# Complete Convergence for Weighted Sums ofρ∗-Mixing Random Variables

Authors: Soo Hak Sung
Journal: Discrete Dynamics in Nature and Society
 (2010)

---

## Abstract

We obtain the complete convergence for weighted sums ofρ∗-mixing random variables. Our result extends the result of Peligrad and Gut (1999) on unweighted average to a weighted average under a mild condition of weights. Our result also generalizes and sharpens the result of An and Yuan (2008).


---

## Body

## 1. Introduction

In many stochastic models, the assumption that random variables are independent is not plausible. So it is of interest to extend the concept of independence to dependence cases. One of these dependence structures isρ*-mixing.Let{Xn,n≥1} be a sequence of random variables defined on a probability space (Ω,ℱ,P), and let ℱnm denote the σ-algebra generated by the random variables Xn,Xn+1,…,Xm. For any S⊂N, define ℱS=σ(Xi,i∈S). Given two σ-algebras 𝒜,ℬ in ℱ, put(1.1)ρ(𝒜,ℬ)=sup{corr(X,Y);X∈L2(𝒜),Y∈L2(ℬ)},
where corr(X,Y)=(EXY-EXEY)/var(X)var(Y). Define the ρ*-mixing coefficients by(1.2)ρn*=sup{ρ(ℱS,ℱT);S,T⊂Nwithdist(S,T)≥n}.
Obviously, 0≤ρn+1*≤ρn*≤ρ0*=1. The sequence {Xn,n≥1} is called ρ*-mixing (or ρ̃-mixing) if there exists k∈N such that ρk*<1. Note that if {Xn,n≥1} is a sequence of independent random variables, then ρn*=0 for all n≥1.A number of limit results forρ*-mixing sequences of random variables have been established by many authors. We refer to Bradley [1] for the central limit theorem, Bryc and Smoleński [2], Peligrad and Gut [3], and Utev and Peligrad [4] for moment inequalities, Gan [5], Kuczmaszewska [6], and Wu and Jiang [7] for almost sure convergence, and An and Yuan [8], Cai [9], Gan [5], Kuczmaszewska [10], Peligrad and Gut [3], and Zhu [11] for complete convergence.The concept of complete convergence of a sequence of random variables was introduced by Hsu and Robbins [12]. A sequence {Xn,n≥1} of random variables converges completely to the constant θ if(1.3)∑n=1∞P(|Xn-θ|>ϵ)<∞∀ϵ>0.
In view of the Borel-Cantelli lemma, this implies that Xn→θ almost surely. Therefore, the complete convergence is a very important tool in establishing almost sure convergence of summation of random variables as well as weighted sums of random variables. Hsu and Robbins [12] proved that the sequence of arithmetic means of independent and identically distributed random variables converges completely to the expected value if the variance of the summands is finite. Erdös [13] proved the converse. The result of Hsu-Robbins-Erdös is a fundamental theorem in probability theory and has been generalized and extended in several directions by many authors. One of the most important generalizations is Baum and Katz [14] strong law of large numbers.Theorem 1.1 (Baum and Katz [14]).
Letp≥1/α and 1/2<α≤1. Let {Xn,n≥1} be a sequence of independent and identically distributed random variables with EX1=0. Then the following statements are equivalent: (i)
E|X1|p<∞;(ii)
∑n=1∞npα-2P(max1≤j≤n|∑i=1jXi|>ϵnα)<∞ for all ϵ>0,
Peligrad and Gut [3] extended the result of Baum and Katz [14] to ρ*-mixing random variables.Theorem 1.2 (Peligrad and Gut [3]).
Letp>1/α and 1/2<α≤1. Let {Xn,n≥1} be a sequence of identically distributed ρ*-mixing random variables with EX1=0. Then the following statements are equivalent: (i)
E|X1|p<∞;(ii)
∑n=1∞npα-2P(max1≤j≤n|∑i=1jXi|>ϵnα)<∞ for all ϵ>0.
Cai [9] complemented Theorem 1.2 when p=1/α.
Recently, An and Yuan [8] obtained a complete convergence result for weighted sums of identically distributed ρ*-mixing random variables.Theorem 1.3 (An and Yuan [8]).
Letp>1/α and 1/2<α≤1. Let {Xn,n≥1} be a sequence of identically distributed ρ*-mixing random variables with EX1=0. Assume that {ani,1≤i≤n,n≥1} is an array of real numbers satisfying
(1.4)∑i=1n|ani|p=O(nδ)forsome0<δ<1,(1.5)#Ank=#{1≤i≤n:|ani|p>(k+1)-1}≥ne-1/k∀k≥1,n≥1.
Then the following statements are equivalent: (i)
E|X1|p<∞;(ii)
∑n=1∞npα-2P(max1≤j≤n|∑i=1janiXi|>ϵnα)<∞ for all ϵ>0.Note that the result of An and Yuan [8] is not an extension of Peligrad and Gut's [3] result, since condition (1.4) does not hold for the array with ani=1,1≤i≤n,n≥1. An and Yuan [8] proved the implication (i)⇒(ii) under condition (1.4), and proved the converse under conditions (1.4) and (1.5). However, the array satisfying both (1.4) and (1.5) does not exist. Noting that #Ank/(k+1)≤∑i=1n|ani|p≤O(nδ), we have that ne-1/k≤#Ank≤(k+1)O(nδ). But, this does not hold when k is fixed and n is large enough.In this paper, we obtain a new complete convergence result for weighted sums of identically distributedρ*-mixing random variables. Our result extends the result of Peligrad and Gut [3], and generalizes and sharpens the result of An and Yuan [8].Throughout this paper, the symbolC denotes a positive constant which is not necessarily the same one in each appearance, [x] denotes the integer part of x, and a⋀b=min{a,b}.


