# Complete Convergence for END Random Variables under Sublinear Expectations

Authors: Qunying Wu
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

In this paper, the complete convergence theorems of partial sums and weighted sums for extended negatively dependent random variables in sublinear expectation spaces have been studied and established. Our results extend the corresponding results of classical probability spaces to the case of sublinear expectation spaces.


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## Body

## 1. Introduction

The classical probability and statistics modeling is based on the linear basis of expectation and probability, and their important theoretical basis is the additivity of probability and expectation. However, in many application fields, because many uncertain factors can not be well solved, in order to better simulate this uncertainty phenomenon, we need new tools. The sublinear expectation space established by Peng ([1], [2]) belongs to this kind of tool. Sublinear expectation space is very useful in the simulation of uncertainty. They are widely used in statistics, finance, and insurance and have been studied by many probability statisticians since it came into being. For example, Peng ([1–5]) gave the basic structure, basic properties, and central limit theorem of sublinear expected space. Denis and Martini ([6]) studied a theoretical framework for the pricing of contingent claims in the presence of model uncertainty. Limit theorems for sublinear expectations have raised a large number of issues of interest recently. Marinacci ([7]) and Zhang (2016a [8], 2016b [9], 2016c [10]) established some inequalities and almost sure convergence theorems. Hu (2016 [11]), Chen (2016 [12]), Wu and Jiang (2018 [13]), and Wu and Lu (2020 [14]) also studied strong limit theorem and law of the iterated logarithm.The concept of complete convergence of sequence of random variables is introduced by Hsu and Robbins (1947 [15]). In view of Borel–Cantelli lemma, complete convergence means almost sure convergence. Therefore, complete convergence is one of the most important problems in probability limit theory. Many of its related results have been obtained in the probability space, for example, Wu (2010 [16]), Hu et al. (2012 [17]), Wang et al. (2014 [18]), Wang et al. (2015 [19]), Liu et al. (2015 [20]), Chen and Sung (2016 [21]), Tan et al. (2016 [22]), Wu and Jiang (2016 [23]), and Shen et al. (2017 [24]). Recently, Yu and Wu (2018 [25]), Wang and Wu (2019 [26]), and Deng and Wang (2020 [27]) obtained complete convergence theorem under sublinear expectations. In this paper, we study the complete convergence for extended negatively dependent random variables under sublinear expectation. As a result, the corresponding complete convergence theorems of the probability space have been generalized to the sublinear expectation space context.Because sublinear expectation space relaxes the additivity of probability and expectation of traditional probability space and expands its application scope, and it is of great theoretical and practical significance to extend the limit theorem of traditional probability to space to sublinear expectation space. At the same time, because the sublinear expectation is nonadditive, the study of its limit theory becomes very complex and difficult.


## 2. Related Concepts and Lemmas

In the sublinear expectation space, many basic concepts need to be defined, such as sublinear expectation, capacity, independence, and identical distribution. This section gives a brief introduction to these concepts, which can be seen in detail in the works of Peng (1997 [1], 1999 [2]).LetΩ,ℱ be a given measurable space and let ℋ be a linear space of real functions defined on Ω,ℱ such that if X1,…,Xn∈ℋ, then φX1,…,Xn∈ℋ for each φ∈Cl,Lipℝn, where Cl,Lipℝn denotes the linear space of local Lipschitz functions.Definition 1.
A sublinear expectationE^ on ℋ is a function E^:ℋ⟶−∞,∞ satisfying the following properties: for all X,Y∈ℋ, we have(i)
Monotonicity: ifX≥Y, then E^X≥E^Y(ii)
Constant preserving:E^c=c(iii)
Subadditivity:E^X+Y≤E^X+E^Y(iv)
Positive homogeneity:E^λX=λE^X,λ≥0
The tripleΩ,ℋ,E^ is called a sublinear expectation space.
The conjugate expectationε^ of E^ by ε^X≔−E^−X,∀X∈ℋ.Definition 2.
LetG⊂ℱ. A function V:G⟶0,1 is called a capacity if(1)V∅=0,VΩ=1,VA≤VB,for∀A⊆B,A,B∈G.
