# Equivalent Conditions of Completepth Moment Convergence for Weighted Sums of I. I. D. Random Variables under Sublinear Expectations

Authors: Mingzhou Xu; Kun Cheng
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

We investigate the completepth moment convergence for weighted sums of independent, identically distributed random variables under sublinear expectations space. Using moment inequality and truncation methods, we prove the equivalent conditions of complete pth moment convergence of weighted sums of independent, identically distributed random variables under sublinear expectations space, which complement the corresponding results obtained in Guo and Shan (2020).


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## Body

## 1. Introduction

Peng [1, 2] presented the concept of the sublinear expectation space to study the uncertainty of probability and distribution. The seminal work of Peng [1, 2] encourages people to study limit theorems under sublinear expectations space. Zhang [3–5] proved results including exponential inequalities, Rosenthal’s inequalities, and Donsker’s invariance principle under sublinear expectations. Wu [6] obtained precise asymptotics for complete integral convergence. Xu and Cheng [7] studied precise asymptotics in the law of iterated logarithm under sublinear expectations. The interested reader could refer to Xu and Zhang [8, 9], Chen [10], Gao and Xu [11], Fang et al. [12], Hu et al. [13], Hu and Yang [14], Huang and Wu [15], Kuczmaszewska [16], Ma and Wu [17], Wang and Wu [18], Wu and Jiang [19], Yu and Wu [20], Zhang [21], Zhong and Wu [22], and references therein for more limit theorems under sublinear expectations.Recently, Guo and Shan [23] studied equivalent conditions of complete qth moment convergence for weighted sums of sequences of negatively orthant dependent random variables. Xu and Cheng [24] obtained equivalent conditions of complete convergence for weighted sums of sequences of i. i. d. random variables under sublinear expectations. Motivated by the work of Guo and Shan [23] and Xu and Cheng [24], here we try to prove the equivalent conditions of complete pth moment convergence of weighted sums of independent, identically distributed random variables under sublinear expectations space, which complement the corresponding results obtained in Guo and Shan [23], also extend results in Xu and Cheng [24] from complete convergence to complete pth moment convergence.We organized the rest of this paper as follows. In Section2, we give necessary basic notions, concepts, and relevant properties and present necessary lemmas under sublinear expectations. In Section 3, we give our main results, Theorems 1–4, whose proofs are presented in Section 4.


## 2. Preliminaries

As in Xu and Cheng [24], we adopt similar notations as in the work by Peng [2] and Chen [10]. Suppose that Ω,ℱ is a given measurable space. We assume that ℋ is a subset of all random variables on Ω,ℱ such that IA∈ℋ (cf. [10]), where A∈ℱ, and X1,…,Xn∈ℋ implies φX1,…,Xn∈ℋ, for each φ∈Cl,Lipℝn, where Cl,Lipℝn denotes the linear space of (local Lipschitz) function φ satisfying(1)φx−φy≤C1+xm+ymx−y,∀x,y∈ℝn,for some C>0 and m∈ℕ depending on φ.Definition 1.
