空间计量经济学是在区域经济模型中处理由于空间因素导致的特殊性质的一系列方法^[1].空间计量经济学可应用于区域科学、地理经济学、城市经济学等领域, 具体来说, 如研究区域经济、房屋价值、人均收入等问题.空间面板数据模型是空间计量经济学中经典模型之一, 它同时考虑了空间相关性和时间依赖性, 将传统的时间序列方法、横截面数据方法以及普通面板数据模型进行综合, 学者们对此模型也进行了诸多研究.如Elhorst^[2]研究了空间面板数据模型的分类和估计问题; Kapoor等^[3]研究了误差项是空间自相关的面板数据模型的参数估计; Yu等^[4]研究了带固定效应空间动态面板数据模型的拟极大似然估计(QMLE); Lee和Yu^[5]进一步研究了带固定效应的空间自回归面板数据模型的QMLE, 并提出正交转换的间接估计方法; 文利霞^[6]研究了空间误差面板数据模型的拟极大似然估计及其渐近性质; 戴晓文等^[7]基于工具变量法研究了含有个体固定效应的空间误差面板数据模型参数的分位回归估计, 并与均值回归方法做比较, 结果表明在处理空间面板数据时工具变量回归估计方法优于均值回归方法; Qin^[8]研究了空间误差模型的经验似然估计.由于人为或者客观原因, 我们得到的数据不一定是完整的; Wang和Lee^[9]研究了数据随机缺失情形下空间面板数据模型的广义矩估计、非线性最小二乘估计和两阶段最小二乘估计估计, 并对这三种估计方法得到的结果进行了比较; 于力超和金勇进^[10]研究了数据缺失机制为非随机缺失情形的面板数据参数估计方法.

对空间计量模型进行估计, 使用比较多的估计方法是(拟)极大似然估计法(如文献[4, 5, 11])、广义矩估计法(如文献[9, 13])、两阶段最小二乘法(如文献[9, 13])等.空间计量模型的理论和应用研究均比较深入且成熟.经验似然是Owen^[14]于1988年在完全样本下提出的一种非参数统计推断方法, 其是在一定约束条件下将参数似然比极大化, 有类似于Bootstrap的抽样特性.经验似然方法应用广泛, 可应用于各种统计模型, 学者们对经验似然也进行了诸多研究. Owen^{[15, 16]}将经验似然应用到线性回归模型的统计推断; 石坚^[17]运用经验似然方法修正了线性相关模型中误差方差的传统最小二乘估计, 修正后的估计的渐近方差比传统估计的更小; Cui和Chen^[18]将经验似然应用到线性变量含误差模型; Kostov^[19]研究了空间分位数回归模型的经验似然推断; 陈燕红^[20]研究了时间序列模型的经验似然推断.据我们所知, 目前尚没有研究面板数据空间计量经济模型的文献报导.于是, 本文探讨空间面板数据模型的经验似然推断, 在正则条件下构造了空间面板数据模型的经验似然统计量, 证明了该统计量的极限分布为卡方分布.

一般的空间面板数据模型^[5]

$ \begin{equation} \left\{ \begin{aligned} &Y_{nt}= \rho M_{n} Y_{nt} + X_{nt} \beta + U_{nt} + V_{nt}, \\ & V_{nt}=\lambda W_{n} V_{nt} + \varepsilon_{nt}, ~t=1, \cdots, T, \end{aligned} \right. \end{equation} $

(2.1)

其中$ Y_{nt}=( y_{1t}, \cdots, y_{nt} )'$为$ n \times 1$维被解释变量; $ X_{nt}$为$ n \times k$维非随机解释变量向量; $ \beta$为$ k \times 1$维系数向量.$ M_{n}$和$ W_{n}$均为$ n \times n$维空间权重矩阵, 其中$ m_{ij}$为矩阵$ M_{n} $的$(i, j)$元素, $ m_{ii}=0$; $ w_{ij}$为矩阵$ W_{n} $的$(i, j)$元素, $ w_{ii}=0$.$ U_{nt}$为$ n \times 1$维固定效应向量.$ V_{nt}$为自相关误差项.$ \varepsilon_{nt}=( \varepsilon_{1t}, \varepsilon_{2t}, \cdots, \varepsilon_{nt} )'$为$n \times 1$维随机扰动项, 其分量的均值为$0$, 协方差为$ \sigma^{2} $, 且独立同分布.$ \rho$和$ \lambda$为空间相关系数.在本文中, “$'$”表示矩阵转置.

由于$ t=1, \cdots, T$, 模型(2.1)可改写为^[1]

$ \begin{equation} \left\{ \begin{aligned} & Y=\rho (I_{T}\otimes M_{n})y + X\beta + U + V, \\ & V=\lambda(I_{T}\otimes W_{n})V + \varepsilon, \end{aligned} \right. \end{equation} $

(2.2)

其中$Y= (Y_{n1}' , \cdots, Y_{nT}')' $为$nT \times 1$维因变量向量, $Y_{nt}=(y_{1t}, \cdots, y_{nt})'$为$n \times 1$维向量.$ X= (X_{n1}', \cdots, X_{nT}')'$为$ nT \times k$维解释变量向量, $ X_{nt}= (X^{*}_{1t}, X^{*}_{2t}, \cdots, X^{*}_{nt})'$为$n \times k$维向量.$ \beta$为$ k \times 1$维系数向量.$ U$为$ nT \times 1$维固定效应向量.$ I_{T}$表示$ T \times T$维单位矩阵.$ M_{n}$和$ W_{n}$均表示$ n \times n$维空间权重矩阵, 其中$ m_{ij}$和$ w_{ij}$分别表示$ M_{n} $和$ W_{n} $的$(i, j)$元素.$V$为自相关误差项.$\varepsilon= (\varepsilon_{n1}', \cdots, \varepsilon_{nT}')'$为$nT \times 1$维随机扰动项, 其均值为$ 0 $, 协方差阵为$ \sigma^{2} I_{nT}$, $ \varepsilon_{nt}= (\varepsilon_{1t}, \varepsilon_{2t}, \cdots, \varepsilon_{nt})'$为$n \times 1$维向量.$\rho $和$\lambda $为空间相关系数. $\otimes $表示矩阵的Kronecker积.空间面板数据模型(2.2)分为空间滞后面板数据模型和空间误差面板数据模型两个大类.

(1) 当$\lambda=0 $时, 模型(2.2)即为空间滞后面板数据模型

$ \begin{equation} Y = \rho(I_{T} \otimes M_{n}) Y + X \beta + U + \varepsilon. \end{equation} $

(2.3)

(2) 当$\rho=0 $时, 模型(2.2)即为空间误差面板数据模型

$ \begin{equation} \left\{ \begin{aligned} & Y = X \beta + U + V, \\ & V=\lambda(I_{T} \otimes W_{n})V + \varepsilon. \end{aligned} \right. \end{equation} $

(2.4)

为了完整性, 我们重述空间误差面板数据模型

$ \begin{equation} \left\{ \begin{aligned} & Y = X \beta + U + V, \\ & V=\lambda(I_{T} \otimes W_{n})V + \varepsilon, \end{aligned} \right. \end{equation} $

(3.1)

其中$Y= (Y_{n1}' , \cdots, Y_{nT}')' $为$nT \times 1$维因变量向量, $Y_{nt}=(y_{1t}, \cdots, y_{nt})'$为$n \times 1$维向量.$ X= (X_{n1}', \cdots, X_{nT}')'$为$ nT \times k$维解释变量向量, $ X_{nt}= (X^{*}_{1t}, X^{*}_{2t}, \cdots, X^{*}_{nt})'$为$n \times k$维向量.$ \beta$为$ k \times 1$维系数向量.$ U$为$ nT \times 1$维固定效应向量. $ I_{T}$表示$ T \times T$维单位矩阵.$ W_{n}$表示$ n \times n$维空间权重矩阵, $ w_{ij}$表示矩阵$ W_{n} $的$(i, j)$元素.$V$为自相关误差项. $ \varepsilon= (\varepsilon_{n1}', \cdots, \varepsilon_{nT}')'$为$ nT \times 1$维随机扰动项, 其均值为$ 0 $, 协方差阵为$ \sigma^{2} I_{nT}$, $ \varepsilon_{nt}= (\varepsilon_{1t}, \varepsilon_{2t}, \cdots, \varepsilon_{nt})'$为$n \times 1$维向量.$\lambda $为空间相关系数. $\otimes $表示矩阵的Kronecker积.