It is called to be subadditive ifVA∪B≤VA+VB for all A,B∈G with A∪B∈G. In the sublinear space Ω,ℋ,E^, we denote a pair V,ν of capacities by(2)VA≔infE^ξ;IA≤ξ,ξ∈ℋ,νA≔1−VAc,∀A∈ℱ,where I⋅ denotes an indicator function and Ac is the complement set of A. By definition of V and ν, it is obvious that V is subadditive, and νA≤VA for any A∈ℱ.
The corresponding Choquet integrals/expectationsCV,Cν are defined by(3)CVX≔∫0∞VX>xdx+∫−∞0VX>x−1dx,where V is replaced by V and ν, respectively.
Some basic useful properties of the sublinear expected space are given below, which are described in detail in the works of Wu and Jiang (2018 [13]).Proposition 1.
(i)
For allX,Y∈ℋ,(4)ε^X≤E^X,E^X+c=E^X+c,E^X−Y≤E^X−Y,E^X−Y≥E^X−E^Y.(ii)
IfE^Y=ε^Y, then E^X+aY=E^X+aE^Y for any a∈ℝ.(iii)
(5)E^f≤VA≤E^g,ε^f≤νA≤ε^g,iff≤IA≤g,f,g∈ℋ.(iv)
Markov inequality:∀X∈ℋ:(6)VX≥x≤E^Xpxp,∀x>0,p>0.(v)
Ho..lder inequality: ∀X,Y∈ℋ,p,q>1, satisfying p−1+q−1=1:(7)E^XY≤E^Xp1/pE^Yq1/q.(vi)
Jensen inequality:∀X∈ℋ:(8)E^Xr1/r≤E^Xs1/sfor0<r≤s.Definition 3 (see [2, 8]).
(i)
Identical distribution: letX1 and X2 be two random variables defined, respectively, in sublinear expectation spaces Ω1,ℋ1,E^1 and Ω2,ℋ2,E^2. They are called identically distributed if(9)E^1φX1=E^2φX2,∀φ∈Cl,Lipℝ,whenever the subexpectations are finite. A sequenceXn;n≥1 of random variables is said to be identically distributed if, for each i≥1, Xi and X1 are identically distributed.(ii)
Extended negatively dependent: a sequence of random variablesXn;n≥1 is said to be upper (resp. lower) extended negatively dependent if there is some dominating constant K≥1 such that(10)E^∏i=1nφiXi≤K∏i=1nE^φiXi,∀n≥2,whenever the nonnegative functionsφix∈Cl,Lipℝ,i=1,2,…, are all nondecreasing (resp. all nonincreasing). They are called extended negatively dependent if they are both upper extended negatively dependent and lower extended negatively dependent.
It is obvious that ifXn;n≥1 is a sequence of upper (resp. lower) extended negatively dependent random variables and functions f1x,f2x,…∈Cl,Lipℝ are all nondecreasing (resp. all nonincreasing), then fnXn;n≥1 is also a sequence of upper (resp. lower) extended negatively dependent random variables.
In the following, letXn;n≥1 be a sequence of random variables in Ω,ℋ,E^, and Sn=∑i=1nXi. The symbol c stands for a generic positive constant which may differ from one place to another. Let an≪bn denote that there exists a constant c>0 such that an≤cbn for sufficiently large n.
To prove our results, we need the following two lemmas.Lemma 1 (see Theorem 3.1 in [8]).
LetXk;k≥1 be a sequence of upper extended negatively dependent random variables in Ω,ℋ,E^ with E^Xk≤0. Then, for any x>0,(11)VSn≥x≤1+Ke∑k=1nE^Xk2x2.Lemma 2.
SupposeX∈ℋ,α>0.(i)
Then, for anyc>0,(12)CVX2/α<∞⇔∑n=1∞VX>cnα/2<∞⇔∑n=1∞nVX>cnα<∞.(ii)
IfCVX2/α<∞, then, for any constant c>0 and integer θ>1,(13)∑k=1∞θ2kVX>c⋅θαk<∞,(14)limn⟶∞nVX>cnα/2=0,limn⟶∞n4/3VX>cn2α/3=0.Proof.