A sublinear expectationE on ℋ is a functional E:ℋ↦ℝ¯≔−∞,∞ satisfying the following properties: for all X,Y∈ℋ, we have(a)
Monotonicity: ifX≥Y, then EX≥EY(b)
Constant preserving:Ec=c, ∀c∈ℝ(c)
Positive homogeneity:EλX=λEX, ∀λ≥0(d)
Subadditivity:EX+Y≤EX+EY whenever EX+EY is not of the form ∞−∞ or −∞+∞A set functionV:ℱ↦0,1 is said to be a capacity if it obeys(a)
V∅=0 and VΩ=1.(b)
VA≤VB, A⊂B and A,B∈ℱ.Moreover, ifV is continuous, then V should satisfy(c)
VAn↑VA if An↑A.(d)
VAn↓VA if An↓A.A capacityV is called subadditive if VA+B≤VA+VB, A,B∈ℱ.In this paper, given a sublinear expectation spaceΩ,ℋ,E, set VA≔infEξ:IA≤ξ,ξ∈ℋ=EIA, ∀A∈ℱ (see (8) and the definitions of V above (8) in [4]). Clearly, V is a subadditive capacity. Denote the Choquet expectations CV by(2)CVX≔∫0∞VX>xdx+∫−∞0VX>x−1dx.Suppose thatX=X1,…,Xm, Xi∈ℋ, and Y=Y1,…,Yn, Yi∈ℋ, are two random vectors on Ω,ℋ,E. Y is called to be independent of X if, for each Borel-measurable function ψ on ℝm×ℝn with ψX,Y,ψx,Y∈ℋ; for each x∈ℝm, we have EψX,Y=EEψx,Y|x=X whenever ψ¯x≔Eψx,Y<∞ for each x and Eψ¯X<∞ (see Definition 2.5 in [10]). Xnn=1∞ is called a sequence of independent random variables if Xn+1 is independent of X1,…,Xn, for each n≥1.Assume thatX1 and X2 are two n-dimensional random vectors defined, respectively, in sublinear expectation spaces Ω1,ℋ1,E1 and Ω2,ℋ2,E2. They are called identically distributed if, for every Borel-measurable function ψ such that ψX1∈ℋ1,ψX2∈ℋ2,(3)E1ψX1=E2ψX2,whenever the sublinear expectations are finite. Xnn=1∞ is called to be identically distributed if, for each i≥1, Xi and X1 are identically distributed.In the sequel, we suppose thatE is countably subadditive, i.e., EX≤∑n=1∞EXn, whenever X≤∑n=1∞Xn, X,Xn∈ℋ, and X≥0, Xn≥0, n=1,2,…. Let C denote a positive constant which may differ from line to line. IA or IA represents the indicator function of A, an≪bn means that there exists a constant C>0 such that an≤Cbn for n large sufficiently, and an≈bn means that an≪bn and bn≪an. We use logx for lnmaxe,x.We first present several necessary lemmas to prove our main results. By using Corollary 2.2, Theorem 2.3 in [5], the proofs of Theorem 3 in [25], and Minkowski’s inequality under sublinear expectations, we see that the following lemma holds.Lemma 1 (cf. Lemma 2.3 in [24]).
Suppose thatXi,1≤i≤n is a sequence of independent random variables under sublinear expectation space Ω,ℋ,E with EXi≤0, EXiM<∞, 1≤i≤n, and M≥2. Then,(4)Emax1≤j≤n∑i=1jXiM≤ClogMn∑i=1nEXiM+∑i=1nEXi2M/2,where C depends on M only.Proof.
For readers’ convenience, we give complete proofs here. First, forb≥0 and n≥1, set Sb,n=∑k=b+1b+nXi and Lb,n=max1≤k≤nSb,k, and(5)gFb,n=CM∑i=b+1b+nEXiM+∑i=b+1b+nEXi2M/2,where CM depends on M determined as in (2.8) of Corollary 2.2 (b) in [5]. By Corollary 2.2 in [5], we know that, for all b≥0 and n≥1,(6)ESb,nM≤gFb,n.
Obviously, forb≥0 and 1≤k≤k+l,(7)gFb,k+gFb+k,l≤gFb,k+l.
SetΛ1=1, and for n≥2, Λn=1+Λm−1, where m is the integer part of 1/2n+2. 1+log2m−1≤log2n implies that Λn≤log2n. Now, it is enough to prove that, for b≥0 and n≥1,(8)ELb,nM≤ΛnMgFb,n.
As in the proof of Theorem 4 of Móricz [25], let n>1 be given. Obviously, n=2m−1 or 2m−2. For b≥0 and m≤k≤n, we see that(9)Sb,k≤Sb,m+Sb+m,k−m,whence, for such k’s,(10)Sb,k≤Sb,m+Lb+m,n−m.
Since, for1≤k<m, we have Sb,k≤Lb,m−1, hence, for 1≤k≤n,(11)Sb,k≤Sb,m+Lb,m−1M+Lb+m,n−mM1/M.
Thus,(12)Lb,n≤Sb,m+Lb,m−1M+Lb+m,n−mM1/M,and by Minkowski’s inequality under sublinear expectations (see (4.10) in Proposition 4.2 of Chapter I of [2]),(13)ELb,nM1/M≤ESb,mM1/M+ELb,m−1M+ELb+m,n−mM1/M.