为了描述方便, 我们重新记$ nT \times k$维解释变量向量

$ \begin{equation*} \begin{aligned} X = (X_{n1}', \cdots, X_{nT}')' = (X^{*}_{11}, \cdots, X^{*}_{n1}, \cdots, X^{*}_{1T}, \cdots, X^{*}_{nT})' ~ \hat{=}~ (\widetilde{X}_{1} , \widetilde{X}_{2}, \cdots, \widetilde{X}_{nT})', \end{aligned} \end{equation*} $

其中$\widetilde{X}'_{i}=( \widetilde{x}_{i1}, \cdots, \widetilde{x}_{ik}) $为$ 1 \times k$维行向量.$ nT \times 1$维随机扰动项

$ \begin{equation*} \begin{aligned} \varepsilon = (\varepsilon_{n1}', \cdots, \varepsilon_{nT}')' = (\varepsilon_{11}, \cdots, \varepsilon_{n1}, \cdots, \varepsilon_{1T}, \cdots, \varepsilon_{nT})' ~ \hat{=} ~(\widetilde{\varepsilon}_{1}, \widetilde{\varepsilon}_{2}, \cdots, \widetilde{\varepsilon}_{nT})'. \end{aligned} \end{equation*} $

记$R_{nT}(\lambda)=I_{nT} - \lambda(I_{T}\otimes W_{n})$, 且$ R_{nT}(\lambda)=( \widetilde{R}_{1}, \widetilde{R}_{2}, \cdots, \widetilde{R}_{nT} )'$为$ nT \times nT$维向量, $\widetilde{R}'_{i}=( R_{i1}, \cdots, R_{i, nT}) $为$ 1 \times nT$维向量.假设$R_{nT}(\lambda) $可逆, 则模型(3.1)可改写为

$ \begin{equation} Y= X\beta + U + R_{nT}^{-1}(\lambda)\varepsilon, \end{equation} $

(3.2)

其中$\varepsilon=R(\lambda)(Y - X\beta - U)$.被解释变量$Y$的拟对数似然函数

$ \begin{equation} L = -\frac{nT}{2} \ln(2\pi) - \frac{nT}{2}\ln(\sigma^{2}) + \ln \left | R_{nT}(\lambda) \right | - \frac{1}{2\sigma^{2}}\varepsilon'\varepsilon, \end{equation} $

(3.3)

其中$\varepsilon = R_{nT} (\lambda)(Y - X\beta - U)$.

令$G_{nT}=(I_{T}\otimes W_{n})R^{-1}(\lambda)$且$\widetilde{G}_{nT}=\frac{1}{2}(G_{nT} + G_{nT}')$. $ G_{nT}$和$ \widetilde{G}_{nT}$均为$ nT \times nT$维矩阵, 且$\widetilde{G}_{nT}$是对称矩阵.由文献[1]可得

$ \begin{eqnarray*} \frac{\partial L}{\partial \beta} &=& \frac{1}{ \sigma^{2} } \varepsilon' R(\lambda) X, \qquad \frac{\partial L}{\partial U} = \frac{1}{\sigma^{2}} \varepsilon' R (\lambda), \\ \frac{\partial L}{\partial \lambda} &=& -\text{tr}( (I_{T} \otimes W_{n}) R_{nT}^{-1}(\lambda) ) + \frac{1}{ \sigma^{2} } \varepsilon' R_{nT}^{-1}(\lambda) \varepsilon = \frac{1}{\sigma^{2}}(\varepsilon'G_{nT}\varepsilon - \sigma^{2}\text{tr}(G_{nT}))\\ &=& \frac{1}{\sigma^{2}}( \varepsilon' \widetilde{G}_{nT}\varepsilon - \sigma^{2}\text{tr}(\widetilde{G}_{nT}) ), \\ \frac{\partial L}{\partial \sigma^{2}}&=&- \frac{nT}{2\sigma^{2}} + \frac{1}{2\sigma^{4}}\varepsilon'\varepsilon =\frac{1}{2\sigma^{4}}(\varepsilon'\varepsilon - nT\sigma^{2}). \end{eqnarray*} $

令以上偏导数等于$0$, 我们有以下式子

$ \begin{eqnarray} &&X'R'(\lambda)\varepsilon=0, \qquad R'(\lambda)\varepsilon=0, \nonumber\\ && \varepsilon' \widetilde{G}_{nT} \varepsilon - \sigma^{2}\text{tr}(\widetilde{G}_{nT})=0, \qquad \varepsilon'\varepsilon - nT\sigma^{2}=0. \end{eqnarray} $

(3.4)

分别用$g_{ij}$和$\widetilde{g}_{ij}$表示矩阵$G_{nT}$和$\widetilde{G}_{nT}$的$(i, j)$元素, 用$b_{i}$表示矩阵$X'R'(\lambda)$的第$i$列, 用$r_{i}$表示矩阵$R_{nT}'(\lambda)$的第$i$列.由于$X'R'(\lambda)= (R(\lambda)X)'$, 根据$X$及$R(\lambda)$的表达式可知, $b_{i}$和$ r_{i}$分别为

$ \begin{equation} b_{i} = ( \sum\limits_{j=1}^{nT} R_{ij} \widetilde{x}_{j1}, \cdots, \sum\limits_{j=1}^{nT} R_{ij} \widetilde{x}_{jk} )', ~ r_{i} = ( R_{i1} , R_{i2}, \cdots, R_{i, nT} )', \end{equation} $

(3.5)

其中$i=1, 2, \cdots, nT$.

我们规定, 当求和符号的上标等于$0$时我们令该和为$0$.为将$(3.4)$式的二次型转化为线性形式, 引入一个鞅差序列^[21].定义$\sigma$-域: $\mathscr{F}=\{\emptyset, \Omega \}$, $\mathscr{F}_{i}=\sigma( \widetilde{\varepsilon}_{1}, \widetilde{\varepsilon}_{2}, \cdots, \widetilde{\varepsilon}_{i})$, $1\leq i \leq nT$.令

$ \begin{equation} \widetilde{Y}_{i, nT} = \widetilde{g}_{ii} ( \widetilde{\varepsilon}_{i}^{2} - \sigma^{2} ) + 2 \widetilde{\varepsilon}_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde{\varepsilon}_{j}, \end{equation} $

(3.6)

则$\mathscr{F}_{i-1} \subseteq \mathscr{F}_{i}$, $\widetilde{Y}_{i, nT}$是$\mathscr{F}_{i}$-可测的, 并且$E( \widetilde{Y}_{i, nT} \mid \mathscr{F}_{i-1})=0$.因此$\{\widetilde{Y}_{i, nT}, \mathscr{F}_{i}, 1 \leq i \leq nT \}$构成一个鞅差序列, 且

$ \begin{equation} \varepsilon'\widetilde{G}_{nT}\varepsilon - \sigma^{2} \text{tr}(\widetilde{G}_{nT})=\sum\limits_{i=1}^{nT}\widetilde{Y}_{i, nT}. \end{equation} $

(3.7)

我们提出$\theta=(\beta', U', \lambda, \sigma^{2})' \in R^{k+nT+2}$的经验似然比统计量$ L_{nT}(\theta)=\underset{p_{i}, 1\leq i \leq nT}{\sup} \prod_{i=1}^{nT}(nT p_{i}), $其中$p_{i}$满足

$ \begin{eqnarray*} && p_{i}\geq 0, 1 \leq i \leq nT, \sum\limits_{i=1}^{nT}p_{i}=1, \qquad \sum\limits_{i=1}^{nT}p_{i}b_{i}\widetilde{\varepsilon}_{i} = 0, \qquad \sum\limits_{i=1}^{nT} p_{i}r_{i} \widetilde{\varepsilon}_{i} =0 , \\ &&\sum\limits_{i=1}^{nT} p_{i}( { \widetilde{g}_{ii} ( \widetilde{\varepsilon}_{i}^{2} - \sigma^{2} ) + 2 \widetilde{\varepsilon}_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde{\varepsilon}_{j} })=0, \qquad \sum\limits_{i=1}^{nT}p_{i}( {\widetilde{\varepsilon}_{i}^{2} - \sigma^{2}} )=0. \end{eqnarray*} $

令

$ \begin{equation} \omega_{i}(\theta)= \begin{pmatrix} b_{i}\widetilde {\varepsilon}_{i}\\ r_{i}\widetilde {\varepsilon}_{i}\\ \widetilde{g}_{ii}(\widetilde{ \varepsilon}_{i}^{2} - \sigma^{2}) + 2 \widetilde{\varepsilon}_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde{\varepsilon}_{j}\\ \widetilde{\varepsilon}_{i}^{2} - \sigma^{2} \end{pmatrix}, \end{equation} $

(3.8)

其中$\omega_{i}(\theta)$为$(k+nT+2)\times 1$维, $\widetilde{\varepsilon}_{i}$为$\varepsilon=R_{nT}(\lambda)(Y - X\beta - U )$的第$i$部分.由Owen^[15], 有

$ \begin{equation} l_{nT}(\theta) = -2 L_{nT}(\theta)=2 \sum\limits_{i=1}^{nT} \log \{1 + \gamma'(\theta)\omega_{i}(\theta)\}, \end{equation} $

(3.9)

其中$\gamma(\theta) \in R^{k+nT+2}$是下面等式的解

$ \begin{equation} \frac{1}{nT} \sum\limits_{i=1}^{nT} \frac{\omega_{i}(\theta)}{1 + \gamma'(\theta)\omega_{i}(\theta)}=0. \end{equation} $

(3.10)

在本文中, 记$\mu_{j}=E \widetilde{\varepsilon}_{1}^{j}$, $j=3, 4$.用Vec(diag$ A)$表示由矩阵$A$的主对角线上的元素构成的列向量.$\left \| a \right \|$表示向量$a$的$L_{2}$范数.用$1_{nT}$表示元素均为$1$的$nT$维列向量.给出以下假设条件

(A1) $n$为无限大常数, $T$为有限常数^[6].

