Convergence rate of extremes from Maxwell sample

Abstract

For the partial maximum from a sequence of independent and identically distributed random variables with Maxwell distribution, we establish the uniform convergence rate of its distribution to the extreme value distribution.

MSC:62E20, 60E05, 60F15, 60G15.

1 Introduction

One interesting problem in extreme value theory is to consider the convergence rate of some extremes. For the uniform convergence rate of extremes under the second-order regular variation conditions, see Falk [1], Balkema and de Haan [2], de Haan and Resnick [3] and Cheng and Jiang [4]. For the extreme value distributions and their associated uniform convergence rates for given distributions, see Hall and Wellner [5], Hall [6], Peng et al. [7], Lin and Peng [8] and Lin et al. [9].

In this note, we discuss the uniform convergence rate of extremes from a sequence of independent and identically distributed (iid) random variables with Maxwell distribution (MD). The probability density function of MD is given by

$$f(x)=\sqrt{\frac{2}{\pi }}\frac{x^2}{{\sigma }^3}exp(-\frac{x^2}{2{\sigma }^2}),x>0.$$

(1.1)

The MD and the convergence rate of extremes from Maxwell sample have been widely used in the field of physics. We establish the uniform convergence rate of its distribution to the extreme value distribution and give an improved proof for the pointwise convergence rate of MD.

Throughout this paper, let $({\xi }_n,n\ge 1)$ be a sequence of iid random variables with common distribution $F(x)={\int }_0^{x}f(t)dt$ with a probability density function $f(x)$ given by (1.1), and let $M_n={max}_{1\le k\le n}{\xi }_k$ be the partial maximum. Liu and Fu [10] proved that

$$\underset{n\to \mathrm{\infty }}{lim}P({\alpha }_n^{-1}(M_n-{\beta }_n)\le x)=exp(-exp(-x)):=\mathrm{\Lambda }(x)$$

with the normalizing constants ${\alpha }_n$ and ${\beta }_n$ given by

$${\alpha }_n=\frac{\sigma }{{(2logn)}^{\frac{1}{2}}},{\beta }_n={(2{\sigma }^{2}logn)}^{\frac{1}{2}}+\frac{\sigma log(2logn)+\sigma log\frac{2}{\pi }}{2{(2logn)}^{\frac{1}{2}}}.$$

(1.2)

By arguments similar to those of Hall [6], Peng et al. [7] and Lin et al. [9], the appropriate normalizing constants $a_n$ and $b_n$ can be given by the following equations:

$$a_n={\sigma }^2{b}_n^{-1}$$

(1.3)

and

$$\sqrt{\frac{\pi }{2}}\frac{b_n}{\sigma }exp\left(\frac{b_n^2}{2{\sigma }^2}\right)=n.$$

(1.4)

By arguments similar to those of Example 2 of Resnick [11], we have

$$b_n={(2{\sigma }^{2}logn)}^{\frac{1}{2}}+\frac{\sigma log(2logn)+\sigma log\frac{2}{\pi }}{2{(2logn)}^{\frac{1}{2}}}+o\left({(logn)}^{-1/2}\right).$$

Hence

$${\alpha }_n/a_n\to 1,({\beta }_n-b_n)/a_n\to 0,$$

implying

$$\underset{n\to \mathrm{\infty }}{lim}P(M_n\le a_{n}x+b_n)=\underset{n\to \mathrm{\infty }}{lim}{F}^n(a_{n}x+b_n)=\mathrm{\Lambda }(x),$$

cf. Leadbetter et al. [12] or Resnick [11].

This paper is organized as follows. Section 2 gives some auxiliary results. In Section 3, we present the main result. Related proofs are given in Section 4.

2 Auxiliary results

To establish the uniform convergence of $F^n(a_{n}x+b_n)$ to its extreme value distribution $\mathrm{\Lambda }(x)$, we need some auxiliary results. The first result is the decomposition of $F(x)$, which is the following result.

