由于Davydov ^[8]的工作, 最近很多学者研究了布朗运动和分数布朗运动的弱收敛问题. Stroock ^[13]研究了一维标准的泊松过程与标准的布朗运动之间的如下关系:令$\{N(t), t\geq0\}$是一个标准的泊松过程, 定义

$ y_\varepsilon(t)=\frac{1}{\varepsilon}\int^t_0(-1)^{N(s/\varepsilon^2)}ds, $

则当$\varepsilon\rightarrow0$时, $y_\varepsilon(t)$在连续函数空间$\mathcal{{C}}([0, 1])$中弱收敛到标准的布朗运动$\{B_t, t\geq 0\}$.受到Stroock ^[13]工作的启发, Bardina和Jolis ^[2]证明了当$\varepsilon\rightarrow0$时, 随机过程族

$ y_\varepsilon(s, t)=\int^t_0\int^s_0\frac{1}{\varepsilon^2}\sqrt{xy}(-1) ^{N(x/\varepsilon, y/\varepsilon)}dxdy, \quad \varepsilon>0 $

在$\mathcal{{C}}([0, 1]^2)$空间中弱收敛到标准的布朗单, 其中$\{N(x, y), (x, y)\in\mathbb{R}^2_+\}$是平面上的一个标准泊松过程(关于多维情形的相应结果可参考Bardina, Jolis和Rovira ^[1]). Delgado和Jolis ^[7]证明了可以利用泊松过程逼近具有如下随机积分表示的一类高斯过程

$ X(t)=\int_0^1K(t, r)dB_r, t\in[0, 1], $

其中$K$是满足一定条件的确定的核函数. Bardina, Jolis和Tudor ^[3]把上述结果推广到分数布朗单和其他两参数高斯过程:令$W^{K_1, K_2}=\{W_{s, t}^{K_1, K_2}, (s, t)\in[0, 1]^2\}$使得

$ W_{s, t}^{K_1, K_2}:=\int_0^1\int_0^1K_1(s, u)K_2(t, v)dW_{u, v}, $

其中$W=\{W_{s, t}, (s, t)\in[0, 1]^2\}$是标准布朗单且核函数$K_1, K_2$满足一些条件.令

$ X_\varepsilon(s, t)=\int_0^1\int_0^1K_1(s, u)K_2(t, v)\frac{1}{\varepsilon^2}\sqrt{uv}(-1)^ {N(\frac{u}{\varepsilon}, \frac{v}{\varepsilon})}dudv, $

那么$X_\varepsilon$在连续函数空间$\mathcal{{C}}([0, 1]^2)$中弱收敛到$W^{K_1, K_2}$.更多关于分数布朗运动, 分数布朗单和多参数过程的相关问题可参见Bardina和Florit ^[4], Li和Dai ^[10], Wang等^{[14, 15]}, 徐锐等^[16], Dai ^[6]等文献.

受上述文献的启发, 令$X(t_1, \cdots, t_d)$是具有如下随机积分表示的一类$d$ -维参数高斯过程

$ X(t_1, \cdots, t_d)=\int_{0}^1\cdots\int_{0}^1 K_1(t_1, u_1), \cdots, K_d(t_d, u_d)dW_{u_1, \cdots, u_d}, t_i\in[0, 1], $

其中$W=\{W_t, t\in[0, 1]^d\}$表示$d$ -维布朗运动, $K_i, i=1, \cdots, d$是满足一定条件的确定的核函数.本文证明了

$ X_{n}(t)=n^{\frac{d}{2}}\int^1_{0}\cdots\int^1_{0}K_1(t_1, u_1)\cdots K_d(t_d, u_d)(\prod\limits_{i=1}^du_i)^\frac{d-1}{2}(-1)^{N(u_1n^\frac{1}{d}, \cdots, u_dn^\frac{1}{d})}du_1\cdots du_d $

在连续函数空间$\mathcal{{C}}([0, 1]^d)$中弱收敛到$X$.此结论是将文献[3]中的结论推广到了$d$ -维参数情形, 而在多维参数情形下的计算过程会遇到更多的困难, 因此更多的借用了文献[1]中的不等式来证明定理.本文中常数$C_0$, $C$, $C_d$, $c_{H_i}$, $C_{H_i}$表示常数, 在不同的位置可以表示不同的值.

令$(s, t]=\prod\limits_{i=1}^d(s_i, t_i]\subset\mathbb{R}^d$, 其中$0<s_i\leq t_i<1, i=1, \cdots, d$.令

$ \Delta_sX(t)=\sum\limits_{\varepsilon_1=0, 1}\cdots \sum\limits_{\varepsilon_d=0, 1}(-1)^{d-\sum\limits_{i=1}^d\varepsilon_i} X(s_1+\varepsilon_1(t_1-s_1), \cdots, s_d+\varepsilon_d(t_d-s_d)) $

是$X$在$(s, t]$上的增量(见Bickel和Wichura ^[5]).特别地, 当$d=2$时,

$ \Delta_sX(t)=X(s_1, s_2)-X(t_1, s_2)-X(s_1, t_2)+X(t_1, t_2) $

是常见的二维空间上的增量.

本节考虑$d$ -维参数高斯过程$X=\{X(t), t\in[0, 1]^d\}$,

$ X(t)=X(t_1, \cdots, t_d)=\int_{0}^1\cdots\int_{0}^1K_1(t_1, u_1)\cdots K_d(t_d, u_d)dW_{u_1, u_2, \cdots, u_d}, $

其协方差函数为

$ {\rm Cov}(X(t), X(s))=\prod\limits_{i=1}^d[\int_{0}^1K_i(t_i, u_i)K_i(s_i, u_i)du_i], $

其中$K_i:[0, 1]\times[0, 1]\rightarrow\mathbb{R}$是满足下列条件的函数:

(H1) (ⅰ) 对$i=1, 2, \cdots, d$, $K_i$是可测的且对任意的$u_i\in[0, 1], $ $K_i(0, u_i)=0, $ a.e..

(ⅱ) 对$i=1, 2, \cdots, d$, 存在单增的连续函数$F_i:[0, 1]\rightarrow \mathbb{R}$和$\alpha_i>1$使得对任意$0\leq s_i<t_i\leq1$, 有

$ \int_{0}^1(K_i(t_i, u_i)-K_i(s_i, u_i))^2du_i\leq(F_i(t_i)-F_i(s_i))^{\alpha_i}. $

(H2) (ⅰ) 对$i=1, 2, \cdots, d$, $K_i$是可测的且对任意的$u_i\in[0, 1], $ $K_i(0, u_i)=0, $ a.e..

(ⅱ)'对$i=1, 2, \cdots, d$, 存在单增的连续函数$F_i:[0, 1]\rightarrow \mathbb{R}$和$0<\rho_i\leq1$使得对任意$0\leq s_i<t_i\leq1$, 有

$ \int_{0}^1(K_i(t_i, u_i)-K_i(s_i, u_i))^2du_i\leq(F_i(t_i)-F_i(s_i))^{\rho_ i}. $

(ⅲ) 对$i=1, 2, \cdots, d$, 存在常数$M_i>0$和$\beta_i>0$使得对任意的$0\leq s_i<t_i\leq1$和$0\leq s_{0i}<t_{0i}\leq1$,

$ \int_{s_{0i}}^{t_{0i}}(K_i(t_i, u_i)-K_i(s_i, u_i))^2du_i\leq M_i(t_{0i}-s_{0i})^{\beta_i}. $

注意到条件(ⅱ)较(ⅱ)'强, 在较弱的条件(ⅱ)'下增加条件(ⅲ)是为了证明$X_n$的胎紧性.

