本文研究如下部分线性EV (errors-in-variables)模型

$ \begin{equation} \label{eq:1} y_{i}=x_{i}\beta+g(t_{i})+e_{i}, X_{i}=x_{i}+\varsigma_{i}, i=1, 2, \cdots, n, \end{equation} $

(1.1)

其中 $X_{i}\in R, t_{i}\in R, i=1, 2, \cdots, n$是已知的点列, $\beta$是未知参数, $g(\cdot)$是定义在[0, 1]上的未知函数, 误差 $\{e_{i}, i=1, 2, \cdots, n\}$是平稳的 $\alpha$ -混合随机变量且 $Ee_{i}=0$和 $E(e_{i}^{2})=1$. $\{x_{i}\}$是通过 $X_{i}=x_{i}+\varsigma_{i}$观测到的, $\{\varsigma_{i}\}$是独立同分布的测量误差, 且 $E\varsigma_{i}=0, {\rm Var}(\varsigma_{i})=\sigma_{\varsigma}^{2}$且和 $\{e_{i}\}$是独立的.

定义1^[1]  设 $\{\xi_{i}, i\geq 1\}$是 $\alpha$ -混合的, 如果 $\alpha$ -混合系数

$ \begin{align*} \alpha(n)=\underset{i\geq 1}{\sup}\{\mid P(AB)-P(A)P(B)\mid: A\in F_{n+i}^{\infty}, B\in F_{1}^{i}\}. \end{align*} $

当 $n\rightarrow\infty$时收敛到 $0$, 其中 $F_{l}^{m}=\sigma\{\xi_{l}, \xi_{l+1}, \cdots, \xi_{m}\}$表示包含 $\xi_{l}, \xi_{l+1}, \cdots, \xi_{m}, l\leq m$的 $\sigma$ -代数.

文献[2]用加权的方法研究了异方差 $\alpha$ -混合半参数模型, 得到了估计量的Berry-Esseen界; 文献[3]研究了部分线性变系数EV模型, 得到了估计量的渐近性质; 文献[4]用加权的方法研究了误差独立的半参数回归模型的矩相合性; 文献[5]用加权的方法研究了NA样本下部分线性回归模型的矩相合性; 文献[6]用小波方法研究了鞅差时间序列半参数回归模型的矩收敛速度; 文献[7]用小波方法研究了 $\varphi$ -混合和 $\psi$ -混合的非参数回归模型的矩收敛速度.然而对 $\alpha$ -混合的部分线性EV模型的矩收敛性还没有研究, 因此本文研究模型(1.1) 的矩收敛性.

本文用小波方法研究模型(1.1), 仍采用文献[8]修正后的最小二乘估计, 即

$ \begin{eqnarray*} \label{eq:2} \widehat{\beta}_{n}=(\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}^{2}-n\sigma_{\varsigma}^{2})^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}\widetilde{y}_{i}, \end{eqnarray*} $

(1.2)

其中 $\widetilde{X}_{i}=X_{i}-\sum\limits_{\mathit{j}{\rm{ = 1}}}^\mathit{n} {}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds X_{j}, \widetilde{y}_{i}=y_{i}-\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds y_{j}$.由此可得非参数部分的小波估计为

$ \begin{eqnarray*} \label{eq:3} \widehat{g}_{n}(t)=\underset{i=1}{\overset{n}\sum}(y_{i}-X_{i}\widehat{\beta}_{n})\displaystyle\int_{A_{i}}E_{m}(t, s)ds, \end{eqnarray*} $

(1.3)

其中 $E_{m}(t, s)=2^{m}E_{0}(2^{m}t, 2^{m}s)=2^{m}\sum\limits_{\mathit{k}\in \mathit{Z}}{{}}\phi(2^{m}t-k)\phi(2^{m}s-k), A_{i}=[s_{i-1}, s_{i}], $ $\phi(\cdot)$是Schwartz空间 $S_{l}$的刻度函数, 相伴 $L^{2}(R)$的多解分析为 $\{V_{m}\in Z\}$, 这里的 $R$为实数集合, $Z$是整数集合, $V_{m}$的再生核为 $E_{m}(t, s)$.