Lemma 1 Let $F(x)$ be the Maxwell distribution function. Then, for $x>0$, we have

$$1-F(x)=\sqrt{\frac{2}{\pi }}\frac{x}{\sigma }(1+\frac{{\sigma }^2}{x^2})exp(-\frac{x^2}{2{\sigma }^2})-r(x)$$

(2.1)

with

$$0<r(x)=\sqrt{\frac{2}{\pi }}{\int }_x^{\mathrm{\infty }}\frac{\sigma }{y^2}exp(-\frac{y^2}{2{\sigma }^2})dy<\sqrt{\frac{2}{\pi }}\frac{{\sigma }^3}{x^3}exp(-\frac{x^2}{2{\sigma }^2}).$$

(2.2)

For simplicity, throughout this paper, let C be a generic positive constant whose value may change from line to line, and let $C_i$, $C_{ij}$ ($i\in N$, $j\in N$) be absolute positive constants.

For the normalizing constants $a_n$, $b_n$ defined by (1.3) and (1.4), respectively, let

$$a_n^{\ast }=a_n{r}_n,b_n^{\ast }=b_n+a_n{\delta }_n,$$

(2.3)

where $r_n\to 1$, ${\delta }_n\to 0$, $n\to \mathrm{\infty }$. So, $a_n^{\ast }/a_n\to 1$, $(b_n^{\ast }-b_n)/a_n\to 0$, implying $F^n(a_n^{\ast }x+b_n^{\ast })\to \mathrm{\Lambda }(x)$. For large n, we have the following result.

Lemma 2 Let $a_n^{\ast }$, $b_n^{\ast }$ be defined by (2.3). For fixed $x\in R$ and sufficiently large n, we have

$$\begin{array}{rl}{F}^n(a_n^{\ast }x+b_n^{\ast })-\mathrm{\Lambda }(x)=& \mathrm{\Lambda }(x)e^{-x}((\frac{x^2}{2}-x-1)a_n{b}_n^{-1}+(r_n-1)x\\ +{\delta }_n+O[{\left(a_n{b}_n^{-1}\right)}^2+{(r_n-1)}^2+{\delta }_n^2]).\end{array}$$

(2.4)

Proof Note that $b_n\sim \sigma {(2logn)}^{\frac{1}{2}}$, which means

$$a_n{b}_n^{-1}\sim \frac{1}{2logn}\to 0.$$

For large n, we have

$$\begin{array}{c}\sqrt{\frac{2}{\pi }}\frac{a_n^{\ast }x+b_n^{\ast }}{\sigma }exp(-\frac{{(a_n^{\ast }x+b_n^{\ast })}^2}{2{\sigma }^2})\hfill \\ =\sqrt{\frac{2}{\pi }}\frac{b_n}{\sigma }(1+a_n{b}_n^{-1}(r_{n}x+{\delta }_n))exp(-\frac{b_n^2}{2{\sigma }^2})\hfill \\ \times exp(-\frac{a_n^2(r_n^2{x}^2+{\delta }_n^2+2r_n{\delta }_{n}x)}{2{\sigma }^2}-(r_n-1)x-x-{\delta }_n)\hfill \\ =n^{-1}{e}^{-x}(1-(\frac{x^2}{2}-x)a_n{b}_n^{-1}-(r_n-1)x-{\delta }_n+O[{\left(a_n{b}_n^{-1}\right)}^2+{(r_n-1)}^2+{\delta }_n^2]).\hfill \end{array}$$

Since

$$\frac{{\sigma }^2}{{(a_n^{\ast }x+b_n^{\ast })}^2}=a_n{b}_n^{-1}-2x{\left(a_n{b}_n^{-1}\right)}^2+O\left({\left(a_n{b}_n^{-1}\right)}^3\right),$$

we have

$$\frac{{\sigma }^4}{{(a_n^{\ast }x+b_n^{\ast })}^4}={\left(a_n{b}_n^{-1}\right)}^2+O\left({\left(a_n{b}_n^{-1}\right)}^3\right).$$

Similarly,

$$\frac{{\sigma }^5}{{(a_n^{\ast }x+b_n^{\ast })}^5}exp(-\frac{{(a_n^{\ast }x+b_n^{\ast })}^2}{2{\sigma }^2})=O\left(n^{-1}{\left(a_n{b}_n^{-1}\right)}^2\right).$$

Hence,

$$\begin{array}{rl}1-F(a_n^{\ast }x+b_n^{\ast })=& n^{-1}{e}^{-x}(1-(\frac{x^2}{2}-x-1)a_n{b}_n^{-1}-(r_n-1)x\\ -{\delta }_n+O[{\left(a_n{b}_n^{-1}\right)}^2+{(r_n-1)}^2+{\delta }_n^2]).\end{array}$$

(2.5)