令$\{N(x), x\in\mathbb{R}_{+}^d\}$是定义在概率空间$(\Omega, \mathcal{F}, P)$上的标准泊松过程, $\forall n>0$, $t_i\in[0, 1]$, $i=1, 2, \cdots d$, 定义

$ \begin{align}\label{sec2-eq2.0} X_n(t)=\int_{[0, 1]^d}K_1(t_1, u_1)\cdots K_d(t_d, u_d)n^{\frac{d}{2}}(\prod\limits_{i=1}^du_i)^\frac{d-1}{2}(-1)^{N(u_1n^\frac{1}{d}, \cdots, u_d n^\frac{1}{d})}du_1\cdots du_d. \end{align} $

(2.1)

本节将证明在连续函数空间$\mathcal{C}([0, 1]^d)$中, 当$n\rightarrow\infty$时, $X_n$弱收敛到$X$.为了简化记号, 用$N_n(u)$表示随机变量$N(u_1n^\frac{1}{d}, \cdots, u_d n^\frac{1}{d})$, 则$N_n$是强度为$n$的泊松过程.

为了证明主要结果, 首先给出如下引理.引理2.1说明了过程$X_n$是连续的.

引理2.1  设$\theta\in L^\infty([0, 1]^d)$, 定义

$ Y(t)=\int_{[0, 1]^d}K_1(t_1, u_1)\cdots K_d(t_d, u_d)\theta(u_1, \cdots, u_d)du_1\cdots du_d, $

其中核函数$K_i, i=1, 2, \cdots, d$满足条件(ⅰ)和(ⅱ)', 那么$Y(t)$是一个连续函数.

证  $\forall 0 < s_i\leq t_i < 1, i=1, \cdots, d$, 根据Hölder不等式和条件(ⅱ)'有

$ \begin{align*} |\Delta_s Y(t)|&=|\int_{[0, 1]^d}( K_1(t_1, u_1)-K_1(s_1, u_1))\cdots( K_d(t_d, u_d)-K_d(s_d, u_d)) \theta(u)du_1\cdots du_d|\\ &\leq \|\theta\|_\infty|\int_{[0, 1]^d}( K_1(t_1, u_1)-K_1(s_1, u_1))\cdots( K_d(t_d, u_d)-K_d(s_d, u_d))du_1\cdots du_d|\\ &= \|\theta\|_\infty|\int_0^1( K_1(t_1, u_1)-K_1(s_1, u_1))du_1\cdots\int_0^1( K_d(t_d, u_d)-K_d(s_d, u_d))du_d|\\ &\leq \|\theta\|_\infty(\int_{0}^1(K_1(t_1, u_1)-K_1(s_1, u_1))^2du_1)^\frac{1}{2}\cdots \\&(\int_{0}^1(K_d(t_d, u_d)-K_d(s_d, u_d))^2du_d)^\frac{1}{2}\\ &\leq\|\theta\|_\infty\prod\limits_{i=1}^d(F_i(t_i)-F_i(s_i))^\frac{\rho_i}{2}. \end{align*} $

再由条件(ⅰ)知$Y(t)$是一个连续函数.

引理2.2  对任意函数$ f_i\in L^2([0, 1]), i=1, 2, \cdots, d$, 存在一个正常数$C_d$使得

$ E(\int_{[0, 1]^d}\prod\limits_{i=1}^df_i(x_i)\theta_n(x)dx_1\cdots dx_d)^2\leq C_d\int_{[0, 1]^d}\prod\limits_{i=1}^df_{i}^2(x_i)dx_1\cdots dx_d, $

其中

$ \theta_n(x)=n^\frac{d}{2}(\prod\limits_{i=1}^dx_i)^{\frac{d-1}{2}}(-1)^{N_n(x)}, ~~x=(x_1, \cdots, x_d). $

证 由于(见参考文献[1])

$ E[(-1)^{\sum\limits_{j=1}^2N_n(x^j)}] =\exp[-2n(\prod\limits_{i=1}^dx_i^1+\prod\limits_{i=1}^dx_i^2-2\prod\limits_{i=1}^dx_i^{1\wedge2})] $

和

$ \prod\limits_{i=1}^dx_i^1+\prod\limits_{i=1}^dx_i^2-2\prod\limits_{i=1}^dx_i^{1\wedge2} =\sum\limits_{i=1}^d(\prod\limits_{k=1}^{i-1}x_k^{1\wedge2})|x_i^1-x_i^2|(\prod\limits_{k=i+1}^{d}x_k^{1\vee2(i)}), $

其中$x_k^{1\wedge2}=\min\{x_k^1, x_k^2\}, ~ x_k^{1\vee2}=\max\{x_k^1, x_k^2\}, $对$j\geq i$定义

$ \begin{align*} x_j^{1\vee2(i)}=\left\{ \begin{aligned} x_j^1, \quad x_i^1\geq x_i^2 ;\\ x_j^2, \quad x_i^1<x_i^2. \end{aligned} \right. \end{align*} $

进而有

$ \begin{align}\label{sec2-eq2.1} &E(\int_{[0, 1]^d}\prod\limits_{i=1}^df_i(x_i)\theta_n(x)dx_1\cdots dx_d)^2 \nonumber \\=&E(\int_{[0, 1]^d}\prod\limits_{i=1}^df_i(x_i)n^\frac{d}{2} (\prod\limits_{i=1}^dx_i)^\frac{d-1}{2}(-1)^{N_n(x_1, \cdots, x_d)}dx_1\cdots dx_d)^2 \nonumber \\=&n^d\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d\prod\limits_{j=1}^2f_i(x_i^j)(\prod\limits_{i=1}^d\prod\limits_{j=1}^2x_i^j)^\frac{d-1}{2} E[(-1)^{\sum\limits_{j=1}^2N_n(x_{1}^j, \cdots, x_{d}^j)}]\prod\limits_{j=1}^2dx_{1}^j\cdots dx_{d}^j\nonumber\\ =&n^d\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d\prod\limits_{j=1}^2f_i(x_i^j)(\prod\limits_{i=1}^d\prod\limits_{j=1}^2x_i^j)^\frac{d-1}{2}\nonumber\\ &\exp[-2n\sum\limits_{i=1}^d(\prod\limits_{k=1}^{i-1}x_k^{1\wedge2})|x_i^1-x_i^2|(\prod\limits_{k=i+1}^{d}x_k^{1\vee2(i)})] \prod\limits_{j=1}^2dx_{1}^j\cdots dx_{d}^j\nonumber\\ \leq& 2^dn^d\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d\prod\limits_{j=1}^2|f_i(x_i^j)|(\prod\limits_{i=1}^d\prod\limits_{j=1}^2x_i^j)^\frac{d-1}{2} \nonumber\\ &\times \exp[-2n\sum\limits_{i=1}^d[(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]]\prod\limits_{i=1}^d I_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j. \end{align} $

(2.2)

先把上面的积分区域分成两部分: $A=\{x_i^1\leq x_i^2\leq2x_i^1, i=1, 2, \cdots, d\}$和$A^c$.通过简单的计算, 积分(2.2)在区域$A$上的值不超过

$ \begin{align*} 2^{d-1}n^d(&\int_{[0, 1]^{2d}}\prod\limits_{i=1}^df_i^2(x_i^1)(\prod\limits_{i=1}^dx_i^1)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j \\&+\int_{[0, 1]^{2d}}\prod\limits_{i=1}^df_i^2(x_i^2)(\prod\limits_{i=1}^dx_i^2)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\\&\prod\limits_{k\neq i}\frac{x_k^2}{2}]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j ). \end{align*} $

