Density expansions of extremes from general error distribution with applications

Abstract

In this paper the higher-order expansions of density of normalized maximum with parent following general error distribution are established. The main results are applied to derive the higher-order expansions of the moments of extremes.

1 Introduction

In extreme value theory, the quality of convergence of normalized partial maximum of a sample has been studied in recent literature. For the convergence rate of distribution of normalized maximum, we refer to Smith [1], Leadbetter et al. [2], de Haan and Resnick [3] for general cases, and specific cases were studied by Hall [4], Nair [5], Peng et al. [6] and Jia and Li [7]. Nair [5] derived the higher-order expansions of moments of normalized maximum with parent following normal distribution. Liao et al. [8] and Jia et al. [9] extended Nair’s results to skew-normal distribution and general error distribution, respectively.

The main objective of this paper is to derive the higher-order expansions of density of normalized maximum with parent following the general error distribution. To the best of our knowledge, there are few studies on the rate of convergence of density of normalized maximum except the work of de Haan and Resnick [10] for local limit theorems and Omey [11] for rates of convergence of densities with regular variation with remainders excluding the case we will study in this paper, i.e., the general error distribution.

Let $\{X_{n},n\geq1\}$ be a sequence of independent and identically distributed (i.i.d.) random variables with marginal cumulative distribution function (cdf) $F_{v}$ following the general error distribution ($F_{v} \sim\operatorname{GED}(v)$ for short), and let $M_{n}=\max_{1\leq k\leq n}X_{k}$ denote its partial maximum. The probability density function (pdf) of the $\operatorname{GED}(v)$ is given by

$$ f_{v}(x)=\frac{v\exp(-(1/2)|x/\lambda|^{v})}{\lambda 2^{1+1/v}\Gamma(1/v)},\quad x\in\mathbb{R}, $$

where $v >0$ is the shape parameter, $\lambda={[2^{-2/v}\Gamma(1/v)/\Gamma(3/v)]}^{1/2}$ and $\Gamma(\cdot)$ denotes the gamma function (Nelson [12]). Note that the $\operatorname{GED}(2)$ reduces to the standard normal distribution.

For the $\operatorname{GED}(v)$, the limiting distribution of maximum $M_{n}$ and its associated higher-order expansions are given by Peng et al. [13] and Jia and Li [7]. Peng et al. [6] showed that

$$ \lim_{n\to\infty}\mathrm{P}(M_{n} \leq{a_{n}x+b_{n}}) =\Lambda(x)=\exp\bigl(-\exp(-x)\bigr), \quad x\in\mathbb{R} $$

(1.1)

provided the norming constants $a_{n}$ and $b_{n}$ satisfy the following equations:

$$ 1-F_{v}(b_{n})=n^{-1},\qquad a_{n}=f(b_{n}), $$

(1.2)

where $f(x)=2v^{-1}\lambda^{v} x^{1-v}$. In the sequel, let

$$ g_{n}(x)=na_{n}F^{n-1}_{v}(a_{n}x+b_{n})f_{v}(a_{n}x+b_{n}) $$

(1.3)

denote the density of normalized maximum, and

$$ \Delta_{n}\bigl(g_{n},\Lambda^{\prime};x \bigr)=g_{n}(x)-\Lambda^{\prime}(x) $$

(1.4)

with $\Lambda^{\prime}(x)=e^{-x}\Lambda(x)$. By Proposition 2.5 in Resnick [14], $\Delta_{n}(g_{n},\Lambda^{\prime})\to0$ as $n\to\infty$. For both applications and theoretical analysis, it is of interest to know the convergence rate of (1.4). This paper focuses on this topic and applies the main results to derive the high-order expansions of moments of extremes.

The paper is organized as follows. Section 2 provides the main results and all proofs are deferred to Section 4. Auxiliary lemmas with proofs are given in Section 3.

2 Main results

In this section, we present the asymptotic expansions of density for the normalized maximum formed by the $\operatorname{GED}(v)$ random variables and its applications to the higher-order expansions of moments of extremes.

Theorem 2.1

Let $F_{v}(x)$ denote the cdf of $\operatorname {GED}(v)$ with $v > 0$, then for $v\neq1$, with norming constants $a_{n}$ and $b_{n}$ given by (1.2), we have

$$ b^{v}_{n} \bigl[b^{v}_{n} \Delta_{n}\bigl(g_{n},\Lambda^{\prime};x \bigr)-k_{v}(x)\Lambda '(x) \bigr]\to \omega_{v}(x)\Lambda'(x) $$

(2.1)

as $n\to\infty$, where $k_{v}(x)$ and $\omega_{v}(x)$ are respectively given by

$$\begin{aligned}& k_{v}(x)=k_{v1}(x)+k_{v2}(x), \end{aligned}$$

(2.2)

$$\begin{aligned}& \omega_{v}(x)=\bigl(1-v^{-1}\bigr)\lambda^{2v} \bigl(\omega_{v1}(x)+\omega _{v2}(x)e^{-x}+ \omega_{v3}(x)e^{-2x} \bigr) \end{aligned}$$

(2.3)

with

$$\begin{aligned}& k_{v1}(x)=-\bigl(1-v^{-1}\bigr)\lambda^{v} \bigl(x^{2}+2x\bigr), \\& k_{v2}(x)=\bigl(1-v^{-1}\bigr)\lambda^{v} \bigl(2+2x+\bigl(x^{2}+2x\bigr)e^{-x} \bigr), \\& \omega_{v1}(x)=-4-2\bigl(1-v^{-1}\bigr)x^{2}- \frac{2}{3}\bigl(1-2v^{-1}\bigr)x^{3}+\frac {1}{2} \bigl(1-v^{-1}\bigr)x^{4}, \\& \omega_{v2}(x)=-\frac{4}{v}x-\frac{2}{v}x^{2}- \frac {2}{3}\bigl(5-4v^{-1}\bigr)x^{3}-\frac{3}{2} \bigl(1-v^{-1}\bigr)x^{4}, \\& \omega_{v3}(x)=\frac{1}{2}\bigl(1-v^{-1}\bigr) \bigl(x^{2}+2x\bigr)^{2}. \end{aligned}$$

Remark 2.1

If we choose the norming constants $a_{n}$ and $b_{n}$ such that

$$ \frac{1-F_{v}(b_{n})}{f_{v}(b_{n})}\sim\frac{2\lambda^{v}}{v}b_{n}^{1-v}, \qquad a_{n}=f(b_{n}) $$

(2.4)

with $v\ne1$, then

$$ b^{v}_{n} \bigl[b^{v}_{n} {\Delta_{n}\bigl(g_{n},\Lambda^{\prime};x\bigr)}- \bar{k}_{v}(x)\Lambda'(x) \bigr]\to \bar{ \omega}_{v}(x)\Lambda'(x) $$

(2.5)

as $n\to\infty$, where $\bar{k}_{v}(x)$ and $\bar{\omega}_{v}(x)$ are respectively given by

$$\begin{aligned} \bar{k}_{v}(x) = \bigl(1-v^{-1}\bigr) \lambda^{v} \bigl(-x^{2}+\bigl(2+2x+x^{2} \bigr)e^{-x} \bigr) \end{aligned}$$

