半参数变系数模型在金融, 经济等数据分析凭借良好的解释能力得到广泛应用. 具有非参数模型的灵活和参数模型易解释的优点, 从而成为近年来用以研究变量间非线性关系的最有效模型之一. 经济学、环境科学、地理学、流行病学等领域的数据都存在空间相依性或空间异质性, 这将导致不同空间尺度的响应变量观测值之间的内生相互作用效应. 空间相关性的存在就会让模型变得比较复杂, 所以如何对模型存在的空间相关性进行有效的处理成为后面的考虑对象. 半参数变系数空间自回归模型就是一个非常重要的模型, 由于其灵活性和可解释性, 越来越受到关注. 例如, Wei等^[1]利用剖面极大似然方法研究了半参数变系数空间自回归模型的统计推断；Luo等^[2]提出的半参数变系数空间自回归模型的经验似然推断；Liang等^[3]考虑了在固定效应和时变系数下半参数空间自回归面板数据模型的推断；以及Su^[4]研究的半参数空间自回归模型的GMM估计.

空间误差未被纳入模型时, 可能导致参数估计的偏误和不一致性. 针对此类问题Zhang等^[5]研究了具有空间误差和未知异方差的部分线性可加高阶空间自回归下的统计推断；Su和Jin^[6]提出的半参数滞后回归模型, 在刻画空间滞后因变量引起的空间相关性特征方面意义鲜明；陈建宝^[7]的半参数变系数空间误差回归模型在捕捉非线性特征和处理空间效应两方面的表现都就较为优越. 本文考虑带空间误差的半参数变系数模型：

$ \left\{ {\begin{array}{*{20}{l}} {{Y_i} = Z_i^{\rm T}\beta + X_i^{\rm T}\alpha \left( {{U_i}} \right) + {m_i}, }\\ {{m_i} = {\rho _0}\sum\limits_{j = 1}^n {{w_{i, j}}{m_j} + {\varepsilon _i}} , } \end{array}{\rm{ }}} \right. $

(1)

其中$ Y $是响应变量, $ Z \in {R^P}, X \in {R^q}, U $是协变量, $ \beta = {({\beta _1}, {\beta _2}, \cdot \cdot \cdot , {\beta _p})^{\rm T}} $是$ p $维的未知参数向量, $ \alpha \left( \cdot \right){\rm{ = }}{\left( {{\alpha _{\rm{1}}}\left( \cdot \right), {\alpha _{\rm{2}}}\left( \cdot \right), \cdot \cdot \cdot , {\alpha _q}\left( \cdot \right)} \right)^{\rm T}} $是q维的未知函数向量, 误差$ {\varepsilon _i} $是均值为0, 方差为$ {\sigma ^2} $的独立同分布的随机变量. $ {\rho _0} $是真实的空间相关相关系数, 反映的是误差项的自相关关系. $ {m_i} $是具有空间误差的误差项.

自从$ Owen $^[8]提出经验似然法进行构造置信区间以来, 该方法受到了广泛的关注. 经验似然是一种基于样本数据构建的似然函数, 为样本提供一个权重分配, 使得在样本的约束条件下, 通过最大化似然函数, 能够反映样本的分布特性. 基于经验似然的置信区域不需要对区域形状施加先验约束. 众所周知, 高维数据分析在许多当代统计研究中经常出现. 各种惩罚方法已被开发用于变量选择. Tang和Leng(2010)^[9]首次引入惩罚经验似然(PEL), 用于分析多变量的平均向量具有发散参数的线性模型的分析和回归系数. Chen等^[10]考虑在纵向数据下高维广义线性模型的惩罚经验似然；He等^[11]利用PEL对带固定效应含误差变量半参数高维面板数据模型进行降维和参数估计. Wang等^[12]利用惩罚经验方法研究GINAR(p)模型. 本文在陈建宝^[7]的半参数变系数空间误差回归模型的基础之上构造了参数以及非参数部分的估计, 然后又对参数部分的随机辅助向量构建了经验似然比函数, 最后通过惩罚经验似然的方法进行了变量选择, 并证明在给出的条件下, 惩罚经验似然估计具有Oracle特征且在零假设下服从渐近卡方分布.

将模型(1)写成矩阵形式如下：

$ \left\{ {\begin{array}{*{20}{l}} {Y = {Z^{\rm T}}\beta + {X^{\rm T}}\alpha \left( u \right) + m, }\\ {m = \rho Wm + \varepsilon , } \end{array}{\rm{ }}i = 1, \cdots , n} \right., $

(2)

第一步, 利用局部多项式的方法对变系数函数$ \alpha ({u_i}) $进行估计, 在任给的$ U $邻域内一点$ u $, 有

$ {\alpha _j}\left( {{U_i}} \right) \approx {\alpha _j}(u) + {\alpha '_j}(u)\left( {{U_i} - u} \right) = {a_j} + {b_j}\left( {{U_i} - u} \right), j = 1, 2, \cdots , q $

从而系数函数可以通过极小化下式估计

$ \frac{1}{n}\mathop \sum \limits_{i = 1}^n {\left\{ {{Y_i} - {Z_i}^{\rm T}\beta - X_i^{\rm T}\left[ {{a_j} + {b_j}\left( {{U_i} - u} \right)} \right]} \right\}^2}{K_h}\left( {{U_i} - u} \right), $

(3)

其中$ {K_h}\left( \right) = {{K\left( {{ \mathord{\left/ {\vphantom { h}} \right. } h}} \right)} \mathord{\left/ {\vphantom {{K\left( {{ \mathord{\left/ {\vphantom { h}} \right. } h}} \right)} h}} \right. } h} $是核函数, $ h $是带宽, 为了书写方便, 引入以下记号

$ X = \left( {\begin{array}{*{20}{l}} {{X_1}}&{{{\left( {{U_1} - u} \right)X_1^{\rm T}} \mathord{\left/ {\vphantom {{\left( {{U_1} - u} \right)X_1^{\rm T}} h}} \right. } h}}\\ \vdots & \vdots \\ {{X_n}}&{{{\left( {{U_n} - u} \right)X_n^{\rm T}} \mathord{\left/ {\vphantom {{\left( {{U_n} - u} \right)X_n^{\rm T}} h}} \right. } h}} \end{array}} \right), $

$ K = diag\left( {{K_h}\left( {{U_1} - u} \right), {K_h}\left( {{U_2} - u} \right), \cdots , {K_h}\left( {{U_n} - u} \right)} \right) $, $ M = \left( {{X_1}\alpha \left( {{U_1}} \right), {X_2}\alpha \left( {{U_2}} \right), \cdots , {X_n}\alpha \left( {{U_n}} \right)} \right) $, 则(3)式的解为

$ \left( {\begin{array}{*{20}{c}} {{{\hat a}_j}}\\ {h{{\hat b}_j}} \end{array}} \right) = {\left( {{X^{\rm T}}KX} \right)^{ - 1}}{X^{\rm T}}K\left( {Y - {Z^{\rm T}}\beta } \right), $

记$ S = e_1^{\rm T}{\left( {{X^{\rm T}}KX} \right)^{ - 1}}{X^{\rm T}}K $, $ {e_1} = {\left( {1, 0} \right)^{\rm T}} $则

$ \tilde M = {X^{\rm T}}\hat \alpha \left( u \right) = S\left( {Y - {Z^{\rm T}}\beta } \right), $

(4)

第二步, 估计$ \beta $, 令$ H = {\left( {{H_1}, {H_2}, \cdots , {H_n}} \right)^{\rm T}} $, $ \tilde Y = \left( {I - S} \right)Y $, $ \tilde Z = \left( {I - S} \right)Z $, $ H $是工具变量, 满足$ E\left( {{H_n}^{\rm T}m} \right) = 0 $, 我们使用$ G = \left( {I - S} \right)H $作为$ \tilde Z $的工具变量矩阵, 来定义$ \beta $的最大经验似然比估计

