分枝随机游动是概率论中的一个经典模型. 该模型描述了在空间中分枝过程粒子的分布发展的过程, 从而推广了经典的Galton-Watson过程. 尽管此模型已有相当长的历史, 它现在仍是概率论及应用中的重要研究对象. 它在各种应用学科如生物、计算机科学和人口动力学中是一个普适模型. 同时, 该模型揭示了随机动力系统的根本性质, 因此被用在许多重要的随机模型中, 如乘积瀑布、无穷粒子系统、随机分形等. 关于分枝随机游动的经典理论, 可以查阅参考文献[1, 2].

自Harris^{[3, Chapter Ⅲ, §16]}提出了分枝随机游动的中心极限定理以来, 有许多关于此课题的工作, 参见文献[4-16].

此文的目标是对于一类依时变化的随机环境中的分枝随机游动, 推导其粒子分布的二阶渐近展开表达式. Révész^[2]开启了对$ {\mathbb{Z}}^d $中简单分枝随机游动中的局部极限定理的收敛速度的研究. Chen^[17]在假定二阶矩的条件下, 推出了局部极限定理的精确收敛速度, 这其实相当于其一阶渐近展开. Gao在文献[18]中弱化了文献[17] 中的条件并修正了其结论, 并进一步在文献[19]中得到了局部极限定理的二阶渐近展开. 该文的主要目标是将文献[19] 中的结果推广至依时变化的随机环境中的分枝随机游动的情形. 随机环境的出现使得模型更切合实际, 但同时导致分析非常困难复杂.

当模型中的随机游动服从$ {\mathbb{R}} $上的非格点分布时, 对应的依时随机环境中的分枝随机游动的中心极限定理的展开问题已经在文献[20, 21] 中得以研究. 这种情形下, 一些相关的其它类型的极限定理已在文献[22] 与^[8]中予以研究. 此外文献[23-25] 详细探讨了与此模型相关的Biggins鞅的性质. 在已有文献中引入过各种形式的随机环境中的分枝随机游动, 参见文献[26-35].

随机环境$ \xi=(\xi_n) $是一列给定空间上的独立同分布的随机变量序列. 由$ \xi_n $的取值可以确定两个分布律: $ {\mathbb{N}} $上的分布律$ p(\xi_n)=\{p_k(\xi_n): k\in{\mathbb{N}}\} $, 及$ {\mathbb{Z}}^d $上的分布律$ G_{\xi_n}(\cdot) $.

(1.1) $ \begin{equation} G_{\xi_n}({\bf 0})=r_n, \quad G_{\xi_n}({\bf e}_v)=G_{\xi_n}(-{\bf e}_v)=\frac{1-r_n}{2d}, \quad v=1, 2, \cdots, d, \end{equation} $

其中$ r_n=r(\xi_n)\in (0, 1) $是依赖于$ \xi_n $的数, $ {\bf 0}=(0, 0, \cdots, 0)\in {\mathbb{Z}}^d $, $ {\bf e}_v(1\leq v\leq d) $是$ {\mathbb{R}}^d $中的标准正交基.

不失一般性, 取$ \xi_n $为$ (\Theta, {\cal F}, \tau_0) $的乘积空间$ (\Theta^{\mathbb{N}}, {\cal F}^{\bigotimes{\mathbb{N}}}, \tau) $中的坐标函数, 其在$ \Theta^{\mathbb{N}} $上的平移变换下$ \theta $: $ \theta(\xi_0, \xi_1, \cdots)=(\xi_1, \xi_2, \cdots) $下是平稳遍历的.

