部分线性EV模型如下

$ \begin{equation} y_k = x_k\beta+g(t_k)+\varepsilon_k, X_k = x_k+u_k, k = 1, 2, \cdot\cdot\cdot, n, \end{equation} $

(1.1)

这里$ \{y_k, k = 1, 2, \cdot\cdot\cdot, n\} $是观测值, $ \beta $是一维未知参数, $ g(t) $是光滑的曲线, $ \{t_k\} $为区间$ [0, 1] $内的序列. $ \{\varepsilon_k, k = 1, 2, \cdot\cdot\cdot, n\} $表示随机误差, 变量$ \{x_k\} $不能直接被观测, 只能通过$ X_k = x_k+u_k $来观测, 其中$ \{u_k\} $是测量误差, 且$ E(u_k) = 0, $var$ (u_k) = \sigma_1^2. $假设$ \{u_k\} $为已知独立同分布的随机变量, 而且对于每个$ k $值, $ \{u_k\} $与$ \{\varepsilon_k, k = 1, 2, \cdot\cdot\cdot, n\} $都是相互独立的.假设$ E(\varepsilon_i) = 0 $, var$ (\varepsilon_i) = 1 $ (即假定已知).则自协方差函数为

$ \begin{equation} \rho(k) = E(\varepsilon_1\varepsilon_{k+1}) = k^{-\theta}L(k), k = 1, 2, \cdot\cdot\cdot, n, \end{equation} $

(1.2)

这里$ 0<\theta<1 $是常数, $ L(t), t\in(0, \infty) $是正的缓慢变化函数, 即$ \lim\limits_{k\rightarrow\infty}\frac{L(t_k)}{L(k)} = 1, t\in(0, \infty). $

几十年来, 部分线性EV模型已经被广泛研究, 文献[1-4]用小波估计的方法研究了部分线性回归模型, 在误差序列$ \{\varepsilon_k\} $独立同分布时, 得到估计量的大样本性质; 若其误差为鞅差序列, 如参考文献[5], 则使用近邻估计方法来研究了部分线性EV模型.相依误差的一个重要的特殊情况就是误差为长相依的.这种情况会出现在经济学, 时间序列分析和其他学科领域的运用中, 见文献[6-8].使用小波估计对半参数模型的研究见文献[3, 9].文献[10]是对固定设计下的半参数回归模型使用非参数权函数法和最小二乘法.在一定的正则性条件下, 文献[11]中研究了部分线性模型中参数$ \beta $和函数$ g(\cdot) $的估计的弱相合性, 均方相合性和收敛速度, 得到了这些估计的渐近表示和渐近分布; 文献[12]研究了长相依随机设计下的部分线性EV模型, 得到了参数估计量的渐近表示, 渐近分布和弱收敛速度.本文是在这些文献的基础上, 运用小波估计研究了固定设计下的参数估计量的渐近表示, 渐近分布和弱收敛速度.

本节中, 我们使用小波方法估计未知参数和非参数.令$ A_i = [s_{i-1}, s_i] $表示区间$ [0, 1] $的划分区间, 且$ t_i\in{A_i}, 1\leqslant{i}\leqslant{n}. $由(1.1)式, 得到

$ \begin{equation} y_k = X_k\beta+g(t_k)+v_k, v_k = \varepsilon_k+\beta{u}_k. \end{equation} $

(2.1)

假设Schwartz空间$ S_l $中存在尺度函数$ \phi(x) $, 在伴随Hilbert空间中存在多分辨率分析$ V_m $, 它的再生核定义为$ E_m(t, s) = 2^m{E_0}(2^m{t}, 2^m{s}) = 2^m\sum\limits_{k\in{Z}}\phi(2^m{t}-k)\phi(2^m{s}-k). $将(2.3)式看成一般部分线性回归模型.首先, 假设$ \beta $已知, 定义$ g(t) $的估计量为

$ \begin{equation} \widetilde{g}_{0}(t) = \widetilde{g}_{0}(t, \beta) = \sum\limits_{i = 1}^{n}(y_i-X_i\beta)\int_{A_i}E_m(t, s)ds. \end{equation} $

(2.2)

然后, 通过最小化方法来定义小波估计量

$ \widehat{\beta}_n = \arg\min\limits_\beta\sum\limits_{i = 1}^{n}(y_i-X_i\beta-\widetilde{g}_0(t_i, \beta) )^{2} = (\sum\limits_{i = 1}^{n}\widetilde{X}_i^2 )^{-1}\sum\limits_{i = 1}^{n}\widetilde{X}_i\widetilde{y}_i, $

这里$ \tilde{X}_k = X_k-\sum\limits_{j = 1}^{n}X_j \int_{A_j}E_m(t_k, s)ds, $ $ \tilde{y}_k = y_k-\sum\limits_{j = 1}^{n}y_j\int_{A_j}E_m(t_k, s)ds. $

由于在线性回归或在部分线性回归中, 由测量误差所导致的不一致性可由“衰减效应”所克服.因此使用以下修正的最小二乘估计量

$ \begin{equation} \widetilde\beta_{n} = (\sum\limits_{i = 1}^{n}\widetilde{X}_i^2-n\sigma_1^2 )^{-1}\sum\limits_{i = 1}^{n}\widetilde{X}_i\widetilde{y}_i. \end{equation} $

(2.3)

最后, 定义$ g(t) $的小波估计量为

$ \begin{equation} \widetilde{g}_{n}(t) = \widetilde{g}_{0}(t, \widetilde\beta_n) = \sum\limits_{i = 1}^{n}(y_i-X_i\widetilde\beta_n) \int_{A_i}E_m(t, s)ds. \end{equation} $

(2.4)

为了获得主要结果, 作如下假设条件

(1) 令$ x_j = f(t_j)+\nu_j, j = 1, 2, \cdot\cdot\cdot, n $, 这里$ f(t_j) $是区间$ [0, 1] $中的函数, 对于实数列$ \{\nu_i\}, $有$ \max\limits_i|\nu_i| = O(n^{-\varphi}), 0<\varphi<\frac{1}{2}; $

(2) $ g( \cdot), f( \cdot)\in H^\alpha $(Sobolev空间), $ \alpha>1/2; $

(3) $ g( \cdot), f( \cdot) $都是阶数为$ \gamma>0 $的Lipschitz函数;

(4) $ 2^m = O(n^{-1}), 0<\theta<\min\{2\gamma/{\tilde m}, 2\lambda/{\tilde m}, (1-\lambda)/{\tilde m}\}, 0<\lambda<{1/2}, {\tilde m} $的定义见附录.

注1   条件(2)–(4)在小波估计中常被用到(见文献[4, 10, 17, 19]).条件(1)是研究部分线性模型时常用到的条件, 但是本文条件与他们有所不同.

(1) $ ' $文献[9, 10, 13]均对$ |\nu_i|^{p}(p\geq2) $施加条件, 本文仅对$ |\nu_i| $施加条件;

(2) $ ' $文献[9, 10, 13]中对$ \nu_i $施加的条件较多, 本文对$ \nu_i $仅有一个条件;

(3) $ ' $虽然文献[10, 13]的条件及文献[9]中的条件A1)ⅰ)均可由本文中$ \nu_i $的条件(条件(1))推出(本文中$ \nu_i $的条件比文献[10, 13]的条件及文献[9]中的条件1强), 但本文中$ \nu_i $的条件推不出文献[9]中的条件A1)ⅱ).说明本文$ \nu_i $条件比文献[9]中的条件A1) ⅱ)弱.