$ E\left( {{G^{\rm T}}m} \right) = E\left[ {{G^{\rm T}}\left( {Y - {Z^{\rm T}}\beta - \tilde M} \right)} \right] = E\left[ {{G^{\rm T}}\left( {\tilde Y - {{\tilde Z}^{\rm T}}\beta } \right)} \right] = 0, $

(5)

引入$ \beta $的辅助随机向量

$ {\xi _i}\left( \beta \right) = G\left( \beta \right)\left( {{{\tilde Y}_i} - {{\tilde Z}_i}^{\rm T}\beta } \right), $

(6)

定义$ \beta $的经验对数似然比函数：

$ L\left( \beta \right) = - \max \left\{ {\sum\limits_{i = 1}^n {\log \left( {n{p_i}} \right)\left| {{p_i} \ge 0, \sum\limits_{i = 1}^n {{p_i}{\xi _i}\left( \beta \right)} = 0, \sum\limits_{i = 1}^n {{p_i} = 1} } \right.} } \right\}, $

由Lagrange乘数法$ L\left( \beta \right) $可以表示为

$ L\left( \beta \right) = \sum\limits_{i = 1}^n {\log \left\{ {1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right\}}, $

(7)

其中$ \lambda $是下面方程的解：

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

(8)

为了达到变量选择的目的, 在经验对数似然比的基础上添加惩罚函数, 得到

$ {L_p}\left( \beta \right) = \sum\limits_{i = 1}^n {\log \left\{ {1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right\}} + n\sum\limits_{j = 1}^p {{p_\tau }} \left( {\left| {{\beta _j}} \right|} \right), $

其中$ {p_\tau }\left( \cdot \right) $是惩罚函数, 他的一阶导满足

$ {p_\tau }^\prime \left( t \right) = \tau \left\{ {I\left( {t \le \tau } \right) + \frac{{\left( {a\tau - t} \right)}}{{\left( {a - 1} \right)\tau }}I\left( {t > \tau } \right)} \right\}, $

其中$ t > 0 $, $ a > 2 $且$ \tau > 0 $是调整参数, 本文选择Fan^[13]建议的$ a = 3.7 $.

假设$ {\rm A} = \left\{ {j:{\beta _{0j}} \ne 0} \right\} $代表由真实参数$ {\beta _0} $的非零元素组成的集合. 我们将参数向量划分为$ \beta = {\left( {{\beta _1}^{\rm T}, {\beta _2}^{\rm T}} \right)^{\rm T}} $, 其中$ {\beta _1} \in {R^d}, {\beta _2} \in {R^{p - d}} $代表非零元素与零元素, 那么真实参数$ {\beta _0} = \left( {\beta _{10}^{\rm T}, \beta _{20}^{\rm T}} \right) = {\left( {\beta _{10}^{\rm T}, {0^{\rm T}}} \right)^{\rm T}} $. 类似的, 记$ \hat \beta = {\left( {\hat \beta _1^{\rm T}, \hat \beta _2^{\rm T}} \right)^{\rm T}} $, 其中$ {\hat \beta _1}, {\hat \beta _2} $分别是$ {\beta _1}, {\beta _2} $的惩罚经验似然估计值. 令

$ {\Sigma _1} = E\left[ {{H_i}{H_i}^{\rm T}} \right] - E\left[ {{\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right)\Phi \left( {{U_i}} \right)} \right], $

$ {\Sigma _2} = \left\{ {E\left[ {{H_i}{H_i}^{\rm T}} \right] - E\left[ {{\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right)\Phi \left( {{U_i}} \right)} \right]} \right\}{\sigma ^2}{\left( {I - \rho W} \right)^{ - 2}}, $

其中$ \Gamma \left( u \right) = E\left( {{X_i}{X_i}^{\rm T}\left| {U = u} \right.} \right) $, $ \Phi \left( u \right) = E\left( {{X_i}{H_i}^{\rm T}\left| {U = u} \right.} \right) $, $ \Sigma = \Sigma _1^{ - 1}{\Sigma _2}\Sigma _1^{ - 1} $, 对应的, 将$ \Sigma $划分为矩阵块$ {\Sigma _{ij}}\left( {i = 1, 2;j = 1, 2} \right) $.

第三步, 估计空间系数$ \rho $和误差项的方差$ {\sigma ^{\rm{2}}} $, 记$ {\bar m_n} = {W_n}{m_n}, {{\bar{\bar{m}}}_n} = W_n^2{m_n}, {\bar \varepsilon _n} = {W_n}{\varepsilon _n}, $则有$ {m_n} = \rho {\bar m_n} + {\varepsilon _n} $, $ {\bar m_n} = \rho {\bar m_n} + {\bar \varepsilon _n} $所以有

$ \begin{array}{*{20}{c}} {{m_n} - \lambda {{\bar m}_n} = {\varepsilon _n}}, \quad {{m_n} - \lambda {{{\bar{\bar{m}}}}_n} = {{\bar \varepsilon }_n}}.\end{array} $

(9)

对(9)式平方再求和得

$ \begin{array}{l} \frac{1}{n}\sum\limits_{i = 1}^n {m_n^2} = \frac{{2\rho }}{n}\sum\limits_{i = 1}^n {{m_n}{{\bar m}_n}} - \frac{{{\rho ^2}}}{n}\sum\limits_{i = 1}^n {{{\bar m}^2}_n} + \frac{1}{n}\sum\limits_{i = 1}^n {\varepsilon _n^2}, \\ \frac{1}{n}\sum\limits_{i = 1}^n {\bar m_n^2} = \frac{{2\rho }}{n}\sum\limits_{i = 1}^n {{{\bar m}_n}{{{\bar{\bar{m}}}}_n}} - \frac{{{\rho ^2}}}{n}\sum\limits_{i = 1}^n {{{{\bar{\bar{m}}}}^2}_n} + \frac{1}{n}\sum\limits_{i = 1}^n {\bar \varepsilon _n^2}, \\ \frac{1}{n}\sum\limits_{i = 1}^n {{m_n}{{\bar m}_n}} = \frac{\rho }{n}\sum\limits_{i = 1}^n {\left( {{m_n}{{{\bar{\bar{m}}}}_n} + {{{\bar{\bar{m}}}}^2}_n} \right)} - \frac{{{\rho ^2}}}{n}\sum\limits_{i = 1}^n {{{\bar m}_n}} {{{\bar{\bar{m}}}}_n} + \frac{1}{n}\sum\limits_{i = 1}^n {\varepsilon _n^{}} {{\bar \varepsilon }_n}, \end{array} $

且对应的期望分别是$ E\left( {\frac{1}{n}\sum\limits_{i = 1}^n {\varepsilon _n^2} } \right) = {\sigma ^2}, E\left( {\frac{1}{n}\sum\limits_{i = 1}^n {\bar \varepsilon _n^2} } \right) = \frac{{{\sigma ^2}}}{n}tr\left( {{W_n}^{\rm T}{W_n}} \right), E\left( {\frac{1}{n}\sum\limits_{i = 1}^n {{\varepsilon _n}{{\bar \varepsilon }_n}} } \right) = 0, $令$ \Theta {\rm{ = }}{\left( {\rho , {\rho ^2}, {\sigma ^2}} \right)^{\rm T}} $, 则有

$ {\Lambda _n}\Theta = {C_n}, $

(10)