(A2) $\{\widetilde{\varepsilon}_{i}, ~i=1, 2, \cdots, nT\}$为均值为$0$, 方差$\sigma^{2}$有限的独立同分布随机变量序列, 且存在$\delta > 0$, 使$E \left | \varepsilon \right |^{4+\delta} < \infty$.

(A3) $R_{nT}(\lambda)$为非奇异矩阵.

(A4) 矩阵$W_{n}$, $R_{nT}(\lambda)$元素的绝对值的行和与列和均一致有界.

(A5) $\{\widetilde{X}_{i}, ~1 \leq i \leq nT \}$一致有界.

(A6) 存在常数$C_{1}$, $C_{2}$, 使$0 < C_{1}\leq \lambda_{\text{min}}(\frac{1}{nT}\Sigma_{k+nT+2}) \leq \lambda_{\text{max}}(\frac{1}{nT}\Sigma_{k+nT+2}) \leq C_{2} < \infty$, 其中$\lambda_{\text{min}}(A)$, $\lambda_{\text{max}}(A)$分别表示矩阵$A$的最小和最大特征值. $\Sigma_{k+nT+2}$表示如下

$ \begin{equation} \Sigma_{k+nT+2}=\Sigma_{k+nT+2}'=Cov\{\sum\limits_{i=1}^{nT}\omega_{i}(\theta)\}= \begin{pmatrix} \Sigma _{11} & \Sigma _{12} & \Sigma _{13} & \Sigma _{14}\\ \Sigma _{21} & \Sigma _{22} & \Sigma _{23} & \Sigma _{23}\\ \Sigma _{31} & \Sigma _{32} & \Sigma _{33} & \Sigma _{34}\\ \Sigma _{41} & \Sigma _{42} & \Sigma _{43} & \Sigma _{44} \end{pmatrix}, \end{equation} $

(3.11)

其中$\Sigma _{11}=\sigma^{2}X'R_{nT}'(\lambda)X$, $\Sigma _{12}=2 \sigma^{2}X'R_{nT}'(\lambda)R_{nT}(\lambda)$, $\Sigma _{13}=2 \mu_{3}X' R_{nT}'(\lambda) \text{Vec(diag} \widetilde{G}_{nT}), $ $\Sigma _{14}=2 \mu_{3}X' R_{nT}'(\lambda) 1_{nT}$, $\Sigma _{22}=\sigma^{2}R_{nT}'(\lambda)R_{nT}(\lambda)$, $\Sigma _{23}=2 \mu_{3}R_{nT}'(\lambda) \text{Vec(diag} \widetilde{G}_{nT})$, $\Sigma _{24}=2 \mu_{3} R_{nT}'(\lambda) 1_{nT}$, $\Sigma _{33}=2 \sigma^{4} \text{tr}(\widetilde{G}_{nT}^{2}) + (\mu_{3} - 3 \sigma^{4}) \| \text{Vec(diag} \widetilde{G}_{nT}) \|^{2}$, $\Sigma _{34}= 2 (\mu_{4} - \sigma^{4}) \text{tr}(\widetilde{G}_{nT})$, $\Sigma _{44}= nT(\mu_{4} - \sigma^{4})$.

注 (A1)源于文献[6], 本文考虑$ n $无限而$ T $有限的情形. (A2)和(A3)是空间面板数据模型的常见假设, 如Lee和Yu^[5], Yu等^{[4, 22]}. (A3)保证(3.2)的表示方法是有效的. (A4)源于Kelejian和Prucha^{[21, 23]}, 在Lee^[13]中也有用到. (A5)和(A6)保证了本文的$Q_{nT}$满足假设条件C2.

引理1^[15]令$\xi_{1}, \cdots, \xi_{nT}$是一个平稳随机变量序列, 且对常数$s>0$有$E \left | \xi_{1} \right |^{s} < \infty$, 那么$ \underset{1\leq i \leq nT}{\max} \left | \xi_{i} \right |=o((nT)^{1/s}).$ a.s.

证 见文献[15]中引理3的证明.

引理2的准备 需要用到文献[21]中的定理1, 我们对此定理做以下描述.考虑线性二次型

$ \begin{equation*} \widetilde{Q}_{n} = \epsilon'_{n}A_{n}\epsilon_{n} + b'_{n} \epsilon_{n} = \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{n}a_{nij}\epsilon_{ni} \epsilon_{nj} + \sum\limits_{i=1}^{n} b_{ni} \epsilon_{ni}, \end{equation*} $

其中$\epsilon_{ni}$是(实值)随机变量, $a_{nij}$和$b_{ni}$分别代表二次型和线性形式的(实值)系数.需要以下假设.

(C1) 实值随机变量序列$\{\epsilon_{ni}, 1\leq i\leq n, n\geq 1\}$满足$E(\epsilon_{ni})=0$, 且对每一个$ n\geq 1$随机变量$\epsilon_{n1}$, $\cdots$, $\epsilon_{nn}$完全独立.且存在$\delta_{1} > 0$使$\underset{1\leq i \leq n, n\geq 1}{\sup} E \left | \epsilon_{ni}\right |^{4 + \delta_{1}} < \infty$.

(C2) 对所有$1\leq i, j \leq n, n\geq1$有$a_{nij}=a_{nji}$, $\underset{1\leq j \leq n, n\geq 1} {\sup} \sum\limits_{i=1}^{n} \left | a_{nij}\right | < \infty$, 存在$\delta_{2} > 0$, 有$\underset{n} {\sup} \sum\limits_{i=1}^{n} \left | b_{ni}\right |^{2 + \delta_{2}} < \infty$.

在这些假设下, $\widetilde{Q}_{n}$的均值$\mu_{\widetilde{Q}_{n}}$和方差$\sigma^{2}_{ \widetilde{Q}_{n} }$分别为

$ \begin{eqnarray*} \mu_{\widetilde{Q}_{n}}&=&\sum\limits_{i=1}^{n}a_{ni}\sigma_{ni}^{2}, \\ \sigma^{2}_{ \widetilde{Q}_{n} } &=& 2 \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{n} a_{nij}^{2} \sigma_{ni}^{2} \sigma_{nj}^{2} + \sum\limits_{i=1}^{n} b_{ni}^{2} \sigma_{ni}^{2} + \sum\limits_{i=1}^{n} \{ a_{nii}^{2} (\mu_{ni}^{4} - 3\sigma_{ni}^{4}) + 2 b_{ni} a_{nii} \mu_{ni}^{(3)}\}, \end{eqnarray*} $

其中$\sigma_{ni}^{2}=E(\epsilon_{ni}^{2})$, $\mu_{ni}^{(s)}=E(\epsilon_{ni}^{s})$, $s=3, 4$.

引理2 若假设条件(C1)和(C2)成立, 且存在常数$C>0$满足$n^{-1} \sigma_{ \widetilde{Q}_{n} }^{2}\geq C$, 则

$ \begin{equation*} \frac{ \widetilde{Q}_{n} - \mu_{\widetilde{Q}_{n}} } {\sigma_{\widetilde{Q}_{n}} } \overset{d}{\rightarrow} N(0, 1). \end{equation*} $

证 见文献[21]中的定理1.

引理3 若假设条件(A1)-(A6)满足, 那么当$n \rightarrow \infty$时,

$ Z_{nT}=\underset{1\leq i \leq nT}{\max} \left \| \omega_{i}(\theta) \right \|=o_{p}((nT)^{1/2}), $

(3.12)

$ \Sigma^{- 1/2}_{k + nT + 2} \sum\limits_{i=1}^{nT} \omega_{i}(\theta) \overset{d}{\rightarrow} N(0, I_{k + nT + 2}), $

(3.13)

$ (nT)^{-1} \sum\limits_{i=1}^{nT} \omega_{i}(\theta) \omega_{i}'(\theta)=(nT)^{-1} \Sigma_{k + nT + 2} + o_{p}(1), $

(3.14)

$ \sum\limits_{i=1}^{nT} \left \| \omega_{i}(\theta) \right \|^{3}=O_{p}(nT). $

(3.15)

证 见第四章.