Suppose now that (8) holds for k<n. Then, by the choice of m, we see that(14)ELb,m−1M≤ΛMm−1gFb,m−1,ELb+m,n−mM≤ΛMn−mgFb+m,n−m≤ΛMm−1gFb+m,n−m.
By these two inequalities above and (7), we see that(15)ELb,m−1M+ELb+m,n−mM≤ΛMm−1gFb,n.
Finally, (6) implies(16)ESb,mM≤gFb,m≤gFb,n.
By (13)–(16), we conclude that(17)ELb,nM1/M≤1+Λm−1g1/MFb,n=Λng1/MFb,n,which implies the result. Hence, by (6), the conclusion of (8) is true for n=1. By induction, (8) holds for all n=1,2,…. The proof is complete.Lemma 2 (see Lemma 2.4 in [24, 26]).
LetXn;n≥1 be a sequence of independent random variables under sublinear expectation space Ω,ℋ,E. Then, for all n≥1 and x>0,(18)1−Vmax1≤j≤nXj>x2∑j=1nVXj>x≤4Vmax1≤j≤nXj>x.Remark 1.
In the proofs of Lemma 2.4 in [24, 26], by independence of IAk≔IXk>x,k=1,…,n, EX+c=Ec+X=EX+c for constant c and X∈ℋ (Definition 2.5 in [10]), we see that(19)∑k=1nEIAk=∑k=1n−2EIAk+EIAn−1+EIAn=∑k=1n−2EIAk+Ex+EIAn|x=IAn−1=∑k=1n−2EIAk+EEx+IAn|x=IAn−1=∑k=1n−2EIAk+EIAn−1+IAn=⋯=EIA1+E∑k=2nIAk=E∑k=1nIAk,which implies that Lemma 2 is valid. The difference between the sublinear expectations and linear expectations could be implied by the subtle observation that(20)EIA1−IA2=EEx−IA2|x=IA1=Ex+E−IA2|x=IA1=EIA1+E−IA2,which is not necessarily equal to EIA1−EIA2.Lemma 3.
LetX be a random variable under sublinear expectation space Ω,ℋ,E, q>0, r>0, and p>0. Then, the following are equivalent:(i)
(21)CVXp<∞,forp>rq,CVXr/qlnX<∞,forp=rq,CVXr/q<∞,forp<rq.(ii)
(22)∫1∞dy∫1∞yr−1VX>x1/pyqdx<∞.Proof.
(23)∫1∞dy∫1∞yr−1VX>x1/pyqdx=∫1∞dt∫tq∞tr−1VX>ssp−1t−qpdsSettings=x1/pyq,t=y≈∫1∞VX>ssp−1ds≈∫1∞VX>ssp−1ds≈CVXp,forp>rq,∫1∞VX>ssr/q−1lnsds≈CVXr/qlnX,forp=rq,∫1∞VX>ssp−1+r−pq/qds≈CVXr/q,forp<rq.
The proof is finished.Lemma 4.
LetX be a random variable under sublinear expectation space Ω,ℋ,E, q>0, r>0, and p>0. Then, the following is equivalent:(i)
(24)CVXp<∞,forp>rq,CVXr/qln2X<∞,forp=rq,CVXr/qlnX<∞,forp<rq.(ii)
(25)∫1∞dy∫1∞yr−1lnyVX>x1/pyqdx<∞.Proof.
(26)∫1∞dy∫1∞yr−1lnyVX>x1/pyqdx≈∫1∞dt∫tq∞tr−1lntVX>ssp−1t−qpdsSettings=x1/pyq,t=y≈∫1∞VX>ssp−1ds∫1s1/qtr−1−qplntdt≈∫1∞VX>ssp−1ds≈CVXp,forp>rq,∫1∞VX>ssr/q−1ln2sds≈CVXr/qln2X,forp=rq,∫1∞VX>ssp−1+r−pq/qlnsds≈CVXr/qlnX,forp<rq.
This finishes the proof.


## 3. Main Results

We state our main results, the proofs of which will be given in Section4.Theorem 1.