$ \begin{align*} \begin{gathered} {\Lambda _n} = \frac{1}{n}\left( {\begin{array}{*{20}{c}} {2E\left( {{m_n}{{\bar m}_n}} \right)}&{ - E\left( {\bar m_n^{\rm T}{{\bar m}_n}} \right)}&1\\ {2E\left( {{{\bar m}_n}{{{\bar{\bar{m}}}}_n}} \right)}&{ - E\left( {{\bar{\bar{m}}}_n^{\rm T}{{{\bar{\bar{m}}}}_n}} \right)}&{tr\left( {W_n^{\rm T}{W_n}} \right)}\\ {2E\left( {{m_n}{{{\bar{\bar{m}}}}_n} + {{{\bar{\bar{m}}}}^2}_n} \right)}&{ - E\left( {\bar m_n^{\rm T}{{{\bar{\bar{m}}}}_n}} \right)}&0 \end{array}} \right), {C_n} = \frac{1}{n}\left( {\begin{array}{*{20}{c}} {E\left( {m_n^{\rm T}{m_n}} \right)}\\ {E\left( {\bar m_n^{\rm T}{{\bar m}_n}} \right)}\\ {E\left( {m_n^{\rm T}{{\bar m}_n}} \right)} \end{array}} \right), \end{gathered} \end{align*} $

则$ \Theta $的估计

$ \hat \Theta = \Lambda _n^{ - 1}{C_n}, $

(11)

一般情况下, $ {\Lambda _n}{C_n} $是未知的, 所以用到两步估计方法来估计$ \Theta $, 此时用$ {m_n} $的估计值$ {\tilde m_n} $估计$ \Theta $, 其中$ {\tilde m_n} = {\tilde Y_n} - \tilde Z_n^{\rm T}\beta $, 令$ {\tilde {\bar{m}}_n} = {W_n}{\tilde m_n}, {\tilde {\bar{\bar{m}}}_n} = W_n^2{\tilde m_n} $, 则有

$ \begin{align*} \begin{gathered} {F_n} = \frac{1}{n}\left( {\begin{array}{*{20}{c}} {2E\left( {{{\tilde m}_n}{{\tilde {\bar{m}}}_n}} \right)}&{ - E\left( {\tilde {\bar{m}}_n^{\rm T}{{\tilde {\bar{m}}}_n}} \right)}&1\\ {2E\left( {{{\tilde {\bar{m}}}_n}{{\tilde {\bar{\bar{m}}}}_n}} \right)}&{ - E\left( {\tilde {\bar{\bar{m}}}_n^{\rm T}{{\tilde {\bar{\bar{m}}}}_n}} \right)}&{tr\left( {W_n^{\rm T}{W_n}} \right)}\\ {2E\left( {{{\tilde m}_n}{{\tilde {\bar{\bar{m}}}}_n} + {{\tilde {\bar{\bar{m}}}}^2}_n} \right)}&{ - E\left( {\tilde {\bar{m}}_n^{\rm T}{{\tilde {\bar{\bar{m}}}}_n}} \right)}&0 \end{array}} \right), {d_n} = \frac{1}{n}\left( {\begin{array}{*{20}{c}} {E\left( {\tilde m_n^{\rm T}{{\tilde m}_n}} \right)}\\ {E\left( {\tilde {\bar{m}}_n^{\rm T}{{\tilde {\bar{m}}}_n}} \right)}\\ {E\left( {\tilde m_n^{\rm T}{{\tilde {\bar{m}}}_n}} \right)} \end{array}} \right), \end{gathered} \end{align*} $

根据(11)式, 有

$ {d_n} = {F_n}\Theta + {v_n}, $

(12)

这里$ {v_n} $是估计偏差, 对残差平方和取最小值计算$ \Theta $的估计有

$ \tilde \Theta = \mathop {\arg \min }\limits_\Theta {\left( {{d_n} - {F_n}\Theta } \right)^{\rm T}}\left( {{d_n} - {F_n}\Theta } \right) = {\left( {{F_n}^{\rm T}{F_n}} \right)^{ - 1}}{F_n}^{\rm T}{d_n}, $

(13)

在讨论样本性质前, 先给出一些正则性条件, 记$ {a_n} = {\left( {{p \mathord{\left/ {\vphantom {p n}} \right. } n}} \right)^{{1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}, {c_n} = {\left\{ {{{\log n} \mathord{\left/ {\vphantom {{\log n} {nh}}} \right. } {nh}}} \right\}^{{1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}} + {h^2} $, C1. 当$ \left| \rho \right| < 1 $时, 矩阵$ \left( {I - \rho W} \right) $是非奇异矩阵.

C2. 当$ \left| \rho \right| < 1 $时, 矩阵$ W $和$ {\left( {I - \rho W} \right)^{ - 1}} $的绝对行和与绝对列和一致有界.

C3. 误差序列$ \varepsilon $完全独立, 且满足：$ E\left( \varepsilon \right) = 0, E{\left( \varepsilon \right)^2} = {\sigma ^2} $.

C4. $ \Gamma \left( u \right) $和$ \Phi \left( u \right) $是Lipschitz连续的, $ \Gamma \left( u \right) = E\left( {{X_i}{X_i}^{\rm T}\left| {U = u} \right.} \right) $是非奇异的, 以及$ \Phi \left( u \right) = E\left( {{X_i}{H_i}^{\rm T}\left| {U = u} \right.} \right) $, $ {\mu _j} = \int {{u^j}} K\left( u \right)du $, $ {v_j} = \int {{u^j}} {K^2}\left( u \right)du $.

C5. $ \left\{ {{\alpha _j}\left( \cdot \right), j = 1, 2 \cdots , q} \right\} $具有连续的二阶导数.

C6. 核密度函数$ K\left( \cdot \right) $是具有紧支撑的的对称密度函数且Lipschitz连续, 存在$ d < 2 - {r^{ - 1}} $, 且窗宽$ h $满足：$ n{h^6} \to 0, {{{n{h^3}} \mathord{\left/ {\vphantom {{n{h^3}} {\left( {\log n} \right)}}} \right. } {\left( {\log n} \right)}}^3} \to \infty $和$ {n^{2d - 1}}h \to \infty $.

C7. $ {\Sigma _1} $和$ {\Sigma _2} $为正定矩阵, 且其特征值均有界.

C8. 当$ n \to \infty , \tau {\left( {{n \mathord{\left/ {\vphantom {n p}} \right. } p}} \right)^{{1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}} \to \infty $和$ {\min _{j \in {\rm A}}}{{{\beta _{j0}}} \mathord{\left/ {\vphantom {{{\beta _{j0}}} \tau }} \right. } \tau } \to \infty $.

C9. 惩罚函数$ {p'_\tau }\left( \cdot \right) $满足$ {\max _{j \in {\rm A}}}{p'_\tau }\left( {\left| {{\beta _{j0}}} \right|} \right) = o\left( {{{\left( {np} \right)}^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}} \right) $和$ {\max _{j \in {\rm A}}}{p''_\tau }\left( {\left| {{\beta _{j0}}} \right|} \right) = o\left( {{p^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}} \right) $.

引理1  假设条件C1-C6成立, 那么

$ \begin{array}{l} {X^{\rm T}}KX = nf\left( u \right)\Gamma \left( u \right) \otimes \left( {\begin{array}{*{20}{c}} 1&0\\ 0&{{\mu _2}} \end{array}} \right)\left( {1 + {O_p}\left( {{c_n}} \right)} \right), \\ {X^{\rm T}}KH = nf\left( u \right)\Phi \left( u \right) \otimes {\left( {1, 0} \right)^{\rm T}}\left( {1 + {O_p}\left( {{c_n}} \right)} \right), \end{array} $

证  由Fan^[14]引理A.2可得.

