# Marcinkiewicz–Zygmund Laws of Large Numbers under Sublinear Expectation

Authors: Cheng Hu
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

In this paper, we derive the weak and strong results of Marcinkiewicz–Zygmund laws of large numbers under sublinear expectation. The results are extensions of the Kolmogorov laws of large numbers under sublinear expectation and the classical Marcinkiewicz–Zygmund laws of large numbers.


---

## Body

## 1. Introduction

The theory of sublinear expectation was initiated by Peng [1, 2] to describe the probability uncertainties in statistics, economics, finance, and other fields which are difficult to be handled by the classical probability theory. The classical laws of large numbers (LLNs for short) which reveal the almost sure laws of stabilized partial sum are of great significance in the probability theory. Recently, the LLNs under sublinear expectation got a lot of development, see for example, Marinacci [3]; Maccheroni and Marinacci [4]; Chen et al. [5]; Chen [6]; Zhang [7]; Hu [8]; Chen et al. [9]; and Hu [10].Peng [11]; Chen et al. [9]; and Hu [12] gave three forms of weak LLNs under sublinear expectation under the first moment condition. That is, for any φ∈Cbℝ,(1)limn⟶∞EφSnn=supμ¯≤x≤μ¯φx,for any ε>0,(2)limn⟶∞vμ¯−ε<Snn<μ¯+ε=1,and for any h∈μ¯,μ¯(3)limn⟶∞Vh−ε<Snn<h+ε=1,where E,ℰ denotes the pair of the sublinear expectation and the conjugate expectation and V,v represent the induced capacities, μ¯=EX1 and μ¯=ℰX1.Chen et al. [5] obtained the Kolmogorov strong LLNs under sublinear expectation under the condition of finite 1+αth moments α>0 for sublinear expectation:(4)Vliminfn⟶∞Snn<μ¯∪limsupn⟶∞Snn>μ¯=0.Zhang [7] got the above strong LLNs under the condition of finite first moment for Choquet expectation. Hu [8] improved the above results under a general moment condition for sublinear expectation which is the weakest one for sublinear expectation.The classical Marcinkiewicz–Zygmund strong LLNs generalized the Kolmogorov strong LLNs by extending the convergence rate of partial sum and give the relation between moment conditions and convergence rate. The norming constants becomen1/p0<p≤2 instead of n and the moment conditions depend on p accordingly. In sublinear situation, Feng and Lan [13] obtained the Marcinkiewicz–Zygmund strong LLNs for arrays of row wise independent random variables. Zhang and Lin [14]; Xu and Zhang [15] got the Marcinkiewicz–Zygmund strong LLNs under the condition of finite pth moments for Choquet expectation by different methods. We know that Choquet expectation is larger than sublinear expectation. The purpose of this paper is to generalize the weak and strong LLNs to the Marcinkiewicz–Zygmund LLNs under some moment conditions for sublinear expectation. We discuss the weak results under the condition of p-order uniform integrability for sublinear expectation and study the strong results under the condition a bit stronger than finite pth moments for sublinear expectation.The plan of this paper is as follows. In Section2, we introduce the basic concepts and lemmas under sublinear expectation. In Section 3, we prove some forms of Marcinkiewicz–Zygmund weak LLNs under sublinear expectation and discuss the equivalence relation among them. In Section 4, the Marcinkiewicz–Zygmund strong LLNs under sublinear expectation is given.


