A Law of the Iterated Logarithm for Sub-Linear Expectation Under a General Moment Condition

Abstract

In this paper, we obtain the law of the iterated logarithm under a general moment condition with respect to sub-linear expectation. We present a novel proof by combining the exponential inequality with the subsequence method. Moreover, we show that the general moment condition we indicated is the weakest one for the law of the iterated logarithm under sub-linear expectation.

1. Introduction

The classical law of the iterated logarithm (LIL for short) is one of the foundational core results in probability theory, rooted in the additive properties of probability and expectation. In recent years, domains such as statistics and risk measurement have been confronted with model uncertainties that can no longer be sufficiently characterized within the classical linear framework. Consequently, non-additive probabilities and non-linear expectations have become more popular, as they provide a much more reliable way to deal with these uncertainties (cf. Huber and Strassen (1973), Peng (1997), Chen and Epstein (2002), Maccheroni and Marinacci (2005) [1,2,3,4]).

The framework of sub-linear expectation was introduced by Peng (2010) [5]. Within this setting, several classical limit theorems have been successfully extended. Chen et al. (2013) and Chen (2016) [6,7] obtained the strong law of large numbers for independent random variables under the finite $(1+\alpha )$-th moment condition ($\alpha >0$) with respect to sub-linear expectation. Hu (2016) [8] established the strong law of large numbers for sub-linear expectation under a general moment condition based on a function class ${\mathsf{\Phi }}_c$, and further indicated that such a general moment condition is the weakest moment condition for the strong law of large numbers with respect to sub-linear expectation. Hu (2018) [9] investigated the central limit theorems under sub-linear expectation without the assumption of identical distribution. Xu and Zhang (2020) [10] obtained the law of the logarithm for arrays of row-wise extended negatively dependent random variables under finite fourth moment conditions with respect to sub-linear expectation.

Among various limit theorems under sub-linear expectation, the LIL has been extensively investigated. Chen and Hu (2014) [11] established the LIL for sequences of uniformly bounded random variables within the framework of capacities, generalizing the classical Kolmogorov and Hartman–Wintner laws to non-additive probability spaces. Zhang (2016) [12] generalized the LIL to negatively dependent and identically distributed random variables, establishing the results under the moment condition with respect to Choquet expectation $C_{\mathbb{V}}\left[|X_1{|}^2/\log \log \left|X_1\right|\right]<\infty$, and further showed that this condition is necessary and sufficient for the validity of the LIL. Zhang (2021) [13] established Wittmann’s LIL for independent but not necessarily identically distributed random variables under sub-linear expectation. Zhang (2022) [14] further derived the LIL for extended negatively dependent random variables under the moment condition with respect to sub-linear expectation $\mathbb{E}\left[|X_1{|}^{2+\alpha }\right]<\infty$ for some $\alpha >0$.

However, in contrast to the necessary and sufficient moment condition with respect to Choquet expectation established by Zhang (2016) [12], all moment conditions currently available for the LIL with respect to sub-linear expectation are only sufficient for its validity. Since the Choquet expectation generated by a sub-additive capacity is larger than the corresponding sub-linear expectation, moment conditions formulated with respect to Choquet expectation are typically stronger than those with respect to sub-linear expectation. This motivates us to investigate whether the LIL can be established under a milder sub-linear expectation moment condition that relaxes the prevailing finite $(2+\alpha )$-th moment assumption for independent, not necessarily identically distributed random variables.

To precisely characterize the moment conditions, we follow Hu (2016) [8] and introduce the following function classes. Let ${\mathsf{\Phi }}_c$ (or, respectively, ${\mathsf{\Phi }}_d$) denote the set of functions $\psi \left(x\right)$ defined on $[0,\infty )$ such that $\psi \left(x\right)$ satisfies:

(1)

$\psi \left(x\right)$ is non-negative and non-decreasing on $[0,\infty )$ and positive on $[x_0,\infty )$ for some $x_0\ge 0$. The series ${\sum }_{n=\left[x_0\right]+1}^{\infty }{\displaystyle \frac{1}{n\psi \left(n\right)}}$ converges (respectively, diverges);

(2)

For any fixed $a>0$, there exists $C>0$ such that $\psi (x+a)\le C\psi \left(x\right)$ for any $x\ge x_0$.

The value of $x_0$ is not assumed to be the same for different $\psi \left(x\right)$. The functions $x^{\alpha }$ and ${\left(\ln (1+x)\right)}^{1+\alpha }$ for any $\alpha >0$ are examples of the class of functions ${\mathsf{\Phi }}_c$. The functions $\ln (1+x)$ and $\ln \ln (e\vee x)$ belong to the class ${\mathsf{\Phi }}_d$.

Inspired by Hu (2016) [8], we derive that the LIL still holds under a general moment condition with respect to sub-linear expectation based on the above complementary function classes. Specifically, we establish that the moment condition ${\sup }_{n\ge 1}\mathbb{E}\left[{\displaystyle \frac{|X_n{|}^2\psi \left(\right|X_n\left|\right)}{\log \log |X_n|}}\right]<\infty$ for some $\psi \left(x\right)\in {\mathsf{\Phi }}_c$ is sufficient to guarantee the validity of the LIL, which is weaker than the $(2+\alpha )$-th moment condition required by Zhang (2022) [14]. We present a novel proof strategy that combines the exponential inequality techniques developed by De Acosta (1983) [15] with the classical subsequence method. We also prove that this moment condition is indeed the weakest possible for the LIL with respect to sub-linear expectation. Conversely, if the moment condition fails for all $\psi \in {\mathsf{\Phi }}_c$, then it can at most hold for some $\psi \in {\mathsf{\Phi }}_d$, and we can always construct an explicit counterexample showing that the LIL fails in this case.

The rest of this paper is organized as follows. In Section 2, we recall some basic concepts and lemmas related to sub-linear expectation. In Section 3, we state and prove our main result, that is the LIL under a general moment condition for sub-linear expectation. In Section 4, we illustrate that this condition is the weakest.

Throughout this paper, we define $\log x=\ln (x\vee e)$, $\forall x\in \mathbb{R}$. We say that two functions $f\left(x\right)$ and $g\left(x\right)$ are asymptotically equivalent, denoted by $f\left(x\right)\sim g\left(x\right)$ as $x\to \infty$, if ${\lim }_{x\to \infty }f\left(x\right)/g\left(x\right)=1$. In addition, the symbol C stands for a positive constant whose value may differ at different places.

2. Basic Concepts and Lemmas

We use the definitions and framework proposed by Peng (2010) [5]. Let $(\mathsf{\Omega },\mathcal{F})$ be a given measurable space and let $\mathcal{H}$ be a linear space of real functions defined on $(\mathsf{\Omega },\mathcal{F})$ such that if $X_1,\dots ,X_n\in \mathcal{H}$ then $\phi (X_1,\dots ,X_n)\in \mathcal{H}$ for each $\phi \in C_{l,\mathit{Lip}}\left({\mathbb{R}}^n\right)$, where $C_{l,\mathit{Lip}}\left({\mathbb{R}}^n\right)$ 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 $\phi$. $\mathcal{H}$ is considered as a space of random variables.

As follows we present some definitions and basic properties of sub-linear expectation.

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 $X,Y\in \mathcal{H}$, we have

(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 $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$ is called a sub-linear expectation space. Given a sub-linear expectation $\mathbb{E}$, we denote the conjugate expectation $\mathcal{E}$ of $\mathbb{E}$ by

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

Definition 2.