引理2  假设条件C1-C7成立, 那么

$ \frac{1}{n}{G^{\rm T}}\left( \beta \right)\left( {I - S} \right){Z_i} = {\Sigma _1} + {o_p}\left( 1 \right), $

(14)

$ \frac{1}{n}{G^{\rm T}}\left( \beta \right)\left( {I - S} \right)\left( {Y - Z\beta } \right) = {\Psi _n} + {o_p}\left( {c_n^2} \right), $

(15)

证  由引理1, 可以得到

$ \left( {{X_i}^{\rm T}, {0_{1 \times q}}} \right){\left\{ {{X^{\rm T}}\left( {{U_1}} \right){W_{h{u_1}}}X\left( {{U_1}} \right)} \right\}^{ - 1}}{X^{\rm T}}\left( {{U_1}} \right){W_{h{u_1}}}H = {X_i}^{\rm T}{\Gamma ^{ - 1}}\left( {{U_i}} \right)\Phi \left( {{U_i}} \right)\left\{ {1 + {O_p}\left( {{c_n}} \right)} \right\}, $

所以有以下推导

$ \begin{array}{l} \quad\frac{1}{n}{H^{\rm T}}{\left( {I - S} \right)^{\rm T}}\left( {I - S} \right){Z_i} = \frac{1}{n}{H^{\rm T}}{\left( {I - S} \right)^{\rm T}}\left( {I - S} \right)\left( {H + {o_p}\left( 1 \right)} \right)\\ = \frac{1}{n}{\sum\limits_{i = 1}^n {\left\{ {{H_i} - {\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}\left[ {1 + {O_p}\left( {{c_n}} \right)} \right]} \right\}} ^2}\\ = \frac{1}{n}{\sum\limits_{i = 1}^n {\left\{ {\left[ {{H_i} - {\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}} \right] + {O_p}\left( {{c_n}} \right){\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}} \right\}} ^2}\\ = \frac{1}{n}{\sum\limits_{i = 1}^n {\left\{ {\left[ {{H_i} - {\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}} \right]} \right\}} ^2}\\ \quad + {O_p}\left( {{c_n}} \right)\frac{1}{n}\sum\limits_{i = 1}^n {\left\{ {\left[ {{H_i} - {\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}} \right]{\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}} \right\}} \\ \quad + {O_p}\left( {{c_n}^2} \right)\frac{1}{n}\sum\limits_{i = 1}^n {\left\{ {{\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}{X_i}^{\rm T}{\Gamma ^{ - 1}}\left( {{U_i}} \right)\Phi \left( {{U_i}} \right)} \right\}} = {\Sigma _1} + {o_p}\left( 1 \right), \end{array} $

$ \frac{1}{n}{G^{\rm T}}\left( \beta \right)\left( {I - S} \right)\left( {Y - Z\beta } \right) = \frac{1}{n}{G^{\rm T}}\left( \beta \right)\left( {I - S} \right)\left( {M + m} \right), $

由引理1得到$ \left( {{X_i}^{\rm T}, {0_{1 \times q}}} \right){\left\{ {{X^{\rm T}}\left( {{U_1}} \right){W_{h{u_1}}}X\left( {{U_1}} \right)} \right\}^{ - 1}}{X^{\rm T}}\left( {{U_1}} \right){W_{h{u_1}}}M = {X_i}^{\rm T}\alpha \left( {{U_i}} \right)\left( {1 + {O_p}\left( {{c_n}} \right)} \right), $由此可以得到下面式子

$ \begin{array}{l} \quad \frac{1}{n}{H^{\rm T}}{\left( {I - S} \right)^{\rm T}}\left( {I - S} \right)M\\ = \frac{1}{n}\sum\limits_{i = 1}^n {\left[ {{H_i} - {\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}\left( {1 + {O_p}\left( {{c_n}} \right)} \right)} \right]} \left[ {{X_i}^{\rm T}\alpha \left( {{U_i}} \right) - {X_i}^{\rm T}\alpha \left( {{U_i}} \right)\left( {1 + {O_p}\left( {{c_n}} \right)} \right)} \right]\\ = \frac{1}{n}\sum\limits_{i = 1}^n {\left( {{H_i} - {\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}} \right)} {X_i}^{\rm T}\alpha \left( {{U_i}} \right)\left( {1 + {O_p}\left( {{c_n}} \right)} \right){O_p}\left( {{c_n}} \right)\\ = {O_p}\left( {c_n^2} \right), \end{array} $

同理$ \left( {{X_i}^{\rm T}, {0_{1 \times q}}} \right){\left\{ {{X^{\rm T}}\left( {{U_1}} \right){W_{h{u_1}}}X\left( {{U_1}} \right)} \right\}^{ - 1}}{X^{\rm T}}\left( {{U_1}} \right){W_{h{u_1}}}m = {O_p}\left( {{c_n}} \right), $所以有下式

$ \frac{1}{n}{H^{\rm T}}{\left( {I - S} \right)^{\rm T}}\left( {I - S} \right)m = \frac{1}{n}\sum\limits_{i = 1}^n {\left[ {{H_i} - {\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}} \right]} {m_i}\left( {1 + {O_p}\left( {{c_n}} \right)} \right), $

此处定义$ \psi = \frac{1}{n}\sum\limits_{i = 1}^n {\left[ {{H_i} - {\Phi ^{\rm T}}\left( {{U_i}} \right){\Gamma ^{ - 1}}\left( {{U_i}} \right){X_i}} \right]} \left( {1 + {O_p}\left( {{c_n}} \right)} \right){\left( {I - \rho W} \right)^{ - 1}}{\varepsilon _i} $, 所以有$ \frac{1}{n}{G^{\rm T}}\left( \beta \right) \\ \left( {I - S} \right)\left( {Y - Z\beta } \right) = {\psi _n} + {O_p}\left( {{c_n}^2} \right), $那么(15)式成立.

综合以上式子引理2证明完成.

引理3  假设条件C1-C8成立, 那么

$ {1 \mathord{\left/ {\vphantom {1 {\sqrt n }}} \right. } {\sqrt n }}\sum\limits_{i = 1}^n {\xi \left( \beta \right)} \to N\left( {0, {\Sigma _2}} \right), $

(16)

$ \mathop {\max }\limits_{1 \le i \le n} \left\| {\xi \left( \beta \right)} \right\| = {o_p}\left( {{n^{{1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}} \right), $

(17)

证  

$ \begin{align*} {1 \mathord{\left/ {\vphantom {1 {\sqrt n }}} \right. } {\sqrt n }}\sum\limits_{i = 1}^n {\xi \left( \beta \right)} &= {1 \mathord{\left/ {\vphantom {1 {\sqrt n }}} \right. } {\sqrt n }}\sum\limits_{i = 1}^n {{G_i}\left( \beta \right)} \left( {{{\tilde Y}_i} - {{\tilde Z}_i}^{\rm T}\beta } \right) = {1 \mathord{\left/ {\vphantom {1 {\sqrt n }}} \right. } {\sqrt n }}\sum\limits_{i = 1}^n {{G_i}\left( \beta \right)} \left( {I - S} \right)\left( {Y - {Z^{\rm T}}\beta } \right)\\ &= \sqrt n \left( {{\psi _n} + {O_p}\left( {{c_n}^2} \right)} \right), \end{align*} $

因为$ E\left( {{\psi _n}} \right) = 0 $, $ Cov\left( {{\psi _n}} \right) = \frac{1}{n}{\sigma ^2}{\Sigma _1}{\left( {I - \rho W} \right)^{ - 2}} $, 其中$ {\Sigma _2} = {\sigma ^2}{\Sigma _1}{\left( {I - \rho W} \right)^{ - 2}} $则(16)式成立.

$ \mathop {\max }\limits_{1 \le i \le n} \left\| {\xi \left( \beta \right)} \right\| \le \mathop {\max }\limits_{1 \le i \le n} \left\| {{G_i}\left( \beta \right)} \right\| \cdot \mathop {\max }\limits_{1 \le i \le n} \left\| {{{\tilde Y}_i} - {{\tilde Z}_i}\beta } \right\|, $

由假设$ E{\left\| X \right\|^{2s}} \le \infty , E{\left\| Z \right\|^{2s}} \le \infty , E{\left\| H \right\|^{2s}} \le \infty , E{\left\| U \right\|^{2s}} \le \infty , E{\left\| \varepsilon \right\|^{2s}} \le \infty , $得

$ \mathop {\max }\limits_{1 \le i \le n} \left\| {{G_i}\left( \beta \right)} \right\| = o\left( {{n^{{1 \mathord{\left/ {\vphantom {1 {2s}}} \right. } {2s}}}}} \right), \mathop {\max }\limits_{1 \le i \le n} \left\| {{{\tilde Y}_i} - {{\tilde Z}_i}\beta } \right\| = o\left( {{n^{{1 \mathord{\left/ {\vphantom {1 {2s}}} \right. } {2s}}}}} \right) $

(17) 式成立.