So,

$$\begin{array}{c}{F}^n(a_n^{\ast }x+b_n^{\ast })-\mathrm{\Lambda }(x)\hfill \\ =(1-n^{-1}{e}^{-x}(1-(\frac{x^2}{2}-x-1)a_n{b}_n^{-1}-(r_n-1)x\hfill \\ {-{\delta }_n+O[{\left(a_n{b}_n^{-1}\right)}^2+{(r_n-1)}^2+{\delta }_n^2]))}^n-\mathrm{\Lambda }(x)\hfill \\ =\mathrm{\Lambda }(x)e^{-x}((\frac{x^2}{2}-x-1)a_n{b}_n^{-1}+(r_n-1)x+{\delta }_n+O[{\left(a_n{b}_n^{-1}\right)}^2+{(r_n-1)}^2+{\delta }_n^2]),\hfill \end{array}$$

which is the desired result. □

3 Main results

In this section we present the pointwise convergence rate and the uniform convergence rate of $F^n(\cdot )$ to its extreme value distribution under different normalizing constants. The first result is the pointwise convergence of extremes under the normalizing constants given by (1.2).

Theorem 1 Let $\{{\xi }_n,n\ge 1\}$ be a sequence of independent identically distributed random variables with common distribution MD. Then

$$F^n({\alpha }_{n}x+{\beta }_n)-\mathrm{\Lambda }(x)\sim \mathrm{\Lambda }(x)e^{-x}\frac{{(log(2logn))}^2}{16logn},$$

(3.1)

for large n, where ${\alpha }_n$, ${\beta }_n$ are defined in (1.2).

Recently Liu and Fu [10] proved the result, we present an improved proof for the pointwise convergence rate in Section 4.

The following is the uniform convergence rate of extremes under the appropriate normalizing constants $a_n$ and $b_n$ given by (1.3) and (1.4), which shows that the optimal convergence rate is proportional to $1/logn$.

Theorem 2 Let $({\xi }_n,n\ge 1)$ be a sequence of independent identically distributed random variables with common distribution MD. For large n, there exist absolute constants $0<d_1<d_2$ such that

$$\frac{d_1}{logn}<\underset{x\in R}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|<\frac{d_2}{logn},$$

(3.2)

where $a_n$ and $b_n$ are defined by (1.3) and (1.4), respectively.

4 Proofs

Proof of Theorem 1 Firstly, we derive the following asymptotic expansions of $b_n$ defined by (1.4)

$$b_n={\beta }_n+o\left({(logn)}^{-\frac{1}{2}}\right)$$

(4.1)

and

$$\begin{array}{rcl}{b}_n& =& {\beta }_n-\frac{\sigma {(log(2logn)+log\frac{2}{\pi })}^2}{16\sqrt{2}{(logn)}^{\frac{3}{2}}}\\ +(\sigma log\frac{4logn+log(2logn)+log\frac{2}{\pi }}{4logn})/{(2logn)}^{\frac{1}{2}}+O\left(\frac{{(log(2logn))}^2}{{(logn)}^{\frac{5}{2}}}\right).\end{array}$$

(4.2)

Setting $b_n={\beta }_n+{\theta }_n$ and substituting into (1.4), we obtain by taking logarithms that

$$log\frac{\pi }{2}+log\sigma -log({\beta }_n+{\theta }_n)+\frac{{\beta }_n^2}{2{\sigma }^2}+\frac{{\beta }_n{\theta }_n}{{\sigma }^2}+\frac{{\theta }_n^2}{{\sigma }^2}=logn.$$

So,

$$\begin{array}{r}\frac{{(log(2logn)+log\frac{2}{\pi })}^2}{16logn}-log\frac{4logn+log(2logn)+log\frac{2}{\pi }}{4logn}\\ +\frac{{\beta }_n{\theta }_n}{{\sigma }^2}+\frac{{\theta }_n^2}{{\sigma }^2}-log(1+\frac{{\theta }_n}{{\beta }_n})=0,\end{array}$$

(4.3)

therefore

$$\frac{{\beta }_n{\theta }_n}{{\sigma }^2}\sim -\frac{{(log(2logn)+log\frac{2}{\pi })}^2}{16logn}+log\frac{4logn+log(2logn)+log\frac{2}{\pi }}{4logn},$$

