Traffic Transformer: Transformer-based framework for temporal traffic accident prediction

1.   Introduction

Let $\{X, X_n; n\geq1\}$ be a sequence of independent and identically distributed (i.i.d.) random variables. Complete convergence first established by Hsu and Robbins ^[1] (for the sufficiency) and Erdős ^[2,3] (for the necessity) proceeds as follows:

if and only if $EX = 0$ and $EX^2 < \infty$. Baum and Katz ^[4] extended the above result and obtained the following theorem:

if and only if $E|X|^r < \infty$, and when $r\geq1$, $EX = 0$.

There are several extensions of the research on complete convergence. One of them is the study of the convergence rate of complete convergence. The first work was the convergence rate, achieved by Heyde ^[5]. He got the result of $\lim\limits_{\epsilon\rightarrow0}\epsilon^{2}\sum_{n = 1}^\infty P(|S_{n}|\geq\epsilon n) = EX^2$ under the conditions $EX = 0$ and $EX^2 < \infty$. For more results on the convergence rate, see Chen ^[6], Sp${\rm \breve{a}}$taru ^[7], Gut and Sp${\rm \breve{a}}$taru ^[8], Sp${\rm \breve{a}}$tarut and Gut ^[9], Gut and Steinebach ^[10], He and Xie ^[11], Kong and Dai ^[12], etc.

But (1.1) does not hold for $p = 2$. However, by replacing $n^{1/p}$ by $\sqrt{n\ln n}$ and $\sqrt{n\ln\ln n}$, Gut and Sp${\rm \breve{a}}$taru ^[8] and Sp${\rm \breve{a}}$tarut and Gut ^[9] established the following results called the convergence rate of the law of the (iterated) logarithm. Supposing that $\{X, X_n; n\geq1\}$ is a sequence of i.i.d. random variables with $EX = 0$ and $EX^2 = \sigma^2 < \infty$, Gut and Sp${\rm \breve{a}}$taru ^[8] and Sp${\rm \breve{a}}$tarut and Gut ^[9] obtained the following results respectively:

where $\mathcal{N}$ is the standard normal distribution, and

Motivated by the above results, the purpose of this paper is to extend (1.2) and (1.3) to sub-linear expectation space (to be introduced in Section 2), which was introduced by Peng ^[13,14], and to study the necessary conditions of (1.2).

Under the theoretical framework of the traditional probability space, in order to infer the model, all statistical models must assume that the error (and thus the response variable) is subject to a certain uniquely determined probability distribution, that is, the distribution of the model is deterministic. Classical statistical modeling and statistical inference are based on such distribution certainty or model certainty. "Distribution certainty" modeling has yielded a set of mature theories and methods. However, the real complex data in economic, financial and other fields often have essential and non negligible probability and distribution uncertainty. The probability distribution of the response variable to be studied is uncertain and does not meet the assumptions of classical statistical modeling. Therefore, classical probability statistical modeling methods cannot be used for this type of data modeling. Driven by uncertainty issues, Peng ^[14,15] established a theoretical framework for sub-linear expectation spaces from the perspective of expectations. Sub-linear expectation has a wide range of application backgrounds and prospects. In recent years, a series of research achievements on limit theory in sub-linear expectation spaces has been established. See Peng ^[14,15], Zhang ^[16,17,18], Hu ^[19], Wu and Jiang ^[20,21], Wu et al. ^[22], Wu and Lu ^[23], etc. Wu ^[24], Liu and Zhang ^[25], Ding ^[26] and Liu and Zhang ^[27] obtained the convergence rate for complete moment convergence. However, the convergence rate results for the (iterative) logarithmic law have not been reported yet. The main difficulty in studying it is that the sub-linear expectation and capacity are not additive, which makes many traditional probability space tools and methods no longer effective; thus, it is much more complex and difficult to study it.

In Section 2, we will provide the relevant definitions of sub-linear expectation space, the basic properties and the lemmas that need to be used in this paper.

