0 引言

对函数型数据的分析和处理是统计学的一个热门问题, 被广泛应用到计量经济学、生物医学、心理学及其他领域.考虑半函数部分线性模型

$ Y = {\mathit{\boldsymbol{X}}^{\rm{T}}}\mathit{\boldsymbol{\beta }} + m\left( T \right) + \varepsilon , $

(1)

其中：Y是实值响应变量；X是取值于R^p上的随机向量；m(·)是未知的光滑实函数；T是取值于抽象无穷维空间H上的函数型协变量；ε是随机误差, 期望为零, 方差有限, 且与X和T是独立的.

文献[1]引入了函数线性模型, 文献[2-3]利用函数主成分分析法研究了函数型线性回归模型的估计及预测问题.为了更好地拟合数据, 一些学者开始研究函数型数据半参数模型.文献[4]提出了半函数部分线性模型(SFPLM), 利用函数核光滑方法并结合最小二乘法给出了参数分量和非参数分量的估计, 得到了参数分量的渐近正态性和非参数分量的收敛速度.文献[5]进一步将SFPLM推广应用于时间序列预测问题.文献[6]基于惩罚函数对SFPLM的高维线性回归部分进行了变量选择, 得到变量选择的oracle性质及非参数分量的非参收敛速度.文献[7]利用主成分基函数展开及最小二乘法研究了部分函数线性模型的均方预测误差的收敛速度.文献[8]考虑了纵向函数型数据变系数模型, 给出了模型中系数函数和历史指标函数的估计, 证明了它们的渐近性质.文献[9]研究了纵向函数型数据单指标模型, 将经典单指标模型的最小平均方差估计(MAVE)方法推广到函数型数据情形.文献[10]研究了部分函数变系数模型, 利用主成分基函数展开及局部光滑法得到了系数函数的估计.

文献[11-12]提出的经验似然方法在参数置信域构造方面优于正态逼近方法, 一是它没有涉及方差估计, 二是它没有对置信域的形状施加约束, 而是由数据自行决定形状.文献[13-16]把经验似然方法应用到各种模型.本文利用经验似然方法研究了SFPLM的估计, 构造了模型(1)中未知参数分量β的经验似然比统计量, 证明了此统计量具有渐近χ²分布, 同时, 给出了非参数分量m(·)的估计及其收敛速度.

1 估计方法

假设数据{(Y_i, X_i1, …, X_ip), T_i}_i=1ⁿ是模型(1)中的一组独立同分布的可观测随机样本，即

$ {Y_i} = \mathit{\boldsymbol{X}}_i^{\rm{T}}{\mathit{\boldsymbol{\beta }}_i} + m\left( {{T_i}} \right) + {\varepsilon _i}, $

(2)

其中ε_i是相互独立的模型误差, 且E(ε_i|X_i, T_i)=0, Var(ε_i|X_i, T_i)=σ² < ∞, i=1, 2, …, n.T_i∈H₁, H₁是H的紧子集.X=(X₁, …, X_n)^T, X_i=(X_i1, …, X_ip)^T, Y=(Y₁, …, Y_n)^T.

令d(·, ·)表示定义在无穷空间H上的半度量^[17].因为T为函数型协变量, 我们引入函数型模型的Nadaraya-Watson-type权重${{W}_{nh}}\left( t, {{T}_{i}} \right)=K\left( d\left( t, {{T}_{i}} \right)/h \right)/\sum\limits_{j=1}^{n}{K\left( d\left( t, {{T}_{j}} \right)/h \right)}$，其中：K是定义于R上的核函数；h是窗宽.在式(2)两边求给定T_i的条件期望, 有E(Y_i|T_i)=E(X_i^T|T_i)β+m(T_i), 与式(1)两边分别相减可得

$ {Y_i} - E\left( {{Y_i}\left| {{T_i}} \right.} \right) = {\left[ {{\mathit{\boldsymbol{X}}_i} - E\left( {{\mathit{\boldsymbol{X}}_i}\left| {{T_i}} \right.} \right)} \right]^{\rm{T}}}\mathit{\boldsymbol{\beta }} + {\varepsilon _i}. $