事实上, 将$ \max\limits_i|\nu_i| = O(n^{-\varphi}), 0<\varphi<\frac{1}{2}. $代入文献[9]中的A1) ⅱ)式, 得

$ \limsup\limits_{n\rightarrow\infty} \frac{1}{\sqrt{n}\log n}n^{1-\varphi} = (\log n)^{-1}n^{\frac{1}{2}-\varphi} = \infty. $

因此文献[9]中条件A1)ⅱ)不成立.

注2   与文献[12]类似, 文献[12]是随机变量, 本文是固定设计.

定理1  假设条件(1)–(4)均成立, 则对任意的$ \alpha\geq3/2, $

$ \begin{align} \mu_n(\tilde\beta_n-\beta) = \lambda_n \frac{J(\tilde m)}{\tilde m!} \sum\limits_{k = 1}^nu_kH_{\tilde m}(e_k)+o_p(1) \end{align} $

(3.1)

且有

$ \begin{align} \tilde\beta_n-\beta = O_p (n^{-\tilde m\theta/2}L^{\tilde m/2}(n) ), n\rightarrow\infty, \end{align} $

(3.2)

这里$ \mu_n = S_n\Gamma_n^{-1}, \lambda_n = (S_n\Gamma_n)^{-1}, \Gamma_n = n^{(1-\tilde n\theta)/2}L^{\tilde m/2}(n), \tilde x_k = x_k-\sum\limits_{j = 1}^nx_j \int_{A_j}E_m(t_k, s)ds $, $ S_n^2 = \sum\limits_{k = 1}^n\tilde x_k^2. $记号$ J( \cdot) $和$ H_{\tilde m}( \cdot) $见附录.

假设定理1的条件及结论成立, 则有以下推论, 其中推论1和推论2与文献[12]中推论相同.

推论1   若$ \{e_k, k = 1, 2, \cdot\cdot\cdot\} $是独立随机变量, $ G(s) = s $, 则对$ E|u_k|^{4+2\delta}<\infty, $ $ E|H_{\tilde m}(e_k)|^{4+2\delta}<\infty, $ $ \forall \delta>0, $有$ n^{1/2}(\tilde\beta-\beta)\overset D\longrightarrow N (0, \sigma_1^2J^2(\tilde m)/\tilde m!), $ $ n\rightarrow\infty. $

推论2   在定理1的条件下, 若$ u_k = 0 $, 则$ n^{1/2}(\tilde\beta-\beta)\overset D\longrightarrow N (0, \nu_k^2J^2(\tilde m)/\tilde m! ), n\rightarrow\infty. $

定理2   若定理1的条件成立, 则对$ \lambda/\tilde m<\theta, $

$ \begin{align} n^{(\tilde m\theta-\lambda)/2}L^{-\tilde m/2} (\tilde g_n(t)-g(t) )\overset{D}\rightarrow C_{\tilde m, \theta}(t)Z_{\tilde m} \end{align} $

(3.3)

且有

$ \begin{align} \tilde g_n(t)-g(t) = O_p (n^{(\lambda-\tilde m)/2}L^{\tilde m/2}(n) ), \; n\rightarrow\infty, \end{align} $

(3.4)

这里$ C_{\tilde m, \theta}(t) = \int_{A_{\tilde m}}E_m(t, s)dsC_{\tilde m, \theta}, C_{\tilde m, \theta} = (2\Gamma(\theta)\cos(\pi\theta/2) )^{-\tilde m/2}J(\tilde m)/{\tilde m!}. $

为了证明主要结果, 首先介绍一些有用的结论及引理.

首先介绍Hermite秩的表示和Hermite级数的一些基本结论.令$ e $表示$ N(0, 1) $上的随机变量, 且$ \mathbf{G} = \{G:E (G(e) ) = 0, E(G^2(e))<\infty \}, $则$ G $是

$ \{L^2(R.\Phi(s)): \{G: \int_{-\infty}^{+\infty}G^2(s)\Phi(s)ds<\infty \} \} $

的一个子集, 这里$ \Phi(s) = \exp(-s^2/2)/\sqrt{2\pi} $. Hermite多项式

$ H_q(s) = (-1)^q\exp(s^2/2)\frac{d^q}{ds^q}\exp(-s^2/2), s\in R, q = 0, 1, 2, \cdot\cdot\cdot $

构成了$ L^2(R, \Phi(s)) $中函数的完整正交系统, 并满足$ E (H_l(e)H_q(e) ) = \delta_{lq}q!. $

对任意的$ G\in\mathbf{G} $, 令$ J(q) = EG(e)H_q(e). $定义$ \tilde m = \min\limits_{q\geq0}\{q:J(q)\neq0\} $, 且为$ G $的Hermite秩.由于$ J(0) = E(G) = 0 $, 故$ \tilde m $通常是正数.

级数$ \sum\limits_{q = 0}^n J(q)/{q!} \cdot H_q(s) $收敛到$ L^2 (R, \Phi(s) ) $中的$ G(s) $ (见文献[17]), 即秩为$ \tilde m $的$ G(s) $的Hermite展式为$ \sum\limits_{q = 0}^n J(q)/{q!} \cdot H_q(s). $

令$ \{e_k, k = 1, 2, \cdot\cdot\cdot, n\} $和$ \rho(k) $如上所述, $ \tilde m>0 $为固定常数, 则以下结果成立^[18].

(F₁) $ \sum\limits_{i = 1}^n\sum\limits_{j = 1}^n |\rho(i-j) |^{\tilde m} = \left\{\begin{array}{ll} O (n^{2-\tilde m\theta}L^{\tilde m}(n) ), & \tilde m\theta<1, \\ O (nL_o(n) ), & \tilde m\theta = 1, \\ O(n), & \tilde m\theta>1, \end{array} \right. $其中$ L_o(n) $是缓慢变化函数.

(F₂)对任意Hermite秩为$ \tilde m<1/\theta, $可测函数$ G\in L^2(R, \Phi(s)), $

$ \hbox {Var} (\sum\limits_{i = 1}^nG(e_i) )\approx\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n |\rho(i-j) |^{\tilde m}\approx n^{2-\tilde m\theta}L^{\tilde m}(n). $

特别的,

$ \hbox {Var} (\sum\limits_{i = 1}^n\tilde G(e_i) )\approx\sum\limits_{i = 1}^n\sum\limits_{j = 1}^na_ia_j\rho^{\tilde m}(i-j)\approx\max\limits_{1\leq i\leq n} a_i^2n^{2-\tilde m\theta}L^{\tilde m}(n). $

(F₃)对任意缓慢变化函数$ L(n) $而言, 有$ n^\kappa L(n)\overset{n\rightarrow\infty}\longrightarrow\left\{\begin{array}{ll} \infty, & \kappa>0, \\ 0, & \kappa<0. \end{array} \right. $

(F₄) $ \{H_q(e_i)\} $是$ L^2 (R, \Phi(s) ) $中的一列随机变量, 且满足

$ H_o(x) = 1, EH_q(e) = 0(q\geq1), E (H_q(e_j)H_q(e_k) ) = q!\rho^q(j-k), \forall j, k. $

引理1^[3]  假设条件(3)成立, 则

(1) $ \sup\limits_t |E_m(t, s) | = O(2^m); $ (2) $ \sup\limits_t \int_0^1 |E_m(t, s) |ds\leq C $; (3) $ \int_0^1 |E_m(t, s) |ds\leq1, n\rightarrow\infty. $

引理2   假设条件(1)和(3)成立, 则$ x_i $有界, $ \tilde{x_i} $有界, 且$ S_n^2 = \sum\limits_{i = 1}^{n}\tilde x_i^2 = O(n) $.