2.   Preliminaries

Let $(\Omega, \mathcal{F})$ be a measurable space and let $\mathcal{H}$ be a linear space of random variables on $(\Omega, \mathcal{F})$ such that if $X_1, \ldots, X_n\in \mathcal{H}$ then $\varphi(X_1, \ldots, X_n)\in \mathcal{H}$ for each $\varphi\in C_{l, Lip}(\mathbb{R}_n)$, where $C_{l, Lip}(\mathbb{R}_n)$ denotes the set of local Lipschitz functions on $\mathbb{R}_n$. In this case, for $X\in\mathcal{H}$, $X$ is called a random variable.

Definition 2.1. A sub-linear expectation $\hat{\mathbb{E}}$ on $\mathcal{H}$ is a function: $\mathcal{H}\rightarrow \mathbb{R}$ satisfying the following for all $X, Y\in \mathcal{H}$:

(a) Monotonicity: If $X\geq Y$ then $\hat{\mathbb{E}}X\geq\hat{\mathbb{E}}Y$;

(b) Constant preservation: $\hat{\mathbb{E}}c = c$;

(c) Sub-additivity: $\hat{\mathbb{E}}(X+Y)\leq \hat{\mathbb{E}}X+\hat{\mathbb{E}}Y$;

(d) Positive homogeneity: $\hat{\mathbb{E}}(\lambda X) = \lambda\hat{\mathbb{E}}X, \lambda\geq0$.

The triple $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$ is called a sub-linear expectation space. The conjugate expectation $\hat{\mathbb{\varepsilon}}$ of $\hat{\mathbb{E}}$ is defined by

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

The upper and lower capacities $(\mathbb{V}, \nu)$ corresponding to $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$ are respectively defined as

The Choquet integrals is defined by

From all of the definitions above, it is easy to obtain the following Proposition 2.1.

Proposition 2.1. Let $X, Y\in \mathcal{H}$ and $A, B\in\mathcal{F}$.

$({\rm{i}})$ $\hat{\varepsilon} X\leq\hat{\mathbb{E}}X, \quad \hat{\mathbb{E}}(X+a) = \hat{\mathbb{E}}X+a, \quad \forall a\in \mathbb{R}$;

$({\rm{ii}})$ $|\hat{\mathbb{E}}(X-Y)|\leq\hat{\mathbb{E}}|X-Y|, \quad \hat{\mathbb{E}}(X-Y)\geq\hat{\mathbb{E}}X-\hat{\mathbb{E}}Y$;

$({\rm{iii}})$ $\nu(A)\leq\mathbb{V}(A), \quad \mathbb{V}(A\cup B)\leq\mathbb{V}(A)+\mathbb{V}(B), \quad \nu(A\cup B)\leq\nu(A)+\mathbb{V}(B);$

$({\rm{iv}})$ If $f\leq I(A)\leq g, f, g\in\mathcal{H}$, then

$({\rm{v}})$(Lemma 4.5 (iii) in Zhang ^[16]) For any $c > 0$,

where, here and hereafter, $a\wedge b: = \min(a, b),$ and $a\vee b: = \max(a, b)$ for any $a, b \in \mathbb{R}$.

$({\rm{vi}})$Markov inequality: $\mathbb{V}(|X|\geq x)\leq \hat{\mathbb{E}}(|X|^p)/x^p, \quad \forall \quad x > 0, p > 0;$

Jensen inequality: $\left(\hat{\mathbb{E}}(|X|^r)\right)^{1/r}\leq\left(\hat{\mathbb{E}}(|X|^s)\right)^{1/s} \quad {\rm for} \quad 0 < r\leq s.$

Definition 2.2. (Peng ^[14,15])

(ⅰ) (Identical distribution) Let $X_1$ and $X_2$ be two random variables on $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$. They are called identically distributed, denoted by $X_1\stackrel{d}{ = }X_2$, if

A sequence $\{X_n; n\geq1\}$ of random variables is said to be identically distributed if for each $i\geq1$, $X_i\stackrel{d}{ = }X_1$.