Let $\mathcal{G}\subset \mathcal{F}$. A function $V:\mathcal{G}\to [0,1]$ is called a capacity if

$$V(\varnothing )=0,V\left(\mathsf{\Omega }\right)=1\mathit{and}V\left(A\right)\le V\left(B\right),\forall A\subset B,A,B\in \mathcal{G}.$$

The capacity V is said to be sub-additive if $V(A\cup B)\le V\left(A\right)+V\left(B\right)$ for all $A,B\in \mathcal{G}$ such that $A\cup B\in \mathcal{G}$. We denote a pair $(\mathbb{V},\mathcal{V})$ of capacities by

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

(1)

where $A^c$ is the complement set of A. Then

$$\begin{array}{c}\mathbb{V}\left(A\right):=\mathbb{E}\left[I_A\right],\mathcal{V}\left(A\right):=\mathcal{E}\left[I_A\right],\mathit{if}{I}_A\in \mathcal{H},\\ \mathbb{E}\left[f\right]\le \mathbb{V}\left(A\right)\le \mathbb{E}\left[g\right],\mathcal{E}\left[f\right]\le \mathcal{V}\left(A\right)\le \mathcal{E}\left[g\right],\mathrm{if}f\le I_A\le g,f,g\in \mathcal{H}.\end{array}$$

(2)

It is obvious that $\mathbb{V}$ is sub-additive, but $\mathcal{V}$ and $\mathcal{E}$ are not.

Also, we define the Choquet integrals/expectations ($C_{\mathbb{V}},C_{\mathcal{V}}$) by

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

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

Definition 3.

A sub-linear expectation $\mathbb{E}:\mathcal{H}\to \mathbb{R}$ is said to be continuous if it satisfies:

(1) 

Lower-continuity: if $X_n\uparrow X$, then $\mathbb{E}\left[X_n\right]\uparrow \mathbb{E}\left[X\right]$, where $0\le X_n,X\in \mathcal{H}$.

(2) 

Upper-continuity: if $X_n\downarrow X$, then $\mathbb{E}\left[X_n\right]\downarrow \mathbb{E}\left[X\right]$, where $0\le X_n,X\in \mathcal{H}$.

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

(1) 

Lower-continuity: if $A_n\uparrow A$, then $V\left(A_n\right)\uparrow V\left(A\right)$, where $A_n,A\in \mathcal{F}$.

(2) 

Upper-continuity: if $A_n\downarrow A$, then $V\left(A_n\right)\downarrow V\left(A\right)$, where $A_n,A\in \mathcal{F}$.

Definition 4.

(Independence). Let $X=(X_1,\dots ,X_m),X_i\in \mathcal{H}$ and $Y=(Y_1,\dots ,Y_n),Y_i\in \mathcal{H}$ be two random variables on $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$. $Y$ is said to be independent of X, if for each test function $\phi \in C_{l,\mathit{Lip}}({\mathbb{R}}^m\times {\mathbb{R}}^n)$ we have $\mathbb{E}\left[\phi \right(X,Y\left)\right]=\mathbb{E}\left[\mathbb{E}\right[\phi (x,Y)|x=X]]$ whenever $\overline{\phi }\left(x\right):=\mathbb{E}\left[\right|\phi (x,Y)\left|\right]<\infty$ for all x and $\mathbb{E}\left[\right|\overline{\phi }\left(X\right)\left|\right]<\infty$. ${\left\{X_n\right\}}_{n=1}^{\infty }$ is said to be a sequence of independent random variables, if $X_{n+1}$ is independent of $(X_1,\dots ,X_n)$ for each $n\ge 1$.

Definition 5.

(Negative dependence). On a sub-linear expectation space $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$, the random vector $\mathit{Y}=(Y_1,\dots ,Y_n)$ is said to be negatively dependent (ND) to another random vector $\mathit{X}=(X_1,\dots ,X_m)$ if for each pair of test functions ${\phi }_1\in C_{l,\mathit{Lip}}\left({\mathbb{R}}^m\right)$ and ${\phi }_2\in C_{l,\mathit{Lip}}\left({\mathbb{R}}^n\right)$, 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 ${\phi }_1$ and ${\phi }_2$ are both coordinatewise non-decreasing (or both coordinatewise non-increasing) with ${\phi }_1\left(\mathit{X}\right)⩾0$, $\mathbb{E}\left[{\phi }_2\left(\mathit{Y}\right)\right]⩾0$, and all relevant expectations are finite.

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

Remark 1.

It follows from Definitions 4 and 5 that negative dependence is a general extension of independence.

The following lemma provides the Borel-Cantelli Lemma under sub-linear expectation whose proof can be found in Chen et al. (2013) [6].

Lemma 1 (Borel-Cantelli Lemma).

Let ${\left\{A_n\right\}}_{n=1}^{\infty }$ be a sequence of events in $\mathcal{F}$ and $\mathbb{V}$ be the capacity induced by lower-continuous sub-linear expectation $\mathbb{E}$. If ${\sum }_{n=1}^{\infty }\mathbb{V}\left(A_n\right)<\infty$, then

$$\mathbb{V}\left({\displaystyle \bigcap _{n=1}^{\infty }}{\displaystyle \bigcup _{i=n}^{\infty }}{A}_i\right)=0.$$

The next lemma gives the Chebyshev’s inequality under sub-linear expectation.

Lemma 2 (Chebyshev’s Inequality).

Let X be a random variable defined on $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$ and function $f\in C_{l,Lip}\left(\mathbb{R}\right)$ be non-decreasing and positive, we have:

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

Proof. 

By the monotonicity of $f\left(x\right)$ and $f\left(x\right)>0$, 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 $f\in C_{l,Lip}\left(\mathbb{R}\right)$, ${\displaystyle \frac{f\left(X\right)}{f\left(x\right)}}$ belongs to $\mathcal{H}$. Then by (2) we have:

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

□

The next lemma refines a key convergence property of ${\mathsf{\Phi }}_c$ and its proof can be found in Cui and Hu (2026) [16].

Lemma 3.

Assume that $\psi \left(x\right)\in {\mathsf{\Phi }}_c$, then

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n\psi \left(n^{1/\beta }\right)}}<\infty ,\forall \beta >1.$$

The following lemma establishes the exponential inequality for negatively dependent random variables under sub-linear expectation and the proof can be found in Zhang (2016) [12].

Lemma 4.

Let $\{X_i;i\ge 1\}$ be a sequence of negatively dependent random variables on $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$ with $\mathbb{E}\left[X_i\right]\le 0$. Denote $S_n={\sum }_{i=1}^n{X}_i$. Then for any $p\ge 2$, there exists a constant $C_p\ge 1$ such that for any $x>0$ and $0<\eta ⩽1$,

$$\mathbb{V}(S_n⩾x)⩽C_p{\eta }^{-2p}{\displaystyle \frac{{\sum }_{i=1}^n\mathbb{E}\left[\right|X_i{|}^p]}{x^p}}+\exp \left\{-{\displaystyle \frac{x^2}{2(1+\eta ){\sum }_{i=1}^n\mathbb{E}\left[X_i^2\right]}}\right\}.$$

(3)

The next lemma is an exponential maximal inequality under sub-linear expectation. This lemma is stated in Hu et al. (2024) [17], while this result occurs first in Zhang (2016) [18].

Lemma 5.

Suppose that $\{X_n;n\ge 1\}$ is a sequence of random variables defined on $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$. Then for any $\lambda \ge 0$,

$$\mathbb{E}\exp \left\{\lambda \underset{k\le n}{\max }{\displaystyle \sum _{i=1}^k}(X_i-\mathcal{E}\left[X_i\right])\right\}\le 4\mathbb{E}\exp \left\{\lambda {\displaystyle \sum _{i=1}^n}(X_i-\mathcal{E}\left[X_i\right])\right\}.$$

(4)

3. Main Results

Starting from this section, we consider the following general moment condition:

$$\underset{n\ge 1}{\sup }\mathbb{E}\left[{\displaystyle \frac{|X_n{|}^2\psi \left(\right|X_n\left|\right)}{\log \log |X_n|}}\right]<\infty \mathrm{for}\mathrm{some}\psi \left(x\right)\in {\mathsf{\Phi }}_c,$$

(5)

which will be used in all subsequent lemmas and main results. We state our main results and their proofs in what follows. The subsequent two lemmas are essential to the proofs of our theorems.

Lemma 6.

For any $\epsilon >0$ and $p>2$, under the moment condition (5), there holds

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[\right(|X_n|\wedge \epsilon \sqrt{2n\log \log n}{)}^p]}{{(\sqrt{2n\log \log n})}^p}}<\infty .$$

Proof. 

