定义1 称随机变量序列$\{{{\eta }_{i}};i\ge 1\}~$为$\phi$-混合序列, 如果当$n\rightarrow\infty$时, 有

$\phi (n) = \mathop {\sup }\limits_{k \ge 1} \left\{ {\left| {\frac{{P(AB)}}{{P(A)}} - P(B)} \right|:A \in F_1^k, B \in F_{k + n}^\infty, P(A) > 0} \right\} \downarrow 0, $

其中${\cal{F}}^t_s$表示由$\{\eta_i, ~s\leq i\leq t\}$生成的$\sigma$ -域, 且称$\phi(n)$为$\phi$ -混合系数.

1959年, Ibragimov^[1]首次提出了$\phi$ -混合条件, 同时Cogburn^[2]进行了相关研究, 发现$\phi$ -混合的概念可作为弱相关的衡量尺度应用于时间序列的研究中. Bradley^[3]给出了一个较好的$\phi$ -混合条件以及其它常用混合条件的综述.由于$\phi$ -混合序列应用广泛, Utev^[4], Chen^[5], Herrndorf^[6], Peligrad^[7], Sen^{[8, 9]}, Shao^[10]和Wang^[11]等都对$\phi$ -混合随机变量序列的极限理论进行了深入研究.

1964年, Nadaraya^[12]首先提出了分位数的核估计. Yamato^[13], Parzen^[14]和Ralescu^[15]等对分位数核估计进行了深入研究, 但他们研究的都是独立情形.

本文研究$\phi$ -混合样本下总体的有限个分位数核估计的联合渐近分布, 主要结果在第二部分, 一些相关引理及其证明在第三部分, 主要结果的证明在第四部分.

设$X_1, X_2, \cdots, X_n$是一列$\phi$ -混合随机变量, 混合系数为$\phi$且该样本对应的总体的分布函数为$F$.分布函数$F(x)$在$x\in{R}$的光滑核估计为$\hat{F}_n(x)=\frac{1}{n}\sum\limits_{j=1}^nK(\frac{x-X_j}{h_n}), $其中$K$为核函数, $h=h_n$为窗宽.

样本分位数$\xi_\gamma=F^{-1}(\gamma)=\inf\{x\in{R}:F(x)\geq{\gamma}\} (0<\gamma<1)$的光滑核估计为

$\hat{\xi}_{\gamma n}=\hat{F}_n^{-1}(\gamma)=\inf\{x\in{R}, \hat{F}_n(x)\geq{\gamma}\}.$

设$p\leq n, q\leq n$为正整数序列, $k=[n/(p+q)]$ (此处[]表示取整).为了获得有限个分位数$\xi_{\gamma_i}(1\leq{i}\leq{k})$的核估计的联合渐近分布, 需要以下假设条件:

(A1) (i) $X_1, X_2, \cdots, X_n$是强平稳的随机变量序列, 该样本对应的总体的分布函数为$F$, 密度函数$f$有界.

(ii) $\{X_i, i\geq1\}$是$\phi$ -混合的且$\sum\limits_{n=1}^{\infty}\phi^{1/2}(n)<\infty$.

(iii) 设$f_{1, j}$为随机变量$X_1, X_{j+1}$的联合分布函数, 则对任意$u, v\in R$都有

$|f_{1, j}(u, v)-f(u)f(v)|\leq C (j>1).$

(iv)  $F^{\prime\prime}(x)$是有界的且$f(\xi_\gamma)>0(0<\gamma<1)$.

(A2) 核函数$K$是有界的且$\int_RudK(u)=0, \int_Ru^2dK(u)<\infty.$

(A3) 窗宽序列$h=h_n ( n\geq1 )$满足$0<h\rightarrow0, nh\rightarrow\infty $且$nh^4\rightarrow0$.

(A4) $p, q$和$k$满足

(i)$pk/n\to 1$; (ii) $ph\to \infty$; (iii) $q\to \infty$且$k\to \infty$; (iv) $k\phi(q)\to 0$.

注1 假设条件和下文中的所有极限都是在$n\to \infty$的条件下取得的.

注2 由$pk/n\to 1$可推出$qk/n\to 0$和$q/p\to 0$.由$0\leq n-k(p+q)\leq p+q$和$q/p\to 0$可推出$0\leq n-k(p+q)\leq Cp$.下文将会用到这些结论, 不再特别说明.