(ⅱ) (Independence) In a sub-linear expectation space $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$, a random vector ${\bf Y} = (Y_1, \ldots, Y_n)$, $Y_i\in \mathcal{H}$ is said to be independent of another random vector ${\bf X} = (X_1, \ldots, X_m), X_i\in \mathcal{H}$ under $\hat{\mathbb{E}}$ if for each $\varphi\in C_{l, Lip}(\mathbb{R}_m\times\mathbb{R}_n)$, there is $\hat{\mathbb{E}}(\varphi({\bf X}, {\bf Y})) = \hat{\mathbb{E}}[\hat{\mathbb{E}}(\varphi({\bf x}, {\bf Y}))|_{{\bf x} = {\bf X}}]$.

(ⅲ) (Independent and identically distributed) A sequence $\{X_n; n\geq1\}$ of random variables is said to be i.i.d., if $X_{i+1}$ is independent of $(X_1, \ldots, X_i)$ and $X_i\stackrel{d}{ = }X_1$ for each $i\geq1$.

From Definition 2.2 (ⅱ), it can be verified that if $Y$ is independent of $X$, and $X\geq0, \hat{\mathbb{E}}Y\geq0$, then $\hat{\mathbb{E}}(XY) = \hat{\mathbb{E}}(X)\hat{\mathbb{E}}(Y)$. Further, if $Y$ is independent of $X$ and $X, Y\geq0$, then

For convenience, in all subsequent parts of this article, let $\{X, X_n; n\geq1\}$ be a sequence of random variables in $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$, and $S_n = \sum_{i = 1}^nX_i$. For any $X\in\mathcal{H}$ and $c > 0$, set $X^{(c)}: = (-c)\vee X\wedge c$. The symbol $c$ represents a positive constant that does not depend on $n$. Let $a_x\sim b_x$ denote $\lim_{x\rightarrow \infty}a_x/b_x = 1$, $a_x\ll b_x$ denote that there exists a constant $c > 0$ such that $a_x\leq cb_x$ for sufficiently large $x$, $[x]$ denote the largest integer not exceeding $x$, and $I(\cdot)$ denote an indicator function.

To prove the main results of this article, the following three lemmas are required.

Lemma 2.1. (Theorem 3.1 (a) and Corollary 3.2 (b) in Zhang ^[16]) Let $\{X_k; k\geq1\}$ be a sequence of independent random variables in $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$.

(i) If $\hat{\mathbb{E}}X_{k}\leq0$, then for any $x, y > 0$,

(ii) If $\hat{\varepsilon}X_{k}\leq0$, then there exists a constant $c > 0$ such that for any $x > 0$,

where $B_n = \sum_{k = 1}^n\hat{\mathbb{E}}X^2_k$.

Here we give the notations of a $G$-normal distribution which was introduced by Peng ^[14].

Definition 2.3. ($G$-normal random variable) For $0\leq\underline{\sigma}^2\leq \bar{\sigma}^2 < \infty$, a random variable $\xi$ in $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$ is called a $G$-normal $\mathcal{N}(0, [\underline{\sigma}^2, \bar{\sigma}^2])$ distributed random variable (write $\xi\sim\mathcal{N}(0, [\underline{\sigma}^2, \bar{\sigma}^2])$ under $\hat{\mathbb{E}}$), if for any $\varphi\in C_{l, Lip}(\mathbb{R})$, the function $u(x, t) = \hat{\mathbb{E}}\left(\varphi(x+\sqrt{t}\xi)\right)$ ($x\in \mathbb{R}, t\geq0$) is the unique viscosity solution of the following heat equation:

where $G(\alpha) = (\bar{\sigma}^2\alpha^+-\underline{\sigma}^2\alpha^-)/2.$

From Peng ^[14], if $\xi\sim\mathcal{N}(0, [\underline{\sigma}^2, \bar{\sigma}^2])$ under $\hat{\mathbb{E}}$, then for each convex function $\varphi$,

If $\sigma = \bar{\sigma} = \underline{\sigma}$, then $\mathcal{N}(0, [\underline{\sigma}^2, \bar{\sigma}^2]) = \mathcal{N}(0, \sigma^2)$ which is a classical normal distribution.