LetXn,n≥1 be a sequence of independent random variables, identically distributed as X under sublinear expectation space Ω,ℋ,E. Assume that r>1, q>1/2, and β>−q/r, and suppose that EX=−E−X=0 for 1/2<q≤1. Suppose that ani≈i/nβ1/nq,1≤i≤n,n≥1 is a triangular array of real numbers. Then, the following is equivalent:(i)
(27)CVXp<∞,forp>rq,CVXr/qlnX<∞,forp=rq,CVXr/q<∞,forp<rq.(ii)
(28)∑n=1∞nr−2CVmax1≤k≤n∑i=1kaniXip−ɛ+<∞,∀ɛ>0.Remark 2.
IfΩ,ℋ,E is classic probability space, then Theorem 1 recovers Theorem 7 in [23] in case in which Xn,n≥1 is a sequence of independent random variables, identically distributed as X. As pointed in Hossein and Nezakati [27], why we need pth moment convergence under sublinear expectations, the complete moment covergence is a more general expression than complete convergence in theory and practice, and the interested reader also could refer to the first study in complete moment convergence by Chow [28].Remark 3.
Under the same conditions in Theorem1, if (28) holds, by the same proof of Remark 3.2 of Hossein and Nezakati [27], we see that, for all ϵ>0,(29)∑n=1∞nr−2Vmax1≤k≤n∑i=1kaniXip>∈<∞.
Moreover, ifV is continuous, by (29), we see that(30)∑i=1naniXi⟶0,a.s.V,asn⟶∞.
In (30), if 1/2<q≤1, β=0, we conclude that ∑i=1nXi/nq⟶0a.s.V,asn⟶∞, which is similar to Theorem 1 of Zhang and Lin [29].Theorem 2.
LetXn,n≥1 be a sequence of independent random variables, identically distributed as X under sublinear expectation space Ω,ℋ,E. Assume that r>1, q>1/2, and β=−q/r<0, and suppose that EX=−E−X=0 for 1/2<q≤1. Suppose that ani≈i/nβ1/nq,1≤i≤n,n≥1 is a triangular array of real numbers. Then, (28) is equivalent to(31)CVXp<∞,forp>rq,CVXr/qln2X<∞,forp=rq,CVXr/qlnX<∞,forp<rq.Theorem 3.
LetXn,n≥1 be a sequence of independent random variables, identically distributed as X under sublinear expectation space Ω,ℋ,E. Assume that r>1, q>1/2, and −q<β<−q/r<0, and suppose that EX=−E−X=0 for 1/2<q≤1. Suppose that ani≈i/nβ1/nq,1≤i≤n,n≥1 is a triangular array of real numbers. Then, (28) is equivalent to(32)CVXp<∞,forp>r−1q+β,CVXr−1/q+βlnX<∞,forp=r−1q+β,CVXr−1/q+β<∞,forp<r−1q+β.As in the proofs of Theorems1–3, we can get the following corollary.Corollary 1.
LetXn,n≥1 be a sequence of independent random variables, identically distributed as X under sublinear expectation space Ω,ℋ,E. Assume that r>1, q>1/2, and β>−q, and suppose that EX=−E−X=0, for 1/2<q≤1. Suppose that ani≈n−i/nβ1/nq,0≤i≤n−1,n≥1 is a triangular array of real numbers. Then,(i)
(27) is equivalent to(33)∑n=1∞nr−2Emax0≤k≤n−1∑i=0kaniXip−ϵ+<∞,β>−qr.(ii)
(31) is equivalent to (33) when β=−q/r.(iii)
(32) is equivalent to (33) when −q<β<−q/r.The following theorem is completepth moment convergence on Cesàro summation of independent, identically distributed random variables underΩ,ℋ,E.Theorem 4.