定理2.1  设$\{X_n, n\geq1\}$是$\phi$ -混合随机样本序列且满足条件(A1)--(A4), $ 0< \gamma_1<\gamma_2<\cdots<\gamma_k<1$.则当$n\to \infty$时, 有

$\sqrt{n}( f(\xi_{\gamma_1})(\hat{\xi}_{\gamma_1n}-\xi_{\gamma_1}), f(\xi_{\gamma_2})(\hat{\xi}_{\gamma_2n}-\xi_{\gamma_2}), \cdots, f(\xi_{\gamma_k})(\hat{\xi}_{\gamma_kn}-\xi_{\gamma_k}))^\prime \stackrel{d}{\longrightarrow} N_k(0, V), $

其中$V=(\sigma_{(st)}) (s, t=1, 2, \cdots, k)$,

${\sigma _{(ss)}}\; = \;{\gamma _s}(1 - {\gamma _s}) + \sum\limits_{j = 1}^{ + \infty } {\{ 2P(} {X_1} < {\xi _{{\gamma _s}}}, {X_{j + 1}} < {\xi _{{\gamma _s}}}) - 2\gamma _s^2\} (s = 1, 2, \cdots, k), \\{\sigma _{(st)}}\; = \;{\sigma _{(ts)}} = {\gamma _s}(1 - {\gamma _t}) + \sum\limits_{j = 1}^{ + \infty } {\{ P(} {X_1} < {\xi _{{\gamma _s}}}, {X_{j + 1}} < {\xi _{{\gamma _t}}})\\\quad \quad + P({X_1} < {\xi _{{\gamma _t}}}, {X_{j + 1}} < {\xi _{{\gamma _s}}}) - 2{\gamma _s}{\gamma _t}\} (1 \le s < t \le k).$

为了证明主要结果, 需要以下引理.在以下引理中用$C$表示一个不依赖于$n$的充分大的数, 在不同情况下可以取不同值.

引理3.1 假设$\{\eta_j: j\geq 1\}$是$\phi$ -混合序列, 混合系数为$\phi(n)$, ${\cal{F}}^t_s$表示由$\{\eta_i, s\leq i\leq t\}$ $(s\leq t)$序列生成的$\sigma$ -域.若$\{f_j(\cdot), j\geq 1\}$都为可测函数, $\{\xi_i(\cdot), i\geq 1\}$是${\cal{F}}^{j_1}_{i_1}, \cdots, {\cal{F}}^{j_n}_{i_n}, \cdots$可测的, 分别满足$1\leq i_1<j_1<i_2\cdots<j_n<\cdots$, 则$\{f_j(\eta_j): j\geq 1\}$是$\phi$ -混合序列, 且混合系数满足$\phi_1(n)\leq \phi(n)$, $\{\xi_j: j\geq 1\}$也是一个$\phi$ -混合序列, 其混合系数满足$\phi_2(n)\leq \phi(n)$.

证 通过$\phi$ -混合随机变量序列的定义可直接证明.

引理3.2 ^[11] 假设$\{\eta_j: j\geq 1\}$是$\phi$ -混合序列, 其混合系数为$\phi(n)$, ${\cal{F}}^t_s$表示由$\{\eta_i, s\leq i\leq t\}$ $(s\leq t)$序列生成的$\sigma$ -域.如果$\xi_1 \in {\cal{F}}^k_1$, $\xi_2 \in {\cal{F}}^{\infty}_{k+n}$且$k\geq 1$, $E|\xi_1|^{p_1}<\infty$, $E|\xi_2|^{p_2}<\infty$, 其中$p_1$和$q_1$满足$p_1>1, q_1>1, 1/p_1+1/q_1=1$.则

$|E(\xi_1\xi_2)-(E\xi_1)(E\xi_2)|\leq 2\phi^{1/p_1}(n)(E|\xi_1|^{p_1})^{1/p_1}(E|\xi_2|^{q_1})^{1/q_1}.$

证见文献Wang^[11]的引理1.2.

