# Complete Convergence for Maximal Sums of Negatively Associated Random Variables

Authors: Victor M. Kruglov
Journal: Journal of Probability and Statistics
 (2010)

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## Abstract

Necessary and sufficient conditions are given for the complete convergence of maximal sums of identically distributed negatively associated random variables. The conditions are expressed in terms of integrability of random variables. Proofs are based on new maximal inequalities for sums of bounded negatively associated random variables.


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## Body

## 1. Introduction

The paper by Hsu and Robbins [1] initiated a great interest to the complete convergence of sums of independent random variables. Their research was continued by Erdös [2, 3], Spitzer [4], and Baum and Katz [5]. Kruglov et al. [6] proved two general theorems that provide sufficient conditions for the complete convergence for sums of arrays of row-wise independent random variables. In the paper of Kruglov and Volodin [7], a criterion was proved for the complete convergence of sums of independent identically distributed random variable in a rather general setting. Taylor et al. [8] and Chen et al. [9, 10] demonstrated that many known sufficient conditions for complete convergence of sums of independent random variables can be transformed to sufficient conditions for the complete convergence of sums of negatively associated random variables. Here we give necessary and sufficient conditions for the complete convergence of maximal sums of negatively associated identically distributed random variables. They resemble the criterions presented by Baum and Katz [5] and by Kruglov and Volodin [7] for the complete convergence of sums of independent identically distributed random variables. Theorems 2.3 and 2.5 are our main results. Theorem 2.3 is new even for independent random variables.In what follows we assume that all random variables under consideration are defined on a probability space(Ω,ℱ,P). We use standard notations, in particular, IA denotes the indicator function of a set A⊆Ω. Recall the notion of negatively associated random variables and some properties of such random variables.Definition 1.1.
Random variablesX1,…,Xn are called negatively associated if
(1.1)Cov(f(Xi1,…,Xik),g(Xj1,…,Xjm))≤0
for any pair of nonempty disjoint subsets A={i1,…,ik} and B={j1,…,jm},k+m≤n, of the set {1,…,n} and for any bounded coordinate-wise increasing real functions f(xi1,…,xik) and g(xj1,…,xjm),x1,…,xn∈ℝ=(-∞,∞). Random variables Xn,n∈ℕ={1,2,…,} are negatively associated if for any n∈ℕ random variables X1,…,Xn are negatively associated.
In this definition the coordinate-wise increasing functionsf and g may be replaced by coordinate-wise decreasing functions. Indeed, if f and g are coordinate-wise decreasing functions, then -f and -g are coordinate-wise increasing functions and the covariance (1.1) coincides with the covariance for -f and -g.Theorem A.
LetXn,n∈ℕ, be negatively associated random variables. Then for every  an,bn∈ℝ,an≤bn, the random variables  Yn=anI(-∞,an)(Xn)+XnI[an,bn](Xn)+bnI(bn,∞)(Xn),n∈ℕ, are negatively associated. For every  n∈ℕ and  x1,…,xn∈ℝ, the inequalities
(1.2)P{X1≤x1,…,Xn≤xn}≤∏k=1nP{Xk≤xk},P{X1≥x1,…,Xn≥xn}≤∏k=1nP{Xk≥xk}
hold.Proof.
It can be found in Taylor et al. [8].Theorem B.
LetXn be negatively associated random variables. Let  Xn*,n∈ℕ, be independent random variables such that  Xk and  Xk* are identically distributed for every  k=1,…,n. Then
(1.3)Eexp{X1+⋯+Xn}≤Eexp{X1*+⋯+Xn*}.
If E|Xk|p<∞ and EXk=0 for all k=1,…,n and for some p≥1, then
(1.4)E|max1≤r≤n∑k=1rXk|p≤2E|∑k=1n‍Xk*|p,n∈ℕ.Proof.
It can be found in Qi-Man Shao [11].


## 2. Main Results

Our basic theorems will be stated in terms of special functions. They were introduced in Kruglov and Volodin [7].Definition 2.1.
A nonnegative functionh(x),x∈[0,∞), belongs to the class ℋq for some q∈(0,2) if it is nondecreasing, is not equal to zero identically, and
(2.1)limsupx→∞h(2x)h(x)<∞,(2.2)limsupy→∞y2/q-1h(y)∫y∞h(x)x-2/qdx<∞.