证  由于$ \{\nu_i\} $是实数列, $ f(\cdot) $是阶数为$ \gamma>0 $的Lipschitz函数, 故$ x_j = f(t_j)+\nu_j, $ $ j = 1, 2, \cdots, n $有界.由$ \tilde x_i = {x_i}-\sum\limits_{j = 1}^{n}{x_j} \int_{A_j}E_m(t_i, s)ds $和引理1 (2), 有

$ \begin{align*} |\tilde{x_i}|& = |{x_i}- \sum\limits_{j = 1}^{n}{x_j} \int_{A_j}E_m(t_i, s)ds |\leq|{x_i}|+ | \sum\limits_{j = 1}^{n}{x_j} \int_{A_j}E_m(t_i, s)ds |\\ &\leq|{x_i}|+ \sum\limits_{j = 1}^{n}|{x_j}| \int_{A_j} |E_m(t_i, s) |ds\leq|{x_i}|+\max\limits_{j}|{x_j}|\sum\limits_{j = 1}^{n} \int_{A_j} |E_m(t_i, s) |ds\\ & = |{x_i}|+\max\limits_{j}|{x_j}| \int_0^1 |E_m(t_i, s) |ds\leq|{x_i}|+C\max\limits_{j}|{x_j}|. \end{align*} $

即$ \tilde{x_i} $有界, 得证.

注意到$ S_n^2 = \sum\limits_{i = 1}^{n}\tilde x_i^2 = O(n) $是显然的.

引理3   令$ B_n = S_n^2 (\sum\limits_{i = 1}^{n}\tilde{X_i}^{2}-n\sigma_1^2 )^{-1} $, 则有$ B_n\overset{p}{\longrightarrow}1, n\rightarrow\infty. $

证   为了证明此引理, 只需证明$ \sum\limits_{i = 1}^n\tilde X_i^2-n\sigma_1^2-\sum\limits_{i = 1}^n\tilde x_i^2\overset p\longrightarrow0\ \ (n\rightarrow\infty). $注意到

$ \sum\limits_{i = 1}^n\tilde X_i^2-n\sigma_1^2-\sum\limits_{i = 1}^n\tilde x_i^2 = \sum\limits_{i = 1}^n(\tilde u_i^2-\sigma_1^2)+2\sum\limits_{i = 1}^n\tilde x_i\tilde u_i = D_1+2D_2. $

由$ C_r $不等式及引理2, 有

$ \begin{align*} ED_2^2& = E(\sum\limits_{i = 1}^n\tilde x_i\tilde u_i)^2 = E (\sum\limits_{i = 1}^n\tilde x_i (u_i-\sum\limits_{j = 1}^n \int_{A_j}E_m(t_i, s)ds ) )^2\\ &\leq2\sum\limits_{i = 1}^n\tilde x_i^2Eu_i^2+2\sum\limits_{i = 1}^n\tilde x_i^2E (\sum\limits_{j = 1}^n u_j \int_{A_j}E_m(t_i, s)ds )^2\\ &\leq CS_n^2+C\sum\limits_{i = 1}^n\tilde x_i^2\sum\limits_{j = 1}^n Eu_j^2 ( \int_{A_j}E_m(t_i, s)ds )^2\\ &\leq CS_n^2+ CS_n^2n^{-1}2^m\leq CS_n^2 = O(n), \\ ED_1^2 = &E (\sum\limits_{i = 1}^n(u_i^2-\sigma_1^2)-2\sum\limits_{i = 1}^nu_i\sum\limits_{j = 1}^nu_j \int_{A_j}E_m(t_i, s)ds+\sum\limits_{i = 1}^n (\sum\limits_{j = 1}^nu_j \int_{A_j}E_m(t_i, s)ds )^2 )^2\\ \leq& CE (\sum\limits_{i = 1}^n(u_i^2-\sigma_1^2) )^2+CE (\sum\limits_{i = 1}^nu_i\sum\limits_{j = 1}^nu_j \int_{A_j}E_m(t_i, s)ds )^2\\ &+CE (\sum\limits_{i = 1}^n (\sum\limits_{j = 1}^nu_j \int_{A_j}E_m(t_i, s)ds )^2 )^2\\ = &D_{11}+D_{12}+D_{13}. \end{align*} $

由$ u_i $的独立性得

$ D_{11} = CE (\sum\limits_{i = 1}^n(u_i^2-\sigma_1^2) )^2\leq Cn. $

由引理1及$ u_i $的独立性得

$ \begin{align*} D_{12} = &CE (\sum\limits_{1\leq i, k\leq n}u_iu_k (\sum\limits_{j = 1}^nu_j \int_{A_j}E_m(t_i, s)ds ) (\sum\limits_{j = 1}^nu_j \int_{A_j}E_m(t_i, s)ds ) )\\ = &C\sum\limits_{i = 1}^nE (u_i^2\sum\limits_{j = 1}^nu_j^2 \int_{A_j}E_m(t_i, s)ds^2 )\\ &+C\sum\limits_{i\neq k}E (u_iu_k\sum\limits_{j\neq l}u_ju_l\int_{A_j}E_m(t_i, s)ds \int_{A_l}E_m(t_k, s)ds) \\ = &C\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n\sigma_1^4 ( \int_{A_j}E_m(t_i, s)ds )^2\\ &+C\sum\limits_{i\neq k}E (u_i^2u_k^2 \int_{A_i}E_m(t_i, s)ds \int_{A_k}E_m(t_k, s)ds )\\ \leq& Cnn^{-1}2^m+Cnn^{-1}2^m = C2^m = Cn^\lambda, \\ D_{13} = &CE (\sum\limits_{1\leq i, k\leq n} (\sum\limits_{j = 1}^nu_j \int_{A_j}E_m(t_i, s)ds )^2 (\sum\limits_{j = 1}^nu_j \int_{A_j}E_m(t_k, s)ds )^2 )\\ \leq& C\sum\limits_{1\leq i, k\leq n} (\sum\limits_{j = 1}^nEu_j^4 ( \int_{A_j}E_m(t_i, s)ds )^2 ( \int_{A_j}E_m(t_k, s) )^2\\ &+\sum\limits_{j = 1}^n\sigma^4 ( \int_{A_j}E_m(t_i, s)ds )^2 ( \int_{A_j}E_m(t_k, s)ds )^2 )\\ \leq& C\sum\limits_{1\leq i, k\leq n}(n^{-1}2^m)^3 = n^{-1}2^{3m} = n^{3\lambda-1}, \end{align*} $

故$ ED_1^2 = O(n). $

综上, 由Chebyshev不等式可得$ B_n = S_n^2 (\sum\limits_{i = 1}^{n}\tilde{X_i}^{2}-n\sigma_1^2 )^{-1}\overset{p}{\longrightarrow}1, n\rightarrow\infty. $