(2.6)

and

$$\begin{aligned} \bar{\omega}_{v}(x) = \lambda^{2v}\bigl(1-v^{-1} \bigr) \bigl(\bar{\omega}_{v1}(x)+\bar{\omega }_{v2}(x)e^{-x}+ \bar{\omega}_{v3}(x)e^{-2x} \bigr) \end{aligned}$$

(2.7)

with

$$\begin{aligned}& \bar{\omega}_{v1}(x) = \frac{2(2v^{-1}-1)}{3}x^{3}+ \frac{1-v^{-1}}{2}x^{4}, \\& \bar{\omega}_{v2}(x) = 4\bigl(v^{-1}-2\bigr)+4 \bigl(v^{-1}-2\bigr)x +2\bigl(2v^{-1}-3\bigr)x^{2}+ \frac{2}{3}\bigl(4v^{-1}-5\bigr)x^{3}+\frac{3}{2} \bigl(v^{-1}-1\bigr)x^{4}, \\& \bar{\omega}_{v3}(x) = 2\bigl(1-v^{-1}\bigr) \biggl(1+x+ \frac{x^{2}}{2} \biggr)^{2}. \end{aligned}$$

For the case of $v=1$, we have the following results.

Theorem 2.2

For $v = 1$, with norming constants $a_{n}=2^{-1/2}$ and $b_{n}=2^{-1/2}\log(n/2)$, we have

$$ n \bigl[n\Delta_{n}\bigl(g_{n}, \Lambda^{\prime};x\bigr)-k_{1}(x)\Lambda'(x) \bigr] \to \biggl(\omega_{1}(x)+\frac{k^{2}_{1}(x)}{2} \biggr) \Lambda'(x) $$

(2.8)

as $n\to\infty$, where $k_{1}(x)$ and $\omega_{1}(x)$ are respectively given by

$$ k_{1}(x)=-\frac{1}{2}e^{-2x},\qquad \omega_{1}(x)=-\frac{1}{3}e^{-3x}. $$

(2.9)

To end this section, we apply the higher-order expansions of densities to derive the asymptotic expansions of the moments of extremes. Methods used here are different from those in Nair [5] and Jia et al. [9].

In the sequel, for nonnegative integers r, let

$$ m_{r}(n)=\int_{x\in\mathbb{R}}x^{r}g_{n}(x) \,dx,\qquad m_{r}=\int_{x\in \mathbb{R}}x^{r} \Lambda'(x)\,dx $$

denote respectively the rth moments of $(M_{n}-b_{n})/a_{n}$ and its limits.

Theorem 2.3

Let $\{X_{n},n\geq1\}$ be an iid sequence with marginal distribution $F_{v} \sim\operatorname{GED}(v)$, then

• (i) for $v \neq1$, with norming constants $a_{n}$ and $b_{n}$ given by (1.2), we have

  $$\begin{aligned}& b_{n}^{v} \bigl[b_{n}^{v} \bigl(m_{r}(n)-m_{r}\bigr)+\bigl(1-v^{-1}\bigr) \lambda^{v}r(m_{r+1}+2m_{r}) \bigr] \\& \quad \to 2r\lambda^{2v}\bigl(1-v^{-1}\bigr) \biggl[\bigl( \bigl(1-v^{-1}\bigr) (r+1)+2\bigr)m_{r}+\bigl( \bigl(1-v^{-1}\bigr) (r+1)+1\bigr)m_{r+1} \\& \qquad {} + \biggl(\frac{1}{4}\bigl(1-v^{-1}\bigr) (r-1)+ \frac{1}{3}\bigl(2-v^{-1}\bigr) \biggr)m_{r+2} \biggr] \end{aligned}$$

  (2.10)

  as $n\to\infty$;

• (ii) for $v = 1$, with norming constants $a_{n}=2^{-1/2}$ and $b_{n}=2^{-1/2}\log(n/2)$, we have

  $$ n \biggl[n\bigl(m_{r}(n)-m_{r}\bigr)+ (-1)^{r}\frac{r}{2}\Gamma^{(r-1)}(2) \biggr] \to(-1)^{r-1}\frac {r}{24} \bigl[8\Gamma^{(r-1)}(3)-3 \Gamma^{(r-1)}(4) \bigr] $$

  (2.11)

  as $n\to\infty$, where $\Gamma^{(r-1)}(t)$ denote the $(r-1)$ th derivative of the gamma function at $x=t$.

3 Auxiliary lemmas

In this section we provide auxiliary lemmas which are needed to prove the main results.

Lemma 3.1

Let $F_{v}(x)$ and $f_{v}(x)$ respectively denote the cdf and pdf of $\operatorname{GED}(v)$ with $v\neq1$, for large x, we have

$$\begin{aligned}& 1-F_{v}(x) \\& \quad = f_{v}(x)\frac{2\lambda^{v}}{v}x^{1-v} \bigl[1+2 \bigl(v^{-1}-1\bigr)\lambda^{v} x^{-v} +4 \bigl(v^{-1}-1\bigr) \bigl(v^{-1}-2\bigr)\lambda^{2v}x^{-2v}+O \bigl(x^{-3v}\bigr) \bigr]. \end{aligned}$$

(3.1)

Furthermore, with the norming constants $a_{n}$ and $b_{n}$ given by (1.2), we have

• (i) for $v\neq1$,

  $$ b^{v}_{n} \bigl[b^{v}_{n} \bigl(F^{n}_{v}(a_{n}x+b_{n})- \Lambda(x) \bigr)-\tilde {k}_{v}(x)\Lambda(x) \bigr] \to \biggl(\tilde{ \omega}_{v}(x)+\frac{\tilde{k}^{2}_{v}(x)}{2} \biggr)\Lambda(x) $$

  (3.2)

  as $n\to\infty$, where $\tilde{k}_{v}(x)$ and $\tilde{\omega}_{v}(x)$ are respectively given by

  $$\begin{aligned}& \tilde{k}_{v}(x)=\bigl(1-v^{-1}\bigr) \lambda^{v}\bigl(x^{2}+2x\bigr)e^{-x}, \end{aligned}$$

  (3.3)
  $$\begin{aligned}& \tilde{\omega}_{v}(x)=\bigl(v^{-1}-1\bigr) \lambda^{2v} \biggl(4x+2x^{2}+\frac {2}{3} \bigl(2-v^{-1}\bigr)x^{3} +\frac{1}{2} \bigl(1-v^{-1}\bigr)x^{4} \biggr)e^{-x}; \end{aligned}$$

  (3.4)

• (ii) for $v = 1$, with norming constants $a_{n}=2^{-1/2}$ and $b_{n}=2^{-1/2}\log(n/2)$, we have

  $$ n \bigl[n \bigl(F^{n}_{1}(a_{n}x+b_{n})- \Lambda(x) \bigr)-{k}_{1}(x)\Lambda (x) \bigr] \to \biggl({ \omega}_{1}(x)+\frac{{k}^{2}_{1}(x)}{2} \biggr)\Lambda(x) $$

  (3.5)

  as $n\to\infty$, where ${k}_{1}(x)$ and ${\omega}_{1}(x)$ are those given by (2.9).