又有

$ \begin{align*} &\int_{[0, 1]^{2d}}\prod\limits_{i=1}^df_i^2(x_i^1)(\prod\limits_{i=1}^dx_i^1)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j\\ =&\int_{[0, 1]^d}\prod\limits_{i=1}^df_i^2(x_i^1)(\prod\limits_{i=1}^dx_i^1)^{d-1}\int_{x_d^1}^1\cdots\int_{x_1^1}^1\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]dx_1^2\cdots dx_d^2dx_1^1\cdots dx_d^1\\ =&\int_{[0, 1]^d}\prod\limits_{i=1}^df_i^2(x_i^1)(\prod\limits_{i=1}^dx_i^1)^{d-1}\int_{x_d^1}^1\cdots\int_{x_2^1}^1\prod\limits_{i=2}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]\\ &\int_{x_1^1}^1\exp[-2n(x_1^2-x_1^1)\prod\limits_{k=2}x_k^1]dx_1^2dx_2^2\cdots dx_d^2dx_1^1\cdots dx_d^1, \end{align*} $

其中

$ \begin{align*} &\int_{x_1^1}^1\exp[-2n(x_1^2-x_1^1)x_2^1\cdots x_d^1]dx_1^2\\ =&\exp[2nx_1^1x_2^1\cdots x_d^1]\int_{x_1^1}^1\exp[-2nx_1^2x_2^1\cdots x_d^1]dx_1^2\\ =&\exp[2nx_1^1x_2^1\cdots x_d^1]\frac{1}{-2nx_2^1x_3^1\cdots x_d^1}\exp[-2nx_1^2x_2^1\cdots x_d^1]\mid_{x_1^1}^1\\ =&\frac{1}{-2nx_2^1x_3^1\cdots x_d^1}\exp[2nx_1^1x_2^1\cdots x_d^1] (\exp[-2nx_2^1x_3^1\cdots x_d^1]-\exp[-2nx_1^1x_2^1\cdots x_d^1])\\ =&\frac{1}{-2nx_2^1x_3^1\cdots x_d^1}(\exp[-2n(1-x_1^1)x_2^1\cdots x_d^1]-1)\\ =&\frac{1}{2nx_2^1x_3^1\cdots x_d^1}-\frac{1}{2nx_2^1x_3^1\cdots x_d^1}\exp[-2n(1-x_1^1)x_2^1\cdots x_d^1]\\ \leq&\frac{1}{2nx_2^1x_3^1\cdots x_d^1}. \end{align*} $

类似的依次对$x_2^2, x_3^2, \cdots, x_d^2$求积分计算可得

$ \begin{align*} \int_{x_d^1}^1\cdots\int_{x_1^1}^1\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]dx_1^2\cdots dx_d^2\leq(2n)^{-d}(\prod\limits_{i=1}^dx_i^1)^{1-d}. \end{align*} $

故

$ \begin{align*} &\int_{[0, 1]^{2d}}\prod\limits_{i=1}^df_i^2(x_i^1)(\prod\limits_{i=1}^dx_i^1)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j\\ \leq&(2n)^{-d}\int_{[0, 1]^d}\prod\limits_{i=1}^df_i^2(x_i^1)dx_1^1\cdots dx_d^1. \end{align*} $

同理可计算

$ \begin{align*} &\int_{[0, 1]^{2d}}\prod\limits_{i=1}^df_i^2(x_i^2)(\prod\limits_{i=1}^dx_i^2)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}\frac{x_k^2}{2}]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j ) \\\leq& (\frac{1}{2^{2-d}n})^d\int_{[0, 1]^{d}}\prod\limits_{i=1}^df_i^2(x_i^2)dx_1^2\cdots dx_d^2. \end{align*} $

综上可得

$ \begin{align*} &2^{d-1}n^d(\int_{[0, 1]^{2d}}\prod\limits_{i=1}^df_i^2(x_i^1)(\prod\limits_{i=1}^dx_i^1)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j \\&+\int_{[0, 1]^{2d}}\prod\limits_{i=1}^df_i^2(x_i^2)(\prod\limits_{i=1}^dx_i^2)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}\frac{x_k^2}{2}]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j )\\ \leq& C_d \int_{[0, 1]^{d}}\prod\limits_{i=1}^df_i^2(x_i)dx_1\cdots dx_d. \end{align*} $

下面计算表达式(2.2)在区域$A^c$上的积分. $A^c$可分解成如下区域的并

$ B_1=\{x_i^1\leq x_i^2, i=1, 2, \cdots, d, x_1^2>2x_1^1\}, \\B_2=\{x_i^1\leq x_i^2, i=1, 2, \cdots, d, x_2^2>2x_2^1\}, \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\vdots\\B_d=\{x_i^1\leq x_i^2, i=1, 2, \cdots, d, x_d^2>2x_d^1\}. $

仅计算在$B_2$上的积分, 其它区域上的积分可以类似计算.在$B_2$上有不等式

$ \begin{align*} &2(x_1^2-x_1^1)x_2^1\cdots x_d^1+2(x_2^2-x_2^1)x_1^1x_3^1\cdots x_d^1+\cdots+2(x_d^2-x_d^1)x_1^1x_2^1\cdots x_{d-1}^1\\ \geq &2(x_1^2-x_1^1)x_2^1\cdots x_d^1+(2x_2^1-x_2^1)x_1^1x_3^1\cdots x_d^1+(x_2^2-x_2^1)x_1^1x_3^1\cdots x_d^1+\cdots\\ &+2(x_d^2-x_d^1)x_1^1x_2^1\cdots x_{d-1}^1 \\ =&2(x_1^2-x_1^1)x_2^1\cdots x_d^1+x_1^1x_2^1x_3^1\cdots x_d^1+(x_2^2-x_2^1)x_1^1x_3^1\cdots x_d^1+\cdots+2(x_d^2-x_d^1)x_1^1x_2^1\cdots x_{d-1}^1\\ \geq &\frac{1}{d}(x_1^2-x_1^1)x_2^1\cdots x_d^1+x_1^1x_2^1x_3^1\cdots x_d^1+\frac{1}{d}(x_2^2-x_2^1)x_1^1x_3^1\cdots \\&x_d^1+\cdots+\frac{1}{d}(x_d^2-x_d^1)x_1^1x_2^1\cdots x_{d-1}^1\\ = &\frac{1}{d}(x_1^2x_2^1\cdots x_d^1+x_1^1x_2^2\cdots x_d^1+\cdots+x_1^1x_2^1\cdots x_d^2). \end{align*} $

因此表达式(2.2)在$B_2$上的积分就可以用下面的式子来控制

$ \begin{align}\label{sec2-eq2.2} &2^dn^d(\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d(f_i^2(x_i^1))(\prod\limits_{i=1}^dx_i^1)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j\nonumber \\&+\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d(f_i^2(x_i^2))(\prod\limits_{i=1}^dx_i^2)^{d-1}\prod\limits_{i=1}^d \exp[-2n(x_i^2-x_i^1)\prod\limits_{k\neq i}x_k^1]\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j)\nonumber \\\leq& 2^dn^d(\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d(f_i^2(x_i^1))(\prod\limits_{i=1}^dx_i^1)^{d-1} \nonumber \\&\times \exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1+x_1^1x_2^2\cdots x_d^1+\cdots+x_1^1x_2^1\cdots x_d^2)\}}\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j\nonumber \\&+\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d(f_i^2(x_i^2))(\prod\limits_{i=1}^dx_i^2)^{d-1} \nonumber \\&\times\exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1+x_1^1x_2^2\cdots x_d^1+\cdots+x_1^1x_2^1\cdots x_d^2)\}}\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j)\nonumber \\\leq& C_d\int_{[0, 1]^d}\prod\limits_{i=1}^df_i^2(x_i)dx_1\cdots dx_d. \end{align} $