引理4  假设条件C1-C7成立, 则$ \left\| \lambda \right\| = {O_p}\left( {{a_n}} \right) $.

证   设$ \lambda = \alpha \delta , $其中$ \alpha \ge 0\delta \in {R^p} $且$ \left\| \delta \right\|{\rm{ = 1}} $, 并且令$ S\left( \beta \right) = \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}\left( \beta \right)} {\xi _i}^{\rm T}\left( \beta \right) $, $ {\bar \xi _i}\left( \beta \right) = \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}\left( \beta \right)} $, $ {\xi _i}^ * \left( \beta \right) = \mathop {\max }\limits_{1 \le i \le n} \left\| {{\xi _i}\left( \beta \right)} \right\| $, 由(17)式$ {\xi _i}^ * \left( \beta \right){\rm{ = }}{O_p}\left( {{a_n}} \right) $.

$ \begin{align*} 0 &= \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\xi _i}\left( \beta \right)}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\xi _i}\left( \beta \right)}}{{1 + \alpha {\delta ^{\rm T}}{\xi _i}\left( \beta \right)}}} = \frac{1}{n}\sum\limits_{i = 1}^n {{\delta ^{\rm T}}{\xi _i}\left( \beta \right) - } \alpha \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{{\left[ {{\delta ^{\rm T}}{\xi _i}\left( \beta \right)} \right]}^2}}}{{1 + \alpha {\delta ^{\rm T}}{\xi _i}\left( \beta \right)}}} \\ &\le {\delta ^{\rm T}}\bar \xi \left( \beta \right) - \frac{\alpha }{{1 + \alpha {\xi ^ * }\left( \beta \right)}}{\delta ^{\rm T}}S\left( \beta \right)\delta , \end{align*} $

因此$ \alpha \left[ {{\delta ^{\rm T}}S\left( \beta \right)\delta - {\xi ^ * }\left( \beta \right){\delta ^{\rm T}}\bar \xi \left( \beta \right)} \right] \le {\delta ^{\rm T}}\bar \xi \left( \beta \right) $, 由Owen^[8]的(3.36)和引理3得到$ \left\| \lambda \right\| = {O_p}\left( {{a_n}} \right) $.

引理5  假设条件C1-C9成立, 那么在$ {D_n} = \left\{ {\beta :\left\| {\beta - {\beta _0}} \right\| \le c{a_n}} \right\} $这个领域内, $ {L_p}\left( \beta \right) $在$ {D_n} $内有最小值.

证  当$ \beta \in {D_n} $, 有$ \mathop {\max }\limits_{1 \le i \le n} \left\| {{\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right\| = {o_p}\left( 1 \right), $将$ {Q_{1n}}\left( {\beta , \lambda } \right) = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\xi _i}\left( \beta \right)}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} = 0 $泰勒展开得到

$ 0 = \bar \xi \left( \beta \right) - S\left( \beta \right)\lambda + {\gamma _n}, $

其中$ {\gamma _n} = \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}\left( \beta \right)\frac{{{{\left[ {{\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right]}^2}}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} $且$ \left| {{\varsigma _i}} \right| \le \left| {{\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right| $, 所以$ \lambda = {S^{ - 1}}\left( \beta \right)\bar \xi \left( \beta \right) + {S^{ - 1}}\left( \beta \right){\gamma _n} $代入$ L\left( \beta \right) $得到

$ \begin{align*} 2L\left( \beta \right) &= 2\sum\limits_{i = 1}^n {\ln \left( {1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right)} \\ &= n{{\bar \xi }^{\rm T}}\left( \beta \right){S^{ - 1}}\left( \beta \right)\bar \xi \left( \beta \right) - n{\gamma _n}^{\rm T}{S^{ - 1}}\left( \beta \right){\gamma _n} + \frac{2}{3}{\sum\limits_{i = 1}^n {\left[ {1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right]} ^3}{\left( {1 + {\varsigma _i}} \right)^{ - 4}}, \end{align*} $

取$ {\varsigma _i} $并且满足$ \left| {{\varsigma _i}} \right| \le \left| {{\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right| $, $ \beta \in \partial {D_n} $这里$ \partial {D_n} $代表$ {D_n} $得边界, $ \beta = {\beta _0} + c{a_n}{\theta _\beta } $, $ {\theta _\beta } $是单位向量

$ \begin{align*} 2L\left( \beta \right) = &n{{\bar \xi }^{\rm T}}\left( \beta \right){S^{ - 1}}\left( \beta \right)\bar \xi \left( \beta \right) + n\left[ {\bar \xi \left( \beta \right) - \bar \xi \left( {{\beta _0}} \right)} \right]{S^{ - 1}}\left( \beta \right)\left[ {\bar \xi \left( \beta \right) - \bar \xi \left( {{\beta _0}} \right)} \right]\\ &+ n\bar \xi \left( {{\beta _0}} \right)\left[ {{S^{ - 1}}\left( \beta \right) - {S^{ - 1}}\left( {{\beta _0}} \right)} \right]\bar \xi \left( {{\beta _0}} \right) + 2n\bar \xi \left( {{\beta _0}} \right){S^{ - 1}}\left( \beta \right)\left[ {\bar \xi \left( \beta \right) - \bar \xi \left( {{\beta _0}} \right)} \right]\\ &- n{\gamma _n}^{\rm T}{S^{ - 1}}\left( \beta \right){\gamma _n} + \frac{2}{3}{\sum\limits_{i = 1}^n {\left[ {1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right]} ^3}{\left( {1 + {\varsigma _i}} \right)^{ - 4}}\\ =& {T_0} + {T_1} + {T_2}, \end{align*} $

当$ n \to \infty $时, 由引理3

$ \begin{align*} \bar \xi \left( \beta \right) - \bar \xi \left( {{\beta _0}} \right) &= \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}\left( \beta \right) - } \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}\left( {{\beta _0}} \right)} = \frac{1}{n}\sum\limits_{i = 1}^n {\left[ {{\xi _i}\left( \beta \right) - {\xi _i}\left( {{\beta _0}} \right)} \right]} \\ &= - \frac{1}{n}G\left( {I - S} \right)Z\left( {\beta - {\beta _0}} \right) + {o_p}\left( 1 \right), \end{align*} $

因此$ {T_1} = n{\Sigma _1}^2{a_n}^2{\Sigma _2} $, $ {T_2}/{T_1}\stackrel{p}\longrightarrow0 $, 且$ 2L\left( {{\beta _0}} \right) - {T_1} = {o_p}\left( 1 \right) $, 这就意味着, 当$ n \to \infty $, $ P\left\{ {2L\left( \beta \right) - 2L\left( {{\beta _0}} \right) > c} \right\} \to 1, c \in R $. 此外, 对于$ n $足够大, C(9)和SCAD惩罚得无偏性意味着$ {p_\tau }\left( {\left| {{\beta _i}} \right|} \right) = {p_\tau }\left( {\left| {{\beta _{i0}}} \right|} \right), i \in {\rm A} $, 因此当$ n $足够大有

$ \begin{array}{l} {L_p}\left( \beta \right) - {L_p}\left( {{\beta _0}} \right) = L\left( \beta \right) - L\left( {{\beta _0}} \right) + n\sum\limits_{i = 1}^p {\left[ {{p_\tau }\left( {\left| {{\beta _i}} \right|} \right) - {p_\tau }\left( {\left| {{\beta _{i0}}} \right|} \right)} \right]} \\ \qquad\qquad\qquad\quad \ge L\left( \beta \right) - L\left( {{\beta _0}} \right) + n\sum\limits_{i \in {\rm A}}^p {\left[ {{p_\tau }\left( {\left| {{\beta _i}} \right|} \right) - {p_\tau }\left( {\left| {{\beta _{i0}}} \right|} \right)} \right]} \ge L\left( \beta \right) - L\left( {{\beta _0}} \right), \end{array} $

因此$ P\left\{ {{L_p}\left( \beta \right) - {L_p}\left( {{\beta _0}} \right)} \right\} \to 1 $对于$ \beta \in \partial {D_n} $成立有个最小值.