引理4^[14]  假设条件(1)–(4)成立, 则

$ \sup\limits_t |f(t)-\sum\limits_{k = 1}^n ( \int_{A_k}E_m(t, s)ds )f(t_k) | = O(n^{-\gamma})+O(\tau_m), $

且有$ \sup\limits_t |g(t)-\sum\limits_{k = 1}^n ( \int_{A_k}E_m(t, s)ds )g(t_k) | = O(n^{-\gamma})+O(\tau_m), $其中

$ \tau_m = \left\{\begin{array}{ll} 2^{-m(\theta-1/2)}, & 1/2<\theta<3/2, \\ \sqrt{m}\cdot2^{-m}, & \theta = 3/2, \\ 2^{-m}, & \theta>3/2. \end{array} \right. $

引理5^[11]  假设条件(4)成立, 则对每个实数$ a_1, a_2, \cdot\cdot\cdot, a_n $, 下列分解成立

$ \sum\limits_{j = 1}^na_j\varepsilon_j = \frac{J(\tilde m)}{\tilde m!}\sum\limits_{j = 1}^na_jH_{\tilde m}(e_j)+\sum\limits_{j = 1}^na_j (G(e_j)-\frac{J(\tilde m)}{\tilde m!}H_{\tilde m}(e_j) ): = T_{l1}+T_{l2}, $

这里$ T_{l1}, T_{l2} $满足

$ ET_{l1}^2\leq\max\limits_{1\leq j\leq n}|a_j|^2O (n^{2-\tilde m\theta}L^{\tilde m}(n) ), \; ET_{l2}^2\leq\max\limits_{1\leq j\leq n}|a_j|^2O (n^{2-\tilde m\theta}L^{\tilde m}(n) ). $

引理6^[13]  令$ \tilde m $表示函数$ G\in\mathbf{G} $的Hermite秩, $ a_i $表示有界的非负实数, 则$ \tilde G(e_i) = a_iG(e_i)\in\mathbf{G} $, 它的Hermite秩也是$ \tilde m $, 且$ \tilde G(e_i) = \sum\limits_{q = \tilde m}^\infty\frac{J(\tilde q)}{\tilde q!}\cdot H_q(e_i), \tilde J(q) = a_iJ(q). $

引理7^[15]  若$ G(e_i)\in\mathbf{G} $的Hermite秩是$ \tilde m $, 则

$ \sum\limits_{j = 1}^nG(e_j) / ({\hbox{Var}} (\sum\limits_{j = 1}^nG(e_j) ) )^{1/2}\overset{D}\longrightarrow C_{\tilde m, \theta}Z_{\tilde m}\ \ (n\rightarrow\infty), $

这里$ C_{\tilde m, \theta} = (2\Gamma(\theta)\cos(\pi\theta/2))^{-\tilde m/2}J(\tilde m)/{\tilde m!}, $

$ Z_m = \int\cdot\cdot\cdot\int\exp(i\sum\limits_{j = 1}^{\tilde m}t_j) ((\exp(i\sum\limits_{j = 1}^{\tilde m}t_j)-1)/i\sum\limits_{j = 1}^{\tilde m}t_j )\prod\limits_{j = 1}^{\tilde m}|t_j|^{(\theta-1)/2}\cdot dW(t_1)\cdot\cdot\cdot dW(t_{\tilde m}). $

定理1的证明  对$ \mu(\tilde\beta_n-\beta) $, 由引理2进行如下分解

$ \begin{align*} \mu(\tilde\beta_n-\beta) = &\lambda_n (S_n^2(\sum\limits_{k = 1}^n\tilde{X}_k^2-n\sigma_1^2)^{-1}\sum\limits_{k = 1}^n\tilde X_ky_k-S_n^2\beta )\\ = &\lambda_n (\sum\limits_{k = 1}^n\tilde X_ky_k-S_n^2\beta )\\ = &\lambda_n (\sum\limits_{k = 1}^n\tilde x_k\varepsilon_k-\sum\limits_{k = 1}^n\tilde x_k\bar\varepsilon_k+\sum\limits_{k = 1}^n\tilde x_k\tilde g_k+\sum\limits_{k = 1}^n\tilde u_k\varepsilon_k )\\ &+\lambda_n (-\sum\limits_{k = 1}^n\tilde u_k\bar\varepsilon_k+\sum\limits_{k = 1}^n\tilde u_k\tilde g_k+\beta\sum\limits_{k = 1}^n\tilde x_k\tilde u_k )\\ = &\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^nu_kH_{\tilde m}(e_k)+\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n\nu_kH_{\tilde m}(e_k)\\ &+\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n\tilde f_kH_{\tilde m}(e_k)+\lambda_n\sum\limits_{k = 1}^n\tilde x_k (\varepsilon_k-\frac{J(\tilde m)}{\tilde m!}H_{\tilde m}(e_k) )\\ &-\lambda_n\sum\limits_{k = 1}^n\tilde x_k\bar\varepsilon_k+\lambda_n\sum\limits_{k = 1}^n\tilde x_k\tilde g_k-\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n\bar u_kH_{\tilde m}(e_k)\\ &+\lambda_n\sum\limits_{k = 1}^n\tilde u_k (\varepsilon_k-\frac{J(\tilde m)}{\tilde m!}H_{\tilde m}(e_k) )\\ &+\lambda_n (-\sum\limits_{k = 1}^n\tilde u_k\bar\varepsilon_k+\sum\limits_{k = 1}^n\tilde u_k\tilde g_k+\beta\sum\limits_{k = 1}^n\tilde x_k\tilde u_k )\\ : = &\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^nu_kH_{\tilde m}(e_k)+\sum\limits_{l = 1}^{10}T_{nl}+o_p(1). \end{align*} $

由引理6及(F4)有

$ \begin{align} ET_{n1}^2& = E (\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n\nu_kH_{\tilde m}(e_k) )^2 = n^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{(\tilde m!)^2} (\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n\nu_i\nu_jE (H_{\tilde m}(e_i)H_{\tilde m}(e_j) ) )\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{(\tilde m!)^2}\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n|\nu_i\nu_j| |E (H_{\tilde m}(e_i)H_{\tilde m}(e_j) ) |\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{\tilde m!}\max\limits_{i, j}|\nu_i\nu_j|\sum\limits_{i = 1}^n\sum\limits_{j = 1}^n |\rho(i-j) |^{\tilde m}\\ & = n^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{\tilde m!}o(1)n^{2-\tilde m\theta}L^{\tilde m}(n) = o_p(1). \end{align} $

(5.1)

由引理2及(F4)有

$ \begin{align} ET_{n2}^2& = E (\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n\tilde f_kH_{\tilde m}(e_k) )^2 = n^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{(\tilde m!)^2} (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\tilde f_k\tilde f_jE (H_{\tilde m}(e_k)H_{\tilde m}(e_j) ) )\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{(\tilde m!)^2}\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n|\tilde f_k\tilde f_j| |E (H_{\tilde m}(e_i)H_{\tilde m}(e_j) ) |\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{\tilde m!}\sup\limits_{k, j}|\tilde f_k\tilde f_j|\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n |\rho(i-j) |^{\tilde m}\\ & = n^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{\tilde m!} (O(n^{-\gamma})+O(\tau_m) )n^{2-\tilde m\theta}L^{\tilde m}(n) = o_p(1). \end{align} $