The classℋq contains all nondecreasing nonnegative functions slowly varying at infinity which are not equal to zero identically, and in particular, lnβ(1+x) with β>0. The functions xα and xαlnβ(1+x) with α∈[0,2/q-1) and β>0 are also in ℋq.Remark 2.2.
If a nonnegative functionh(x),x∈[0,∞), is nondecreasing and satisfies condition (2.1), then
(2.3)h(x)≤Δxl
for all x greater than some x0≥1 and for some Δ>0 and l>0.Proof.
We may assume thath(x)>0 for all x greater than some x0≥1. From condition (2.1), it follows that supx≥x0h(2x)/h(x)=d<∞. Choose a number l>0 such that d≤2l. If x≥2x0, then 2nx0≤x<2n+1x0 for some n∈ℕ and h(x)≤h(2n+1x0)≤dn+1h(x0)≤2l(n+1)h(x0)≤2lxlh(x0). Inequality (2.3) holds for all x≥x0 with Δ=2lh(x0).Theorem 2.3.
LetXn,n∈ℕ, be negatively associated identically distributed random variables,  Sn=X1+⋯+Xn,0<q<2,r>1. Let  h(x),x∈[0,∞), be a function which is nondecreasing, is not equal to zero identically, and satisfies condition (2.1). Then the following conditions are equivalent:
(2.4)E(|X1|rqh(|X1|q))<∞,EX1=0,forq≥1,(2.5)∑n=1∞nr-2h(n)P{max1≤k≤n|Sk|>εn1/q}<∞,∀ε>0,(2.6)∑n=1∞nr-2h(n)P{supk≥n|k-1/qSk|>ε}<∞,∀ε>0.Corollary 2.4.
LetXn,n∈ℕ, be negatively associated identically distributed random variables,  Sn=X1+⋯+Xn,0<q<2,r>1,β≥0. The following conditions are equivalent:
(2.7)E(|X1|rq(ln(1+|X1|))βq)<∞,EX1=0,forq≥1,∑n=1∞nr-2(lnn)βP{max1≤k≤n|Sk|>εn1/q}<∞,∀ε>0,∑n=1∞nr-2(lnn)βP{supk≥n|k-1/qSk|>ε}<∞,∀ε>0.A part of Theorem2.3 can be generalized to a larger range of r,r≥1, under additional restrictions on functions h(x),x∈[0,∞).Theorem 2.5.
LetXn,n∈ℕ, be negatively associated identically distributed random variables,  Sn=X1+⋯+Xn,h∈ℋq,0<q<2,r≥1. Then the following conditions are equivalent:
(2.8)E(|X1|rqh(|X1|q))<∞,EX1=0,forq≥1,(2.9)∑n=1∞nr-2h(n)P{max1≤k≤n|Sk|>εn1/q}<∞,∀ε>0.Corollary 2.6.
LetXn,n∈ℕ, be negatively associated identically distributed random variables,  Sn=X1+⋯+Xn,0<q<2,β≥0. The following conditions are equivalent:
(2.10)E(|X1|q(ln(1+|X1|))βq)<∞,EX1=0,forq≥1,∑n=1∞(lnn)βnP{max1≤k≤n|Sk|>εn1/q}<∞,∀ε>0.Proof of Theorem2.3.
The theorem is obvious if the random variableX1 is degenerate, that is, P{X1=const.}=1. From now on we suppose that the random variable X1 is not degenerate. Denote
(2.11)δ={r-1r+lif0<rq<1,r-14(r+l)if1≤rq≤2,2-q4q(r+l)ifrq>2,
where l is the same as in (2.3). Define the function f(x),x∈ℝ,(2.12)f(x)={εn1/qδifx>εn1/qδ,xif|x|≤εn1/qδ,-εn1/qδifx<-εn1/qδ,
where ε>0 is a fixed number. By Theorem A the random variables Yk=f(Xk),k=1,…,n, are negatively associated. Put Sk′=Y1+⋯+Yk,k=1,…,n. Note that |Y1|≤|X1| and Y1→X1 as n→∞. If E|X1|rq<∞, then by the dominated convergence theorem we have
(2.13)limn→∞E|Y1|rq=E|X1|rq,limn→∞E|Y1-EY1|rq=E|X1-EX1|rqifrq≥1.