给定环境$ \xi=(\xi_n) $, 拟研究过程可以描述如下. 在时刻$ 0 $, 一个初始粒子位于$ S_\varnothing={\bf 0}\in{\mathbb{Z}}^d $; 在时刻$ 1 $, 该粒子被$ N=N_\varnothing $个$ 1 $代的新粒子$ i=\varnothing{i}(1\leq{i}\leq{N}) $替代, 分别位于$ S_i=L_i(\varnothing)\in{\mathbb{Z}}^d $, 其中$ N, L_1(\varnothing), L_2(\varnothing), \cdots $是相互独立的. $ N $服从分布$ p(\xi_0) $, $ L_i(\varnothing) $服从分布$ G_{\xi_0} $. 一般地, 每一个$ n $代粒子在$ n+1 $时刻被$ N_u $个$ n+1 $代的新粒子$ ui(1\leq{i}\leq{N_u}) $所取代, 且每个$ ui $位置为$ S_{ui}=S_u+L_i(u) $, 其中$ N_u, L_1(u), L_2(u) , \cdots $相互独立, $ N_u $服从分布$ p(\xi_n) $, $ L_{i}(u) (i=1, 2, \cdots) $服从分布$ G_{\xi_n} $由定义, 给定环境$ \xi $, 随机变量$ N_u $和$ L_i(u) (1\leq i\leq N_u) $（指标$ u $是正整数的有限序列）是彼此相互独立的.

给定环境$ \xi\in\Theta^{\mathbb{N}} $, 令$ (\Gamma, {\cal G}, {\mathbb{P}}_\xi) $是概率空间, 在其上定义上述过程. 概率$ {\mathbb{P}}_\xi $常称为quenched分布.全概率空间可定义为乘积空间$ (\Theta^{\mathbb{N}}\times\Gamma, \epsilon^{\mathbb{N}}\bigotimes{\cal G}, {\mathbb{P}}) $, 其中$ {\mathbb{P}}={\mathbb{E}}(\delta_\xi\bigotimes{{\mathbb{P}}_\xi}) $, $ \delta_\xi $在$ \xi $处的Dirac测度, $ {\mathbb{E}} $是关于随机变量$ \xi $的期望. 因此对于所有定义在$ \Theta^{\mathbb{N}}\times\Gamma $的可测函数$ g $, 有

$ \begin{eqnarray} \int_{\Theta^{\mathbb{N}}\times\Gamma}g(x, y){\rm d}{\mathbb{P}}(x, y)={\mathbb{E}}\int_\Gamma{g(\xi, y)}{\rm d}{\mathbb{P}}_\xi(y). \end{eqnarray} $

全概率$ {\mathbb{P}} $通常称之为annealed分布. 而quenched概率$ {\mathbb{P}}_\xi $可以看做给定$ \xi $时$ {\mathbb{P}} $的条件概率. 关于$ {\mathbb{P}} $的期望记为$ {\mathbb{E}} $. 关于$ {\mathbb{P}}_\xi $的期望记为$ {\mathbb{E}}_\xi $.

令$ {\mathbb{T}} $表示以$ \{N_u\} $为组成元素的系谱树. 根据定义, 可得: (a). $ \varnothing\in{\mathbb{T}} $; (b). $ ui\in{\mathbb{T}} $蕴含了$ u\in{\mathbb{T}} $; (c).若$ u\in{\mathbb{T}} $, 那么$ ui\in{\mathbb{T}} $当且仅当$ 1\leq{i}\leq{N_u} $. 令

$ \begin{eqnarray} {\mathbb{T}}_n=\{u\in{\mathbb{T}}: |u|=n\} \end{eqnarray} $

为$ n $代粒子的集合, 其中$ |u| $表示序列$ u $的长度, 也即$ u $所属的代数.

令$ Z_n(\cdot) $表示$ n $粒子的计数测度: 对$ B\subset{\mathbb{Z}}^d $, 有

$ \begin{eqnarray} Z_n(B)=\sum\limits_{u\in{\mathbb{T}}_n}{\bf 1}_B(S_u). \end{eqnarray} $

特别地, 通常将$ Z_n(\{z\}) $简记为$ Z_n(z) $, 意指位于$ z\in {\mathbb{Z}}^d $处的第$ n $代粒子的个数.