定理1 在(A1)-(A6)假设条件下, 当$n \rightarrow \infty$时, 有

$ \begin{equation} l_{nT}(\theta) \overset{d}{\rightarrow}\chi^{2}_{(k+nT+2)}, \end{equation} $

(3.16)

其中$ \chi^{2}_{(k+nT+2)} $表示自由度为$k+nT+2 $的卡方分布.

证 令$\gamma(\theta)=\gamma$, $\rho_{0}= \| \gamma \|$, $\gamma=\rho_{0}\eta_{0}$.下面证明$\| \gamma \| = O_{p}( (nT)^{-1/2})$.由(3.10)式, 有

$ \begin{equation*} \frac{1}{nT} \sum\limits_{j=1}^{nT} \frac{ \eta_{0}'\omega_{j}(\theta)(1+\gamma'\omega_{j}(\theta)) - \eta_{0}'\omega_{j}\gamma'\omega_{j}(\theta)} {1+\gamma'\omega_{j}(\theta)}=0, \end{equation*} $

即

$ \begin{equation*} \frac{\eta_{0}'}{nT}\sum\limits_{j=1}^{nT}\omega_{j}(\theta) - \frac{\rho_{0}}{nT} \sum\limits_{j=1}^{nT} \frac{ ( \eta_{0}'\omega_{j}(\theta))^{2}}{ 1+\gamma'\omega_{j}(\theta)}=0, \end{equation*} $

则有

$ \begin{equation} \left | \eta_{0}'\frac{1}{nT}\sum\limits_{j=1}^{nT}\omega_{j}(\theta) \right | = \left | \frac{\rho_{0}}{nT}\sum\limits_{j=1}^{nT} \frac{ ( \eta_{0}' \omega_{j}(\theta))^{2}}{ 1+\gamma'\omega_{j}(\theta)} \right |. \end{equation} $

(3.17)

根据(3.12)式有

$ \begin{equation} \begin{aligned} | 1+\gamma'\omega_{j}(\theta) | \leq 1 + \left \| \gamma \right \| \left \| \omega_{j}(\theta) \right \| \leq 1 + \rho_{0} \underset{1 \leq j \leq nT}{\max} \left \| \omega_{j}(\theta) \right \| =1 + \rho_{0}Z_{nT}. \end{aligned} \end{equation} $

(3.18)

记$\bar{\omega}= \frac{1}{nT}\sum\limits_{j=1}^{nT}\omega_{j}(\theta)$, $ S_{0}=\frac{1}{nT}\sum\limits_{j=1}^{nT}\omega_{j}(\theta) \omega_{j}'(\theta)$, 由(3.17), (3.18)式及二次型的极值理论^[24], 有

$ \begin{equation*} \begin{aligned} | \eta_{0}' \bar \omega | & \geq \frac{\rho_{0}}{nT}\sum\limits_{j=1}^{nT} \frac{ ( \eta_{0}'\omega_{j}(\theta))^{2}}{ 1+\rho_{0}Z_{nT}} = \frac{ \rho_{0} } {1+\rho_{0}Z_{nT}} \eta_{0}' \frac{1}{nT} \{ \sum\limits_{j=1}^{nT}\omega_{j}(\theta)\omega'_{j}(\theta) \} \eta_{0}\\ &= \frac{ \rho_{0} } {1+\rho_{0}Z_{nT}} \eta_{0}' S_{0} \eta_{0} \geq \frac{ \rho_{0} } {1+\rho_{0}Z_{nT}} \lambda_{\min}(S_{0}) \eta_{0}'\eta_{0} = \frac{ \rho_{0} } {1+\rho_{0}Z_{nT}} \lambda_{\min}(S_{0}), \end{aligned} \end{equation*} $

即$| \eta_{0}' \bar \omega | \geq \frac{ \rho_{0} } {1+\rho_{0}Z_{n}} \lambda_{\min}(S_{0})$, 即

$ \begin{equation} | \eta_{0}' \Sigma_{k+nT+2}^{1/2} \Sigma_{k+nT+2}^{-1/2} \bar{\omega}| \geq \frac{ \rho_{0} } {1+\rho_{0}Z_{nT}} \lambda_{\min}(S_{0}). \end{equation} $

(3.19)

而

$ \begin{equation} \begin{aligned} | \eta_{0}' \Sigma^{1/2}_{k+nT+2} \Sigma^{-1/2}_{k+nT+2} \bar{\omega}| &= | ( \Sigma_{k+nT+2}^{1/2} \eta_{0})' \Sigma_{k+nT+2}^{-1/2} \bar {\omega}| \leq \left \| \Sigma_{k+nT+2}^{1/2} \eta_{0} \right \| \left \| \Sigma_{k+nT+2}^{-1/2} \bar {\omega} \right \| \\ &= \sqrt{\eta_{0}'\Sigma_{k+nT+2} \eta_{0} } \left \| \Sigma_{k+nT+2}^{-1/2} \bar {\omega} \right \| \leq \sqrt{ \lambda_{\max}(\Sigma_{k+nT+2})\eta_{0}'\eta_{0} } \left \| \Sigma_{k+nT+2}^{-1/2} \bar {\omega} \right \| \\ &= \lambda_{\max}(\Sigma^{1/2}_{k+nT+2}) \left \| \Sigma^{-1/2}_{k+nT+2} \bar {\omega} \right \|, \end{aligned} \end{equation} $

(3.20)

由(3.19)和(3.20)二式有

$ \begin{equation*} \lambda_{\max}(\Sigma_{k+nT+2}^{1/2}) \left \| \Sigma_{k+nT+2}^{-1/2} \bar {\omega} \right \| \geq \frac{ \rho_{0} } {1+\rho_{0}Z_{nT}} \lambda_{\min}(S_{0}), \end{equation*} $

进而

$ \begin{equation} \frac{ \rho_{0} } {1+\rho_{0}Z_{nT}} \leq \frac{1} {\lambda_{\min}(S_{0})}\lambda_{\max}(\Sigma_{k+nT+2}^{1/2}) \left \| \Sigma_{k+nT+2}^{-1/2} \bar {\omega} \right \|. \end{equation} $

(3.21)

由条件(A6)及引理3, $ S_{0} \overset {P}{\rightarrow} (nT)^{-1}\Sigma_{k+nT+2}, $所以

$ \begin{equation*} \lambda_{\min}(S_{0}) \overset {P}{\rightarrow} \lambda_{\min}((nT)^{-1}\Sigma_{k+nT+2}) , ~ \frac{1}{\lambda_{\min}(S_{0})} \overset {P}{\rightarrow} \frac{1}{\lambda_{\min}((nT)^{-1}\Sigma_{k+nT+2})} , \end{equation*} $

故

$ \begin{equation} \frac{1}{\lambda_{\min}(S_{0})}=O_{p}(1). \end{equation} $

(3.22)

另外

$ \begin{equation} \begin{aligned} (nT)^{-1/2}\lambda_{\max}(\Sigma_{k+nT+2}^{1/2}) = (nT)^{\frac{1}{2}}(\lambda_{\max}(\Sigma_{k+nT+2}))^{\frac{1}{2}} = \lambda_{\max}( \frac{1}{nT}\Sigma_{k+nT+2} ) ^{\frac{1}{2}} \leq C^{\frac{1}{2}}, \end{aligned} \end{equation} $

(3.23)

$ \begin{equation} \| \Sigma_{k+nT+2}^{-1/2} \bar{\omega} \| = \| \frac{1}{nT} \Sigma_{k+nT+2}^{-1/2} \sum\limits_{j=1}^{nT}\omega_{i}(\theta) \| = \frac{1}{nT} \| \Sigma_{k+nT+2}^{-1/2} \sum\limits_{j=1}^{nT}\omega_{i}(\theta) \| = \frac{1}{nT} O_{p}(1), \end{equation} $

(3.24)

综合(3.21)-(3.24)几个式子, 得

$ \begin{equation*} \frac{ \rho_{0} } {1+\rho_{0}Z_{nT}}=O_{p}((nT)^{-1/2}), \end{equation*} $

由(3.12)得

$ \begin{equation} \rho_{0} = O_{p}((nT)^{-1/2}). \end{equation} $

(3.25)

令$\xi_{i}=\gamma'\omega_{i}(\theta)$, 由(3.25)式及(3.12)式得到

$ \begin{equation} \underset{1 \leq i \leq nT}{\max} | \xi_{i} | = O_{p}((nT)^{1/2}) O_{p}((nT)^{-1/2}) = o_{p}(1). \end{equation} $

(3.26)