下面是本文的基本假设.

(1) $x_{i}=f(t_{i})+\eta_{i}, i=1, 2, \cdots, n$, 其中 $f(\cdot)$是定义于[0, 1]的函数, $\{\eta_{i}\}~{\rm i.i.d}$且 $E\eta_{i}=0, {\rm Var}(\eta_{i})=\sigma_{\eta}$, 而且 $\{\eta_{i}\}$和 $\{(e_{i}, \varsigma_{i})\}$是相互独立的.

(2) $g(\cdot)$和 $f(\cdot)\in H^{\alpha}$ (Sobolev空间), $\alpha>\frac{1}{2}$.

(3) $g(\cdot), f(\cdot)$满足 $\kappa$阶Lipschitz条件, $\kappa>0$.

(4) $\phi(\cdot)\in S_{\iota}$(阶为 $\iota$的Schwartz空间, $\iota\geq\alpha$), $\phi$满足1阶Lipschitz条件且具有紧支撑, 当 $\xi\rightarrow 0$时, $\mid\widehat{\phi}(\xi)-1\mid=o(\xi)$, 其中 $\widehat{\phi}$为 $\phi$的Fourier变换.

(5) $\underset{1\leq i\leq n}{\max}(s_{i}-s_{i-1})=o(n^{-1}), 2^{m}=o(n^{\frac{1}{3}})$.

注1  条件(1) 是文献[8]的特殊情形, 条件(2)-(5) 是小波估计中经常用到的(如文献[9-11]等).由此可见本文的假设条件是相当一般的.

引理1^[12]  若条件(1)-(5) 成立, 则

$ \begin{align*} \underset{0\leq t\leq 1}{\sup}\mid f(t)-\underset{k=1}{\overset{n}\sum}(\displaystyle\int_{A_{k}}E_{m}(t, s)ds)f(t_{k})\mid&=o(n^{-\kappa})+o(\tau_{m}), \\ \underset{0\leq t\leq 1}{\sup}\mid g(t)-\underset{k=1}{\overset{n}\sum}(\displaystyle\int_{A_{k}}E_{m}(t, s)ds)g(t_{k})\mid&=o(n^{-\kappa})+o(\tau_{m}). \end{align*} $

引理2^[13]  若条件(5) 成立, 其中 $k\in N, C_{k}$只与 $k$有关的实数, 则有

(a) $\mid E_{0}(t, s)\mid\leq\frac{C_{k}}{(1+\mid t-s \mid)^{k}}$, $\mid E_{m}(t, s)\mid\leq\frac{2^{m}C_{k}}{(1+2^{m}\mid t-s \mid)^{k}}$;

(b) $\underset{0\leq s\leq 1}{\sup}\mid E_{m}(t, s)\mid=o(2^{m})$;

(c) $\underset{0\leq t\leq 1}{\sup}\displaystyle\int_{0}^{1}\mid E_{m}(t, s)\mid ds\leq C$;

(d) $\displaystyle\int_{0}^{1}E_{m}(t, s)ds\rightarrow 1, n\rightarrow\infty$.

引理3  若条件(1)-(5) 成立, 则 $n^{-1}(\sum\limits_{\mathit{i}{\rm{ = 1}}}^\mathit{n} {} \widetilde{X}^{2}_{i}-n\sigma_{\varsigma}^{2})\rightarrow\sigma_{\eta}^{2}~~ {\rm a.s.}.$