根据上述记号定义, $ \{Z_n({\mathbb{Z}}^d)\} $组成了随机环境中的分枝过程^[36-38]. 对$ n\geq0 $, 令$ \widehat{N}_n $和$ \widehat{L}_n $是在$ {\mathbb{P}_\xi} $下的条件分布律分别为$ p(\xi_n) $和$ G_{\xi_n} $的随机变量, 定义

$ \begin{eqnarray} m_n=m(\xi_n)={\mathbb{E}}_\xi\widehat{N}_n, \qquad\Pi_n=m_0\cdots{m}_{n-1}, \qquad\Pi_0=1. \end{eqnarray} $

已知正则化序列

(1.2) $ \begin{equation} W_n=\frac{1}{\Pi_n}Z_n({\mathbb{Z}}^d), \qquad n\geq1 \end{equation} $

是适应于$ \sigma $代数族$ ({\cal F}_n) $的鞅, 其中

$ \begin{eqnarray} {\cal F}_0=\{\emptyset, \Omega\}, \qquad{\cal F}_n=\sigma(\xi, N_u : |u|<n), \qquad \mbox{对} \ \ n\geq1. \end{eqnarray} $

下列$ \sigma $代数族$ \{ {\cal D}_n\} $也将会在后面用到

$ \begin{eqnarray} {\cal D}_0=\{\emptyset, \Omega\}, \quad {\cal D}_n=\sigma(\xi, N_u, L_i(u): \ \ i\geq{1}, |u|<n) \ \ \mbox{对} \ \ n\geq{1}. \end{eqnarray} $

此文将始终假设下面条件成立

(1.3) $ \begin{equation} {\mathbb{E}}\log{m_0}>0 \quad \mbox{ 且 } \quad {\Bbb E}\left[\frac{1}{m_0}\widehat{N}_0(\log^+{\widehat{N}_0})^{\lambda+1}\right]<\infty, \end{equation} $

其中$ \lambda>0 $的具体限制条件将在定理中给出. 在这些条件下, 过程$ \big\{Z_n({\mathbb{Z}}^d)\big\} $是上临界的, 意指以正概率有$ Z_n\big({\mathbb{Z}}^d\big)\rightarrow \infty $, 并且极限$ W=\lim_nW_n $满足$ {\mathbb{E}}W=1 $, $ W>0 $在$ \{Z_n\big({\mathbb{Z}}^d\big)\rightarrow \infty\} $几乎必然(简记为a.s.)成立, 参见文献[37, 39].

对$ x=(x_1, \cdots, x_d)\in{\mathbb{R}}^d $和$ y=(y_1, \cdots, y_d)\in{\mathbb{R}}^d $, 定义

$ \begin{eqnarray} \langle{x, y}\rangle=x_1y_1+x_2y_2+\cdots+x_dy_d, \quad \|x\|=\sqrt{x_1^2+x_2^2+\cdots+x_d^2}. \end{eqnarray} $

令

(1.4) $ \begin{equation} {b}_{n}=\sum\limits_{i=0}^{n-1} (1-r_i), \quad {b}_{2, n} = \sum\limits_{i=0}^{n-1} (1-r_i)^2, \quad {b}_{3, n}= \sum\limits_{i=0}^{n-1} (1-r_i)^3. \end{equation} $

易见$ {b}_{n}\leq n, {b}_{2, n}\leq n, {b}_{3, n}\leq n. $由大数定律得: 当$ n\rightarrow \infty $时, 有

(1.5) $ \begin{equation} \frac{ {b}_{n} }{ n} \mathop{\longrightarrow}\limits^{{\rm a.s. }} {\mathbb{E}} (1-r_0) , \; \; \frac{ {b}_{2, n} }{ n}\mathop{\longrightarrow}\limits^{{\rm a.s. }} {\mathbb{E}} (1-r_0)^2 , \; \; \frac{ {b}_{3, n} }{ n} \mathop{\longrightarrow}\limits^{{\rm a.s. }} {\mathbb{E}} (1-r_0)^3 . \end{equation} $

除了已知的鞅$ \{ W_n\} $, 该文还将需要下面的序列

$ N_{1, n} = \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n} S_u, N_{2, n} = \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n} \Big( {\left\Vert {S_u} \right\Vert} ^2- {b}_n \Big ), $

$ N_{3, n} = \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n} \Big( {\left\Vert {S_u} \right\Vert} ^2S_u- (1+2/d) {b}_nS_u \Big ), $

$ N_{4, n} = \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n} \Big[{\left\Vert {S_u} \right\Vert}^4- (2+4/d) {b}_n{\left\Vert {S_u} \right\Vert}^2+ (1+2/d)( {b}_n^2+ {b}_{2, n}) - {b}_n\Big], $

$ N_{2, n}^z= \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n} \Big( \langle {{z}}, {{S_u}}\rangle^2- {b}_{ n} {\left\Vert {z} \right\Vert}^2/d\Big), $

这些都是适应于$ \{ {\cal D}_n\} $的鞅, 其极限在主要定理中用到. 这些鞅的具体性质和收敛速度由下面结论给出.