Proof

See Lemma 1 and Theorem 1 in Jia and Li [7]. □

Lemma 3.2

Let $F_{v}(x)$ denote the cdf of $\operatorname {GED}(v)$ with $v > 0$, then

• (i) for $v\neq1$, with norming constants given by (1.2), we have

  $$ F^{n-1}_{v}(a_{n}x+b_{n}) = \biggl(1+\tilde{k}_{v}(x)b^{-v}_{n}+ \biggl( \tilde{\omega}_{v}(x)+\frac {\tilde{k}^{2}_{v}(x)}{2} \biggr) b^{-2v}_{n} \bigl(1+o(1)\bigr) \biggr)\Lambda(x) $$

  (3.6)

  as $n\to\infty$, where $\tilde{k}_{v}(x)$ and $\tilde{\omega}_{v}(x)$ are respectively given by (3.3) and (3.4);

• (ii) for $v=1$, with norming constants $a_{n}=2^{-1/2}$ and $b_{n}=2^{-1/2}\log(n/2)$, we have

  $$ F^{n-1}_{1}(a_{n}x+b_{n}) = \biggl(1+\frac{1}{n}{k}_{1}(x)+\frac{1}{n^{2}} \biggl({ \omega}_{1}(x)+\frac {{k}^{2}_{1}(x)}{2} \biggr) \bigl(1+o(1)\bigr) \biggr) \Lambda(x) $$

  (3.7)

  as $n\to\infty$, where ${k}_{1}(x)$ and ${\omega}_{1}(x)$ are given by (2.9).

Proof

(i) It follows from (3.2) and (3.3) that

$$\begin{aligned} F^{n}_{v}(a_{n}x+b_{n}) =& b^{-v}_{n} \biggl(b^{-v}_{n} \biggl(\tilde{\omega}_{v}(x)+\frac{\tilde {k}^{2}_{v}(x)}{2} \biggr) \bigl(1+o(1) \bigr) +\tilde{k}_{v}(x) \biggr)\Lambda(x)+\Lambda(x) \\ =& \biggl(1+\tilde{k}_{v}(x)b^{-v}_{n}+ \biggl( \tilde{\omega}_{v}(x)+\frac {\tilde{k}^{2}_{v}(x)}{2} \biggr) b^{-2v}_{n} \bigl(1+o(1)\bigr) \biggr)\Lambda(x). \end{aligned}$$

(3.8)

Noting that

$$ F^{n}_{v}(a_{n}x+b_{n})\to \Lambda(x)=\exp\bigl(-\exp(-x)\bigr), $$

by taking logarithms, we have

$$n\bigl(1- F_{v}(a_{n}x+b_{n})\bigr)\to e^{-x}. $$

Thus

$$ 1- F_{v}(a_{n}x+b_{n})\sim n^{-1}e^{-x}=o \bigl(b^{-3v}_{n}\bigr) $$

since $b_{n}\sim2^{1/v}\lambda(\log n)^{1/v}$, which implies

$$ \frac{1}{F_{v}(a_{n}x+b_{n})} = 1+ \bigl(1-F_{v}(a_{n}x+b_{n}) \bigr) \bigl(1+o(1)\bigr)=1+o\bigl(b^{-3v}_{n}\bigr). $$

(3.9)

The desired result (3.6) follows by (3.8) and (3.9). The proof of (ii) is similar and details are omitted here. □

Lemma 3.3

Let $f_{v}(x)$ denote the pdf of $\operatorname {GED}(v)$ with $v \neq1$, then

$$ f_{v}(x) = \bigl(1-F_{v}(x)\bigr) \frac{v}{2\lambda^{v}}x^{v-1} \bigl(1+2\bigl(1-v^{-1}\bigr) \lambda^{v} x^{-v}-4\bigl(1-v^{-1}\bigr)\lambda ^{2v}x^{-2v}+O\bigl(x^{-3v}\bigr) \bigr) $$

(3.10)

for large x, and

$$\begin{aligned} C_{n}(x) :=& \frac{a_{n}f_{v}(a_{n}x+b_{n})}{1-F_{v}(a_{n}x+b_{n})} \\ =& 1+\bigl(1-v^{-1}\bigr)2\lambda^{v}(x+1)b_{n}^{-v} +4\lambda^{2v}\bigl(1-v^{-1}\bigr) \biggl(-1- \frac{x}{v}+\frac {1-2v^{-1}}{2}x^{2} \biggr)b_{n}^{-2v} \\ &{}+O\bigl(b_{n}^{-3v}\bigr) \end{aligned}$$

(3.11)

as $n\to\infty$.

Proof

The desired results follow directly by (3.1). □

Lemma 3.4

Let

$$H_{v}(b_{n};x)=\frac{1-F_{v}(a_{n}x+b_{n})}{1-F_{v}(b_{n})}e^{x}-1 $$

and the norming constants $a_{n}$ and $b_{n}$ be given by (1.2), then

$$ b^{v}_{n} \bigl(b^{v}_{n} H_{v}(b_{n};x)- k_{v1}(x) \bigr) \to \omega^{\circ}_{v}(x) $$

(3.12)

as $n\to\infty$, where $k_{v1}(x)$ is given by (2.2) and $\omega^{\circ}_{v}(x)$ is given by

$$ \omega^{\circ}_{v}(x)=\bigl(1-v^{-1}\bigr) \lambda^{2v} \biggl(4x+2x^{2}+\frac {2}{3} \bigl(2-v^{-1}\bigr)x^{3} +\frac{1}{2} \bigl(1-v^{-1}\bigr)x^{4} \biggr). $$

Proof

Let

$$ B_{n}(x)=\frac{1+2(v^{-1}-1)\lambda^{v} (a_{n}x+b_{n})^{-v}+ 4(v^{-1}-1)(v^{-1}-2)\lambda^{2v}(a_{n}x+b_{n})^{-2v}+O((a_{n}x+b_{n})^{-3v})}{ 1+2(v^{-1}-1)\lambda^{v} (b_{n})^{-v}+ 4(v^{-1}-1)(v^{-1}-2)\lambda^{2v}(b_{n})^{-2v}+O((b_{n})^{-3v})}, $$

it is easy to check that $\lim_{n\to\infty}B_{n}(x)=1$ and

$$ B_{n}(x)-1=\bigl(1+o(1)\bigr) \bigl[-4\bigl(v^{-1}-1\bigr) \lambda ^{2v}b_{n}^{-2v}x+O\bigl(b_{n}^{-3v} \bigr) \bigr]. $$

Hence,

$$ \lim_{n\to\infty}b_{n}^{v} \bigl(B_{n}(x)-1 \bigr)=0 $$

(3.13)

and

$$ \lim_{n\to\infty}b_{n}^{2v} \bigl(B_{v}(x)-1 \bigr)=-4\bigl(v^{-1}-1\bigr) \lambda^{2v}x. $$

(3.14)