Prove that (2.4)⇒(2.5). Assume that 0<rq<1. The probability P{max1≤k≤n|Sk|>εn1/q} can be estimated as follows:
(2.14)P{max1≤k≤n|Sk|>εn1/q}=P{{max1≤k≤n|Sk′|>εn1/q}∩⋂k=1n{Xk=Yk}}+P{{max1≤k≤n|Sk|>εn1/q}∩(⋃k=1n{Xk≠Yk})}≤P{max1≤k≤n|Sk′|>εn1/q}+P{max1≤k≤n|Xk|>εn1/qδ}.
We intend to use Lemma 3.1 from the third part of the paper. Put γ=rq,x=εn1/q,c=εn1/q(r-1)/(r+l). From (2.13) it follows that the inequality E|Y1|γ<2E|X1|γ holds for all n∈ℕ greater than some n0. By inequality (3.1), we get, for all n>n0,(2.15)P{max1≤k≤n|Sk′|>εn1/q}≤2exp{xc-xcln(xcγ-1nE|Y1|γ+1)}≤2exp{r+lr-1-r+lr-1ln(εrq(r-1)rq-12E|X1|rq(r+l)rq-1nr-1+1)}≤2exp{r+lr-1}(εrq(r-1)rq-12E|X1|rq(r+l)rq-1)-(r+l)/(r-1)n-(r+l)=Cn-(r+l).
These inequalities and (2.3) imply that
(2.16)∑n=n0+1∞nr-2h(n)P{max1≤k≤n|Sk′|>εn1/q}≤ΔC∑n=n0+1∞nr+l-2n-(r+l)<∞.
Since P{max1≤k≤n|Xk|>εn1/qδ}≤nP{|X1|>εn1/qδ}, we obtain
(2.17)∑n=n0+1∞nr-2h(n)P{max1≤k≤n|Xk|>εn1/qδ}≤∑n=n0+1∞nr-1h(n)P{|X1|>εn1/qδ}≤E(|X1εδ|rqh(|X1εδ|q)).
The last inequality holds by Lemma 3.2. From (2.1) it follows that
(2.18)E(|X1εδ|rqh(|X1εδ|q))≤const.E(|X1|rqh(|X1|q)).
Condition (2.4) and inequalities (2.14)–(2.17) imply (2.5) for 0<rq<1.
Now assume thatrq≥1. First we consider the case 1≤rq≤2. Note that
(2.19)P{max1≤k≤n|Sk|>εn1/q}≤P{max1≤k≤n|Sk-ESn′|>εn1/q-max1≤k≤n|ESk′|}.
By Lemma3.4, we have max1≤k≤nE|Sk′|n-1/q→0 as n→∞, and hence
(2.20)P{max1≤k≤n|Sk|>εn1/q}≤P{max1≤k≤n|Sk-ESn′|>εn1/q2}
for all n∈ℕ greater than some n0′. The probability on the right-hand side can be estimated as follows:
(2.21)P{max1≤k≤n|Sk-ESk′|>εn1/q2}=P{{max1≤k≤n|Sk′-ESk′|>εn1/q2}∩⋂k=1n{Xk=Yk}}+P{{max1≤k≤n|Sk-ESk′|>εn1/q2}∩(⋃k=1n{Xk≠Yk})}≤P{max1≤k≤n|Sk′-ESk′|>εn1/q2}+P{max1≤k≤n|Xk|>εn1/qδ}.
In order to apply Lemma 3.1, put γ=rq,x=εn1/q/2,c=εn1/q(r-1)/(4(r+l)). From (2.13) it follows that the inequality E|Y1-EY1|γ<2E|X1-EX1|γ holds for all n∈ℕ greater than some n0′′≥n0′. By inequality (3.2) we get, for n>n0′′,(2.22)P{max1≤k≤n|Sk′-ESk′|>εn1/q2}≤4exp{x2γ-1c-x2cln(xcγ-1nE|Y1-EY1|γ+1)}≤4exp{2(r+l)r-1-r+lr-1ln(εrq(r-1)rq-12E|X1-EX1|rq(4(r+l))rq-1nr-1+1)}≤4exp{2(r+l)r-1}(εrq(r-1)rq-12E|X1-EX1|rq(4(r+l))rq-1)-(r+l)/(r-1)n-(r+l)=C1n-(r+l).