Noticing that $\log \log (\epsilon \sqrt{2n\log \log n})\sim \log \log n$, as $n\to \infty$, we have

$$\begin{array}{cc}& {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[\right(|X_n|\wedge \epsilon \sqrt{2n\log \log n}{)}^p]}{{(\sqrt{2n\log \log n})}^p}}\hfill \\ & \le {\displaystyle \sum _{n=1}^{\infty }}\mathbb{E}\left[{\displaystyle \frac{|X_n{|}^2\psi \left(\right|X_n\left|\right)}{\log \log \left(\right|X_n\left|\right)}}\right]{\displaystyle \frac{{(\epsilon \sqrt{2n\log \log n})}^{p-2}\log \log (\epsilon \sqrt{2n\log \log n})}{\psi (\epsilon \sqrt{2n\log \log n}){(\sqrt{2n\log \log n})}^p}}\hfill \\ & \le \underset{n\ge 1}{\sup }\mathbb{E}\left[{\displaystyle \frac{|X_n{|}^2\psi \left(\right|X_n\left|\right)}{\log \log \left(\right|X_n\left|\right)}}\right]{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\log \log (\epsilon \sqrt{2n\log \log n})}{\psi (\epsilon \sqrt{2n\log \log n})2n\log \log n}}\hfill \\ & \le C{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n\psi (\epsilon \sqrt{2n\log \log n})}}\hfill \\ & \le C{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n\psi \left(\sqrt{n}\right)}},\hfill \end{array}$$

where the last series converges by Lemma 3. □

Lemma 7.

Let X be a random variable defined on $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$. Assume that the moment condition (5) holds. Then

$$\underset{n\ge 1}{\sup }\mathbb{E}[{\left(X_n^2-c\right)}^{+}]\to 0,\mathit{as}c\to \infty .$$

Proof. 

Since the convergence of ${\sum }_{n=\left[x_0\right]+1}^{\infty }{\displaystyle \frac{1}{n\psi \left(n\right)}}$ implies ${\lim }_{x\to \infty }\psi \left(x\right)/\log \log x=\infty$, we have

$$\underset{n\ge 1}{\sup }\mathbb{E}\left[{(X_n^2-c)}^{+}\right]\le {\displaystyle \frac{\log \log \sqrt{c}}{\psi \left(\sqrt{c}\right)}}\underset{n\ge 1}{\sup }\mathbb{E}\left[{\displaystyle \frac{|X_n{|}^2\psi \left(\right|X_n\left|\right)}{\log \log |X_n|}}\right]\to 0,$$

as $c\to \infty$.

□

The next theorem states the main result of this paper which is the LIL under sub-linear expectation.

Theorem 1.

Let $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$ be a sub-linear expectation space where $\mathbb{E}$ is lower-continuous and $\mathbb{V}$ is the induced capacity. Let ${\left\{X_n\right\}}_{n=1}^{\infty }$ be a sequence of ND random variables satisfying the moment condition (5). Assume that $\mathbb{E}\left[X_n\right]=\mathcal{E}\left[X_n\right]\equiv \mu ,\mathbb{E}\left[{(X_n-\mu )}^2\right]\equiv {\overline{\sigma }}^2$ for each $n\ge 1$. Denote $S_n={\sum }_{i=1}^n{X}_i$. Then

$$\begin{array}{cc}\hfill & \mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{S_n-n\mu }{\sqrt{2n\log \log n}}}>\overline{\sigma }\right)=0,\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill & \mathbb{V}\left(\underset{n\to \infty }{\lim \inf }{\displaystyle \frac{S_n-n\mu }{\sqrt{2n\log \log n}}}<-\overline{\sigma }\right)=0.\hfill \end{array}$$

(7)

Remark 2.

To investigate the limit properties of truncated random variables, Zhang (2021) [13] introduced the notation for the limit of the truncated sub-linear expectation, which is defined as

$$\breve{\mathbb{E}}\left[X\right]=\underset{c\to \infty }{\lim }\mathbb{E}\left[(-c)\vee (X\wedge c)\right]$$

It can be verified that if ${\lim }_{c\to \infty }\mathbb{E}\left[{\left(\right|X|-c)}^{+}\right]=0$, then $\breve{\mathbb{E}}\left[X\right]=\mathbb{E}\left[X\right]$ holds. So in Theorem 1, under the moment condition (5), by Lemma 7 there holds for each $n\ge 1$,

$$\breve{\mathbb{E}}\left[X_n\right]=-\breve{\mathbb{E}}[-X_n]=\mathbb{E}\left[X_n\right]=\mathcal{E}\left[X_n\right]\equiv \mu .$$

Then we have

$$\breve{\mathbb{E}}\left[{(X_n-\breve{\mathbb{E}}\left[X_n\right])}^2\right]=\breve{\mathbb{E}}\left[{(X_n+\breve{\mathbb{E}}[-X_n])}^2\right]=\mathbb{E}\left[{(X_n-\mu )}^2\right]\equiv {\overline{\sigma }}^2,$$

Thus the moment condition (5) ensures that the asymptotic analysis of variances under sub-linear expectation can be reduced to the original sub-linear expectation setting without any loss of information.

Proof. 

If (6) holds, then replacing $\left\{X_n\right\}$ with $\{-X_n\}$ in (6), we have (7). Therefore, it suffices to prove (6). Without loss of generality, we may assume that $\mu =0$. To establish (6), we only need to show that

$$\mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{S_n}{\sqrt{2n\log \log n}}}\ge (1+7\epsilon )\overline{\sigma }\right)=0,\forall \epsilon >0.$$

(8)

Denote $a_n=\sqrt{2n\log \log n},b_n=\tau \sqrt{{\displaystyle \frac{n}{\log \log n}}},$ where $\tau$ is a positive constant to be specified subsequently. Define

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

Thus we have

$$X_n=Y_n+X_n^{\left(1\right)}+X_n^{\left(2\right)}+X_n^{\left(3\right)}.$$

Moreover, for any $\epsilon >0$, it suffices to prove the following four statements:

$$\begin{array}{cc}\hfill & \mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=1}^n{Y}_i}{a_n}}\ge (1+4\epsilon )\overline{\sigma }\right)=0,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & \mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=1}^n{X}_i^{\left(1\right)}}{a_n}}\ge \epsilon \overline{\sigma }\right)=0,\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill & \mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=1}^n{X}_i^{\left(2\right)}}{a_n}}\ge \epsilon \overline{\sigma }\right)=0,\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill & \mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=1}^n{X}_i^{\left(3\right)}}{a_n}}\ge \epsilon \overline{\sigma }\right)=0.\hfill \end{array}$$

(12)

• Step 1. Convergence of the far-tail truncated term: $X_n^{\left(3\right)}$

For (12), noting that $\log \log \left(\sqrt{2n\log \log n}\right)\sim \log \log n$ as $n\to \infty$, by the moment condition (5) and Lemma 3, we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}(X_n^{\left(3\right)}\ne 0)& \le {\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}\left(\right|X_n|>a_n)\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[|X_n{|}^2\psi \left(\right|X_n\left|\right)/\log \log \left(\right|X_n\left|\right)\right]}{a_n^2\psi \left(a_n\right)/\log \log \left(a_n\right)}}\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[|X_n{|}^2\psi \left(\right|X_n\left|\right)/\log \log \left(\right|X_n\left|\right)\right]}{2n\log \log n\psi (\sqrt{2n\log \log n})/\log \log (\sqrt{2n\log \log n})}}\hfill \\ \hfill & \le C\underset{n\ge 1}{\sup }\mathbb{E}\left[{\displaystyle \frac{|X_n{|}^2\psi \left(\right|X_n\left|\right)}{\log \log \left(\right|X_n\left|\right)}}\right]·{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{2n\psi (\sqrt{2n\log \log n})}}\hfill \\ \hfill & \le C{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n\psi \left(\sqrt{n}\right)}}<\infty .\hfill \end{array}$$

(13)

Hence, $\mathbb{V}(\{X_n^{\left(3\right)}\ne 0\}\mathrm{i}.\mathrm{o}.)=0$ by the Borel-Cantelli lemma. Thus we obtain (12).