证  注意到

$ \begin{align}\label{eq:4} n^{-1}(\underset{i=1}{\overset{n}\sum}\widetilde{X}^{2}_{i}-n\sigma_{\varsigma}^{2}) &=n^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{f}_{i}^{2} +n^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{\eta}_{i}^{2} +(n^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{\varsigma}_{i}^{2}-\sigma_{\varsigma}^{2})\nonumber\\ &~~~~~~+2n^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{f}_{i}\widetilde{\eta}_{i} +2n^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{f}_{i}\widetilde{\varsigma}_{i} +2n^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{\varsigma}_{i}\widetilde{\eta}_{i}\\ %\end{align*} %\begin{align} \label{eq:4} &=\underset{i=1}{\overset{6}\sum}U_{i}.\nonumber \end{align} % \end{equation} $

(2.1)

由引理1可得 $U_{1}\rightarrow 0$.由强大数定理和

$ \begin{eqnarray*} \underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\eta_{j}=o(n^{-1/3}\log n)~~ {\rm a.s..} \end{eqnarray*} $

(见文献[14]), 有

$ \begin{equation} \label{eq:5} U_{2}\rightarrow\sigma_{\eta}^{2}~~ {\rm a.s.} \end{equation} $

(2.2)

同理, 很容易证明 $U_{3}\rightarrow 0~~ {\rm a.s.}$.使用Cauchy-Schwarz不等式, 有

$ \begin{eqnarray} \label{eq:6} &&U_{4}\leq 2(U_{1}U_{2})^{1/2}\rightarrow 0~~ {\rm a.s.}, ~~ U_{5}\leq 2(U_{1}n^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{\varsigma}_{i}^{2})^{1/2}\rightarrow 0~~ {\rm a.s.}, \end{eqnarray} $

(2.3)

$ \begin{eqnarray} \label{eq:7} &&U_{6}\leq 2(U_{2}n^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{\varsigma}_{i}^{2})^{1/2}\rightarrow 0~~ {\rm a.s.}, \end{eqnarray} $

(2.4)

由(2.1)-(2.4) 式即得引理3.

引理4^[13]  (1) 存在 $\delta>0$, 使得 $Ee_{i}=0$且 $E\mid e_{i}\mid^{2+\delta}<\infty$, 则

$ \begin{align*} E(\underset{i=1}{\overset{n}\sum}e_{i})^{2}\leq(1+16\underset{i=1}{\overset{n}\sum}\alpha^{2/(2+\delta)}(\iota))\underset{i=1}{\overset{n}\sum}\parallel e_{i}\parallel^{2}_{2+\delta}, \end{align*} $

其中 $\parallel e_{i}\parallel_{2+\delta}=(E\mid e_{i}\mid^{2+\delta})^{1/(2+\delta)}$.

(2) 存在 $r>2, \delta>0, \lambda>\frac{r(r+\delta)}{2\delta}$和 $\alpha(n)=o(n^{-\lambda})$, 使得 $Ee_{i}=0$且 $E\mid e_{i}\mid^{r+\delta}<\infty$, 则 $\forall \varepsilon>0$, 存在正整数 $c=c(k, r, \delta, \lambda)$, 有

$ \begin{align*} E(\underset{1\leq m\leq n}{\max}\mid \underset{i=1}{\overset{n}\sum}e_{i}\mid^{r})\leq c(n^{\varepsilon}\underset{i=1}{\overset{n}\sum}E\mid e_{i}\mid^{r}+(\underset{i=1}{\overset{n}\sum}\parallel e_{i}\parallel^{2}_{r+\delta})^{r/2}). \end{align*} $

引理5^[15]  如果 $\{X_{k}\}$是数学期望为 $0$的独立r.v.序列, 那么对 $r\geq 2$,

$ \begin{align*} E\mid \Sigma X_{k}\mid^{r}\leq C_{r}((\Sigma EX_{k}^{2})^{r/2}+\Sigma E\mid X_{k}\mid^{r}). \end{align*} $

引理6  如果 $\{\xi_{i}, i\geq 1\}$是 $\alpha$ -混合的随机变量, $\{\zeta_{i}, i\geq 1\}$是独立的随机变量, 那么 $\{\xi_{i}\zeta_{i}, i\geq 1\}$也是 $\alpha$ -混合的随机变量.