These inequalities and (2.3) imply
(2.23)∑n=n0′′+1∞nr-2h(n)P{max1≤k≤n|Sk′-ESk′|>εn1/q2}≤ΔC1∑n=n0′′+1∞nr+l-2n-(r+l)<∞.
From this inequality and (2.17) with n0′′ instead of n0, (2.20) and (2.21), it follows that (2.5) holds for 1≤rq≤2. The case rq>2 can be considered in the same way. In order to apply Lemma 3.1, put γ=2,x=εn1/q/2,c=εn1/q(2-q)/(4q(r+l)). From (2.13) it follows that the inequality E|Y1-EY1|2<2E|X1-EX1|2 holds for all n∈ℕ greater than some n0′′′≥n0′. By inequality (3.2), we get, for n>n0′′′,(2.24)P{max1≤k≤n|Sk′-ESk′|>εn1/q2}≤4exp{x2c-x2cln(xcnE|Y1-EY1|2+1)}≤4exp{q(r+l)2-q-q(r+l)2-qln(ε2(2-q)16E|X1-EX1|2q(r+l)n2/q-1+1)}≤4exp{q(r+l)2-q}(ε2(2-q)16E|X1-EX1|2q(r+l))-q(r+l)/(2-q)n-(r+l)=C2n-(r+l).
These inequalities and (2.3) imply
(2.25)∑n=n0′′′+1∞nr-2h(n)P{max1≤k≤n|Sk′-ESk′|>εn1/q2}≤ΔC2∑n=n0′′′+1∞nr+l-2n-(r+l)<∞.
From this inequality and (2.17) with n0′′′ instead of n0, (2.20) and (2.21), it follows that (2.5) holds for rq>2.
Prove that (2.5)⇒(2.6). Note that(2.26)∑n=3∞nr-2h(n)P{supk≥n|k-1/qSk|>ε}=∑j=1∞∑n=2j+12j+1nr-2h(n)P{supk≥n|k-1/qSk|>ε}≤∑j=1∞2(j+1)(r-1)h(2j+1)P{supk≥2j|k-1/qSk|>ε}≤∑j=1∞2(j+1)(r-1)h(2j+1)∑i=j∞P{max2i≤k≤2i+1|k-1/qSk|>ε}=∑i=1∞∑j=1i2(j+1)(r-1)h(2j+1)P{max2i≤k≤2i+1|k-1/qSk|>ε}≤2r-12r-1-1∑i=1∞2(i+1)(r-1)h(2i+1)P{max2i≤k≤2i+1|Sk|>ε2i/q}.
The last series can be estimated as follows:
(2.27)∑i=1∞2(i+1)(r-1)h(2i+1)P{max2i≤k≤2i+1|Sk|>ε2i/q}≤25r-3h(25)+max{1,22-r}∑i=1∞∑n=2i+1+12i+2nr-2h(n)P{max1≤k≤n|Sk|>ε2-2/qn1/q}≤25r-3h(25)+max{1,22-r}∑n=1∞nr-2h(n)P{max1≤k≤n|Sk|>ε2-2/qn1/q}.
Condition (2.5) and these inequalities imply (2.6).
Prove that (2.6)⇒(2.4). The sequence {P{supk≥n|k-1/qSk|>ε}}n≥1 decreases to zero for any ε>0. Indeed, if the sequence converges to a number a>0, then
(2.28)∞=a∑n=1∞nr-2h(n)≤∑n=1∞nr-2h(n)P{supk≥n|k-1/qSk|>ε}<∞.
Note that P{maxn≤k≤2n|Sk|>ε(2n)1/q}≤P{supk≥n|k-1/qSk|>ε}→0 as n→∞.
With the help of (1.2), we obtain
(2.29)P{maxn<k≤2nXk>n1/q}=1-P{maxn<k≤2nXk≤n1/q}≥1-∏k=n+12nP{Xk≤n1/q}=1-(1-P{X1>n1/q})n,P{maxn<k≤2n(-Xk)>n1/q}=1-P{maxn<k≤2n(-Xk)≤n1/q}≥1-∏k=n+12nP{Xk≥-n1/q}=1-(1-P{-X1>n1/q})n.