(2.3)

事实上

$ \begin{align*} &\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d(f_i^2(x_i^1))(\prod\limits_{i=1}^dx_i^1)^{d-1} \\&\times \exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1+x_1^1x_2^2\cdots x_d^1+\cdots+x_1^1x_2^1\cdots x_d^2)\}}\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j\\ =&\int_{[0, 1]^{d}}\prod\limits_{i=1}^d(f_i^2(x_i^1))(\prod\limits_{i=1}^dx_i^1)^{d-1} \\&\quad\times \int_{x_d^1}^1\cdots\int_{x_1^1}^1\exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1+x_1^1x_2^2\cdots x_d^1+\cdots+x_1^1x_2^1\cdots x_d^2)\}}\\&dx_1^2\cdots dx_d^2dx_1^1\cdots dx_d^1\\ =&\int_{[0, 1]^{d}}\prod\limits_{i=1}^d(f_i^2(x_i^1))(\prod\limits_{i=1}^dx_i^1)^{d-1} \int_{x_d^1}^1\cdots\int_{x_2^1}^1 \exp{\{-\frac{n}{d}(x_1^1x_2^2\cdots x_d^1+\cdots+x_1^1x_2^1\cdots x_d^2)\}} \\&\quad\times \int_{x_1^1}^1\exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1)\}}dx_1^2dx_2^2\cdots dx_d^2dx_1^1\cdots dx_d^1. \end{align*} $

先对$x_1^2$求积分有

$ \begin{align*} &\int_{x_1^1}^1\exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1)\}}dx_1^2 =-\frac{d}{n}\frac{1}{x_2^1\cdots x_d^1}\exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1)\}}\mid_{x_1^1}^1\\ =&-\frac{d}{n}\frac{1}{x_2^1\cdots x_d^1}(\exp{\{-\frac{n}{d}(x_2^1\cdots x_d^1)\}}-\exp{\{-\frac{n}{d}(x_1^1x_2^1\cdots x_d^1)\}}) \\\leq& \frac{d}{n}\frac{1}{x_2^1\cdots x_d^1}\exp{\{-\frac{n}{d}(x_1^1x_2^1\cdots x_d^1)\}} \leq \frac{d}{n}\frac{1}{x_2^1\cdots x_d^1}. \end{align*} $

再依次对$x_2^2$, $\cdots, x_d^2$求积分有

$ \begin{align*} &\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d(f_i^2(x_i^1))(\prod\limits_{i=1}^dx_i^1)^{d-1} \\&\times \exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1+x_1^1x_2^2\cdots x_d^1+\cdots+x_1^1x_2^1\cdots x_d^2)\}}\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j\\ \leq&(\frac{d}{n})^d\int_{[0, 1]^{d}}\prod\limits_{i=1}^d(f_i^2(x_i^1))dx_1^1\cdots dx_d^1. \end{align*} $

同理可计算

$ \begin{align*} &\int_{[0, 1]^{2d}}\prod\limits_{i=1}^d(f_i^2(x_i^2))(\prod\limits_{i=1}^dx_i^2)^{d-1} \\&\times\exp{\{-\frac{n}{d}(x_1^2x_2^1\cdots x_d^1+x_1^1x_2^2\cdots x_d^1+\cdots+x_1^1x_2^1\cdots x_d^2)\}}\prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j\\ \leq&(\frac{d}{n})^d\int_{[0, 1]^{d}}\prod\limits_{i=1}^d(f_i^2(x_i^2))dx_1^2\cdots dx_d^2, \end{align*} $

故(2.3)式成立.综上, 可以得到

$ E(\int_{[0, 1]^d}\prod\limits_{i=1}^df_i(x_i)\theta_n(x)dx_1\cdots dx_d)^2\leq C_d\int_{[0, 1]^d}\prod\limits_{i=1}^df_{i}^2(x_i)dx_1\cdots dx_d. $

引理2.3  设$Y=\{Y(s), s\in[0, 1]^d\}$是一个连续过程.假设对固定的偶数$m\in\mathbb{N}$和$\delta_i\in(0, 1)$, $i=1, 2, \cdots, d$存在常数$L>0$使得对任意的$ 0<s_i<t_i<2s_i, ~ i=1, 2, \cdots, d$, 有

$ E(\Delta_s(Y(t)))^m\leq L\prod\limits_{i=1}^d(t_i-s_i)^{m\delta_i} $

(2.4)

成立.那么存在一个仅依赖于$m$和$\delta_i, i=1, 2, \cdots, d$的常数$C$使得对任意的$ 0\leq s_i<t_i\leq1, i=1, 2, \cdots, d$, 有$E(\Delta_s(Y(t)))^m\leq CL\prod\limits_{i=1}^d(t_i-s_i)^{m\delta_i}$成立.

证 易证$\forall 0<s_i<t_i<2^\frac{1}{\delta_i}s_i$, 过程$Y$满足

$ E(\Delta_s(Y(t)))^m\leq C_0L\prod\limits_{i=1}^d(t_i-s_i)^{m\delta_i}, $

(2.5)

其中$C_0$是依赖于$m$和$\delta_i, i=1, 2, \cdots, d$的常数.

下面把区间$(s_i, t_i]$分解成可数个互不相交的小区间的和

$ (s_i, t_i]=\bigcup\limits_{j_i=0}^\infty(a_{j_i+1}^i, a_{j_i}^i], i=1, 2, \cdots, d, $

其中$a_{j_i}^i=\frac{t_i+(2^\frac{j_i}{\delta_i}-1)s_i}{2^\frac{j_i}{\delta_i}}, i=1, 2, \cdots d$.

根据$Y$的连续性有

$ \Delta_{s_1, \cdots, s_d}Y(t_1, \cdots, t_d)=\sum\limits_{j_1, \cdots, j_d=0}^\infty\Delta_{a_{j_1+1}^1, \cdots, a_{j_d+1}^d}Y(a_{j_1}^1, \cdots, a_{j_d}^d), $

且上式右边的每个增量均满足(2.5)式.所以

$ \begin{align*} &E[\Delta_{s_1, \cdots, s_d}Y(t_1, \cdots, t_d)]^m\\ \leq &E\sum\limits_{j_1, \cdots, j_d=0}^\infty2^{(m-1)(j_1+1)}2^{(m-1)(j_2+1)}\cdots2^{(m-1)(j_d+1)} [\Delta_{a_{j_1+1}^1, \cdots, a_{j_d+1}^d}Y(a_{j_1}^1, \cdots, a_{j_d}^d)]^m\\ \leq &\sum\limits_{j_1, \cdots, j_d=0}^\infty2^{(m-1)(j_1+1)}2^{(m-1)(j_2+1)}\cdots2^{(m-1)(j_d+1)} C_0L\prod\limits_{i=1}^d(a_{j_i}^i-a_{j_i+1}^i)^{m\delta_i}\\ =&(\prod\limits_{i=1}^d(t_i-s_i)^{m\delta_i})C_0L(\prod\limits_{i=1}^d(2^\frac{1}{\delta_i}-1)^{m\delta_i}) \sum\limits_{j_1, \cdots, j_d=0}^\infty(2^{j_1+1}2^{j_2+2}\cdots2^{j_d+1})^{-1}\\ \leq &CL\prod\limits_{i=1}^d(t_i-s_i)^{m\delta_i}, \end{align*} $

