On the Convergence Rates Under Sub-Linear Expectation

Abstract

In this paper, we establish the convergence rates for the law of large numbers and the law of the logarithm for identically distributed random variables with either negative dependence or independence in the framework of sub-linear expectation. We propose a novel subsequence partitioning method to address the convergence of the series of capacities. This method allows us to verify the convergence of capacity series for weighted sum sequences under suitable moment conditions in the sub-linear expectation setting. Under the assumption of countable sub-additivity of the upper capacity, we derive sufficient Choquet moment and tail conditions for the above problems, which yield convergence rates of order $1/n$ and $\left(\log n\right)/n$. Our results generalize the corresponding conclusions in classical probability theory.

1. Introduction

Limit theorems constitute the core theoretical framework of probability theory and mathematical statistics, whose essence is to reveal the long-term asymptotic behavior of random variables. In recent years, many uncertain phenomena have appeared in fields including statistics, risk measurement, financial mathematics and other related areas. Classical probability theory is not effective enough to model such uncertain phenomena, and researchers have found that non-additive probabilities and non-linear expectations are appropriate for describing these phenomena in the above fields (cf. [1,2,3]).

Peng (2007) [4] introduced the definition and theoretical framework of sub-linear expectation, which serves as a canonical example of non-linear expectation. Under this framework, several classical limit theorems have been extended to the sub-linear expectation setting. Zhang (2016) [5] obtained the strong law of large numbers for sub-linear expectation under the Choquet moment condition. Hu (2018) [6] studied the strong law of large numbers for sub-linear expectation under the controlled moment condition. Hu et al. (2024) [7] established the strong law of large numbers for delayed sums under sub-linear expectation. Zhang (2016) [8] proved the law of the iterated logarithm under sub-linear expectation. Xu and Zhang (2020) [9] established the law of the logarithm for arrays of row-wise extended negatively dependent random variables under sub-linear expectation.

Convergence rates in the sense of complete moment convergence constitute a fundamental component of limit theorems by quantifying the speed at which random variables converge to their limits. There have been numerous studies on convergence rates or complete moment convergence within the classical probability framework. Li et al. (1995) [10] established complete convergence results for weighted sums of independent random variables. Chen and Wang (2008) [11] investigated the convergence rates and refined convergence rates for probabilities of moderate deviations of moving average processes. Liu et al. (2015) [12] derived convergence rates via the invariance principle. Zhang (2019) [13] studied the convergence rates which extended and generalized the known Davis–Gut Law and Lai Law. Under sub-linear expectation, Wu et al. (2025) [14] established the complete convergence theorems for arrays of row-wise widely orthant-dependent random variables under some mild conditions.

Let $\{X_n,n\ge 1\}$ be a sequence of independent and identically distributed random variables with the partial sums $S_n={\sum }_{k=1}^n{X}_k$, $n\ge 1$. Baum and Katz (1965) [15] established the convergence rates for the law of large numbers, i.e.,

For $0<t<1$,

$$E\left[\right|X_k{|}^t]<\infty \iff {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}P\left\{|S_n|>n^{1/t}\epsilon \right\}<\infty ,\forall \epsilon >0.$$

(1)

For $1\le t<2$,

$$E\left[\right|X_k{|}^t]<\infty \iff {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}P\left\{|S_n-nE\left[X_k\right]|>n^{1/t}\epsilon \right\}<\infty ,\forall \epsilon >0.$$

(2)

Davis (1968) [16] derived the convergence rates for the law of the logarithm, i.e.,

$$E{X}_1=0\mathrm{and}E\left[X_1^2\right]<\infty \iff {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}P\left\{|S_n|>\epsilon \sqrt{n\log n}\right\}<\infty ,\forall \epsilon >0.$$

(3)

The aim of this paper is to extend the classical convergence rates to the sub-linear expectation framework. We consider convergence rates of order $1/n$ and $\left(\log n\right)/n$, and then we derive the corresponding sufficient moment conditions. The results show that under sub-linear expectation, the Choquet moment condition for the convergence rate of order $1/n$ for the law of large numbers is consistent with the classical results. For the law of the logarithm, the Choquet moment condition required for the convergence rate of order $\left(\log n\right)/n$ coincides with the classical finite second moment condition, while the moment condition corresponding to the rate of order $1/n$ is weaker than the finite second moment condition.

Furthermore, to establish the convergence of the series of capacities, we propose a novel proof framework. A key method, which holds for sequences of non-negative or negatively dependent random variables, is derived for the convergence of the series of capacities for weighted partial sums with exponentially increasing weights. This method, which incorporates the subsequence partitioning of series, serves as a critical tool to justify both the far-tail and intermediate truncation terms in the partial sum truncation analysis for our main theorems.

This paper is organized as follows. Section 2 provides the basic concepts and lemmas. Section 3 contains our results on convergence rates for the law of large numbers and the law of the logarithm under sub-linear expectation as well as the corresponding proofs.

In what follows, C denotes a positive constant, $C_{\alpha }$ a positive constant depending on $\alpha$, and $C_{\alpha ,\beta }$ a positive constant depending on $\alpha$ and $\beta$ with other positive constants defined similarly. All such positive constants may take different values from line to line. For real numbers x and y, we denote $x^{+}=\max \{x,0\}$, $x^{-}=\max \{-x,0\}$, $x\vee y=\max (x,y)$, $x\wedge y=\min (x,y)$ and $\log x=\ln (x\vee e)$. The notation $a_n$∼$b_n$ as $n\to$∞ means that the limit of the ratio $a_n/b_n$ equals 1. $\left[x\right]$ denotes the greatest integer less than or equal to x.

2. Basic Concepts and Lemmas

This paper is based on the framework of sub-linear expectation proposed by Peng (2007) [4]. Let $(\mathrm{\Omega },\mathcal{F})$ be a given measurable space and $\mathcal{H}$ be a linear space of real functions defined on $\mathrm{\Omega }$ such that if $X_1,X_2,\dots ,X_n\in \mathcal{H}$ then $\phi (X_1,X_2,\dots ,X_n)\in \mathcal{H}$ for each $\phi \in C_{l,\mathrm{Lip}}\left({\mathbb{R}}^n\right)$ where $C_{l,\mathrm{Lip}}\left({\mathbb{R}}^n\right)$ denotes the linear space of (local Lipschitz) functions $\phi$ satisfying

$$|\phi \left(\mathit{x}\right)-\phi \left(\mathit{y}\right)|\le {C(1+|\mathit{x}|}^m+{\left|\mathit{y}\right|}^m\left)\right|\mathit{x}-\mathit{y}|,\forall \mathit{x},\mathit{y}\in {\mathbb{R}}^n,$$

for some $C>0,m\in \mathbb{N}$ depending on $\phi$.

Next, we present several fundamental definitions and properties of sub-linear expectation. For further properties, the reader may refer to [3,4,17,18].

Definition 1. 

A sub-linear expectation $\mathbb{E}$ on $\mathcal{H}$ is a functional $\mathbb{E}:\mathcal{H}\to \overline{\mathbb{R}}:=[-\infty ,\infty ]$ satisfying the following properties: for all $X,Y\in \mathcal{H}$,

(a) 

Monotonicity: If $X\ge Y$ then $\mathbb{E}\left[X\right]\ge \mathbb{E}\left[Y\right].$

(b) 

Constant preserving: $\mathbb{E}\left[c\right]=c,\forall c\in \mathbb{R}.$

(c) 

Positive homogeneity: $\mathbb{E}\left[\lambda X\right]=\lambda \mathbb{E}\left[X\right],\forall \lambda \ge 0.$

(d) 

Sub-additivity: $\mathbb{E}[X+Y]\le \mathbb{E}\left[X\right]+\mathbb{E}\left[Y\right]$ whenever $\mathbb{E}\left[X\right]+\mathbb{E}\left[Y\right]$ is not of the form $+\infty -\infty$ or $-\infty +\infty$.

The triple $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$ is called a sub-linear expectation space.

In addition, we give the definition of the conjugate expectation $\mathcal{E}$ of sub-linear expectation $\mathbb{E}$, which is a functional $\mathcal{E}:\mathcal{H}\to \overline{\mathbb{R}}$ satisfying

$$\mathcal{E}\left[X\right]:=-\mathbb{E}[-X],\forall X\in \mathcal{H}.$$

Definition 2. 

A set function $V:\mathcal{F}\to [0,1]$ is called a capacity if it satisfies

(i)

$V(\varnothing )=0$, $V\left(\mathrm{\Omega }\right)=1$.

(ii)

For $A\subset B$, where $A,B\in \mathcal{F}$, we have $V\left(A\right)\le V\left(B\right)$.

Let $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$ be a sub-linear expectation space. We define the pair of induced capacities $(\mathbb{V},v)$ by

$$\mathbb{V}\left(A\right):=\inf \left\{\mathbb{E}\left[\xi \right]:I_A\le \xi ,\xi \in \mathcal{H}\right\},v\left(A\right):=1-\mathbb{V}\left(A^c\right),\forall A\in \mathcal{F},$$

(4)

where $A^c$ is the complement set of A. We conventionally refer to $\mathbb{V}$ as the upper capacity and v as the lower capacity. Then

$$\mathbb{V}\left(A\right):=\mathbb{E}\left[I_A\right],v\left(A\right):=\mathcal{E}\left[I_A\right],\mathrm{if}{I}_A\in \mathcal{H}.$$

$$\mathbb{E}\left[f\right]\le \mathbb{V}\left(A\right)\le \mathbb{E}\left[g\right],\mathcal{E}\left[f\right]\le v\left(A\right)\le \mathcal{E}\left[g\right],\mathrm{if}f\le I_A\le g,f,g\in \mathcal{H}.$$

Definition 3. 

A pair $(C_{\mathbb{V}},C_v)$ of the Choquet integrals/expectations is defined by

$$C_V\left[X\right]={\int }_0^{\infty }V(X\ge t)dt+{\int }_{-\infty }^0\left[V(X\ge t)-1\right]dt,$$

with V being replaced by $\mathbb{V}$ and v, respectively.

Definition 4. 

(i) A sub-linear expectation $\mathbb{E}$ is said to be countably sub-additive if for $X,X_n\in \mathcal{H}$, $n=1,2,\dots$

$$\mathbb{E}\left[X\right]\le {\displaystyle \sum _{n=1}^{\infty }}\mathbb{E}\left[X_n\right],\mathit{whenever}X\le {\displaystyle \sum _{n=1}^{\infty }}{X}_n,X\ge 0,X_n\ge 0.$$

(ii) 

A capacity V is said to be sub-additive if $V(A\cup B)\le V\left(A\right)+V\left(B\right)$ for all $A,B\in \mathcal{F}$. A capacity V is said to be countably sub-additive if

$$V\left({\displaystyle \bigcup _{n=1}^{\infty }}{A}_n\right)\le {\displaystyle \sum _{n=1}^{\infty }}V\left(A_n\right),\forall A_n\in \mathcal{F}.$$

Because $\mathbb{V}$ may be not countably sub-additive in general, its countably sub-additive extension ${\mathbb{V}}^{*}$ is defined by

$${\mathbb{V}}^{*}\left(A\right)=\inf \left\{{\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}\left(A_n\right):A\subset {\displaystyle \bigcup _{n=1}^{\infty }}{A}_n\right\},v^{*}\left(A\right)=1-{\mathbb{V}}^{*}\left(A^c\right),A\in \mathcal{F}.$$

Then, it can be shown that ${\mathbb{V}}^{*}\left(A\right)$ is a countably sub-additive capacity with ${\mathbb{V}}^{*}\left(A\right)\le \mathbb{V}\left(A\right)$ and the following properties:

(a)

If $\mathbb{V}$ is countably sub-additive, then ${\mathbb{V}}^{*}\equiv \mathbb{V}$.

(b)

If $I_A\le g,g\in \mathcal{H}$, then ${\mathbb{V}}^{*}\left(A\right)\le \mathbb{E}\left[g\right]$. Furthermore, if $\mathbb{E}$ is countably sub-additive, then

$$\mathbb{E}\left[f\right]\le {\mathbb{V}}^{*}\left(A\right)\le \mathbb{V}\left(A\right)\le \mathbb{E}\left[g\right],\forall f\le I_A\le g,f,g\in \mathcal{H}.$$

It can be verified that if $\mathbb{E}$ is countably sub-additive or $\mathbb{E}\left[\right|X\left|\right]={\lim }_{c\to \infty }\mathbb{E}\left[\left|X\right|\wedge c\right]$, then $\mathbb{E}\left[\right|X\left|\right]\le C_{\mathbb{V}}\left[\left|X\right|\right]$.

Remark 1. 

In this paper, all the convergence rate results are derived for the capacity $\mathbb{V}$ under the assumption of the countable sub-additivity of $\mathbb{V}$. From the properties above, it follows that these results also hold for ${\mathbb{V}}^{*}$ without requiring the assumption of countable sub-additivity.