LetXn,n≥1 be a sequence of independent random variables, identically distributed as X under sublinear expectation space Ω,ℋ,E. Assume that r>1, q>1/2, 0<α≤1, and p>0, and suppose that EX=−E−X=0, for 1/2<q≤1. Suppose Anα=α+1α+2,…,α+n/n!, n=1,2,… and A0α=1. Then,(i)
(27) is equivalent to(34)∑n=1∞nr−2Anα−pEmax0≤k≤n−1∑i=0kAn−iα−1Xip−ϵAnαp+<∞,∀ϵ>0,when1−1/r<α≤1.(ii)
(31) is equivalent to (34) when α=1−1/r.(iii)
(34) is equivalent to(35)CVXp<∞,forp>r−1qα,CVXr−1/qαlnX<∞,forp=r−1qα,CVXr−1/qα<∞,forp<r−1qα,when 0<α<1−1/r.Remark 4.
IfΩ,ℋ,E is classic probability space, then Theorems 1–3 and Corollary 1 recover, respectively, Theorems 10, 11, and 14 and Corollary 13 in Guo and Shan [23] in case in which Xn,n≥1 is a sequence of independent random variables, identically distributed as X.


## 4. Proofs of the Main Results

Proof.
of Theorem 1. We first prove that (27) implies (28). Notice that(36)∑n=1∞nr−2CVmax1≤k≤n∑i=1kaniXip−ϵ+=∑n=1∞nr−2∫0∞Vmax1≤k≤n∑i=1kaniXip>ϵ+xdx=∑n=1∞nr−2∫ϵ1Vmax1≤k≤n∑i=1kaniXip>xdx+∑n=1∞nr−2∫1∞Vmax1≤k≤n∑i=1kaniXip>xdx≤∑n=1∞nr−2Vmax1≤k≤n∑i=1kaniXi>ϵ1/p+∑n=1∞nr−2∫1∞Vmax1≤k≤n∑i=1kaniXi>x1/pdx≔I+II.
From Xu and Cheng [24] and the fact that (27) implies CVXr/q<∞, we see that I<∞. We next establish II<∞. Choose 0<α<1/p, and δ>0 is small sufficiently and integer K is large enough. For every 1≤i≤n, n≥1, we note the fact that n is large sufficiently to guarantee xαn−δ<x1/p/4K. Since the first finite terms of the series do not affect the convergence of the series, without loss of restictions, we could assume the definitions of Xnij below are meaningful. Write(37)Xni1=−xαn−δIaniXi<−xαn−δ+aniXiIaniXi≤xαn−δ+xαn−δIaniXi>xαn−δ,Xni2=aniXi−xαn−δIxαn−δ<aniXi<x1/p4K,Xni3=aniXi+xαn−δI−x1/p4K<aniXi<−xαn−δ,Xni4=aniXi+xαn−δIaniXi≤−x1/p4K+aniXi−xαn−δIaniXi≥x1/p4K.
Observe that∑i=1kaniXi=∑i=1kXni1+∑i=1kXni2+∑i=1kXni3+∑i=1kXni4. Notice that(38)max1≤k≤n∑i=1kaniXi>x1/p⊂∪j=14max1≤k≤n∑i=1kXnij>x1/p4.
Consequently, to prove (28), we only need to prove that(39)Jj≔∑n=1∞nr−2∫1∞Vmax1≤k≤n∑i=1kXnij>x1/p4dx<∞,j=1,2,3,4.
From the definition ofXni4, we deduce that(40)max1≤k≤n∑i=1kXni4>x1/p4⊂max1≤i≤naniXni>x1/p4K.
Observe thatβ>−q/r implies βr−1/q+β>−1. Therefore,(41)∫1s1/qtβr−1/q+βdt≈s1/q+βr−1/qq+β.
Fromani≈i/nβ1/nq, we see that(42)J4≤∑n=1∞nr−2∑i=1n∫1∞VaniXni>x1/p4Kdx≈∑n=1∞nr−2∑i=1n∫1∞VX>x1/p4CKnq+βi−βdx≈∫1∞dx∫1∞ur−2du∫1uVX>x1/p4CKuq+βv−βdvSettings=uq+βv−β,t=v≈∫1∞dx∫1∞ds∫1s1/qsr−1/q+β−1tβr−1/q+βVX>x1/p4CKsdt≈∫1∞dx∫1∞sr/q−1VX>x1/p4CKsds.