其中第一步使用了不等式

$ [\sum\limits_{j_1, \cdots, j_d=0}^\infty a_{j_1, \cdots, j_d}]^m\leq \sum\limits_{j_1, \cdots, j_d=0}^\infty2^{(m-1)(j_1+1)}2^{(m-1)(j_2+1)}\cdots2^{(m-1)(j_d+1)}(a_{j_1, \cdots, j_d})^m. $

引理2.4  设$X_n$是(2.1)式中定义的随机过程, 核函数$K_i$, $i=1, 2, \cdots, d$满足条件(H2), 那么对任意的偶数$m\in\mathbb{N}$, 存在仅依赖于$m$和条件(ⅲ)中出现的参数的常数$C$, 使得对任意$0\leq s_i<t_i\leq1, i=1, 2, \cdots, d, $ $\sup_nE[\Delta_sX_n(t)]^m\leq C\prod\limits_{i=1}^d(F_i(t_i)-F_i(s_i))^{\frac{m\rho_i}{4}}.$

证

$ \begin{align} &E[\Delta_sX_n(t)]^m\nonumber\\ =&E[\int_{[0, 1]^d}(K_1(t_1, x_1)-K_1(s_1, x_1))\cdots(K_d(t_d, x_d)-K_d(s_d, x_d))\theta_n(x_1, \cdots, x_d)dx_1\cdots dx_d]^m\nonumber \\=&E[\int_{[0, 1]^d}(K_1(t_1, x_1)-K_1(s_1, x_1))\cdots(K_d(t_d, x_d)-K_d(s_d, x_d))\nonumber \\&\times n^\frac{d}{2}(\prod\limits_{i=1}^dx_i)^\frac{d-1}{2}(-1)^{N_n(x_1, \cdots, x_d)}dx_1\cdots dx_d]^m\nonumber \\=&E[\Delta_0Y(1)]^m\nonumber, \end{align} $

其中

$ \begin{align*} Y(s_0):=&n^\frac{d}{2}\int_{\prod\limits_{i=1}^d[0, s_{0i}]} (K_1(t_1, x_1)-K_1(s_1, x_1))\cdots(K_d(t_d, x_d)-K_d(s_d, x_d)) \\&\times(\prod\limits_{i=1}^dx_i)^\frac{d-1}{2}(-1)^{N_n(x_1, \cdots, x_d)}dx_1\cdots dx_d. \end{align*} $

根据引理2.3, 只需要证明对任意的$0<s_{0i}<t_{0i}<2s_{0i}, ~i=1, 2, \cdots, d$, 有

$ \begin{align}\label{sec2-eq2.5} E[\Delta_{s_0}Y(t_0)]^m\leq C_{M, m}\prod\limits_{i=1}^d (F_i(t_i)-F_i(s_i))^{\frac{m\rho_i}{4}}\prod\limits_{i=1}^d(t_{0i}-s_{0i})^{m\gamma}, \end{align} $

(2.6)

其中$\gamma=\frac{1}{4}\inf\{\beta_1, \beta_2, \cdots, \beta_d\}$.

事实上

$ \begin{align} E[\Delta_{s_0}Y(t_0)]^m\nonumber=&n^{\frac{md}{2}}E\int_{[0, 1]^{md}}\prod\limits_{i=1}^d\prod\limits_{j=1}^mI_{(s_{0i}, t_{0i})}\\&(x_i^j) \prod\limits_{i=1}^d\prod\limits_{j=1}^m(K_i^j(t_i^j, x_i^j)-K_i^j(s_i^j, x_i^j))(\prod\limits_{i=1}^d\prod\limits_{j=1}^mx_i^j)^\frac{d-1}{2}\nonumber \\&\times (-1)^{\sum\limits_{j=1}^mN_n(x_1^j, \cdots, x_d^j)}\prod\limits_{j=1}^mdx_1^j\cdots dx_d^j\nonumber \\=&n^{\frac{md}{2}}\int_{[0, 1]^{md}}\prod\limits_{i=1}^d\prod\limits_{j=1}^mI_{(s_{0i}, t_{0i})}(x_i^j) \prod\limits_{i=1}^d\prod\limits_{j=1}^m(K_i^j(t_i^j, x_i^j)-K_i^j(s_i^j, x_i^j)) (\prod\limits_{i=1}^d\prod\limits_{j=1}^mx_i^j)^\frac{d-1}{2}\nonumber \\&\times E[(-1)^{\sum\limits_{j=1}^mN_n(x_1^j, \cdots, x_d^j)}]\prod\limits_{j=1}^mdx_1^j\cdots dx_d^j.\nonumber \end{align} $

注意到

$ \begin{align}\label{sec2-eq2.6} E[(-1)^{\sum\limits_{j=1}^mN_n(x_1^j, \cdots, x_d^j)}]\leq \prod\limits_{i=1}^d|E((-1)^{\sum\limits_{j=1}^m\Delta_{(0, \cdots, 0, s_i, 0, \cdots, 0)}N_n(s_1, \cdots, s_{i-1}, x_i^j, s_{i+1}, \cdots, s_d)})|. \end{align} $

(2.7)

事实上, 对$x\in\prod\limits_{i=1}^d(s_i, t_i]$有$[0, x]=A_x\bigcup B_x$, 其中

$ A_x:=\bigcup\limits_{i=1}^d[(0, \cdots, 0, s_i, 0, \cdots, 0), (s_1, \cdots, s_{i-1}, x_i^j, s_{i+1}, \cdots, s_d)]\subset G, $

这里

$ G:=\bigcup\limits_{i=1}^d[(0, \cdots, 0, s_i, 0, \cdots, 0), (s_1, \cdots, s_{i-1}, t_i, s_{i+1}, \cdots, s_d)]. $

进一步$B_x\subset G^c$.给定$x^j, j=1, 2, \cdots, m, $那么$\bigcup\limits_{j=1}^mA_{x^j}$和$\bigcup\limits_{j=1}^mB_{x^j}$是互不相交的.所以相关的增量也是独立的, 故易得到(2.7)式.

现假设$x_i^1\leq x_i^2\leq\cdots\leq x_i^m$对所有$i=1, 2, \cdots, d$.考虑具有独立增量的泊松过程, 若$Z\sim P(\lambda)$, 则$E[(-1)^Z]=\exp(-2\lambda)$.故

$ \begin{align*} &|E((-1)^{\sum\limits_{j=1}^m\Delta_{(0, \cdots, 0, s_i, 0, \cdots, 0)}N_n(s_1, \cdots, s_{i-1}, x_i^j, s_{i+1}, \cdots, s_d)})| \\=&\exp{[-2n\sum\limits_{j=1}^m(x_i^j-s_i)\prod\limits_{k\neq i}s_k]} \leq \exp[-2n\sum\limits_{l=1}^\frac{m}{2}(x_i^{2l}-x_i^{2l-1})\prod\limits_{k\neq i}s_k]. \end{align*} $