证  令 $F_{\iota}^{m}=\sigma\{\xi_{\iota}, \xi_{\iota+1}, \cdots, \xi_{m}\}$表示包含 $\xi_{\iota}, \xi_{\iota+1}, \cdots, \xi_{m}, \iota\leq m$的 $\sigma$ -代数;

$ F'{\iota}^{m}=\sigma\{\zeta_{\iota}, \zeta_{\iota+1}, \cdots, \zeta_{m}\} $

表示包含 $\zeta_{\iota}, \zeta_{\iota+1}, \cdots, \zeta_{m}, \iota\leq m$的 $\sigma$ -代数;

$ F''{\iota}^{m}=\sigma\{\xi_{\iota}\zeta_{\iota}, \xi_{\iota+1}\zeta_{\iota+1}, \cdots, \xi_{m}\zeta_{m}\} $

表示包含 $\iota^{m}=\sigma\{\xi_{\iota}\zeta_{\iota}, \xi_{\iota+1}\zeta_{\iota+1}, \cdots, \xi_{m}\zeta_{m}\}, \iota\leq m$的 $\sigma$ -代数. $\forall A_{1}\in F_{n+i}^{\infty}, \forall A_{2}\in F_{1}^{i}, \forall B_{1}\in F^{'\infty}_{n+i}, \forall B_{2}\in F_{1}^{'i}, \forall C_{1}\in F^{''\infty}_{n+i}, \forall C_{2}\in F_{1}^{''i}$, 有

$ \begin{align*} \mid P(C_{1}C_{2})-P(C_{1})P(C_{2})\mid &=\mid P(B_{1})P(B_{2})P(A_{1}A_{2})-P(B_{1})P(B_{2})P(A_{1})P(A_{2})\mid\\ &=P(B_{1})P(B_{2})\mid P(A_{1}A_{2})-P(A_{1})P(A_{2})\mid. \end{align*} $

因为 $\{\xi_{i}, i\geq 1\}$是 $\alpha$ -混合的随机变量, 所以 $\{\xi_{i}\zeta_{i}, i\geq 1\}$也是 $\alpha$ -混合的随机变量.

定理1  若本文假设(1)-(5) 成立, 且存在 $r>2, \delta>0, \lambda>\frac{r(r+\delta)}{2\delta}$和 $\alpha(n)=o(n^{-\lambda})$, 使得 $\underset{i}{\sup}E\mid e_{i}\mid^{r+\delta}<\infty$, 且满足

$ E\mid \varsigma_{i}\mid^{2r\vee(r+\delta)}<\infty, E\mid \eta_{i}\mid^{2r\vee(r+\delta)}<\infty, 0<\varepsilon<\frac{r}{3}-\frac{2}{3}. $

则 $E\mid \widehat{\beta}_{n}-\beta\mid^{r}=o(n^{-\frac{r}{3}})+o(n^{-2\kappa r})+o(\tau_{m}^{2r}), $其中

$ \tau_{m}=\left\{ \begin{array}{ll} 2^{-m(\alpha-\frac{1}{2})},&\hbox{$\frac{1}{2}<\alpha<\frac{3}{2}$, } \\ \frac{\sqrt{m}}{2^{m}},&\hbox{$\alpha=\frac{3}{2}$, } \\ 2^{-m},&\hbox{$\alpha>\frac{3}{2}$.} \end{array} \right. $

证  注意到 $\widetilde{y}_{i}=\widetilde{x}_{i}\beta+\widetilde{g}(t_{i})+\widetilde{e}_{i}$, 易得