## 2. Main Result

To prove our main result, we need the following lemma which is a Rosenthal-type inequality forρ*-mixing random variables.Lemma 2.1 (Utev and Peligrad [4]).
Let{Xn,n≥1} be a sequence of ρ*-mixing random variables with EXn=0 and E|Xn|r<∞ for some r≥2 and all n≥1. Then there exists a constant D=D(r,k,ρk*) depending only on r,k, and ρk* such that for any n≥1,(2.1)E(max1≤j≤n|∑i=1jXi|r)≤D{∑i=1nE|Xi|r+(∑i=1nEXi2)r/2},
where ρk*<1.Now we state the main result of this paper.Theorem 2.2.
Letp>1/α and 1/2<α≤1. Let {Xn,n≥1} be a sequence of identically distributed ρ*-mixing random variables with EX1=0. Assume that {ani,1≤i≤n,n≥1} is an array of real numbers satisfying
(2.2)∑i=1n|ani|q=O(n)forsomeq>p.
If E|X1|p<∞, then
(2.3)∑n=1∞npα-2P(max1≤j≤n|∑i=1janiXi|>ϵnα)<∞∀ϵ>0.
Conversely, if (2.3) holds for any array {ani} satisfying (2.2), then E|X1|p<∞.To prove Theorem2.2, we first prove the following lemma which is the sufficiency of Theorem 2.2 when the array is bounded.Lemma 2.3.
Let{Xn,n≥1} be a sequence of identically distributed ρ*-mixing random variables with EX1=0 and E|X1|p<∞ for some p>1/α and 1/2<α≤1. Assume that {ani,1≤i≤n,n≥1} is an array of real numbers satisfying |ani|≤1 for 1≤i≤n and n≥1. Then (2.3) holds.Proof.
For1≤i≤n and n≥1, define Xni'=XiI(|Xi|≤nα). Since EXi=0 and ∑i=1n|ani|≤n, we have that
(2.4)n-αmax1≤j≤n|∑i=1janiEXni′|=n-αmax1≤j≤n|∑i=1janiEXiI(|Xi|>nα)|≤n-α∑i=1n|ani|E|X1|I(|X1|>nα)≤n1-αE|X1|I(|X1|>nα)≤n1-pαE|X1|pI(|X1|>nα)→0
as n→∞. Hence for n large enough, we have
(2.5)n-αmax1≤j≤n|∑i=1janiEXni′|<ϵ2.
It follows that
(2.6)∑n=1∞npα-2P(max1≤j≤n|∑i=1janiXi|>ϵnα)≤∑n=1∞npα-2∑i=1nP(|Xi|>nα)+∑n=1∞npα-2P(max1≤j≤n|∑i=1janiXni′|>ϵnα)≤∑n=1∞npα-1P(|X1|>nα)+C∑n=1∞npα-2P(max1≤j≤n|∑i=1jani(Xni′-EXni′)|>ϵnα2)=:I+CJ.
Noting that ∑n=1∞npα-1P(|X1|>nα)≤CE|X1|p<∞, we have I<∞. Thus, it remains to show that J<∞.
We have by Markov's inequality and Lemma2.1 that for any r≥2,(2.7)J≤(2ϵ)r∑n=1∞npα-rα-2Emax1≤j≤n|∑i=1jani(Xni′-EXni′)|r≤C∑n=1∞npα-rα-2{(∑i=1nani2E|Xni′|2)r/2+∑i=1n|ani|rE|Xni′|r}≤C∑n=1∞npα-rα-2+r/2(E|X1|2I(|X1|≤nα))r/2+C∑n=1∞npα-rα-1E|X1|rI(|X1|≤nα)=:CJ1+CJ2.