为构造β的经验似然比函数, 引入辅助随机变量${\eta _i}\left( \mathit{\boldsymbol{\beta }} \right) = {{\mathit{\boldsymbol{\tilde X}}}_i}\left( {{{\tilde Y}_i}-\mathit{\boldsymbol{\tilde X}}_i^{\rm{T}}\mathit{\boldsymbol{\beta }}} \right)$, 其中：${{\mathit{\boldsymbol{\tilde X}}}_i} = {\mathit{\boldsymbol{X}}_i}-g\left( {{T_i}} \right);{{\tilde Y}_i} = {Y_i}-\mu \left( {{T_i}} \right);$$g\left( \cdot \right) = E\left( {{\mathit{\boldsymbol{X}}_i}|{T_i}} \right);\mu \left( \cdot \right){\rm{ = }}E\left( {{Y_i}|{T_i}} \right)$.显见E[η_i(β)]=0, 利用这一信息，能够定义一个经验似然比函数${\mathit{\boldsymbol{\hat R}}_i}\left( \mathit{\boldsymbol{\beta }} \right)$.如果β是真参数，则可证明η_i(β)渐近于自由度为p的χ²分布.然而η_i(β)不能直接用于统计推断，因为它包含两个未知函数μ(·)和g(·).

要解决这一问题可在η_i(β)中用两个估计量分别代替它们.利用函数型核估计方法分别定义它们的估计量，$\hat g\left( {{T_i}} \right) = \sum\limits_{j = 1}^n {{W_{nh}}\left( {{T_i}, {T_j}} \right){\mathit{\boldsymbol{X}}_j}, \hat \mu } \left( {{T_i}} \right) = \sum\limits_{j = 1}^n {{W_{nh}}\left( {{T_i}, {T_j}} \right)} {Y_j}$.因此, 可以得到η_i(β)的一个估计量，即${\mathit{\boldsymbol{\hat \eta }}_i}\left( \mathit{\boldsymbol{\beta }} \right) = {\mathit{\boldsymbol{\hat X}}_i}\left( {{{\hat Y}_i}-\mathit{\boldsymbol{\hat X}}_i^{\rm{T}}\mathit{\boldsymbol{\beta }}} \right)$, 其中：${{\mathit{\boldsymbol{\hat X}}}_i}{\rm{ = }}{\mathit{\boldsymbol{X}}_i}-\hat g\left( {{T_i}} \right);{{\hat Y}_i} = {Y_i}-\hat \mu \left( {{T_i}} \right)$.那么，定义β的经验对数似然比函数为$\hat R\left( \mathit{\boldsymbol{\beta }} \right) =-2\max \left\{ {\sum\limits_{i = 1}^n {\log \left( {n{p_i}} \right)|{p_i} \ge 0, \sum\limits_{i = 1}^n {{p_i} = 1}, \sum\limits_{i = 1}^n {{p_i}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} = 0} } \right\}$如果0点在$\left( {{{\mathit{\boldsymbol{\hat \eta }}}_1}\left( \mathit{\boldsymbol{\beta }} \right), \cdots, {{\mathit{\boldsymbol{\hat \eta }}}_n}\left( \mathit{\boldsymbol{\beta }} \right)} \right)$的凸集内部^[11-12], 那么$\hat R\left( \mathit{\boldsymbol{\beta }} \right)$存在唯一的解.利用Lagrange乘子法，可得

$ \hat R\left( \mathit{\boldsymbol{\beta }} \right) = 2\sum\limits_{i = 1}^n {\log \left( {1 + {\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} \right)} , $

(3)

其中λ为Lagrange乘子, 满足

$ \frac{1}{n}\sum\limits_{i = 1}^n {{{\hat \eta }_i}\left( \mathit{\boldsymbol{\beta }} \right)/\left( {1 + {\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right) = 0} \right)} , $

(4)

极大化$\hat R\left( \mathit{\boldsymbol{\beta }} \right)$，可得β的极大经验似然估计${\mathit{\boldsymbol{\hat \beta }}}$, 由${\mathit{\boldsymbol{\hat \beta }}}$可得m(T)的估计为$\hat m\left( T \right)-\sum\limits_{i = 1}^n {{W_{nh}}\left( {T, {T_i}} \right)\left( {{Y_i}-\mathit{\boldsymbol{X}}_i^{\rm{T}}\mathit{\boldsymbol{\hat \beta }}} \right)} $.