Definition 5. 

(i) (Identical distribution) Let ${\mathit{X}}_1$ and ${\mathit{X}}_2$ be two n-dimensional random vectors defined respectively in sub-linear expectation spaces $({\mathrm{\Omega }}_1,{\mathcal{H}}_1,{\mathbb{E}}_1)$ and $({\mathrm{\Omega }}_2,{\mathcal{H}}_2,{\mathbb{E}}_2)$. They are called identically distributed, denoted by ${\mathit{X}}_1\stackrel{d}{=}{\mathit{X}}_2$, if

$${\mathbb{E}}_1\left[\phi \left({\mathit{X}}_1\right)\right]={\mathbb{E}}_2\left[\phi \left({\mathit{X}}_2\right)\right],\forall \phi \in C_{l,\mathrm{Lip}}\left({\mathbb{R}}^n\right),$$

whenever the sub-linear expectations are finite. A sequence $\{X_n;n\ge 1\}$ of random variables is said to be identically distributed if $X_i\stackrel{d}{=}{X}_1$ for each $i\ge 1$.

(ii) 

(Independence) In a sub-linear expectation space $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$, a random vector $\mathit{Y}=(Y_1,\dots ,Y_n),Y_i\in \mathcal{H}$ is said to be independent to another random vector $\mathit{X}=(X_1,\dots ,X_m),$ $X_i\in \mathcal{H}$ under $\mathbb{E}$ if for each test function $\phi \in C_{l,\mathit{Lip}}({\mathbb{R}}^m\times {\mathbb{R}}^n)$, we have

$$\mathbb{E}\left[\phi (\mathit{X},\mathit{Y})\right]=\mathbb{E}\left[\mathbb{E}\left[\phi (\mathit{x},\mathit{Y})\right]{|}_{\mathit{x}=\mathit{X}}\right],$$

whenever $\overline{\phi }\left(\mathit{x}\right):=\mathbb{E}\left[\right|\phi (\mathit{x},\mathit{Y})\left|\right]<\infty$ for all $\mathit{x}$ and $\mathbb{E}\left[\right|\overline{\phi }\left(\mathit{X}\right)\left|\right]<\infty$.

(iii) 

(IID random variables) A sequence $\{X_n;n\ge 1\}$ of random variables is said to be independent and identically distributed (IID) if $X_n\stackrel{d}{=}{X}_1$ and $X_{n+1}$ is independent to $(X_1,\dots ,X_n)$ for each $n\ge 1$.

From the definition of independence, it is easily seen that if Y is independent to X and $X\ge 0$, $\mathbb{E}\left[Y\right]\ge 0$, then

$$\mathbb{E}\left[XY\right]=\mathbb{E}\left[X\right]\mathbb{E}\left[Y\right].$$

Furthermore, if Y is independent to X and $X\ge 0$, $Y\ge 0$, then

$$\mathbb{E}\left[XY\right]=\mathbb{E}\left[X\right]\mathbb{E}\left[Y\right],\mathcal{E}\left[XY\right]=\mathcal{E}\left[X\right]\mathcal{E}\left[Y\right].$$

Definition 6 

(Negative dependence). In a sub-linear expectation space $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$, a random vector $\mathit{Y}=(Y_1,\dots ,Y_n)$ with $Y_i\in \mathcal{H}$ is said to be negatively dependent (ND) to another random vector $\mathit{X}=(X_1,\dots ,X_m)$ with $X_i\in \mathcal{H}$ under $\mathbb{E}$ if for each pair of test functions ${\phi }_1\in C_{l,\mathrm{Lip}}\left({\mathbb{R}}^m\right)$ and ${\phi }_2\in C_{l,\mathrm{Lip}}\left({\mathbb{R}}^n\right)$, we have

$$\mathbb{E}\left[{\phi }_1\left(\mathit{X}\right){\phi }_2\left(\mathit{Y}\right)\right]\le \mathbb{E}\left[{\phi }_1\left(\mathit{X}\right)\right]\mathbb{E}\left[{\phi }_2\left(\mathit{Y}\right)\right],$$

whenever either ${\phi }_1$ and ${\phi }_2$ are coordinate-wise non-decreasing or ${\phi }_1$ and ${\phi }_2$ are coordinate-wise non-increasing with ${\phi }_1\left(\mathit{X}\right)\ge 0$, $\mathbb{E}\left[{\phi }_2\left(\mathit{Y}\right)\right]\ge 0$, $\mathbb{E}\left[|{\phi }_1\left(\mathit{X}\right){\phi }_2\left(\mathit{Y}\right)|\right]<\infty$, $\mathbb{E}\left[|{\phi }_1\left(\mathit{X}\right)|\right]<\infty$, $\mathbb{E}\left[|{\phi }_2\left(\mathit{Y}\right)|\right]<\infty$.

A sequence $\{X_n;n\ge 1\}$ of random variables is said to be negatively dependent if $X_{n+1}$ is negatively dependent to $(X_1,\dots ,X_n)$ for each $n\ge 1$.

It is easily seen that if $\{X_n;n\ge 1\}$ is a sequence of independent random variables and $f_1\left(x\right),f_2\left(x\right),\dots \in C_{l,\mathrm{Lip}}\left(\mathbb{R}\right)$, then $\{f_n\left(X_n\right);n\ge 1\}$ is also a sequence of independent random variables; if $\{X_n;n\ge 1\}$ is a sequence of negatively dependent random variables and $f_1\left(x\right),f_2\left(x\right),\dots \in C_{l,\mathrm{Lip}}\left(\mathbb{R}\right)$ are non-decreasing (resp. non-increasing) functions, then $\{f_n\left(X_n\right);n\ge 1\}$ is also negatively dependent.

It is obvious that if $\mathit{Y}$ is independent to $\mathit{X}$, then $\mathit{Y}$ is negatively dependent to $\mathit{X}$. The following is the classical example of a sub-linear expectation space, along with sequences of negatively dependent random variables defined on this space, as introduced by Huber and Strassen (1973) [1].

Example 1. 

Let $\mathcal{P}$ be a family of probability measures defined on $(\mathrm{\Omega },\mathcal{F})$. For any random variable ξ, we denote the upper expectation by $\mathbb{E}\left[\xi \right]={\sup }_{Q\in \mathcal{P}}{E}_Q\left[\xi \right]$. Then, $\mathbb{E}[·]$ is a sub-linear expectation. Moreover, if $\mathit{X}$ and $\mathit{Y}$ are independent under each $Q\in \mathcal{P}$, then $\mathit{Y}$ is negatively dependent to $\mathit{X}$ under $\mathbb{E}$. In fact,

$$\mathbb{E}\left[{\phi }_1\left(\mathit{X}\right){\phi }_2\left(\mathit{Y}\right)\right]=\underset{Q\in \mathcal{P}}{\sup }{E}_Q\left[{\phi }_1\left(\mathit{X}\right){\phi }_2\left(\mathit{Y}\right)\right]=\underset{Q\in \mathcal{P}}{\sup }{E}_Q\left[{\phi }_1\left(\mathit{X}\right)\right]E_Q\left[{\phi }_2\left(\mathit{Y}\right)\right]$$

$$⩽\underset{Q\in \mathcal{P}}{\sup }{E}_Q\left[{\phi }_1\left(\mathit{X}\right)\right]\underset{Q\in \mathcal{P}}{\sup }{E}_Q\left[{\phi }_2\left(\mathit{Y}\right)\right]=\mathbb{E}\left[{\phi }_1\left(\mathit{X}\right)\right]\mathbb{E}\left[{\phi }_2\left(\mathit{Y}\right)\right],$$

whenever ${\phi }_1\left(\mathit{X}\right)⩾0$ and $\mathbb{E}\left[{\phi }_2\left(\mathit{Y}\right)\right]⩾0$.

However, $\mathit{Y}$ may be not independent to $\mathit{X}$ in Example 1. With a similar argument, we can show that $\mathit{Y}$ is negatively dependent to $\mathit{X}$ under $\mathbb{E}$ if $\mathit{X}$ and $\mathit{Y}$ are negatively dependent under each $Q\in \mathcal{P}$.

In the following example, we construct a non-trivial sub-linear expectation space, built on the G-expectation framework introduced by Peng (2007) [4], equipped with a sequence of independent and identically distributed random variables.

Example 2. 

G-normal distribution.   A one-dimensional random variable ξ in a sub-linear expectation space $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$ is said to be G-normal distributed, denoted by $\xi \sim N(0,[{\underline{\sigma }}^2,{\overline{\sigma }}^2])$, where $\mathbb{E}\left[{\xi }^2\right]={\overline{\sigma }}^2$, $-\mathbb{E}[-{\xi }^2]={\underline{\sigma }}^2$, if for any $\phi \in C_{l,\mathit{Lip}}\left(\mathbb{R}\right)$, the function

$$u(t,x)=\mathbb{E}\left[\phi \left(x+\sqrt{t}\xi \right)\right]$$

is the unique viscosity solution of the G-heat equation:

$$\left\{\begin{array}{cc}{\partial }_{t}u-G\left({\partial }_{xx}u\right)=0,\hfill & t>0,x\in \mathbb{R},\hfill \\ u(0,x)=\phi \left(x\right),\hfill & \end{array}\right.$$

where $G\left(a\right)={\displaystyle \frac{1}{2}}{\overline{\sigma }}^2{a}^{+}-{\displaystyle \frac{1}{2}}{\underline{\sigma }}^2{a}^{-},a\in \mathbb{R}$.

It can be verified that the G-normal distributed random variable ξ satisfies: $a\xi +b\overline{\xi }\stackrel{d}{=}\sqrt{a^2+b^2}\xi ,\forall a,b\ge 0$, where $\overline{\xi }\stackrel{d}{=}\xi$ and $\overline{\xi }$ is independent to ξ. This implies $\mathbb{E}\left[\xi \right]=\mathbb{E}[-\xi ]=0$.

G-expectation.   Let $\tilde{\mathrm{\Omega }}=C[0,\infty )$ denote the space of all real-valued continuous paths ${\left({\omega }_t\right)}_{t\ge 0}$ with initial value ${\omega }_0=0$, which are equipped with the metric

$$\rho ({\omega }^1,{\omega }^2)={\displaystyle \sum _{i=1}^{\infty }}{2}^{-i}\left[\left(\underset{t\in [0,i]}{\max }|{\omega }_t^1-{\omega }_t^2|\right)\wedge 1\right].$$

Let $W_t\left(\omega \right)={\omega }_t$ be the canonical coordinate process on $\tilde{\mathrm{\Omega }}$. Define the space of cylindrical functionals

$$L_{ip}\left(\tilde{\mathrm{\Omega }}\right)=\left\{\phi (W_{t_1},\dots ,W_{t_k}):k\in \mathbb{N},0\le t_1<\cdots <t_k<\infty ,\phi \in C_{b,\mathrm{Lip}}\left({\mathbb{R}}^k\right)\right\},$$

where $C_{b,\mathit{Lip}}\left({\mathbb{R}}^k\right)$ stands for the space of bounded Lipschitz continuous functions on ${\mathbb{R}}^k$.

The G-expectation $\tilde{\mathbb{E}}$ is defined on $L_{ip}\left(\tilde{\mathrm{\Omega }}\right)$: for any cylindrical functional $\phi (W_{t_1},\dots ,W_{t_n})$ with $0\le t_1<\cdots <t_n$, set

$$\tilde{\mathbb{E}}\left[\phi (W_{t_1},\dots ,W_{t_{n-1}},W_{t_n}-W_{t_{n-1}})\right]=\tilde{\mathbb{E}}\left[\psi (W_{t_1},\dots ,W_{t_{n-1}})\right],$$

where $\psi (x_1,\dots ,x_{n-1})=\tilde{\mathbb{E}}\left[\phi \left(x_1,\dots ,x_{n-1},\sqrt{t_n-t_{n-1}}\xi \right)\right]$ with $\xi \sim N(0,[{\underline{\sigma }}^2,{\overline{\sigma }}^2])$.

For any $p\ge 1$, let $L_G^p\left(\tilde{\mathrm{\Omega }}\right)$ be the completion of $L_{ip}\left(\tilde{\mathrm{\Omega }}\right)$ under the norm

$${\parallel X\parallel }_{L_G^p}={\left(\tilde{\mathbb{E}}\left[{\left|X\right|}^p\right]\right)}^{1/p}.$$

The G-expectation $\tilde{\mathbb{E}}$ can be continuously extended to $L_G^p\left(\tilde{\mathrm{\Omega }}\right)$, and $(\tilde{\mathrm{\Omega }},L_G^p\left(\tilde{\mathrm{\Omega }}\right),\tilde{\mathbb{E}})$ forms a valid sub-linear expectation space.