## 2. Preliminaries

LetΩ,ℱ be a measurable space and ℋ be a linear space of real functions defined on Ω,ℱ such that if X1,X2,…,Xn∈ℋ then φX1,X2,…,Xn∈ℋ for each φ∈Cl,Lipℝn where Cl,Lipℝn denotes the linear space of local Lipschitz continuous functions φ satisfying(5)φx−φy≤C1+xm+ymx−y,∀x,y∈ℝn,for some C>0, m∈ℕ depending on φ. ℋ contains all IA where A∈ℱ. We also denote Cb,Lipℝn as the linear space of bounded Lipschitz continuous functions φ satisfying(6)φx−φy≤Cx−y,∀x,y∈ℝn,for some C>0.Definition 1.
A functionE: ℋ⟶ℝ¯ is said to be a sublinear expectation if it satisfies for ∀X,Y∈ℋ,(a)
Monotonicity:X≥Y implies 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 −∞+∞
The tripleΩ,ℋ,E is called the sublinear expectation space.Remark 1.
By Definition 1, we can obtain two properties of sublinear expectationE:(1)
EX+c=EX+c,∀c∈ℝ(2)
EX−Y≥EX−EYRemark 2.
LetP be a family of probability measures defined on Ω,ℱ. For any random variable X∈ℱ, the upper expectation defined by E^X:=supQ∈PEQX is a sublinear expectation. So, the results in this paper can also be applied to upper expectation.
The conjugate expectation ofE is defined by(7)ℰX≔−E−X,∀X∈ℋ.
Obviously, for allX∈ℋ, ℰX≤EX.Definition 2 (see [16]).
A set functionV:ℱ⟶0,1 is called a capacity if it satisfies(a)
V∅=0,VΩ=1(b)
VA≤VB,A⊂B,A,B∈ℱ
In this paper, we consider the capacities induced by sublinear expectation and the conjugate expectation:VA:=EIA, vA:=ℰIA=1−VAc, ∀A∈ℱ.
The continuity from above and continuity from below of sublinear expectation and capacity can also be defined similar to the classical probability theory (see [7]).Definition 3.
(independence).Y=Y1,…,YnYi∈ℋ is said to be independent of X=X1,…,XmXi∈ℋ, if for each test function φ∈Cl,Lipℝm×ℝn(8)EφX,Y=EEφx,Yx=X,whenever the sublinear expectations on both sides are finite.
X
n
n
=
1
∞ is said to be a sequence of independent random variables, if Xn+1 is independent of X1,…,Xn for each n≥1.Remark 3.
Peng [2] also gave the definition of identical distribution under the sublinear expectation space. The results in this paper do not need the random variables to be identically distributed.
To prove our main results, we need the following lemmas. The proofs of Lemma 1, Lemma 2, and Lemma 3 can be found in [5] and [8].Lemma 1 (Borel–Cantelli lemma).
If sublinear expectationE is continuous from below and ∑n=1∞VAn<∞, then(9)V∩n=1∞∪i=n∞Ai=0.Lemma 2 (Chebyshev’s inequality).
Letfx>0 be a nondecreasing function on ℝ. Then, for any real x,(10)VX≥x≤EfXfx,vX≥x≤ℰfXfx.Lemma 3.
IfEX<∞, then vX<∞=1.In the following sections, we consider the sequenceXnn=1∞ of independent random variables defined on a sublinear expectation space Ω,ℋ,E with EXn=μ¯, ℰXn=μ¯ for each n≥1. Denote Sn=∑i=1nXi, S0=0.