由(3.10)式, 也有

$ \begin{equation*} \begin{aligned} \frac{1}{nT} \sum\limits_{j=1}^{nT}\frac{\omega_{j}(\theta)}{1+\gamma'\omega_{j}(\theta)} &= \frac{1}{nT} \sum\limits_{j=1}^{nT}\frac{\omega_{j}(\theta)(1+\gamma'\omega_{j}(\theta)) - \omega_{j}(\theta)\gamma'\omega_{j}(\theta)}{1+\gamma'\omega_{j}(\theta)}\\ &= \frac{1}{nT} \sum\limits_{j=1}^{nT}\omega_{j}(\theta) - \frac{1}{nT} \sum\limits_{j=1}^{nT} \frac{\omega_{j}(\theta)\gamma'\omega_{j}(\theta)}{1+\gamma'\omega_{j}(\theta)}\\ &= \frac{1}{nT} \sum\limits_{j=1}^{nT}\omega_{j}(\theta) - \frac{1}{nT} \sum\limits_{j=1}^{nT} \frac{\omega_{j}(\theta)\gamma'\omega_{j}(\theta)(1+\gamma'\omega_{j}(\theta)) - \omega_{j}(\theta)(\gamma'\omega_{j}(\theta))^{2}} {1+\gamma'\omega_{j}(\theta)}\\ &= \frac{1}{nT} \sum\limits_{j=1}^{nT}\omega_{j}(\theta) - \frac{1}{nT} \sum\limits_{j=1}^{nT}\omega_{j}(\theta)\gamma'\omega_{j}(\theta) + \frac{1}{nT} \sum\limits_{j=1}^{nT} \frac{ \omega_{j}(\theta)(\gamma'\omega_{j}(\theta))^{2}} {1+\gamma'\omega_{j}(\theta)}\\ &= \frac{1}{nT} \sum\limits_{j=1}^{nT}\omega_{j}(\theta) - (\frac{1}{nT} \sum\limits_{j=1}^{nT}\omega_{j}\omega_{j}')\gamma + \frac{1}{nT} \sum\limits_{j=1}^{nT} \frac{ \omega_{j}(\theta)(\gamma'\omega_{j}(\theta) )^{2}} {1+\gamma'\omega_{j}(\theta)}\\ &= \bar{\omega} - S_{0}\gamma + \frac{1}{nT} \sum\limits_{j=1}^{nT} \frac{ \omega_{j}(\theta)\xi_{j}^{2} } {1+\xi_{j}}=0, \end{aligned} \end{equation*} $

所以

$ \begin{equation} \gamma=S_{0}^{-1}\bar{\omega} + S_{0}^{-1}\frac{1}{nT} \sum\limits_{j=1}^{nT} \frac{ \omega_{j}(\theta)\xi_{j}^{2}} {1+\xi_{j}}\hat{=}S_{0}^{-1}\bar{\omega} +\zeta. \end{equation} $

(3.27)

由于$ (nT)^{-1} \sum\limits_{j=1}^{nT} \left \| \omega_{j}(\theta) \right \|^{3} \left \| \xi \right \|^{2} =(nT)^{-1} O_{p}(nT)O_{p}((nT)^{-1})= O_{p}((nT)^{-1}), $所以$\zeta$是有界的.利用泰勒展开式, 有

$ \begin{equation} \log (1 + \gamma'(\theta)\omega_{i}(\theta)) = \log (1 + \xi_{i})=\xi_{i} - \frac{1}{2}\xi_{i}^{2} + d_{i}, \end{equation} $

(3.28)

其中$d_{i} = \frac{1}{3(1+h)^{3}}\xi_{i}^{3} \hat{=}B\xi_{i}^{3}$, $h$介于$0$与$\xi_{i}$之间, $0 < B < \infty$.由(3.25)式及引理3, 当$ n \rightarrow \infty$时,

$ \begin{equation} \| \sum\limits_{i=1}^{nT}d_{i} \| \leq B \sum\limits_{i=1}^{nT} | \xi_{i}|^{3} \leq B \rho_{0}^{3} \sum\limits_{i=1}^{nT} \| \omega_{i}(\theta) \|^{3} = O_{p}((nT)^{-1/2}) = o_{p}(1). \end{equation} $

(3.29)

另外, 由引理3得$\lambda_{\max}(S_{0}) \overset{p}{\rightarrow} \lambda_{\max}((nT)^{-1} \Sigma_{k+nT+2}) \leq C_{2}$, 即$\lambda_{\max}(S_{0}) =O_{p}(1)$, 所以

$ \begin{equation} nT \zeta' S_{0}^{-1} \zeta \leq nT \cdot \lambda_{\max}(S_{0}) \zeta' \zeta = nT \cdot \lambda_{\max}(S_{0}) \left \| \zeta \right \|^{2}=O_{p}((nT)^{-1}) = o_{p}(1). \end{equation} $

(3.30)

根据(3.27)及(3.28)二式,

$ \begin{equation*} \begin{aligned} l_{n}(\theta) =& 2 \sum\limits_{i=1}^{nT} \log (1+ \gamma'(\theta)\omega_{i}(\theta) ) = 2 \sum\limits_{i=1}^{nT} ( \xi_{i} - \frac{1}{2}\xi_{i}^{2} + d_{i} ) = 2 \sum\limits_{i=1}^{nT}\xi_{i} - \sum\limits_{i=1}^{nT}\xi_{i}^{2} + 2 \sum\limits_{i=1}^{nT}d_{i}\\ =& 2 \gamma' {\sum\limits_{i=1}^{nT} \omega_{i}(\theta)} - \gamma' \{\sum\limits_{i=1}^{nT} \omega_{i}'(\theta)\omega_{i}(\theta)\} \gamma + 2 \sum\limits_{i=1}^{nT}d_{i} = 2 nT \gamma' \bar{\omega} - nT \gamma' S_{0}\gamma + 2 \sum\limits_{i=1}^{nT}d_{i}\\ =& 2 nT (S_{0}^{-1}\bar{\omega} +\zeta)' \bar{\omega} - nT (S_{0}^{-1}\bar{\omega} +\zeta)' S_{0}(S_{0}^{-1}\bar{\omega} +\zeta)+ 2 \sum\limits_{i=1}^{nT}d_{i}\\ = & 2 nT (S_{0}^{-1}\bar{\omega} )' \bar{\omega} + 2 nT \zeta' \bar{\omega} - ( nT \bar{\omega}' S_{0}^{-1} \bar{\omega}+ 2 nT \bar{\omega}' \zeta + nT \zeta' S_{0}^{-1} \zeta) + 2 \sum\limits_{i=1}^{nT}d_{i}\\ =& nT \cdot \bar{\omega}' S_{0}^{-1} \bar{\omega} - nT \cdot \zeta' S_{0}^{-1} \zeta + o_{p}(1)\\ =& nT \cdot \bar{\omega}' \Sigma^{-1/2}_{k+nT+2} (\Sigma^{-1/2}_{k+nT+2})^{-1}S_{0}^{-1} (\Sigma^{-1/2}_{k+nT+2})^{-1} \Sigma^{-1/2}_{k+nT+2} \bar{\omega} - nT \cdot \zeta' S_{0}^{-1} \zeta + 2 \sum\limits_{i=1}^{nT}d_{i}\\ =& ( nT \Sigma^{-1/2}_{k+nT+2} \bar{\omega} )' ( nT \Sigma^{-1/2}_{k+nT+2} S_{0} \Sigma^{-1/2}_{k+nT+2} )^{-1} ( nT \Sigma^{-1/2}_{k+nT+2} \bar{\omega} ) - nT \cdot \zeta' S_{0}^{-1} \zeta + 2 \sum\limits_{i=1}^{nT}d_{i}.\\ \end{aligned} \end{equation*} $

由假设条件A(6)及引理3, 得

$ \begin{equation} ( nT \Sigma^{-1/2}_{k+nT+2} \bar{\omega} )' ( nT \Sigma^{-1/2}_{k+nT+2} S_{0} \Sigma^{-1/2}_{k+nT+2} )^{-1} ( nT \Sigma^{-1/2}_{k+nT+2} \bar{\omega} ) \overset{d}{\rightarrow} \chi^{2}_{k+nT+2}. \end{equation} $

(3.31)

由(3.29)-(3.31)三个式子就完成了定理1的证明.