(4.4)

which implies

$$\begin{array}{rl}{\theta }_n\sim & -\frac{\sigma {(log(2logn)+log\frac{2}{\pi })}^2}{16\sqrt{2}{(logn)}^{\frac{3}{2}}}\\ +(\sigma log\frac{4logn+log(2logn)+log\frac{2}{\pi }}{4logn})/{(2logn)}^{\frac{1}{2}}.\end{array}$$

(4.5)

Once again let

$${\theta }_n=-\frac{\sigma {(log(2logn)+log\frac{2}{\pi })}^2}{16\sqrt{2}{(logn)}^{\frac{3}{2}}}+(\sigma log\frac{4logn+log(2logn)+log\frac{2}{\pi }}{4logn})/{(2logn)}^{\frac{1}{2}}+{\nu }_n,$$

where ${\nu }_n=o(\frac{{(log(2logn))}^2}{{(logn)}^{\frac{3}{2}}})$. By similar arguments, we can obtain (4.2).

Note that $a_n=\frac{{\sigma }^2}{b_n}$, we have

$$r_n-1=\frac{{\alpha }_n}{a_n}-1\sim \frac{log(2logn)}{4logn},{\delta }_n=\frac{{\beta }_n-b_n}{a_n}\sim \frac{{(log(2logn))}^2}{16logn}.$$

Noting $a_n{b}_n^{-1}\sim \frac{1}{2logn}$, by Lemma 2, we have

$$F^n({\alpha }_{n}x+{\beta }_n)-\mathrm{\Lambda }(x)\sim \mathrm{\Lambda }(x)e^{-x}\frac{{(log(2logn))}^2}{16logn}.$$

The proof is complete. □

Proof of Theorem 2 Letting $r_n=1$, ${\delta }_n=0$ in (2.3) and noting $a_n{b}_n^{-1}\sim \frac{1}{2logn}$, and by Lemma 2, there exists an absolute constant $d_1>0$ such that

$$\underset{x\in R}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|>\frac{d_1}{logn}.$$

(4.6)

Thus, in order to obtain the upper bound, we need to prove

$$(\mathrm{a})\underset{-c_n\le x<0}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|<{\mathbb{D}}_1{a}_n{b}_n^{-1},$$

(4.7)

$$(\mathrm{b})\underset{0\le x\le d_n}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|<{\mathbb{D}}_2{a}_n{b}_n^{-1},$$

(4.8)

$$(\mathrm{c})\underset{d_n\le x<\mathrm{\infty }}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|<{\mathbb{D}}_3{a}_n{b}_n^{-1},$$

(4.9)

$$(\mathrm{d})\underset{-\mathrm{\infty }<x\le -c_n}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|<{\mathbb{D}}_4{a}_n{b}_n^{-1},$$

(4.10)

where ${\mathbb{D}}_i>0$ ($i=1,2,3,4$), and

$$c_n=:loglog\frac{b_n^2}{{\sigma }^2}>0,d_n=:-loglog\frac{b_n^2}{b_n^2-{\sigma }^2}>0.$$

Obviously,

$$\sigma {(2logn)}^{\frac{1}{2}}<b_n<\sigma {(2logn)}^{\frac{1}{2}}(1+C_0)$$

and

$$b_n-a_n{c}_n=b_n(1-\frac{{\sigma }^2}{b_n^2}{c}_n)=b_n(1-\frac{{\sigma }^2}{b_n^2}loglog\frac{b_n^2}{{\sigma }^2})>0.$$

Define ${\mathrm{\Psi }}_n(x)=1-F(a_{n}x+b_n)$, then

$$\begin{array}{rl}nlog(1-F(a_{n}x+b_n))& =-n{\mathrm{\Psi }}_n(x)+n{\mathrm{\Psi }}_n(x)+nlog(1-{\mathrm{\Psi }}_n(x))\\ =-n{\mathrm{\Psi }}_n(x)-R_n(x).\end{array}$$

(4.11)

By the following inequality

$$-x-\frac{x^2}{2(1-x)}<log(1-x)<-x(0<x<1),$$

we have

$$0<R_n(x)=-(n{\mathrm{\Psi }}_n(x)+nlog(1-{\mathrm{\Psi }}_n(x)))<\frac{n{\mathrm{\Psi }}_n^2(x)}{2(1-{\mathrm{\Psi }}_n(x))}.$$