## 3. The Marcinkiewicz–Zygmund Weak LLNs

Theorem 1.
(1)
Iflimn⟶∞supm≥1EXmpIXm>n=0 for some 1≤p<2, then for any ε>0,
(11)
lim
n
⟶
∞
V
S
n
−
n
μ
¯
n
1
/
p
≤
−
ε
∪
S
n
−
n
μ
¯
n
1
/
p
≥
ε
=
0
,and for any h∈μ¯,μ¯,(12)limn⟶∞V−ε<Sn−nhn1/p<ε=1.(2)
Iflimn⟶∞supm≥1EXmpIXm>n=0 for some 0<p<1, then(13)limn⟶∞VSnn1/p≥ε=0.Proof.
(1)
For the case1≤p<2, we only give the proof of (11). The proof of (12) is similar to Theorem 3.1 of Hu [10], so we omit it.
For any fixedε>0, define a nondecreasing function ϕx∈Cb2ℝ such that ϕx=0 when x≤ε/2, 0≤ϕx≤1 when ε/2<x≤ε, and ϕx=1 when x>ε. It is obvious that Ix≥ε≤ϕx and ϕ0=0. It follows that(14)VSn−nμ¯n1/p≥ε≤EϕSn−nμ¯n1/p=EϕSn−nμ¯n1/p−ϕ0=∑m=1nEϕSm−mμ¯n1/p−EϕSm−1−m−1μ¯n1/p.
Letfx=Eϕx+Xm/n1/p. By the independence of Xmm=1n, we have(15)EϕSm−mμ¯n1/p=EEϕx+Xmn1/px=Sm−1−mμ¯/n1/p=EfSm−1−mμ¯n1/p.
Therefore,(16)EϕSm−mμ¯n1/p−EϕSm−1−m−1μ¯n1/p=EfSm−1−mμ¯n1/p−EϕSm−1−m−1μ¯n1/p≤EfSm−1−mμ¯n1/p−ϕSm−1−m−1μ¯n1/p≤supx∈ℝfx−ϕx+μ¯n1/p=supx∈ℝEϕx+Xmn1/p−ϕx+μ¯n1/p=supx∈ℝEϕx+Xm−μ¯n1/p−ϕx.
For any1≤m≤n, there exist two random variables θm,θ¯m∈0,1 satisfying(17)ϕx+Xm−μ¯n1/p−ϕx=ϕ′xXm−μ¯n1/p+ϕ′x+θmXm−μ¯n1/p−ϕ′xXm−μ¯n1/p,ϕ′x+θmXm−μ¯n1/p−ϕ′x=ϕ′′x+θmθ¯mXm−μ¯n1/p⋅θmXm−μ¯n1/p.
Thus, byϕ′x≥0, for any δ>0, we have(18)VSn−nμ¯n1/p≥ε≤∑m=1nsupx∈ℝEϕx+Xm−μ¯n1/p−ϕx≤∑m=1nsupx∈ℝEϕ′xXm−μ¯n1/p+Eϕ′x+θmXm−μ¯n1/p−ϕ′x⋅Xm−μ¯n1/p≤∑m=1nsupx∈ℝϕ′xμ¯−μ¯n1/p+Eϕ′x+θmXm−μ¯n1/p−ϕ′xXm−μ¯n1/p⋅IXm>n1/pδ+ϕ′′x+θmθ¯mXm−μ¯n1/p⋅θmXm−μ¯2n2/pIXm≤n1/pδ.
LetM=maxsupx∈ℝϕx,supx∈ℝϕ′x,supx∈ℝϕ′′x. It follows that(19)VSn−nμ¯n1/p≥ε≤2Mn1/p∑m=1nEXm−μ¯IXm>n1/pδ+Mn2/p∑m=1nEXm−μ¯2IXm≤n1/pδ≤2Mn1/p∑m=1nEXmIXm>n1/pδ+2Mμ¯n1/p∑m=1nVXm>n1/pδ+2Mn2/p∑m=1nEXm2IXm≤n1/pδ+2Mμ¯2n2/p∑m=1nVXm≤n1/pδ≤2Mn1/pnp−1/pδp−1∑m=1nEXmpIXm>n1/pδ+2Mμ¯n1/pnp/pδp∑m=1nEXmpIXm>n1/pδ+2Mn2−p/pδ2−pn2/p∑m=1nEXmpIXm≤n1/pδ+2Mμ¯2n2/p∑m=1n1≤2Mnδp−1∑m=1nEXmpIXm>n1/pδ+2Mμ¯n1/pδpsupm≤nEXmp+2Mδ2−psupm≤nEXmp+2Mμ¯2n2/p−1⟶2Mδ2−psupm≥1EXmp,asn⟶∞.
Then, by the arbitrariness ofδ, we obtain