(i)
For anyc>0, it can be obtained by the definition of CVX2/α and the substitution of integral variables that(15)CVX2/α=∫0∞VX>xα/2dx=c2/α∫0∞VX>cyα/2dylety=cα/2x=2c2/α∫0∞zVX>czαdz.letz=y1/2.BecauseVX>cyα/2 is nonnegative and monotonically decreasing, according to the integral convergence method, we can obtain(16)CVX2/α<∞⇔∑n=1∞VX>cnα/2<∞.Also, note that(17)∫0∞zVX>czαdz=∑n=1∞∫n−1nzVX>czαdz.This implies that(18)12∑n=1∞nVX>cnα≤∫0∞zVX>czαdz≤1+2∑n=1∞nVX>cnα.Hence,(19)∫0∞zVX>czαdz<∞⇔∑n=1∞nVX>cnα<∞.From combination (16), we get (12) setup.(ii)
Note that, for any integerθ>1,(20)∑n=1∞nVX>cnα≥∑k=1∞∑θk−1≤n<θkθk−1VX>cθαk=θ−2θ−1∑k=1∞θ2kVX>cθαk.
Hence, by (12), CVX2/α<∞ implies (13).
By∫0∞VX>cxα/2dx<∞ and VX>cxα/2↓, we have(21)limx⟶∞xVX>cxα/2=0.
This is equivalent to(22)limx⟶∞x4/3VX>cx2α/3=0.
Therefore, (14) holds.


## 3. Complete Convergence Theorems

In the sublinear expectation space, we define thatXn⟶X completely converges as follows.Definition 4.
A sequence of random variablesXn;n≥1 is said to completely converge to X, denoted by Xn⟶cX as n⟶∞ if ∑n=1∞VXn−X>ε<∞ for any ϵ>0.
The purpose of this paper is to extend corresponding results from the probability space to sublinear expectation space. Our results are as follows.Theorem 1.
LetXn;n≥1 be a sequence of upper extended negatively dependent and identically distributed random variables. Let, for some 0<α≤1,(23)E^X12/α≤CVX12/α<∞,(24)0<ank≤cn−αfork≤nand some0<c<∞,ank=0fork>n,(25)cn≔∑k=1nank2=olog−1n.
Then,(26)∑n=1∞V∑k=1nankXk−E^Xk>ϵ<∞for anyϵ>0.
Furthermore, ifXn;n≥1 is extended negatively dependent, then(27)∑n=1∞V∑k=1nankXk−ε^Xk<−ϵ<∞for anyϵ>0.
In particular, ifXn;n≥1 is extended negatively dependent and E^Xk=ε^Xk, then(28)∑k=1nankXk−E^Xk⟶c0.Remark 1.
It is known by Lemma 4.5 (iii) in [9] that if limc⟶∞E^X2/α−c+=0 or E^ is countably subadditive, then E^X12/α≤CVX12/α.Theorem 2.
LetXn;n≥1 be a sequence of extended negatively dependent and identically distributed random variables. Let, for some α>1,(29)CVX12/α<∞.
Then,(30)Snnα⟶c0.Remark 2.
Theorems1 and 2 extend the corresponding complete convergence results from the probability space to sublinear expectation space.Proof.
of Theorem 1. By the Jensen inequality,E^X12≤E^X12/αα<∞ for 0<α≤1. Without loss of generality, we can assume that E^X12=1.
For upper extended negatively dependent random variablesXn;n≥1, in order to ensure that the truncated random variables are also upper extended negatively dependent, we need that truncated functions belong to Cl,Lip and are nondecreasing. Let fcx≔xIx≤c+cIx>c for any c>0; for any 1≤k≤n,n≥1, let(31)Xnk1≔fn−α/3ank−1Xk=XkIXk≤n−α/3ank−1+n−α/3ank−1IXk>n−α/3ank−1,Xnk2≔Xk−n−α/3ank−1IXk>ϵank−14,Xnk3≔Xk−Xnk1−Xnk2=Xk−n−α/3ank−1In−α/3ank−1<Xk<ϵank−14.
Then,Xnk1;1≤k≤n,n≥1 is also a sequence of upper extended negatively dependent random variables by fcx∈Cl,Lip and fcx being nondecreasing. Define(32)Tni≔∑k=1nankXnki,i=1,2,3,T˜n1≔∑k=1nankXnk1−E^Xnk1.
Then,(33)∑k=1nankXk−E^Xk=T˜n1+Tn2+Tn3+Jn,where Jn=∑k=1nankE^Xnk1−E^Xk.
Therefore, in order to prove (26), it suffices to verify that, for any ϵ>0,(34)∑n=1∞VT˜n1>ϵ<∞,(35)∑n=1∞VTni>ε<∞,i=2,3,(36)Jn⟶0asn⟶∞.