引理3.3^[11] 假定$\{\eta_j: j\geq 1\}$是$\phi$ -混合序列, 混合系数为$\phi(n)$, 且$\sum^{\infty}\limits_{n=1}\phi^{1/2}(n)<\infty$, $E\eta_j=0 (j\geq 1)$.假设$E|\eta_j|^{r_0}<\infty$ $(r_0\geq 2, j\geq 1)$.则对于任意$n\geq 1$且$a\geq 0$, 有以下式子成立:

$E{(\sum\limits_{i = a + 1}^{a + n} {{\eta _i}} )^2} \le C\sum\limits_{i = a + 1}^{a + n} E \eta _i^2, $

(3.1)

$E|\sum\limits_{i = a + 1}^{a + n} {{\eta _i}} {|^{{r_0}}} \le C\{ {(\sum\limits_{i = a + 1}^{a + n} E \eta _i^2)^{{r_0}/2}} + \sum\limits_{i = a + 1}^{a + n} E |{\eta _i}{|^{{r_0}}}\} .$

(3.2)

证 不等式(3.1) 来自于Wang^[11]的定理2.1.

由$\max\limits_{1\leq j\leq n}|\sum\limits^{a+j}_{i=a+1}\eta_i|^{r_0}\geq {|\sum\limits^{a+n}_{i=a+1}\eta_i|^{r_0}}, \max\limits_{1\leq i\leq n}|\eta_{a+i}|^{r_0}\leq{\sum\limits^{a+n}_{i=a+1}|\eta_i|^{r_0}}$, 不等式(3.1) 和Wang^[11]的引理1.3知不等式(3.2) 成立.

引理3.4^[17] 设$\{\eta_i, i\geq 1\}$是$\phi$ -混合序列, ${\cal{F}}^t_s$表示由$\{\eta_i, s\leq i\leq t\}$ $(s\leq t)$生成的$\sigma$ -域.如果$\{\xi_i, 1\leq i\leq n\}$为${\cal{F}}^{j_1}_{i_1}, \cdots, {\cal{F}}^{j_n}_{i_n}$可测的, 且分别满足$1\leq i_1<j_1<i_2\cdots<j_n, i_{l+1}-j_l\geq m$和$|\xi_l|\leq 1$ $(l=1, \cdots, n)$.则

$|E(\prod\limits_{l = 1}^n {{\xi _l}} ) - \prod\limits_{l = 1}^n E ({\xi _l})| \le 8(n - 1)\phi (m).$

(3.3)

证 这结论在$\alpha$ -混合的情形下是成立的, 参见$\mathrm{Volkonkii}$和 Rozanov^[17]的引理1.1, 易证此结论在$\phi$ -混合条件下也成立.

引理3.5 设$\{X_n, n\geq1\}$是$\phi$ -混合随机样本序列且满足条件(A1)--(A4), $0<\gamma_1<\gamma_2<\cdots<\gamma_k<1$, $y_i\in R, x_{ni}=\xi_{\gamma_i}+\frac{\sigma(\xi_{\gamma_i})y_i}{f(\xi_{\gamma_i})\sqrt{n}} (i=1, 2, \cdots, k)$.则当$n\to \infty$时, 有

$\sqrt{n}(\hat{F_n}(x_{n1})-E\hat{F_n}(x_{n1}), \hat{F_n}(x_{n2})-E\hat{F_n}(x_{n2}), \cdots, \hat{F_n}(x_{nk})-E\hat{F_n}(x_{nk}) )^\prime\stackrel{d}{\longrightarrow} N_k(0, V), $

其中$V=(\sigma_{(st)}) (s, t=1, 2, \cdots, k)$,

${\sigma _{(ss)}}\; = \;{\gamma _s}(1 - {\gamma _s}) + \sum\limits_{j = 1}^{ + \infty } {\{ 2P(} {X_1} < {\xi _{{\gamma _s}}}, {X_{j + 1}} < {\xi _{{\gamma _s}}}) - 2\gamma _s^2\} (s = 1, 2), \\{\sigma _{(st)}}\; = \;{\sigma _{(ts)}} = {\gamma _s}(1 - {\gamma _t}) + \sum\limits_{j = 1}^{ + \infty } {\{ P(} {X_1} < {\xi _{{\gamma _s}}}, {X_{j + 1}} < {\xi _{{\gamma _t}}})\\{\rm{ }}\;\; + P({X_1} < {\xi _{{\gamma _t}}}, {X_{j + 1}} < {\xi _{{\gamma _s}}}) - 2{\gamma _s}{\gamma _t}\} (1 \le s < t \le 2).$