In particular, notice that $\varphi(x) = |x|^p, p\geq1$ is a convex function, taking $\varphi(x) = |x|^p, p\geq1$ in (2.4), we get

Equation (2.5) implies that

Lemma 2.2. (Theorem 4.2 in Zhang ^[17], Corollary 2.1 in Zhang ^[18]) Let $\{X, X_n; n\geq1\}$ be a sequence of i.i.d. random variables in $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$. Suppose that

(i)$\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^2\wedge c)$ is finite;

(ii)$x^2\mathbb{V}(|X|\geq x)\rightarrow0$ as $x\rightarrow \infty$;

(iii)$\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^{(c)}) = \lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}((-X)^{(c)}) = 0$.

Then for any bounded continuous function $\varphi$,

and if $F(x): = \mathbb{V}(|\xi|\geq x)$, then

where $\xi\sim\mathcal{N}(0, [\underline{\sigma}^2, \bar{\sigma}^2])$ under $\hat{\mathbb{E}}$, $\bar{\sigma}^2 = \lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^2\wedge c)$ and $\underline{\sigma}^2 = \lim\limits_{c\rightarrow \infty}\hat{\varepsilon}(X^2\wedge c)$.

Lemma 2.3. (Lemma 2.1 in Zhang ^[17]) Let $\{X_n; n\geq1\}$ be a sequence of independent random variables in $(\Omega, \mathcal{H}, \hat{\mathbb{E}})$, and $0 < \alpha < 1$ be a real number. If there exist real constants $\beta_{n, k}$ such that

then

3.   The main results and the proofs

The results of this article are as follows.

Theorem 3.1. Let $\{X, X_{n}; n\geq1\}$ be a sequence of i.i.d. random variables in $(\Omega, \mathcal{H}, \hat{\mathbb{E}}).$ Suppose that

Then for $0\leq\delta\leq1$,

where, here and hereafter, $\xi\sim\mathcal{N}(0, [\underline{\sigma}^2, \bar{\sigma}^2])$ under $\hat{\mathbb{E}}$, $\bar{\sigma}^2 = \lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^2\wedge c)$ and $\underline{\sigma}^2 = \lim\limits_{c\rightarrow \infty}\hat{\varepsilon}(X^2\wedge c)$.

Conversely, if (3.2) holds for $\delta = 1$, then (3.1) holds.

Theorem 3.2. Under the conditions of Theorem 3.1,

Remark 3.1. Theorems 3.1 and 3.2 not only extend Theorem 3 in ^[8] and Theorem 2 in ^[9], respectively, from the probability space to sub-linear expectation space, but they also study and obtain necessary conditions for Theorem 3.1.

Remark 3.2. Under the condition $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(|X|-c)^+ = 0$ ($\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^2-c)^+ = 0\Rightarrow \lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(|X|-c)^+ = 0)$, it is easy to verify that $\hat{\mathbb{E}}(\pm X) = \lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}((\pm X)^{(c)})$. So, Corollary 3.9 in Ding ^[26] has two more conditions than Theorem 3.2: $\hat{\mathbb{E}}$ is continuous and $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^2-c)^+ = 0$. Therefore, Corollary 3.9 in Ding ^[26] and Theorem 3.2 cannot be inferred from each other.

Proof of the direct part of Theorem 3.1.. Note that

Hence, in order to establish (3.2), it suffices to prove that

and

Given that $\frac{\ln^\delta n}{n}$ and $\mathbb{V}(|\xi|\geq \varepsilon\sqrt{\ln n})$ is monotonically decreasing with respect to $n$, it holds that

and

Therefore, (3.4) follows from

Let $M\geq40$; write $A_{M, \epsilon}: = \exp(M\epsilon^{-2})$.