由于$x_i^j\leq t_i\leq2s_i$对所有$i=1, 2, \cdots, d, j=1, 2, \cdots, m$成立, 那么有

$ \begin{align}\label{sec2-eq2.11} E[(-1)^{\sum\limits_{j=1}^mN_n(x_1^j, \cdots, x_d^j)}] \leq&\prod\limits_{i=1}^d \exp[-2n\sum\limits_{l=1}^\frac{m}{2}(x_i^{2l}-x_i^{2l-1})\prod\limits_{k\neq i}s_k]\nonumber\\ \leq &\prod\limits_{i=1}^d\exp[-2^{2-d}n(\sum\limits_{l=1}^\frac{m}{2}(x_i^{2l}-x_i^{2l-1})\prod\limits_{k\neq i}x_k^{2l-1})]. \end{align} $

(2.8)

因此根据(2.8)式可得

$ \begin{align*} &E[\Delta_{s_0}Y(t_0)]^m\\ =&n^{\frac{md}{2}}(m!)^d\int_{[0, 1]^{md}}\prod\limits_{i=1}^d\prod\limits_{j=1}^mI_{(s_{0i}^j, t_{0i}^j)}(x_i^j) \prod\limits_{i=1}^d\prod\limits_{j=1}^m(K_i^j(t_i^j, x_i^j)-K_i^j(s_i^j, x_i^j)) (\prod\limits_{i=1}^d\prod\limits_{j=1}^mx_i^j)^\frac{d-1}{2} \\&\times E[(-1)^{\sum\limits_{j=1}^mN_n(x_1^j, \cdots, x_d^j)}] \prod\limits_{i=1}^dI_{\{x_i^1\leq\cdots\leq x_i^m\}}\prod\limits_{j=1}^mdx_1^j\cdots dx_d^j\\ \leq &n^{\frac{md}{2}}(m!)^d\int_{[0, 1]^{md}}\prod\limits_{i=1}^d\prod\limits_{j=1}^mI_{(s_{0i}^j, t_{0i}^j)}(x_i^j) \prod\limits_{i=1}^d\prod\limits_{j=1}^m|K_i^j(t_i^j, x_i^j)-K_i^j(s_i^j, x_i^j)| (\prod\limits_{i=1}^d\prod\limits_{j=1}^mx_i^j)^\frac{d-1}{2} \\&\times \prod\limits_{i=1}^d\exp[-2^{2-d}n(\sum\limits_{l=1}^\frac{m}{2}(x_i^{2l}-x_i^{2l-1})\prod\limits_{k\neq i}x_k^{2l-1})] \prod\limits_{i=1}^dI_{\{x_i^1\leq\cdots\leq x_i^m\}}\prod\limits_{j=1}^mdx_1^j\cdots dx_d^j \\\leq&C_m(n^d\int_{[0, 1]^{d}}\prod\limits_{i=1}^d\prod\limits_{j=1}^2I_{(s_{0i}^j, t_{0i}^j)}(x_i^j) \prod\limits_{i=1}^d\prod\limits_{j=1}^2|K_i^j(t_i^j, x_i^j)-K_i^j(s_i^j, x_i^j)| (\prod\limits_{i=1}^d\prod\limits_{j=1}^2x_i^j)^\frac{d-1}{2} \\&\times\prod\limits_{i=1}^d\exp[-2^{2-d}n((x_i^{2}-x_i^{1})\prod\limits_{k\neq i}x_k^{1})] \prod\limits_{i=1}^dI_{\{x_i^1\leq x_i^2\}}\prod\limits_{j=1}^2dx_1^j\cdots dx_d^j)^\frac{m}{2}. \end{align*} $

由引理2.2的计算过程可类似计算上式不超过

$ \begin{align*} &C_m(\int_{[0, 1]^d}\prod\limits_{i=1}^d(K_i(t_i, x_i)-K_i(s_i, x_i))^2 \prod\limits_{i=1}^dI_{(s_{0i}, t_{0i})}(x_i)dx_1\cdots dx_d)^\frac{m}{2} \\\leq&C_m(\int_{[0, 1]^d}\prod\limits_{i=1}^d(K_i(t_i, x_i)-K_i(s_i, x_i))^2 \prod\limits_{i=1}^dI_{(s_{0i}, t_{0i})}(x_i)dx_1\cdots dx_d)^\frac{m}{4} \\&\times (\int_{[0, 1]^d}\prod\limits_{i=1}^d(K_i(t_i, x_i)-K_i(s_i, x_i))^2dx_1\cdots dx_d)^\frac{m}{4}\\\leq &C_m\prod\limits_{i=1}^d(M_i^{\frac{m}{4}}(t_{0i}-s_{0i})^{\frac{m\beta_i}{4}}(F_i(t_i)-F_i(s_i))^\frac{m\rho_i}{4}) \\\leq &C_{M, m}\prod\limits_{i=1}^d(t_{0i}-s_{0i})^{\gamma m}\prod\limits_{i=1}^d(F_i(t_i)-F_i(s_i))^\frac{m\rho_i}{4}, \end{align*} $

其中第二个不等式使用了条件(H2), $\gamma=\frac{1}{4}\inf\{\beta_1, \beta_2, \cdots, \beta_d\}$.所以不等式(2.6)成立, 进而引理成立.

定理2.5  在(H1)或(H2)的条件下, 当$n\rightarrow\infty$时, $(2.1)$式定义的随机过程族$\{X_n(t), t\in[0, 1]^d\}$在连续函数空间$\mathcal{C}([0, 1]^d)$中弱收敛到$X$.

证 首先, 证明随机过程族$\{X_n\}$的胎紧性.由于$\{X_n\}$在坐标轴上取值为零, 利用Bickel和Wichura ^[5]中建立的关于多参数随机过程的胎紧性准则, 只需证明对某些$m\geq2$存在常数$C>0$和$\eta>1$, 以及单增的连续函数$F_i, ~i=1, 2, \cdots, d$, 使得对$0\leq s_i\leq t_i\leq1, i=1, 2, \cdots, d$,

$ \sup\limits_nE[\Delta_sX_n(t)]^m\leq C(\prod\limits_{i=1}^d(F_i(t_i)-F_i(s_i)))^\eta $

(2.9)

成立即可.

假设条件(H1)成立, 根据引理2.2和条件(ⅱ)有

$ \begin{align*} &E[\Delta_sX_n(t)]^2\\ =&E(\int_{[0, 1]^d}(K_1(t_1, x_1)-K_1(s_1, x_1))\cdots(K_d(t_d, x_d)-K_d(s_d, x_d)) \theta_n(x_1, \cdots, x_d)dx_1\cdots dx_d)^2\\ \leq &C\int_{[0, 1]^d}\prod\limits_{i=1}^d(K_i(t_i, x_i)-K_i(s_i, x_i))^2dx_1\cdots dx_d \leq C\prod\limits_{i=1}^d(F_i(t_i)-F_i(s_i))^{\alpha_i}, \end{align*} $

其中第一个不等式使用了引理2.2, $\alpha_i>1, i=1, 2, \cdots, d$.即(2.9)式成立.

假设条件(H2)成立.根据引理2.4, 对任意的偶数$m\in\mathbb{N}$存在一个常数$C$使得对$0\leq s_i\leq t_i\leq1, i=1, 2, \cdots, d$时, 有$\sup_nE[\Delta_sX_n(t)]^m\leq C(\prod\limits_{i=1}^d(F_i(t_i)-F_i(s_i)))^\frac{m\rho_i}{4}$成立, 即(2.9)式成立.

其次, 证明随机过程$X_n$依有限维分布收敛到$X$.事实上, 只要证明$\forall k\geq1$, $a_1, \cdots, a_k\in\mathbb{R}$且$(t_1^j, \cdots, t_d^j)\in[0, 1]^d$, 当$n\rightarrow\infty$时, 随机变量列$S_n=\sum\limits_{j=1}^ka_jX_n(t_1^j, \cdots, t_d^j)$依分布收敛到$S=\sum\limits_{j=1}^ka_jX(t_1^j, \cdots, t_d^j)$.等价的只要证明相应的特征函数收敛即可.