证为了证明引理3.5, 只需证明当$k=2$时成立; 当$k>2$时同理可证.下证

$y_n\triangleq\sqrt{n}(\hat{F}_n(x_{n1})-E\hat{F}_n(x_{n1}), \hat{F}_n(x_{n2})-E\hat{F}_n(x_{n2}))^\prime \stackrel{d}{\longrightarrow} N_2(0, V), $

(3.4)

其中$V=(\sigma_{(st)}) (s, t=1, 2)$,

${\sigma _{(ss)}}\; = \;{\gamma _s}(1 - {\gamma _s}) + \sum\limits_{j = 1}^{ + \infty } {\{ 2P(} {X_1} < {\xi _{{\gamma _s}}}, {X_{j + 1}} < {\xi _{{\gamma _s}}}) - 2\gamma _s^2\} (s = 1, 2), \\{\sigma _{(st)}}\; = \;{\sigma _{(ts)}} = {\gamma _s}(1 - {\gamma _t}) + \sum\limits_{j = 1}^{ + \infty } {\{ P(} {X_1} < {\xi _{{\gamma _s}}}, {X_{j + 1}} < {\xi _{{\gamma _t}}})\\\quad \quad + P({X_1} < {\xi _{{\gamma _t}}}, {X_{j + 1}} < {\xi _{{\gamma _s}}}) - 2{\gamma _s}{\gamma _t}\} (1 \le s < t \le 2).$

为证式(3.4), 只需证$\forall a=(a_1, a_2)'\in R^{2}$都有

$a'y_n\stackrel{d}{\longrightarrow} N(0, a'Va).$

(3.5)

令

${S_n} \buildrel \Delta \over = {a^\prime }{y_n} = {a_1}\sqrt n ({\hat F_n}({x_{n1}}) - E{\hat F_n}({x_{n1}})) + {a_2}\sqrt n ({\hat F_n}({x_{n2}}) - E{\hat F_n}({x_{n2}})){\rm{ }}\\ = \frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n \{ {a_1}[K(\frac{{{x_{n1}}-{X_i}}}{h})-EK(\frac{{{x_{n1}}-{X_i}}}{h})] + {a_2}[K(\frac{{{x_{n2}}-{X_i}}}{h})-EK(\frac{{{x_{n2}}-{X_i}}}{h})]\} \\ = \frac{1}{{\sqrt n }}\sum\limits_{i = 1}^n {({a_1}{Z_{i1}} + {a_2}{Z_{i2}})}, {\rm{ }}$

其中$Z_{i1}\triangleq K(\frac{x_{n1}-X_i}{h})-EK(\frac{x_{n1}-X_i}{h}), Z_{i2}\triangleq K(\frac{x_{n2}-X_i}{h})-EK(\frac{x_{n2}-X_i}{h})$.

再将$S_n$进行分块, 令

$S_n=S_n^\prime+S_n^{\prime\prime}+S_n^{\prime\prime\prime}, S_n^\prime =\sum\limits_{m=1}^k e_{nm}, S_n^{\prime\prime}=\sum\limits_{m=1}^k e_{nm}^\prime, S_n^{\prime\prime\prime}=e_{nm}^{\prime\prime}, \\ e_{nm}=\frac{1}{\sqrt{n}}\sum\limits_{i=r_m}^{r_m+p-1}(a_1Z_{i1}+a_2Z_{i2}), \\ e_{nm}^\prime=\frac{1}{\sqrt{n}}\sum\limits_{i=l_m}^{l_m+q-1}(a_1Z_{i1}+a_2Z_{i2}), e_{nm}^{\prime\prime}=\frac{1}{\sqrt{n}}\sum\limits_{i=k(p+q)+1}^{n}(a_1Z_{i1}+a_2Z_{i2}), $

其中$r_m=(m-1)(p+q)+1, l_m=(m-1)(p+q)+p+1 (m=1, 2, \cdots, k).$

要证式(3.5) 成立, 只需证

$S_n^\prime\stackrel{d}{\longrightarrow} N(0, a^\prime Va), )$

(3.6)

$S_n^{\prime\prime}=o_p(1), $

(3.7)

和

$S_n^{\prime\prime\prime}=o_p(1), $

(3.8)

同时成立.先证

${\rm{Var}}(S_n^\prime ) = \sum\limits_{m = 1}^k {{\rm{Var}}} ({e_{nm}}) + 2\sum\limits_{1 \le i < j \le k}^{} {{\rm{Cov}}({e_{ni}}, {e_{nj}})} \to {a^\prime }Va.$

(3.9)