• Step 2. Convergence of the intermediate truncated term: $X_n^{\left(1\right)}$, $X_n^{\left(2\right)}$

For (10), by the definition of $X_n^{\left(1\right)}$, we have $X_n^{\left(1\right)}\ge 0$. It is sufficient to show that

$$\mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=2^n+1}^{2^{n+1}}{X}_i^{\left(1\right)}}{a_{2^{n+1}}}}\ge \epsilon \overline{\sigma }\right)=0.$$

(14)

Let $Z_i=X_i^{\left(1\right)}-\mathbb{E}\left[X_i^{\left(1\right)}\right]$. Then for any $\epsilon >0$, by (3), we have

$$\mathbb{V}\left({\displaystyle \frac{{\sum }_{i=2^n+1}^{2^{n+1}}{Z}_i}{a_{2^{n+1}}\overline{\sigma }}}\ge \epsilon \right)\le C{\displaystyle \sum _{i=2^n+1}^{2^{n+1}}}{\displaystyle \frac{\mathbb{E}\left[\right|Z_i{|}^p]}{a_i^p{\overline{\sigma }}^p}}+\exp \left\{-{\displaystyle \frac{{\epsilon }^2{a}_{2^{n+1}}^2{\overline{\sigma }}^2}{4{\sum }_{i=2^n+1}^{2^{n+1}}\mathbb{E}{Z}_i^2}}\right\}.$$

Note that $\mathbb{E}\left[Z_i^2\right]\le 4\mathbb{E}[{(X_i^{\left(1\right)})}^2]\le 4\mathbb{E}[{(X_i^2-b_i^2)}^{+}]$. By Lemma 7, we have $\underset{2^n<i\le 2^{n+1}}{\sup }\mathbb{E}\left[Z_i^2\right]<{\displaystyle \frac{1}{2}}{\epsilon }^2{\overline{\sigma }}^2$ for all sufficiently large n. Thus, applying Lemma 6, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{{\sum }_{i=2^n+1}^{2^{n+1}}{Z}_i}{a_{2^{n+1}}\overline{\sigma }}}\ge \epsilon \right)& \le C{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \sum _{i=2^n+1}^{2^{n+1}}}{\displaystyle \frac{\mathbb{E}\left[\right|Z_i{|}^p]}{a_i^p{\overline{\sigma }}^p}}+{\displaystyle \sum _{n=1}^{\infty }}\exp \left\{-{\displaystyle \frac{{\epsilon }^2{a}_{2^{n+1}}^2{\overline{\sigma }}^2}{4{\sum }_{i=2^n+1}^{2^{n+1}}\mathbb{E}{Z}_i^2}}\right\}\hfill \\ \hfill & \le C{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[\left(\right|X_i{|\wedge a_i)}^p\right]}{a_i^p}}+{\displaystyle \sum _{n=1}^{\infty }}\exp \left\{-{\displaystyle \frac{2^{n+2}\log \log \left(2^{n+1}\right){\epsilon }^2{\overline{\sigma }}^2}{2^{n+2}\underset{2^n<i\le 2^{n+1}}{\sup }\mathbb{E}\left[Z_i^2\right]}}\right\}\hfill \\ \hfill & \le C{\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{\mathbb{E}\left[\left(\right|X_i{|\wedge a_i)}^p\right]}{a_i^p}}+C{\displaystyle \sum _{n=1}^{\infty }}\exp \left\{-2\log \log 2^{n+1}\right\}<\infty .\hfill \end{array}$$

Hence, by the Borel-Cantelli lemma, we have

$$\mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=2^n+1}^{2^{n+1}}{Z}_i}{a_{2^{n+1}}}}\ge \epsilon \overline{\sigma }\right)=0.$$

(15)

On the other hand,

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{a_{2^{n+1}}\overline{\sigma }}}{\displaystyle \sum _{i=2^n+1}^{2^{n+1}}}\mathbb{E}\left[X_i^{\left(1\right)}\right]& \le {\displaystyle \frac{1}{a_{2^{n+1}}\overline{\sigma }}}{\displaystyle \sum _{i=2^n+1}^{2^{n+1}}}\mathbb{E}\left[{\displaystyle \frac{|X_i{|}^2\psi \left(\right|X_i\left|\right)}{\log \log \left(\right|X_i\left|\right)}}\right]{\displaystyle \frac{\log \log b_i}{b_i\psi \left(b_i\right)}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{a_{2^{n+1}}\overline{\sigma }}}\underset{i\ge 1}{\sup }\mathbb{E}\left[{\displaystyle \frac{|X_i{|}^2\psi \left(\right|X_i\left|\right)}{\log \log \left(\right|X_i\left|\right)}}\right]{\displaystyle \sum _{i=2^n+1}^{2^{n+1}}}{\displaystyle \frac{\log \log \tau \sqrt{\frac{i}{\log \log i}}}{\tau \sqrt{\frac{i}{\log \log i}}\psi (\tau \sqrt{\frac{i}{\log \log i}})}}\hfill \\ \hfill & \le {\displaystyle \frac{C}{\sqrt{2^{n+2}\log \log 2^{n+1}}}}·{\displaystyle \frac{2^n\sqrt{\log \log 2^{n+1}}\log \log \tau \sqrt{\frac{2^{n+1}}{\log \log 2^{n+1}}}}{\sqrt{2^n}\psi \left(\tau \sqrt{\frac{2^{n+1}}{\log \log 2^{n+1}}}\right)}}\hfill \\ \hfill & \le {\displaystyle \frac{C\log \log \tau \sqrt{\frac{2^{n+1}}{\log \log 2^{n+1}}}}{\psi \left(\tau \sqrt{\frac{2^{n+1}}{\log \log 2^{n+1}}}\right)}}\to 0,asn\to \infty .\hfill \end{array}$$

(16)

Combining (16) with (15), (14) is proved. So (10) holds. Since $X_n^{\left(2\right)}\le 0$ for all $n\ge 1$, (11) holds immediately.

• Step 3. Convergence of the near-truncated term: $Y_n$

Fix the arbitrarily small positive constant $\epsilon$ mentioned above. By Remark 2 we have

$$\mathbb{E}\left[{(Y_i-\mathbb{E}\left[Y_i\right])}^2\right]\to {\overline{\sigma }}^2\mathrm{as}i\to \infty .$$

Choose $\delta >0$ sufficiently small such that ${(1+\epsilon )}^2(1-\delta )>1$. Then, there exists a positive integer N such that for all $i>N$, we have

$$1-\delta \le {\displaystyle \frac{{\overline{\sigma }}^2}{\underset{j>N}{\sup }\mathbb{E}\left[{(Y_j-\mathbb{E}\left[Y_j\right])}^2\right]}}\le 1+\delta .$$

(17)

Thus, we have

$$\begin{array}{c}\hfill \mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=1}^n{Y}_i}{a_n}}\ge (1+4\epsilon )\overline{\sigma }\right)\le \mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=1}^N{Y}_i}{a_n}}\ge \epsilon \overline{\sigma }\right)+\mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=N+1}^n{Y}_i}{a_n}}\ge (1+3\epsilon )\overline{\sigma }\right).\end{array}$$

It is obvious that

$$\mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=1}^N{Y}_i}{a_n}}\ge \epsilon \overline{\sigma }\right)=0.$$

Therefore, it suffices to prove

$$\mathbb{V}\left(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{{\sum }_{i=N+1}^n{Y}_i}{a_n}}\ge (1+3\epsilon )\overline{\sigma }\right)=0.$$

(18)

Let $n_k=\left[e^{k^{1-\alpha }}\right]$, where $\alpha >0$ satisfies ${(1+\epsilon )}^2(1-\delta )(1-\alpha )>1$. Then it can be verified that $n_{k+1}/n_k\to 1$ and ${\displaystyle \frac{n_{k+1}-n_k}{n_k}}\sim {\displaystyle \frac{C}{k^{\alpha }}}$, as $k\to \infty$. Consider k sufficiently large such that $n_k>N$. Then, for $n_k<n\le n_{k+1}$, we have

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

(19)