(3.12)式的证明

$ \begin{equation*} \begin{aligned} Z_{nT} & \leq \underset{1\leq i \leq nT}{\max} \| b_{i}\widetilde{ \varepsilon }_{i} \| + \underset{1\leq i \leq nT}{\max} \| r_{i} \widetilde{\varepsilon}_{i} \| + \underset{1\leq i \leq nT}{\max} | \widetilde{g}_{ii} (\widetilde{\varepsilon}_{i}^{2} - \sigma^{2}) + 2 \widetilde{\varepsilon}_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde{\varepsilon}_{j}| + \underset{1\leq i \leq nT}{\max} \| \widetilde{ \varepsilon }_{i}^{2} - \sigma^{2} \|\\ & \leq \underset{1\leq i \leq nT}{\max} \left \| b_{i}\widetilde{ \varepsilon}_{i} \right \| + \underset{1\leq i \leq nT}{\max} \left \| r_{i}\widetilde{ \varepsilon}_{i} \right \| + \underset{1\leq i \leq nT}{\max} | \widetilde{g}_{ii} (\widetilde{ \varepsilon }_{i}^{2} - \sigma^{2}) | + \underset{1\leq i \leq nT}{\max} | 2 \widetilde{ \varepsilon }_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde{ \varepsilon }_{j} | + \underset{1\leq i \leq nT}{\max} \left \| \widetilde{ \varepsilon }_{i}^{2} - \sigma^{2} \right \|, \end{aligned} \end{equation*} $

由条件(A1)-(A5)及引理1, 有

$ \underset{1\leq i \leq nT}{\max} \| b_{i}\widetilde{ \varepsilon }_{i} \| = \underset{1\leq i \leq nT}{\max} \left \| b_{i} \right \| o_{p}( (nT)^{1/4} ) = o_{p}((nT)^{1/4}), $

(4.1)

$ \underset{1\leq i \leq nT}{\max} \left \| r_{i} \widetilde{ \varepsilon }_{i} \right \| = \underset{1\leq i \leq nT}{\max} \left \| r_{i} \right \| o_{p}((nT)^{1/4})=o_{p}((nT)^{1/4}), $

(4.2)

$ \underset{1\leq i \leq nT}{\max} \left | \widetilde{g}_{ii} ( \widetilde{ \varepsilon }_{i}^{2} - \sigma^{2}) \right | = \underset{1\leq i \leq nT}{\max} \left | \widetilde{g}_{ii} \right | o_{p}((nT)^{1/2})=o_{p}((nT)^{1/2}), $

(4.3)

$ \underset{1\leq i \leq nT}{\max} | \widetilde{ \varepsilon }_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde{ \varepsilon }_{j}| = (\underset{1\leq i \leq nT}{\max} \left | \widetilde{\varepsilon}_{i} \right |)^{2} + \underset{1\leq i \leq nT}{\max} | \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} |) = o_{p}((nT)^{1/2}), $

(4.4)

$ \underset{1\leq i \leq nT}{\max} \left \| \widetilde{ \varepsilon }_{i}^{2} - \sigma^{2} \right \|=o_{p}((nT)^{1/4}), $

(4.5)

由(4.1)-(4.5)式可知$Z_{nT}=o_{p}((nT)^{1/2})$, (3.12)式得证.

(3.13)式的证明 对任意给定的$l=(l_{1}', ~l_{2}', ~l_{3}, ~l_{4})' \in R^{k+nT+2}$, $\left \| l \right \|=1$, 其中$l_{1} = (l_{11}, \cdots, l_{1k})' \in R^{k}$, $l_{2}=(l_{21}, \cdots, l_{2, nT})' \in R^{nT}$, $l_{3} \in R^{1}$, $l_{4} \in R^{1}$.则

$ \begin{eqnarray*} l'\omega_{i}(\theta) &=& l_{1}'b_{i} \widetilde{ \varepsilon }_{i} + l_{2}'r_{i} \widetilde{ \varepsilon }_{i} + l_{3}\{ \widetilde{g}_{ii} ( \widetilde{ \varepsilon }_{i}^{2} - \sigma^{2}) + 2 \widetilde{ \varepsilon}_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde{\varepsilon}_{j} \} + l_{4}( \widetilde{\varepsilon}_{i}^{2} - \sigma^{2} )\\ &=& ( l_{3} \widetilde{g}_{ii} + l_{4} )( \widetilde{ \varepsilon }_{i}^{2} - \sigma^{2} ) + 2 \widetilde{ \varepsilon }_{i} \sum\limits_{j=1}^{i-1} l_{3} \widetilde{g}_{ij} \widetilde{ \varepsilon }_{j} + l_{1}'b_{i} \widetilde{\varepsilon }_{i} + l_{2}'r_{i} \widetilde{\varepsilon }_{i}\\ &=& (l_{3}\widetilde{g}_{ii} + l_{4})( \widetilde{\varepsilon }_{i}^{2} - \sigma^{2}) + 2 \widetilde{ \varepsilon }_{i} \sum\limits_{j=1}^{i-1} l_{3} \widetilde{g}_{ij} \widetilde{ \varepsilon }_{j} + ( l_{1}'b_{i} + l_{2}'r_{i} ) \widetilde{ \varepsilon }_{i}, \end{eqnarray*} $

其中$\widetilde{g}_{ij}$, $b_{i}$, $r_{i}$如第3节所述, 所以

$ \begin{equation*} \sum\limits_{i=1}^{nT}l'\omega_{i}(\theta) = \sum\limits_{i=1}^{nT} (l_{3}\widetilde{g}_{ii} + l_{4})( \widetilde{\varepsilon }_{i}^{2} - \sigma^{2}) + 2 \sum\limits_{i=1}^{nT} \sum\limits_{j=1}^{i-1} l_{3} \widetilde{g}_{ij} \widetilde{ \varepsilon }_{i} \widetilde{ \varepsilon }_{j} + \sum\limits_{i=1}^{nT} (l_{1}'b_{i} + l_{2}'r_{i}) \widetilde{ \varepsilon}_{i}. \end{equation*} $

令$ Q_{nT}=\sum\limits_{i=1}^{nT}\sum\limits_{j=1}^{nT}u_{ij} \widetilde{ \varepsilon }_{i} \widetilde{ \varepsilon }_{j} + \sum\limits_{i=1}^{nT}v_{i}\widetilde{ \varepsilon }_{i}, $其中$u_{ii}=l_{3}\widetilde{g}_{ii} + l_{4}$, $u_{ij}=l_{3} \widetilde{g}_{ij}(i\neq j)$, $v_{i}=l_{1}'b_{i} + l_{2}'r_{i}$, 则

$ \begin{equation*} Q_{nT} = \sum\limits_{i=1}^{nT}l'w_{i}(\theta)= \sum\limits_{i=1}^{nT}\{ u_{ii}( \widetilde{ \varepsilon }_{i}^{2} - \sigma^{2} ) + 2 \sum\limits_{j=1}^{i-1} u_{ij} \widetilde{ \varepsilon }_{i} \widetilde{ \epsilon }_{j} + v_{i} \widetilde{ \varepsilon }_{i} \}. \end{equation*} $

接下来检验$Q_{nT}$是否满足条件(C2).由假设条件(A4), 有

$ \begin{equation} \sum\limits_{i=1}^{nT} \left | u_{ij} \right | \leq \left | l_{3} \right | \sum\limits_{i=1}^{nT} \left | \widetilde{g}_{ij} \right | + \left | l_{4} \right |. \end{equation} $

(4.6)

接下来证$(nT)^{-1}\sum_{i=1}^{nT} \left | v_{i} \right |^{3}\leq C$.由$\left | v_{i} \right |^{3}=\left | l_{1}'b_{i} + l_{2}'r_{i} \right |^{3} \leq C(\left | l_{1}'b_{i} \right |^{3} + \left | l_{2}'r_{i} \right |^{3})$, 有

$ \begin{equation*} \begin{aligned} (nT)^{-1}\sum\limits_{i=1}^{nT} \left | v_{i} \right |^{3} \leq (nT)^{-1}\sum\limits_{i=1}^{nT} C(\left | l_{1}'b_{i} \right |^{3} + \left | l_{2}'r_{i} \right |^{3}) \leq (nT)^{-1}\sum\limits_{i=1}^{nT} C\left | l_{1}'b_{i} \right |^{3} + (nT)^{-1}\sum\limits_{i=1}^{nT} C \left | l_{2}'r_{i} \right |^{3}. \end{aligned} \end{equation*} $

根据假设条件(A4)和$b_{i}$的表达式, 得

$ \begin{align*} \left |l_{1}'b_{i}\right | =& | l_{11}\sum\limits_{j=1}^{nT} R_{ij} \widetilde{x}_{j1} + \cdots + l_{1k}\sum\limits_{j=1}^{nT} R_{ij} \widetilde{x}_{jk} | \leq C |R_{i1}| \left \| \widetilde{X}_{1}\right \| + C |R_{i2}| \| \widetilde{X}_{2} \| +\cdots + C |R_{i, nT}| \| \widetilde{X}_{nT} \| \\ \leq & C \underset {1 \leq i \leq nT}{\max} \| \widetilde{X}_{i} \| (\sum\limits_{j=1}^{nT} |R_{ij} |) \leq C \underset {1 \leq i \leq nT}{\max} \| \widetilde{X}_{i} \| , \end{align*} $

即$ \left |l_{1}'b_{i}\right | \leq C \underset {1 \leq i \leq nT}{\max} \| \widetilde{X}_{i} \|$.同理

$ \begin{equation*} \left |l_{2}'r_{i}\right | =\left |l_{21}R_{i1} + l_{22}R_{i2} + \cdots + l_{2, nT}R_{i, nT}\right |\leq C \sum\limits_{j=1}^{nT} \left |R_{ij}\right | \leq C , \end{equation*} $

再由假设条件(A5), 有

$ \begin{equation} (nT)^{-1} \sum\limits_{i=1}^{nT} C | l_{1}'b_{i} |^{3} \leq C (nT)^{-1} \sum\limits_{i=1}^{nT}\{ C \underset{1 \leq i \leq nT}{\max} \| \widetilde{X}_{i} \| \}^{3} \leq C \underset{1 \leq i \leq nT}{\max} \| \widetilde{X}_{i} \|^{3} \leq C, \end{equation} $

(4.7)

因此$(nT)^{-1}\sum\limits_{i=1}^{nT} \left | v_{i} \right |^{3}\leq C$.故$Q_{nT}$满足条件(C2).