$ \begin{align}\label{eq:8} &(\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}^{2}-n\sigma_{\varsigma}^{2})^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}\widetilde{y}_{i}-\beta\nonumber\\ =&(\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}^{2}-n\sigma_{\varsigma}^{2})^{-1}\underset{i=1}{\overset{n}\sum}(\sigma_{\varsigma}^{2}- \widetilde{X}_{i}\widetilde{\varsigma}_{i})\beta +(\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}^{2}-n\sigma_{\varsigma}^{2})^{-1} \underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}\widetilde{g}(t_{i})\\ &+(\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}^{2}-n\sigma_{\varsigma}^{2})^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}e_{i} -(\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}^{2}-n\sigma_{\varsigma}^{2})^{-1}\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i} \underset{j=1}{\overset{n}\sum}e_{j}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\nonumber\\ =&T_{1}+T_{2}+T_{3}-T_{4}, \nonumber \end{align} $

(3.1)

则

$ \begin{align}\label{eq:9} T_{1}&=(n^{-1}(\underset{i=1}{\overset{n}\sum}\widetilde{X}_{i}^{2}-n\sigma_{\varsigma}^{2}))^{-1} (n^{-1}\underset{i=1}{\overset{n}\sum}(\sigma_{\varsigma}^{2}- \widetilde{\varsigma}_{i}^{2})\beta-n^{-1}\underset{i=1}{\overset{n}\sum}(\widetilde{f}(t_{i})\widetilde{\varsigma}_{i}\beta)- n^{-1}\underset{i=1}{\overset{n}\sum}(\widetilde{\eta}_{i}\widetilde{\varsigma}_{i}\beta))\nonumber\\ &=T_{11}+T_{12}+T_{13}. \end{align} $

(3.2)

由 $C_{r}$不等式, 引理3, 引理5和Cauchy-Schwarz不等式有

$ \begin{align}\label{eq:10} E\mid T_{11}\mid^{r}&\leq cn^{-r}E\mid \underset{i=1}{\overset{n}\sum}\{(\sigma_{\varsigma}^{2}- \varsigma_{i}^{2})\beta+2\varsigma_{i}\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\varsigma_{j}-(\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)\\ &ds\varsigma_{j})^{2}\}\mid^{r}\nonumber\\ &\leq cn^{-r}E\mid\underset{i=1}{\overset{n}\sum}(\sigma_{\varsigma}^{2}- \varsigma_{i}^{2})\beta\mid^{r}+cn^{-1}\underset{i=1}{\overset{n}\sum}\{E\mid \underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)\\ &ds\varsigma_{j}\mid^{2r}\}^{\frac{1}{2}}(E\mid \varsigma_{i}\mid^{2r})^{\frac{1}{2}}\nonumber\\ &~~~~~~+cE\mid\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\varsigma_{j}\mid^{2r}\nonumber\\ &=o(n^{-\frac{r}{3}}). \end{align} $

(3.3)

由 $C_{r}$不等式, 引理1, 引理3和引理5, 有

$ \begin{align}\label{eq:11} E\mid T_{12}\mid^{r}&\leq cn^{-r}\underset{1\leq i\leq n}{\max}\mid \widetilde{f}(t_{i})\mid^{r}(E\mid\underset{i=1}{\overset{n}\sum}\varsigma_{i}\mid^{r}+E\mid\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\varsigma_{j}\mid^{r})\nonumber\\ &=o(n^{-\frac{r}{2}}(n^{-\kappa r}+\tau_{m}^{r})). \end{align} $

(3.4)

由 $C_{r}$不等式, 引理3和引理5, 有

$ \begin{align}\label{eq:12} E\mid T_{13}\mid^{r}&\leq cn^{-r}E\mid\underset{i=1}{\overset{n}\sum}\eta_{i}\varsigma_{i}\mid^{r}+cn^{-r}E\mid\underset{i=1}{\overset{n}\sum} \underset{k=1}{\overset{n}\sum}\displaystyle\int_{A_{k}}E_{m}(t_{i}, s)ds\varsigma_{k}\eta_{i}\mid^{r}\nonumber\\ &~~~~~~+cn^{-r}E\mid\underset{i=1}{\overset{n}\sum} \underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\eta_{j}\varsigma_{i}\mid^{r}\nonumber\\ &~~~~~~+cn^{-r}E\mid\underset{i=1}{\overset{n}\sum} \underset{k=1}{\overset{n}\sum}\displaystyle\int_{A_{k}}E_{m}(t_{i}, s)ds\varsigma_{k}\eta_{i}\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\eta_{j}\varsigma_{i}\mid^{r}\nonumber\\ &=o(n^{-\frac{r}{3}}). \end{align} $