Without loss of generality, based on the earlier assumption $\mathbb{E}\left[X_i\right]=0$ for each $i\ge 1$, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=N+1}^{n_k}}\left|\mathbb{E}\left[Y_i\right]\right|& \le {\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}\left|\mathbb{E}\left[Y_i\right]\right|\hfill \\ \hfill & ={\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}\left|\mathbb{E}\left[Y_i\right]-\mathbb{E}\left[X_i\right]\right|\le {\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}\mathbb{E}\left[\right(|X_i|-b_i{)}^{+}]\hfill \\ \hfill & \le {\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\mathbb{E}\left[|X_i{|}^2\psi \left(\right|X_i\left|\right)/\log \log \left(\right|X_i\left|\right)\right]}{b_i\psi \left(b_i\right)/\log \log \left(b_i\right)}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{a_n}}\underset{i\ge 1}{\sup }\mathbb{E}\left[{\displaystyle \frac{|X_i{|}^2\psi \left(\right|X_i\left|\right)}{\log \log \left(\right|X_i\left|\right)}}\right]{\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\log \log \tau \sqrt{\frac{i}{\log \log i}}}{\tau \sqrt{\frac{i}{\log \log i}}\psi (\tau \sqrt{\frac{i}{\log \log i}})}}\hfill \\ \hfill & \le {\displaystyle \frac{C}{\sqrt{2n\log \log n}}}{\displaystyle \frac{n\log \log \tau \sqrt{\frac{n}{\log \log n}}}{\tau \sqrt{\frac{n}{\log \log n}}\psi (\tau \sqrt{\frac{n}{\log \log n}})}}\hfill \\ \hfill & \le C{\displaystyle \frac{\log \log \tau \sqrt{\frac{n}{\log \log n}}}{\psi (\tau \sqrt{\frac{n}{\log \log n}})}}\to 0,\mathrm{as}n\to \infty .\hfill \end{array}$$

(20)

By (19) and (20), it follows

$$\begin{array}{cc}\hfill \mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=N+1}^n}{Y}_i\ge (1+3\epsilon )\overline{\sigma }\right)\le & \mathbb{V}\left({\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=N+1}^{n_k}}(Y_i-\mathbb{E}\left[Y_i\right])\ge (1+\epsilon )\overline{\sigma }\right)\hfill \\ \hfill & +\mathbb{V}\left(\underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \frac{1}{a_n}}{\displaystyle \sum _{i=n_k+1}^n}{Y}_i\ge 2\epsilon \overline{\sigma }\right)\hfill \\ \hfill :=& I_1+I_2.\hfill \end{array}$$

(21)

Hence, to obtain (18), by the Borel-Cantelli lemma, we only need to prove ${\sum }_{n=1}^{\infty }{I}_1<\infty$ and ${\sum }_{n=1}^{\infty }{I}_2<\infty$.

For $I_1$, by the Chebyshev’s inequality, we have

$$\begin{array}{cc}\hfill I_1& \le \mathbb{V}\left({\displaystyle \frac{1}{a_{n_k}}}{\displaystyle \sum _{i=N+1}^{n_k}}(Y_i-\mathbb{E}\left[Y_i\right])\ge (1+\epsilon )\overline{\sigma }\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{\exp \left\{\lambda (1+\epsilon )\overline{\sigma }\right\}}}\mathbb{E}\left[\exp \left\{{\displaystyle \frac{\lambda }{a_{n_k}}}{\displaystyle \sum _{i=N+1}^{n_k}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right].\hfill \end{array}$$

where $\lambda >0$ depends on n and will be determined later. By the elementary exponential inequality

$$e^x\le 1+x+{\displaystyle \frac{x^2}{2}}{e}^{\left|x\right|},\forall x\in \mathbb{R},$$

and noting that for any $i\le n_k$,

$${\displaystyle \frac{\lambda }{a_{n_k}}}|Y_i-\mathbb{E}\left[Y_i\right]|\le {\displaystyle \frac{2\lambda b_i}{a_{n_k}}}\le {\displaystyle \frac{2\lambda b_{n_k}}{a_{n_k}}}={\displaystyle \frac{\sqrt{2}\lambda \tau }{\log \log n_k}},$$

it follows that

$$\begin{array}{cc}& \exp \left\{{\displaystyle \frac{\lambda }{a_{n_k}}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)\right\}\hfill \\ & \le 1+{\displaystyle \frac{\lambda }{a_{n_k}}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)+{\displaystyle \frac{{\lambda }^2}{2a_{n_k}^2}}{\left(Y_i-\mathbb{E}\left[Y_i\right]\right)}^2\exp \left\{{\displaystyle \frac{\lambda }{a_{n_k}}}|Y_i-\mathbb{E}\left[Y_i\right]|\right\}\hfill \\ & \le 1+{\displaystyle \frac{\lambda }{a_{n_k}}}\left(Y_i-\mathbb{E}\left[Y_i\right]\right)+{\displaystyle \frac{{\lambda }^2}{4n_k\log \log n_k}}{\left(Y_i-\mathbb{E}\left[Y_i\right]\right)}^2\exp \left\{{\displaystyle \frac{\sqrt{2}\lambda \tau }{\log \log n_k}}\right\}.\hfill \end{array}$$

Taking expectations on both sides, for $i\le n_k$, we have

$$\begin{array}{cc}\hfill \mathbb{E}\left[\exp \left\{{\displaystyle \frac{\lambda }{a_{n_k}}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right]& \le 1+{\displaystyle \frac{{\lambda }^2}{4n_k\log \log n_k}}\underset{j>N}{\sup }\mathbb{E}\left[{(Y_j-\mathbb{E}\left[Y_j\right])}^2\right]\exp \left\{{\displaystyle \frac{\sqrt{2}\lambda \tau }{\log \log n_k}}\right\}\hfill \\ & \le \exp \left\{{\displaystyle \frac{{\lambda }^2}{4n_k\log \log n_k}}\underset{j>N}{\sup }\mathbb{E}\left[{(Y_j-\mathbb{E}\left[Y_j\right])}^2\right]\exp \left\{{\displaystyle \frac{\sqrt{2}\lambda \tau }{\log \log n_k}}\right\}\right\},\hfill \end{array}$$

where the last inequality holds since $1+x\le e^x$ for all $x\ge 0$.

Consequently,

$$\begin{array}{cc}\hfill \mathbb{E}& \left[\exp \left\{{\displaystyle \frac{\lambda }{a_{n_k}}}{\displaystyle \sum _{i=N+1}^{n_k}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right]\le {\displaystyle \prod _{i=N+1}^{n_k}}\mathbb{E}\left[\exp \left\{{\displaystyle \frac{\lambda }{a_{n_k}}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right]\hfill \\ \hfill & \le \exp \left\{{\displaystyle \frac{{\lambda }^2}{4\log \log n_k}}\underset{j>N}{\sup }\mathbb{E}\left[{(Y_j-\mathbb{E}\left[Y_j\right])}^2\right]\exp \left\{{\displaystyle \frac{\sqrt{2}\lambda \tau }{\log \log n_k}}\right\}\right\},\hfill \end{array}$$

(22)

where the first inequality follows from the negative dependence of $Y_i$ and the monotonicity of the exponential function. Let

$$\lambda ={\displaystyle \frac{2(1+\epsilon )\overline{\sigma }\log \log n_k}{\underset{j>N}{\sup }\mathbb{E}\left[{(Y_j-\mathbb{E}\left[Y_j\right])}^2\right]}}.$$

By (17), we have

$$\begin{array}{cc}\hfill I_1& \le \exp \left\{-\lambda (1+\epsilon )\overline{\sigma }+{\displaystyle \frac{{\lambda }^2}{4\log \log n_k}}\underset{i>N}{\sup }\mathbb{E}\left[{(Y_i-\mathbb{E}\left[Y_i\right])}^2\right]\exp \left\{{\displaystyle \frac{\sqrt{2}\lambda \tau }{\log \log n_k}}\right\}\right\}\hfill \\ & \le \exp \left\{-{(1+\epsilon )}^2(1-\delta )\log \log n_k\left[2-\exp \left\{{\displaystyle \frac{2\sqrt{2}\tau (1+\epsilon )\overline{\sigma }}{\underset{i>N}{\sup }\mathbb{E}\left[{(Y_i-\mathbb{E}\left[Y_i\right])}^2\right]}}\right\}\right]\right\}.\hfill \end{array}$$