(20)
lim
n
⟶
∞
V
S
n
−
n
μ
¯
n
1
/
p
≥
ε
=
0.
Considering−Xii=1∞, applying the above consequence, we have
(21)
lim
n
⟶
∞
V
−
S
n
−
n
E
−
X
n
n
1
/
p
≥
ε
=
0.
Equivalently,
(22)
lim
n
⟶
∞
V
S
n
−
n
μ
¯
n
1
/
p
≤
−
ε
=
0.
Noting thatVA∪B≤VA+VB, we obtain (11).
(2)
For the case0<p<1, applying the function ϕx∈Cb2ℝ in (1) (actually, we only need ϕx∈Cb1ℝ here), we have(23)VSnn1/p≥ε≤EϕSnn1/p=∑m=1nEϕSmn1/p−EϕSm−1n1/p.
Also, byfx=Eϕx+Xm/n1/p, we get(24)EϕSmn1/p=EEϕx+Xmn1/px=Sm−1/n1/p=EfSm−1n1/pEϕSmn1/p−EϕSm−1n1/p≤supx∈ℝEϕx+Xmn1/p−ϕx.
For any1≤m≤n, there exists some random variables λm∈0,1 satisfying(25)ϕx+Xmn1/p−ϕx=ϕ′x+λmXmn1/pXmn1/p.
So, for anyδ>0,(26)VSnn1/p≥ε≤∑m=1nsupx∈ℝEϕx+Xmn1/p−ϕx≤∑m=1nsupx∈ℝEϕx+Xmn1/p−ϕxIXm>n1/pδ+Eϕ′x+λmXmn1/pXmn1/pIXm≤n1/pδ≤2Mnδp∑m=1nEXmpIXm>n1/pδ+Mn1−p/pδ1−pn1/p⋅nsupm≤nEXmp⟶Mδ1−psupm≥1EXmp,asn⟶∞.
By the arbitrariness ofδ, we obtain(27)limn⟶∞VSnn1/p≥ε=0.
Considering−Xii=1∞, we have(28)limn⟶∞VSnn1/p≤−ε=0.Theorem 2.
For any0<p<2, the following two statements are equivalent:(1)
For anyε>0,(29)limn⟶∞VSnn1/p≥ε=0.(2)
For anyφ∈Cbℝ,(30)limn⟶∞EφSnn1/p=φ0.Proof.
(1) ⟹ (2) : Let L=supxφx, then for any ε>0,(31)EφSnn1/p=EφSnn1/pISnn1/p<ε+φSnn1/pISnn1/p≥ε≤sup−ε<x<εφx+LVSnn1/p≥ε⟶sup−ε<x<εφx,asn⟶∞,EφSnn1/p≥inf−ε<x<εφx−LVSnn1/p≥ε⟶inf−ε<x<εφx,asn⟶∞.
By the arbitrariness ofε and the continuity of φ, we obtain expression (30).
(2) ⟹ (1) : For any ε>0, define a function ϕx∈Cbℝ as follows:(32)ϕx=1,x<−εorx>ε,xε,−ε≤x≤ε.
We can easily obtain thatIx≥ε≤ϕx and ϕ0=0. Then,(33)VSnn1/p>ε≤EϕSnn1/p⟶ϕ0=0,asn⟶∞.
Theorem1 together with Theorem 2 gives birth to the following result.Corollary 1.
Iflimn⟶∞supm≥1EXmpIXm>n=0, for some 0<p<2, EXn=ℰXn=0, for each n≥1 when 1≤p<2. Then, for any φ∈Cbℝ,(34)limn⟶∞EφSnn1/p=φ0.Remark 4.
If sublinear expectationE coincides with the classical expectation EP where P is classical probability, (11)–(13) and (34) all can be reduced to the classical forms.