(3.5)

由 $C_{r}$不等式, 引理1, 引理3和引理5, 有

$ \begin{align}\label{eq:13} E\mid T_{2}\mid^{r}\leq& cn^{-r}\underset{1\leq i\leq n}{\max}\mid\widetilde{g}(t_{i})\mid^{r}\mid\underset{i=1}{\overset{n}\sum}\widetilde{f}(t_{i})\mid^{r} +cn^{-r}\underset{1\leq i\leq n}{\max}\mid\widetilde{g}(t_{i})\mid^{r}E\mid\underset{i=1}{\overset{n}\sum}\widetilde{\varsigma}_{i}\mid^{r}\nonumber\\ &+cn^{-r}\underset{1\leq i\leq n}{\max}\mid\widetilde{g}(t_{i})\mid^{r}E\mid\underset{i=1}{\overset{n}\sum}\widetilde{\eta}_{i}\mid^{r}\nonumber\\ \leq& cn^{-r}\underset{1\leq i\leq n}{\max}\mid\widetilde{g}(t_{i})\mid^{r}\mid\underset{i=1}{\overset{n}\sum}\widetilde{f}(t_{i})\mid^{r}\nonumber\\ &+cn^{-r}\underset{1\leq i\leq n}{\max}\mid\widetilde{g}(t_{i})\mid^{r}(E\mid\underset{i=1}{\overset{n}\sum}\varsigma_{i}\mid^{r}+ E\mid\underset{i=1}{\overset{n}\sum}\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\varsigma_{j}\mid^{r})\nonumber\\ &+cn^{-r}\underset{1\leq i\leq n}{\max}\mid\widetilde{g}(t_{i})\mid^{r}(E\mid\underset{i=1}{\overset{n}\sum}\eta_{i}\mid^{r}+ E\mid\underset{i=1}{\overset{n}\sum}\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\eta_{j}\mid^{r})\nonumber\\ =&o(n^{-\frac{r}{3}}(n^{-\kappa r}+\tau_{m}^{r}))+o(n^{-2\kappa r})+o(\tau_{m}^{2r}). \end{align} $

(3.6)

由 $C_{r}$不等式, 引理3, 引理4, 引理5和引理6, 有

$ \begin{align}\label{eq:14} E\mid T_{3}\mid^{r}&\leq cn^{-r}E\mid \underset{i=1}{\overset{n}\sum}\widetilde{f}(t_{i})e_{i}\mid^{r}+ cn^{-r}E\mid \underset{i=1}{\overset{n}\sum}\varsigma_{i}e_{i}\mid^{r}+cn^{-r}E\mid \underset{i=1}{\overset{n}\sum}\\ &(\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\varsigma_{j})e_{i}\mid^{r}\nonumber\\ &~~~~~~+cn^{-r}E\mid \underset{i=1}{\overset{n}\sum}\eta_{i}e_{i}\mid^{r} +cn^{-r}E\mid \underset{i=1}{\overset{n}\sum}(\underset{j=1}{\overset{n}\sum}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\eta_{j})e_{i}\mid^{r}\\ &=o(n^{-\frac{r}{3}}). \end{align} $

(3.7)