Let $\gamma$ be a fixed small positive constant satisfying ${(1+\epsilon )}^2(1-\delta )(1-\alpha )(1-\gamma )>1$. Choose $\tau >0$ sufficiently small such that

$$2-\exp \left\{{\displaystyle \frac{2\sqrt{2}\tau (1+\epsilon )\overline{\sigma }}{{\sup }_{i>N}\mathbb{E}\left[{(Y_i-\mathbb{E}\left[Y_i\right])}^2\right]}}\right\}\ge 1-\gamma .$$

Thus we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{k=1}^{\infty }}{I}_1& \le {\displaystyle \sum _{k=1}^{\infty }}\exp \left\{-{(1+\epsilon )}^2(1-\delta )(1-\gamma )\log \log n_k\right\}\hfill \\ \hfill & \le {\displaystyle \sum _{k=1}^{\infty }}{k}^{-{(1+\epsilon )}^2(1-\delta )(1-\alpha )(1-\gamma )}<\infty .\hfill \end{array}$$

(23)

Finally, we consider $I_2$. By $a_{n_{k+1}}/a_{n_k}\to 1$, as $k\to \infty$, we have

$$\begin{array}{cc}\hfill I_2& \le \mathbb{V}\left(\underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \frac{a_{n_{k+1}}}{a_{n_k}}}·{\displaystyle \frac{1}{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^n}{Y}_i\ge 2\epsilon \overline{\sigma }\right)\hfill \\ \hfill & \le \mathbb{V}\left(\underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \frac{1}{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^n}{Y}_i\ge \epsilon \overline{\sigma }\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{\exp \left\{\rho \epsilon \overline{\sigma }\right\}}}\mathbb{E}\left[\exp \left\{\rho \underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \frac{1}{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^n}{Y}_i\right\}\right],\hfill \end{array}$$

(24)

where $\rho >0$ depends on n and will be determined later. By (4), we have

$$\begin{array}{cc}\hfill & \mathbb{E}\left[\exp \left\{\rho \underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \frac{1}{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^n}{Y}_i\right\}\right]\hfill \\ \hfill & \le \exp \left\{\rho \underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \frac{1}{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^n}\mathcal{E}\left[Y_i\right]\right\}·\mathbb{E}\left[\exp \left\{\rho \underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \frac{1}{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^n}(Y_i-\mathcal{E}\left[Y_i\right])\right\}\right]\hfill \\ \hfill & \le \exp \left\{\rho \underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \frac{1}{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^n}\mathcal{E}\left[Y_i\right]\right\}·4\mathbb{E}\left[\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}(Y_i-\mathcal{E}\left[Y_i\right])\right\}\right]\hfill \\ \hfill & =4\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}\underset{n_k<n\le n_{k+1}}{\max }{\displaystyle \sum _{i=n}^{n_{k+1}}}(-\mathcal{E}\left[Y_i\right])+{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\mathbb{E}\left[Y_i\right]\right\}·\hfill \\ \hfill & \mathbb{E}\left[\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right]\hfill \\ \hfill & \le 4\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\left(\right|\mathcal{E}\left[Y_i\right]|+|\mathbb{E}\left[Y_i\right]\left|\right)\right\}·\mathbb{E}\left[\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right]\hfill \\ \hfill & =4\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\left(\right|\mathcal{E}\left[Y_i\right]-\mathcal{E}\left[X_i\right]|+|\mathbb{E}\left[Y_i\right]-\mathbb{E}\left[X_i\right]\left|\right)\right\}·\hfill \\ \hfill & \mathbb{E}\left[\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right]\hfill \\ \hfill & \le 4\exp \left\{{\displaystyle \frac{2\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\mathbb{E}|X_i-Y_i|\right\}·\mathbb{E}\left[\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right],\hfill \end{array}$$

(25)

where the second-to-last equality follows from the earlier assumption $\mathbb{E}\left[X_i\right]=\mathcal{E}\left[X_i\right]=0$ for each $i\ge 1$. Similar to the proof of (22), we obtain

$$\begin{array}{cc}\hfill & \mathbb{E}\left[\exp \left\{{\displaystyle \frac{\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}(Y_i-\mathbb{E}\left[Y_i\right])\right\}\right]\hfill \\ \hfill & \le \exp \left\{{\displaystyle \frac{{\rho }^2\underset{j>N}{\sup }\mathbb{E}\left[{(Y_j-\mathbb{E}\left[Y_j\right])}^2\right]}{4\log \log n_{k+1}}}{\displaystyle \frac{n_{k+1}-n_k}{n_{k+1}}}\exp \left\{{\displaystyle \frac{\sqrt{2}\rho \tau }{\log \log n_{k+1}}}\right\}\right\}.\hfill \end{array}$$

(26)

Hence, combing (24)–(26), we have

$$\begin{array}{cc}\hfill I_2& \le 4\exp \left\{{\displaystyle \frac{2\rho }{a_{n_{k+1}}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\mathbb{E}|X_i-Y_i|\right\}\hfill \\ \hfill & ·\exp \left\{-\rho \epsilon \overline{\sigma }+{\displaystyle \frac{{\rho }^2\underset{i>N}{\sup }\mathbb{E}\left[{(Y_i-\mathbb{E}\left[Y_i\right])}^2\right]}{4\log \log n_{k+1}}}{\displaystyle \frac{n_{k+1}-n_k}{n_{k+1}}}\exp \left\{{\displaystyle \frac{\sqrt{2}\rho \tau }{\log \log n_{k+1}}}\right\}\right\}\hfill \\ \hfill & :=4I_{21}·I_{22}.\hfill \end{array}$$

(27)

Choose

$$\rho ={\displaystyle \frac{\overline{\sigma }}{\underset{j>N}{\sup }\mathbb{E}\left[{(Y_j-\mathbb{E}\left[Y_j\right])}^2\right]}}\log \log n_{k+1}(\alpha \log k-\log \log k+\log \epsilon -\log \overline{\sigma }).$$

For $I_{21}$, by (20) we have

$${\displaystyle \frac{1}{a_{n_{k+1}-n_k}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\mathbb{E}|X_i-Y_i|\le {\displaystyle \frac{1}{a_{n_{k+1}-n_k}}}{\displaystyle \sum _{i=n_k+1}^{n_{k+1}}}\mathbb{E}\left[\right(|X_i|-b_i{)}^{+}]\to 0.$$

(28)

On the other hand

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\rho a_{n_{k+1}-n_k}}{a_{n_{k+1}}}}\hfill \\ \hfill & \le C{\displaystyle \frac{\log \log n_{k+1}(\alpha \log k-\log \log k+\log \epsilon -\log \overline{\sigma })}{\sqrt{2n_{k+1}\log \log n_{k+1}}}}\sqrt{2(n_{k+1}-n_k)\log \log (n_{k+1}-n_k)}\hfill \\ \hfill & \le C\log k\sqrt{{\displaystyle \frac{\log \log n_{k+1}}{n_{k+1}}}}\sqrt{(n_{k+1}-n_k)\log \log (n_{k+1}-n_k)}\hfill \\ \hfill & \le C\log k\sqrt{{\displaystyle \frac{\log \log n_{k+1}·\log \log (n_{k+1}-n_k)}{k^{\alpha }}}}\hfill \\ \hfill & \le C\log k·\log (k+1)\sqrt{{\displaystyle \frac{1}{k^{\alpha }}}}\to 0,\hfill \end{array}$$

(29)

as $k\to \infty$. Combining (28) with (29), we obtain $I_{21}\to 1$ as $k\to \infty$.