由假设条件(A1)(i)知

$\sum\limits_{m=1}^k \mathrm{Var}(e_{nm})=\frac{kp}{n}\mathrm{Var}(a_1Z_{11}+a_2Z_{12})+2\frac{k}{n}\sum\limits_{j=1}^{p-1}(p-j)\mathrm{Cov}(a_1Z_{11}+a_2Z_{12}, a_1Z_{j+1, 1}+a_2Z_{j+1, 2}), $

由$0<\gamma_1<\gamma_2<1$和假设条件(A1)(iv)知$\xi_{\gamma_1}<\xi_{\gamma_2}$.又因

$x_{ni}=\xi_{\gamma_i}+\frac{\sigma(\xi_{\gamma_i})y_i}{f(\xi_{\gamma_i})\sqrt{n}}\to \xi_{\gamma_i} (i=1, 2).$

故当$n$充分大时, 有$x_{n1}<x_{n2}$.由假设条件(A1)(i), (A2), (A3), (A4)(i)和文献$\mathrm{Cai}$和Roussas^[18]中引理3.1知

$\begin{array}{l} \frac{{kp}}{n}{\rm{Var}}({a_1}{Z_{11}} + {a_2}{Z_{12}}) = \frac{{kp}}{n}[a_1^2{\rm{Var}}({Z_{11}}) + 2{a_1}{a_2}{\rm{Cov}}({Z_{11}},{Z_{12}}) + a_2^2{\rm{Var}}({Z_{12}})]\\ \to a_1^2{\gamma _1}(1 - {\gamma _1}) + 2{a_1}{a_2}{\gamma _1}(1 - {\gamma _2}) + a_2^2{\gamma _2}(1 - {\gamma _2}). \end{array}$

由(A4)(i)(ii), 引理3.2和$\mathrm{Cai}$和Roussas^[18]引理3.3中式(3.6) 和(3.7) 知

$\quad \quad 2\frac{{kp}}{n}\sum\limits_{j = 1}^{p - 1} {{\rm{Cov}}} ({a_1}{Z_{11}} + {a_2}{Z_{12}}, {a_1}{Z_{j + 1, 1}} + {a_2}{Z_{j + 1, 2}})\\\quad = \quad 2\frac{{kp}}{n}\sum\limits_{j = 1}^{p - 1} {\{ a_1^2{\rm{Cov}}(} {a_1}{Z_{11}}, {a_1}{Z_{j + 1, 1}}) + {a_1}{a_2}{\rm{Cov}}({a_1}{Z_{11}}, {a_1}{Z_{j + 1, 2}})\\\quad \quad + {a_1}{a_2}{\rm{Cov}}({a_1}{Z_{12}}, {a_1}{Z_{j + 1, 1}}) + a_2^2{\rm{Cov}}({a_1}{Z_{12}}, {a_1}{Z_{j + 1, 2}})\} \\ \to \quad 2{a_1}{a_2}\sum\limits_{i = 1}^\infty {[P({X_1}} < {\xi _{\gamma 1}}, {X_{j + 1}} < {\xi _{\gamma 1}})-\gamma _1^2] + 2{a_1}{a_2}\sum\limits_{i = 1}^\infty {[P({X_1}} < {\xi _{\gamma 1}}, {X_{j + 1}} < {\xi _{\gamma 2}})-\gamma 1\gamma 2]\\\quad \quad + 2a_1^2\sum\limits_{i = 1}^\infty {[P({X_1}} < {\xi _{\gamma 2}}, {X_{j + 1}} < {\xi _{\gamma 1}})-\gamma 1\gamma 2] + 2a_2^2\sum\limits_{i = 1}^\infty {[P({X_1}} < {\xi _{\gamma 2}}, {X_{j + 1}} < {\xi _{\gamma 2}})- \gamma _2^2$

且$\frac{k}{n}\sum\limits_{j=1}^{p-1}j\mathrm{Cov}(a_1Z_{11}+a_2Z_{12}, a_1Z_{j+1, 1}+a_2Z_{j+1, 2})\longrightarrow 0$.故

$\sum\limits_{m = 1}^k {{\rm{Var}}} ({e_{nm}}) \to {a^\prime }Va, $

(3.10)