设$K^*(x_1, \cdots, x_d)=\sum\limits_{j=1}^ka_j\prod\limits_{i=1}^dK_i(t_i^j, x_i)$, 定义

$ S_n=\int_{[0, 1]^d}K^*(x_1, \cdots, x_d)\theta_n(x_1, \cdots, x_d)dx_1\cdots dx_d $

和

$ S=\int_{[0, 1]^d}K^*(x_1, \cdots, x_d)dW_{x_1, \cdots, x_d}. $

由于函数$K_i(t_i^j, x_i)\in L^2([0, 1]^2)$, $i=1, 2, \cdots, d$, 所以可选取简单函数$\gamma_i^{j, l}(x_i)=\sum\limits_{k^i=0}^{P_l^i-1}K_{k^i}^lI_{(x_{k^i}^l, x_{k^i+1}^l]}(x_i), $其中$0=x_0^l<x_1^l<\cdots<x_{P_l^i}^l=1$, 使得$\displaystyle\int_0^1(\gamma_i^{j, l}-K_i(t_i^j, x_i))^2dx_i\leq\frac{1}{l}, ~~ i=1, 2, \cdots, d, $所以

$ \int_{[0, 1]^d}(K^l(x_1, \cdots, x_d)-K^*(x_1, \cdots, x_d))^2dx_1\cdots dx_d\leq C\frac{1}{l}, $

(2.10)

其中$K^l(x_1, \cdots, x_d)=\sum\limits_{j=1}^ka_j\prod\limits_{i=1}^d\gamma_i^{jl}(x_i).$

定义

$ S_n^l=\int_{[0, 1]^d}K^l(x_1, \cdots, x_d)\theta_n(x_1, \cdots, x_d)dx_1\cdots dx_d $

和$S^l=\displaystyle\int_{[0, 1]^d}K^l(x_1, \cdots, x_d)dW_{x_1, \cdots, x_d}.$因此

$ E|S_n^l-S_n|^2\leq C\int_{[0, 1]^d}(K^l(x_1, \cdots, x_d)-K^*(x_1, \cdots, x_d))^2dx_1\cdots dx_d\leq C\frac{1}{l}. $

(2.11)

另一方面, 根据Bardina, Jolis和Rovira ^[1], 当$n\rightarrow\infty$时,

$ S_n^l=\sum\limits_{j=1}^ka_j\prod\limits_{j=1}^d(\sum\limits_{k^i=0}^{P_l^i-1}K_{k^i}^l) \int_{\prod\limits_{i=1}^d[x_{k^i}^l, x_{k^i+1}^l]}\theta_n(x_1, \cdots, x_d)dx_1\cdots dx_d $

依分布收敛到

$ \begin{align} &\sum\limits_{j=1}^ka_j\prod\limits_{j=1}^d(\sum\limits_{k^i=0}^{P_l^i-1}K_{k^i}^l) \int_{\prod\limits_{i=1}^d[x_{k^i}^l, x_{k^i+1}^l]}dW_{x_1, \cdots, x_d}\nonumber \\=&\int_{[0, 1]^d}\sum\limits_{j=1}^ka_j\prod\limits_{j=1}^d(\sum\limits_{k^i=0}^{P_l^i-1}K_{k^i}^l) \prod\limits_{i=1}^dI_{(x_{k^i}^l, x_{k^i+1}^l]}(x_i)dW_{x_1, \cdots, x_d}\nonumber \\=&\int_{[0, 1]^d}K^l(x_1, \cdots, x_d)dW_{x_1, \cdots, x_d}\nonumber =S^l\nonumber. \end{align} $

因此$\forall x\in\mathbb{R}$和$l\in\mathbb{N}$, 有

$ E(e^{ixS_n^l})\rightarrow E(e^{ixS^l}), \quad n\rightarrow\infty. $

(2.12)

最后$\forall x\in\mathbb{R}$和$l, n\in\mathbb{N}, $ $|E[e^{ixS_n}]-E[e^{ixS}]|\leq \eta_1+\eta_2+\eta_3, $其中

$ \eta_1=|E[e^{ixS_n}]-E[e^{ixS_n^l}]|, ~~\eta_2=|E[e^{ixS_n^l}]-E[e^{ixS^l}]|, ~~\eta_3=|E[e^{ixS^l}]-E[e^{ixS}]|\nonumber. $

联立(2.11)式以及对$t\in\mathbb{R}$, $|\frac{e^{ita}-e^{itb}}{it}|=|\displaystyle\int_a^be^{ity}dy|\leq|a-b|, $可以得到

$ \eta_1\leq E|e^{ixS_n}-e^{ixS_n^l}|\leq |x|E|S_n-S_n^l|\leq |x|[E(S_n-S_n^l)^2]^\frac{1}{2}\leq C|x|\frac{1}{\sqrt{l}}\rightarrow0, \quad l\rightarrow\infty. $

根据(2.12)式, 当$n\rightarrow\infty$时, $\eta_2\rightarrow0$.根据(2.10)式和多重随机积分的性质有

$ \eta_3\leq E|e^{ixS^l}-e^{ixS}|\leq |x|E|S^l-S|\leq |x|(E(S^l-S)^2)^\frac{1}{2}\leq C|x|\frac{1}{\sqrt{l}}\rightarrow0 , \quad l\rightarrow\infty. $

综上, 定理得证.

基于Mandelbrot和Van Ness ^[11]的工作, 分数布朗运动$B^\alpha=\{B_t^\alpha, t\in\mathbb{R}_+\}, \alpha\in(0, 1)$受到越来越多学者的关注, 现已广泛应用于金融、通讯等领域, 它是具有自相似、长相依、平稳增量的高斯过程, 其协方差函数为

$ R(t, s)=E[B_t^\alpha, B_s^\alpha]=\frac{1}{2}(t^{2\alpha}+s^{2\alpha}-|t-s|^{2\alpha}), \nonumber $

且具有如下积分表示$B_t^\alpha=\displaystyle\int_0^tK_\alpha(t, s)dB_s, $其中$B$是一个标准的布朗运动且核函数$K_\alpha$为

$ K_\alpha(t, s)=[c_\alpha(t-s)^{\alpha-\frac{1}{2}}+c_\alpha(\frac{1}{2}-\alpha)\int_s^t(u-s)^{\alpha-\frac{3}{2}}(1-(\frac{s}{u})^{\frac{1}{2}-\alpha})du]I_{(0, t)}(s), $

其中

$ c_\alpha=(\frac{2\alpha\Gamma(\frac{3}{2}-\alpha)}{\Gamma(\alpha+\frac{1}{2})\Gamma(2-2\alpha)})^\frac{1}{2} $

是一个标准化的常数.

作为分数布朗运动的扩张过程, $d$维分数布朗单$B^H=\{B_t^H, t\in\mathbb{R}_+^d\}$是定义在概率空间$(\Omega, \digamma, P)$上的中心高斯过程且协方差函数$E[B_t^H B_s^H]=\prod\limits_{k=1}^d\frac{1}{2}[s_k^{2H_k}+t_k^{2H_k}-|t_k-s_k|^{2H_k}], $其中$H=(H_1, H_2, \cdots, H_d)\in(0, 1)^d$.当$H_1=H_2=\cdots=H_d=\frac{1}{2}$, $B^H=\{B_t^H, t\in\mathbb{R}_+^d\}$是标准的$d$维布朗单$W=\{W_t, t\in\mathbb{R}_+^d\}$.它有连续的样本轨道且在坐标轴上为零, 具有如下积分表示

$ \int_0^{t_d}\cdots\int_0^{t_1}K_{H_1}(t_1, u_1)K_{H_2}(t_2, u_2)\cdots K_{H_d}(t_d, u_d)dW_u. $

以下验证$d$维分数布朗单满足本文中的假设条件.