令$e_{i}$为坐标轴$i$方向上的单位向量, 下面求$Q_{nT}$的方差.由$u_{ij}$和$v_{i}$的表达式, 有

$ \begin{equation*} \begin{aligned} \sum\limits_{i=1}^{nT}\sum\limits_{j=1}^{nT} u_{ij}^{2} =&\sum\limits_{i=1}^{nT}\{(l_{3} \widetilde{g}_{ii} + l_{4})^{2} + \underset{i\neq j}{\sum} (l_{3}\widetilde{g}_{ij})^{2}\} = \sum\limits_{i=1}^{nT}\{ (l_{3} \widetilde{g}_{ii})^{2} + 2 l_{3}l_{4} \widetilde{g}_{ii} + l_{4}^{2} + \underset{i\neq j}{\sum} (l_{3}\widetilde{g}_{ij})^{2}\}\\ =&2 l_{3}l_{4} \sum\limits_{i=1}^{nT} \widetilde{g}_{ii} + nT \cdot l_{4}^{2} + \sum\limits_{i=1}^{nT} \sum\limits_{j=1}^{nT} (l_{3} \widetilde{g}_{ij})^{2} = 2 l_{3}l_{4} \text{tr}(\widetilde{G}_{nT}) + nT \cdot l_{4}^{2} + l_{3}^{2}\text{tr}(\widetilde{G}_{nT}^{2}), \\ \sum\limits_{i=1}^{nT} u_{ii}^{2} =&\sum\limits_{i=1}^{nT}\{ l_{3} \widetilde{g}_{ii} + l_{4} \}^{2} = \sum\limits_{i=1}^{nT}\{ (l_{3} \widetilde{g}_{ii})^{2} + 2 l_{3}l_{4}\widetilde{g}_{ii} + l_{4}^{2} \} \\ =& l_{3}^{2} \sum\limits_{i=1}^{nT} \widetilde{g}_{ii}^{2} + 2 l_{3}l_{4}\text{tr}(\widetilde{G}_{nT}) + nT \cdot l_{4}^{2}\\ =&l_{3}^{2} \| \text{Vec ( diag} \widetilde{G}_{nT} ) \|^{2} + 2 l_{3}l_{4} \text{tr}(\widetilde{G}_{nT}) + nT \cdot l_{4}^{2}, \\ \sum\limits_{i=1}^{nT} v_{i}^{2} =& \sum\limits_{i=1}^{nT} ( l_{1}' b_{i} + l_{2}' r_{i})^{2} = \sum\limits_{i=1}^{nT} ( l_{1}' b_{i}b_{i}' l_{1} + l_{2}' r_{i}r_{i}'l_{2} + l_{1}' b_{i}r_{i}'l_{2} + l_{2}' r_{i}b_{i}' l_{1} )\\ =& l_{1}' \{ \sum\limits_{i=1}^{nT} b_{i}b_{i}' \} l_{1} + l_{2}' \{ \sum\limits_{i=1}^{nT} r_{i}r_{i}' \} l_{2} + 2 l_{1}' \{ \sum\limits_{i=1}^{nT} b_{i}r_{i}' \} l_{2}\\ =& l_{1}' \{ \sum\limits_{i=1}^{nT} X' R_{nT}'(\lambda) e_{i} e_{i}' R_{nT}(\lambda) X \} l_{1} + l_{2}' \{ \sum\limits_{i=1}^{nT} R_{nT}'(\lambda) e_{i} e_{i}' R_{nT}(\lambda) \} l_{2}\\ &+ 2 l_{1}' \{ \sum\limits_{i=1}^{nT} X' R_{nT}'(\lambda) e_{i} e_{i}' R_{nT}(\lambda) \} l_{2}\\ =& l_{1}' X' R_{nT}'(\lambda) R_{nT}(\lambda) X l_{1} + l_{2}' R_{nT}'(\lambda) R_{nT}(\lambda) l_{2} + 2 l_{1}' X' R_{nT}'(\lambda) R_{nT}(\lambda) l_{2}, \\ \sum\limits_{i=1}^{nT} u_{ii} v_{i} =& \sum\limits_{i=1}^{nT} ( l_{3} \widetilde{g}_{ii} + l_{4} )( l_{1}' b_{i} + l_{2}' r_{i}) = \sum\limits_{i=1}^{nT} ( l_{3} \widetilde{g}_{ii} l_{1}' b_{i}) + \sum\limits_{i=1}^{nT} ( l_{3} \widetilde{g}_{ii} l_{2}' r_{i} ) + \sum\limits_{i=1}^{nT} ( l_{4} l_{1}' b_{i}) + \sum\limits_{i=1}^{nT} ( l_{4} l_{2}' r_{i})\\ =& l_{1}' \{ \sum\limits_{i=1}^{nT} \widetilde{g}_{ii} b_{i} \} l_{3} + l_{2}' \{ \sum\limits_{i=1}^{nT} \widetilde{g}_{ii} r_{i} \} l_{3} + l_{1}' \{ \sum\limits_{i=1}^{nT} b_{i} \} l_{4} + l_{2}' \{ \sum\limits_{i=1}^{nT} r_{i} \} l_{4}\\ =& l_{1}' \{ \sum\limits_{i=1}^{nT} \widetilde{g}_{ii} X'R_{nT}'(\lambda) e_{i} \} l_{3} + l_{2}' \{ \sum\limits_{i=1}^{nT} \widetilde{g}_{ii} R_{nT}'(\lambda) e_{i} \} l_{3} + l_{1}' \{ \sum\limits_{i=1}^{nT} X'R'_{nT}(\lambda) e_{i} \} l_{4} + l_{2}' \{ \sum\limits_{i=1}^{nT} R'_{nT}(\lambda) e_{i} \} l_{4}\\ =& l_{1}' X' R_{nT}'(\lambda)( \sum\limits_{i=1}^{nT} \widetilde{g}_{ii} e_{i} ) l_{3} + l_{2}' R_{nT}'(\lambda)( \sum\limits_{i=1}^{nT} \widetilde{g}_{ii} e_{i} ) l_{3} + l_{1}' X' R_{nT}'(\lambda)( \sum\limits_{i=1}^{nT} e_{i} ) l_{4} + l_{2}' R_{nT}'(\lambda)( \sum\limits_{i=1}^{nT} e_{i} ) l_{4}\\ =& l_{1}' X' R_{nT}'(\lambda)\text{Vec(diag} \widetilde{G}_{nT}) l_{3} + l_{2}' R_{nT}'(\lambda) \text{Vec(diag} \widetilde{G}_{nT}) l_{3} + l_{1}' X' R_{nT}'(\lambda)1_{nT} l_{4} + l_{2}' R_{nT}'(\lambda)1_{nT} l_{4}, \end{aligned} \end{equation*} $