G-Brownian motion.   The canonical process ${\left\{W_t\right\}}_{t\ge 0}$ is a standard G-Brownian motion under $\tilde{\mathbb{E}}$, which satisfies:

(i) 

$W_0=0$.

(ii) 

Independent increments: for any $0\le t_1<\cdots <t_n<t$, the increment $W_t-W_{t_n}$ is independent to $(W_{t_1},\dots ,W_{t_n})$.

(iii) 

Stationary increments: for any $0\le s<t$, $W_t-W_s\stackrel{d}{=}{W}_{t-s}$.

(iv) 

$W_1\sim N(0,[{\underline{\sigma }}^2,{\overline{\sigma }}^2])$.

IID sequence.   We now construct the sequence of random variables

$$X_n=W_n-W_{n-1},n\ge 1.$$

By the stationary increment property, $X_n\stackrel{d}{=}{W}_1$ for all $n\ge 1$, so $\{X_n;n\ge 1\}$ is an identically distributed sequence of G-normal random variables. By the independent increment property, $X_{n+1}$ is independent to $(X_1,\dots ,X_n)$ for each $n\ge 1$. Hence, $\{X_n;n\ge 1\}$ is an IID sequence under $\tilde{\mathbb{E}}$. Since independence implies negative dependence, the sequence also satisfies the negative dependence.

The following three lemmas play an important role in our proofs. Lemma 1 presents Rosenthal’s inequalities under sub-linear expectation for negatively dependent or independent random variables (see Zhang (2016) [5], Theorem 2.1, Corollary 2.2). Lemma 2 gives the upper bounds of the exponential inequalities under sub-linear expectation for negatively dependent random variables (see Zhang (2016) [8], Theorem 3.1). Lemma 3 is the Bernstein-type inequality under sub-linear expectation (see Zhang (2016) [19], Lemma 4.5).

Lemma 1. 

Let ${\left\{X_k\right\}}_{k=1}^n$ be a sequence of random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$.

(i) 

Suppose that $X_{k+1}$ is negatively dependent to $(X_1,\dots ,X_k)$ for each $k=1,\dots ,n-1$, and $\mathbb{E}\left[X_k\right]\le 0$. Then,

$$\mathbb{E}\left[{\left|\underset{1\le k\le n}{\max }\left({\displaystyle \sum _{i=k+1}^n}{X}_i\right)\right|}^p\right]\le 2^{2-p}{\displaystyle \sum _{k=1}^n}\mathbb{E}\left[{\left|X_k\right|}^p\right],1\le p\le 2$$

and

$$\mathbb{E}\left[{\left|\underset{1\le k\le n}{\max }\left({\displaystyle \sum _{i=k+1}^n}{X}_i\right)\right|}^p\right]\le C_p{n}^{p/2-1}{\displaystyle \sum _{k=1}^n}\mathbb{E}\left[{\left|X_k\right|}^p\right],p\ge 2.$$

In particular,

$$\mathbb{E}\left[{\left({\left({\displaystyle \sum _{k=1}^n}{X}_k\right)}^{+}\right)}^p\right]\le \left\{\begin{array}{cc}{2}^{2-p}{\displaystyle {\sum }_{k=1}^n}\mathbb{E}\left[{\left|X_k\right|}^p\right],\hfill & 1\le p\le 2,\hfill \\ C_p{n}^{p/2-1}{\displaystyle {\sum }_{k=1}^n}\mathbb{E}\left[{\left|X_k\right|}^p\right],\hfill & p\ge 2.\hfill \end{array}\right.$$

(ii) 

Suppose that $X_{k+1}$ is independent to $(X_1,\dots ,X_k)$ for each $k=1,\dots ,n-1$, and $\mathbb{E}\left[X_k\right]\le 0$. Then

$$\begin{array}{c}\hfill \mathbb{E}\left[{\left({\left({\displaystyle \sum _{k=1}^n}{X}_k\right)}^{+}\right)}^p\right]\le C_p\left\{{\displaystyle \sum _{k=1}^n}\mathbb{E}\left[{\left|X_k\right|}^p\right]+{\left({\displaystyle \sum _{k=1}^n}\mathbb{E}\left[{\left|X_k\right|}^2\right]\right)}^{p/2}\right\},\mathrm{for}p\ge 2.\end{array}$$

(iii) 

Suppose that $X_{k+1}$ is negatively dependent to $(X_1,\dots ,X_k)$ for each $k=1,\dots ,n-1$. Then,

$$\begin{array}{cc}\hfill \mathbb{E}\left[\underset{k\le n}{\max }{\left|{\displaystyle \sum _{i=1}^k}{X}_i\right|}^p\right]& \le C_p\left\{{\displaystyle \sum _{k=1}^n}\mathbb{E}\left[{\left|X_k\right|}^p\right]+{\left({\displaystyle \sum _{k=1}^n}\mathbb{E}\left[{\left|X_k\right|}^2\right]\right)}^{p/2}\right.\hfill \\ & \left.+{\left({\displaystyle \sum _{k=1}^n}\left[{\left(\mathcal{E}\left[X_k\right]\right)}^{-}+{\left(\mathbb{E}\left[X_k\right]\right)}^{+}\right]\right)}^p\right\},\mathrm{for}p\ge 2.\hfill \end{array}$$

Lemma 2. 

Let ${\left\{X_k\right\}}_{k=1}^n$ be a sequence of negatively dependent random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$ with $\mathbb{E}\left[X_k\right]\le 0$. Then, for any $p\ge 2$, there exists a constant $C_p\ge 1$ such that

$$\mathbb{V}\left(S_n\ge x\right)\le C_p{\delta }^{-2p}{\displaystyle \frac{{\sum }_{k=1}^n\mathbb{E}\left[{\left|X_k\right|}^p\right]}{x^p}}+\exp \left\{-{\displaystyle \frac{x^2}{2{\sum }_{k=1}^n\mathbb{E}\left[X_k^2\right](1+\delta )}}\right\},\forall x>0,\mathrm{and}0<\delta \le 1.$$

Lemma 3. 

Let ${\left\{X_k\right\}}_{k=1}^n$ be a sequence of independent random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$ with $\left|X_k\right|\le a$, $k=1,2,\dots ,n$, for some constant $a>0$. Then,

$$\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(X_i-\mathbb{E}\left[X_i\right]\right)\ge x\right)\le \exp \left\{-{\displaystyle \frac{x^2}{2\left(2ax+{\sum }_{i=1}^n\mathbb{E}\left[{(X_i-\mathbb{E}\left[X_i\right])}^2\right]\right)}}\right\}.$$

Chebyshev’s inequality is a fundamental tool in classical probability theory. The following lemma establishes the Chebyshev’s inequality under sub-linear expectation.

Lemma 4. 

Let X be a random variable in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$ and $f\left(x\right)\in C_{l,\mathit{Lip}}\left(\mathbb{R}\right)$ be a non-decreasing and positive function. Then, for any $x\in \mathbb{R}$,

$$\mathbb{V}\left(X\ge x\right)\le {\displaystyle \frac{\mathbb{E}\left[f\left(X\right)\right]}{f\left(x\right)}}.$$

Proof. 

By the monotonicity of $f\left(x\right)$, we have

$$I_{\{X\ge x\}}\le {\displaystyle \frac{f\left(X\right)}{f\left(x\right)}}{I}_{\{X\ge x\}}\le {\displaystyle \frac{f\left(X\right)}{f\left(x\right)}}.$$

Since $f\left(x\right)\in C_{l,\mathrm{Lip}}\left(\mathbb{R}\right)$ and $f\left(x\right)>0$, ${\displaystyle \frac{f\left(X\right)}{f\left(x\right)}}$ belongs to $\mathcal{H}$. It follows that

$$\mathbb{V}(X\ge x)\le \mathbb{E}\left[{\displaystyle \frac{f\left(X\right)}{f\left(x\right)}}\right]={\displaystyle \frac{\mathbb{E}\left[f\right(X\left)\right]}{f\left(x\right)}}.$$

□

3. Main Results and Proofs

Before stating our main results, we first present a lemma that forms the core of our proof framework and constitutes a key contribution of this paper. It provides sufficient conditions for the convergence of capacity series for weighted partial sums with exponentially growing weights, and it holds under a range of moment conditions for both non-negative and negatively dependent random sequences. As a unified technical tool, it will be invoked repeatedly in subsequent proofs to verify the convergence of both far-tail and intermediate truncation terms in our main theorems.

Lemma 5. 

Let ${\left\{Y_n\right\}}_{n=1}^{\infty }$ be a sequence of random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$. Let ${\left\{{\lambda }_n\right\}}_{n=1}^{\infty }$ be a positive sequence such that for some $N\ge 1$ and $r>1$, we have ${\lambda }_{n+1}\ge r{\lambda }_n$ for each $n>N$. Then, for any $\epsilon >0$, we have

(i) 

Suppose that $Y_1,Y_2,\dots$ are non-negative, ${\sum }_{n=1}^{\infty }\mathbb{E}\left[Y_n^p\right]<\infty$ for some $0<p\le 1$; then,

$${\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_n}}{\displaystyle \sum _{k=1}^n}{\lambda }_k{Y}_k\ge \epsilon \right)<\infty .$$

(5)

(ii) 

Suppose that $Y_1,Y_2,\dots$ are non-negative, ${\sum }_{n=1}^{\infty }n\mathbb{E}\left[Y_n^p\right]<\infty$ for some $0<p\le 1$; then,

$${\displaystyle \sum _{n=1}^{\infty }}n\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_n}}{\displaystyle \sum _{k=1}^n}{\lambda }_k{Y}_k\ge \epsilon \right)<\infty .$$

(6)

(iii) 

Suppose that $Y_1,Y_2,\dots$ are negatively dependent, $\mathbb{E}\left[Y_n\right]\le 0$ for each $n\ge 1$, and ${\sum }_{n=1}^{\infty }\mathbb{E}\left[{\left(Y_n^{+}\right)}^p\right]<\infty$ for some $1<p\le 2$; then, (5) holds.

(iv) 

Suppose that $Y_1,Y_2,\dots$ are negatively dependent, $\mathbb{E}\left[Y_n\right]\le 0$ for each $n\ge 1$, $\mathbb{E}\left[{\left(Y_n^{+}\right)}^2\right]=O\left({\displaystyle \frac{1}{n^{\alpha }}}\right)$ for some $\alpha >0$, and ${\sum }_{n=1}^{\infty }\mathbb{E}\left[{\left(Y_n^{+}\right)}^p\right]<\infty$ for some $p>2$; then, (5) holds.

(v) 

Suppose that $Y_1,Y_2,\dots$ are negatively dependent, $\mathbb{E}\left[Y_n\right]\le 0$ for each $n\ge 1$, $\mathbb{E}\left[{\left(Y_n^{+}\right)}^2\right]=O\left({\displaystyle \frac{1}{n^{\alpha }}}\right)$ for some $\alpha >0$, and ${\sum }_{n=1}^{\infty }n\mathbb{E}\left[{\left(Y_n^{+}\right)}^p\right]<\infty$ for some $p>2$; then, (6) holds.

Proof. 

Noting that ${\lambda }_n\ge r^{n-k}{\lambda }_k$, for $n\ge k>N$, for any $p>0$, we have

$${\displaystyle \sum _{n=k}^{\infty }}{\displaystyle \frac{1}{{\lambda }_n^p}}\le {\displaystyle \frac{1}{{\lambda }_k^p}}{\displaystyle \sum _{n=k}^{\infty }}{r}^{-p(n-k)}\le {\displaystyle \frac{C_{r,p}}{{\lambda }_k^p}},$$

and

$${\displaystyle \sum _{n=k}^{\infty }}{\displaystyle \frac{n}{{\lambda }_n^p}}\le {\displaystyle \frac{1}{{\lambda }_k^p}}{\displaystyle \sum _{n=k}^{\infty }}n{r}^{-p(n-k)}\le {\displaystyle \frac{C_{r,p}k}{{\lambda }_k^p}},$$

(i)

Noticing that

$${\left({\displaystyle \sum _{i=1}^n}{x}_i\right)}^p\le {\displaystyle \sum _{i=1}^n}{x}_i^p,$$

for $x_1,x_2,\dots ,x_n>0$ and $0<p\le 1$, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_n}}{\displaystyle \sum _{k=1}^n}{\lambda }_k{Y}_k\ge \epsilon \right)& \le {\displaystyle \frac{1}{{\epsilon }^p}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{{\lambda }_n^p}}\mathbb{E}\left[{\left({\displaystyle \sum _{k=1}^n}{\lambda }_k{Y}_k\right)}^p\right]\hfill \\ \hfill & \le {\displaystyle \frac{1}{{\epsilon }^p}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{{\lambda }_n^p}}{\displaystyle \sum _{k=1}^n}{\lambda }_k^p\mathbb{E}\left[Y_k^p\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{{\epsilon }^p}}{\displaystyle \sum _{k=1}^{\infty }}{\lambda }_k^p\mathbb{E}\left[Y_k^p\right]{\displaystyle \sum _{n=k}^{\infty }}{\displaystyle \frac{1}{{\lambda }_n^p}}\hfill \\ \hfill & \le C_{\epsilon ,p,r}{\displaystyle \sum _{k=1}^{\infty }}\mathbb{E}\left[Y_k^p\right]<\infty .\hfill \end{array}$$

(ii)

The proof is similar to (i), so we omit it.