Let us first estimate $I_{21}(\epsilon)$. For any $\beta > \epsilon^2$,

By (2.2), $\hat{\mathbb{E}}(X^2\wedge c)\leq\int_{0}^c\mathbb{V}(X^2\geq x){\rm d}x$; also, notice that $\mathbb{V}(X^2\geq x)$ is a decreasing function of $x$. So, $C_\mathbb{V}(X^2) = \int_{0}^\infty\mathbb{V}(X^2\geq x){\rm d}x < \infty$ implies that $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^2\wedge c)$ is finite and $\lim\limits_{x\rightarrow \infty}x^2\mathbb{V}(|X|\geq x) = \lim\limits_{x\rightarrow \infty}x\mathbb{V}(X^2\geq x) = 0$. Therefore, (3.1) implies the conditions of Lemma 2.2. From (2.6),

Note that $F(y)$ is a monotonically decreasing function, so its discontinuous points are countable. Hence (3.8) holds for each $y$, except on a set with the null Lebesgue measure. Combining $y^{2\delta+1}\sup\limits_{n\geq A_{\beta, \epsilon}}\left|\mathbb{V}\left(\frac{|S_{n}|}{\sqrt{n}}\geq y\right)-F(y)\right|\leq2M^{\delta+1/2}$ for any $0\leq y\leq\sqrt{M}$, by the Lebesgue bounded convergence theorem, (3.8) leads to the following:

Let $\epsilon\rightarrow0$ first, then let $\beta\rightarrow0$; from (3.7) and (3.9), we get

Next, we estimate that $I_{22}(\epsilon)$. For $0 < \mu < 1$, let $\varphi_\mu(x)\in C_{l, Lip}(\mathbb{R})$ be an even function such that $0\leq \varphi_\mu(x)\leq1$ for all $x$ and $\varphi_\mu(x) = 0$ if $|x|\leq\mu$ and $\varphi_\mu(x) = 1$ if $|x| > 1$. Then

Given (2.1) and (3.11), and that $X, X_{i}$ are identically distributed, for any $x > 0$ and $0 < \mu < 1$, we get

Without loss of generality, we assume that $\bar{\sigma} = 1$. For $n\geq\exp(M\epsilon^{-2})\geq\exp(40\epsilon^{-2})$, set $b_n: = \epsilon\sqrt{n\ln n}/20$; from Proposition 2.1 (ii) and the condition that $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^{(c)}) = 0$,

Using Lemma 2.1 for $\{X_i^{(b_n)}-\hat{\mathbb{E}}X_i^{(b_n)}; 1\leq i\leq n\}$, and taking $x = \epsilon\sqrt{n\ln n}/2$ and $y = 2b_n = \epsilon \sqrt{n\ln n}/10$ in Lemma 2.1 (ⅰ), by Proposition 2.1 (ⅰ), $\hat{\mathbb{E}}(X_i^{(b_n)}-\hat{\mathbb{E}}X_i^{(b_n)}) = 0$, and noting that $|X_i^{(b_n)}-\hat{\mathbb{E}}X_i^{(b_n)}|\leq y$, $B_n = \sum\limits_{i = 1}^n\hat{\mathbb{E}}(X_i^{(b_n)}-\hat{\mathbb{E}}X_i^{(b_n)})^2 \leq4n\hat{\mathbb{E}}(X_i^{(b_n)})^2\leq4n$; combining this with (3.12) we get

from $\frac{\epsilon^2n\ln n}{4(\epsilon^2n\ln n/20+4n)}\left\{1+\frac{2}{3}\ln\left(1+\frac{\epsilon^2\ln n}{80}\right)\right\} \geq3\ln\left(\frac{\epsilon^2\ln n}{80}\right)$.