By (3.1), we have

$$\begin{aligned}& \frac{1-F_{v}(a_{n}x+b_{n})}{1-F_{v}(b_{n})}e^{x} \\& \quad = B_{n}(x) \exp \biggl[-\int^{x}_{0} \biggl(\frac{(v-1)a_{n}}{b_{n}+a_{n}t}+\frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt \biggr] \\& \quad = B_{n}(x) \biggl\{ 1-\int^{x}_{0} \biggl(\frac{(v-1)a_{n}}{b_{n}+a_{n}t}+\frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt \\& \qquad {} +\frac{1}{2} \biggl[\int^{x}_{0} \biggl(\frac{(v-1)a_{n}}{b_{n}+a_{n}t}+\frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt \biggr]^{2} \bigl(1+o(1)\bigr) \biggr\} . \end{aligned}$$

(3.15)

Combining (3.13)-(3.15) together, we have

$$\begin{aligned}& \lim_{n\to\infty}b^{v}_{n} H_{v}(b_{n};x) \\& \quad = \lim_{n\to\infty}b^{v}_{n} \biggl( {B_{n}(x) \biggl\{ 1-\int^{x}_{0} \biggl(\frac{(v-1)a_{n}}{b_{n}+a_{n}t}+\frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda ^{v}}-1 \biggr)\,dt\bigl(1+o(1)\bigr) \biggr\} -1} \biggr) \\& \quad = \lim_{n\to\infty} \biggl(b^{v}_{n} \bigl({B_{n}(x)-1} \bigr)-b^{v}_{n}{B_{n}(x) \int^{x}_{0} \biggl(\frac{(v-1)a_{n}}{b_{n}+a_{n}t}+ \frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt} \biggr) \\& \quad = -\lim_{n\to\infty}\int^{x}_{0}b_{n}^{v} \biggl(\frac {(v-1)a_{n}}{b_{n}+a_{n}t}+\frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt \\& \quad = -\bigl(1-v^{-1}\bigr)\lambda^{v} \bigl(x^{2}+2x\bigr) \\& \quad = k_{v1}(x). \end{aligned}$$

Hence,

$$\begin{aligned}& \lim_{n\to\infty}b^{v}_{n} \bigl(b^{v}_{n} H_{v}(b_{n};x)- k_{v1}(x) \bigr) \\& \quad = \lim_{n\to\infty}\biggl[b^{2v}_{n} \bigl(B_{n}(x)-1\bigr)-B_{n}(x)b^{v}_{n} \\& \qquad {}\times \biggl(\int^{x}_{0} b_{n}^{v} \biggl(\frac{(v-1)a_{n}}{b_{n}+a_{n}t}+\frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt+k_{v1}(x) \biggr) \\& \qquad {}+\frac{1}{2}\bigl(1+o(1)\bigr)B_{n}(x)b^{2v}_{n} \biggl(\int^{x}_{0} \biggl(\frac {(v-1)a_{n}}{b_{n}+a_{n}t}+ \frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt \biggr)^{2}\biggr] \\& \quad = -4\bigl(v^{-1}-1\bigr)\lambda^{2v}x+ \frac{2\lambda^{2v}(1-v^{-1})}{v}x^{2} -\frac{2}{3}\bigl(1-v^{-1}\bigr) \bigl(1-2v^{-1}\bigr)\lambda^{2v}x^{3} \\& \qquad {}+ \frac {(1-v^{-1})^{2}}{2}\lambda^{2v}\bigl(x^{2}+2x \bigr)^{2} \\& \quad = \omega^{\circ}_{v}(x). \end{aligned}$$

The proof is complete. □

Lemma 3.5

Let $C_{n}(x)$ be given by (3.11) and $D_{n}(x)$ be denoted by

$$D_{n}(x)= 1+\tilde{k}_{v}(x)b^{-v}_{n}+ \biggl(\tilde{\omega}_{v}(x)+\frac {\tilde{k}^{2}_{v}(x)}{2} \biggr) b^{-2v}_{n}\bigl(1+o(1)\bigr). $$

For $v\neq1$ and $-d\log b_{n}< x< cb_{n}^{\frac{3}{v}}$ with $0< c, d<1$, we have

$$\begin{aligned}& \bigl\vert C_{n}(x)D_{n}(x)\bigr\vert < 2, \\& \bigl\vert b_{n}^{v}\bigl(C_{n}(x)D_{n}(x)-1 \bigr)\bigr\vert \leq 1+\bigl(1+v^{-1}\bigr)\lambda^{v} \bigl(2+2|x|+\bigl(x^{2}+2\vert x\vert \bigr)e^{-x} \bigr), \\& \bigl\vert b_{n}^{v}\bigl[b_{n}^{v} \bigl(C_{n}(x)D_{n}(x)-1\bigr)-k_{v2}(x)\bigr] \bigr\vert \\& \quad \leq 1+\bigl(1+v^{-1}\bigr)\lambda^{2v} \biggl[4+ \frac{4}{v}|x|+\bigl(2+4v^{-1}\bigr)x^{2} + \biggl( \frac{4}{v}|x|+\bigl(4+6v^{-1}\bigr)x^{2} \\& \qquad {}+\frac{2}{3}\bigl(1+2v^{-1}\bigr)|x|^{3}+ \frac{1+v^{-1}}{2}x^{4} \biggr)e^{-x}+\frac{1}{2} \bigl(1+v^{-1}\bigr) \bigl(x^{2}+2\vert x\vert \bigr)^{2}e^{-2x} \biggr] \end{aligned}$$

for large n, where $k_{v1}(x)$ and $k_{v2}(x)$ are given by (2.2).

Proof

The desired results follow from Lemmas 3.2 and 3.3. □

The following Mills’ inequalities are from the $\operatorname{GED}(v)$ in Jia et al. [9], which will be used later.

Lemma 3.6

Let $F_{v}(x)$ and $f_{v}(x)$ denote the cdf and pdf of $\operatorname{GED}(v)$, respectively. Then

• (i) for $v>1$ and all $x>0$, we have

  $$ \frac{2\lambda^{v}}{v}x^{1-v} \biggl( 1+ \frac{2(v-1)\lambda ^{v}}{v}x^{-v} \biggr)^{-1} < \frac{1-F_{v}(x)}{f_{v}(x)} < \frac{2\lambda^{v}}{v}x^{1-v}; $$

  (3.16)

• (ii) for $0< v<1$ and all $x>\lambda[2(1/v-1)]^{1/v}$, we have

  $$ \frac{2\lambda^{v}}{v}x^{1-v} < \frac{1-F_{v}(x)}{f_{v}(x)} < \frac{2\lambda^{v}}{v}x^{1-v} \biggl( 1+\frac{2(v-1)\lambda ^{v}}{v}x^{-v} \biggr)^{-1}. $$

  (3.17)

Lemma 3.7

Let the norming constant $b_{n}$ be given by (1.2), for any constant $0< c<1$ and arbitrary nonnegative integers i, j and k, we have

$$\begin{aligned}& \lim_{n\to\infty}b_{n}^{i}\int _{cb_{n}^{\frac{v}{3}}}^{\infty} |x|^{j}e^{-kx} \Lambda(x)\,dx = 0, \end{aligned}$$

(3.18)

$$\begin{aligned}& \lim_{n\to\infty}b_{n}^{i}\int _{cb_{n}^{\frac{v}{3}}}^{\infty} |x|^{j}g_{n}(x)\,dx = 0 \end{aligned}$$

(3.19)

if $v\neq1$.