(iii)

By Lemma 1 (i), we obtain

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_n}}{\displaystyle \sum _{k=1}^n}{\lambda }_k{Y}_k\ge \epsilon \right)& \le {\displaystyle \frac{1}{{\epsilon }^p}}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{{\lambda }_n^p}}\mathbb{E}\left[{\left({\left({\displaystyle \sum _{k=1}^n}{\lambda }_k{Y}_k^{+}\right)}^{+}\right)}^p\right]\hfill \\ \hfill & \le C_{\epsilon ,p}{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{{\lambda }_n^p}}{\displaystyle \sum _{k=1}^n}{\lambda }_k^p\mathbb{E}\left[{\left(Y_k^{+}\right)}^p\right]\hfill \\ \hfill & =C_{\epsilon ,p}{\displaystyle \sum _{k=1}^{\infty }}{\lambda }_k^p\mathbb{E}\left[{\left(Y_k^{+}\right)}^p\right]{\displaystyle \sum _{n=k}^{\infty }}{\displaystyle \frac{1}{{\lambda }_n^p}}\hfill \\ \hfill & \le C_{\epsilon ,p,r}{\displaystyle \sum _{k=1}^{\infty }}\mathbb{E}\left[{\left(Y_k^{+}\right)}^p\right]<\infty .\hfill \end{array}$$

(iv)

For any $p>2$ and $0<\delta \le 1$, by Lemma 2, it follows

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}& \left({\displaystyle \frac{1}{{\lambda }_n}}{\displaystyle \sum _{k=1}^n}{\lambda }_k{Y}_k\ge \epsilon \right)\le {\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}\left({\displaystyle \sum _{k=1}^n}{\lambda }_k{Y}_k^{+}\ge {\lambda }_n\epsilon \right)\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}\left[{\displaystyle \frac{C_{p,\delta ,\epsilon }}{{\lambda }_n^p}}{\displaystyle \sum _{k=1}^n}\mathbb{E}\left[{\left({\lambda }_k{Y}_k^{+}\right)}^p\right]+\exp \left\{-{\displaystyle \frac{{\epsilon }^2{\lambda }_n^2}{2(1+\delta ){\sum }_{k=1}^n\mathbb{E}\left[{\left({\lambda }_k{Y}_k^{+}\right)}^2\right]}}\right\}\right]\hfill \\ \hfill & \le C_{p,\delta ,\epsilon }{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{{\lambda }_n^p}}{\displaystyle \sum _{k=1}^n}{\lambda }_k^p\mathbb{E}\left[{\left(Y_k^{+}\right)}^p\right]+{\displaystyle \sum _{n=1}^{\infty }}\exp \left\{-{\displaystyle \frac{{\epsilon }^2{\lambda }_n^2}{2(1+\delta ){\sum }_{k=1}^n{\lambda }_k^2\mathbb{E}\left[{\left(Y_k^{+}\right)}^2\right]}}\right\}\hfill \\ \hfill & :=I+II.\hfill \end{array}$$

For I, by the similar method of (i), we can easily obtain $I<\infty$.

For $II$, noting that $\mathbb{E}\left[{\left(Y_k^{+}\right)}^2\right]\le {\displaystyle \frac{C}{k^{\alpha }}}$ and ${\lambda }_k\le r^{k-n}{\lambda }_n$ for $n\ge k>N$, we have

$${\displaystyle \sum _{k=1}^n}{\lambda }_k^2\mathbb{E}\left[{\left(Y_k^{+}\right)}^2\right]\le C{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{\lambda }_k^2}{k^{\alpha }}}\le C{\lambda }_n^2{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{r^{2(k-n)}}{k^{\alpha }}}\le {\displaystyle \frac{C_{r,\alpha }{\lambda }_n^2}{n^{\alpha }}}.$$

Therefore,

$$II\le {\displaystyle \sum _{n=1}^{\infty }}\exp \left\{-{\displaystyle \frac{{\epsilon }^2}{2C_{r,\alpha }(1+\delta )}}·n^{\alpha }\right\}<\infty .$$

(v)

The proof is similar to (iv), so we omit it.

□

Remark 2. 

The auxiliary sequence ${\left\{{\lambda }_n\right\}}_{n=1}^{\infty }$ introduced in Lemma 5 is required to grow at least exponentially for sufficiently large n. Examples satisfying ${\lambda }_{n+1}\ge r{\lambda }_n$ for all $n>N$ ($r>1$) include ${\lambda }_n=r^n$, ${\lambda }_n=n^{\alpha }{r}^n$ and ${\lambda }_n=r^{2^n}$ with $r>1,\alpha >0$. We point out that polynomial sequences ${\lambda }_n=n^{\alpha }$ with $\alpha >0$ fail this condition.

Next, we investigate the convergence rates for the law of large numbers under sub-linear expectation, starting with a supporting lemma.

Lemma 6. 

Let ${\left\{X_n\right\}}_{n=1}^{\infty }$ be a sequence of random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$. If $C_{\mathbb{V}}\left[{\left|X_1\right|}^t\right]<\infty$ for some $t>0$, then for any $\epsilon >0$, the following statements hold:

(i) 

For any $0<p<t$,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left({\left(\left|X_1\right|-\epsilon n^{1/t}\right)}^{+}\right)}^p\right]}{n^{p/t}}}<\infty .$$

(7)

(ii) 

For any $p>t$,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon n^{1/t}\right)\right)}^p\right]}{n^{p/t}}}<\infty .$$

(8)

(iii) 

For any $0<p\le t$,

$$\mathbb{E}\left[{\left({\left(|X_1|-c\right)}^{+}\right)}^p\right]=o\left(c^{p-t}\right),asc\to \infty .$$

(9)

Proof. 

(i) For any $0<p<t$, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left({\left(|X_1|-\epsilon n^{1/t}\right)}^{+}\right)}^p\right]}{n^{p/t}}}& \le {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{C_{\mathbb{V}}\left[{\left({\left(|X_1|-\epsilon n^{1/t}\right)}^{+}\right)}^p\right]}{n^{p/t}}}\hfill \\ \hfill & \le C_p{\displaystyle \sum _{n=1}^{\infty }}{n}^{-p/t}{\int }_{\epsilon n^{1/t}}^{\infty }{x}^{p-1}\mathbb{V}\left(\right|X_1|>x)dx\hfill \\ \hfill & \le C_{p,t}{\displaystyle \sum _{n=1}^{\infty }}{n}^{-p/t}{\int }_{{\epsilon }^{t}n}^{\infty }{y}^{{\displaystyle \frac{p}{t}}-1}\mathbb{V}\left(|X_1{|}^t>y\right)dy\hfill \\ \hfill & \le C_{p,t}{\int }_1^{\infty }{z}^{-p/t}{\int }_{{\epsilon }^{t}z}^{\infty }{y}^{{\displaystyle \frac{p}{t}}-1}\mathbb{V}\left(|X_1{|}^t>y\right)dydz\hfill \\ \hfill & \le C_{p,t}{\int }_{{\epsilon }^t}^{\infty }{y}^{{\displaystyle \frac{p}{t}}-1}\mathbb{V}\left(|X_1{|}^t>y\right){\int }_1^{y/{\epsilon }^t}{z}^{-p/t}dzdy\hfill \\ \hfill & \le C_{p,t,\epsilon }{\int }_0^{\infty }\mathbb{V}\left(|X_1{|}^t>y\right)dy=C_{p,t,\epsilon }{C}_{\mathbb{V}}\left[{\left|X_1\right|}^t\right]<\infty ,\hfill \end{array}$$

where the first inequality follows from $\mathbb{E}\left[\left|X\right|\right]\le C_{\mathbb{V}}\left(\left|X\right|\right)$. Thus, we obtain (7).

(ii)

For any $p>t$, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon n^{1/t}\right)\right)}^p\right]}{n^{p/t}}}& \le C_p{\displaystyle \sum _{n=1}^{\infty }}{n}^{-p/t}{\int }_{\epsilon }^{\epsilon n^{1/t}}{x}^{p-1}\mathbb{V}\left(\left|X_1\right|>x\right)dx+C_{\epsilon ,p}{\displaystyle \sum _{n=1}^{\infty }}{n}^{-p/t}\hfill \\ \hfill & \le C_{p,t}{\int }_1^{\infty }{z}^{-p/t}{\int }_{{\epsilon }^t}^{{\epsilon }^{t}z}{y}^{{\displaystyle \frac{p}{t}}-1}\mathbb{V}\left({\left|X_1\right|}^t>y\right)dydz+C_{\epsilon ,p,t}\hfill \\ \hfill & \le C_{p,t}{\int }_{{\epsilon }^t}^{\infty }{y}^{{\displaystyle \frac{p}{t}}-1}\mathbb{V}\left({\left|X_1\right|}^t>y\right){\int }_{y/{\epsilon }^t}^{\infty }{z}^{-p/t}dzdy+C_{\epsilon ,p,t}\hfill \\ \hfill & \le C_{p,t,\epsilon }{\int }_0^{\infty }\mathbb{V}\left({\left|X_1\right|}^t>y\right)dy+C_{\epsilon ,p,t}\hfill \\ \hfill & =C_{p,t,\epsilon }{C}_{\mathbb{V}}\left[{\left|X_1\right|}^t\right]+C_{\epsilon ,p,t}<\infty .\hfill \end{array}$$

(iii)

For any $0<p\le t$, we have

$$\begin{array}{cc}\hfill \mathbb{E}\left[{\left({\left(\left|X_1\right|-c\right)}^{+}\right)}^p\right]& \le C_p{\int }_c^{\infty }{x}^{p-1}\mathbb{V}\left(|X_1|>x\right)dx\le C_{p,t}{\int }_{c^t}^{\infty }{y}^{{\displaystyle \frac{p}{t}}-1}\mathbb{V}\left(|X_1{|}^t>y\right)dy\hfill \\ \hfill & \le C_{p,t}·c^{p-t}{\int }_{c^t}^{\infty }\mathbb{V}\left(|X_1{|}^t>y\right)dy=o\left(c^{p-t}\right).\hfill \end{array}$$

□

Theorem 1. 

Let ${\left\{X_n\right\}}_{n=1}^{\infty }$ be a sequence of negatively dependent and identically distributed random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$. Suppose that $\mathbb{V}$ is countably sub-additive. Set $S_n={\sum }_{k=1}^n{X}_k$, $\mathbb{E}\left[X_1\right]=\overline{\mu }$, $\mathcal{E}\left[X_1\right]=\underline{\mu }$. Then, for any $\epsilon >0$, we have

(i) 

If $C_{\mathbb{V}}\left[{\left|X_1\right|}^t\right]<\infty$ for some $0<t<1$ and ${\lim }_{c\to \infty }\mathbb{E}\left[{\left(|X_1|-c\right)}^{+}\right]=0$, then

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left(\left|S_n\right|>n^{1/t}\epsilon \right)<\infty ,$$

(10)

(ii) 

If $C_{\mathbb{V}}\left[{\left|X_1\right|}^t\right]<\infty$ for some $1\le t<2$, then

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left(S_n-n\overline{\mu }>n^{1/t}\epsilon \right)<\infty ,$$

(11)

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left(S_n-n\underline{\mu }<-n^{1/t}\epsilon \right)<\infty .$$

(12)

Proof. 