Denote an+=P{X1>n1/q} and an-=P{-X1>n1/q}. Note that P{|X1|>n1/q}=an++an- and 1-(1-an±)n≤P{maxn<k≤2n|Xk|>n1/q}≤P{maxn≤k≤2n|Sk|>n1/q/2}. Since P{maxn≤k≤2n|Sk|>n1/q/2}→0 and an±≤P{|X1|>n1/q}→0 as n→∞, then nln(1-an±)>-ln2 and 0≤an++an-<1 for all n∈ℕ greater than some n0. By the inequalities -ln(1-x)≥x for x∈[0,1) and 1-ex≥ex|x| for x≤0, we obtain
(2.30)nan±≤-nln(1-an±)≤2(1-enln(1-an±))=2(1-(1-an±)n)≤2P{max1≤k≤n|Xk|>n1/q}
for all n>n0, and
(2.31)∑n=n0+1∞nr-1h(n)P{|X1|>n1/q}=∑n=n0+1nr-1h(n)(an++an-)≤4∑n=n0+1∞nr-2h(n)P{maxn<k≤2n|Xk|>n1/q}≤4∑n=1∞nr-2h(n)P{maxn≤k≤2n|Sk|>n1/q2}≤4∑n=1∞nr-2h(n)P{maxn≤k≤2n|k-1/qSk|>2-1/q2}≤4∑n=1∞nr-2h(n)P{supk≥n|k-1/qSk|>2-1/q2}<∞.
It follows by Lemma 3.3 that E(|X1|rqh(|X1|q))<∞.
Now we will prove thata=EX1=0 provided that 1≤q<2. Assume that a≠0. Since n|a|≤|Sn-na|+|Sn|, then
(2.32)1=P{|Sn-na|+|Sn|>|a|n1/q2}≤P{|Sn-na|>|a|n1/q4}+P{|Sn|>|a|n1/q4},
and hence
(2.33)∞=∑n=1∞nr-2h(n)≤∑n=1∞nr-2h(n)P{|Sn-na|>|a|n1/q4}+∑n=1∞nr-2h(n)P{|Sn|>|a|n1/q4}.
But this contradicts the convergence of the series on the right-hand side of the inequality. The equality EX1=0 is proved.Proof of Theorem2.5.
Both theorems overlap. We need to consider only the caser=1. Prove that (2.8)⇒(2.9). Define random variables Zn=-εn1/qI(-∞,-εn1/q)(Xn)+XnI[-εn1/q,εn1/q](Xn)+εn1/qI(εn1/q,∞)(Xn),n∈ℕ. By Theorem A the random variables Zn,n∈ℕ, as well as Zn-EZn,n∈ℕ, are negatively associated. Denote Sk′′=(Z1-EZ1)+⋯+(Zk-EZk). By Theorem B there exist independent random variables X1*,…,Xn* such that the random variables Zk-EZk and Xk* are identically distributed for all k=1,…,n, and E(max1≤k≤n|Sk′′|2)≤2E(X1*+⋯+Xn*)2. Similarly to (2.21), we can prove that
(2.34)P{max1≤k≤n|Sk|>εn1/q}≤P{max1≤k≤n|Sk′′-ESk′′|>εn1/q2}+P{max1≤k≤n|Xk|>εn1/q}
for all n∈ℕ grater than some n0. In the same way as (2.17), one can prove that
(2.35)∑n=n0+1∞h(n)nP{max1≤k≤n|Xk|>εn1/q}≤∑n=n0+1∞h(n)P{|X1|>εn1/q}≤E(|X1ε|qh(|X1ε|q))<∞.
With the help of the Markov inequality, we obtain
(2.36)P{max1≤k≤n|Sk′′-ESk′′|>εn1/q2}≤4ε-2n-2/qE(max1≤k≤n|Sk′′-ESk′′|2)≤8ε-2n-2/qE(X1*+⋯+Xn*)2=8ε-2n1-2/qE|Z1-ZY1|2.
From (2.34) and (2.35), it follows that (2.9) holds if the series ∑n=1∞h(n)n-2/qE|Z1-EZ1|2 converges. This series can be estimated as follows:
(2.37)∑n=1∞h(n)n2/qE|Z1-EZ1|2≤∑n=1∞h(n)n2/qE|Z1|2=∑n=1∞h(n)n2/qE(X12I{|X1|≤εn1/q})+ε2∑k=1∞h(n)P{|X1|>εn1/q}.