Hence, from Lemma3 and (27), we obtain J4<∞. From the definition of Xni2, we conclude that Xni2≥0. From the subadditivity of capacity and Definition 2.5 in [10], it follows that(43)Vmax1≤k≤n∑i=1kXni2>x1/p4=V∑i=1nXni2>x1/p4≤Vthere are at leastKindicesi∈1,nsuch thataniXni>xαn−δ≤∑1≤i1<i2<⋯<iK≤nVanijXij>xαn−δ,for all1≤j≤KIn Definition2.5in Chen10,we setX=ani1Xi1,…,aniK−1XiK−1,Y=aniKXiKφX,Yn=Iani1Xi1>xαn−δ,…,aniK−1XiK−1>xαn−δ,aniKXiK>xαn−δ=∑1≤i1<i2<⋯<iK≤nEIani1Xi1>xαn−δ,…,aniK−1XiK−1>xαn−δVaniKXiK>xαn−δ=⋯=∑1≤i1<i2<⋯<iK≤n∏j=1KVanijXij>xαn−δ≤∑j=1nVanjX>xαn−δK.
Fromβ>−q/r, we obtain ∑i=1nanir/q≈∑i=1nn−rq+β/qiβr/q≈n1−r. Since (27) implies EXr/q<∞, by Markov’s inequality under sublinear expectations (cf. (9) in Hu and Wu [30]) and (43), we conclude that(44)J2≤∑n=1∞nr−2∫1∞∑j=1nVanjX>xαn−δKdx≤∑n=1∞nr−2∫1∞∑j=1nx−rα/qnrδ/qanjr/qEXr/qKdx≈∑n=1∞nr−2−Kr−1−rδ/q∫1∞x−rKα/qdx.
Notice thatr>1 and α>0; we could choose δ small sufficiently and integer K large enough such that r−2−Kr−1−rδ/q<−1 and −rKα/q<−1. Hence, from (44), we obtain J2<∞. Similarly, we can get J3<∞. In order to estimate J1, we first prove that(45)supx≥11x1/pmax1≤k≤n∑i=1kEXni1⟶0,asn⟶∞.
Observe that (27), Lemma 4.5 in Zhang [4], and Hölder’s inequality under sublinear expectations imply that EXr/q<∞ and EX1/p<∞. When q>1, observe that Xni1≤xαn−δ and Xni1≤aniXi, by Hölder’s inequality, we see that(46)max1≤k≤n∑i=1kEXni1≤∑i=1nEXni1≤xα1−1/qn−δ1−1/q∑i=1nEaniXi1/q≪xα1−1/qn−δ1−1/q∑i=1nani1/q≤xα1−1/qn−δ1−1/qnr−1/r∑i=1nanir/q1/r≈xα1−1/qn−δ1−1/q.
Observing thatα1−1/q<α<1/p, by (46), for any x≥1, we obtain(47)supx≥11x1/pmax1≤k≤n∑i=1kEXni1≪n−δ1−1/q⟶0,asn⟶∞.
When1/2<q≤1, noticing that EX=E−X=0 and by choosing δ small sufficiently such that −δ1−r/q+1−r<0, we see that(48)max1≤k≤n∑i=1kEXni1≤2∑i=1nEaniXiIaniXi>xαn−δ≤2xα1−r/qn−δ1−r/q∑i=1nEaniXir/q≤2xα1−r/qn−δ1−r/q∑i=1nanir/q≈xα1−r/qn−δ1−r/q+1−r.
Observing that1−r/q<0, we have(49)supx≥11x1/pmax1≤k≤n∑i=1kEXni1≪n−δ1−r/q+1−r⟶0,asn⟶∞.
Thus, to establishJ1<∞, we only need to prove that(50)J1∗≔∑n=1∞nr−2∫1∞Vmax1≤k≤n∑i=1kXni1−EXni1>x1/p8dx<∞.
Observe thatXni1,1≤i≤n,n≥1 is independent, identically distributed under Ω,ℋ,E. By Markov’s inequality under sublinear expectations, Cr’s inequality, and Lemma 1, we conclude that, for a suitably large M, which will be determined later:(51)Vmax1≤k≤n∑i=1kXni1−EXni1>x1/p8≪x−M/plnnM∑i=1nEXni1M+∑i=1nEXni12M/2.