Without loss of generality, assume $\overline{\mu }=0$. Denote $b_n=n^{1/t}$, define

$$\begin{array}{cc}\hfill & Y_n=(-\epsilon b_n)\vee (X_n\wedge \epsilon b_n),\hfill \\ \hfill & X_n^{\left(1\right)}=0\vee (X_n-\epsilon b_n),\hfill \\ \hfill & X_n^{\left(2\right)}=0\wedge (X_n+\epsilon b_n).\hfill \end{array}$$

Thus, we have

$$X_n=Y_n+X_n^{\left(1\right)}+X_n^{\left(2\right)}.$$

For $X_n^{\left(1\right)}$, we need to show

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{3}}{b}_n\right)<\infty .$$

(13)

Noting that $b_{2^k}/b_{2^{k-1}}=2^{1/t}>1$ for each $k\ge 1$, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}& {\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{3}}{b}_n\right)={\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \sum _{n=2^{k-1}}^{2^k-1}}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \frac{1}{b_n}}{\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{3}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \sum _{n=2^{k-1}}^{2^k-1}}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \frac{1}{b_{2^{k-1}}}}{\displaystyle \sum _{m=1}^k}{\displaystyle \sum _{i=2^{m-1}}^{2^m-1}}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{3}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{k=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{b_{2^{k-1}}}}{\displaystyle \sum _{m=1}^k}\left(\left({\displaystyle \frac{1}{b_{2^{m-1}}}}{\displaystyle \sum _{i=2^{m-1}}^{2^m-1}}{X}_i^{\left(1\right)}\right)b_{2^{m-1}}\right)>{\displaystyle \frac{\epsilon }{3}}\right){\displaystyle \sum _{n=2^{k-1}}^{2^k-1}}{\displaystyle \frac{1}{n}}\hfill \\ \hfill & \le C{\displaystyle \sum _{k=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{b_{2^{k-1}}}}{\displaystyle \sum _{m=1}^k}\left(\left({\displaystyle \frac{1}{b_{2^{m-1}}}}{\displaystyle \sum _{i=2^{m-1}}^{2^m-1}}{X}_i^{\left(1\right)}\right)b_{2^{m-1}}\right)>{\displaystyle \frac{\epsilon }{3}}\right).\hfill \end{array}$$

(14)

Denoting $Z_k={\displaystyle \frac{1}{b_{2^{k-1}}}}{\sum }_{i=2^{k-1}}^{2^k-1}{X}_i^{\left(1\right)}$ and ${\lambda }_k=b_{2^{k-1}}$, (14) becomes

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{3}}{b}_n\right)\le C{\displaystyle \sum _{k=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_k}}{\displaystyle \sum _{m=1}^k}{\lambda }_m{Z}_m>{\displaystyle \frac{\epsilon }{3}}\right).\end{array}$$

By the definition of $X_n^{\left(1\right)}$, for any $0<p\le 1$, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=1}^{\infty }}\mathbb{E}\left[Z_k^p\right]& ={\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{1}{b_{2^{k-1}}^p}}\mathbb{E}\left[{\left({\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}{X}_i^{\left(1\right)}\right)}^p\right]\hfill \\ \hfill & \le {\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{1}{b_{2^{k-1}}^p}}{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}\mathbb{E}\left[{\left(X_i^{\left(1\right)}\right)}^p\right]\hfill \\ \hfill & \le C{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left({\left(|X_1|-\epsilon b_i\right)}^{+}\right)}^p\right]}{b_i^p}}.\hfill \end{array}$$

(15)

In the case $0<t<1$, taking $p=1$ in (15), noting that ${\lim }_{i\to \infty }\mathbb{E}\left[{\left(|X_1|-\epsilon b_i\right)}^{+}\right]=0$ and ${\sum }_{i=1}^{\infty }1/b_i<\infty$, we have

$${\displaystyle \sum _{k=1}^{\infty }}\mathbb{E}\left[Z_k\right]\le o\left(1\right){\displaystyle \sum _{i=1}^{\infty }}1/b_i<\infty .$$

In the case $1<t<2$, taking $p=1$ in (15), by Lemma 6 (i), we have

$$\begin{array}{c}\hfill {\displaystyle \sum _{k=1}^{\infty }}\mathbb{E}\left[Z_k\right]\le C{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left(|X_1|-\epsilon b_i\right)}^{+}\right]}{b_i}}<\infty .\end{array}$$

In the case $t=1$, taking $0<p<1$ in (15), by Lemma 6 (i), we have

$$\begin{array}{c}\hfill {\displaystyle \sum _{k=1}^{\infty }}\mathbb{E}\left[Z_k^p\right]\le C{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left({\left(|X_1|-\epsilon b_i\right)}^{+}\right)}^p\right]}{b_i^p}}<\infty .\end{array}$$

Noting that $Z_k\ge 0$ for each $k\ge 1$, by Lemma 5 (i), (13) holds for $0<t<2$.

For $X_n^{\left(2\right)}$, by noting that $X_n^{\left(2\right)}\le 0$, it is obvious that for $0<t<2$,

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(2\right)}>{\displaystyle \frac{\epsilon }{3}}{b}_n\right)<\infty .\end{array}$$

(16)

For $Y_n$, let $n_k=\left[{\theta }^k\right]$ where $\theta >1$. For each $n\ge 1$, there exists k such that $n_k<n\le n_{k+1}$. Then, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{i=1}^n}{Y}_i& \le {\displaystyle \sum _{i=1}^{n_k}}{Y}_i+\underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \sum _{i=n_k+1}^n}{Y}_i\hfill \\ \hfill & \le {\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}+{\displaystyle \sum _{i=1}^{n_k}}\left|\mathbb{E}\left[Y_i\right]\right|+\underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \sum _{i=n_k+1}^n}{Y}_i\hfill \\ \hfill & :=I+II+III.\hfill \end{array}$$

For I, note that $\{Y_i-\mathbb{E}\left[Y_i\right]$, $i=1,2,\dots \}$ are negatively dependent.

In the case $0<t<1$, by Lemma 1 (i), we obtain

$$\begin{array}{cc}\hfill \mathbb{V}\left({\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}>{\displaystyle \frac{\epsilon }{9}}{b}_{n_k}\right)& \le C_{\epsilon }{\displaystyle \frac{{\sum }_{i=1}^{n_k}\mathbb{E}\left[\left|Y_i-\mathbb{E}\left[Y_i\right]\right|\right]}{b_{n_k}}}\hfill \\ \hfill & =C_{\epsilon }{\displaystyle \frac{{\sum }_{i=1}^{n_k}\mathbb{E}\left[\left|X_i\right|\wedge \epsilon b_i\right]}{b_{n_k}}}.\hfill \end{array}$$

Therefore, by the fact that ${\sum }_{n=n_k}^{n_{k+1}-1}{\displaystyle \frac{1}{n}}\sim C_{\theta }$ and ${\sum }_{\begin{array}{c}{n}_k\ge i\end{array}}{b_{n_k}}^{-1}\sim C_{\theta ,t}{b_i}^{-1}$, as k → ∞, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}>{\displaystyle \frac{\epsilon }{9}}{b}_n\right)& ={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \sum _{n=n_k}^{n_{k+1}-1}}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}>{\displaystyle \frac{\epsilon }{9}}{b}_n\right)\hfill \\ \hfill & \le {\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \sum _{n=n_k}^{n_{k+1}-1}}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}>{\displaystyle \frac{\epsilon }{9}}{b}_{n_k}\right)\hfill \\ \hfill & ={\displaystyle \sum _{k=0}^{\infty }}\mathbb{V}\left({\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}>{\displaystyle \frac{\epsilon }{9}}{b}_{n_k}\right){\displaystyle \sum _{n=n_k}^{n_{k+1}-1}}{\displaystyle \frac{1}{n}}\hfill \\ \hfill & \le C_{\theta }{\displaystyle \sum _{k=0}^{\infty }}\mathbb{V}\left({\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}>{\displaystyle \frac{\epsilon }{9}}{b}_{n_k}\right)\hfill \\ \hfill & \le C_{\theta ,\epsilon }{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\sum }_{i=1}^{n_k}\mathbb{E}\left[\left|X_1\right|\wedge \left(\epsilon b_i\right)\right]}{b_{n_k}}}\hfill \\ \hfill & \le C_{\theta ,\epsilon }{\displaystyle \sum _{i=1}^{\infty }}\mathbb{E}\left[\left|X_1\right|\wedge \left(\epsilon b_i\right)\right]\sum _{\begin{array}{c}{n}_k\ge i\end{array}}{b}_{n_k}\hfill \\ \hfill & \le C_{\theta ,\epsilon ,t}{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[\left|X_1\right|\wedge \left(\epsilon b_i\right)\right]}{b_i}}<\infty ,\hfill \end{array}$$

(17)

where the last inequality follows from (8).

In the case $1\le t<2$, note that ${\sum }_{\begin{array}{c}{n}_k\ge i\end{array}}{b_{n_k}}^{-2}\sim C_{\theta ,t}{b_i}^{-2}$, as k → ∞. By the similar method, we can also obtain that

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}>{\displaystyle \frac{\epsilon }{9}}{b}_n\right)& \le C_{\theta }{\displaystyle \sum _{k=0}^{\infty }}\mathbb{V}\left({\left({\displaystyle \sum _{i=1}^{n_k}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right)}^{+}>{\displaystyle \frac{\epsilon }{9}}{b}_{n_k}\right)\hfill \\ \hfill & \le C_{\theta ,\epsilon }{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\sum }_{i=1}^{n_k}\mathbb{E}\left[{\left|Y_i-\mathbb{E}\left[Y_i\right]\right|}^2\right]}{b_{n_k}^2}}\hfill \\ \hfill & \le C_{\theta ,\epsilon }{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{\sum }_{i=1}^{n_k}\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon b_i\right)\right)}^2\right]}{b_{n_k}^2}}\hfill \\ \hfill & \le C_{\theta ,\epsilon }{\displaystyle \sum _{i=1}^{\infty }}\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon b_i\right)\right)}^2\right]\sum _{\begin{array}{c}{n}_k\ge i\end{array}}{b}_{n_k}^{-2}\hfill \\ \hfill & \le C_{\theta ,\epsilon ,t}{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon b_i\right)\right)}^2\right]}{b_i^2}}<\infty ,\hfill \end{array}$$

(18)

where the last inequality follows from (8).

For $II$, in the case $0<t<1$, from the fact that $\overline{\mu }=0$, $\left|\mathbb{E}\left[Y_i\right]\right|=\left|\mathbb{E}\left[Y_i\right]-\overline{\mu }\right|\le \mathbb{E}\left[{\left(|X_1|-\epsilon b_i\right)}^{+}\right]=o\left(1\right)$ as $i\to \infty$, we have

$$\begin{array}{c}\hfill {\displaystyle \frac{{\sum }_{i=1}^{n_k}\left|\mathbb{E}\left[Y_i\right]\right|}{b_{n_k}}}\le o\left(1\right)·n_k^{1-1/t}\to 0,ask\to \infty .\end{array}$$

(19)

In the case $1\le t<2$, by (9), we have $\left|\mathbb{E}\left[Y_i\right]\right|\le \mathbb{E}\left[{\left(|X_1|-\epsilon b_i\right)}^{+}\right]=o\left({\left(\epsilon b_i\right)}^{1-t}\right)$ as $i\to \infty$. Noting that ${\sum }_{i=1}^{n_k}{\left(\epsilon b_i\right)}^{1-t}\le C_{\epsilon ,t}{b}_{n_k}$, we have

$$\begin{array}{c}\hfill {\displaystyle \frac{{\sum }_{i=1}^{n_k}\left|\mathbb{E}\left[Y_i\right]\right|}{b_{n_k}}}\le o\left(1\right)·{\displaystyle \frac{{\sum }_{i=1}^{n_k}{\left(\epsilon b_i\right)}^{1-t}}{b_{n_k}}}\to 0,ask\to \infty .\end{array}$$

(20)

For $III$, we have

$$\begin{array}{cc}\hfill \underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \sum _{i=n_k+1}^n}{Y}_i& \le {\left({\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\left(Y_i^{+}-\mathbb{E}\left[Y_i^{+}\right]\right)\right)}^{+}\hfill \\ \hfill & +{\left({\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\left(Y_i^{-}-\mathbb{E}\left[Y_i^{-}\right]\right)\right)}^{+}+2{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\left|\mathbb{E}\left[Y_i\right]\right|\hfill \\ \hfill & :=II{I}_1+II{I}_2+II{I}_3.\hfill \end{array}$$

The proofs of the convergence of the series for $II{I}_1$ and $II{I}_2$ are analogous to that of I, and the proof of the convergence of the series for $II{I}_3$ is similar to that of $II$. Thus, we obtain

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left(\underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \sum _{i=n_k+1}^n}{Y}_i>{\displaystyle \frac{\epsilon }{9}}{b}_n\right)<\infty .$$

(21)

Therefore, from (17)–(21), we have for any $0<t<2$,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{Y}_i>{\displaystyle \frac{\epsilon }{3}}{b}_n\right)<\infty .$$

(22)

Combining (13), (16) and (22), we obtain

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left(S_n>\epsilon n^{1/t}\right)& \le {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{3}}{n}^{1/t}\right)+{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(2\right)}>{\displaystyle \frac{\epsilon }{3}}{n}^{1/t}\right)\hfill \\ \hfill & +{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{Y}_i>{\displaystyle \frac{\epsilon }{3}}{n}^{1/t}\right)<\infty .\hfill \end{array}$$

(23)

Thus, (11) holds. Replacing $\{X_n;n\ge 1\}$ with $\{-X_n;n\ge 1\}$ in (23), we immediately obtain (12).