First, suppose that $x\ge -c_n$. By (2.1), we have

$$\begin{array}{rl}{\mathrm{\Psi }}_n(x)& \le {\mathrm{\Psi }}_n(-c_n)=1-F(b_n-a_n{c}_n)\\ <\sqrt{\frac{2}{\pi }}\frac{b_n-a_n{c}_n}{\sigma }(1+\frac{{\sigma }^2}{{(b_n-a_n{c}_n)}^2})exp(-\frac{{(b_n-a_n{c}_n)}^2}{2{\sigma }^2})\\ <2\sqrt{\frac{2}{\pi }}\frac{b_n}{\sigma }(1-a_n{b}_n^{-1}{c}_n)exp(-\frac{{(b_n)}^2}{2{\sigma }^2}+c_n-\frac{a_n{b}_n^{-1}{c}_n^2}{2})\\ <2n^{-1}{e}^{c_n}=2n^{-1}log\frac{b_n^2}{{\sigma }^2}\\ <\underset{n\ge n_0}{sup}\frac{2log(C_{1}logn)}{n}<{\mathfrak{C}}_{11}<1\end{array}$$

with $C_1=2{(1+C_0)}^2$, implying

$$\underset{x\ge -c_n}{inf}(1-{\mathrm{\Psi }}_n(x))>1-{\mathfrak{C}}_{11}>0.$$

Therefore,

$$\begin{array}{rcl}0& <& R_n(x)\le \frac{n{\mathrm{\Psi }}_n^2(x)}{2(1-{\mathfrak{C}}_{11})}\le \frac{n{\mathrm{\Psi }}_n^2(-c_n)}{2(1-{\mathfrak{C}}_{11})}\\ <& \frac{n^{-1}{(log(C_{1}logn))}^2}{2(1-{\mathfrak{C}}_{11})}=\frac{n^{-1}{(log(C_{1}logn))}^2{a}_n{b}_n^{-1}}{2(1-{\mathfrak{C}}_{11})a_n{b}_n^{-1}}\\ <& \frac{n^{-1}{(log(C_{1}logn))}^2}{4(1-{\mathfrak{C}}_{11})logn}{a}_n{b}_n^{-1}\\ <& C{a}_n{b}_n^{-1}.\end{array}$$

By $1-e^{-x}<x$, $x>0$, we have

$$|exp(-R_n(x))-1|<R_n(x)<C{a}_n{b}_n^{-1}.$$

Setting $A_n(x)=exp(-n{\mathrm{\Psi }}_n(x)+e^{-x})$, $B_n(x)=exp(-R_n(x))$, we obtain

$$\begin{array}{rcl}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|& =& \mathrm{\Lambda }(x)|A_n(x)B_n(x)-1|\\ =& \mathrm{\Lambda }(x)|A_n(x)B_n(x)-B_n(x)+B_n(x)-1|\\ \le & \mathrm{\Lambda }(x)|A_n(x)-1|+|B_n(x)-1|\\ <& \mathrm{\Lambda }(x)|A_n(x)-1|+C{a}_n{b}_n^{-1}.\end{array}$$

(4.12)

By (2.1) and (2.2), we have

$$\begin{array}{rcl}-n{\mathrm{\Psi }}_n(x)+e^{-x}& =& -n[\sqrt{\frac{2}{\pi }}\frac{b_n+a_{n}x}{\sigma }(1+\frac{{\sigma }^2}{{(b_n+a_{n}x)}^2})exp(-\frac{{(b_n+a_{n}x)}^2}{2{\sigma }^2})\\ -r(a_{n}x+b_n)]+e^{-x}\\ =& (1+a_n{b}_n^{-1}x)e^{-x}{C}_n(x),\end{array}$$

where

$$C_n(x)=(-1-\frac{{\sigma }^2}{{(b_n+a_{n}x)}^2}+\frac{{\sigma }^4}{{(b_n+a_{n}x)}^4}{\delta }_n(a_{n}x+b_n))exp(-\frac{a_n{b}_n^{-1}{x}^2}{2})+{(1+a_n{b}_n^{-1}x)}^{-1}$$

with $0<{\delta }_n(a_{n}x+b_n)<1$. To prove (4.7), we consider the case of $-c_n\le x<0$. By $e^{-x}>1-x$, $x>0$, we have