由假设条件(A1)(i), (A4)(i)和引理3.2知

$\quad \quad \sum\limits_{1 \le i < j \le k} | {\rm{Cov}}({e_{ni}}, {e_{nj}})| = \sum\limits_{l = 1}^{k- 1} | (k- i){\rm{Cov}}({e_{n1}}, {e_{n, l + 1}})| \le k\sum\limits_{l = 1}^{k - 1} | {\rm{Cov}}({e_{n1}}, {e_{n, l + 1}})|\\{\rm{ }}\quad \le \quad \frac{{kp}}{n}\sum\limits_{l = 1}^{k - 1} {\sum\limits_{s = l(p + q) - p}^{l(p + q) + p} | } {\rm{Cov}}({a_1}{Z_{11}} + {a_2}{Z_{12}}, {a_1}{Z_{s + 1, 1}} + {a_2}{Z_{s + 1, 2}})| \le \frac{{Ckp}}{n}\sum\limits_{s = q}^\infty {{\phi ^{1/2}}} (s) \to 0.$

由以上两式知(3.9) 式成立.

当$i\neq j$时, 由假设条件(A1)(iii)知

$|{\rm{Cov}}({Z_{i1}}, {Z_{j2}})|\quad = \quad {h^2}|\int_{{R^2}} K (u)K(v)\{ {f_{1, j}}({x_{n1}} - hu, {x_{n2}} - hv)\\\quad \quad - f({x_{n1}} - hu)f({x_{n2}} - hv)\} dudv| \le C{h^2}.{\rm{ }}$

同理可证

$\sum\limits_{m = 1}^k {{\rm{Var}}} (e_{nm}^\prime )\;\; = \;\;\frac{{kq}}{n}{\rm{Var}}({a_1}{Z_{11}} + {a_2}{Z_{12}}) + \frac{{2k}}{n}\sum\limits_{1 \le i < j \le q} {{\rm{Cov}}({a_1}{Z_{i1}} + {a_2}{Z_{i2}}, {a_1}{Z_{j1}} + {a_2}{Z_{j2}})} \\ \le \;\;\frac{{Ckq}}{n} + \frac{{Ck{q^2}{h^2}}}{n} \to 0, $

且

$\;\;\;\;\sum\limits_{1 \le i < j \le k} {|{\rm{Cov}}(e_{ni}^\prime, e_{nj}^\prime )|} = \sum\limits_{l = 1}^{k - 1} {(k - l)} |{\rm{Cov}}(e_{n1}^\prime, e_{n, i + 1}^\prime )| \le k\sum\limits_{i = 1}^{k - 1} | {\rm{Cov}}(e_{n1}^\prime, e_{n, i + 1}^\prime )|\\{\rm{ }}\;\; \le \;\;\frac{{kq}}{n}\sum\limits_{l = 1}^{k - 1} {\sum\limits_{s = l(p + q) - (q - 1)}^{l(p + q) + (q - 1)} | } {\rm{Cov}}({a_1}{Z_{11}} + {a_2}{Z_{12}}, {a_1}{Z_{s + 1, 1}} + {a_2}{Z_{s + 1, 2}})| \le \frac{{Ckq}}{n}\sum\limits_{s = p}^\infty {{\phi ^{1/2}}} (s) \to 0.{\rm{ }}$

由以上两式知

$E(S_n^{\prime\prime})^2\longrightarrow0, $

(3.11)

则式(3.7) 成立.

由引理3.3知

$E(S_n^{\prime\prime\prime})^2\leq \frac{C[n-k(p+q)]}{n}E(a_1Z_{11}+a_2Z_{12})^2 \longrightarrow0, $

(3.12)

则式(3.8) 成立.

由(A4)(iv)和引理3.4知

$|Ee^{it\sum\limits_{m=1}^{k}e_{nm}}-\prod\limits_{m=1}^k Ee^{ite_{nm}}|\leq {Ck\phi(q)}\longrightarrow0, $

(3.13)

故$\{e_{nm}\}$是渐近独立的.设$\{E_{nm};m=1, 2, \cdots, k\}$是独立随机变量序列, 其中$E_{nm}$的分布函数和$e_{nm}$的分布函数相同$(m=1, 2, \cdots, k)$.为证式(3.6) 成立, 只需证下式成立

$\sum\limits_{m = 1}^k {{E_{nm}}} \mathop \to \limits^d N(0, {a^\prime }Va).$

(3.14)

又设$s_n^2=k\mathrm{Var}E_{n1}$和$F_{nm}=E_{nm}/s_n$, $1\leq m\leq k$由式(3.9) 的证明过程知$s_n^2\rightarrow a^\prime Va$.故式(3.14) 等价于