Hence, (10) can be directly derived from the following expression.

$$\mathbb{V}\left(\left|S_n\right|>n^{1/t}\epsilon \right)\le \mathbb{V}\left(S_n>n^{1/t}\epsilon \right)+\mathbb{V}\left(S_n<-n^{1/t}\epsilon \right).$$

□

Remark 3. 

Theorem 1 reveals that for the law of large numbers under sub-linear expectation, the moment condition corresponding to the convergence rates of order $1/n$ coincides precisely with its counterpart in the classical probability framework, as specified in (1) and (2).

In what follows, we establish the convergence rates for the law of the logarithm under sub-linear expectation. We first state a lemma that serves a purpose similar to Lemma 6.

Lemma 7. 

Let ${\left\{X_n\right\}}_{n=1}^{\infty }$ be a sequence of random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$. If $C_{\mathbb{V}}\left[{\displaystyle \frac{X_1^2}{\log |X_1|}}\right]<\infty$, then for any $\epsilon >0$, the following statements hold:

(i) 

For any $0<p<2$,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left({\left(|X_1|-\epsilon \sqrt{n\log n}\right)}^{+}\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}<\infty .$$

(24)

(ii) 

For any $p>2$,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left(|X_1|\wedge \left(\epsilon \sqrt{n\log n}\right)\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}<\infty .$$

(25)

Proof. 

(i) We start with the proof of (24). Let $f\left(x\right)$ be the inverse function of $\epsilon \sqrt{x\log x}$.

For $0<p<2$, we have

• $$\begin{array}{cc}\hfill {\displaystyle \sum _{n=4}^{\infty }}& {\displaystyle \frac{\mathbb{E}\left[{\left({\left(|X_1|-\epsilon \sqrt{n\log n}\right)}^{+}\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}\le {\displaystyle \sum _{n=4}^{\infty }}{\displaystyle \frac{C_{\mathbb{V}}\left[{\left(\left(\right|X_1{|-\epsilon \sqrt{n\log n})}^{+}\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}\hfill \\ \hfill & \le C_p{\displaystyle \sum _{n=4}^{\infty }}{\left(n\log n\right)}^{-p/2}{\int }_{\epsilon \sqrt{n\log n}}^{\infty }{x}^{p-1}\mathbb{V}\left(\left|X_1\right|>x\right)dx\hfill \\ \hfill & \le C_{p,\epsilon }{\displaystyle \sum _{n=4}^{\infty }}{\left(n\log n\right)}^{-p/2}{\int }_n^{\infty }{\left(\log y\right)}^{p/2}{y}^{p/2-1}\mathbb{V}\left(f\right(|X_1\left|\right)>y)dy\hfill \\ \hfill & \le C_{p,\epsilon }{\int }_4^{\infty }{\left(z\log z\right)}^{-p/2}{\int }_z^{\infty }{\left(\log y\right)}^{p/2}{y}^{p/2-1}\mathbb{V}\left(f\right(|X_1\left|\right)>y)dydz\hfill \\ \hfill & \le C_{p,\epsilon }{\int }_4^{\infty }{\left(\log y\right)}^{p/2}{y}^{p/2-1}\mathbb{V}\left(f\right(|X_1\left|\right)>y){\int }_4^y{\left(z\log z\right)}^{-p/2}dzdy\hfill \\ \hfill & \le C_{p,\epsilon }{\int }_0^{\infty }\mathbb{V}\left(f\right(|X_1\left|\right)>y)dy\hfill \\ \hfill & =C_{p,\epsilon }{\int }_0^{\infty }\mathbb{V}({\displaystyle \frac{X_1^2}{\log |X_1|}}>{\displaystyle \frac{{\left(\epsilon \sqrt{y\log y}\right)}^2}{\log \left(\epsilon \sqrt{y\log y}\right)}})dy\hfill \\ \hfill & =C_{p,\epsilon }{\int }_0^{\infty }\mathbb{V}({\displaystyle \frac{X_1^2}{\log |X_1|}}>{\displaystyle \frac{{\epsilon }^{2}y\log y}{\log \epsilon +(\log y+\log \log y)/2}})dy.\hfill \end{array}$$

  Note that there exist some $M>0$ such that for any $y\ge M$,

  $${\displaystyle \frac{{\epsilon }^{2}y\log y}{\log \epsilon +(\log y+\log \log y)/2}}\ge C_{\epsilon }y.$$

  Then, we have

  $${\int }_M^{\infty }\mathbb{V}({\displaystyle \frac{X_1^2}{\log |X_1|}}>{\displaystyle \frac{{\epsilon }^{2}y\log y}{\log \epsilon +(\log y+\log \log y)/2}})dy\le C_{\epsilon }{C}_{\mathbb{V}}\left[{\displaystyle \frac{X_1^2}{\log |X_1|}}\right]<\infty .$$

  Combining this with

  $${\int }_0^M\mathbb{V}({\displaystyle \frac{X_1^2}{\log |X_1|}}>{\displaystyle \frac{{\epsilon }^{2}y\log y}{\log \epsilon +(\log y+\log \log y)/2}})dy\le M,$$

  we obtain (24).

(ii)

By a similar argument as in (i), we can also obtain for $p>2$,

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=4}^{\infty }}& {\displaystyle \frac{\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon \sqrt{n\log n}\right)\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}\le {\displaystyle \sum _{n=4}^{\infty }}{\displaystyle \frac{C_{\mathbb{V}}\left[{\left(\left|X_1\right|\wedge \left(\epsilon \sqrt{n\log n}\right)\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}\hfill \\ \hfill & \le C_p{\displaystyle \sum _{n=4}^{\infty }}{\left(n\log n\right)}^{-p/2}{\int }_{\epsilon \sqrt{4\log 4}}^{\epsilon \sqrt{n\log n}}{x}^{p-1}\mathbb{V}\left(\left|X_1\right|>x\right)dx+C_{\epsilon ,p}{\displaystyle \sum _{n=4}^{\infty }}{\displaystyle \frac{1}{{\left(\sqrt{n\log n}\right)}^p}}\hfill \\ \hfill & \le C_{p,\epsilon }{\int }_4^{\infty }{\left(z\log z\right)}^{-p/2}{\int }_4^z{\left(\log y\right)}^{p/2}{y}^{p/2-1}\mathbb{V}\left(f\right(|X_1\left|\right)>y)dydz+C_{\epsilon ,p}\hfill \\ \hfill & \le C_{p,\epsilon }{\int }_4^{\infty }{\left(\log y\right)}^{p/2}{y}^{p/2-1}\mathbb{V}\left(f\right(|X_1\left|\right)>y){\int }_y^{\infty }{\left(z\log z\right)}^{-p/2}dzdy+C_{\epsilon ,p}\hfill \\ \hfill & \le C_{p,\epsilon }{\int }_0^{\infty }\mathbb{V}\left(f\right(|X_1\left|\right)>y)dy+C_{\epsilon ,p}\hfill \\ \hfill & <\infty .\hfill \end{array}$$

The derivation of the final inequality is identical to the corresponding part in (i) and is thus omitted.

  □

Theorem 2. 

Let ${\left\{X_n\right\}}_{n=1}^{\infty }$ be a sequence of independent and identically distributed random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$. Suppose that $\mathbb{V}$ is countably sub-additive. Set $S_n={\sum }_{k=1}^n{X}_k$, $\mathbb{E}\left[X_1\right]=\overline{\mu }$, $\mathcal{E}\left[X_1\right]=\underline{\mu }$, $\mathbb{E}\left[{(X_1-\mathbb{E}\left[X_1\right])}^2\right]={\overline{\sigma }}^2$. If $C_{\mathbb{V}}\left[{\displaystyle \frac{X_1^2}{\log |X_1|}}\right]<\infty$ and ${\lim }_{c\to \infty }\mathbb{E}\left[\left(\right|X_1{{|}^2-c)}^{+}\right]=0$, then for any $\epsilon >0$, we have

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left(S_n-n\overline{\mu }>\epsilon \sqrt{n\log n}\right)<\infty ,$$

(26)

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left(S_n-n\underline{\mu }<-\epsilon \sqrt{n\log n}\right)<\infty .$$

(27)

Proof. 

Without loss of generality, we assume $\overline{\mu }=0$, ${\overline{\sigma }}^2=1$. Denote $a_n=\sqrt{n\log n}$, $b_n={\displaystyle \frac{5}{4}}\sqrt{n/\log n}$. We define

$$\begin{array}{cc}\hfill & Y_n=(-\epsilon b_n)\vee (X_n\wedge \epsilon b_n),\hfill \\ \hfill & X_n^{\left(1\right)}=\left(0\vee (X_n-\epsilon b_n)\right)\wedge (\epsilon a_n-\epsilon b_n),\hfill \\ \hfill & X_n^{\left(2\right)}=\left(0\wedge (X_n+\epsilon b_n)\right)\vee (\epsilon b_n-\epsilon a_n),\hfill \\ \hfill & X_n^{\left(3\right)}=0\vee (X_n-\epsilon a_n),\hfill \\ \hfill & X_n^{\left(4\right)}=0\wedge (X_n+\epsilon a_n).\hfill \end{array}$$

Thus, we have

$$\begin{array}{cc}\hfill & X_n=Y_n+X_n^{\left(1\right)}+X_n^{\left(2\right)}+X_n^{\left(3\right)}+X_n^{\left(4\right)}.\hfill \end{array}$$

We first show (26). For $X_n^{\left(1\right)}$, from the definition of $X_n^{\left(1\right)}$, we have

$$\left|\mathbb{E}\left[X_i^{\left(1\right)}\right]\right|\le \mathbb{E}\left[{\left(|X_1|-\epsilon b_i\right)}^{+}\right]\le {\left(\epsilon b_i\right)}^{-1}\mathbb{E}\left[{\left(X_1^2-{\left(\epsilon b_i\right)}^2\right)}^{+}\right].$$

(28)

Note that ${\sum }_{i=1}^n{b}_i^{-1}\sim C{a}_n$ as $n\to \infty$. Hence, by ${\lim }_{c\to \infty }\mathbb{E}\left[\left(\right|X_1{{|}^2-c)}^{+}\right]=0$,

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\sum }_{i=1}^n\left|\mathbb{E}\left[X_i^{\left(1\right)}\right]\right|}{a_n}}& \le {\displaystyle \frac{{\sum }_{i=1}^n{\left(\epsilon b_i\right)}^{-1}\mathbb{E}\left[{\left(X_1^2-{\left(\epsilon b_i\right)}^2\right)}^{+}\right]}{a_n}}\hfill \\ \hfill & =o\left(1\right)·{\displaystyle \frac{{\sum }_{i=1}^n{\left(\epsilon b_i\right)}^{-1}}{a_n}}\to 0,\mathrm{as}n\to \infty .\hfill \end{array}$$

(29)

Following the argument of (14) in Theorem 1, we have

$$\begin{array}{cc}& {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)\hfill \\ \hfill \le & C{\displaystyle \sum _{k=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{a_{2^{k-1}}}}{\displaystyle \sum _{m=1}^k}\left(\left({\displaystyle \frac{1}{a_{2^{m-1}}}}{\displaystyle \sum _{i=2^{m-1}}^{2^m-1}}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)\right)a_{2^{m-1}}\right)>{\displaystyle \frac{\epsilon }{5}}\right).\hfill \end{array}$$

Denoting $Z_k={\displaystyle \frac{1}{a_{2^{k-1}}}}{\sum }_{i=2^{k-1}}^{2^k-1}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)$ and ${\lambda }_k=a_{2^{k-1}}$, we have

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)\le C{\displaystyle \sum _{k=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_k}}{\displaystyle \sum _{m=1}^k}{\lambda }_m{Z}_m>{\displaystyle \frac{\epsilon }{5}}\right).$$

Since $\mathbb{E}\left[{\left(X_i^{\left(1\right)}\right)}^2\right]\le \mathbb{E}\left[{\left(X_1^2-{\left(\epsilon b_i\right)}^2\right)}^{+}\right]\to 0$ as $i\to \infty$, we have ${\sum }_{i=2^{k-1}}^{2^k-1}\mathbb{E}\left[{\left(X_i^{\left(1\right)}\right)}^2\right]\le 2^{k-1}\epsilon$ for sufficiently large k. Thus, by Lemma 1 (i), we have

$$\mathbb{E}\left[{\left(Z_k^{+}\right)}^2\right]\le C{\displaystyle \frac{{\sum }_{i=2^{k-1}}^{2^k-1}\mathbb{E}\left[{\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)}^2\right]}{a_{2^{k-1}}^2}}\le {\displaystyle \frac{C_{\epsilon }}{k}},\mathrm{for}\mathrm{sufficiently}\mathrm{large}k.$$