Rewrite the first summand on the right-hand side in the following way:
(2.38)∑n=1∞h(n)n2/qE(|X1|2I{|X1|≤εn1/q})=∑n=1∞h(n)n2/q∑k=1nE(X12I{ε(k-1)1/q<|X1|≤εk1/q})=∑k=1∞∑n=k∞h(n)n2/qE(X12I{ε(k-1)1/q<|X1|≤εk1/q}).
From (2.1) and (2.2), it follows that there exist numbers k0∈ℕ,k0>2, and C>0 such that
(2.39)h(k)≤Ch(k2),∫k∞h(x)x-2/qdx≤Ch(k)k1-2/q,∀k≥k0.
For any k∈ℕ, we have
(2.40)∑n=k∞h(n)n2/q≤22/q∑n=k∞∫nn+1h(x)x2/qdx=22/q∫k∞h(x)x2/qdx<∞.
If k>k0, then
(2.41)∑n=k∞h(n)n2/q≤22/q∫k∞h(x)x2/qdx≤22/qCh(k)k1-2/q≤22/qC2h(k-1)(k-1)1-2/q.
With the help of these estimates, we get
(2.42)∑n=1∞h(n)n2/qE(|X1|2I{|X1|≤εn1/q})≤22/q∑k=1k0∫k∞h(x)x2/qdx+22/qC2∑k=k0+1∞h(k-1)(k-1)2/q-1E(X12I{ε(k-1)1/q<|X1|≤εk1/q}).
The last series can be estimated as follows:
(2.43)∑k=k0+1∞h(k-1)(k-1)2/q-1E(X12I{ε(k-1)1/q<|X1|≤εk1/q})≤ε222/q-1∑k=k0+1∞E(|ε-1X|qh(|ε-1X1|q)I{ε(k-1)1/q<|X1|≤εk1/q})≤ε222/q-1E(|ε-1X1|qh(|ε-1X1|q))<∞.
As a result we get that ∑n=1∞h(n)n-2/qE(|X1|2I{|X1|≤εn1/q})<∞. Taking account of (2.35) and (2.37), we see that ∑n=1∞h(n)n-2/qE|Z1-EZ1|2<∞.
Prove that (2.9)⇒(2.8) for r=1. Note that(2.44)∑n=1∞h(2n)P{max2n<k≤2n+1|Xk|>2n/q}≤∑n=1∞h(2n)P{max2n≤k≤2n+1|Sk|>2n/q2}≤2∑n=1∞∑j=2n+12n+2-1h(j)jP{max2n≤k≤j|Sk|>2-2/qj1/q2}≤2∑n=1∞h(n)nP{max1≤k≤n|Sk|>2-2/qn1/q2}<∞.
Denote bn+=P{X1>2n/q} and bn-=P{-X1>2n/q}. Note that P{|X1|>2n/q}=bn++bn-. With the help of (1.2), one can prove the inequality 1-(1-bn±)2n≤P{max2n+1<k≤2n+2|Xk|>2n/q}. Since P{max2n<k≤2n+1|Xk|>2n/q}→0 and bn±≤P{|X1|>2n/q}→0 as n→∞, then 2nln(1-bn±)>-ln2 and 0≤bn++bn-<1 for all n∈ℕ greater than some n0. By the inequalities -ln(1-x)≥x for x∈[0,1) and 1-ex≥ex|x| for x≤0, we obtain
(2.45)2nbn±≤-2nln(1-bn±)≤2(1-e2nln(1-bn±))=2(1-(1-bn±)2n)≤2P{max2n+1<k≤2n+2|Xk|>2n/q},∑n=1∞h(2n)(2n+1-2n)P{|X1|>2n/q}=∑n=1∞h(2n)2n(bn++bn-)≤4∑n=1∞h(2n)P{max2n<k≤2n+1|Xk|>2n/q}<∞.
By Lemma 3.3, we have that E(|X1|qh(|X1|q))<∞. In the same way as in the proof of the previous theorem, one can prove that EX1=0 if q≥1.


## 3. Auxiliary Results

LetX1,…,Xn be random variables. Denote Sk=X1+⋯+Xk for k=1,…,n and An,γ=E|X1|γ+⋯+E|Xn|γ,Bn,γ=E|X1-EX1|γ+⋯+E|Xn-EXn|γ for some γ∈(0,2],coshx=(ex+e-x)/2,x∈ℝ=(-∞,∞).Lemma 3.1.