命题1.1  序列$ \{ N_{q, n} \} (q=1, 2, 3, 4) $及$ \{N_{2, n}^z\} $是适应于$ \{ {\cal D}_{n}\}_{n\geq0} $的鞅. 进而, 假设对某个$ \lambda>4 $, 条件(1.3)成立. 那么存在随机变量$ {\cal V}_q (q=1, 2, 3, 4) $及$ {\cal V}_2^z $, 使得

(1.6) $ \begin{equation} N_{q, n}- {\cal V}_q=o(n^{-(\lambda-q)})\ {\rm a.s.}, \quad N_{2, n}^z - {\cal V}_2^z = o(n^{-(\lambda-2)}) \ {\rm a.s.}. \end{equation} $

该命题将在第2节证明.

利用上面定义的记号, 该文的主要结论可叙述如下.

定理1.1  假设对$ \lambda>3(d+6) $, 式(0.3)成立, 及对$ \iota>0 $, 有$ {\mathbb{E}}m_0^{-\iota}<\infty $成立. 那么对任意$ z=(z_1, \cdots, z_d)\in{\mathbb{Z}}^d $, 当$ n $趋于无穷时, 有

(1.7) $ \begin{eqnarray} \frac{1}{\Pi_n}Z_n(z)&=& \Big (\frac{ d}{2\pi {b}_{ n} } \Big)^{d/2}\Big \{ W + \frac{d}{ {b}_n} \Big [ F_{1}(z)-\frac{1}{8} \frac{ {b}_{2, n} }{ {b}_n} (d+2) W\Big ] {}\\ && + \frac{d^2}{ {b}_n^2} \Big[F_{2}(z) + \frac{ {b}_{2, n}}{ {b}_n} F_3(z) + Q_d( {b}_n, {b}_{2, n}, {b}_{3, n}) W\Big]+ \frac{1}{n^2}o(1)\Big\} \ {\rm a.s.} , \end{eqnarray} $

其中

(1.8) $ \begin{equation} F_{1}(z)= \Big(\frac{1}{8} d-\frac{1}{2}{\left\Vert {z} \right\Vert}^2 \Big)W+\langle {{z}}, {{ {\cal V}_1}}\rangle-\frac{1}{2} {\cal V}_2, \end{equation} $

(1.9) $ \begin{eqnarray} F_{2 }(z)&=& \frac{1}{8} {\cal V}_4 - \frac{1}{2}\langle {{z}}, {{ {\cal V}_3}}\rangle+\frac{1}{2} {\cal V}^z_2 + \Big( \frac{1}{4} {\left\Vert {z} \right\Vert}^2-\frac{d}{16}-\frac{1}{4}\Big) {\cal V}_{2} + \Big( \frac{d}{8}+\frac{1}{2} -\frac{1}{2}{\left\Vert {z} \right\Vert}^2\Big)\langle {{z}}, {{ {\cal V}_1}}\rangle \\ && +\Big[ \frac{1}{8}{\left\Vert {z} \right\Vert}^4 - \frac{1}{16} (d+4) {\left\Vert {z} \right\Vert}^2 \Big]W, \end{eqnarray} $

(1.10) $ \begin{equation} F_3(z)= (d+4)\Big(1+\frac{2}{d}\Big)\Big(\frac{1}{16} {\cal V}_2-\frac{1}{8}\langle {{z}}, {{ {\cal V}_1}}\rangle + \frac{1}{16}{\left\Vert {z} \right\Vert}^2 W \Big), \end{equation} $

(1.11) $ \begin{equation} Q_d(u, v, w) = (d+4)\Big(1+\frac{2}{d}\Big) \Big[ \frac{1}{128} ( d+6 ) \Big(\frac{v}{u}\Big)^2 - \frac{d}{64} \frac{v}{u}- \frac{1}{24} \frac{w}{u}\Big] + \frac{d}{16}+\frac{1}{128}d^2. \end{equation} $

为便于比较, 在下面的推论中, 给出了常环境中的简单分枝随机游动的相应结果. 其中的结论(Ⅰ) 是定理1.1的直接推论, 结论(Ⅱ) 已在文献[19]中证得.