Proof

By arguments similar to Lemma 3.3 in Jia et al. [9], we can get (3.18). The rest is to prove (3.19). By (3.16) and (3.17), and Lemma 3.5, we have

$$\begin{aligned}& b_{n}^{i}\int_{cb_{n}^{\frac{v}{3}}}^{\infty} |x|^{j}g_{n}(x)\,dx \\& \quad = b_{n}^{i}\int_{cb_{n}^{\frac{v}{3}}}^{\infty} |x|^{j}na_{n}F^{n-1}_{v}(a_{n}x+b_{n})f_{v}(a_{n}x+b_{n}) \,dx \\& \quad = b_{n}^{i}\int_{cb_{n}^{\frac{v}{3}}}^{\infty} |x|^{j}\frac {1-F_{v}(a_{n}x+b_{n})}{1-F_{v}(b_{n})}e^{x} C_{n}(x)D_{n}(x) \Lambda'(x)\,dx \\& \quad \leq 4b_{n}^{i}\int_{cb_{n}^{\frac{v}{3}}}^{\infty} |x|^{j} e^{-x}\Lambda(x)\,dx \\& \quad \to 0 \end{aligned}$$

as $n\to\infty$. The proof is complete. □

Lemma 3.8

Assume that the shape parameter $v\neq1$, then for any constant $0< d<1$ and arbitrary nonnegative integers i, j and k, we have

$$\begin{aligned}& \lim_{n\to\infty}b_{n}^{i}\int _{-\infty}^{-d\log b_{n}}|x|^{j}e^{-kx} \Lambda(x)\,dx = 0, \end{aligned}$$

(3.20)

$$\begin{aligned}& \lim_{n\to\infty}b_{n}^{i}\int _{-\infty}^{-d\log b_{n}}|x|^{j}g_{n}(x)\,dx = 0. \end{aligned}$$

(3.21)

Proof

By arguments similar to that of Lemma 3.7, we have

$$\begin{aligned}& b_{n}^{i}\int_{-\infty}^{-d\log b_{n}} |x|^{j}e^{-kx}\Lambda(x)\,dx \\& \quad \leq b_{n}^{i} \exp\biggl(-\frac{b_{n}^{d}}{2}\biggr) \int_{1}^{\infty} x^{j}e^{kx}\exp \biggl(-\frac {e^{x}}{2}\biggr)\,dx \\& \quad \to 0 \end{aligned}$$

as $n\to\infty$ since $\int_{1}^{\infty} x^{j}e^{kx}\exp(-\frac{e^{x}}{2})\,dx<\infty$.

For assertion (3.21), we only consider the case of $v>1$ since the proof of the case of $0< v<1$ is similar. Rewrite

$$\begin{aligned} b_{n}^{i}\int_{-\infty}^{-d\log b_{n}}|x|^{j}g_{n}(x) \,dx =& b_{n}^{i}\int_{-\infty}^{-\frac{3v\lambda^{-v}b_{n}^{v}}{2}}|x|^{j}g_{n}(x) \,dx+ b_{n}^{i}\int_{-\frac{3v\lambda^{-v}b_{n}^{v}}{2}}^{-\frac{v\lambda ^{-v}b_{n}^{\frac{v-1}{2}}}{2}}|x|^{j}g_{n}(x) \,dx \\ &{} +b_{n}^{i}\int_{-\frac{v\lambda^{-v}b_{n}^{\frac{v-1}{2}}}{2}}^{-d\log b_{n}}|x|^{j}g_{n}(x) \,dx \\ =& I_{n}+\mathit{II}_{n}+\mathit{III}_{n}. \end{aligned}$$

First note that $\int_{\mathbb{R}}|x|^{j}f_{v}(x)\,dx<\infty$ and the symmetry of $f_{v}$ implies $F_{v}(-x)+F_{v}(x)=1$. By using (1.2) and (3.16) we have

$$\begin{aligned} I_{n} < & 2^{j}v\lambda^{-v}b_{n}^{i+j+v} \bigl(1-F_{v}(2b_{n}) \bigr)^{n-2} \\ < & 2^{j}v\lambda^{-v}b_{n}^{i+j+v}\exp \biggl((n-2) \biggl(c-(v-1)\log b_{n}-\frac{(2b_{n})^{v}}{2\lambda^{v}} \biggr) \biggr) \\ \to& 0 \end{aligned}$$

(3.22)

as $n\to\infty$, where $c=(1-v-1/v)\log2+(v-1)\log\lambda-\log \Gamma(1/v)$.

To show $\mathit{II}_{n}\to0$ and $\mathit{III}_{n}\to0$, we consider the case of $v>2$ first. By using the following inequalities

$$ 1-vx < (1-x)^{v} < 1-vx + \frac{v(v-1)}{2}x^{2}, \quad 0< x< \frac{1}{2}, v>2, $$

(3.23)

we can get

$$na_{n}f_{v}\bigl(b_{n}-b_{n}^{\frac{1-v}{2}} \bigr) < 2\exp\biggl(\frac{v}{2\lambda^{v}}b_{n}^{\frac{v-1}{2}}\biggr) $$

and

$$\begin{aligned}& \frac{1-F_{v}(b_{n}-b_{n}^{\frac{1-v}{2}})}{1-F_{v}(b_{n})} \\& \quad > \frac{\frac{2\lambda^{v}}{v}(b_{n}-b_{n}^{\frac{1-v}{2}})^{1-v}(1+\frac {2(v-1)\lambda^{v}}{v}(b_{n}-b_{n}^{\frac{1-v}{2}})^{-v})^{-1} f_{v}(b_{n}-b_{n}^{\frac{1-v}{2}})}{\frac{2\lambda ^{v}}{v}b_{n}^{1-v}f_{v}(b_{n})} \\& \quad > \frac{(1-b_{n}^{-\frac{1+v}{2}})^{1-v}}{1+\frac{2(v-1)\lambda ^{v}}{v}(b_{n}-b_{n}^{\frac{1-v}{2}})^{-v}} \exp \biggl( \frac{v}{2\lambda^{v}}b_{n}^{\frac{v-1}{2}}- \frac {v(v-1)}{4\lambda^{v}} b_{n}^{-1} \biggr) \\& \quad > \frac{1}{2}\exp \biggl(\frac{v}{2\lambda^{v}}b_{n}^{\frac {v-1}{2}} \biggr) \end{aligned}$$

for large n, which implies that

$$\begin{aligned} \mathit{II}_{n} < & b_{n}^{i} \biggl(\frac{3v\lambda^{-v}b_{n}^{v}}{2}\biggr)^{j}na_{n}f_{v} \biggl(-a_{n}\frac{v\lambda ^{-v}b_{n}^{\frac{v-1}{2}}}{2}+b_{n}\biggr) F_{v}^{n-1}\biggl(-a_{n}\frac{v\lambda^{-v}b_{n}^{\frac{v-1}{2}}}{2}+b_{n} \biggr)\frac {3v\lambda^{-v}b_{n}^{v}}{2} \\ < & b_{n}^{i}\biggl(\frac{3v\lambda^{-v}b_{n}^{v}}{2}\biggr)^{j+1}na_{n}f_{v} \biggl(-a_{n}\frac{v\lambda ^{-v}b_{n}^{\frac{v-1}{2}}}{2}+b_{n}\biggr) \exp \bigl( -(n-1) \bigl(1-F_{v} \bigl( b_{n}-b_{n}^{\frac{1-v}{2}} \bigr) \bigr) \bigr) \\ =& 2b_{n}^{i}\biggl(\frac{3v\lambda^{-v}b_{n}^{v}}{2} \biggr)^{j+1}\frac{\lambda2^{1+\frac {1}{v}}\Gamma(\frac{1}{v})}{v} \exp \biggl(\frac{b_{n}^{v}+vb_{n}^{\frac{v-1}{2}}}{2\lambda^{v}}- \frac {1}{2}\exp\biggl(\frac{v}{2\lambda^{v}}b_{n}^{\frac{v-1}{2}} \biggr) \biggr) \\ \to& 0 \end{aligned}$$