If negatively associated random variablesX1,…,Xn are bounded by a constant  c>0, then
(3.1)P{max1≤k≤n|Sk|>x}≤2exp{xc-xcln(xcγ-1An,γ+1)},0<γ≤1,(3.2)P{max1≤k≤n|Sk-ESk|>x}≤4exp{x2γ-1c-x2cln(xcγ-1Bn,γ+1)},1≤γ≤2,
for any number x>0.Proof.
Prove Inequality (3.1). It is easily verified that {max1≤k≤n|Sk|>x}⊆{∑k=1nXk+>x}∪{∑k=1nXk->x} where Xk+=max{0,Xk} and Xk-=max{0,-Xk}. With the help of Markov inequality, we get
(3.3)P{max1≤k≤n|Sk|>x}≤P{∑k=1nXk+∣>x}+P{∑k=1nXk->x}≤e-hxEexp{h∑k=1nXk+}+e-hxEexp{h∑k=1nXk-}
for any h>0. By Theorem A, random variables Xk+,k=1,…,n, as well as Xk-,k=1,…,n, are negatively associated. By Theorem B, the inequality
(3.4)Eexp{h(X1±+⋯+Xn±)}≤Eexp{h((X1*)±+⋯+(Xn*)±)}
holds where random variables X1*,…,Xn* are independent, and for any k=1,…,n, random variables Xk and Xk* are identically distributed. It follows that
(3.5)P{max1≤k≤n|Sk|>x}≤e-hx∏k=1nEeh(Xk*)++e-hx∏k=1nEeh(Xk*)-=e-hx∏k=1nEehXk++e-hx∏k=1nEehXk-≤2e-hx∏k=1nEeh|Xk|.
Further we can proceed as in Prokhorov [12], Fuk and Nagaev [13], and Kruglov [14]. Assume that 0<γ≤1. The function (ehx-1)/xγ,0≤x≤c, increases and hence (ehx-1)/xγ≤(ehc-1)/cγ. With the help of this inequality, we obtain
(3.6)∏k=1nEeh|Xk|=∏k=1nE(1+|Xk|-γ(eh|Xk|-1)|Xk|γ)≤∏k=1n(1+c-γ(ehc-1)E|Xk|γ)=exp{c-γ(ehc-1)An,γ}.
From this inequality and from (3.5), it follows
(3.7)P{max1≤k≤n|Sk|>x}≤2exp{-hx+c-γ(ehc-1)An,γ}.
Put h=c-1ln(xcγ-1/An,γ+1). As a result we obtain (3.1).
Now we assume that1≤γ≤2. By the Markov inequality we get(3.8)P{max1≤k≤n|Sk-ESk|>x}≤Ecosh(hmax1≤k≤n|Sk-ESk|)cosh(hx)
for any h>0. By Theorem B, the inequality
(3.9)E(max1≤k≤n|Sk-ESk|)r≤2E|(X1*-EX1*)+⋯+(Xn*-EXn*)|r
holds for any r>1. Denote Sn*=X1*+⋯+Xn*. Note that cosh(|x|)=cosh(x) for any x∈ℝ. With these remarks, we obtain
(3.10)Ecosh(hmax1≤k≤n|Sk-ESk|)=1+∑r=1∞E(hmax1≤k≤n|Sk-ESk|2r)(2r)!≤1+2∑r=1∞E|h(Sn*-ESn*)|2r(2r)!≤2Ecosh(h|Sn*-ESn*|)=2Ecosh(h(Sn*-ESn*)),
and hence
(3.11)P{max1≤k≤n|Sk-ESk|>x}≤2Ecosh(h(Sn*-ESn*))cosh(hx)≤2e-hx(Eeh(Sn*-ESn*)+Ee-h(Sn*-ESn*)).
By the inequalities α≤eα-1 and eα-1-α≤coshα-1 for α∈ℝ, we obtain
(3.12)Eeh(Sn*-ESn*)=∏k=1nEeh(Xk*-EXk*)≤∏k=1nexp{E(eh(Xk*-EXk*)-1)}=exp{∑k=1nE(eh(Xk*-EXk*)-1-h(Xk*-EXk*))}≤exp{2∑k=1nE(cosh(h(Xk*-EXk*))-1)}.