## 4. The Marcinkiewicz–Zygmund Strong LLNs

To study the moment conditions for strong LLNs, we need the following concept of function class which was introduced by Petrov [17]. Let Φc (or, respectively, Φd) denote the set of nonnegative functions ψx defined on 0,∞ satisfying(1)
ψx is positive and nondecreasing on 0,∞. The series ∑n=1∞1/nψn converges (respectively, diverges).(2)
For any fixeda>0, there exists C>0 such that ψx+a≤Cψx for any x>0.Functionsxα and ln1+x1+αα>0 are examples of Φc. The functions ln1+x and lnlne∨x belong to Φd.Before we give the main results, we need the following two lemmas which are used to cope with the truncated parts.Lemma 4.
Suppose thatψx∈Φc. Then, for any 0<p<2,(35)∑n=1∞1nψn1/p/ln1+n<∞.Proof.
Sinceψx∈Φc, we have ψn1/p/ln1+n≥ψn for each n≥1. The rest of the proof is similar to Lemma 2.5 of Hu [8].Lemma 5.
Suppose thatXn−μ¯≤2n1/p/ln1+n for some 0<p<2, and(36)supn≥1EXnplnp−11+XnψXn<∞,for  someψ∈Φcwhen1≤p<2,supn≥1EXnpψXn<∞,for  someψ∈Φcwhen0<p<1.
Then, for anym>1, we have(37)supn≥1Eexpmln1+nn1/p∑i=1nXi−μ¯<∞.Proof.
Sinceex≤1+x+x2/2ex for all real x, and for i≤n,(38)mln1+nn1/pXi−μ¯≤mln1+nn1/p⋅2i1/pln1+i≤2m,we have for any m>1,(39)expmln1+nn1/pXi−μ¯≤1+mln1+nn1/pXi−μ¯+m2ln21+n2n2/pXi−μ¯2expmln1+nn1/pXi−μ¯.
Taking expectations, fori≤n,(40)Eexpmln1+nn1/pXi−μ¯≤1+m2ln21+n2n2/pEXi−μ¯2expmln1+nn1/pXi−μ¯≤expm2e2mln21+n2n2/pEXi−μ¯2.
Sinceψ∈Φc, the function fx=ψx/ln1+x converges to ∞ as x⟶∞. Then, for any 0<a<1,(41)ln21+nn2/p−1sup1≤i≤nEXi−μ¯2=ln21+nn2/p−1sup1≤i≤nEXi−μ¯2IXi−μ¯≤na+ln21+nn2/p−1sup1≤i≤nEXi−μ¯2Ina≤Xi−μ¯≤2i1/pln1+i.
For1≤p<2, by noting that(42)supn≥1EXn−μ¯plnp−11+Xn−μ¯ψXn−μ¯≤supn≥1EXn+μ¯plnp−11+Xn+μ¯ψXn+μ¯≤Csupn≥1EXnp+μ¯plnp−11+XnψXn<∞,we have(43)ln21+nn2/p−1sup1≤i≤nEXi−μ¯2≤ln21+nn2/p−1⋅n2a+ln21+nn2/p−1⋅22−pi2−p/pln2−p1+i⋅1lnp−11+naψna⋅sup1≤i≤nEXi−μ¯plnp−11+Xi−μ¯ψXi−μ¯≤ln21+nn2/p−1−2a+ln21+nn2/p−1⋅22−pn2−p/pln2−p1+n⋅1lnp−11+naln1+nafna⋅supi≥1EXi−μ¯plnp−11+Xi−μ¯ψXi−μ¯=ln21+nn2/p−1−2a+O1fna.
For0<p<1, by noting that(44)supn≥1EXn−μ¯pψXn−μ¯≤supn≥1EXn+μ¯pψXn+μ¯≤Csupn≥1EXnp+μ¯pψXn<∞,we have(45)ln21+nn2/p−1sup1≤i≤nEXi−μ¯2≤ln21+nn2/p−1⋅n2a+ln21+nn2/p−1⋅22−pi2−p/pln2−p1+i⋅1ψna⋅sup1≤i≤nEXi−μ¯pψXi−μ¯≤ln21+nn2/p−1−2a+ln21+nn2/p−1⋅22−pn2−p/pln2−p1+n⋅1ln1+nafna⋅supi≥1EXi−μ¯pψXi−μ¯=ln21+nn2/p−1−2a+O1ln1−p1+nfna.
We seta such that 2/p−1−2a>0, i.e., a<1/p−1/2. So, for 0<p<2, we have(46)limn⟶∞ln21+nn2/p−1sup1≤i≤nEXi−μ¯2=0.