定理1  在C1-C9的假设条件下, 有$ \sqrt n \left( {\hat \beta - \beta } \right)\stackrel{D}\longrightarrow N\left( {0, {\Sigma _1}{\sigma ^2}{{\left( {I - \rho W} \right)}^{ - 2}}} \right) $其中$ \stackrel{D}\longrightarrow $表示依分布收敛.

证  将$ {Q_{1n}}\left( {\beta , \lambda } \right) = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\xi _i}\left( \beta \right)}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} $, $ {Q_{2n}}\left( {\beta , \lambda } \right) = \frac{\lambda }{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}{\left( {\frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial \beta }}} \right)^{\rm T}} $在$ \left( {\beta , 0} \right) $处泰勒展开得到下式：

$ {Q_{in}}\left( {\hat \beta , \hat \lambda } \right) = {Q_{in}}\left( {\beta , 0} \right) + \frac{{\partial {Q_{in}}\left( {\beta , 0} \right)}}{{\partial \beta }}\left( {\hat \beta - \beta } \right) + \frac{{\partial {Q_{in}}\left( {\beta , 0} \right)}}{{\partial {\lambda ^{\rm T}}}}\left( {\hat \lambda - 0} \right) + {o_p}\left( {{n^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}} \right), $

写成矩阵形式

$ \left( {\begin{array}{*{20}{c}} {\hat \lambda }\\ {\hat \beta - \beta } \end{array}} \right) = {K_{^n}}^{ - 1}\left( {\begin{array}{*{20}{c}} {{Q_{1n}}\left( {\beta , 0} \right) + {o_p}\left( {{n^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}} \right)}\\ {{o_p}\left( {{n^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}} \right)} \end{array}} \right), $

$ {K_n} = \left( {\begin{array}{*{20}{c}} {\frac{{\partial {Q_{1n}}\left( {\beta , 0} \right)}}{{\partial {\lambda ^{\rm T}}}}}&{\frac{{\partial {Q_{1n}}\left( {\beta , 0} \right)}}{{\partial \beta }}}\\ {\frac{{\partial {Q_{2n}}\left( {\beta , 0} \right)}}{{\partial {\lambda ^{\rm T}}}}}&{\frac{{\partial {Q_{2n}}\left( {\beta , 0} \right)}}{{\partial \beta }}} \end{array}} \right) \to K = \left( {\begin{array}{*{20}{c}} {{K_{11}}}&{{K_{12}}}\\ {{K_{21}}}&{{K_{22}}} \end{array}} \right), $

$ {K^{ - 1}} = \left( {\begin{array}{*{20}{c}} {{K_{11}}^{ - 1} + {K_{11}}^{ - 1}{K_{21}}{K_{22.1}}^{ - 1}{K_{11}}^{ - 1}{K_{12}}^{ - 1}}&{ - {K_{11}}^{ - 1}{K_{12}}{K_{22.1}}^{ - 1}}\\ { - {K_{11}}^{ - 1}{K_{12}}{K_{22.1}}^{ - 1}}&{{K_{22.1}}^{ - 1}} \end{array}} \right), $

其中$ {K_{22.1}} = {K_{22}} - {K_{21}}{K_{11}}^{ - 1}{K_{12}} $, 那么他的逆矩阵$ {K_{22.1}}^{ - 1} = - {\left( {{K_{21}}{K_{11}}^{ - 1}{K_{12}}} \right)^{ - 1}} $. 由Owen^[8]得

$ \frac{{\partial {Q_{1n}}\left( {\beta , 0} \right)}}{{\partial \beta }} = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial \beta }}} , \frac{{\partial {Q_{1n}}\left( {\beta , 0} \right)}}{{\partial {\lambda ^{\rm T}}}} = - \frac{1}{n}{\xi _i}\left( \beta \right){\xi _i}{\left( \beta \right)^{\rm T}}, $

$ \frac{{\partial {Q_{2n}}\left( {\beta , 0} \right)}}{{\partial \beta }} = 0, \frac{{\partial {Q_{2n}}\left( {\beta , 0} \right)}}{{\partial {\lambda ^{\rm T}}}} = \frac{1}{n}{\sum\limits_{i = 1}^n {\left( {\frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial \beta }}} \right)} ^{\rm T}}, $

所以$ \sqrt n \left( {\hat \beta - \beta } \right) = {K_{22.1}}^{ - 1}{K_{21}}{K_{11}}^{ - 1}\sqrt n {Q_{1n}}\left( {\beta , 0} \right) + {o_p}\left( 1 \right) \stackrel{D}\longrightarrow N\left( {0, {\Sigma _1}{\sigma ^2}{{\left( {I - \rho W} \right)}^{ - 2}}} \right) $, 定理1证明完成.

定理2  假设条件C1-C9成立, 那么

$ - 2L\left( \beta \right) = 2\sum\limits_{i = 1}^n {\log \left\{ {1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right\}} \stackrel{D}\longrightarrow \chi _{p + 1}^2, $

其中$ \chi _{p + 1}^2 $表示具有$ p + 1 $个自由度的卡方分布.

证  假设$ \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}^{\rm T}\left( \beta \right)} {\xi _i}\left( \beta \right) $是正定的, 且$ \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\xi _i}\left( \beta \right)}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} = 0 $, 使用Owen^[8]的(2.14)同样的方法结合引理4, 得

$ \begin{align*} \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\xi _i}\left( \beta \right)}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} &= \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}\left( \beta \right)} \left[ {1 - {\lambda ^{\rm T}}{\xi _i}\left( \beta \right) + \frac{{{{\left[ {{\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right]}^2}}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} \right]\\ &= \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}\left( \beta \right)} - \frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}^{\rm T}\left( \beta \right)} {\xi _i}\left( \beta \right){\lambda ^{\rm T}} + \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\xi _i}\left( \beta \right){{\left[ {{\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right]}^2}}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} = 0, \end{align*} $

因此$ \lambda = {S^{ - 1}}\left( \beta \right)\bar \xi \left( \beta \right) + {o_p}\left( {{a_n}} \right) $通过泰勒展开得：

$ 2L\left( \beta \right) = 2n{\lambda ^{\rm T}}\bar \xi \left( \beta \right) - n{\lambda ^{\rm T}}S\left( \beta \right)\lambda + {o_p}\left( 1 \right) = n\bar \xi \left( \beta \right){S^{ - 1}}\left( \beta \right)\bar \xi \left( \beta \right) + {o_p}\left( 1 \right), $

结合引理4, 我们可以证明当$ n \to \infty $时,

$ {\left\{ {\frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n {{\xi _i}\left( \beta \right)} } \right\}^{\rm T}}{\left\{ {\frac{1}{n}\sum\limits_{i = 1}^n {{\xi _i}^{\rm T}\left( \beta \right)} {\xi _i}\left( \beta \right)} \right\}^{ - 1}}\left\{ {\frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n {{\xi _i}\left( \beta \right)} } \right\}\stackrel{D}\longrightarrow \chi _{p + 1}^2, $

定理2证明完成.