2 主要结果

为叙述方便, 始终假设c表示一不依赖于n的正的常数, c每次出现可以取不同的值.引入记号:

$ \mathit{\boldsymbol{B}}\left( {t,h} \right) = \left\{ {t' \in H:d\left( {t,t'} \right) \le h} \right\},{q_{ij}} = {X_{ij}} - E\left( {{X_{ij}}\left| {{T_i}} \right.} \right),{\mathit{\boldsymbol{q}}_i} = {\left( {{q_{i1}}, \cdots ,{q_{ip}}} \right)^{\rm{T}}}, $

$ {g_j}\left( t \right) = E\left( {{X_{ij}}\left| {{T_i}} \right. = t} \right),{{\bar g}_j}\left( t \right) = {g_j}\left( t \right) - {{\hat g}_j}\left( t \right),{{\hat g}_j}\left( t \right) = \sum\limits_{i = 1}^n {{W_{nh}}\left( {t,{T_i}} \right){X_{ij}},} $

$ {{\tilde g}_j}\left( {{T_i}} \right) = E\left( {{X_{ij}}\left| {{T_i}} \right.} \right) - \sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right)E\left( {{X_{kj}}\left| {{T_k}} \right.} \right)} ,{{\tilde g}_h}\left( {{T_i}} \right) = \\E\left( {{\mathit{\boldsymbol{X}}_i}\left| {{T_i}} \right.} \right) - \sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right)E\left( {{\mathit{\boldsymbol{X}}_k}\left| {{T_k}} \right.} \right),} $

$ {{\tilde m}_h}\left( T \right) = m\left( T \right) - \sum\limits_{j = 1}^n {{W_{nh}}\left( {T,{T_j}} \right)m\left( {{T_j}} \right)} ,{{\bar \varepsilon }_j} = \sum\limits_{j = 1}^n {{W_{nh}}\left( {T,{T_j}} \right){\varepsilon _j}} . $

为了得到$\hat R\left( \mathit{\boldsymbol{\beta }} \right)$的渐近分布，需要以下条件：

(C1) K满足Lipschitz条件，支撑为[0, 1]，并且$\exists c$满足∀u∈[0, 1], -K′(u)>c > 0.

(C2)存在一个(0，∞)上的正值函数ϕ和正值c₀、c₁、c₂使得

$ \int_0^1 {\phi \left( {hs} \right){\rm{d}}s} > {c_0}\phi \left( h \right),{c_1}\phi \left( h \right) \le p\left( {T \in \mathit{\boldsymbol{B}}\left( {t,h} \right)} \right) \le {c_2}\phi \left( h \right). $

(C3)存在常数c > 0，α > 0，$\forall f \in \left\{ {m, {g_1}, \cdots, {g_p}} \right\}$, 都有|f(u)-f(v)|≤cd^α(u, v).

(C4) E|ε₁|^r+E|q₁₁|^r+…+E|q_1p|^r < ∞, r≥3.

(C5)σ²=Var(ε)>0，B=E(q₁q₁^T)是正定阵.

条件(C1)和(C2)是研究函数型非参数模型的常见条件^[17]，条件(C3)~(C5)是研究部分线性模型的常见条件^[18].下面的定理给出了经验对数似然比渐近于χ²分布.

定理1  假设条件(C1)~(C5)成立, nh^4α→0, 当n→∞, 并且当n足够大时，存在常数b > 0, 使得(2/r)+b > 1/2, 有ϕ(h)≥n^(2/r)+b-1/(log n)², 则$\hat R\left( \mathit{\boldsymbol{\beta }} \right)\xrightarrow{L}\chi _{p}^{2}$, 其中“$\xrightarrow{L}$”表示依分布收敛.

基于定理1, 可得参数分量β的具有渐近置信水平1-α的置信域I_α(β), 即对任意0 < α < 1, 存在C_α使得P(χ_P² > C_α)=α, 则${I_\alpha }\left( \mathit{\boldsymbol{\beta }} \right) = \left\{ {\mathit{\boldsymbol{\beta }} \in {{\bf{R}}^p}|\hat R\left( \mathit{\boldsymbol{\beta }} \right) \le {C_\alpha }} \right\}$.极大化$\hat R\left( \mathit{\boldsymbol{\beta }} \right)$，可得β的极大经验似然估计${\mathit{\boldsymbol{\hat \beta }}}$，下面给出${\mathit{\boldsymbol{\hat \beta }}}$的渐近正态性.

定理2  假设定理1中的条件成立，则$ \sqrt{n}\left( {\mathit{\boldsymbol{\hat \beta }}-\mathit{\boldsymbol{\beta }}} \right)\xrightarrow{L}N\left( {0, {\sigma ^2}{\mathit{\boldsymbol{B}}^{-1}}} \right)$.进一步可给出估计量$\hat m\left( T \right)$的收敛速度.