Observe that ${\displaystyle \frac{{\lambda }_{k+1}}{{\lambda }_k}}\to \sqrt{2}$ as $k\to \infty$, $\mathbb{E}\left[Z_k\right]\le 0$ for each $k\ge 1$. For any $p>2$, by Lemma 1 (ii), we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=1}^{\infty }}\mathbb{E}\left[{\left(Z_k^{+}\right)}^p\right]\le {\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{1}{a_{2^{k-1}}^p}}\mathbb{E}\left[{\left({\left({\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)\right)}^{+}\right)}^p\right]\hfill \\ \hfill \le & {\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{C_p}{a_{2^{k-1}}^p}}\left\{{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}\mathbb{E}\left[{\left|X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right|}^p\right]+{\left({\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}\mathbb{E}\left[{\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)}^2\right]\right)}^{{\displaystyle \frac{p}{2}}}\right\}\hfill \\ \hfill \le & C_p{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}{\displaystyle \frac{\mathbb{E}\left[{\left(X_i^{\left(1\right)}\right)}^p\right]}{a_i^p}}+C_{p,\epsilon }{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{{\left(2^{k-1}\right)}^{\frac{p}{2}}}{{\left(2^{k-1}·k\right)}^{\frac{p}{2}}}}\hfill \\ \hfill \le & C_p{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon a_i\right)\right)}^p\right]}{a_i^p}}+C_{p,\epsilon }{\displaystyle \sum _{k=1}^{\infty }}{k}^{-{\displaystyle \frac{p}{2}}}<\infty ,\hfill \end{array}$$

(30)

where the last inequality follows from (25). By Lemma 5 (iv), we have

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)<\infty .$$

(31)

By (29) and (31), there holds

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)<\infty .$$

(32)

For $X_n^{\left(3\right)}$, also by the method similar to (14) in Theorem 1, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(3\right)}>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)& \le C{\displaystyle \sum _{k=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{a_{2^{k-1}}}}{\displaystyle \sum _{m=1}^k}\left(\left({\displaystyle \frac{1}{a_{2^{m-1}}}}{\displaystyle \sum _{i=2^{m-1}}^{2^m-1}}{X}_i^{\left(3\right)}\right)a_{2^{m-1}}\right)>{\displaystyle \frac{\epsilon }{5}}\right).\hfill \end{array}$$

Denoting $Z_k^{\prime }={\displaystyle \frac{1}{a_{2^{k-1}}}}{\sum }_{i=2^{k-1}}^{2^k-1}{X}_i^{\left(3\right)}$ and ${\lambda }_k^{\prime }=a_{2^{k-1}}$, we have

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(3\right)}>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)\le C{\displaystyle \sum _{k=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_k^{\prime }}}{\displaystyle \sum _{m=1}^k}{\lambda }_m^{\prime }{Z}_m^{\prime }>{\displaystyle \frac{\epsilon }{5}}\right).\end{array}$$

Note that $Z_k^{\prime }\ge 0$ for each $k\ge 1$. According to the definition of $X_n^{\left(3\right)}$, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=1}^{\infty }}\mathbb{E}\left[Z_k^{\prime }\right]& \le C{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{1}{a_{2^{k-1}}}}{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}\mathbb{E}\left[X_i^{\left(3\right)}\right]\hfill \\ \hfill & \le C{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}{\displaystyle \frac{\mathbb{E}\left[X_i^{\left(3\right)}\right]}{a_i}}\le C{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[{\left(\left|X_1\right|-\epsilon a_i\right)}^{+}\right]}{a_i}}<\infty ,\hfill \end{array}$$

where the last inequality follows from (24). Hence, by Lemma 5 (i),

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}{X}_i^{\left(3\right)}>{\displaystyle \frac{\epsilon }{5}}\right)<\infty .$$

(33)

For $X_n^{\left(2\right)}$ and $X_n^{\left(4\right)}$, since $X_n^{\left(2\right)},X_n^{\left(4\right)}\le 0$, we immediately obtain

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}{X}_i^{\left(2\right)}>{\displaystyle \frac{\epsilon }{5}}\right)<\infty ,{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}{X}_i^{\left(4\right)}>{\displaystyle \frac{\epsilon }{5}}\right)<\infty .$$

(34)

For $Y_n$, note the fact that $\overline{\mu }=0$, $\left|\mathbb{E}\left[Y_i\right]\right|=\left|\mathbb{E}\left[Y_i\right]-\overline{\mu }\right|\le \mathbb{E}\left[{\left(|X_1|-\epsilon b_i\right)}^{+}\right]\le {\left(\epsilon b_i\right)}^{-1}\mathbb{E}\left[{\left(X_1^2-{\left(\epsilon b_i\right)}^2\right)}^{+}\right]$. Hence, by ${\sum }_{i=1}^n{\left(b_i\right)}^{-1}\sim C{a}_n$ as $n\to \infty$,

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\sum }_{i=1}^n\left|\mathbb{E}\left[Y_i\right]\right|}{a_n}}& \le {\displaystyle \frac{{\sum }_{i=1}^n{\left(\epsilon b_i\right)}^{-1}\mathbb{E}\left[{\left(X_1^2-{\left(\epsilon b_i\right)}^2\right)}^{+}\right]}{a_n}}\hfill \\ \hfill & =o\left(1\right)·{\displaystyle \frac{{\sum }_{i=1}^n{\left(\epsilon b_i\right)}^{-1}}{a_n}}\to 0,\mathrm{as}n\to \infty .\hfill \end{array}$$

(35)

On the other hand, observe that

$$\left|{\displaystyle \frac{Y_i}{\sqrt{n/\log n}}}\right|\le {\displaystyle \frac{5\epsilon }{4}},1\le i\le n.$$

For any $0<\epsilon <1$, taking $a={\displaystyle \frac{5\epsilon }{4}}$ and $x={\displaystyle \frac{\epsilon }{5}}\log n$ in Lemma 3, by noting that $2ax={\displaystyle \frac{1}{2}}{\epsilon }^2\log n<{\displaystyle \frac{\epsilon }{2}}\log n$, we have

$$\mathbb{V}\left({\displaystyle \frac{{\sum }_{i=1}^n\left(Y_i-\mathbb{E}\left[Y_i\right]\right)}{\sqrt{n/\log n}}}>{\displaystyle \frac{\epsilon }{5}}\log n\right)\le \exp \left\{-{\displaystyle \frac{{\left(\frac{\epsilon }{5}\right)}^2{\left(\log n\right)}^2}{\epsilon \log n+\frac{2\log n}{n}{\sum }_{i=1}^n\mathbb{E}\left[{\left(Y_i-\mathbb{E}\left[Y_i\right]\right)}^2\right]}}\right\}.$$

(36)

Moreover, $\left|\mathbb{E}\left[X_i^2\right]-\mathbb{E}\left[Y_i^2\right]\right|\le \mathbb{E}\left|X_i^2-Y_i^2\right|\le \mathbb{E}\left[{\left(X_1^2-{\left(\epsilon b_i\right)}^2\right)}^{+}\right]\to 0$, and $\left|\mathbb{E}\left[Y_i\right]\right|\to 0$ as $i\to \infty$, we have ${\sum }_{i=1}^n\mathbb{E}\left[{\left(Y_i-\mathbb{E}\left[Y_i\right]\right)}^2\right]\le \left(1+{\displaystyle \frac{\epsilon }{2}}\right)n$. Combining this with (36), we obtain

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)\hfill \\ \hfill =& {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \frac{{\sum }_{i=1}^n\left(Y_i-\mathbb{E}\left[Y_i\right]\right)}{\sqrt{n/\log n}}}>{\displaystyle \frac{\epsilon }{5}}\log n\right)\hfill \\ \hfill \le & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\exp \left\{-{\displaystyle \frac{{\left(\frac{\epsilon }{5}\right)}^2{\left(\log n\right)}^2}{\epsilon \log n+\frac{2\log n}{n}{\sum }_{i=1}^n\mathbb{E}\left[{\left(Y_i-\mathbb{E}\left[Y_i\right]\right)}^2\right]}}\right\}\hfill \\ \hfill \le & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\exp \left\{-{\displaystyle \frac{1}{50}}·{\displaystyle \frac{{\epsilon }^2\log n·n}{\frac{\epsilon }{2}n+\left(1+\frac{\epsilon }{2}\right)n}}\right\}\hfill \\ \hfill =& {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\exp \left\{-{\displaystyle \frac{{\epsilon }^2}{50(1+\epsilon )}}\log n\right\}\hfill \\ \hfill =& {\displaystyle \sum _{n=1}^{\infty }}{n}^{-\left(1+{\displaystyle \frac{{\epsilon }^2}{50(1+\epsilon )}}\right)}<\infty .\hfill \end{array}$$

(37)

From (35) and (37), it follows that

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{Y}_i>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)<\infty .$$

(38)

Combining (38) with (32)–(34), we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left(S_n>\epsilon \sqrt{n\log n}\right)\hfill \\ \hfill \le & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)+{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(2\right)}>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)\hfill \\ \hfill +& {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(3\right)}>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)+{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(4\right)}>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)\hfill \\ \hfill +& {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{Y}_i>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)<\infty .\hfill \end{array}$$

Hence, (26) holds. And (27) follows immediately by taking $\{-X_n;n\ge 1\}$ instead of $\{X_n;n\ge 1\}$ in (26).    □

Remark 4. 

As established in Theorem 2, for the law of the logarithm under sub-linear expectation, the convergence rate of order $1/n$ derived here is of a lower order than its counterpart $\left(\log n\right)/n$ as specified in (3), while the required moment condition is also milder than the second-moment existence assumption therein.

We proceed to establish the convergence rate of order $\left(\log n\right)/n$ for the law of the logarithm under sub-linear expectation. The proofs of the main lemma and theorem are analogous to those of Lemma 7 and Theorem 2; hence, we only present the key steps.

Lemma 8. 

Let ${\left\{X_n\right\}}_{n=1}^{\infty }$ be a sequence of random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$. If $C_{\mathbb{V}}\left[X_1^2\right]<\infty$, then for any $\epsilon >0$, the following statements hold:

(i) 

For any $0<p<2$,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n\mathbb{E}\left[{\left({\left(|X_1|-\epsilon \sqrt{n\log n}\right)}^{+}\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}<\infty .$$

(ii) 

For any $p>2$,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n\mathbb{E}\left[{\left(|X_1|\wedge \left(\epsilon \sqrt{n\log n}\right)\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}<\infty .$$

(iii) 

$$\underset{c\to \infty }{\lim }\mathbb{E}\left[\left(\right|X_1{{|}^2-c)}^{+}\right]=0.$$

Proof. 

Let $f\left(x\right)$ be the inverse function of $\epsilon \sqrt{x\log x}$.

(i)

For any $0<p<2$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=4}^{\infty }}{\displaystyle \frac{\log n\mathbb{E}\left[{\left({\left(|X_1|-\epsilon \sqrt{n\log n}\right)}^{+}\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}\hfill \\ \hfill \le & C_p{\displaystyle \sum _{n=4}^{\infty }}{n}^{-p/2}{\left(\log n\right)}^{1-p/2}{\int }_{\epsilon \sqrt{n\log n}}^{\infty }{x}^{p-1}\mathbb{V}\left(\left|X_1\right|>x\right)dx\hfill \\ \hfill \le & C_p{\int }_4^{\infty }{z}^{-p/2}{\left(\log z\right)}^{1-p/2}{\int }_{{\epsilon }^{2}z\log z}^{\infty }{y}^{p/2-1}\mathbb{V}\left(\right|X_1{|}^2>y)dydz\hfill \\ \hfill \le & C_p{\int }_{{\epsilon }^{2}4\log 4}^{\infty }{y}^{p/2-1}\mathbb{V}\left(\right|X_1{|}^2>y){\int }_4^{f\left(\sqrt{y}\right)}{z}^{-p/2}{\left(\log z\right)}^{1-p/2}dzdy\hfill \\ \hfill \le & C_{p,\epsilon }{\int }_0^{\infty }\mathbb{V}\left(\right|X_1{|}^2>y)dy\hfill \\ \hfill =& C_{p,\epsilon }{C}_{\mathbb{V}}\left[{X_1}^2\right]<\infty .\hfill \end{array}$$

(ii)

For any $p>2$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=4}^{\infty }}{\displaystyle \frac{\log n\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon \sqrt{n\log n}\right)\right)}^p\right]}{{\left(\sqrt{n\log n}\right)}^p}}\hfill \\ \hfill \le & C_p{\displaystyle \sum _{n=4}^{\infty }}{n}^{-p/2}{\left(\log n\right)}^{1-p/2}{\int }_{\epsilon \sqrt{4\log 4}}^{\epsilon \sqrt{n\log n}}{x}^{p-1}\mathbb{V}\left(\left|X_1\right|>x\right)dx\hfill \\ \hfill & +C_{\epsilon ,p}{\displaystyle \sum _{n=4}^{\infty }}{\displaystyle \frac{\log n}{{\left(\sqrt{n\log n}\right)}^p}}\hfill \\ \hfill \le & C_p{\int }_{{\epsilon }^{2}4\log 4}^{\infty }{y}^{p/2-1}\mathbb{V}\left(\right|X_1{|}^2>y){\int }_{f\left(\sqrt{y}\right)}^{\infty }{z}^{-p/2}{\left(\log z\right)}^{1-p/2}dzdy+C_{\epsilon ,p}\hfill \\ \hfill \le & C_{p,\epsilon }{\int }_0^{\infty }\mathbb{V}\left(\right|X_1{|}^2>y)dy+C_{\epsilon ,p}\hfill \\ \hfill =& C_{p,\epsilon }{C}_{\mathbb{V}}\left[{X_1}^2\right]+C_{\epsilon ,p}<\infty .\hfill \end{array}$$

(iii)

This result can be obtained immediately by noting that

$$\mathbb{E}\left[\left(\right|X_1{{|}^2-c)}^{+}\right]\le {\int }_c^{\infty }\mathbb{V}\left(\right|X_1{|}^2>x)dx\to 0,asc\to \infty .$$

□

Theorem 3. 