Since $\{-X, -X_i\}$ also satisfies the (3.1), we can replace the $\{X, X_i\}$ with $\{-X, -X_i\}$ in the upper form

Therefore

This implies the following from Markov's inequality and (2.5),

Let $\epsilon\rightarrow0$ first, then let $M\rightarrow \infty$; we get

Combining this with (3.10) and (3.6), (3.5) is established. □

Proof of the converse part of Theorem 3.1. If (3.2) holds for $\delta = 1$, then

Take $\xi$ as defined by Lemma 2.2 ($\hat{\mathbb{E}}|\xi| < \infty$ from (2.5)) and the bounded continuous function $\psi$ such that $I(x > q\hat{\mathbb{E}}|\xi|+1)\leq\psi(x)\leq I(x > q\hat{\mathbb{E}}|\xi|)$ for any fixed $q > 0$. Therefore, for any $\epsilon > 0, q > 0$ and $n\geq\exp\left(\frac{q\hat{\mathbb{E}}|\xi|+1}{\epsilon}\right)^2$, according to (2.1), Lemma 2.2 and the Markov inequality, one has

From the arbitrariness of $q$, letting $q\rightarrow \infty$, we get the following for any $\epsilon > 0$,

So, there is an $n_0$ such that $\mathbb{V}(|S_{n}|\geq\epsilon \sqrt{n\ln n}) < 1/4$ for $n\geq n_0$. Now for $n\geq 2n_0$, if $k\leq n/2$, then $n-k\geq n/2\geq n_0$, and, combining this with (2.1), (3.11) and (3.12), we get that,

Also, if $n/2 < k\leq n$, then $n, k\geq n/2\geq n_0$; thus,

Taking $\alpha = 1/2, \beta_{n, k} = 0$ in Lemma 2.3, for $n\geq2n_0$,

Since $\max_{k\leq n}|X_k|\leq2\max_{k\leq n}|S_k|$, it follows that for $n\geq2n_0$

Let $Y_k = \varphi_{8/9}\left(\frac{X_k}{9\epsilon\sqrt{n\ln n}}\right)$. Then,

Since $\{X_k; k\geq1\}$ is a sequence of i.i.d. random variables, $\{1-Y_k; k\geq1\}$ is also a sequence of i.i.d. random variables, and $1-Y_k\geq0$; given (2.1), (2.3) and $\hat{\mathbb{E}}(-X) = -\hat{\mathbb{\varepsilon}}(X)$, it can be concluded that,

Hence, by (3.15) and (3.13)

On the other hand,

Hence,

Next, we prove that $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^{(c)}) = \lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}((-X)^{(c)}) = 0$. For $c_1 > c_2 > 0$, by (2.2) and (3.16),

This implies that

By the Cauchy criterion, $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^{(c)})$ and $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}((-X)^{(c)})$ exist and are finite. It follows that $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^{(c)}) = \lim\limits_{n\rightarrow \infty}\hat{\mathbb{E}}(X^{(n)}): = a$. So, for any $\epsilon > 0$, when $n$ is large enough, $|\hat{\mathbb{E}}(X^{(n)})-a| < \epsilon$; by Proposition 2.1 (iii), Lemma 2.1 (ii), $\hat{\mathbb{E}}(-X_k^{(n)}+\hat{\mathbb{E}}X_k^{(n)})^2\leq4\hat{\mathbb{E}}(X_k^{(n)})^2\leq4C_\mathbb{V}(X^2)$ and (3.16),

It is concluded that,

If $a > 0$, taking $\epsilon < a/2$, then $\epsilon_1: = a-2\epsilon > 0$, and

On the other hand, by (3.14),

which contradicts (3.17). It follows that $a\leq0$. Similarly, we can prove that $b: = \lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}((-X)^{(c)})\leq0$. From $(-X)^{(c)} = -X^{(c)}$ and

we conclude that $a = b = 0$, i.e., $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^{(c)}) = \lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}((-X)^{(c)}) = 0$. This completes the proof of Theorem 3.1. □

Proof of Theorem 3.2. Note that

Hence, in order to establish (3.3), it suffices to prove that

and

Obviously, (3.18) follows from

Let $M\geq32$; write $B_{M, \epsilon}: = \exp(\exp(M\epsilon^{-2}))$.