Choosing large sufficientlyM such that −M1/p−α−rα/q<−1 and −1−M−r/qδ<−1, we have(52)∑n=1∞nr−2lnnM∑i=1n∫1∞x−M/pEXni1Mdx≪∑n=1∞nr−2n−δM−r/qlnnM∑i=1nanir/q∫1∞x−M1/p−α−rα/qdx≪∑n=1∞n−1−δM−r/qlnnM<∞.
Whenr/q≥2, (27) implies that EX2<∞. Observing that q>1/2, we could choose M large enough such that −M/p<−1, r−2−2q−1M/2<−1. Then, by β>−q/r≥−1/2,(53)∑n=1∞nr−2lnnM∫1∞x−M/p∑i=1nEXni12M/2dx≪∑n=1∞nr−2lnnM∑i=1nani2M/2∫1∞x−M/pdx≪∑n=1∞nr−2lnnM∑i=1ni2βn−2q+βM/2≪∑n=1∞nr−2−2q−1M/2lnnM<∞.
Whenr/q<2, choosing M large enough such that −M1/p−α+αr/2q<−1 and r−2−δ2−r/q+r−1M/2<−1, we have(54)∑n=1∞nr−2lnnM∫1∞x−M/p∑i=1nEXni12M/2dx≪∑n=1∞nr−2n−δ2−r/qM/2lnnM∑i=1nanir/qM/2∫1∞x−M1/p−α+αr/2qdx≪∑n=1∞nr−2−δ2−r/q+r−1M/2lnnM<∞.
Consequently, by (51)–(54), we get J1∗<∞. Now, we prove that (28) implies (27). By Markov’s inequality under sublinear expectations (see (8) of Wu [6]), (28), and Lemma 4.5 (iii) in [4],(55)∑n=1∞nr−2Vmax1≤k≤n∑i=1kaniXi>ϵ=∑n=1∞nr−2E1⋅Imax1≤k≤n∑i=1kaniXi>ϵ≤∑n=1∞nr−2Emax1≤k≤n∑i=1kaniXip−ϵ/2p+ϵ/2pImax1≤k≤n∑i=1kaniXi>ϵ≤∑n=1∞nr−2Emax1≤k≤n∑i=1kaniXip−ϵ/2p+ϵ/2p≤∑n=1∞nr−2Emax1≤k≤n∑i=1kaniXip−ϵ/2p+ϵ/2p<∞.
Since(56)max1≤k≤nankXk≤2max1≤k≤n∑i=1kaniXi,by (55) and similar proofs of (3.17) in [31], we conclude that(57)Vmax1≤k≤nankXk>ϵ⟶0,asn⟶∞.
Therefore, by (57) and Lemma 2, we obtain(58)∑i=1nVaniXi>ϵ≪Vmax1≤k≤nankXk>ϵ,for allϵ>0.
Now, combining (58) with (28) gives(59)∑n=1∞nr−2∫ϵ∞∑i=1nVaniXi>x1/pdx<∞.
From the proof of (42), we conclude that (59) is equivalent to (27).Proof.
of Theorem 2. As in the proof of Theorem 1, with Lemma4 in place of Lemma 3, and using(60)∑i=1nanir/q≈∑i=1nn−rq+β/qiβr/q≈n−rq+β/q/qlnn,∫1s1/qtβr−1/q+βdt=∫1s1/qt−1dt≈lns,when β=−q/r, we could prove Theorem 2. Hence, the proof is omitted.Proof.
of Theorem 3. As in the proof of Theorem1, using(61)∑i=1nanir/q≈∑i=1nn−rq+β/qiβr/q≈n−rq+β/q,ifβ<−qr,and with(62)∫1s1/qtβr−1/q+βdt≈C,ifβ<−qr,in place of (41), we could prove Theorem 3. Thus, the proof is omitted.Proof.
of Theorem 4. Settingani=An−iα−1/Anαq, 0≤i≤n and n≥1, we observe that ani≈n−iqα−1n−qα, 0≤i<n, n≥1, and ann≈n−qα. As in the proof of Theorem 14 in [23], letting β=qα−1 in Corollary 1, we finish the proof of Theorem 4.



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