Let ${\left\{X_n\right\}}_{n=1}^{\infty }$ be a sequence of independent and identically distributed random variables in $(\mathrm{\Omega },\mathcal{H},\mathbb{E})$. Suppose that $\mathbb{V}$ is countably sub-additive. Set $S_n={\sum }_{k=1}^n{X}_k$, $\mathbb{E}\left[X_1\right]=\overline{\mu }$, $\mathcal{E}\left[X_1\right]=\underline{\mu }$. If $C_{\mathbb{V}}\left[X_1^2\right]<\infty$; then, for any $\epsilon >0$, we have

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left(S_n-n\overline{\mu }>\epsilon \sqrt{n\log n}\right)<\infty ,$$

(39)

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left(S_n-n\underline{\mu }<-\epsilon \sqrt{n\log n}\right)<\infty .$$

(40)

Proof. 

Noting that $\mathbb{E}\left[X_1^2\right]\le C_{\mathbb{V}}\left[X_1^2\right]<\infty$, without loss of generality, we assume $\overline{\mu }=0$, $\mathbb{E}\left[X_1^2\right]=1$. We still define $Y_n,X_n^{\left(1\right)},X_n^{\left(2\right)},X_n^{\left(3\right)}$ and $X_n^{\left(4\right)}$ as in the proof of Theorem 2. We first show (39).

For $X_n^{\left(1\right)}$, by the same argument as in the proof of (28) in Theorem 2, we also have

$${\displaystyle \frac{{\sum }_{i=1}^n\left|\mathbb{E}\left[X_i^{\left(1\right)}\right]\right|}{a_n}}\to 0,\mathrm{as}n\to \infty .$$

Noting that ${\displaystyle {\displaystyle \sum _{n=2^{k-1}}^{2^k-1}}\frac{\log n}{n}\sim Ck\mathrm{as}k\to \infty }$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)\hfill \\ \hfill & ={\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \sum _{n=2^{k-1}}^{2^k-1}}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)>{\displaystyle \frac{\epsilon }{5}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \sum _{n=2^{k-1}}^{2^k-1}}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \frac{1}{a_{2^{k-1}}}}{\displaystyle \sum _{m=1}^k}{\displaystyle \sum _{i=2^{m-1}}^{2^m-1}}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)>{\displaystyle \frac{\epsilon }{5}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{k=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{1}{a_{2^{k-1}}}}{\displaystyle \sum _{m=1}^k}\left(\left({\displaystyle \frac{1}{a_{2^{m-1}}}}{\displaystyle \sum _{i=2^{m-1}}^{2^m-1}}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)\right)a_{2^{m-1}}\right)>{\displaystyle \frac{\epsilon }{5}}\right){\displaystyle \sum _{n=2^{k-1}}^{2^k-1}}{\displaystyle \frac{\log n}{n}}\hfill \\ \hfill & \le C{\displaystyle \sum _{k=1}^{\infty }}k\mathbb{V}\left({\displaystyle \frac{1}{a_{2^{k-1}}}}{\displaystyle \sum _{m=1}^k}\left(\left({\displaystyle \frac{1}{a_{2^{m-1}}}}{\displaystyle \sum _{i=2^{m-1}}^{2^m-1}}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)\right)a_{2^{m-1}}\right)>{\displaystyle \frac{\epsilon }{5}}\right).\hfill \end{array}$$

(41)

Denoting $Z_k={\displaystyle \frac{1}{a_{2^{k-1}}}}{\sum }_{i=2^{k-1}}^{2^k-1}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)$ and ${\lambda }_k=a_{2^{k-1}}$, we have

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)\le C{\displaystyle \sum _{k=1}^{\infty }}k\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_k}}{\displaystyle \sum _{m=1}^k}{\lambda }_m{Z}_m>{\displaystyle \frac{\epsilon }{5}}\right).$$

As in Theorem 2, we can easily have ${\displaystyle \frac{{\lambda }_{k+1}}{{\lambda }_k}}\to \sqrt{2}$ as $k\to \infty$, $\mathbb{E}\left[Z_k\right]\le 0$ and $\mathbb{E}\left[{\left(Z_k^{+}\right)}^2\right]\le C_{\epsilon }/k$ for sufficiently large k. By Lemma 1 (ii), setting $p>4$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{k=1}^{\infty }}k\mathbb{E}\left[{\left(Z_k^{+}\right)}^p\right]\hfill \\ \hfill \le & {\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{C_{p}k}{a_{2^{k-1}}^p}}\left\{{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}\mathbb{E}\left[{\left|X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right|}^p\right]+{\left({\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}\mathbb{E}\left[{\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)}^2\right]\right)}^{{\displaystyle \frac{p}{2}}}\right\}\hfill \\ \hfill \le & C_p{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}{\displaystyle \frac{\log i\mathbb{E}\left[{\left(X_i^{\left(1\right)}\right)}^p\right]}{a_i^p}}+C_{p,\epsilon }{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{k·{\left(2^{k-1}\right)}^{\frac{p}{2}}}{{\left(2^{k-1}·k\right)}^{\frac{p}{2}}}}\hfill \\ \hfill \le & C_p{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\log i\mathbb{E}\left[{\left(\left|X_1\right|\wedge \left(\epsilon a_i\right)\right)}^p\right]}{a_i^p}}+C_{p,\epsilon }{\displaystyle \sum _{k=1}^{\infty }}{k}^{1-{\displaystyle \frac{p}{2}}}<\infty ,\hfill \end{array}$$

where the last inequality follows from Lemma 8 (ii). By Lemma 5 (v), we have

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]\right)>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)<\infty .\end{array}$$

Hence,

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)<\infty .$$

(42)

For $X_n^{\left(3\right)}$, denote $Z_k^{\prime }={\displaystyle \frac{1}{a_{2^{k-1}}}}{\sum }_{i=2^{k-1}}^{2^k-1}{X}_i^{\left(1\right)}$ and ${\lambda }_k^{\prime }=a_{2^{k-1}}$. By a method similar to (41), we obtain

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(3\right)}>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)\le C{\displaystyle \sum _{k=1}^{\infty }}k\mathbb{V}\left({\displaystyle \frac{1}{{\lambda }_k^{\prime }}}{\displaystyle \sum _{m=1}^k}{\lambda }_m^{\prime }{Z}_m^{\prime }>{\displaystyle \frac{\epsilon }{5}}\right).\end{array}$$

Note that $Z_k^{\prime }\ge 0$ for each $k\ge 1$. According to the definition of $X_n^{\left(3\right)}$, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=1}^{\infty }}k\mathbb{E}\left[Z_k^{\prime }\right]& \le {\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{k}{a_{2^{k-1}}}}{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}\mathbb{E}\left[X_i^{\left(3\right)}\right]\hfill \\ \hfill & \le C{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \sum _{i=2^{k-1}}^{2^k-1}}{\displaystyle \frac{\log i\mathbb{E}\left[X_i^{\left(3\right)}\right]}{a_i}}\le C{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\log i\mathbb{E}\left[{\left(\left|X_1\right|-\epsilon a_i\right)}^{+}\right]}{a_i}}<\infty ,\hfill \end{array}$$

where the last inequality follows from Lemma 8 (i). Hence, by Lemma 5 (ii),

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}{X}_i^{\left(3\right)}>{\displaystyle \frac{\epsilon }{5}}\right)<\infty .$$

(43)

For $X_n^{\left(2\right)}$ and $X_n^{\left(4\right)}$, since $X_n^{\left(2\right)},X_n^{\left(4\right)}\le 0$, we still have

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}{X}_i^{\left(2\right)}>{\displaystyle \frac{\epsilon }{5}}\right)<\infty ,{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}{X}_i^{\left(4\right)}>{\displaystyle \frac{\epsilon }{5}}\right)<\infty .$$

(44)

For $Y_n$, by the same argument as in the proof of (35) in Theorem 2 and Lemma 8 (iii), we also obtain

$${\displaystyle \frac{{\sum }_{i=1}^n\left|\mathbb{E}\left[Y_i\right]\right|}{a_n}}\to 0,\mathrm{as}n\to \infty .$$

(45)

On the other hand,

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)\hfill \\ \hfill \le & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\exp \left\{-{\displaystyle \frac{{\left(\frac{\epsilon }{5}\right)}^2{\left(\log n\right)}^2}{\epsilon \log n+\frac{2\log n}{n}{\sum }_{i=1}^n\mathbb{E}\left[{\left(Y_i-\mathbb{E}\left[Y_i\right]\right)}^2\right]}}\right\}\hfill \\ \hfill \le & {\displaystyle \sum _{n=1}^{\infty }}\log n·n^{-\left(1+{\displaystyle \frac{{\epsilon }^2}{50(1+\epsilon )}}\right)}<\infty .\hfill \end{array}$$

(46)

From (45) and (46), it follows that

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{Y}_i>{\displaystyle \frac{\epsilon }{5}}{a}_n\right)<\infty .$$

(47)

Combining (47) with (42)–(44), we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left(S_n>\epsilon \sqrt{n\log n}\right)\hfill \\ \hfill \le & {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(1\right)}>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)+{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(2\right)}>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)\hfill \\ \hfill +& {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(3\right)}>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)+{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{X}_i^{\left(4\right)}>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)\hfill \\ \hfill +& {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log n}{n}}\mathbb{V}\left({\displaystyle \sum _{i=1}^n}{Y}_i>{\displaystyle \frac{\epsilon }{5}}\sqrt{n\log n}\right)<\infty .\hfill \end{array}$$

Hence, (39) holds. And (40) follows immediately by taking $\{-X_n;n\ge 1\}$ instead of $\{X_n;n\ge 1\}$ in (39).    □

Remark 5. 

As established in Theorem 3, for the convergence rate of order $\left(\log n\right)/n$ for the law of the logarithm under sub-linear expectation, the required moment condition is consistent with the second-moment existence assumption therein as specified in (3).

Remark 6. 

As demonstrated in Example 2, the G-expectation space forms a canonical non-trivial sub-linear expectation space, on which the IID sequence $\left\{X_n\right\}$ of G-normal random variables is well defined. Let $\tilde{\mathbb{V}}$ be the upper capacity induced by $\tilde{\mathbb{E}}$ defined as (4). By the representation theorem for G-expectation (Denis et al. (2011) [20]), there exists a weakly compact family of probability measures $\mathcal{P}$ on $(\tilde{\mathrm{\Omega }},\mathcal{B}\left(\tilde{\mathrm{\Omega }}\right))$ such that

$$\tilde{\mathbb{V}}\left(A\right)=\underset{P\in \mathcal{P}}{\sup }P\left(A\right),\forall A\in \mathcal{B}\left(\tilde{\mathrm{\Omega }}\right).$$

It follows immediately that $\tilde{\mathbb{V}}$ is countably sub-additive, which satisfies the capacity assumption in Theorems 1–3. For the G-normal random variable $X_1\stackrel{d}{=}{W}_1$, Meng and Wu (2026) [21] established that its Choquet moments of all positive orders are finite, i.e., $C_{\tilde{\mathbb{V}}}\left[\right|X_1{|}^p]<\infty$ for all $p>0$. Then, the corresponding tail conditions in Theorems 1–3 also hold automatically.

Consequently, all the convergence rate results of Theorems 1–3 apply directly to the IID G-normal sequence constructed in Example 2.

Remark 7. 

We note that probabilistic metric spaces provide an alternative framework for describing inherent randomness in convergence problems. As a recent development, Achtoun et al. (2024) [22] proposed the controlled strong probabilistic b-metric space and obtained related fixed point results. Extending our convergence rate conclusions to such a probabilistic metric setting is a worthwhile topic for further research.