(5.2)

对$ T_{n3} $求二阶矩

$ \begin{align} ET_{n3}^2& = E (\lambda_n\sum\limits_{k = 1}^n\tilde x_k\tilde G(e_k) )^2 = E (\lambda_n\sum\limits_{k = 1}^n(\tilde f_k+\nu_k)\tilde G(e_k) )^2\\ &\leq C (E (\lambda_n^2 (\sum\limits_{k = 1}^n\tilde f_k\tilde G(e_k) ) )^2+E (\lambda_n^2 (\sum\limits_{k = 1}^n\nu_k\tilde G(e_k) )^2 ) )\\ &: = T_{n3}^{(1)}+T_{n3}^{(2)}, \end{align} $

由引理2及(F2)有

$ \begin{align} T_{n3}^{(1)}& = CE (\lambda_n\sum\limits_{k = 1}^n\tilde f_k\tilde G(e_k) )^2\\ & = Cn^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\tilde f_k\tilde f_j\tilde G(e_k)\tilde G(e_j) )\\ &\leq Cn^{\tilde m\theta-2}L^{-\tilde m}(n)\sup\limits_{k, j}|\tilde f_k\tilde f_j|E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\tilde G(e_k)\tilde G(e_j) )\\ &\leq Cn^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{\tilde m!} (O(n^{-2\gamma})+O(\tau_m^2) )Var (\sum\limits_{j = 1}^n\tilde G(e_j) )\\ &\leq Cn^{\tilde m\theta-2}L^{-\tilde m}(n)\frac{J^2(\tilde m)}{\tilde m!} (O(n^{-2\gamma})+O(\tau_m^2) )n^{2-\tilde m\theta}L^{\tilde m}(n) = o_p(1). \end{align} $

(5.3)

由引理6及(F2)有

$ \begin{align} T_{n3}^{(2)}&\leq CE(\lambda_n^2)E (\sum\limits_{k = 1}^n\nu_k\tilde G(e_k) )^2 = n^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{j = 1}^n\sum\limits_{k = 1}^n\nu_j\nu_k\tilde G(e_j)\tilde G(e_k) )\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\max\limits_{j, k}|\nu_j\nu_k|E (\sum\limits_{j = 1}^n\sum\limits_{k = 1}^n\tilde G(e_j)\tilde G(e_k) )\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\max\limits_{j, k}|\nu_j\nu_k|\text {Var} (\sum\limits_{j = 1}^n\tilde G(e_j) )\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)o(1)n^{2-\tilde m\theta}L^{\tilde m}(n) = o_p(1). \end{align} $

(5.4)

由(5.3)–(5.4)式可得

$ \begin{align} ET_{n3}^2\leq o_p(1). \end{align} $

(5.5)

对$ T_{n4} $求二阶矩

$ \begin{align} ET_{n4}^2& = E (\lambda_n\sum\limits_{k = 1}^n\tilde x_k\bar\varepsilon_k )^2 = E (\lambda_n\sum\limits_{k = 1}^n(\tilde f_k+\nu_k)\bar\varepsilon_k )^2\\ & = E (\lambda_n^2(\sum\limits_{k = 1}^n\tilde f_k\bar\varepsilon_k)^2+\lambda_n^2(\sum\limits_{k = 1}^n\nu_k\bar\varepsilon_k)^2 ): = T_{n4}^{(1)}+T_{n4}^{(2)}. \end{align} $

由引理2及(F2)有

$ \begin{align} T_{n4}^{(1)}&\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\sup\limits_{k}|\tilde f_k|^2E (\sum\limits_{k = 1}^n\bar\varepsilon_k^2 )\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\sup\limits_{k}|\tilde f_k|^2n\max\limits_kE (\sum\limits_{k = 1}^n\int_{A_j}E_m(t_k, s)dsG(e_j) )^2\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\sup\limits_{k}|\tilde f_k|^2nn^{2\lambda-2}\max\limits_kE (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^nG(e_k)G(e_j) )\\ & = n^{2\lambda-3+\tilde m\theta}L^{-\tilde m}(n) (O(n^{-2\gamma})+O(\tau_m^2) )\text {Var} (\sum\limits_{k = 1}^nG(e_k) ) = n^{2\lambda-1}o(1) = o_p(1). \end{align} $

(5.6)

由引理6及(F2)有

$ \begin{align} T_{n4}^{(2)}&\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\max\limits_{k, j}|\nu_k\nu_j|E (\sum\limits_{k = 1}^n\bar\varepsilon_k^2 )\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\max\limits_{k, j}|\nu_k\nu_j|n\max\limits_kE (\sum\limits_{k = 1}^n\int_{A_j}E_m(t_k, s)dsG(e_j) )^ 2\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\max\limits_{k, j}|\nu_k\nu_j|nn^{2\lambda-2}\max\limits_kE (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^nG(e_k)G(e_j) )\\ & = n^{2\lambda-3+\tilde m\theta}L^{-\tilde m}(n)o(1){\hbox{Var}} (\sum\limits_{k = 1}^nG(e_k) ) = n^{2\lambda-1}o(1) = o_p(1). \end{align} $

(5.7)

由(5.6)–(5.7)式可得

$ \begin{align} ET_{n4}^2\leq o_p(1). \end{align} $

(5.8)

对$ T_{n5} $求二阶矩

$ \begin{align} ET_{n5}^2& = E (\lambda_n\sum\limits_{k = 1}^n\tilde x_k\tilde g_k )^2 = E (\lambda_n\sum\limits_{k = 1}^n(\tilde f_k+\nu_k)\tilde g_k )^2 = E (\lambda_n^2(\sum\limits_{k = 1}^n\tilde f_k\tilde g_k)^2 )+E (\lambda_n^2(\sum\limits_{k = 1}^n\nu_k\tilde g_k)^2 )\\& : = T_{n5}^{(1)}+T_{n5}^{(2)}. \end{align} $

由引理2有

$ \begin{align} T_{n5}^{(1)}&\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\sup\limits_{k, j}|\tilde f_k\tilde f_j|\sup\limits_{k, j}|\tilde g_k\tilde g_j|\\ & = n^{\tilde m\theta-2}L^{-\tilde m}(n)n^2 (O(n^{-2\gamma})+O(\tau_m^2) )^2\\ & = n^{\tilde m\theta-4\gamma}L^{-\tilde m}(n) = o_p(1). \end{align} $

(5.9)

由引理2及引理6有

$ \begin{align} T_{n5}^{(1)}&\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\max\limits_{k, j}|\nu_k\nu_j|\sup\limits_{k, j}|\tilde g_k\tilde g_j|\\ & = n^{\tilde m\theta-2}L^{-\tilde m}(n)n^2o(1) (O(n^{-2\gamma})+O(\tau_m^2) ) = o_p(1). \end{align} $

(5.10)

由(5.9)–(5.10)式可得

$ \begin{align} ET_{n5}^2\leq o_p(1). \end{align} $

(5.11)