It shall be noted that, in the probability space, there is an equality:EIX≤a=PX≤a; however, in the sublinear expectation space, E^ is defined through continuous functions in Cl,Lip and the indicator function Ix≤a may be discontinuous. Therefore, the expression E^IX≤a may not exist. This needs to modify the indicator function by functions in Cl,Lip. To this end, we define the function gx∈Cl,Lipℝ as follows.
For0<μ<1, let gx∈Cl,Lipℝ be a nonincreasing function such that 0≤gx≤1 for all x and gx=1 if x≤μ and gx=0 if x>1. Then,(37)Ix≤μ≤gx≤Ix≤1,Ix>1≤1−gx≤Ix>μ.
Note that(38)Xk−Xnk1=Xk−n−α/3ank−1IXk>n−α/3ank−1≤XkIXk>n−α/3ank−1≥cn2α/3≪Xk2/αn2α2/α−1/3.
By (23), (24) and (36) follow that(39)Jn≤∑k=1nankE^Xk−Xnk1≤∑k=1nankE^X1α/2n4/3−2α/3≪n−α/3−1/3⟶0,n⟶∞.
Now, we prove that (34). Let X˜nk1=Xnk1−E^Xnk1. Then, T˜n1=∑k=1nankX˜nk1. Fix n≥1. Let u=minϵ/8cn,nα/3/2. By Xnk1≤n−α/3ank−1,Xnk1≤Xk, (24), and Jensen inequality, we obtain(40)uankX˜nk1≤uankXnk1+uankE^Xnk1≤12+uankE^X121/2≤12+cn−2α/32≤1.
It follows that(41)expuankX˜nk1=expuankX˜nk1IuankX˜nk1<−1+uankX˜nk1IuankX˜nk1≤1=expuankX˜nk1IuankX˜nk1<−1expuankX˜nk1IuankX˜nk1≤1≤1,uankX˜nk1<−1,expuankX˜nk1,uankX˜nk1≤1,=1+uankX˜nk1+uankX˜nk1,uankX˜nk1<−1,1+∑j=1∞uankX˜nk1jj!,uankX˜nk1≤1,≤1+uankX˜nk1+u2ank2X˜nk12,uankX˜nk1<−1,1+uankX˜nk1+uankX˜nk1221+2∑j=3∞1j!,uankX˜nk1≤1,≤1+uankX˜nk1+u2ank2X˜nk12.
Hence, byE^X˜nk1=0, and E^X˜nk12≤2E^Xnk12+E^Xnk12≤4E^Xnk12≤4E^X12=4, we obtain(42)E^expuankX˜nk1≤1+4u2ank2≤exp4u2ank2.
SinceuankX˜nk1,n≥1,k≤n is also upper extended negatively dependent, it follows from (25) that(43)E^expuT˜n1=E^∏k=1nexpuankX˜nk1≤K∏k=1nE^expuankX˜nk1≪∏k=1nexp4u2ank2=exp4u2cn.
By the Markov inequality,(44)VT˜n1>ε≤exp−ϵuE^expuT˜n1≪exp−ϵu+4u2cn.
Letℕ1≔n;ϵ/4cn>nα/3 and ℕ2≔n;ϵ/4cn≤nα/3.
Ifn∈ℕ1, i.e., ϵ/4cn>nα/3, then, by the definition of u, we have u=nα/3/2 and −ϵu+4u2cn≤−ϵnα/3/4. Hence,(45)∑n∈ℕ1VT˜n1>ϵ≪∑n=1∞exp−ϵnα/34<∞.
Ifn∈ℕ2, i.e., ϵ/4cn≤nα/3, then, by the definition of u, we have u=ϵ/8cn and −ϵu+4u2cn=−ϵ2/16cn. Note that cn<ϵ2/32logn for sufficiently large n from cn=olog−1n. Hence,(46)∑n∈ℕ2VT˜n1>ϵ≤∑n=1∞exp−ϵ216cn≪∑n=1∞1n2<∞.
That is that (34) holds from(47)∑n=1∞VT˜n1>ϵ=∑n∈ℕ1VT˜n1>+∑n∈ℕ2VT˜n1>ϵ<∞.