(3.24)

as $n\to\infty$.

Similarly, for $-\frac{v\lambda^{-v}b_{n}^{\frac {v-1}{2}}}{2}< x<-d\log b_{n}$, we have

$$\begin{aligned}& na_{n}f_{v}(a_{n}x+b_{n}) \\& \quad < \biggl(1+\frac{2(v-1)\lambda^{v}}{v}b_{n}^{-v}\biggr) \frac{f_{v}(a_{n}x+b_{n})}{f_{v}(b_{n})} \\& \quad < \biggl(1+\frac{2(v-1)\lambda^{v}}{v}b_{n}^{-v}\biggr)\exp \biggl(-\frac {b_{n}^{v}}{2\lambda^{v}}2\lambda^{v}b_{n}^{-v}x \biggr) \\& \quad < 2e^{-x} \end{aligned}$$

and

$$\begin{aligned}& \frac{1-F_{v}(a_{n}x+b_{n})}{1-F_{v}(b_{n})} \\& \quad > \frac{\frac{2\lambda^{v}}{v}(a_{n}x+b_{n})^{1-v}(1+\frac{2(v-1)\lambda ^{v}}{v}(a_{n}x+b_{n})^{-v})^{-1} f_{v}(a_{n}x+b_{n})}{\frac{2\lambda^{v}}{v}b_{n}^{1-v}f_{v}(b_{n})} \\& \quad > \frac{(1+\frac{2\lambda^{v}}{v}b_{n}^{-v}x)^{1-v}}{1+\frac {2(v-1)\lambda^{v}}{v}(a_{n}x+b_{n})^{-v}} \exp\biggl(-x-\frac{v-1}{v}\lambda^{v}b_{n}^{-v}x^{2} \biggr) \\& \quad > \frac{1}{2}e^{-x} \end{aligned}$$

for large n by using (3.23). Then

$$\begin{aligned} \mathit{III}_{n} < & 2b_{n}^{i} \int_{-\frac{v\lambda^{-v}b_{n}^{\frac{v-1}{2}}}{2}}^{-d\log b_{n}}|x|^{j}e^{-x}\exp \biggl(-\frac{1}{2}e^{-x}\biggr)\,dx \\ < & 2b_{n}^{i}\exp\biggl(-\frac{e^{d\log b_{n}}}{4}\biggr)\int _{-\frac{v\lambda ^{-v}b_{n}^{\frac{v-1}{2}}}{2}}^{-d\log b_{n}}|x|^{j}e^{-x}\exp \biggl(-\frac {1}{4}e^{-x}\biggr)\,dx \\ \to& 0 \end{aligned}$$

(3.25)

as $n\to\infty$.

Combining with (3.22)-(3.25), the assertion (3.21) is derived for $v>2$. Similar proofs for the case of $1< v\leq2$ and details are omitted here. The proof is complete. □

Lemma 3.9

Let $\alpha=\min(1,v)$ as $v\neq1$. For large n and $-d\log b_{n}< x< cb_{n}^{\frac{v}{3}}$, both $x^{r}b_{n}^{v}\Delta_{n}(g_{n}, \Lambda^{\prime};x)$ and $x^{r}b_{n}^{v}[b_{n}^{v}\Delta_{n}(g_{n},\Lambda^{\prime};x)-k_{v}(x)\Lambda'(x)]$ are bounded by integrable functions independent of n, with $r>0$, $0< c<1$ and $0< d<\alpha$, where $a_{n}$ and $b_{n}$ are given by (1.2), and $k_{v}(x)$ is given by (2.2).

Proof

We only consider the case of $v>1$. For the case of $0< v<1$, the proofs are similar and details are omitted here. Rewrite

$$b_{n}^{v} \Delta_{n}\bigl(g_{n}, \Lambda^{\prime};x\bigr)= b_{n}^{v} {H_{v}(b_{n};x)}e^{-x}\Lambda(x)+b_{n}^{v} \bigl(C_{n}(x)D_{n}(x)-1\bigr)e^{-x}\Lambda(x), $$

where $C_{n}(x)$ is given by (3.11), $H_{v}(b_{n};x)$ and $D_{n}(x)$ are respectively defined in Lemma 3.4 and Lemma 3.5. Note that $\int_{-\infty}^{\infty}x^{k}e^{-tx}\exp(-e^{-x})\,dx=(-1)^{k}\Gamma^{(k)}(t)$ is finite for $t>0$ and nonnegative integers k. Lemma 3.5 shows that $b_{n}^{v}(C_{n}(x)D_{n}(x)-1)e^{-x}\Lambda(x)$ is bounded by integrable function independent of n. The rest is to prove that $b_{n}^{v} {H_{v}(b_{n};x)}$ is bounded by $m(x)$, where $m(x)$ is a polynomial on x. Rewrite

$$ b_{n}^{v} {H_{v}(b_{n};x)} = I_{n}(x)+J_{n}(x), $$

(3.26)

where

$$\begin{aligned}& I_{n}(x) = b_{n}^{v}\bigl(B_{n}(x)-1 \bigr), \\& J_{n}(x) = b_{n}^{v}B_{n}(x)\int ^{x}_{0} \biggl(\frac{(v-1)a_{n}}{b_{n}+a_{n}t}+ \frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt. \end{aligned}$$

For $-d\log b_{n}< x< cb_{n}^{\frac{v}{3}}$, from Lemma 3.4 it follows that

$$ \bigl\vert I_{n}(x)\bigr\vert < 1 $$

(3.27)

and

$$ \bigl\vert J_{n}(x)\bigr\vert < 1+ \biggl[ \frac{v^{-1}}{2^{-1}v\lambda^{-v}-d}|x|+\bigl(1-v^{-1}\bigr)\lambda ^{v}x^{2}+ \frac{4}{3}\bigl(1-v^{-1}\bigr)\bigl\vert 1-2v^{-1} \bigr\vert \lambda^{2v}|x|^{3} \biggr]. $$

(3.28)

Hence, the desired result (3.26) follows by combining (3.16), (3.27) and (3.28) together.