$$\begin{array}{rl}{C}_n(x)& <(1-\frac{a_n{b}_n^{-1}{x}^2}{2})((-1+\frac{{\sigma }^4}{{(b_n+a_{n}x)}^4}){\delta }_n(a_{n}x+b_n))+{(1+a_n{b}_n^{-1}x)}^{-1}\\ <(1-\frac{a_n{b}_n^{-1}{x}^2}{2})\{-1+{\left(a_n{b}_n^{-1}\right)}^2{(1+a_n{b}_n^{-1}x)}^{-4}\}+{(1+a_n{b}_n^{-1}x)}^{-1}\\ =({(1+a_n{b}_n^{-1}x)}^{-4}+\frac{x^2}{2}-x{(1+a_n{b}_n^{-1}x)}^{-1})a_n{b}_n^{-1}\end{array}$$

(4.13)

and

$$\begin{array}{rcl}{C}_n(x)& >& (-1-\frac{{\sigma }^2}{{(b_n+a_{n}x)}^2})exp(-\frac{a_n{b}_n^{-1}{x}^2}{2})+{(1+a_n{b}_n^{-1}x)}^{-1}\\ >& (-1-\frac{{\sigma }^2}{{(b_n+a_{n}x)}^2})+{(1+a_n{b}_n^{-1}x)}^{-1}\\ >& (-{(1+a_n{b}_n^{-1}x)}^{-2}-x{(1+a_n{b}_n^{-1}x)}^{-1})a_n{b}_n^{-1}\\ >& -{(1+a_n{b}_n^{-1}x)}^{-2}.\end{array}$$

(4.14)

Hence, for $-c_n\le x<0$, by combining (4.13) and (4.14) together, we have

$$\begin{array}{rcl}|C_n(x)|& <& ({(1+a_n{b}_n^{-1}x)}^{-4}+\frac{x^2}{2}-x{(1+a_n{b}_n^{-1}x)}^{-1}+{(1+a_n{b}_n^{-1}x)}^{-2})a_n{b}_n^{-1}\\ <& ({(1-a_n{b}_n^{-1}{c}_n)}^{-4}+\frac{c_n^2}{2}+c_n{(1-a_n{b}_n^{-1}{c}_n)}^{-1}+{(1-a_n{b}_n^{-1}{c}_n)}^{-2})a_n{b}_n^{-1}\\ <& C_{21}.\end{array}$$

Furthermore, for $-c_n\le x<0$, we have

$$\begin{array}{rcl}|-n{\mathrm{\Psi }}_n(x)+e^{-x}|& <& (1+a_n{b}_n^{-1}x)e^{-x}|C_n(x)|\\ <& ({(1+a_n{b}_n^{-1}x)}^{-4}+\frac{x^2}{2}-x{(1+a_n{b}_n^{-1}x)}^{-1}+{(1+a_n{b}_n^{-1}x)}^{-2})e^{-x}{a}_n{b}_n^{-1}\\ <& ({(1-a_n{b}_n^{-1}{c}_n)}^{-4}+\frac{c_n^2}{2}+c_n{(1-a_n{b}_n^{-1}{c}_n)}^{-1}\\ +{(1-a_n{b}_n^{-1}{c}_n)}^{-2})e^{c_n}{a}_n{b}_n^{-1}\\ <& C_{22}.\end{array}$$

Noting that $0<|e^x-1|<|x|(e^x+1)$, $x\in R$ and $e^{-x}>1-x+x^2/2$ for $-c_n\le x<0$, we have

$$\begin{array}{rcl}\mathrm{\Lambda }(x)|A_n(x)-1|& =& \mathrm{\Lambda }(x)|exp(-n{\mathrm{\Psi }}_n(x)+e^{-x})-1|\\ <& \mathrm{\Lambda }(x)|-n{\mathrm{\Psi }}_n(x)+e^{-x}|(exp(-n{\mathrm{\Psi }}_n(x)+e^{-x})+1)\\ <& (e^{C_{22}}+1)({(1+a_n{b}_n^{-1}x)}^{-4}+\frac{x^2}{2}-x{(1+a_n{b}_n^{-1}x)}^{-1}+{(1+a_n{b}_n^{-1}x)}^{-2})\\ \times a_n{b}_n^{-1}exp(-e^{-x}-x)\\ <& C_{23}{a}_n{b}_n^{-1}.\end{array}$$

Together with (4.12), we establish (4.7).