In the last inequality, we used the fact that |ani|≤1 for 1≤i≤n and n≥1.
Ifp≥2, then we take large enough r such that r>max{(pα-1)/(α-1/2),p}. Since r>(pα-1)/(α-1/2), we get
(2.8)J1≤C∑n=1∞npα-rα-2+r/2<∞.
Since r>p, we also get
(2.9)J2=∑n=1∞npα-rα-1∑i=1nE|X1|rI((i-1)α<|X1|≤iα)=∑i=1∞E|X1|rI((i-1)α<|X1|≤iα)∑n=i∞npα-rα-1≤C∑i=1∞E|X1|rI((i-1)α<|X1|≤iα)ipα-rα≤CE|X1|p<∞.
If p<2, then we take r=2. Since r>p, (2.9) still holds, and so J1=J2<∞.We next prove the sufficiency of Theorem2.2 when the array is unbounded.Lemma 2.4.
Let{Xn,n≥1} be a sequence of identically distributed ρ*-mixing random variables with EX1=0 and E|X1|p<∞ for some p>1/α and 1/2<α≤1. Assume that {ani,1≤i≤n,n≥1} is an array of real numbers satisfying ani=0 or |ani|>1, and
(2.10)∑i=1n|ani|q≤nforsomeq>p.
Then (2.3) holds.Proof.
Ifp<2, then we can take δ>0 such that p<p+δ<min{2,q}. Since ani=0 or |ani|>1, we have that ∑i=1n|ani|p+δ≤∑i=1n|ani|q≤n. Thus we may assume that (2.10) holds for some p<q<2 when p<2.
LetSnj′=∑i=1janiXiI(|aniXi|≤nα) for 1≤j≤n and n≥1. In view of EXi=0, we get
(2.11)n-αmax1≤j≤n|ESnj′|=n-αmax1≤j≤n|∑i=1janiEXiI(|aniXi|>nα)|≤n-α∑i=1nE|aniXi|I(|aniXi|>nα)≤n-pα∑i=1nE|aniXi|pI(|aniXi|>nα)≤n-pα∑i=1n|ani|pE|X1|p≤n-pα(∑i=1n|ani|q)p/qn1-p/qE|X1|p≤n1-pαE|X1|p→0,
since pα>1. Hence for n large enough, we have that n-αmax1≤j≤n|ESnj′|<ϵ/2. It follows that
(2.12)∑n=1∞npα-2P(max1≤j≤n|∑i=1janiXi|>ϵnα)≤∑n=1∞npα-2P(max1≤i≤n|aniXi|>nα)+∑n=1∞npα-2P(max1≤j≤n|Snj′|>ϵnα)≤∑n=1∞npα-2∑i=1nP(|aniXi|>nα)+C∑n=1∞npα-2P(max1≤j≤n|Snj′-ESnj′|>ϵnα2)=:I+CJ.
For 1≤j≤n-1 and n≥2, let
(2.13)Inj={1≤i≤n:n1/q(j+1)-1/q<|ani|≤n1/qj-1/q}.
Then {Inj,1≤j≤n-1} are disjoint, ⋃j=1n-1Inj={1≤i≤n:ani≠0}, and ∑j=1k#Inj≤k+1 for 1≤k≤n-1, since
(2.14)n≥∑{1≤i≤n:ani≠0}|ani|q=∑j=1n-1∑i∈Inj|ani|q≥n∑j=1k1j+1#Inj≥n(k+1)∑j=1k#Inj.
For convenience of notation, let t=1/(α-1/q). Since ani=0 or |ani|>1, and ∑i=1n|ani|q≤n, we have a11=0. It follows that
(2.15)I=∑n=2∞npα-2∑j=1n-1∑i∈InjP(|aniXi|>nα)≤∑n=2∞npα-2∑j=1n-1P(|X1|t>njt/q)#Inj≤∑n=2∞npα-2∑j=1n-1#Inj∑k≥[njt/q]P(k<|X1|t≤k+1)≤∑n=2∞npα-2∑k=n∞P(k<|X1|t≤k+1)∑j=1(n-1)⋀[((k+1)/n)q/t]#Inj≤∑n=2∞npα-2∑k=n∞P(k<|X1|t≤k+1)n⋀([(k+1n)q/t]+1)≤∑n=1∞npα-2∑k=n[n1+t/q]P(k<|X1|t≤k+1)([(k+1n)q/t]+1)+∑n=1∞npα-1∑k=[n1+t/q]+1∞P(k<|X1|t≤k+1)=:I1+I2.