Then, there exists someM>0 such that for any n≥1,(47)ln21+nn2/psup1≤i≤nEXi−μ¯2≤Mn.
By the independence ofXnn=1∞, we have(48)Eexpmln1+nn1/p∑i=1nXi−μ¯=∏i=1nEexpmln1+nn1/pXi−μ¯≤expMm2e2m2nn=eMm2e2m/2<∞.Theorem 3.
Suppose thatE is continuous from below.(1)
Ifsupn≥1EXnplnp−11+XnψXn<∞ for some ψ∈Φc and 1≤p<2, then(49)Vliminfn⟶∞Sn−nμ¯n1/p<0∪limsupn⟶∞Sn−nμ¯n1/p>0=0.(2)
Ifsupn≥1EXnplnp1+XnψXn<∞ for some ψ∈Φc and 0<p<1, then(50)Vlimn⟶∞Snn1/p>0=0.Proof.
(1)
For1≤p<2, define fnx=−n1/p/ln1+n∨x∧n1/p/ln1+n and f^nx=x−fnx. Then, fn⋅,f^n⋅∈Cl,Lipℝ. Let(51)Yn=fnXn−μ¯−EfnXn−μ¯+μ¯,
and denoteS¯n=∑i=1nYi. Then,
(52)
X
n
−
μ
¯
=
Y
n
−
μ
¯
+
f
^
n
X
n
−
μ
¯
+
E
f
n
X
n
−
μ
¯
,
1
n
1
/
p
S
n
−
n
μ
¯
=
1
n
1
/
p
S
¯
n
−
n
μ
¯
+
1
n
1
/
p
∑
i
=
1
n
f
^
i
X
i
−
μ
¯
+
1
n
1
/
p
∑
i
=
1
n
E
f
i
X
i
−
μ
¯
.
Note that
(53)
E
f
i
X
i
−
μ
¯
=
E
X
i
−
μ
¯
−
f
^
i
X
i
−
μ
¯
≤
E
X
i
−
μ
¯
+
E
−
f
^
i
X
i
−
μ
¯
≤
E
f
^
i
X
i
−
μ
¯
.
Therefore,(54)1n1/pSn−nμ¯≤1n1/pS¯n−nμ¯+1n1/p∑i=1nf^iXi−μ¯+1n1/p∑i=1nEf^iXi−μ¯.
We can easily obtain thatYnn=1∞ are independent; Yn−μ¯≤2n1/p/ln1+n, EYn=μ¯ and supn≥1EYnplnp−11+YnψYn<∞.
For anyε>0, we set m>1/ε. Then, by Chebyshev’s inequality,(55)VS¯n−nμ¯n1/p≥ε=Vmln1+nn1/p∑i=1nYi−μ¯≥εmln1+n≤e−εmln1+nEexpmln1+nn1/p∑i=1nYi−μ¯≤11+nεmsupn≥1Eexpmln1+nn1/p∑i=1nYi−μ¯.
By Lemma 5 and∑n=1∞1/1+nεm<∞, we have(56)∑n=1∞VS¯n−nμ¯n1/p≥ε<∞.
By Borel–Cantelli lemma, we have(57)Vlimsupn⟶∞S¯n−nμ¯n1/p≥ε=0.
The continuity from below ofV can be deduced by the continuity from below of E. So, we have(58)Vlimsupn⟶∞S¯n−nμ¯n1/p>0=0.
Moreover,(59)∑i=1∞Ef^iXi−μ¯i1/p≤∑i=1∞EXi−μ¯IXi−μ¯>i1/p/ln1+ii1/p≤∑i=1∞EXi−μ¯plnp−11+Xi−μ¯ψXi−μ¯i1/pi1/p/ln1+ip−1lnp−11+i1/p/ln1+iψi1/p/ln1+i,
where(60)i1/pi1/pln1+ip−1lnp−11+i1/pln1+iψi1/pln1+i=iln1−p1+ilnp−11+i1/pln1+iψi1/pln1+i≥iln1−p1+i1plni−lnln1+ip−1ψi1/pln1+i≥iln1−p1+ilnp−1i2pp−1ψi1/pln1+i≥Ciψi1/pln1+i.
By Lemma 4, we have(61)∑i=1∞Ef^iXi−μ¯i1/p<∞.
By Kronecker lemma, we have(62)limn⟶∞1n1/p∑i=1nEf^iXi−μ¯=0.
On the other hand, by the continuity from below ofE,(63)E∑i=1∞f^iXi−μ¯i1/p≤∑i=1∞Ef^iXi−μ¯i1/p<∞.
Then, by Lemma 3, we have(64)v∑i=1∞f^iXi−μ¯i1/p<∞=1,
which implies(65)vlimn⟶∞1n1/p∑i=1nf^iXi−μ¯=0=1.
Takinglimsupn⟶∞ on both sides of (54), and by (58), (62), and (65), we have(66)vlimsupn⟶∞Sn−nμ¯n1/p≤0=1.
Equivalently,(67)Vlimsupn⟶∞Sn−nμ¯n1/p>0=0.
Considering−Xnn=1∞ with E−Xi=−μ¯, we have(68)Vlimsupn⟶∞−Sn−n−μ¯n1/p>0=0.