Second, we prove (4.8). Note that

$$\begin{array}{rcl}{C}_n(x)& <& (-1+\frac{{\sigma }^4}{{(b_n+a_{n}x)}^4}{\delta }_n(a_{n}x+b_n))(1-\frac{a_n{b}_n^{-1}{x}^2}{2})+{(1+a_n{b}_n^{-1}x)}^{-1}\\ <& {\left(a_n{b}_n^{-1}\right)}^2{(1+a_n{b}_n^{-1}x)}^{-4}+\frac{x^2}{2}{a}_n{b}_n^{-1}-a_n{b}_n^{-1}x{(1+a_n{b}_n^{-1}x)}^{-1}\\ <& {\left(a_n{b}_n^{-1}\right)}^2{(1+a_n{b}_n^{-1}x)}^{-4}+\frac{x^2}{2}{a}_n{b}_n^{-1}\\ <& (1+\frac{x^2}{2})a_n{b}_n^{-1}\end{array}$$

(4.15)

and

$$C_n(x)>(-{(1+a_n{b}_n^{-1}x)}^{-2}-x{(1+a_n{b}_n^{-1}x)}^{-1})a_n{b}_n^{-1}.$$

(4.16)

By (4.15) and (4.16), for $0\le x<d_n$, we have

$$\begin{array}{rcl}|C_n(x)|& <& (1+\frac{x^2}{2}+{(1+a_n{b}_n^{-1}x)}^{-2}+x{(1+a_n{b}_n^{-1}x)}^{-1})a_n{b}_n^{-1}\\ <& (2+x+\frac{x^2}{2})a_n{b}_n^{-1}.\end{array}$$

Hence,

$$\begin{array}{rcl}|-n{\mathrm{\Psi }}_n(x)+e^{-x}|& <& (1+a_n{b}_n^{-1}x)e^{-x}|C_n(x)|\\ <& (1+a_n{b}_n^{-1}x)e^{-x}(2+x+\frac{x^2}{2})a_n{b}_n^{-1}\\ <& C_{31}{a}_n{b}_n^{-1}<C_{32}.\end{array}$$

Therefore

$$\begin{array}{rcl}\mathrm{\Lambda }(x)|A_n(x)-1|& <& \mathrm{\Lambda }(x)|-n{\mathrm{\Psi }}_n(x)+e^{-x}|(exp(-n{\mathrm{\Psi }}_n(x)+e^{-x})+1)\\ <& C_{31}(e^{C_{32}}+1)\mathrm{\Lambda }(d_n)a_n{b}_n^{-1}\\ <& C_{33}{a}_n{b}_n^{-1}.\end{array}$$

(4.17)

Combining (4.12) and (4.17) together, we can derive that

$$\underset{0\le x\le d_n}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|<(C_{12}+C_{33})a_n{b}_n^{-1}=:{\mathbb{D}}_2{a}_n{b}_n^{-1}.$$

Hence (4.8) is proved.

Third, for $x\ge d_n$, we have

$$\underset{x\ge d_n}{sup}(1-\mathrm{\Lambda }(x))\le 1-\mathrm{\Lambda }(d_n)=a_n{b}_n^{-1}.$$

(4.18)

By $1-e^x<-x$, $x\in R$, we have

$$\begin{array}{rcl}1-F^n(a_n{d}_n+b_n)& =& 1-exp(nlogF(a_n{d}_n+b_n))\\ <& -nlogF(a_n{d}_n+b_n)\\ =& n{\mathrm{\Psi }}_n(d_n)+R_n(d_n).\end{array}$$

(4.19)

By (2.1) and $log(1+x)<x$, $0<x<1$, we have

$$\begin{array}{rcl}n{\mathrm{\Psi }}_n(d_n)& =& n(1-F(a_n{d}_n+b_n))\\ <& (1+a_n{b}_n^{-1}{d}_n)e^{-d_n}(1+a_n{b}_n^{-1}{(1+a_n{b}_n^{-1}{d}_n)}^{-2})\\ <& 2(1+a_n{b}_n^{-1}{d}_n)log\frac{b_n^2}{b_n^2-{\sigma }^2}\\ <& 2(d_n+a_n^{-1}{b}_n)\frac{{\sigma }^2}{b_n^2-{\sigma }^2}{a}_n{b}_n^{-1}\\ =& (1+\frac{{\sigma }^2}{b_n^2-{\sigma }^2}-\frac{{\sigma }^2}{b_n^2-{\sigma }^2}loglog\left(\frac{{\sigma }^2}{b_n^2-{\sigma }^2}\right))a_n{b}_n^{-1}\\ <& C_{41}{a}_n{b}_n^{-1}.\end{array}$$