定理3  假设条件C1-C9成立, 且$ {{{p^5}} \mathord{\left/ {\vphantom {{{p^5}} n}} \right. } n} \to 0 $, 当$ n \to \infty $有

1. 稀疏性：$ {\hat \beta _2} = 0 $.

2. 渐近正态性：$ \sqrt n {B_n}{\Sigma _p}^{ - \frac{1}{2}}\left( {{{\hat \beta }_1} - {{\hat \beta }_{10}}} \right)\stackrel{D}\longrightarrow N\left( {0, g} \right) $, 其中$ {B_n} $是$ s \times d $矩阵且$ {B_n}{B_n}^{\rm T} \to g $, $ {\Sigma _n} = {\Sigma _{11}} - {\Sigma _{12}}\Sigma _{22}^{ - 1}{\Sigma _{21}} $. 根据引理5, 当$ \beta \in {D_n} $时, $ j = 1, 2, \cdots , p $有

$ \frac{1}{n}\frac{{\partial {L_p}\left( \beta \right)}}{{\partial {\beta _j}}} = \frac{1}{n}\sum\limits_{i = 1}^n {{\lambda ^{\rm T}}\frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial {\beta _j}}}} \left\{ {1 + op\left( 1 \right)} \right\} + {p'_\tau }\left( {\left| {{\beta _j}} \right|} \right)sign\left( {{\beta _j}} \right) = {\rm I} + {\rm I}{\rm I}, $

结合假设条件条件C1-C9有

$ \mathop {\max }\limits_{j \in {\rm A}} \left| {\rm I} \right| \le \mathop {\max }\limits_{j \in {\rm A}} \left| {{\lambda ^{\rm T}}} \right|\left| {\frac{1}{n}\sum\limits_{i = 1}^n {\frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial {\beta _j}}}} } \right| = {O_p}\left( {{a_n}} \right), $

当$ \tau {\left( {{n \mathord{\left/ {\vphantom {n p}} \right. } p}} \right)^{\frac{1}{2}}} \to \infty $时, 对于任意, $ j \in {\rm A} $, $ {p'_\tau }\left( {\left| {{\beta _j}} \right|} \right)sign\left( {{\beta _j}} \right) $渐进的主导$ \frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial {\beta _j}}} $的符号, 因此对于任意$ j \notin {\rm A} $, 当$ n \to \infty $时, 依概率趋近于1, 则

$ \frac{{\partial {L_p}\left( \beta \right)}}{{\partial {\beta _j}}} > 0, {\beta _j} \in \left( {0, c{a_n}} \right), \frac{{\partial {L_p}\left( \beta \right)}}{{\partial {\beta _j}}} < 0, {\beta _j} \in \left( { - c{a_n}, 0} \right), $

故对于任意$ j \in {\rm A} $, 依概率趋近于1有$ {\hat \beta _j} = 0 $, 则稀疏性: $ {\hat \beta _2} = 0 $得证.

取$ {\psi _1} $和$ {\psi _2} $满足$ {\psi _1}\beta = {\beta _1} $和$ {\psi _2}\beta = {\beta _2} $, 由拉格朗日乘法得到$ {L_p}\left( \beta \right) $的极小值,

$ \frac{1}{n}\sum\limits_{i = 1}^n {\log \left\{ {1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)} \right\}} \sum\limits_{j = 1}^p {{{p'}_\tau }\left( {\left| {{\beta _j}} \right|} \right)} + {\nu ^{\rm T}}{\psi _2}\beta, $

其中, $ \nu $表示$ \left( {p - s} \right) $维拉格朗日算子, 定义

$ \begin{array}{c} {{\tilde Q}_{1n}}\left( {\beta , \lambda , \nu } \right) = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{{\xi _i}\left( \beta \right)}}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} , \\ {{\tilde Q}_{2n}}\left( {\beta , \lambda , \nu } \right) = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{1}{{1 + {\lambda ^{\rm T}}{\xi _i}\left( \beta \right)}}} {\left( {\frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial {\beta ^{\rm T}}}}} \right)^{\rm T}}\lambda + b\left( \beta \right) + {\psi _2}^{\rm T}\nu , \\ {{\tilde Q}_{3n}}\left( {\beta , \lambda , \nu } \right) = {\psi _2}\beta , \end{array} $

其中$ b\left( \beta \right) = {\left\{ {{{p'}_\tau }\left( {\left| {{\beta _1}} \right|} \right)sign\left( {{\beta _1}} \right), {{p'}_\tau }\left( {\left| {{\beta _2}} \right|} \right)sign\left( {{\beta _2}} \right), \cdots , {{p'}_\tau }\left( {\left| {{\beta _p}} \right|} \right)sign\left( {{\beta _p}} \right), 0 \cdots , 0} \right\}^{\rm T}} $, 关于$ \left( {\hat \beta , \hat \lambda , \hat \nu } \right) $的最小值满足$ 0 = {\tilde Q_{jn}}\left( {\hat \beta , \hat \lambda , \nu } \right) $, $ \left( {j = 1, 2, 3} \right) $. 因为$ \left\| \lambda \right\| = {O_p}\left( {{a_n}} \right) $对于$ \beta \in {D_n} $, 可以从$ 0 = {\tilde Q_{2n}}\left( {\hat \beta , \hat \lambda , \hat \nu } \right) $知$ \left\| \nu \right\| = {O_p}\left( {{a_n}} \right) $, 将$ {\tilde Q_{jn}}\left( {\hat \beta , \hat \lambda , \nu } \right) $在$ \left( {{\beta _0}, 0, 0} \right) $处泰勒展开, 给出以下偏导

$ \begin{array}{c} \frac{{\partial {{\tilde Q}_{1n}}\left( {\beta , 0, 0} \right)}}{{\partial \lambda }} = - S\left( {{\beta _0}} \right), \frac{{\partial {{\tilde Q}_{1n}}\left( {\beta , 0, 0} \right)}}{{\partial \beta }} = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{\partial {\xi _i}\left( {{\beta _0}} \right)}}{{\partial \beta }}} , \frac{{\partial {{\tilde Q}_{1n}}\left( {\beta , 0, 0} \right)}}{{\partial \hat \nu }} = 0, \\ \frac{{\partial {{\tilde Q}_{2n}}\left( {\beta , 0, 0} \right)}}{{\partial \lambda }} = \frac{1}{n}\sum\limits_{i = 1}^n {\frac{{\partial {\xi _i}\left( {{\beta _0}} \right)}}{{\partial \beta }}} , \frac{{\partial {{\tilde Q}_{2n}}\left( {\beta , 0, 0} \right)}}{{\partial \beta }} = b'\left( \beta \right), \frac{{\partial {{\tilde Q}_{2n}}\left( {\beta , 0, 0} \right)}}{{\partial \hat \nu }} = {\psi _2}, \\ \frac{{\partial {{\tilde Q}_{3n}}\left( {\beta , 0, 0} \right)}}{{\partial \lambda }} = 0, \frac{{\partial {{\tilde Q}_{3n}}\left( {\beta , 0, 0} \right)}}{{\partial \beta }} = {\psi _2}, \frac{{\partial {{\tilde Q}_{3n}}\left( {\beta , 0, 0} \right)}}{{\partial \hat \nu }} = 0, \end{array} $

通过泰勒展开写成矩阵形式

$ \left( {\begin{array}{*{20}{c}} {{{\tilde Q}_{1n}}\left( {{\beta _0}, 0, 0} \right)}\\ 0\\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{l}} { - {\Sigma _2}}&{{\Sigma _1}}&0\\ {{\Sigma _1}}&0&{{\psi _2}^{\rm T}}\\ 0&{{\psi _2}}&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {\hat \lambda }\\ {\hat \beta - {\beta _0}}\\ {\hat \nu } \end{array}} \right) + {R_n}, $