由(F4)有

$ \begin{align} ET_{n6}^2& = E (\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n\bar u_kH_{\tilde m}(e_k) )^2\\ & = \frac{J^2(\tilde m)}{(\tilde m!)^2}n^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\bar u_k\bar u_jH_{\tilde m}(e_k)H_{\tilde m}(e_j) )\\ &\leq \frac{J^2(\tilde m)}{(\tilde m!)^2}n^{\tilde m\theta-2}L^{-\tilde m}(n)\sum\limits_{1\leq j, k\leq n} (E(\bar u_k\bar u_j)\tilde m!\rho^{\tilde m}(j-k) )\\ &\leq \frac{J^2(\tilde m)}{\tilde m!}n^{\tilde m\theta-2}L^{-\tilde m}(n)\max\limits_{k, j} |E(\bar u_k\bar u_j) |\sum\limits_{1\leq j, k\leq n}\rho^{\tilde m}(j-k), \\ E(\bar u_k\bar u_j)^2&\leq (\sup\limits_s |E_m(t_k, s) | )^4E ((n^{-1}\sum\limits_{i = 1}^nu_i)^2(n^{-1}\sum\limits_{l = 1}^nu_l)^2 )\\ &\leq (\sup\limits_s |E_m(t_k, s) | )^4E (n^{-1}\sum\limits_{i = 1}^nu_i )^4\\ &\leq C (2^{2m}n^{-1} )^2 (n^{-1}\sup\limits_iE^4{u_i} ) = o_p(1), \end{align} $

(5.12)

$ \begin{align} ET_{n6}^2&\leq\frac{J^2(\tilde m)}{\tilde m!}n^{\tilde m\theta-1}L^{-\tilde m}(n)n^{-1}o_p(1)n^{2-\tilde m\theta}L^{\tilde m}(n) = o_p(1). \end{align} $

(5.13)

对$ T_{n7} $求二阶矩

$ \begin{align} ET_{n7}^2& = E (\lambda_n\sum\limits_{k = 1}^n\tilde u_k\tilde G(e_k) )^2 = E (\lambda_n\sum\limits_{k = 1}^n(u_k-\bar u_k)\tilde G(e_k) )^2\\ &\leq CE (\lambda_n^2 (\sum\limits_{k = 1}^nu_k\tilde G(e_k) )^2 )+CE (\lambda_n^2 (\sum\limits_{k = 1}^n\bar u_k\tilde G(e_k) )^2 )\\ &: = T_{n7}^{(1)}+T_{n7}^{(2)}. \end{align} $

由(F2)有

$ \begin{align} T_{n7}^{(1)}& = CE (\lambda_n^2 (\sum\limits_{k = 1}^nu_k\tilde G(e_k) )^2 ) = Cn^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^nu_ku_j\tilde G(e_k)\tilde G(e_j) )\\ & = Cn^{\tilde m\theta-2}L^{-\tilde m}(n)n^{-1}\max\limits_jE(u_j^2)E (\sum\limits_{k = 1}^n\tilde G(e_k) )^2\\ &\leq Cn^{\tilde m\theta-2}L^{-\tilde m}(n)n^{-1}n^{2-\tilde m\theta}L^{\tilde m}(n) = o_p(1). \end{align} $

(5.14)

由(5.12)式及(F2)有

$ \begin{align} T_{n7}^{(2)}& = CE (\lambda_n^2 (\sum\limits_{k = 1}^n\bar u_k\tilde G(e_k) )^2 ) = Cn^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\bar u_k\bar u_j\tilde G(e_k)\tilde G(e_j) )\\ & = Cn^{\tilde m\theta-2}L^{-\tilde m}(n)\max\limits_jE(\bar u_k\bar u_j)E (\sum\limits_{k = 1}^n\tilde G(e_k) )^2\\ &\leq Cn^{\tilde m\theta-2}L^{-\tilde m}(n)o_p(1)n^{2-\tilde m\theta}L^{\tilde m}(n) = o_p(1). \end{align} $

(5.15)

由(5.14)–(5.15)式可得

$ \begin{align} ET_{n7}^2\leq o_p(1). \end{align} $

(5.16)

对$ T_{n8} $求二阶矩

$ \begin{align} ET_{n8}^2& = E (\lambda_n\sum\limits_{k = 1}^n\tilde u_k\bar\varepsilon_k )^2 = E (\lambda_n\sum\limits_{k = 1}^n(u_k-\bar u_k)\bar\varepsilon_k )^2\\ & = E (\lambda_n^2(\sum\limits_{k = 1}^nu_k\bar\varepsilon_k)^2 )+E (\lambda_n^2(\sum\limits_{k = 1}^n\bar u_k\bar\varepsilon_k)^2 ) : = T_{n8}^{(1)}+T_{n8}^{(2)}. \end{align} $

由(F2)有

$ \begin{align} T_{n8}^{(1)}& = n^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^nu_ku_j\bar\varepsilon_k\bar\varepsilon_j ) \leq n^{\tilde m\theta-2}L^{-\tilde m}(n)n\max\limits_jE(u_j^2)E(\bar\varepsilon_j^2)\\ & = n^{\tilde m\theta-2}L^{-\tilde m}(n)n\max\limits_jE(u_j^2)E (\sum\limits_{k = 1}^n\int_{A_j}E_m(t_k, s)dsG(e_j) )^2\\ &\leq Cn^{\tilde m\theta-2}L^{-\tilde m}(n)nn^{2\lambda-2}\max\limits_{k, j}E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^nG(e_k)G(e_j) )\\ & = Cn^{2\lambda-1} = o_p(1). \end{align} $

(5.17)

由(5.12)式及(F2)有

$ \begin{align} T_{n8}^{(2)}& = n^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\bar u_k\bar u_j\bar\varepsilon_k\bar\varepsilon_j ) \leq n^{\tilde m\theta-2}L^{-\tilde m}(n)n\max\limits_jE(\bar u_j^2)E(\bar\varepsilon_j^2)\\ & = n^{\tilde m\theta-2}L^{-\tilde m}(n)n\max\limits_jE(\bar u_j^2)E (\sum\limits_{k = 1}^n\int_{A_j}E_m(t_k, s)dsG(e_j) )^2\\ &\leq Cn^{\tilde m\theta-2}L^{-\tilde m}(n)no_p(1)n^{2\lambda-2}\max\limits_{k, j}E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^nG(e_k)G(e_j) )\\ & = Cn^{2\lambda-1}o_p(1) = o_p(1). \end{align} $

(5.18)

由(5.17)–(5.18)式可得

$ \begin{align} ET_{n8}^2\leq o_p(1). \end{align} $

(5.19)

对$ T_{n9} $求二阶矩

$ \begin{align} ET_{n9}^2& = E (\lambda_n\sum\limits_{k = 1}^n\tilde u_k\tilde g_k )^2 = E (\lambda_n\sum\limits_{k = 1}^n(u_k-\bar u_k)\tilde g_k )^2\\ & = E (\lambda_n^2(\sum\limits_{k = 1}^nu_k\tilde g_k)^2 )+E (\lambda_n^2(\sum\limits_{k = 1}^n\bar u_k\tilde g_k)^2 )\\ &: = ET_{n9}^{(1)}+ET_{n9}^{(2)}. \end{align} $