(4.20)

Noting that $R_n(d_n)<C_{12}{a}_n{b}_n^{-1}$, and combining (4.18), (4.19), (4.20) and (4.14) together, we have

$$\begin{array}{rcl}\underset{x\ge d_n}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|& <& \underset{x\ge d_n}{sup}(1-F^n(a_{n}x+b_n))+\underset{x\ge d_n}{sup}(1-\mathrm{\Lambda }(x))\\ <& n{\mathrm{\Psi }}_n(d_n)+R_n(d_n)+a_n{b}_n^{-1}\\ <& (C_{41}+C_{12}+1)a_n{b}_n^{-1}=:{\mathbb{D}}_3{a}_n{b}_n^{-1},\end{array}$$

which is (4.9).

Finally, consider the case of $-\mathrm{\infty }<x<-c_n$. If $a_{n}x+b_n\le 0$, then $F^n(a_{n}x+b_n)=0$. By $\mathrm{\Lambda }(-x)<\frac{1}{x}$, $x>1$, we have

$$\underset{x\le -b_n/a_n}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|=\underset{x\le -b_n/a_n}{sup}\mathrm{\Lambda }(x)\le \mathrm{\Lambda }(-b_n^2/{\sigma }^2)<\frac{{\sigma }^2}{b_n^2}=a_n{b}_n^{-1}.$$

So, we only need to consider the case of $a_{n}x+b_n>0$. By using the monotonicity of $\mathrm{\Lambda }(x)$, we have

$$\underset{x\le -c_n}{sup}\mathrm{\Lambda }(x)\le \mathrm{\Lambda }(-c_n)=a_n{b}_n^{-1}.$$

(4.21)

Noting $log(1-x)<-x$, $0<x<1$ and $e^{-x}>1-x$, $x\in R$, and combining (2.1) and (2.2) together, we have

$$\begin{array}{c}\underset{x\le -c_n}{sup}{F}^n(a_{n}x+b_n)\hfill \\ \le F^n(b_n-a_n{c}_n)\hfill \\ <{(1-n^{-1}(1-a_n{b}_n^{-1}{c}_n)(1-{\left(a_n{b}_n^{-1}\right)}^2{(1-a_n{b}_n^{-1}{c}_n)}^{-4})exp(c_n-\frac{a_n{b}_n^{-1}{c}_n^2}{2}))}^n\hfill \\ <exp(-e^{c_n}(1-a_n{b}_n^{-1}{c}_n)(1-{\left(a_n{b}_n^{-1}\right)}^2{(1-a_n{b}_n^{-1}{c}_n)}^{-4})exp(-\frac{a_n{b}_n^{-1}{c}_n^2}{2}))\hfill \\ <exp(-e^{c_n}(1-(a_n{b}_n^{-1}{c}_n+{\left(a_n{b}_n^{-1}\right)}^2{(1-a_n{b}_n^{-1}{c}_n)}^{-3}+\frac{a_n{b}_n^{-1}{c}_n^2}{2})))\hfill \\ <exp(-e^{c_n})exp\left((a_n{b}_n^{-1}{c}_n+{\left(a_n{b}_n^{-1}\right)}^2{(1-a_n{b}_n^{-1}{c}_n)}^{-3}+\frac{a_n{b}_n^{-1}{c}_n^2}{2})e^{c_n}\right)\hfill \\ <C_{51}{a}_n{b}_n^{-1}.\hfill \end{array}$$

Together with (4.21), we have

$$\begin{array}{rcl}\underset{-\mathrm{\infty }<x\le -c_n}{sup}|F^n(a_{n}x+b_n)-\mathrm{\Lambda }(x)|& \le & \underset{-\mathrm{\infty }<x<-c_n}{sup}{F}^n(a_{n}x+b_n)+\underset{-\mathrm{\infty }<x<-c_n}{sup}\mathrm{\Lambda }(x)\\ \le & F^n(b_n-a_n{c}_n)+\mathrm{\Lambda }(-c_n)\\ <& (C_{12}+1)a_n{b}_n^{-1}=:{\mathbb{D}}_4{a}_n{b}_n^{-1}.\end{array}$$

This is (4.10). The proof is complete. □