Next, we prove that∑n=1∞VTn2>ϵ<∞. By (24),(48)Tn2>ϵ⊂∪k=1nXk>ϵank−14⊂∪k=1nXk>ϵnα4c−1≔bnα,where b=ϵ4c−1>0. It shall be noted that the identical distribution is defined under E^, not under V (see Definition 3). Xi identical distribution implies E^fXi=E^fX1 for f⋅∈Cl,Lipℝ, but does not imply VfXi∈A=VfX1∈A. Therefore, in the calculation of VfXi∈A, we need to convert V to E^. Thus, by (5) and (37),(49)VTn2>ϵ≤∑k=1nVXk>bnα≤∑k=1nE^1−gXkbnα=∑k=1nE^1−gX1bnα≤∑k=1nVX1>μbnα≤nVX1>μbnα.
Therefore, from (12) and (23),(50)∑n=1∞VTn2>ε≤∑n=1∞nVX1>μbnα<∞.
Lastly, we prove that∑n=1∞VTn3>ϵ<∞. Since(51)Tn3≔∑k=1nankXk−n−α/3ank−1In−α/3ank−1<Xk<ϵank−14>ε⊂∑k=1nankXkIn−α/3ank−1<Xk<ϵank−14>ϵ⊂there exist at least4indicesksuch thatXk>ank−1n−α/3.
The last inclusion relation above holds that if there are only less than 4 indicesk, such that Xk>ank−1n−α/3, then, in the sum ∑k=1nankXkIn−α/3ank−1<Xk<ϵank−1/4, there are only less than 4 indices k, such that ankXkIn−α/3ank−1<Xk<ϵank−1/4≠0, and each term ankXkIn−α/3ank−1<Xk<ϵank−1/4<ϵ/4, so ∑k=1nankXkIn−α/3ank−1<Xk<ϵank−1/4<ϵ. This is in contradiction with ∑k=1nankXkIn−α/3ank−1<Xk<ϵank−1/4>ϵ.
Therefore, by the definition of extended negatively dependent (5), (24), and (37), we have(52)VTn3>ε≤Vthere exist at least4indicesksuch thatankXk>n−α/3≤∑1≤i1<⋯<i4≤nVani1Xi1>n−α/3,ani2Xi2>n−α/3,ani3Xi3>n−α/3,ani4Xi4>n−α/3≤∑1≤i1<⋯<i4≤nE^∏j=141−ganijnα/3Xij≤∑1≤i1<⋯<i4≤nK∏j=14E^1−ganijnα/3Xij≤∑1≤i1<⋯<i4≤nK∏j=14E^1−ganijnα/3Xij≪∑1≤i1<…<i4≤nE^1−gcn−2α/3X14≤Cn4V4X1>μc−1n2α/3≤n4V4X1>μc−1n2α/3.
Thus,(53)∑n=1∞VTn3>ϵ≪∑n=1∞n−4/3n4/3VX1>μc−1n2α/34≪∑n=1∞n−4/3<∞,from (14). Together with (50), we obtain (35). This completes the proof of (26).
Furthermore, ifXn;n≥1 is extended negatively dependent, then −Xn;n≥1; also, satisfying the conditions of Theorem 1 and considering −Xn;n≥1 instead of Xn;n≥1 in (26), for any ϵ>0, we can obtain by E^−Xk=−ε^Xk,(54)∑n=1∞V∑k=1nank−Xk−E^−Xk>ϵ=∑n=1∞V−∑k=1nankXk−ε^Xk>ϵ<∞.
That is, (27) is established.Proof.
of Theorem2. We first prove that(55)∑n=1∞VSn>ϵnα<∞,∀ϵ>0.
For anyc>0, let hcx≔−cIx<−c+xIx≤c+cIx>c, for all n≥1,1≤k≤n, and let(56)Ynk≔hnαXk=−nαIXk<−nα+XkIXk≤nα+nαIXk>nα,Znk≔Xk−Ynk=Xk+nαIXk<−nα+Xk−nαIXk>nα.
Then,Ynk;1≤k≤n,n≥1 is also a sequence of upper extended negatively dependent random variables by hcx∈Cl,Lip and hcx being nondecreasing. Note that(57)Sn=∑k=1nYnk−E^Ynk+∑k=1nZnk+∑k=1nE^Ynk≔Un1+Un2+Un3.
In order to prove (55), it suffices to verify that(58)∑n=1∞VUni>ϵnα<∞for anyϵ>0,i=1,2.(59)Un3nα⟶0asn⟶∞.