定理3  在定理1的条件下, 有$\mathop {\sup }\limits_{T \in {H_1}} \left| {\hat m\left( T \right)-m\left( T \right)} \right| = O\left( {{h^\alpha }} \right) + O\left( {\sqrt {\frac{{\log \;n}}{{n\phi \left( h \right)}}} } \right)$.

3 定理证明

引理1  假设定理1中的条件成立, 若β是参数的真值, 则有$\frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} \xrightarrow{\rm{L}}N\left( {0, {\sigma ^2}\mathit{\boldsymbol{B}}} \right)$.

证明  由于${\hat \eta _i}\left( \mathit{\boldsymbol{\beta }} \right) = {\mathit{\boldsymbol{\hat X}}_i}\left( {{{\hat Y}_i}-\mathit{\boldsymbol{\hat X}}_i^{\rm{T}}\mathit{\boldsymbol{\beta }}} \right)$, 经计算可得

$ \begin{array}{l} \frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} = \frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat X}}}_i}\left( {{{\hat Y}_i} - {{\mathit{\boldsymbol{\hat X}}}_i}\mathit{\boldsymbol{\beta }} } \right)} =\\ \frac{1}{{\sqrt n }}\left( {{{\mathit{\boldsymbol{\hat X}}}_1}, \cdots ,{{\mathit{\boldsymbol{\hat X}}}_n}} \right)\left( {\begin{array}{*{20}{c}} {{{\hat Y}_1} - \mathit{\boldsymbol{\hat X}}_1^{\rm{T}}\beta }\\ \vdots \\ {{{\hat Y}_n} - \mathit{\boldsymbol{\hat X}}_n^{\rm{T}}\mathit{\boldsymbol{\beta }} } \end{array}} \right) =\\\frac{1}{{\sqrt n }}{{\mathit{\boldsymbol{\hat X}}}^{\rm{T}}}\left( {\hat Y - \mathit{\boldsymbol{\hat X}}\mathit{\boldsymbol{\beta }} } \right) = \\ {n^{ - 1/2}}\left( {{{\mathit{\boldsymbol{\hat X}}}^{\rm{T}}}\hat Y - {{\mathit{\boldsymbol{\hat X}}}^{\rm{T}}}\mathit{\boldsymbol{\hat X\mathit{\boldsymbol{\beta }} }}} \right) = \\{n^{ - 1/2}}\left\{ {\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat X}}}_i}{{\tilde m}_h}\left( {{T_i}} \right)} - \sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat X}}}_i}\left[ {\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){\varepsilon _k}} } \right]} + \sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat X}}}_i}{\varepsilon _i}} } \right\} \equiv \\ {n^{ - 1/2}}\left( {{\mathit{\boldsymbol{S}}_{n1}} - {\mathit{\boldsymbol{S}}_{n2}} + {\mathit{\boldsymbol{S}}_{n3}}} \right), \end{array} $

其中：${\tilde m_h}\left( {{T_i}} \right) = m\left( {{T_i}} \right)-\sum\limits_{j = 1}^n {{W_{nh}}\left( {{T_i}, {T_j}} \right)m\left( {{T_j}} \right);{{\hat X}_{ij}} = {{\tilde g}_j}\left( {{T_i}} \right) + {q_{ij}}-\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i}, {T_k}} \right){q_{kj}}.} }$$i = 1, \cdots , n, j = 1, \cdots, p$.令S_n1j表示S_n1的第j个分量, 则有

$ {\mathit{\boldsymbol{S}}_{n1j}} = \sum\limits_{i = 1}^n {{{\hat X}_{ij}}{{\tilde m}_h}\left( {{T_i}} \right)} = \sum\limits_{i = 1}^n {{{\tilde g}_j}\left( {{T_i}} \right){{\tilde m}_h}\left( {{T_i}} \right)} + \sum\limits_{i = 1}^n {{q_{ij}}{{\tilde m}_h}\left( {{T_i}} \right)} - \sum\limits_{i = 1}^n \\{\left[ {\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){q_{kj}}} } \right]{{\tilde m}_h}\left( {{T_i}} \right)} . $