由引理2有

$ \begin{align} T_{n9}^{(1)}& = n^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^nu_ku_j\tilde g_k\tilde g_j ) \leq n^{\tilde m\theta-2}L^{-\tilde m}(n)n\max\limits_jE(u_j^2)E(\tilde g_j^2)\\ & = n^{\tilde m\theta-2}L^{-\tilde m}(n)n\max\limits_jE(u_j^2)\sup\limits_j|\tilde g_j|^2\\ &\leq Cn^{\tilde m\theta-2}L^{-\tilde m}(n)n (O(n^{-2\gamma})+O(\tau_m^2) )\\ & = Cn^{\tilde m\theta-1}L^{-\tilde m}(n)o_p(1) = o_p(1). \end{align} $

(5.20)

由(5.12)式及引理2有

$ \begin{align} T_{n9}^{(2)}& = n^{\tilde m\theta-2}L^{-\tilde m}(n)E (\sum\limits_{k = 1}^n\sum\limits_{j = 1}^n\bar u_k\bar u_j\tilde g_k\tilde g_j ) \leq n^{\tilde m\theta-2}L^{-\tilde m}(n)n\max\limits_jE(\bar u_j^2)E(\tilde g_j^2)\\ & = n^{\tilde m\theta-2}L^{-\tilde m}(n)n\max\limits_jE(\bar u_j^2)\sup\limits_j|\tilde g_j|^2\\ &\leq n^{\tilde m\theta-2}L^{-\tilde m}(n)no_p(1) (O(n^{-2\gamma})+O(\tau_m^2) )\\ & = n^{\tilde m\theta-1}L^{-\tilde m}(n)o_p(1) = o_p(1). \end{align} $

(5.21)

由(5.20)–(5.21)式可得

$ \begin{align} ET_{n9}^2\leq o_p(1). \end{align} $

(5.22)

对$ T_{n10} $求二阶矩

$ \begin{align*} ET_{n10}^2& = E (\beta\lambda_n\sum\limits_{k = 1}^n\tilde x_k\tilde u_k )^2 = E (\beta\lambda_n\sum\limits_{k = 1}^n\tilde x_k(u_k-\bar u_k) )^2\nonumber\\ &\leq E (\beta\lambda_n\sum\limits_{k = 1}^n\tilde x_ku_k )^2+E (\beta\lambda_n\sum\limits_{k = 1}^n\tilde x_k\bar u_k )^2\nonumber\\ &: = T_{n10}^{(1)}+T_{n10}^{(2)}.\nonumber\\ \end{align*} $

对于$ T_{n10}^{(1)} $有

$ \begin{align} T_{n10}^{(1)}& = E (\beta\lambda_n\sum\limits_{k = 1}^n\tilde x_ku_k )^2\leq\beta^2\Gamma^{-2}E (S_n^{-2}\sum\limits_{k = 1}^n\tilde x_k )^2\max\limits_kE(u_k^2)\\ &\leq Cn^{\tilde m\theta-1}L^{-\tilde m}(n) = o_p(1). \end{align} $

(5.23)

对于$ T_{n10}^{(2)} $, 由(5.12)式有

$ \begin{align} T_{n10}^{(2)}& = E (\beta\lambda_n\sum\limits_{k = 1}^n\tilde x_k\bar u_k )^2\leq\beta^2\Gamma^{-2}E (S_n^{-2}\sum\limits_{k = 1}^n\tilde x_k )^2\max\limits_kE(\bar u_k^2)\\ &\leq Cn^{\tilde m\theta-1}L^{-\tilde m}(n)o_p(1) = o_p(1). \end{align} $

(5.24)

由(5.12)式, 有

$ \begin{align} E (\lambda_n\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n u_kH_{\tilde m}(e_k) )^2 = o_p(1). \end{align} $

(5.25)

由(5.23)–(5.24)式可得

$ \begin{align} & ET_{n10}^2\leq o_p(1), \end{align} $

(5.26)

$ \begin{align} & E(\mu_n^{-1})^2 = \Gamma_n^2n^{-1}E(nS_n^{-2}) = O(1)n^{-\tilde m\theta}L^{\tilde m}(n) = O(1). \end{align} $

(5.27)

由(5.12)和Chebyshev不等式可得

$ \begin{align} \mu_n^{-1} = O_p(n^{-\tilde m\theta/2}L^{\tilde m/2}). \end{align} $

(5.28)

定理1证毕.

推论1的证明  由(3.1)和(5.28)式, 有

$ \begin{align} \tilde\beta_n-\beta = n^{-1}\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n (u_k+\nu_k)H_{\tilde m}(e_k)+o_p(1). \end{align} $

(5.29)

令$ \xi_k = n^{-1}\frac{J(\tilde m)}{\tilde m!} (u_k+\nu_k)H_{\tilde m}(e_k). $则$ \{\xi_k, k = 1, 2, \cdot\cdot\cdot, n\} $是独立随机变量, 且$ E\xi_k = 0 $,

$ \text {Var} (\sum\limits_{k = 1}^n\xi_k ) = \sum\limits_{k = 1}^n\text {Var}(\xi_k) = \sum\limits_{k = 1}^n n^{-2} (\frac{J(\tilde m)}{\tilde m!} )^2\sigma_1^2E (H_{\tilde m}^2(e_k) ) = n^{-1}\sigma_1^2\frac{J^2(\tilde m)}{\tilde m!}. $

因为$ B^{-\frac{2+\delta}{2}}: = (\sum\limits_{k = 1}^nE\xi_k^2 )^{-(2+\delta)/2} = n^{(2+\delta)/2} (C\frac{J^2(\tilde m)}{\tilde m!} )^{-(2+\delta)/2}, $且

$ \sum\limits_{k = 1}^nE|\xi_k|^{2+\delta}\leq|\frac{J(\tilde m)}{\tilde m!}|^{2+\delta}n^{-2-\delta}\sum\limits_{k = 1}^n (|\nu_k|^{4+2\delta}E|H_{\tilde m}(e_k)|^{4+2\delta} )^{1/2} = O(n^{-1-\delta}), $

则有$ B^{-(2+\delta)/2}\sum\limits_{k = 1}^nE|\varsigma_k|^{2+\delta} = O(n^{\delta/2}) = o(1). $因此Lindeberg条件成立, 由中心极限定理有

$ \begin{align} \sum\limits_{k = 1}^n\varsigma_k / ({\hbox{Var}} (\sum\limits_{k = 1}^n\varsigma_k ) )^{1/2}\overset{D}\rightarrow N(0, 1). \end{align} $

(5.30)

因此由(5.29)–(5.30)式得推论1成立.

