡(1+Y)]<∞,if0<r<1,C8E[Ylog⁡2(1+Y)]<∞,ifr=1,C9E[Yrlog⁡(1+Y)]<∞,ifr>1,forp>r,{C10EY<∞,if0<p<1,C11E[Ylog⁡(1+Y)]<∞,ifp=1,C12EYp<∞,ifp>1.
Since{Ỹsj,1≤j<∞} is a mean zero AANA sequence and EỸsj2≤EYsj2, Ysj2=Yj2I(|Yj|≤s1/p)+s2/pI(|Yj|>s1/p), j≥1, similar to the proof of (3.11), we obtain that
(3.39)H2≤C1∑k=1∞2-kp/r∫2kp/r∞s-2/pE{max⁡1≤n≤2k(∑i=1∞ai∑j=i+1i+nỸsj)2}ds≤C1∑k=1∞2-kp/r∫2kp/r∞s-2/p(∑i=1∞|ai|)2sup⁡i≥1E{max⁡1≤n≤2k(∑j=i+1i+nỸsj)2}ds≤C2∑k=1∞2-kp/r∫2kp/r∞s-2/psup⁡i≥1∑j=i+1i+2kEỸsj2ds≤C3∑k=1∞2-kp/r+k∫2kp/r∞s-2/pE[Y2I(Y≤s1/p)]ds+C4∑k=1∞2-kp/r+k∫2kp/r∞P(Y>s1/p)ds=:C3H21+C4H22.
Similar to the proof of Theorem 1.1 of Chen and Gan [7], by p<2 and the conditions of Theorem 2.3, we have that
(3.40)H21=∑k=1∞2-kp/r+k∑m=k∞∫2mp/r2(m+1)p/rs-2/pE[Y2I(Y≤s1/p)]ds≤∑k=1∞2-kp/r+k∑m=k∞2mp/r-2m/rE[Y2I(Y≤2(m+1)/r)]=∑m=1∞2m(p-2)/rE[Y2I(Y≤2(m+1)/r)]∑k=1m2k(1-p/r)≤{C1∑m=1∞2m(r-2)/rE[Y2I(Y≤2(m+1)/r)],ifp<r,C2∑m=1∞m2m(r-2)/rE[Y2I(Y≤2(m+1)/r)],ifp=r,C3∑m=1∞2m(p-2)/rE[Y2I(Y≤2(m+1)/r)],ifp>r≤{C4EYr<∞,ifp<r,C5E[Yrlog⁡(1+Y)]<∞,ifp=r,C6EYp<∞,ifp>r.
On the other hand, by the proof of (3.28) and (3.38), it follows
(3.41)H22≤∑k=1∞2k-kp/r∫2kp/r∞s-1/pE[YI(Y>s1/p)]ds≤{forp<r,{C1EY<∞,if0<r<1,C2E[Ylog⁡(1+Y)]<∞,ifr=1,C3EYr<∞,ifr>1,forp=r,{C4E[Ylog⁡(1+Y)]<∞,if0<r<1,C5E[Ylog⁡2(1+Y)]<∞,ifr=1,C6E[Yrlog⁡(1+Y)]<∞,ifr>1,forp>r,{C7EY<∞,if0<p<1,C8E[Ylog⁡(1+Y)]<∞,ifp=1,C9EYp<∞,ifp>1.
Similar to the proof of (3.23), by the property EYj=0 and the proofs of (3.28) and (3.38), one has
(3.42)H3≤4∑k=1∞2-kp/r∫2kp/r∞s-1/p(max⁡1≤n≤2k|∑i=1∞ai∑j=i+1i+nEYsj|)ds≤8∑k=1∞2-kp/r∫2kp/r∞s-1/p(max⁡1≤n≤2k∑i=1∞|ai|∑j=i+1i+nE[|Yj|I(|Yj|>s1/p)])ds≤C1∑k=1∞2k-kp/r∫2kp/r∞s-1/pE[YI(Y>s1/p)]ds≤{forp<r,{C1EY<∞,if0<r<1,C2E[Ylog⁡(1+Y)]<∞,ifr=1,C3EYr<∞,ifr>1,forp=r,{C4E[Ylog⁡(1+Y)]<∞,if0<r<1,C5E[Ylog⁡2(1+Y)]<∞,ifr=1,C6E[Yrlog⁡(1+Y)]<∞,ifr>1,forp>r,{C7EY<∞,if0<p<1,C8E[Ylog⁡(1+Y)]<∞,ifp=1,C9EYp<∞,ifp>1.
Consequently, by (3.25), (3.27), (3.28), (3.38), (3.39), (3.40), (3.41), and (3.42), we finally obtain (2.6).Remark 3.1.
Zhou and Lin [17] obtained the result (2.6) for partial sums of moving average process under φ-mixing sequence. But there is one problem in their proof. On page 694 of Zhou and Lin [17], they presented that
(3.43)I1≤⋯≤{C∑m=1∞2m-m/rE[|Y1|I(|Y1|>2m/r)],ifp<r,C∑m=1∞m2m-m/rE[|Y1|I(|Y1|>2m/r)],ifp=r,C∑m=1∞2mp/r-m/rE[|Y1|I(|Y1|>2m/r)],ifp>r,≤{CE|Y1|r<∞,ifp<r,CE[|Y1|rlog⁡(1+|Y1|)]<∞,ifp=r,CE|Y1|p<∞,ifp>r,
where 1≤r<2 and p>0. For the case p<r, by taking r=1, we cannot get that
(3.44)∑m=1∞2m-m/rE[|Y1|I(|Y1|>2m/r)]=∑m=1∞E[|Y1|I(|Y1|>2m)]≤CE|Y1|<∞.
Here, we give a counter example to illustrate this problem. Assume that the density function of nonnegative random variable Y is
(3.45)f(y)=Cy2ln⁡2y,y>2,C=[∫2∞1y2ln⁡2ydy]-1.
Obviously, it can be found that
(3.46)EY=C∫2∞1yln⁡2ydy<∞.
But for r=1,
(3.47)∑m=1∞2m-m/rE[YI(Y>2m/r)]=∑m=1∞∑n=m∞E[YI(2n<Y≤2n+1)]=∑n=1∞E[YI(2n<Y≤2n+1)]∑m=1n1=C∑n=1∞n∫2n2n+11yln⁡2ydy=Cln⁡2∑n=1∞1n+1=∞.



---

*Source: 863931-2012-05-17.xml*