由 $C_{r}$不等式, 引理1, 引理3和引理4, 有

$ \begin{align}\label{eq:15} E\mid T_{4}\mid^{r}&\leq cn^{-r}E\mid\underset{i=1}{\overset{n}\sum}\widetilde{f}(t_{i})\underset{j=1}{\overset{n}\sum}e_{j}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\mid^{r}+cn^{-r}E\mid\underset{i=1}{\overset{n}\sum}\widetilde{\varsigma}_{i}\\ &\underset{j=1}{\overset{n}\sum}e_{j}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\mid^{r}\nonumber\\ &~~~~~~~+cn^{-r}E\mid\underset{i=1}{\overset{n}\sum}\widetilde{\eta}_{i}\underset{j=1}{\overset{n}\sum}e_{j}\displaystyle\int_{A_{j}}E_{m}(t_{i}, s)ds\mid^{r} =o(n^{-\frac{r}{3}}). \end{align} $

(3.8)

由 $C_{r}$不等式, (3.1)-(3.8) 式, 得

$ \begin{align*} &E\mid \widehat{\beta}_{n}-\beta\mid^{r}\leq E\mid T_{1}\mid^{r}+E\mid T_{2}\mid^{r}+E\mid T_{3}\mid^{r}+E\mid T_{4}\mid^{r}\\ &=o(n^{-\frac{r}{3}})+o(n^{-2\kappa r})+o(\tau_{m}^{2r}), \end{align*} $

定理1得证.

定理2  若本文假设(1)-(5) 成立, 且存在 $r>2, \delta>0, \lambda>\frac{r(r+\delta)}{2\delta}$和 $\alpha(n)=o(n^{-\lambda})$, 使得 $\underset{i}{\sup}E\mid e_{i}\mid^{r+\delta}<\infty$, 且满足

$ E\mid \varsigma_{i}\mid^{2r\vee(r+\delta)}<\infty, E\mid \eta_{i}\mid^{2r\vee(r+\delta)}<\infty, 0<\varepsilon<\frac{r}{3}-\frac{2}{3}, $

则 $E\mid \widehat{g}_{n}(t)-g(t)\mid^{r}=o(n^{-\kappa r})+o(\tau_{m}^{r})+o(n^{-\frac{r}{3}}).$

证  注意到

$ \begin{align}\label{eq:16} \widehat{g}_{n}(t)-g(t) &=\underset{i=1}{\overset{n}\sum}(x_{i}\beta+g(t_{i})+\varepsilon_{i}-X_{i}\widehat{\beta}_{n})\displaystyle\int_{A_{i}}E_{m}(t, s)ds-g(t)\nonumber\\ &=\underset{i=1}{\overset{n}\sum}f(t_{i})(\beta-\widehat{\beta}_{n})\displaystyle\int_{A_{i}}E_{m}(t, s)ds+\underset{i=1}{\overset{n}\sum}(\beta-\widehat{\beta}_{n})\eta_{i}\displaystyle\int_{A_{i}}E_{m}(t, s)ds\nonumber\\ &~~~~~~+(\underset{i=1}{\overset{n}\sum}g(t_{i})\displaystyle\int_{A_{i}}E_{m}(t, s)ds-g(t))+\underset{i=1}{\overset{n}\sum}\varepsilon_{i}\displaystyle\int_{A_{i}}E_{m}(t, s)ds\nonumber\\ &~~~~~~-\underset{i=1}{\overset{n}\sum}\varsigma_{i}(\widehat{\beta}_{n}-\beta)\displaystyle\int_{A_{i}}E_{m}(t, s)ds-\underset{i=1}{\overset{n}\sum}\varsigma_{i}\beta\displaystyle\int_{A_{i}}E_{m}(t, s)ds\nonumber\\ &=\underset{i=1}{\overset{6}\sum}T_{i}. \end{align} $

(3.9)

由引理2, 有

$ \begin{eqnarray} \label{eq:17} E\mid T_{1}\mid^{r}&\leq& c\mid \underset{i=1}{\overset{n}\sum}f(t_{i})\displaystyle\int_{A_{i}}E_{m}(t, s)ds\mid^{r}E\mid \beta-\widehat{\beta}_{n}\mid^{r}\nonumber\\ &\leq& cE\mid \beta-\widehat{\beta}_{n}\mid^{r}. \end{eqnarray} $