由文献[4]可知$\mathop {\max }\limits_{1 \le j \le p} \mathop {\max }\limits_{1 \le i \le n} \left| {{{\tilde g}_j}\left( {{T_i}} \right)} \right| = O\left( {{h^\alpha }} \right) + O\left( {\sqrt {\frac{{\log \;n}}{{n\phi \left( h \right)}}} } \right)a.\;s.$；$\mathop {\max }\limits_{1 \le i \le n} \;{\tilde m_h}\left( {{T_i}} \right) = O\left( {{h^\alpha }} \right) + O\left( {\sqrt {\frac{{\log \;n}}{{n\phi \left( h \right)}}} } \right)a.\;s.$，

由文献[19]可知$\left| {\sum\limits_{i = 1}^n {{q_{ij}}} } \right| = {O_p}\left( {\sqrt n \log \;n} \right)$.故可得：

$ \begin{array}{l} \left| {\sum\limits_{i = 1}^n {{{\tilde g}_j}\left( {{T_i}} \right){{\tilde m}_h}\left( {{T_i}} \right)} } \right| \le n\mathop {\max }\limits_{1 \le j \le p} \mathop {\max }\limits_{1 \le i \le n} \left| {{{\tilde g}_j}\left( {{T_i}} \right)} \right| \cdot \mathop {\max }\limits_{1 \le i \le n} \left| {{{\tilde m}_h}\left( {{T_i}} \right)} \right| = \\n{\left[ {O\left( {{h^\alpha }} \right) + O\left( {\sqrt {\frac{{\log n}}{{n\varphi \left( h \right)}}} } \right)} \right]^2}a.s. = \\ O\left( {2n{h^{2\alpha }} + 2\phi {{\left( h \right)}^{ - 1}}\log n} \right)a.s. = O\left( {{n^{1/2}}} \right)a.s., \end{array} $

(5)

$ \begin{array}{l} \left| {\sum\limits_{i = 1}^n {{q_{ij}}{{\tilde m}_h}\left( {{T_i}} \right)} } \right| \le \mathop {\max }\limits_{1 \le i \le n} \left| {{{\tilde m}_h}\left( {{T_i}} \right)} \right| \cdot \left| {\sum\limits_{i = 1}^n {{q_{ij}}} } \right| = \left[ {O\left( {{h^\alpha }} \right) +\\ O\left( {\sqrt {\frac{{\log n}}{{n\varphi \left( h \right)}}} } \right)} \right] \cdot {O_p}\left( {\sqrt n \log n} \right) = \\ {O_p}\left( {{n^{1/2}}{h^\alpha }\log n + \phi {{\left( h \right)}^{ - 1/2}}\log n} \right) = {O_p}\left( 1 \right). \end{array} $

(6)

由文献[4]可知$\mathop {\max }\limits_{1 \le i \le n} \left| {\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i}, {T_k}} \right){q_{kj}}} } \right| = O\left( {{{\left( {n\phi \left( h \right)} \right)}^{{\rm{-}}1{\rm{/}}2}}\log n} \right)a.\;s.$因此有，

$ \begin{array}{l} \left| {\sum\limits_{i = 1}^n {\left[ {\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){q_{kj}}} } \right]{{\tilde m}_h}\left( {{T_i}} \right)} } \right| \le n\mathop {\max }\limits_{1 \le i \le n} \left| {\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){q_{kj}}} } \right| \cdot \mathop {\max }\limits_{1 \le i \le n} \left| {{{\tilde m}_h}\left( {{T_i}} \right)} \right| = \\ O\left( {n{h^{\alpha }}{{\left( {n\phi \left( h \right)} \right)}^{ - 1/2}}\log n} \right) + O\left( {n{{\left( {n\phi \left( h \right)} \right)}^{ - 1/2}}\log n\sqrt {\frac{{\log n}}{{n\varphi \left( h \right)}}} } \right) = \\ O\left( {{n^{ 1/2}}{h^\alpha }\phi {{\left( h \right)}^{ - 1/2}}\log n + \phi {{\left( h \right)}^{ - 1}}{{\left( {\log n} \right)}^{3/2}}} \right) = O\left( {{n^{1/2}}} \right)a.s., \end{array} $

(7)

由式(5)~(7)可知S_n1j=O_p(n^1/2).