Since pα-2-q/t=-α(q-p)-1<-1, we obtain
(2.16)I1≤C∑n=1∞npα-2-q/t∑k=n[n1+t/q]P(k<|X1|t≤k+1)kq/t≤C∑k=1∞P(k<|X1|t≤k+1)kq/t∑n=[kq/(q+t)]knpα-2-q/t≤C∑k=1∞P(k<|X1|t≤k+1)kq/t-qα(q-p)/(q+t)≤CE|X1|p<∞.
We also obtain
(2.17)I2≤∑k=1∞P(k<|X1|t≤k+1)∑n=1[kq/(t+q)]npα-1≤C∑k=1∞P(k<|X1|t≤k+1)kp(α-1/q)≤CE|X1|p<∞.
From I1<∞ and I2<∞, we have I<∞. Thus, it remains to show that J<∞.
We have by Markov's inequality and Lemma2.1 that for any r≥2,(2.18)J≤C∑n=1∞npα-rα-2Emax1≤j≤n|Snj′-ESnj′|r≤C∑n=1∞npα-rα-2(∑i=1nE|aniXi|2I(|aniXi|≤nα))r/2+C∑n=1∞npα-rα-2∑i=1nE|aniXi|rI(|aniXi|≤nα)=:J1+J2.
Observe that for r≥q and n>m,(2.19)n≥∑j=1n-1∑i∈Inj|ani|q≥n∑j=1n-11j+1#Inj≥n(m+1)r/q-1∑j=mn-1(j+1)-r/q#Inj.
So ∑j=mn-1j-r/q#Inj≤Cm-(r/q-1) for r≥q and n>m.
ForJ1 and J2, we proceed with two cases.
(i) Ifp≥2, then we take r large enough such that r>max{(pα-1)/(α-1/2),q}. Then we obtain that
(2.20)J1≤C∑n=1∞npα-rα-2(∑i=1n|ani|2)r/2≤C∑n=1∞npα-rα-2(∑i=1n|ani|q)r/2≤C∑n=1∞npα-rα-2+r/2<∞.
The second inequality follows by the fact that ani=0 or |ani|>1.
Noting thata11=0, we also obtain that
(2.21)J2=∑n=2∞npα-rα-2∑j=1n-1∑i∈InjE|aniXi|rI(|aniXi|≤nα)≤∑n=2∞npα-rα-2+r/q∑j=1n-1j-r/q#InjE|X1|rI(|X1|t≤n(j+1)t/q)≤∑n=2∞npα-rα-2+r/q∑j=1n-1j-r/q#Inj∑0≤k≤[n(j+1)t/q]E|X1|rI(k<|X1|t≤k+1)=∑n=2∞npα-rα-2+r/q∑j=1n-1j-r/q#Inj∑k=02nE|X1|rI(k<|X1|t≤k+1)+∑n=2∞npα-rα-2+r/q∑j=1n-1j-r/q#Inj∑k=2n+1[n(j+1)t/q]E|X1|rI(k<|X1|t≤k+1)=:J3+J4.
Since pα-rα-2+r/q<qα-rα-2+r/q=-(r-q)(α-1/q)-1<-1 and q>p, we have that
(2.22)J3=∑n=2∞npα-rα-2+r/q∑k=02nE|X1|rI(k<|X1|t≤k+1)∑j=1n-1j-r/q#Inj≤C∑n=2∞npα-rα-2+r/q∑k=02nE|X1|rI(k<|X1|t≤k+1)≤C∑k=1∞E|X1|rI(k<|X1|t≤k+1)∑n=[k/2]∞npα-rα-2+r/q≤C∑k=1∞E|X1|rI(k<|X1|t≤k+1)kpα-rα-1+r/q≤C∑k=1∞P(k<|X1|t≤k+1)kpα-1≤CE|X1|p<∞.
Since 1/t+1/q-α=0 and pα-2-q/t=-α(q-p)-1<-1, we also have that
(2.23)J4≤∑n=2∞npα-rα-2+r/q∑k=2n+1[n(q+t)/q]E|X1|rI(k<|X1|t≤k+1)∑j=[(k/n)q/t]-1n-1j-r/q#Inj≤C∑n=2∞npα-rα-2+r/q∑k=2n+1[n(q+t)/q]E|X1|rI(k<|X1|t≤k+1)([(kn)q/t]-1)-(r/q-1)≤C∑k=5∞E|X1|rI(k<|X1|t≤k+1)k-(r-q)/t∑n=[kq/(q+t)][k/2]npα-2-q/t≤C∑k=5∞E|X1|rI(k<|X1|t≤k+1)k-(r-q)/t-(α-1/q)(q-p)≤CE|X1|p<∞.