For $I_{22}$, to simplify the notation in subsequent proof, denote $\sup \mathbb{E}=\underset{j>N}{\sup }\mathbb{E}\left[{(Y_j-\mathbb{E}\left[Y_j\right])}^2\right]$. Then we have

$$\begin{array}{cc}\hfill I_{22}& \le \exp \{-{\displaystyle \frac{\epsilon {\overline{\sigma }}^2}{\sup \mathbb{E}}}\log \log n_{k+1}(\alpha \log k-\log \log k+\log \epsilon -\log \overline{\sigma })\hfill \\ \hfill & +{\displaystyle \frac{{\overline{\sigma }}^2}{4\sup \mathbb{E}}}\log \log n_{k+1}{(\alpha \log k-\log \log k+\log \epsilon -\log \overline{\sigma })}^2{\displaystyle \frac{C}{k^{\alpha }}}{\displaystyle \frac{k^{\frac{\sqrt{2}\tau \overline{\sigma }\alpha }{\sup \mathbb{E}}}·{\epsilon }^{\frac{\sqrt{2}\tau \overline{\sigma }}{\sup \mathbb{E}}}}{{\left(\log k\right)}^{\frac{\sqrt{2}\tau \overline{\sigma }}{\sup \mathbb{E}}}·{\overline{\sigma }}^{\frac{\sqrt{2}\tau \overline{\sigma }}{\sup \mathbb{E}}}}}\}\hfill \\ \hfill & \le \exp \{{\displaystyle \frac{{\overline{\sigma }}^2}{\sup \mathbb{E}}}\log \log n_{k+1}(\alpha \log k-\log \log k+\log \epsilon -\log \overline{\sigma })·\hfill \\ \hfill & [-\epsilon +{\displaystyle \frac{C}{4}}·k^{({\displaystyle \frac{\sqrt{2}\tau \overline{\sigma }}{\sup \mathbb{E}}}-1)\alpha }·{\left(\log k\right)}^{1-{\displaystyle \frac{\sqrt{2}\tau \overline{\sigma }}{\sup \mathbb{E}}}}]\},\hfill \end{array}$$

where the last inequality follows $\alpha \log k-\log \log k+\log \epsilon -\log \overline{\sigma }\le C\log k$, for k sufficiently large. Taking $\tau >0$ small enough to satisfy ${\displaystyle \frac{\sqrt{2}\tau \overline{\sigma }}{\sup \mathbb{E}}}<1$, let k sufficiently large such that

$${\displaystyle \frac{C}{4}}·k^{({\displaystyle \frac{\sqrt{2}\tau \overline{\sigma }}{\sup \mathbb{E}}}-1)\alpha }·{\left(\log k\right)}^{1-{\displaystyle \frac{\sqrt{2}\tau \overline{\sigma }}{\sup \mathbb{E}}}}\le {\displaystyle \frac{1}{2}}\epsilon .$$

Consequently, by (17) we have

$$\begin{array}{cc}\hfill I_{22}& \le \exp \left\{-{\displaystyle \frac{{\epsilon \overline{\sigma }}^2}{2\sup \mathbb{E}}}\log \log n_{k+1}(\alpha \log k-\log \log k+\log \epsilon -\log \overline{\sigma })\right\}\hfill \\ \hfill & \le \exp \left\{-C\epsilon \log \log n_{k+1}·\log k·(1-\delta )\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{{(k+1)}^{C\epsilon (1-\delta )(1-\alpha )\log k}}}.\hfill \end{array}$$

(30)

Thus by (27), (30) and $I_{21}\to 1$ as $k\to \infty$, we have

$$\begin{array}{c}\hfill {\displaystyle \sum _{k=1}^{\infty }}{I}_2\le C{\displaystyle \sum _{k=1}^{\infty }}{I}_{22}\le C{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{1}{{(k+1)}^{C\epsilon (1-\delta )(1-\alpha )\log k}}}<\infty .\end{array}$$

(31)

Hence, by virtue of (21), (23), (31) and the Borel-Cantelli lemma, (9) holds. Now (9)–(12) are all established. The proof is complete. □

4. The Weakest Condition for Sub-Linear Expectation

In this section, we show that the moment condition (5) is the weakest moment condition with respect to sub-linear expectation to ensure the validity of the LIL. To illustrate this, we recall the following condition introduced by Zhang (2022) [19]:

(CC): The sub-linear expectation $\mathbb{E}$ on ${\mathcal{H}}_b$ satisfies

$$\mathbb{E}\left[X\right]=\underset{P\in \mathcal{P}}{\sup }P\left[X\right],X\in {\mathcal{H}}_b,$$

where ${\mathcal{H}}_b=\{f\in \mathcal{H};f\mathrm{is}\mathrm{bounded}\}$, and $\mathcal{P}$ is a countable-dimensionally weakly compact family of probability measures on $(\mathsf{\Omega },\sigma (\mathcal{H}\left)\right)$. This compactness implies that, for any $X_1,X_2,\dots \in {\mathcal{H}}_b$ and any sequence $\left\{P_n\right\}\subset \mathcal{P}$, there exist a subsequence $\left\{n_k\right\}$ and a probability measure $P\in \mathcal{P}$ such that

$$\underset{k\to \infty }{\lim }{P}_{n_k}\left[\phi (X_1,\dots ,X_d)\right]=P\left[\phi (X_1,\dots ,X_d)\right],\phi \in C_{b,\mathrm{Lip}}\left({\mathbb{R}}^d\right),d\ge 1.$$

According to Zhang (2021) [13], if $\mathsf{\Omega }$ is a complete separable metric space, each element in $\mathcal{H}$ is a continuous function on $\mathsf{\Omega }$. The sub-linear expectation $\mathbb{E}$ is defined as

$$\mathbb{E}\left[X\right]=\underset{P\in \mathcal{P}}{\max }P\left[X\right],X\in {\mathcal{H}}_b,$$

where $\mathcal{P}$ is a weakly compact family of probability measures on the metric space $\mathsf{\Omega }$, then condition (CC) is satisfied.

Zhang (2022) [19] demonstrated that for an independent and identically distributed (i.i.d.) sequence ${\left\{X_n\right\}}_{n=1}^{\infty }$ on $(\mathsf{\Omega },\mathcal{H},\mathbb{E})$, if condition (CC) holds, the necessity of the LIL follows, i.e.,

$$\mathbb{V}(\underset{n\to \infty }{\lim \sup }{\displaystyle \frac{|S_n|}{\sqrt{2n\log \log n}}}=\infty )<1⟹C_{\mathbb{V}}\left({\displaystyle \frac{|X_1{|}^2}{\log \log |X_1|}}\right)<\infty .$$

(32)

It seems that we cannot derive the similar necessary moment condition with respect to sub-linear expectation. However, we may employ the above moment condition with respect to Choquet expectation to prove that condition (5) is the weakest moment condition ensuring the validity of the LIL under sub-linear expectation.

The following counterexample illustrates that for any given $\psi \left(x\right)\in {\mathsf{\Phi }}_d$, there always exists a sequence ${\left\{X_n\right\}}_{n=1}^{\infty }$ of i.i.d. random variables such that ${\sup }_{n\ge 1}\mathbb{E}\left[{\displaystyle \frac{|X_n{|}^2\psi \left(\right|X_n\left|\right)}{\log \log |X_n|}}\right]=\mathbb{E}\left[{\displaystyle \frac{|X_1{|}^2\psi \left(\right|X_1\left|\right)}{\log \log |X_1|}}\right]<\infty$, but $C_{\mathbb{V}}\left({\displaystyle \frac{|X_1{|}^2}{\log \log |X_1|}}\right)=\infty$, which implies that the LIL in Theorem 1 fails to hold.

Example 1.

Let ${\mathsf{\Omega }}_i=\left\{a_0\right\}\cup \{a_k^{+},a_k^{-}|k=1,2,\dots \},i=1,2,\dots$ be a sequence of countable sample spaces, and ${\mathcal{F}}_i$ be the σ-algebra consisting of all subsets of ${\mathsf{\Omega }}_i$. Define the family of probability measures ${\mathcal{P}}_i=\{P_0,P_1,P_2,\dots \}$, $i=1,2,\dots$ on $({\mathsf{\Omega }}_i,{\mathcal{F}}_i)$ as follows:

(i) 

The Dirac measure $P_0$:

$$P_0\left(a_0\right)=1,P_0\left(a_k^{+}\right)=P_0\left(a_k^{-}\right)=0\mathit{for}\mathit{all}k\ge 1;$$

(ii) 

For each integer $i\ge 1$:

$$P_i\left(a_0\right)=1-{\displaystyle \frac{1}{i\psi \left(i\right)}},P_i\left(a_i^{+}\right)=P_i\left(a_i^{-}\right)={\displaystyle \frac{1}{2i\psi \left(i\right)}}\mathit{and}{P}_i\left(a_k^{+}\right)=P_i\left(a_k^{-}\right)=0,\mathit{for}\mathit{all}k\ne i.$$

It is easy to check that ${\left\{P_i\right\}}_{i\ge 1}$ weakly converges to $P_0$ as $i\to \infty$. So each ${\mathcal{P}}_i$, $i=1,2,\dots$ is weakly compact.