$\sum\limits_{m = 1}^k {{F_{nm}}} \mathop \to \limits^d N(0, 1).$

(3.15)

由假设条件(A1)(i)和(A2) 知

$\quad E{({a_1}{Z_{i1}} + {a_2}{Z_{i2}})^2}\\ = \int_R [{a_1}(K(\frac{{{x_{n1}}-x}}{h})-EK(\frac{{{x_{n1}}-x}}{h})) + {a_2}(K(\frac{{{x_{n2}} - x}}{h}) - EK(\frac{{{x_{n2}} - x}}{h})){]^2}dF(x)\\ = h\int_R [{a_1}(K(u)-EK(u)) + {a_2}(K(\frac{{{x_{n2}}-{x_{n1}}}}{h} + u)-EK(\frac{{{x_{n2}} - {x_{n1}}}}{h} + u)){]^2}f({x_{n1}} - hu)du\\ \le Ch.$

同理可得

$E|a_1Z_{i1}+a_2Z_{i2}|^3\leq Ch. $

由引理3.3和以上两式知

$\quad \quad kE|{F_{n1}}{|^3} = ks_n^{ - 3}E|{e_{n1}}{|^3} = ks_n^{ - 3}{n^{ - 3/2}}E|\sum\limits_{i = {r_1}}^{{r_1} + p - 1} {{a_1}} {Z_{i1}} + {a_2}{Z_{i2}}{|^3}\\ \le \;\;Cks_n^{ - 3}{n^{ - 3/2}}\{ \sum\limits_{i = {r_1}}^{{r_1} + p - 1} E |{a_1}{Z_{i1}} + {a_2}{Z_{i2}}{|^3} + {[\sum\limits_{i = {r_1}}^{{r_1} + p-1} E {({a_1}{Z_{i1}} + {a_2}{Z_{i2}})^2}]^{3/2}}\} {\rm{ }}\\ \le \;\;Cks_n^{ - 3}{n^{ - 3/2}}\{ ph + {(ph)^{3/2}}\} \le Cks_n^{ - 3}{n^{ - 3/2}}{(ph)^{3/2}}\\ \le \;\;Cs_n^{ - 3}{(kp/n)^{3/2}}{k^{ - 1/2}}{h^{3/2}} \to 0, $

其中用到了$ph\rightarrow \infty$, $s_n^2\rightarrow a^\prime Va$, $kp/n\leq 1$和$k\rightarrow \infty$.由$\mathrm{Lyapunov}$中心极限定理知式(3.15) 成立, 故式(3.6) 成立.

由式(3.6), (3.7), (3.8) 和Cramer-Wold定理知式(3.5) 成立.

为证明定理2.1, 只需证明当$k=2$时成立; 当$k>2$时同理可证.

令$\sigma(\xi_{\gamma_i})=\sigma(ii) (i=1, 2)$, $x_{n1}=\xi_{\gamma_1}+\frac{\sigma(\xi_{\gamma_1})y_1}{f(\xi_{\gamma_1})\sqrt{n}}, x_{n2}=\xi_{\gamma_2}+\frac{\sigma(\xi_{\gamma_2})y_2}{f(\xi_{\gamma_2})\sqrt{n}}$.易知

$E\hat{F}_n(x_{nj})=\int_R K(\frac{x_{nj}-x}{h})dF(x)=\int_R F(x_{nj}-hu)dK(u) (j=1, 2).$

再将$F(x_{nj}-hu)$在$\xi_{\gamma j}$处$\mathrm{Taylor}$展开得

$F(x_{nj}-hu)=F(\xi_{\gamma_j})+f(\xi_{\gamma_j})\bigg[\frac{\sigma(\xi_{\gamma_j})y_j}{f(\xi_{\gamma_j})\sqrt{n}}-hu\bigg]+\frac{1}{2}f'(u^*)\bigg[\frac{\sigma(\xi_{\gamma_j})y_j}{f(\xi_{\gamma_j})\sqrt{n}}-hu\bigg]^2, $

其中$u^*$满足$|u^*-\xi_{\gamma_j}|\leq |\sigma(\xi_{\gamma_j})y_j/f(\xi_{\gamma_j})\sqrt{n}-hu|$.由以上两式及假设条件知