We first prove that (59). For any 0<a<b, by (37), we have(60)Ia<x≤b=Ix≤b−Ix≤a≤gμxb−gxa,(61)I−b≤x<−a=I−xb≤1−I−xa≤1≤g−μxb−g−xa.
For everyn, there exists a k such that 2k−1≤n<2k; thus, for any r>0, by gx↓, cr inequality, (37) and (60), and (61), we obtain(62)Yn1r≪X1rIX1≤nα+nαrIX1>nα≤1+X1rI1<X1≤2αk+I−2αk≤X1<1+2αrkIX1>2αk−1+I−X1>2αk−1=1+∑j=1kX1rI2αj−1<X1≤2αj+I−2αj≤X1<−2αj−1+2αrkIX1>2αk−1+I−X1>2αk−1≤1+∑j=1kX1rgμX12αj−gX12αj−1+g−μX12αj−g−X12αj−1+2αrk1−gX12αk−1+1−g−X12αk−1.
Hence, by (5) and (37) and gμX1/2αj−gX1/2αj−1≤Iμ2αj−1≤X1≤μ−12αj, g−μX1/2αj−g−X1/2αj−1≤I−μ−12αj≤X1≤−μ2αj−1,(63)E^Yn1r≪1+∑j=1kE^X1rgμX12αj−gX12αj−1+E^X1rg−μX12αj−g−X12αj−1+2αrkE^1−gX12αk−1+E^1−g−X12αk−1≤1+∑j=1kμ−r2αrjVX1≥μ2αj−1+μ−r2αrjVX1≥μ2αj−1+2αrkVX1>μ2αk−1+VX1<−μ2αk−1≪1+∑j=1k2αrjVX1≥μ2−α2αj+2αrkVX1>μ2−α2αk≪∑j=0k2αrjVX1≥μ2−α2αj.
Therefore, byn−α+1↓0 from α>1, for 2k−1≤n<2k, we obtain(64)n−αUn3≤n1−αE^Yn1≤2k−11−αE^Yn1≪21−αk∑j=0k2αjVX1≥μ2−α2αj≔Un31.
Next, we estimateUn31. Noting that, by (13),(65)∑j=0∞2αj2α−1jVX1>2αj−1=∑j=0∞2jVX1>2−α2αj<∞.
It follows that(66)Un31⟶0,asn⟶∞,from 2α−1j↑∞ and the Kronecker lemma. This combination, (59) and (64), was established.
SinceYnk−E^Ynk,n≥1,k≤n is also upper extended negatively dependent with E^Ynk−E^Ynk=0. It follows from Lemma 1, (63), and (13) that(67)∑n=1∞VUn1>ϵnα≪∑n=1∞n−2α∑k=1nE^Ynk2=∑n=1∞n1−2αE^Yn12=∑k=1∞∑2k−1≤n<2kn1−2αE^Yn12≪∑k=1∞∑2k−1≤n<2k2k−11−2α∑j=0k22αjVX1≥μ2−α2αj≪∑k=1∞221−α∑j=0k22αjVX1≥μ2−α2αj≪∑j=0∞22αj∑k=j∞221−αkVX1>μ2−α2αj≪∑j=0∞22jVX1>μ2−α2αj<∞.
By (5), (12), and (37),(68)∑n=1∞VUn2>ϵnα≪∑n=1∞V∑k=1nZk>ϵnα≤∑n=1∞Vthere exists ak∈1,nsuch thatXk>nαorXk<−nα≤∑n=1∞∑k=1nVXk>nα+VXk<−nα≤∑n=1∞∑k=1nE^1−gXknα+E^1−g−Xknα=∑n=1∞nE^1−gX1nα+E^1−g−X1nα≤∑n=1∞nVX1>μnα<∞.
From this combination, (67), we get (58). Therefore, (55) holds.
Because of−Xn;n≥1, also satisfying the conditions of Theorem 2 and considering −Xn;n≥1 instead of Xn;n≥1 in (55), we can obtain(69)∑n=1∞V−Sn>ϵnα=∑n=1∞VSn<−ϵnα<∞.
Hence,(70)∑n=1∞VSn>ϵnα≤∑n=1∞VSn>ϵnα+∑n=1∞VSn<−ϵnα<∞,namely, (30) is established. This completes the Proof of Theorem 2.



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