From J3<∞ and J4<∞, we have J2<∞.
(ii) Ifp<2, then we take r=2. As noted above, we may assume that p<q<2. Since r>q, as in the case p≥2, we have J1=J2≤CE|X1|p<∞.We now prove Theorem2.2 by using Lemmas 2.3 and 2.4.Proof of Theorem2.2.
Sufficiency.
Without loss of generality, we may assume that∑i=1n|ani|q≤n for some q>p. For n≥1, let
(2.24)An={1≤i≤n:|ani|≤1},Bn={1≤i≤n:|ani|>1},
and let ani′=ani if i∈An,ani′=0 otherwise, and ani′′=ani if i∈Bn,ani′′=0 otherwise. Then
(2.25)max1≤j≤n|∑i=1janiXi|≤max1≤j≤n|∑i=1jani′Xi|+max1≤j≤n|∑i=1jani′′Xi|.
It follows that
(2.26)∑n=1∞npα-2P(max1≤j≤n|∑i=1janiXi|>ϵnα)≤∑n=1∞npα-2P(max1≤j≤n|∑i=1jani′Xi|>ϵnα2)+∑n=1∞npα-2P(max1≤j≤n|∑i=1jani′′Xi|>ϵnα2)=:I+J.
By Lemma 2.3, we have I<∞. By Lemma 2.4, we have J<∞. Hence (2.3) holds.Necessity.
Choose, for eachn≥1,an1=⋯=ann=1. Then {ani} satisfies (2.2). By (2.3), we obtain that
(2.27)∑n=1∞npα-2P(max1≤j≤n|∑i=1jXi|>ϵnα)<∞∀ϵ>0,
which implies that
(2.28)∑n=1∞npα-2P(max1≤j≤n|Xj|>ϵnα)<∞∀ϵ>0.
Observe that
(2.29)∞>∑i=1∞∑n=2i-1+12inpα-2P(max1≤j≤n|Xj|>ϵnα)≥{∑i=1∞(2i-1)pα-22i-1P(max1≤j≤2i-1|Xj|>ϵ(2i)α)ifpα≥2,∑i=1∞(2i)pα-22i-1P(max1≤j≤2i-1|Xj|>ϵ(2i)α)if1<pα<2,≥{∑i=1∞P(max1≤j≤2i-1|Xj|>ϵ(2i)α)ifpα≥2,2pα-2∑i=1∞P(max1≤j≤2i-1|Xj|>ϵ(2i)α)if1<pα<2.
Hence we have that for any ϵ>0,P(max1≤j≤2i-1|Xj|>ϵ(2i)α)→0 as i→∞, and so P(max1≤j≤n|Xj|>nα)→0 as n→∞. The rest of the proof is same as that of Peligrad and Gut [3] and is omitted.Remark 2.5.
Takingani=1 for 1≤i≤n and n≥1, we can immediately get Theorem 1.2 from Theorem 2.2. If the array {ani} satisfies (1.4), then it satisfies (2.2): taking q such that p<q<p/δ, we have
(2.30)∑i=1n|ani|q≤max1≤i≤n|ani|q-p∑i=1n|ani|p≤Cnδ(q-p)/pnδ≤Cn.
So the implication (i)⇒(ii) of Theorem 1.3 follows from Theorem 2.2. As noted after Theorem 1.3, the implication (ii)⇒(i) of Theorem 1.3 is not true. Therefore, our result extends the result of Peligrad and Gut [3] to a weighted average, and generalizes and sharpens the result of An and Yuan [8].



---

*Source: 630608-2010-04-12.xml*