Equivalently,(69)Vliminfn⟶∞Sn−nμ¯n1/p<0=0.
Noting thatVA∪B≤VA+VB, we obtain (49).
(2)
For0<p<1, let Zn=fnXn−EfnXn and denote Sn∗=∑i=1nZi. Then,(70)Xn=Zn+f^nXn+EfnXn,1n1/pSn=1n1/pSn∗+1n1/p∑i=1nf^iXi+1n1/p∑i=1nEfiXi.
We can easily obtain thatZnn=1∞ are independent, Zn≤2n1/p/ln1+n, EZn=0, and supn≥1EZnplnp1+ZnψZn<∞.
For anyε>0, we set m>1/ε. Then, by Chebyshev’s inequality and Lemma 5, we have(71)VSn∗n1/p≥ε=Vmln1+nn1/p∑i=1nZi≥εmln1+n≤11+nεmsupn≥1Eexpmln1+nn1/p∑i=1nZi<∞.
By Borel–Cantelli lemma, we have(72)Vlimsupn⟶∞Sn∗n1/p≥ε=0.
Hence, by the continuity from below ofV, we have(73)Vlimsupn⟶∞Sn∗n1/p>0=0.
LetBi:=Xi>i1/p/ln1+i for i≥1. Then,(74)∑i=1∞VBi≤∑i=1∞EXiplnp1+XiψXii/lnp1+ilnp1+i1/p/ln1+iψi1/p/ln1+i≤Csupi≥1EXiplnp1+XiψXi∑i=1∞1iψi1/p/ln1+i<∞.
By Borel–Cantelli lemma, we have(75)V∩n=1∞∪i=n∞Bi=0.
Then, for anyω∈∪n=1∞∩i=n∞Bic, there exists nω∈ℕ∗ such that for any i>nω, Xiω≤i1/p/ln1+i. For these ω,(76)∑i=1∞f^iXii1/p≤∑i=1∞XiIXi>i1/p/ln1+ii1/p=∑i=1nω+∑i=nω+1∞XiIXi>i1/p/ln1+ii1/p<∞.
Hence,(77)v∑i=1∞f^iXii1/p<∞=1,which implies(78)vlimn⟶∞1n1/p∑i=1nf^iXi=0=1.
Furthermore,(79)∑i=1∞EfiXii1/p≤∑i=1∞EXiIXi≤i1/p/ln1+i+i1/p/ln1+iIXi>i1/p/ln1+ii1/p≤∑i=1∞EXipψXii1/pψi1/p/ln1+i⋅i1−p/pln1−p1+i+∑i=1∞VXi>i1/p/ln1+iln1+i≤supi≥1EXipψXi∑i=1∞1iln1−p1+iψi1/p/ln1+i+∑i=1∞VBi<∞,which implies(80)limn⟶∞1n1/p∑i=1nEfiXi=0.
Takinglimsupn⟶∞ on both sides of (70) and by (73), (78), and (80), we have(81)vlimsupn⟶∞Snn1/p≤0=1.
Equivalently,(82)Vlimsupn⟶∞Snn1/p>0=0.
Considering−Xnn=1∞, we have(83)Vliminfn⟶∞Snn1/p<0=0.Remark 5.
The two moment conditions for case1≤p<2 and case 0<p<1 in Theorem 3 are both stronger than supn≥1EXnp<∞ but weaker than supn≥1EXnp+α<∞α>0. Zhang and Lin [14] obtained that the pth moments for Choquet expectation is the necessary and sufficient conditions of the Marcinkiewicz–Zygmund strong LLNs. One can find a counterexample that the pth moments for sublinear expectation is finite but the pth moments for Choquet expectation is not finite (similar to Example 4.1 of Hu [12]. So the pth moments for sublinear expectation cannot maintain the Marcinkiewicz–Zygmund strong LLNs.Remark 6.
If sublinear expectationE coincides with the classical expectation EP where P is classical probability, Theorem 3 can be reduced to the classical Marcinkiewicz–Zygmund strong LLNs:(84)Plimn⟶∞Sn−nan1/p=0=1,where a=0 if 0<p<1 and a=EX1 if 1≤p<2.



---

*Source: 5050973-2020-04-11.xml*