推论2的证明  由(3.1)和(5.28)式, 有

$ \begin{align} \tilde\beta_n-\beta = n^{-1}\frac{J(\tilde m)}{\tilde m!}\sum\limits_{k = 1}^n \nu_kH_{\tilde m}(e_k)+o_p(1). \end{align} $

(5.31)

令$ \varsigma_k = n^{-1}\frac{J(\tilde m)}{\tilde m!} \nu_kH_{\tilde m}(e_k). $则$ \{\varsigma_k, k = 1, 2, \cdot\cdot\cdot, n\} $是独立随机变量, 且$ E\varsigma_k = 0 $,

$ \text {Var} (\sum\limits_{k = 1}^n\varsigma_k ) = \sum\limits_{k = 1}^n\text {Var}(\varsigma_k) = \sum\limits_{k = 1}^n n^{-2} (\frac{J(\tilde m)}{\tilde m!} )^2\nu_k^2E (H_{\tilde m}^2(e_k) ) = n^{-1}\nu_k^2\frac{J^2(\tilde m)}{\tilde m!}. $

因为

$ B^{-\frac{2+\delta}{2}}: = (\sum\limits_{k = 1}^nE\varsigma_k^2 )^{-(2+\delta)/2} = n^{(2+\delta)/2} (C\frac{J^2(\tilde m)}{\tilde m!} )^{-(2+\delta)/2}, $

且

$ \sum\limits_{k = 1}^nE|\varsigma_k|^{2+\delta}\leq|\frac{J(\tilde m)}{\tilde m!}|^{2+\delta}n^{-2-\delta}\sum\limits_{k = 1}^n (|\nu_k|^{4+2\delta}E|H_{\tilde m}(e_k)|^{4+2\delta} )^{1/2} = O(n^{-1-\delta}), $

有$ B^{-(2+\delta)/2}\sum\limits_{k = 1}^nE|\varsigma_k|^{2+\delta} = O(n^{\delta/2}) = o(1). $因此Lindeberg条件成立, 由中心极限定理有

$ \begin{align} \sum\limits_{k = 1}^n\varsigma_k / ({\hbox{Var}} (\sum\limits_{k = 1}^n\varsigma_k ) )^{1/2}\overset{D}\rightarrow N(0, 1). \end{align} $

(5.32)

因此由(5.31)–(5.32)式得推论2成立.

定理2的证明

$ \begin{align} &\tilde g_n(t)-g(t) = \sum\limits_{i = 1}^{n}(y_i-X_i\tilde\beta_n)\int_{A_i}E_m(t, s)ds-g(t)\\ = & (\sum\limits_{i = 1}^{n}g(t_i)\int_{A_i}E_m(t, s)ds-g(t) ) +(\beta-\tilde\beta)\sum\limits_{i = 1}^{n}x_i\int_{A_i}E_m(t, s)ds+\sum\limits_{i = 1}^{n}\varepsilon_i\int_{A_i}E_m(t, s)ds\\ &+(\beta-\tilde\beta)\sum\limits_{i = 1}^{n}u_i\int_{A_i}E_m(t, s)ds-\beta\sum\limits_{i = 1}^{n}u_i\int_{A_i}E_m(t, s)ds\\ : = &T_1^{(1)}+T_2^{(1)}+T_3^{(1)}+T_4^{(1)}+T_5^{(1)}. \end{align} $

注意到

$ \begin{align*} &n^{(\tilde m\theta-\lambda)/2}L^{-\tilde m/2}(n)T_1^{(1)}\\ = &O (n^{(2\gamma+\lambda)/\tilde m-\theta}L(n) )^{-\tilde m/2}+O ( (n^{\lambda/{\tilde m}-\theta}L(n) )^{-\tilde m/2}\tau_m )\\ = &o_p(1). \end{align*} $

因为

$ \begin{align} &E (\sum\limits_{i = 1}^{n}x_i\int_{A_i}E_m(t, s)ds )^2 \end{align} $

(5.33)

$ \begin{align} \leq& (\sum\limits_{i = 1}^nf(t_i)\int_{A_i}E_m(t, s)ds )^2+ (\sum\limits_{i = 1}^n\nu_i\int_{A_i}E_m(t, s)ds )^2\\ \leq&\sup\limits_{i, k} |f(t_i)f(t_k) | (\int_0^1E_m(t, s)ds )^2+\max\limits_{i, k} |\nu_i\nu_k | (\int_0^1E_m(t, s)ds )^2\\ \leq& O_p(1)+o_p(1)\leq O_p(1). \end{align} $

(5.34)

所以有

$ \begin{align} \sum\limits_{i = 1}^{n}x_i\int_{A_i}E_m(t, s)ds = O_p(1), \end{align} $

(5.35)

从而

$ \begin{align} n^{(\tilde m\theta-\lambda)/2}L^{-\tilde m/2}(n)T_2^{(1)} = O_p(n^{-\lambda/2}) = o_p(1). \end{align} $

(5.36)

由上式和引理2有

$ \begin{align} \text {Var}(T_3^{(1)})& = \text {Var} (\sum\limits_{i = 1}^{n}G(e_i)\int_{A_i}E_m(t, s)ds )\\ &\leq ( \int_{A_i}E_m(t, s)ds )^2{\hbox{Var}} (\sum\limits_{i = 1}^{n}G(e_i) )\\ &\leq\max\limits_i | \int_{A_i}E_m(t, s)ds |^2n^{2-\tilde m\theta}L^{\tilde m}(n)\\ & = O_p (n^{\lambda-\tilde m\theta}L^{\tilde m}(n) ). \end{align} $

(5.37)

由引理6和引理7有

$ \begin{align} n^{(\tilde m\theta-\lambda)/2}L^{-\tilde m/2}(n)T_3^{(1)}\overset{D}\rightarrow C_{\tilde m, \theta}(t)Z_{\tilde m}. \end{align} $

(5.38)

由引理1有

$ \begin{align} \text {Var} (T_4^{(1)} )& = \beta^2\sigma_1^2\sum\limits_{i = 1}^{n} ( \int_{A_i}E_m(t, s)ds )^2\leq Cn^{-1}2^m = O_p(n^{\lambda-1}), \end{align} $

(5.39)

因此由Markov不等式和(F$ _3 $)可得

$ \begin{align} n^{(\tilde m\theta-\lambda)/2}L^{-\tilde m/2}(n)T_4^{(1)} = O_p (n^{-(1-\tilde m\theta)}L^{-\tilde m/2}(n) ) = o_p(1). \end{align} $

(5.40)

由引理1, 可得到

$ \begin{align} n^{(\tilde m\theta-\lambda)/2}L^{-\tilde m/2}(n)T_5^{(1)} = o_p(1). \end{align} $

(5.41)

定理2证毕.

为了对本文的证明结果做进一步的解释和验证, 选取2013年5月到2018年8月之间的全国居民消费价格指数, 城市居民消费价格指数和农村居民消费价格指数, 使用Mathmatic来做模拟应用(具体数值见东方财富网《中国居民消费价格指数》).

使用下述的式子

$ \begin{eqnarray*} && \left\{\begin{array}{ll} \tilde\beta_n = (\tilde X^2-n\Sigma_1^2 )^{-1}\tilde X\tilde y, \\ \tilde g_n(t) = (y-X\tilde\beta_n )\int_AE_m(t, s)ds, \end{array} \right. \end{eqnarray*} $

其中设全国居民消费价格指数为$ y $, 城市居民消费价格指数和农村居民消费价格指数为矩阵$ X $, 通过$ X $, 由Mathmatic算得

$ \begin{align*} \tilde\beta_{n} = \begin{pmatrix} 0.725815\\ 0.274299 \end{pmatrix}. \end{align*} $

$ \tilde g_n(t) $的散点图如下图 1

相对误差$ H $由下式算得

$ H = \frac{|y_t-y|}{y}\cdot100\%. $

64个数据中有55个在0.05%以内, 占总数的85.94%;有9个在0.1%以内, 占总数的14.06%, 在一定程度上说明估计出的$ \tilde\beta_{n} $和$ \tilde g_n(t) $是有效的.这个例子也直接说明了本文前面所证明的结论是正确的, 对于数据处理预测也有实际的作用.

具体计算的结果如下表 1.