其中$ {R_n} = \sum\limits_{l = 1}^5 {R_n^{\left( l \right)}} , R_n^{\left( l \right)} = {\left( {R{{_{1n}^{\left( 1 \right)}}^{\rm T}}, R{{_{2n}^{\left( 1 \right)}}^{\rm T}}, 0} \right)^{\rm T}}, R_{jn}^{( 1 )\rm T} \in {R^p}\left( {j = 1, 2} \right), k = 1, 2, \cdots , p $且第$ k $个分量为$ R_{jn, k}^{( 1 )\rm T} = \frac{1}{2}{\left( {\hat \phi - {\phi _0}} \right)^{\rm T}}\frac{{{\partial ^2}{{\tilde Q}_{jn, k}}\left( {\tilde \phi } \right)}}{{\partial \phi \partial {\phi ^{\rm T}}}}\left( {\hat \phi - {\phi _0}} \right) $, 其中$ \phi = {\left( {{\beta ^{\rm T}}, {\lambda ^{\rm T}}} \right)^{\rm T}}, \tilde \phi = {\left( {{{\tilde \beta }^{\rm T}}, {{\tilde \lambda }^{\rm T}}} \right)^{\rm T}} $, 满足$ \left\| {\tilde \phi - {\phi _0}} \right\| \le \left\| {\hat \phi - {\phi _0}} \right\| $.

$ \begin{array}{c} {R_n}^{\left( 2 \right)} = {\left( {0, b\left( {{\beta _0}} \right), 0} \right)^{\rm T}}, \\ {R_n}^{\left( 3 \right)} = {\left[ {0, \left[ {b'\left( {\tilde \beta } \right)\left( {\hat \beta - {\beta _0}} \right)} \right], 0} \right]^{\rm T}}, \\ {R_n}^{\left( 4 \right)} = {\left\{ {{{\left[ {\left( {{\Sigma _2} - S\left( {{\beta _0}} \right)} \right)\hat \lambda } \right]}^{\rm T}} + {{\left[ {\left( {\frac{1}{n}\sum\limits_{i = 1}^n {\frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial \beta }}} - {\Sigma _1}} \right)\left( {\hat \beta - {\beta _0}} \right)} \right]}^{\rm T}}, 0, 0} \right\}^{\rm T}}, \\ {R_n}^{\left( 5 \right)} = {\left\{ {0, {{\left[ {\left( {\frac{1}{n}\sum\limits_{i = 1}^n {\frac{{\partial {\xi _i}\left( \beta \right)}}{{\partial \beta }}} - {\Sigma _1}} \right)\hat \lambda } \right]}^{\rm T}}, 0} \right\}^{\rm T}}, \end{array} $

结合条件得, 对$ l = 1, \cdots , 5.{R_n}^{\left( l \right)} = {o_p}\left( {{n^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. } 2}}}} \right) $, 记$ {\Pi _{11}} = - {\Sigma _2}, {\Pi _{12}} = \left( {{\Sigma _1}, 0} \right), {\Pi _{21}} = {\Pi _{12}}^{\rm T} $,

$ {\Pi _{22}} = \left( {\begin{array}{*{20}{c}} 0&{{\psi _2}^{\rm T}}\\ {{\psi _2}}&0 \end{array}} \right), \Pi = \left( {\begin{array}{*{20}{c}} {{\Pi _{11}}}&{{\Pi _{12}}}\\ {{\Pi _{21}}}&{{\Pi _{22}}} \end{array}} \right), $

设$ \kappa = \left( {{\beta ^{\rm T}}, {\nu ^{\rm T}}} \right) $

$ \left( {\begin{array}{*{20}{c}} {\hat \lambda }\\ {\hat \kappa - {\kappa _0}} \end{array}} \right) = {\Pi ^{ - 1}}\left\{ {\left( {\begin{array}{*{20}{c}} { - {{\tilde Q}_{1n}}\left( {\beta , 0, 0} \right)}\\ 0 \end{array}} \right) + {R_n}} \right\}, $

且$ {\Pi ^{ - 1}} = \left( {\begin{array}{*{20}{c}} {{\Pi _{11}}^{ - 1} + {\Pi _{11}}^{ - 1}{\Pi _{21}}{\Pi _{22.1}}^{ - 1}{\Pi _{11}}^{ - 1}{\Pi _{12}}^{ - 1}}&{ - {\Pi _{11}}^{ - 1}{\Pi _{12}}{\Pi _{22.1}}^{ - 1}}\\ { - {\Pi _{11}}^{ - 1}{\Pi _{12}}{\Pi _{22.1}}^{ - 1}}&{{\Pi _{22.1}}^{ - 1}} \end{array}} \right) $, 其中$ {\Pi _{22.1}} = {\Pi _{22}} - {\Pi _{21}}{\Pi _{11}}^{ - 1}{\Pi _{12}} = \left( {\begin{array}{*{20}{c}} {{\Sigma ^{ - 1}}}&{{\psi _2}^{\rm T}}\\ {{\psi _2}}&0 \end{array}} \right) $, 因此$ \hat \kappa - {\kappa _0} = {\Pi _{22.1}}^{ - 1}{\Pi _{21}}{\Pi _{11}}^{ - 1}{\tilde Q_{1n}}\left( {{\beta _0}, 0, 0} \right) + {o_p}\left( {{n^{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 2}} \right. } 2}}}} \right) $.

$ {\Pi _{22.1}}^{{\rm{ - 1}}}{\rm{ = }}\left( {\begin{array}{*{20}{c}} {\Sigma - \Sigma {\psi _2}^{\rm T}\left( {{\psi _2}\Sigma {\psi _2}^{\rm T}} \right){\psi _2}\Sigma }&{ - \Sigma {\psi _2}^{\rm T}{{\left( {{\psi _2}\Sigma {\psi _2}^{\rm T}} \right)}^{ - 1}}}\\ { - {{\left( {{\psi _2}\Sigma {\psi _2}^{\rm T}} \right)}^{ - 1}}{\psi _2}\Sigma }&{{{\left( {{\psi _2}\Sigma {\psi _2}^{\rm T}} \right)}^{ - 1}}} \end{array}} \right), $

因此$ \hat \beta - {\beta _0} = \left\{ {\Sigma - \Sigma {\psi _2}^{\rm T}{{\left( {{\psi _2}\Sigma {\psi _2}^{\rm T}} \right)}^{ - 1}}{\psi _2}\Sigma } \right\}\left\{ {{\Sigma _1}^{\rm T}{\Sigma _2}^{ - 1}\bar \xi \left( {{\beta _0}} \right) + {o_p}\left( {{n^{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 2}} \right. } 2}}}} \right)} \right\} $, 对$ {\hat \beta _1} $展开得

$ {\hat \beta _1} - {\beta _{10}} = \left\{ {{\psi _1}\Sigma - {\psi _1}\Sigma {\psi _2}^{\rm T}{{\left( {{\psi _2}\Sigma {\psi _2}^{\rm T}} \right)}^{ - 1}}{\psi _2}\Sigma } \right\}\left\{ {{\Sigma _1}^{\rm T}{\Sigma _2}^{ - 1}\bar \xi \left( {{\beta _0}} \right) + {o_p}\left( {{n^{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 2}} \right. } 2}}}} \right)} \right\}, $

其中$ {\psi _1}\Sigma - {\psi _1}\Sigma {\psi _2}^{\rm T}{\left( {{\psi _2}\Sigma {\psi _2}^{\rm T}} \right)^{ - 1}}{\psi _2}\Sigma = \left( {{\Sigma _n}, 0} \right) $则

$ \sqrt n {B_n}{\Sigma _n}^{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 2}} \right. } 2}}\left( {{{\hat \beta }_1} - {\beta _{10}}} \right) \stackrel{D}\longrightarrow N\left( {0, g} \right) $

定理3证明完成.