Let us first estimate $J_{21}(\epsilon)$. For any $\beta > \epsilon^2$,

Similar to (3.9) we have

Therefore, let $\epsilon\rightarrow0$ first, then let $\beta\rightarrow0$; we get

Next, we estimate that $J_{22}(\epsilon)$. Without loss of generality, we still assume that $\bar{\sigma} = 1$. For $n\geq\exp(\exp(M\epsilon^{-2}))\geq\exp(\exp(32\epsilon^{-2}))$, set $a_n: = \epsilon\sqrt{n\ln\ln n}/16$; from Proposition 2.1 (ii) and the condition that $\lim\limits_{c\rightarrow \infty}\hat{\mathbb{E}}(X^{(c)}) = 0$,

Using Lemma 2.1 for $\{X_i^{(a_n)}-\hat{\mathbb{E}}X_i^{(a_n)}; 1\leq i\leq n\}$, and taking $x = \epsilon\sqrt{n\ln\ln n}/2$ and $y = 2a_n = \epsilon \sqrt{n\ln\ln n}/8$ in Lemma 2.1 (i), if we note that $|X_i^{(a_n)}-\hat{\mathbb{E}}X_i^{(a_n)}|\leq y$, and $B_n\leq4n$, combined with (3.12) we get

from $\frac{\epsilon^2n\ln\ln n}{4(\epsilon^2n\ln\ln n/16+4n)}\left\{1+\frac{2}{3}\ln\left(1+\frac{\epsilon^2\ln\ln n}{64}\right)\right\} \geq2\ln\left(\frac{\epsilon^2\ln\ln n}{64}\right)$.

Since $\{-X, -X_i\}$ also satisfies (3.1), we can replace the $\{X, X_i\}$ with $\{-X, -X_i\}$ in the upper form

Therefore

This implies the following from Markov's inequality and (2.5):

Hence

Combining this with (3.20) and (3.21), (3.19) is established. □

4.   Conclusions

Statistical modeling is one of the key and basic topics in statistical theory research and practical application research. Under the theoretical framework of traditional probability space, in order to infer the model, all statistical models must assume that the error (and therefore the response variable) follows a unique and deterministic probability distribution, that is, the distribution of the model is deterministic. However, complex data in the fields of economics, finance, and other fields often have inherent and non negligible probability and distribution uncertainties. The probability distribution of the response variables that need to be studied is uncertain and does not meet the assumptions of classical statistical modeling. Therefore, classical probability statistical modeling methods cannot be used to model these types of data. How to analyze and model uncertain random data has been an unresolved and challenging issue that has long plagued statisticians. Driven by uncertainty issues, Peng ^[13] established a theoretical framework for the sub-linear expectation space from the perspective of expectations, providing a powerful tool for analyzing uncertainty problems. The sub-linear expectation has a wide range of potential applications. In recent years, the limit theory for sub-linear expectation spaces has attracted much attention from statisticians, and a series of research results have been achieved. This article overcomes the problem of many traditional probability space tools and methods no longer being effective due to the non additivity of sub-linear expectations and capacity; it also demonstrates the development of sufficient and necessary conditions for the rate convergence of logarithmic laws in sub-linear expectation spaces.

Use of AI tools declaration

The authors declare that they have not used artificial intelligence tools in the creation of this article.

Acknowledgments

This paper was supported by the National Natural Science Foundation of China (12061028) and Guangxi Colleges and Universities Key Laboratory of Applied Statistics.

Conflict of interest

Regarding this article, the author claims no conflict of interest.