对于核$K_{H_i}, i=1, 2, \cdots, d, H_i>\frac12$, 有

$ \int_0^1(K_{H_i}(t_i, x_i)-K_{H_i}(s_i, x_i))^2dx_i=E(B_{t_i}^{H_i}-B_{s_i}^{H_i})^2=(t_i-s_i)^{2H_i}. $

此时取$F_i(x)=x, \alpha_i=2H_i, $则核函数$K_{H_i}, i=1, 2, \cdots, d$满足条件(H1).

核函数$K_{H_i}, i=1, 2, \cdots, d$也满足条件(H2).事实上只需要验证核函数满足条件(ⅲ).

若$H_i=\frac{1}{2}, i=1, 2, \cdots, d$, 那么$K_{H_i}(t_i, x_i)=I_{(0, t_i)}(x_i)$, 则条件(ⅲ)显然成立.

对任意的$a$, $b$, $z$, $|z|>1$且任意的$c\neq0, -1$定义高斯超几何函数$F(a, b, c, z)$如下(详见参考文献[12])

$ F(a, b, c, z)=\sum\limits_{k=0}^\infty\frac{(a)_k(b)_k}{(c)_kk!}z^k, $

其中$(a)_k=a(a+1)(a+2)\cdots(a+k-1)=\frac{\Gamma(a+k)}{\Gamma(a)}$.因此核函数有如下形式(参见文献[9])

$ \begin{align*} K_{H_i}(t_i, x_i)=\frac{2^{-2H_i}\sqrt{\pi}}{\Gamma(H_i)\sin(\pi H_i)}x_i^{H_i-\frac{1}{2}} +\frac{(t_i-x_i)^{H_i-\frac{1}{2}}}{2\Gamma(H_i+\frac{1}{2})}F(\frac{1}{2}-H_i, 1, 2-2H_i, \frac{x_i}{t_i}), \end{align*} $

又由参考文献[12]知上式中的超几何函数在$H_i>\frac{1}{2}$的情况下可写作

$ F(\frac{1}{2}-H_i, 1, 2-2H_i, \frac{x_i}{t_i})= C_1F(\frac{1}{2}-H_i, 1, H_i+\frac{1}{2}, 1-\frac{x_i}{t_i}) +C_2(1-\frac{x_i}{t_i})^{\frac{1}{2}-H_i}(\frac{x_i}{t_i})^{2H_i-1}, $

其中$C_1$, $C_2$是常数.在超几何函数$F(\frac{1}{2}-H_i, 1, H_i+\frac{1}{2}, 1-\frac{x_i}{t_i})$中因为$H_i+\frac{1}{2}-(\frac{1}{2}-H_i)-1=2H_i-1>0$, 所以$F(\frac{1}{2}-H_i, 1, H_i+\frac{1}{2}, 1-\frac{x_i}{t_i})$在$[0, t_i]$上关于$x_i$是连续的.因此在$H_i>\frac{1}{2}$时, $K_{H_i}(t_i, x_i)x_i^{H_i-\frac{1}{2}}$在$[0, t_i]$上关于$x_i$是连续的, 故在$[0, t_i]$上存在常数$C'$使得$|K_{H_i}(t_i, x_i)x_i^{H_i-\frac{1}{2}}|\leq C'$, 即$|K_{H_i}(t_i, x_i)|\leq C'x_i^{-H_i+\frac{1}{2}}.$同理存在$C''$使得$|K_{H_i}(s_i, x_i)|\leq C''x_i^{-H_i+\frac{1}{2}}$.所以$\forall s_i<t_i, i=1, 2, \cdots, d$, 有$0<|K_{H_i}(t_i, x_i)-K_{H_i}(s_i, x_i)| \leq Cx_i^{\frac{1}{2}-H_i}, $其中$C$取$C'$和$C''$中较大者.因此对$s_{0i}<t_{0i}$, 有

$ \int_{s_{0i}}^{t_{0i}}(K_{H_i}(t_i, x_i)-K_{H_i}(s_i, x_i))^2dx_i\leq C_{H_i}(t_{0i}-s_{0i})^{2-2H_i}, $

其中$C_{H_i}=\frac{C^2}{2-2H_i}$, 此时取$ \beta_i=2-2H_i>0.$

若$H_i<\frac{1}{2}, i=1, 2, \cdots, d$时, 由于超几何函数$F(\frac{1}{2}-H_i, 1, 2-2H_i, \frac{x_i}{t_i})$中$2-2H_i-(\frac{1}{2}-H_i)-1=\frac{1}{2}-H_i>0$, 所以$F(\frac{1}{2}-H_i, 1, 2-2H_i, \frac{x_i}{t_i})$在$[0, t_i]$上关于$x_i$是连续的.因此在$H_i<\frac{1}{2}$时, $K_{H_i}(t_i, x_i)x_i^{-H_i+\frac{1}{2}}(t_i-x_i)^{-H_i+\frac{1}{2}}$在$[0, t_i]$上关于$x_i$是连续的, 故在$[0, t_i]$上存在常数$d'$使得$|K_{H_i}(t_i, x_i)x_i^{-H_i+\frac{1}{2}}(t_i-x_i)^{-H_i+\frac{1}{2}}|\leq d'$, 即$|K_{H_i}(t_i, x_i)|\leq d'x_i^{H_i-\frac{1}{2}}(t_i-x_i)^{H_i-\frac{1}{2}}.$同理存在$d''$使得$|K_{H_i}(s_i, x_i)|\leq d''x_i^{H_i-\frac{1}{2}}(s_i-x_i)^{H_i-\frac{1}{2}}$.则$\forall 0\leq s_i<t_i\leq1$和$\forall x_i\in[0, 1], i=1, 2, \cdots, d$有

$ |K_{H_i}(t_i.x_i)-K_{H_i}(s_i, x_i)|\leq D((s_i-x_i)^{H_i-\frac{1}{2}}I_{[0, s_i)}(x_i)+[x_i^{H_i-\frac{1}{2}}+(t_i-x_i)^{H_i-\frac{1}{2}}]I_{[s_i, t_i)}(x_i)), $

其中$D$取$d'$和$d''$中较大者.所以$\forall s_{0i}<t_{0i}, i=1, 2, \cdots, d, $

$ \begin{align*} &\int_{s_{0i}}^{t_{0i}}(K_{H_i}(t_i.x_i)-K_{H_i}(s_i, x_i))^2dx_i \leq\int_{s_{0i}}^{t_{0i}}[D(x_i^{H_i-\frac{1}{2}}+(t_i-x_i)^{H_i-\frac{1}{2}})]^2dx_i\\ \leq&\int_{s_{0i}}^{t_{0i}}2D^2x_i^{2H_i-1}dx_i+\int_{s_{0i}}^{t_{0i}}2D^2(t_i-x_i)^{2H_i-1}dx_i &\\\leq\int_{s_{0i}}^{t_{0i}}2D^2x_i^{2H_i-1}dx_i \leq C_{H_i}(t_{0i}-s_{0i})^{2H_i}, \end{align*} $

其中$C_{H_i}=\frac{D^2}{H_i}$.