$E\hat{F}_n(x_{nj})=F(\xi_{\gamma_j})+\frac{\sigma(\xi_{\gamma_j})y_j}{\sqrt{n}}+O\bigg(\frac{y_j^2}{n}+h^2\bigg) (j=1, 2), $

易知

$\sqrt{n}[E\hat{F}_n(x_{nj})-\gamma_j]=\sigma(\xi_{\gamma_j})y_j+O\bigg(\frac{y_j^2}{\sqrt{n}}+\sqrt{nh^4}\bigg) (j=1, 2).$

又令$H_n(y_1, y_2)=P\big(\frac{\sqrt{n}f(\xi_{\gamma_1})(\hat{\xi}_{\gamma_1n}-\xi_{\gamma_1})}{\sigma(\xi_{\gamma_1})}\leq y_1, \frac{\sqrt{n}f(\xi_{\gamma_2})(\hat{\xi}_{\gamma_2n}-\xi_{\gamma_2})}{\sigma(\xi_{\gamma_2})}\leq y_2\big)$.故

${H_n}({y_1}, {y_2})\; = \;P({\hat \xi _{{\gamma _1}n}} \le {\xi _{{\gamma _1}}} + \frac{{\sigma ({\xi _{{\gamma _1}}}){y_1}}}{{f({\xi _{{\gamma _1}}})\sqrt n }}, {\hat \xi _{{\gamma _2}n}} \le {\xi _{{\gamma _2}}} + \frac{{\sigma ({\xi _{{\gamma _2}}}){y_2}}}{{f({\xi _{{\gamma _2}}})\sqrt n }}){\rm{ }}\\\quad \quad = \;P({\hat F_n}({x_{n1}}) \ge {\gamma _1}, {\hat F_n}({x_{n2}}) \ge {\gamma _2}){\rm{ }}\\\quad \quad = \;P( - \sqrt n ({\hat F_n}({x_{n1}}) - E{\hat F_n}({x_{n1}})) \le \sqrt n (E{\hat F_n}({x_{n1}}) - {\gamma _1}), \\\quad \quad - \sqrt n ({\hat F_n}({x_{n2}}) - E{\hat F_n}({x_{n2}})) \le \sqrt n (E{\hat F_n}({x_{n2}}) - {\gamma _2}))\\\quad = \;P( - \sqrt n ({\hat F_n}({x_{n1}}) - E{\hat F_n}({x_{n1}}))/\sigma ({\xi _{{\gamma _1}}}) \le {y_1} + O(\frac{{y_1^2}}{{\sqrt n }} + \sqrt {n{h^4}} ), {\rm{ }}\\\quad \quad - \sqrt n ({\hat F_n}({x_{n2}}) - E{\hat F_n}({x_{n2}}))/\sigma ({\xi _{{\gamma _2}}}) \le {y_2} + O(\frac{{y_2^2}}{{\sqrt n }} + \sqrt {n{h^4}} )).{\rm{ }}$

设二元正态分布$(X_1, X_2)^\prime\sim N(0, V)$, 故

$(Y_1, Y_2)^\prime\triangleq (X_1/\sigma(\xi_{\gamma_1}), X_2/\sigma(\xi_{\gamma_2}))^\prime\sim N(0, \hat{V}), $

其中$\hat{V}=(\hat{\sigma}_{(st)}) (s, t=1, 2)$, $\hat{\sigma}_{st}=\hat{\sigma}_{ts}=\sigma_{st}/\sqrt{\sigma_{ss}\sigma_{tt}} (1\leq s\leq t\leq2)$, $(Y_1, Y_2)^\prime$的分布函数为$F(y_1, y_2)=P(Y_1\leq y_1, Y_2\leq y_2)$.

由引理3.5知

$(-\sqrt{n}(\hat{F}_n(x_{n1})-E\hat{F}_n(x_{n1}))/\sigma(\xi_{\gamma_1}), -\sqrt{n}(\hat{F}_n(x_{n2})-E\hat{F}_n(x_{n2}))/\sigma(\xi_{\gamma_2}))^\prime\stackrel{d}{\longrightarrow} N(0, \hat{V}).$

而由假设条件(A3) 知$O(\frac{y_j^2}{\sqrt{n}}+\sqrt{nh^4})\longrightarrow 0 (j=1, 2)$.

综上可得$H_n(y_1, y_2)-F(y_1, y_2)\longrightarrow 0$, 故定理2.1成立.