Rewrite

$$\begin{aligned}& b^{v}_{n} \bigl[b^{v}_{n} \Delta_{n}\bigl(g_{n},\Lambda^{\prime};x \bigr)-k_{v}(x)\Lambda '(x) \bigr] \\& \quad = b^{v}_{n} \biggl[b^{v}_{n} \biggl(\frac{1-F_{v}(a_{n}x+b_{n})}{1-F_{v}(b_{n})}e^{x} C_{n}(x)D_{n}(x)-1 \biggr)-k_{v}(x) \biggr]\Lambda'(x) \\& \quad = b^{v}_{n} \biggl[b^{v}_{n} C_{n}(x)D_{n}(x) \biggl(\frac {1-F_{v}(a_{n}x+b_{n})}{1-F_{v}(b_{n})}e^{x}-1 \biggr) \\& \qquad {}+b^{v}_{n}\bigl(C_{n}(x)D_{n}(x)-1 \bigr)-\bigl(k_{v1}(x)+k_{v2}(x)\bigr) \biggr] \Lambda'(x) \\& \quad = b^{v}_{n} \bigl[b^{v}_{n} C_{n}(x)D_{n}(x) \bigl(H_{v}(b_{n};x)-b^{-v}_{n}k_{v1}(x) \bigr) +b^{v}_{n} \bigl(C_{n}(x)D_{n}(x)-1-b^{-v}_{n}k_{v2}(x) \bigr) \\& \qquad {} + \bigl(C_{n}(x)D_{n}(x)-1 \bigr)k_{v1}(x) \bigr]\Lambda'(x). \end{aligned}$$

By Lemma 3.5, we only need to estimate the bound of $b^{v}_{n} [b^{v}_{n} C_{n}(x)D_{n}(x) (H_{v}(b_{n};x)-b^{-v}_{n}k_{v1}(x) ) ]$. Rewrite

$$ b^{v}_{n} \bigl[b^{v}_{n} C_{n}(x)D_{n}(x) \bigl(H_{v}(b_{n};x)-b^{-v}_{n}k_{v1}(x) \bigr) \bigr] = H_{n}(x)-K_{n}(x)+L_{n}(x), $$

where

$$\begin{aligned}& H_{n}(x) = b_{n}^{2v} \bigl(B_{n}(x)-1 \bigr), \\& K_{n}(x) = b^{2v}_{n} \biggl({B_{n}(x) \int^{x}_{0} \biggl(\frac {(v-1)a_{n}}{b_{n}+a_{n}t}+ \frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt+b_{n}^{-v}k_{v1}(x)} \biggr), \\& L_{n}(x) = \frac{1}{2}\bigl(1+o(1)\bigr)B_{n}(x)b^{2v}_{n} \biggl(\int^{x}_{0} \biggl(\frac {(v-1)a_{n}}{b_{n}+a_{n}t}+ \frac{v a_{n}(b_{n}+a_{n}t)^{v-1}}{2\lambda^{v}}-1 \biggr)\,dt \biggr)^{2}. \end{aligned}$$

For $0< x< cb_{n}^{\frac{v}{3}}$, by using $1-vy<(1+y)^{-v}<1$ for $v>0$ and $y>0$, we have

$$ \bigl\vert H_{n}(x)\bigr\vert < 1+8\lambda^{2v} \bigl(1+8\lambda^{v}\bigr)x $$

due to Lemma 3.4. If $-d\log b_{n} < x<0$, by using $1+vy<(1+y)^{v}<1$ for $v>1$ and $-1< y<0$, Lemma 3.4 shows that

$$ \bigl\vert H_{n}(x)\bigr\vert < 1+16\lambda^{2v} \bigl(1+8\lambda^{v}\bigr)|x| $$

for large n. Similarly,

$$\begin{aligned}& \bigl\vert K_{n}(x)\bigr\vert < 2 \bigl(1-v^{-1}\bigr)\biggl\vert \frac{v}{2\lambda^{v}}-d\biggr\vert ^{-1}\lambda ^{v}x^{2}+\frac{4}{3} \bigl(1-v^{-1}\bigr)\bigl\vert 1-2v^{-1}\bigr\vert \lambda^{2v}|x|^{3}, \end{aligned}$$

(3.29)

$$\begin{aligned}& \bigl\vert L_{n}(x)\bigr\vert < 1+ \frac{1}{2} \biggl[\frac{v^{-1}}{2^{-1}v\lambda ^{-v}-d}|x|+\bigl(1-v^{-1}\bigr) \lambda^{v}x^{2}+\frac{4}{3}\bigl(1-v^{-1} \bigr) \bigl\vert 1-2v^{-1}\bigr\vert \lambda^{2v}|x|^{3} \biggr]^{2} \end{aligned}$$

(3.30)

as $-d\log b_{n}< x< cb_{n}^{\frac{v}{3}}$. Hence, we derive the desired result by combining (3.16), (3.29) and (3.30) together.

The proof is complete. □

For $v=1$, note that the $\operatorname{GED}(1)$ is the Laplace distribution with pdf given by

$$ f_{1}(x)=2^{-\frac{1}{2}}\exp\bigl(-2^{\frac{1}{2}}\vert x\vert \bigr), \quad x\in\mathbb{R}, $$

and its distributional tail can be written as

$$ 1-F_{1}(x)=2^{-\frac{1}{2}}f_{1}(x)=2^{-1}\exp \bigl(-2^{\frac{1}{2}}\bigr)\exp \biggl(-\int_{1}^{x} \frac{1}{f(t)}\,dt \biggr),\quad x>0 $$

with $f(t)=2^{-\frac{1}{2}}$. For the Laplace distribution, similar to the case of $v>1$, we have the following two results.

Lemma 3.10

For $0< d<1$ and an arbitrary nonnegative real number j, we have

$$\begin{aligned}& \lim_{n\to\infty}n^{2}\int _{-\infty}^{-db_{n}^{\frac {1}{2}}}|x|^{j}e^{-kx} \Lambda(x)\,dx = 0 ,\quad k=1,2,\ldots, \end{aligned}$$

(3.31)

$$\begin{aligned}& \lim_{n\to\infty}n^{2}\int _{-\infty}^{-db_{n}^{\frac {1}{2}}}|x|^{j}F_{1}^{n}(a_{n}x+b_{n}) \,dx = 0. \end{aligned}$$

(3.32)

Lemma 3.11

For $x>-db_{n}^{\frac{1}{2}}$, both $x^{r}n ((F_{1}^{n}(a_{n}x+b_{n}))'-\Lambda'(x) )$ and $x^{r}n[n ((F_{1}^{n}(a_{n}x+b_{n}))'-\Lambda'(x) )+\frac {1}{2}e^{-2x}\Lambda'(x)]$ are bounded by integrable functions independent of n, where $r>0$ and $0< d<1$.