Define the product measurable space and product probability family:

$$\mathsf{\Omega }={\displaystyle \prod _{i=1}^{\infty }}{\mathsf{\Omega }}_i,\mathcal{F}={\displaystyle \prod _{i=1}^{\infty }}{\mathcal{F}}_i,\mathcal{P}={\displaystyle \prod _{i=1}^{\infty }}{\mathcal{P}}_i.$$

The sub-linear expectation is induced by the upper expectation: for any bounded random variable $X\in \mathcal{F}$,

$$\mathbb{E}\left[X\right]=\underset{P\in \mathcal{P}}{\sup }{E}_P\left[X\right].$$

Since each ${\mathcal{P}}_i$ is weakly compact, by the diagonal argument for probability sequences, the product family $\mathcal{P}$ is countable-dimensionally weakly compact and thus satisfies condition (CC).

Next, we construct a special sequence of random variables to make the LIL fail to hold. To simplify the notation, denote $l_k=\sqrt{k\log \log k}$, $k=0,1,2,\dots$ Define random variables on each $({\mathsf{\Omega }}_i,{\mathcal{F}}_i)$:

$$Z\left(a_0\right)=0,Z\left(a_k^{+}\right)=l_k+l_{k+1},Z\left(a_k^{-}\right)=-(l_k+l_{k+1}),\mathit{for}\mathit{all}k\ge 1.$$

Then Z satisfies that for any $i,k\ge 1$,

$$P_i\left(\right|Z|>2l_k)=P_i\left(\{a_k^{+},a_k^{-}|k\ge i\}\right)=\left\{\begin{array}{cc}0\hfill & \mathit{if}k>i\hfill \\ {\displaystyle \frac{1}{i\psi \left(i\right)}}\hfill & \mathit{if}k\le i\hfill \end{array}\right..$$

Further define the sequence ${\left\{Y_n\right\}}_{n=1}^{\infty }$ on the product space $(\mathsf{\Omega },\mathcal{F})$ by

$$Y_n\left(\omega \right)=Y_n({\omega }_1,{\omega }_2,\dots )=Z\left({\omega }_n\right),\forall n\ge 1,\omega =({\omega }_1,{\omega }_2,\dots )\in \mathsf{\Omega }.$$

It is easy to check that ${\left\{Y_n\right\}}_{n=1}^{\infty }$ is a sequence of i.i.d. random variables under sub-linear expectation $\mathbb{E}$.

We now show that $\mathbb{E}\left[Y_n\right]=\mathcal{E}\left[Y_n\right]=0$ for all $n\ge 1$. For $P_0$, we have $E_{P_0}\left[Z\right]=Z\left(a_0\right)=0$. For $P_i$, $i\ge 1$, we have

$$\begin{array}{cc}\hfill E_{P_i}\left[Z\right]& =Z\left(a_0\right)P_i\left(a_0\right)+Z\left(a_i^{+}\right)P_i\left(a_i^{+}\right)+Z\left(a_i^{-}\right)P_i\left(a_i^{-}\right)\hfill \\ \hfill & =0·\left(1-{\displaystyle \frac{1}{i\psi \left(i\right)}}\right)+(l_i+l_{i+1})·{\displaystyle \frac{1}{2i\psi \left(i\right)}}-(l_i+l_{i+1})·{\displaystyle \frac{1}{2i\psi \left(i\right)}}=0.\hfill \end{array}$$

Then for any $P\in {\mathcal{P}}_i$, we have $E_P\left[Z\right]\equiv 0$. Hence

$$\mathbb{E}\left[Y_n\right]=\underset{P\in \mathcal{P}}{\sup }{E}_P\left[Y_n\right]=\underset{P\in {\mathcal{P}}_1}{\sup }{E}_P\left[Z\right]=0=\underset{P\in {\mathcal{P}}_1}{\inf }{E}_P\left[Z\right]=\underset{P\in \mathcal{P}}{\inf }{E}_P\left[Y_n\right]=\mathcal{E}\left[Y_n\right].$$

We proceed to discuss the convergence of the moment condition with respect to sub-linear expectation and the moment condition with respect to Choquet expectation for ${\left\{Y_n\right\}}_{n=1}^{\infty }$.

• Note that for any $\psi \left(x\right)\in {\mathsf{\Phi }}_d$,

$$\begin{array}{cc}\hfill \mathbb{E}\left[{\displaystyle \frac{|Y_1{|}^2\psi \left(\right|Y_1\left|\right)}{\log \log |Y_1|}}\right]& =\underset{P\in \mathcal{P}}{\sup }{E}_P\left[{\displaystyle \frac{|Y_1{|}^2\psi \left(\right|Y_1\left|\right)}{\log \log |Y_1|}}\right]\hfill \\ \hfill & =\underset{P\in {\mathcal{P}}_1}{\sup }{E}_P\left[{\displaystyle \frac{{\left|Z\right|}^2\psi \left(\right|Z\left|\right)}{\log \log \left|Z\right|}}\right]\hfill \\ \hfill & \le \underset{k\ge 1}{\sup }\left[{\displaystyle \frac{{(l_k+l_{k+1})}^2\psi (l_k+l_{k+1})}{\log \log (l_k+l_{k+1})k\psi \left(k\right)}}\right]\hfill \\ \hfill & \le \underset{k\ge 1}{\sup }\left[{\displaystyle \frac{C\psi (l_k+l_{k+1})}{\psi \left(k\right)}}\right]<\infty .\hfill \end{array}$$

But

$$\begin{array}{cc}\hfill C_{\mathbb{V}}\left[{\displaystyle \frac{|Y_1{|}^2}{\log \log |Y_1|}}\right]& \ge {\displaystyle \sum _{i=1}^{\infty }}\mathbb{V}\left({\displaystyle \frac{|Y_1{|}^2}{\log \log |Y_1|}}>i\right)={\displaystyle \sum _{i=1}^{\infty }}\underset{P\in {\mathcal{P}}_1}{\sup }P\left({\displaystyle \frac{{\left|Z\right|}^2}{\log \log \left|Z\right|}}>i\right)\hfill \\ & \ge {\displaystyle \sum _{i=1}^{\infty }}{P}_{i+1}\left({\displaystyle \frac{{\left|Z\right|}^2}{\log \log \left|Z\right|}}>i\right)\hfill \\ & ={\displaystyle \sum _{i=1}^{\infty }}{P}_{i+1}\left(\sqrt{{\displaystyle \frac{{\left|Z\right|}^2}{\log \log \left|Z\right|}}·\log \log {\displaystyle \frac{{\left|Z\right|}^2}{\log \log \left|Z\right|}}}>l_i\right)\hfill \\ & \ge {\displaystyle \sum _{i=1}^{\infty }}{P}_{i+1}\left(\right|Z|>2l_i)\hfill \\ & ={\displaystyle \sum _{i=1}^{\infty }}{\displaystyle \frac{1}{(i+1)\psi (i+1)}}=\infty .\hfill \end{array}$$

Therefore by (32), we have successfully constructed a sequence ${\left\{Y_n\right\}}_{n=1}^{\infty }$ of i.i.d. random variables under sub-linear expectation such that ${\sup }_{n\ge 1}\mathbb{E}\left[{\displaystyle \frac{|Y_n{|}^2\psi \left(\right|Y_n\left|\right)}{\log \log |Y_n|}}\right]<\infty$ for any $\psi \left(x\right)\in {\mathsf{\Phi }}_d$ while $C_{\mathbb{V}}\left[{\displaystyle \frac{|Y_1{|}^2}{\log \log |Y_1|}}\right]=\infty$, confirming that the LIL fails to hold under this setting.