令S_n2j表示S_n2的第j个分量, 类似可证S_n2j=O_p(n^1/2)，则，

$ {\mathit{\boldsymbol{S}}_{n3}} = \sum\limits_{i = 1}^n {\left[ {{{\tilde g}_h}\left( {{T_i}} \right)} \right]{\varepsilon _i}} + \sum\limits_{i = 1}^n {{q_i}{\varepsilon _i}} - \sum\limits_{i = 1}^n {\left[ {\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){q_{kj}}} } \right]{\varepsilon _i}} \equiv {\mathit{\boldsymbol{A}}_{n1}} + {\mathit{\boldsymbol{A}}_{n2}} + {\mathit{\boldsymbol{A}}_{n3}}, $

A_nij表示A_ni的第j个分量, 则有

$ \begin{array}{l} {\mathit{\boldsymbol{A}}_{n1j}} = \sum\limits_{i = 1}^n {\left[ {{{\tilde g}_j}\left( {{T_i}} \right)} \right]{\varepsilon _i}} \le \mathop {\max }\limits_{1 \le j \le p} \mathop {\max }\limits_{1 \le i \le n} \left| {{{\tilde g}_j}\left( {{T_i}} \right)} \right| \cdot \left| {\sum\limits_{i = 1}^n {{\varepsilon _i}} } \right| =\\ \left[ {O\left( {{h^\alpha }} \right) + O\left( {\sqrt {\frac{{\log n}}{{n\phi \left( h \right)}}} } \right)} \right] \cdot {O_p}\left( {\sqrt n \log n} \right) = \\ {O_p}\left( {{n^{1/2}}{h^\alpha }\log n} \right) + {O_p}\left( {\phi {{\left( h \right)}^{ - 1/2}}{{\left( {\log n} \right)}^{3/2}}} \right) = {O_p}\left( 1 \right), \end{array} $

$ \begin{array}{l} {\mathit{\boldsymbol{A}}_{n3j}} = \sum\limits_{i = 1}^n {\left[ {\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){q_{kj}}} } \right]{\varepsilon _i}} \le \mathop {\max }\limits_{1 \le i \le n} \left| {\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){q_{kj}}} } \right| \cdot \left| {\sum\limits_{i = 1}^n {{\varepsilon _i}} } \right| = \\ O\left( {{n^{ - 1/2}}\phi {{\left( h \right)}^{ - 1/2}}\log n} \right){O_p}\left( {\sqrt n \log n} \right) = {O_p}\left( {\phi {{\left( h \right)}^{ - 1/2}}{{\left( {\log n} \right)}^{3/2}}} \right) = {O_p}\left( 1 \right), \end{array} $

由中心极限定理可知${n^{-1/2}}{\mathit{\boldsymbol{A}}_{n2}} = {n^{-1/2}}\sum\limits_{i = 1}^n {{\mathit{\boldsymbol{q}}_i}} {\varepsilon _i}\xrightarrow{L}N\left( {0, \sigma _\varepsilon ^2\mathit{\boldsymbol{B}}} \right)$.引理得证.

引理2  假设定理1中的条件成立, 若β是参数的真值, 则有$\frac{1}{n}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)\mathit{\boldsymbol{\hat \eta }}_i^{\rm{T}}} \left( \mathit{\boldsymbol{\beta }} \right)\xrightarrow{p}{\sigma ^2}\mathit{\boldsymbol{B}}$.

证明  由于${\mathit{\boldsymbol{\hat \eta }}_i}\left(\mathit{\boldsymbol{\beta }} \right) = {\mathit{\boldsymbol{\hat X}}_i}\left( {{{\hat Y}_i}- {{\mathit{\boldsymbol{\hat X}}}_i}\mathit{\boldsymbol{\beta }}} \right) = \left[{{\mathit{\boldsymbol{q}}_i} + {{\tilde g}_h}\left( {{T_i}} \right)-\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i}, {T_k}} \right){\mathit{\boldsymbol{q}}_k}} } \right]$$\left[{{\varepsilon _i}-{{\bar \varepsilon }_i} + {{\tilde m}_h}\left( {{T_i}} \right)} \right] = {\mathit{\boldsymbol{q}}_i}{\varepsilon _i} + $$\left\{ {{\mathit{\boldsymbol{q}}_i}\left[{{{\tilde m}_h}\left( {{T_i}} \right)-{{\bar \varepsilon }_i}} \right]} \right.$$+ \left[{{{\tilde g}_h}\left( {{T_i}} \right)-\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i}, {T_k}} \right){\mathit{\boldsymbol{q}}_k}} } \right]{\varepsilon _i}$$+ \left[{{{\tilde g}_h}\left( {{T_i}} \right)-\sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i}, {T_k}} \right){\mathit{\boldsymbol{q}}_k}} } \right]$$\left. {\left[{{{\tilde m}_h}\left( {{T_i}} \right)-{{\bar \varepsilon }_i}} \right]} \right\}$$\equiv {\mathit{\boldsymbol{U}}_i}{\rm{ + }}{\mathit{\boldsymbol{D}}_i}$, 其中: ${\mathit{\boldsymbol{U}}_i} = {\mathit{\boldsymbol{q}}_i}{\varepsilon _i}$;