(3.10)

由Cauchy-Schwarz不等式和引理5, 有

$ \begin{align}\label{eq:18} E\mid T_{2}\mid^{r}&\leq c(E\mid\underset{i=1}{\overset{n}\sum}\eta_{i}\displaystyle\int_{A_{i}}E_{m}(t, s)ds\mid^{2r})^{1/2}(E\mid \beta-\widehat{\beta}_{n}\mid^{2r})^{1/2}\nonumber\\ &=o(n^{-\frac{r}{3}})(E\mid \beta-\widehat{\beta}_{n}\mid^{2r})^{1/2}. \end{align} $

(3.11)

由引理1, 可得

$ \begin{equation} \label{eq:19} E\mid T_{3}\mid^{r}=o(n^{-\kappa r})+o(\tau_{m}^{r}). \end{equation} $

(3.12)

由引理4, 有

$ \begin{eqnarray*} \label{eq:20} &E\mid T_{4}\mid^{r}\leq(n^{\varepsilon}\underset{i=1}{\overset{n}\sum}E\mid \varepsilon_{i}\displaystyle\int_{A_{i}}E_{m}(t, s)ds\mid^{r}+(\underset{i=1}{\overset{n}\sum}\parallel\varepsilon_{i}\displaystyle\int_{A_{i}}E_{m}(t, s)ds\parallel_{r+\delta}^{2})^{r/2})\\ &=o(n^{-\frac{r}{3}}). \end{eqnarray*} $

(3.13)

由Cauchy-Schwarz不等式和引理5, 有

$ \begin{align}\label{eq:21} E\mid T_{5}\mid^{r}&\leq c(E\mid\underset{i=1}{\overset{n}\sum}\varsigma_{i}\displaystyle\int_{A_{i}}E_{m}(t, s)ds\mid^{2r})^{1/2}(E\mid \beta-\widehat{\beta}_{n}\mid^{2r})^{1/2}\nonumber\\ &=o(n^{-\frac{r}{3}})(E\mid \beta-\widehat{\beta}_{n}\mid^{2r})^{1/2}. \end{align} $

(3.14)

由引理5, 有

$ \begin{eqnarray*} \label{eq:22} E\mid T_{6}\mid^{r}=E\mid\underset{i=1}{\overset{n}\sum}\varsigma_{i}\beta\displaystyle\int_{A_{i}}E_{m}(t, s)ds\mid^{r}=o(n^{-\frac{r}{3}}). \end{eqnarray*} $

(3.15)

由 $C_{r}$不等式和(3.10)-(3.15) 式, 有

$ \begin{align*} E\mid \widehat{g}_{n}(t)-g(t)\mid^{r}&\leq c(E\mid T_{1}\mid^{r}+E\mid T_{2}\mid^{r}+E\mid T_{3}\mid^{r}+E\mid T_{4}\mid^{r}+E\mid T_{5}\mid^{r}+E\mid T_{6}\mid^{r})\\ &=o(n^{-\kappa r})+o(\tau_{m}^{r})+o(n^{-\frac{r}{3}}). \end{align*} $

定理2得证.

注2  当 $\varsigma_{i}=0$时, 模型(1.1) 退化为一般的部分线性回归模型, 因此模型(1.1) 是一般的部分线性模型的推广.文献[4]要求 $e_{i}$独立而本文只需 $e_{i}$是 $\alpha$ -混合, 在条件比文献[4]弱的情况下, 由定理1和定理2可以直接得到文献[4]相应的结论.进一步, 当 $\beta=0$时模型(1.1) 退化为非线性模型, 在文献[7]中要求 $e_{i}$是 $\varphi$ -混合, 而模型(1.1) 是 $\alpha$ -混合比文献[7]的条件弱, 因此文献[7]的结论是定理2的推论.