其中$1_{nT}$为元素均为$1$的$nT$维列向量.则$Q_{nT}$的方差为

$ \begin{equation*} \begin{aligned} \sigma_{Q_{nT}}^{2} =& 2 \sum\limits_{i=1}^{nT}\sum\limits_{j=1}^{nT}u_{ij}^{2} \sigma^{4} + \sum\limits_{i=1}^{nT}v_{i}^{2} \sigma^{2} + \sum\limits_{i=1}^{nT}\{ u_{ii}^{2} ( \mu_{4} - 3 \sigma^{4} ) + 2 u_{ii}v_{i}\mu_{3} \}\\ =& 2 \sigma^{4} \{ 2 l_{3}l_{4} \text{tr}(\widetilde{G}_{nT}) + nT l_{4}^{2} + l_{3}^{2} \text{tr}(\widetilde{G}_{nT}^{2})\}\\ &+ \sigma^{2} \{ l_{1}' X' R_{nT}'(\lambda) R_{nT}(\lambda) X l_{1} + l_{2}' R_{nT}'(\lambda) R_{nT}(\lambda) l_{2} + 2 l_{1}' X' R_{nT}'(\lambda) R_{nT}(\lambda) l_{2} \}\\ &+ ( \mu_{4} - 3 \sigma^{4} )\{ l_{3}^{2} \|\text{Vec (diag} \widetilde{G}_{nT}) \|^{2} + 2 l_{3}l_{4} \text{tr}(\widetilde{G}_{nT}) + nT \cdot l_{4}^{2} \}\\ &+ 2 \mu_{3} \{ l_{1}' X' R_{nT}'(\lambda)\text{Vec(diag} \widetilde{G}_{nT}) l_{3} + l_{2}' R_{nT}'(\lambda) \text{Vec(diag} \widetilde{G}_{nT}) l_{3} + l_{1}' X' R_{nT}'(\lambda)1_{nT} l_{4}\\ &+ l_{2}' R_{nT}'(\lambda)1_{nT} l_{4} \}\\ =& l'\Sigma_{k+nT+2}l, \end{aligned} \end{equation*} $

根据假设条件(A6),

$ \begin{equation*} \begin{aligned} (nT)^{-1}\sigma_{Q_{nT}}^{2} &=(nT)^{-1} l'\cdot \Sigma_{k+nT+2} \cdot l \geq \lambda_{\min} ( (nT)^{-1}\Sigma_{k+nT+2})\cdot l'l\\ &=\lambda_{\min} ( (nT)^{-1}\Sigma_{k+nT+2})\geq C_{1} >0, \end{aligned} \end{equation*} $

根据引理2, 有

$ \begin{equation*} \frac{ Q_{nT} - E(Q_{nT}) }{ \sigma_{Q_{nT}} } \overset {d}{\rightarrow} N(0, 1). \end{equation*} $

令$E(Q_{nT})=0$, 得到$(3.13)$式.

(3.14)式的证明 类似文献[8]中引理3中(13)式的证明.略.

(3.15)式的证明 由$\omega_{i}(\theta)$的表达式(3.8), 有

$ \begin{equation*} \begin{aligned} \sum\limits_{i=1}^{nT} E \left \| \omega_{i}(\theta)\right \|^{3} \leq & \sum\limits_{i=1}^{nT} E \left \| b_{i}\varepsilon_{i}\right \|^{3} + \sum\limits_{i=1}^{nT} E \left \| r_{i}\varepsilon_{i}\right \|^{3} + \sum\limits_{i=1}^{nT} E |\widetilde{g}_{ii} ( \varepsilon_{i}^{2} - \sigma^{2})\\& + 2 \varepsilon_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \varepsilon_{j}|^{3} + \sum\limits_{i=1}^{nT} E |\varepsilon_{i}^{2} - \sigma^{2}|^{3}. \end{aligned} \end{equation*} $

根据假设条件(A1)-(A5), $b_{i}$的表达式及独立随机变量和的矩不等式, 有

$ \begin{equation*} \begin{aligned} \| b_{i} \widetilde {\varepsilon }_{i} \| &= \sqrt{\left \| b_{i} \widetilde {\varepsilon }_{i}\right \|^{2} } = \{ (\sum\limits_{j=1}^{nT}R_{ij} \widetilde {x}_{j1})^{2} + \cdots + (\sum\limits_{j=1}^{nT}R_{ij} \widetilde {x}_{jk})^{2} \}^{1/2} \cdot | \widetilde {\varepsilon }_{i}| \\ & \leq | \sum\limits_{j=1}^{nT}R_{ij} \widetilde {x}_{j1} + \cdots + \sum\limits_{j=1}^{nT}R_{ij} \widetilde {x}_{jk} | \cdot | \widetilde {\varepsilon }_{i}| \leq \{ \| \widetilde {X}_{1} \| |R_{i1}| + \cdots + \| \widetilde {X}_{nT} \| |R_{i, nT}| \} \cdot | \varepsilon_{i}|\\ & \leq \underset{1\leq i \leq nT} {\max} \| \widetilde {X}_{i} \| \{\sum\limits_{j=1}^{nT} |R_{ij}| \} \cdot | \widetilde {\varepsilon }_{i}| \leq C \underset{1\leq i \leq nT} {\max} \| \widetilde {X}_{i} \| \cdot | \widetilde {\varepsilon }_{i}|, \end{aligned} \end{equation*} $

进而有

$ \begin{equation} \sum\limits_{j=1}^{nT} E \left \| b_{i} \widetilde {\varepsilon }_{i}\right \|^{3} \leq C nT \underset{1 \leq i \leq nT} {\max} \left \| \widetilde { X }_{i} \right \|^{3} E | \widetilde {\varepsilon }_{1}|^{3}=O(nT). \end{equation} $

(4.8)

同理

$ \begin{equation} \sum\limits_{i=1}^{nT} E \| r_{i} \widetilde {\varepsilon }_{i} \|^{3} \leq \sum\limits_{j=1}^{nT} E \left \{\left \| r_{i} \right \|^{3} | \widetilde { \varepsilon }_{i}|^{3} \right \} \leq nT \underset{1\leq i \leq nT} {\max} \left \| r_{i} \right \|^{3} E | \widetilde {\varepsilon }_{1}|^{3}=O(nT), \end{equation} $

(4.9)

$ \begin{equation} \begin{aligned} & \sum\limits_{i=1}^{nT} E |\widetilde{g}_{ii} ( \widetilde { \varepsilon }_{i}^{2} - \sigma^{2}) + 2 \widetilde { \varepsilon }_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde {\varepsilon }_{j}|^{3} \leq C \sum\limits_{i=1}^{nT} E |\widetilde{g}_{ii} (\widetilde { \varepsilon }_{i}^{2} - \sigma^{2})|^{3} + C \sum\limits_{i=1}^{nT} E |2 \widetilde { \varepsilon }_{i} \sum\limits_{j=1}^{i-1} \widetilde{g}_{ij} \widetilde {\varepsilon }_{j}|^{3}\\ \leq & C \sum\limits_{i=1}^{nT} E |\widetilde{g}_{ii} ( \widetilde {\varepsilon }_{i}^{2} - \sigma^{2})|^{3} + C \sum\limits_{i=1}^{nT} E | \widetilde { \varepsilon }_{i}|^{3} \sum\limits_{j=1}^{i-1}| \widetilde{g}_{ij} \widetilde { \varepsilon }_{j}|^{3} + C \sum\limits_{i=1}^{nT} E | \widetilde { \varepsilon }_{i}|^{3} \sum\limits_{j=1}^{i-1}| \widetilde{g}_{ij} \varepsilon_{j}|^{3}\\ \leq & C \sum\limits_{i=1}^{nT} E |\widetilde{g}_{ii} ( \widetilde { \varepsilon }_{i}^{2} - \sigma^{2})|^{3} + C \sum\limits_{i=1}^{nT} E | \widetilde { \varepsilon }_{i}|^{3} \sum\limits_{j=1}^{i-1}| \widetilde{g}_{ij} \widetilde { \varepsilon }_{j}|^{3} + C \sum\limits_{i=1}^{nT} E | \widetilde { \varepsilon }_{i}|^{3} \{\sum\limits_{j=1}^{i-1}E(\widetilde{g}_{ij} \widetilde { \varepsilon }_{j})^{2} \}^{3/2}=O(nT), \end{aligned} \end{equation} $

(4.10)

$ \begin{equation} \begin{aligned} \sum\limits_{j=1}^{nT} E |\widetilde {\varepsilon }_{i}^{2} - \sigma^{2}|^{3} \leq \sum\limits_{j=1}^{nT} C ( E | \widetilde {\varepsilon }_{i})^{2}|^{3} + \sigma^{6}) \leq nT C, \end{aligned} \end{equation} $

(4.11)

由(4.8)-(4.11), 得$\sum\limits_{i=1}^{nT} E |\varepsilon_{i}^{2} - \sigma^{2}|^{3}=O(nT)$.于是$ \sum\limits_{i=1}^{nT} E \| \omega_{i}(\theta) \| ^{3}=O(nT). $再由Markov不等式, 有

$ \begin{equation*} P \left \{ \frac{1}{nT} \sum\limits_{i=1}^{nT} E \| \omega_{i}(\theta) \| ^{3} > a \right \} \leq \frac{1}{a}E \left \{ \frac{1}{nT} \sum\limits_{i=1}^{nT} \| \omega_{i}(\theta) \| ^{3} \right \} \leq \frac{C}{a}, \end{equation*} $

其中$a>0 $, 故$\sum\limits_{i=1}^{nT} \|\omega_{i}(\theta) \| ^{3}=O(nT)$, (3.15)式得证.由此完成了引理3的证明.