Put f(α)=(coshα-1)|α|-γ for α≠0 and f(0)=1/2 if γ=2, and f(0)=0 if 1≤γ<2. It can be easily verified that the function f is continuous, even, and increases on (0,∞). Note that |Xk*-EXk*|≤2c for all k=1,…,n. With these remarks, we obtain
(3.13)cosh(h(Xk*-EXk*))-1=(cosh(h(Xk*-EXk*))-1)|h(Xk*-EXk*)|-γ|h(Xk*-EXk*)|γ≤(cosh(2hc)-1)(2c)-γE|Xk*-EXk*|γ,
and hence
(3.14)Eeh(Sn*-ESn*)≤exp{(cosh(2hc)-1)(2c)-γBn,γ}.
In the same way, one can prove the inequality
(3.15)Ee-h(Sn*-ESn*)≤exp{(cosh(2hc)-1)(2c)-γBn,γ}.
From these inequalities and from (3.11), it follows that
(3.16)P{max1≤k≤n|Sk-ESk|>x}≤4exp{-hx+(cosh(2hc)-1)(2c)-γBn,γ}≤4exp{-hx+2(e2hc-1)(2c)-γBn,γ}.
Put h=(2c)-1ln(xcγ-1/Bn,γ+1). As a result we obtain (3.2).Lemma 3.2.
Leth(x),x∈[0,∞), be a nondecreasing nonnegative function,  ξ be a nonnegative random variable,  r≥1,andq>0. Then
(3.17)∑n=1∞nr-1h(n)P{ξ>n1/q}≤E(ξrqh(ξq)).Proof.
It can be found in Kruglov and Volodin [7].Lemma 3.3.
Leth(x),x∈[0,∞), be a nondecreasing nonnegative function possessing property  (2.1),  ξ  be a nonnegative random variable,  r≥1,q>0, and  {bn}n≥1 an unbounded nondecreasing sequence of positive numbers such that  bn+1≤bbn for all  n∈ℕ and for some number  b>0,b0=0. Then there exist numbers  k0∈ℕ and  d>0 such that
(3.18)dE(ξrqh(ξq)(ξq))I{ξ≥bk0})≤∑n=1∞bnr-1h(bn)(bn-bn-1)P{ξ>bn1/q}.Proof.
It can be found in Kruglov and Volodin [7].The next lemma was proved in Kruglov and Volodin [7] under an additional restriction.Lemma 3.4.
LetXn,n∈ℕ, be identically distributed random variables such that  E|X1|q<∞ for some  q∈(0,2) and  EX1=0 if  1≤q<2. Define the function  f(x)=-ɛn1/qI(-∞,-ɛn1/q)(x)+xI[-ɛn1/q,ɛn1/q](x)+ɛn1/qI(ɛn1/q,∞)(x),x∈ℝ, where  ɛ>0 is a fixed number. Then
(3.19)limn→∞1n1/qmax1≤k≤n|∑j=1kEf(Xj)|=0.Proof.
Note that(3.20)max1≤k≤n|∑j=1kEf(Xj)|=n|Ef(X1)|≤n|E(X1I{|X1|≤ɛn1/q})|+ɛn1/qP{|X1|>ɛn1/q},limn→∞1n1/qn1/q+1P{|X1|>ɛn1/q}=limn→∞nP{|X1|q>ɛqn}=0.
It suffices to prove that
(3.21)limn→∞nn1/qE(X1I{|X1|≤ɛn1/q})=0.
Suppose that 0<q<1. For any δ>0, there exist n0∈ℕ such that E(|X1|qI{|X1|>ɛn1/q})<δ. We get, for any n≥n0,(3.22)nn1/qE(|X1|I{|X1|≤ɛn1/q})≤nn1/qE(|X1|I{|X1|≤ɛn01/q})+ɛ1-qE(|X1|qI{ɛn01/q<|X1|≤ɛn1/q}),
and hence
(3.23)limsupn→∞nn1/qE(|X1|I{|X1|≤n1/q})≤ɛ1-qδ.
This implies (3.21), since δ>0 can be chosen arbitrarily small.
If1≤q<2 and EX1=0, then we get(3.24)nn1/q|E(X1I{|X1|≤ɛn1/q})|=nn1/q|E(X1I{|X1|>ɛn1/q})|≤ɛ1-qE(|X1|qI{|X1|>ɛn1/q})→0.



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*Source: 764043-2010-06-24.xml*