4 Proofs of the main results

Proof of Theorem 2.1

From Lemma 3.3 and Lemma 3.5 it follows that

$$\begin{aligned}& C_{n}(x)D_{n}(x) \\& \quad = 1+{k_{v2}}(x)b_{n}^{-v}+ \bigl(1-v^{-1}\bigr)\lambda^{2v} \biggl[-4- \frac {4}{v}x+2\bigl(1-2v^{-1}\bigr)x^{2} + \biggl(- \frac{4}{v}x+2\bigl(2-3v^{-1}\bigr)x^{2} \\& \qquad {} +\frac{2(1-2v^{-1})}{3}x^{3}-\frac{1-v^{-1}}{2}x^{4} \biggr)e^{-x} +\frac{(1-v^{-1})}{2}\bigl(x^{2}+2x \bigr)^{2}e^{-2x} \biggr]b_{n}^{-2v} +O \bigl(b_{n}^{-3v}\bigr), \end{aligned}$$

(4.1)

where $k_{v2}(x)$ is given by (2.2). Note that (3.12) shows

$$ \frac{1-F_{v}(a_{n}x+b_{n})}{1-F_{v}(b_{n})}e^{x} = 1+k_{v1}(x)b_{n}^{-v}+ \omega^{\circ}_{v}(x)b_{n}^{-2v}+O \bigl(b_{n}^{-3v}\bigr). $$

(4.2)

Hence,

$$\begin{aligned}& \Delta_{n}\bigl(g_{n},\Lambda^{\prime};x\bigr) \\& \quad = \biggl(\frac{1-F_{v}(a_{n}x+b_{n})}{1-F_{v}(b_{n})}e^{x} C_{n}(x)D_{n}(x)-1 \biggr)\Lambda'(x) \\& \quad = \biggl[\bigl(1-v^{-1}\bigr)\lambda^{v} \bigl(2-x^{2}+\bigl(x^{2}+2x\bigr)e^{-x} \bigr)b_{n}^{-v}+\bigl(1-v^{-1}\bigr) \lambda^{2v} \biggl(-4-2\bigl(1-v^{-1}\bigr)x^{2} \\& \qquad {}-\frac{2(1-2v^{-1})}{3}x^{3}+\frac{1-v^{-1}}{2}x^{4}+ \biggl(- \frac {4}{v}x-\frac{2}{v}x^{2}-\frac{2(5-4v^{-1})}{3}x^{3}- \frac {3(1-v^{-1})}{2}x^{4} \biggr)e^{-x} \\& \qquad {}+\frac{1-v^{-1}}{2}\bigl(x^{2}+2x\bigr)^{2}e^{-2x} \biggr)b_{n}^{-2v}+O\bigl(b_{n}^{-3v} \bigr) \biggr]\Lambda'(x) \\& \quad = \bigl[k_{v}(x)b_{n}^{-v}+ \bigl(1-v^{-1}\bigr)\lambda^{2v} \bigl(\omega _{v1}(x)+\omega_{v2}(x)e^{-x}+\omega_{v3}(x)e^{-2x} \bigr)b_{n}^{-2v}+O\bigl(b_{n}^{-3v} \bigr) \bigr]\Lambda'(x) \\& \quad = \bigl(k_{v}(x)b_{n}^{-v}+ \omega_{v}(x)b_{n}^{-2v}+O\bigl(b_{n}^{-3v} \bigr) \bigr)\Lambda'(x) \end{aligned}$$

implying that

$$\lim_{n\to\infty}b^{v}_{n} \bigl[b^{v}_{n} \Delta_{n}\bigl(g_{n},\Lambda^{\prime};x \bigr)-k_{v}(x)\Lambda'(x) \bigr] = \omega_{v}(x) \Lambda'(x). $$

The proof is complete. □

Proof of Theorem 2.3

For $v\neq1$, by Lemmas 3.5-3.9 and the dominated convergence theorem, we have

$$\begin{aligned}& b_{n}^{v}\bigl(m_{r}(n)-m_{r}\bigr) \\& \quad = b_{n}^{v}\int_{-\infty}^{-d\log b_{n}}x^{r} \Delta_{n}\bigl(g_{n},\Lambda^{\prime};x\bigr)\,dx+ b_{n}^{v}\int_{-d\log b_{n}}^{cb_{n}^{\frac{v}{3}}}x^{r} \Delta_{n}\bigl(g_{n},\Lambda^{\prime};x\bigr)\,dx \\& \qquad {}+ b_{n}^{v}\int_{cb_{n}^{\frac{v}{3}}}^{\infty}x^{r} \Delta_{n}\bigl(g_{n},\Lambda ^{\prime};x\bigr)\,dx \\& \quad \to \int_{-\infty}^{\infty}x^{r}k_{v}(x) \Lambda'(x)\,dx \\& \quad = -\bigl(1-v^{-1}\bigr)\lambda^{v}r(2m_{r}+m_{r+1}) \end{aligned}$$

and

$$\begin{aligned}& b_{n}^{v} \bigl[b_{n}^{v} \bigl(m_{r}(n)-m_{r}\bigr)+\bigl(1-v^{-1}\bigr) \lambda^{v}r(2m_{r}+m_{r+1}) \bigr] \\& \quad = \int_{-\infty}^{\infty}x^{r}b_{n}^{v} \bigl[b_{n}^{v} \Delta_{n}\bigl(g_{n}, \Lambda ^{\prime};x\bigr)-k_{v}(x)\Lambda'(x)\bigr] \,dx \\& \quad = \int_{-\infty}^{-d\log b_{n}}x^{r}b_{n}^{2v} \Delta_{n}\bigl(g_{n},\Lambda^{\prime};x\bigr)\,dx- \int_{-\infty }^{-d\log b_{n}}x^{r}b_{n}^{v}k_{v}(x) \Lambda'(x)\,dx \\& \qquad {}-\int_{cb_{n}^{\frac{v}{3}}}^{\infty }x^{r}b_{n}^{v}k_{v}(x) \Lambda'(x)\,dx+\int_{-d\log b_{n}}^{cb_{n}^{\frac{v}{3}}}x^{r}b_{n}^{v} \bigl[b_{n}^{v} \Delta_{n}\bigl(g_{n}, \Lambda^{\prime};x\bigr) -k_{v}(x)\Lambda'(x)\bigr] \,dx \\& \qquad {} + \int_{cb_{n}^{\frac{v}{3}}}^{\infty}x^{r}b_{n}^{2v} \Delta _{n}\bigl(g_{n},\Lambda^{\prime};x\bigr)\,dx \\& \quad \to \int_{-\infty}^{\infty}x^{r} \omega_{v}(x)\Lambda'(x)\,dx \\& \quad = 2r\lambda^{2v}\bigl(1-v^{-1}\bigr) \biggl[\bigl( \bigl(1-v^{-1}\bigr) (r+1)+2\bigr)m_{r}+\bigl( \bigl(1-v^{-1}\bigr) (r+1)+1\bigr)m_{r+1} \\& \qquad {}+ \biggl(\frac{1}{4}\bigl(1-v^{-1}\bigr) (r-1)+ \frac{1}{3}\bigl(2-v^{-1}\bigr) \biggr)m_{r+2} \biggr] \end{aligned}$$

as $n\to\infty$.

For the case of $v=1$, note that

$$ \int_{-\infty}^{\infty}x^{k}e^{-mx} \Lambda(x)\,dx=(-1)^{k}\Gamma^{(k)}(m). $$

Combining with Lemmas 3.10 and 3.11 and the dominated convergence theorem, we can derive the desired results.

The proof is complete. □