$ \begin{array}{l} {\mathit{\boldsymbol{D}}_i} = {\mathit{\boldsymbol{q}}_i}\left[ {{{\tilde m}_h}\left( {{T_i}} \right) - {{\bar \varepsilon }_i}} \right] + \left[ {{{\tilde g}_h}\left( {{T_i}} \right) - \sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){\mathit{\boldsymbol{q}}_k}} } \right]{\varepsilon _i} +\\\left[ {{{\tilde g}_h}\left( {{T_i}} \right) - \sum\limits_{k = 1}^n {{W_{nh}}\left( {{T_i},{T_k}} \right){\mathit{\boldsymbol{q}}_k}} } \right]\left[ {{{\tilde m}_h}\left( {{T_i}} \right) - {{\bar \varepsilon }_i}} \right];\\ \frac{1}{n}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)\mathit{\boldsymbol{\eta }}_i^{\rm{T}}\left( \mathit{\boldsymbol{\beta }} \right)} = \frac{1}{n}\sum\limits_{i = 1}^n {{\mathit{\boldsymbol{U}}_i}\mathit{\boldsymbol{U}}_i^{\rm{T}}} +\\ \frac{1}{n}\sum\limits_{i = 1}^n {{\mathit{\boldsymbol{D}}_i}\mathit{\boldsymbol{D}}_i^{\rm{T}}} + \frac{1}{n}\sum\limits_{i = 1}^n {{\mathit{\boldsymbol{U}}_i}\mathit{\boldsymbol{D}}_i^{\rm{T}}} + \frac{1}{n}\sum\limits_{i = 1}^n {{\mathit{\boldsymbol{D}}_i}\mathit{\boldsymbol{U}}_i^{\rm{T}}} \equiv {J_1} + {J_2} + {J_3} + {J_4}. \end{array} $

结合引理1, 并由大数定律可得${{J}_{1}}\xrightarrow{P}{{\sigma }^{2}}B$.下面证明${{J}_{1}}\xrightarrow{P}0$.令J_{2, rs}为J₂的第(r, s)个分量, D_ir为D_i的第r个分量, D_is为D_i的第s个分量, 则利用Cauchy-Schwarz不等式可得$\left| {{J}_{2, rs}} \right|\le {{\left( \frac{1}{n}\sum\limits_{i=1}^{n}{D_{ir}^{2}} \right)}^{1/2}}{{\left( \frac{1}{n}\sum\limits_{i=1}^{n}{D_{is}^{2}} \right)}^{1/2}}$.类似引理1的证明可得${{n}^{-1}}\sum\limits_{i=1}^{n}{D_{ir}^{2}}={{O}_{p}}\left( 1 \right)$, ${{n}^{-1}}\sum\limits_{i=1}^{n}{D_{is}^{2}}-{{O}_{p}}\left( 1 \right)$, 进而有${{j}_{2}}\xrightarrow{P}0$, 类似的讨论, 可以证明${{j}_{v}}\xrightarrow{P}0, v=3, 4$.本引理得证.

引理3  在定理1的条件下，若β是真实参数，则$\mathop {\max }\limits_{1 \le i \le n} \left\| {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} \right\| = {O_p}\left( {{n^{1/2}}} \right)$, $\left\| \mathit{\boldsymbol{\lambda }} \right\| = {O_p}\left( {{n^{-1/2}}} \right)$.

证明  利用引理1的证明方法及文献[11]可证得上式.

定理1的证明 由式(4)可得

$ 0 = \sum\limits_{i = 1}^n {\frac{{{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)}}{{1 + {\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)}}} = \sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} - \sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)\mathit{\boldsymbol{\eta }}_i^{\rm{T}}\left( \mathit{\boldsymbol{\beta }} \right)\lambda } + \sum\limits_{i = 1}^n {\frac{{{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right){{\left[ {{\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} \right]}^2}}}{{1 + {\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)}}} . $

由引理1~3可以证得

$ \sum\limits_{i = 1}^n {\left[ {{\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}{{\left( \mathit{\boldsymbol{\beta }} \right)}^2}} \right]} = \sum\limits_{i = 1}^n {{\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right) + {O_p}\left( 1 \right)} ,\mathit{\boldsymbol{\lambda }} =\\ {\left[ {\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)\mathit{\boldsymbol{\eta }}_i^{\rm{T}}\left( \mathit{\boldsymbol{\beta }} \right)} } \right]^{ - 1}}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} + {O_p}\left( {{n^{ - 1/2}}} \right). $

对式(5)进行Taylor展开, 并结合引理1~3可得$- 2\mathit{\boldsymbol{\hat R}}\left( \mathit{\boldsymbol{\beta }} \right) = 2\sum\limits_{i = 1}^n {\left\{ {{\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)- \frac{1}{2}{{\left[{{\mathit{\boldsymbol{\lambda }}^{\rm{T}}}{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} \right]}^2}} \right\} + {O_p}\left( 1 \right)} $.

结合上式, 并经过简单计算得$\mathit{\boldsymbol{\hat R}}\left( \mathit{\boldsymbol{\beta }} \right){\rm{ = }}\frac{{{{\left[{\frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} } \right]}^{\rm{T}}}\left[{\frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} } \right]}}{{{n^{ -1}}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)\mathit{\boldsymbol{\hat \eta }}_i^{\rm{T}}\left( \mathit{\boldsymbol{\beta }} \right)} }}$.

由引理1及2可证明定理1.

定理2的证明 利用文献[20]的证法可得$\mathit{\boldsymbol{\hat \beta }}- \mathit{\boldsymbol{\beta }} = {{\mathit{\boldsymbol{\hat B}}}^{- 1}}\left[{{n^{-1}}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat \eta }}}_i}\left( \mathit{\boldsymbol{\beta }} \right)} } \right] + {O_p}\left( {{n^{ -1/2}}} \right)$.其中$\mathit{\boldsymbol{\hat B}} = {n^{-1}}\sum\limits_{i = 1}^n {{{\mathit{\boldsymbol{\hat X}}}_i}} \mathit{\boldsymbol{\hat X}}_i^{\rm{T}}$, 由文献[4]可知，${\mathit{\boldsymbol{\hat B}}}\xrightarrow{P}\mathit{\boldsymbol{B}}$, 进而由引理1及Slutsky定理, 即可完成本定理的证明.

定理3的证明  计算可得

$ \begin{array}{l} \hat m\left( T \right) = \sum\limits_{i = 1}^n {{W_{nh}}\left( {T,{T_i}} \right)\left[ {{Y_i} - \mathit{\boldsymbol{X}}_i^{\rm{T}}\mathit{\boldsymbol{\hat \beta }}} \right]} = \sum\limits_{i = 1}^n {{W_{nh}}\left( {T,{T_i}} \right)\left( {{\mathit{\boldsymbol{X}}_i}\mathit{\boldsymbol{\beta }} +\\ m\left( {{T_i}} \right) + {\varepsilon _i} - {\mathit{\boldsymbol{X}}_i}\mathit{\boldsymbol{\hat \beta }}} \right)} = \\ \sum\limits_{i = 1}^n {{W_{nh}}\left( {T,{T_i}} \right)\left[ {m\left( {{T_i}} \right) + {\varepsilon _i}} \right]} - \sum\limits_{i = 1}^n {{W_{nh}}\left( {T,{T_i}} \right)\mathit{\boldsymbol{X}}_i^{\rm{T}}\left( {\mathit{\boldsymbol{\hat \beta }} - \mathit{\boldsymbol{\beta }}} \right)} . \end{array} $

因此$\mathop {\sup }\limits_{t \in {H_1}} \left| {\hat m\left( T \right)m\left( T \right)} \right| \le $$\mathop {\sup }\limits_{T \in {H_1}} \left| {\sum\limits_{i = 1}^n {{W_{nh}}\left( {T, {T_i}} \right)\left[{m\left( {{T_i}} \right) + {\varepsilon _i}} \right] -m\left( T \right)} } \right|$$\mathop {\sup }\limits_{T \in {H_1}} \left\| {\sum\limits_{i = 1}^n {{W_{nh}}\left( {T, {T_i}} \right)\mathit{\boldsymbol{X}}_i^{\rm{T}}} } \right\| \cdot \left\| {\mathit{\boldsymbol{\hat \beta }}-\mathit{\boldsymbol{\beta }}} \right\|$.

由文献[4]中定理2可得本定理证明.