推论1.1  在一个简单随机游动中, 后代分布为$ \{p_k \}_ {k\in {\mathbb{N}}} $, 运动机制服从分布律$ {\cal P} $

$ \begin{eqnarray} {\cal P}(0)=r, \cdots {\cal P}(- {\bf e}_s )={\cal P}( {\bf e}_s)=(1-r)/(2d), {\quad} 1\leq s\leq d. \end{eqnarray} $

假定对$ \lambda>3(d+6) $, 成立

$ \begin{eqnarray} m:= \sum\limits_{k=0}^\infty k p_k >1, \quad \sum\limits_{k=0}^\infty k (\ln k)^{1+\lambda} p_k <\infty . \end{eqnarray} $

那么对任意$ z=(z_1, \cdots, z_d)\in{\mathbb{Z}}^d $, 当$ n $趋于无穷时, 有

(Ⅰ) 当$ 1>r>0 $, 有

$ \begin{eqnarray} \frac{1}{m^n} Z_n(z) = \Big(\frac{d }{ 2\pi n(1-r) }\Big)^{d/2} \Big[ W + \frac{1}{n} {{\cal H}_{r, 1}} (z)+\frac{1}{n^2} {{\cal H}_{r, 2}} (z) \Big] + \frac{1}{n^{2+d/2}}o(1)\quad {\rm a.s.}, \end{eqnarray} $

(Ⅱ) 当$ r=0 $, 假设$ n\equiv z_1+z_2+\cdots+z_d ( {\rm mod }\ 2) $, 有

$ \begin{eqnarray} \frac{1}{m^n} Z_n(z) = 2 \Big(\frac{d }{ 2\pi n }\Big)^{d/2} \Big[ W + \frac{1}{n} {{\cal H}_{0, 1}}(z)+\frac{1}{n^2} {{\cal H}_{0, 2}}(z) \Big] + \frac{1}{n^{2+d/2}}o(1)\quad {\rm a.s.}, \end{eqnarray} $

其中$ {\cal V}_1, {\cal V}_2 , { {\cal V}_2^z}, { {\cal V}_3}, { {\cal V}_4} $是一些随机变量, 以及对$ r\in [0, 1) $, 有

$ {{\cal H}_{r, 1}} (z)=\frac{d}{1-r} \Big[ \Big(\frac{1}{8}r(d+2) - \frac{1}{4} -\frac{1}{2}{\left\Vert {z} \right\Vert}^2\Big)W+ \langle {{z}}, {{ {\cal V}_1}}\rangle-\frac{1}{2} { {\cal V}_2} \Big], $

$ \begin{eqnarray} {{\cal H}_{r, 2}} (z)&= &\frac{d^2 }{ (1-r)^2}\Big[ \Big( \frac{1}{8}{\left\Vert {z} \right\Vert}^4+\mu_{r, d} {\left\Vert {z} \right\Vert}^2+\chi_{r, d}\Big)W+\Big(2\mu_{r, d}-\frac{1}{2}{\left\Vert {z} \right\Vert}^2\Big)\langle {{ {\cal V}_1}}, {{z}}\rangle \\ &&+ \Big(\frac{{\left\Vert {z} \right\Vert}^2}{4}-\mu_{r, d}\Big) { {\cal V}_2} +\frac{1}{2} { {\cal V}_2^z}-\frac{1}{2}\langle {{z}}, {{ { {\cal V}_3}}}\rangle +\frac{1}{8} { {\cal V}_4}\Big], \end{eqnarray} $

$ \mu_{r, d}= - \frac{1}{8}\Big(1+\frac{4}{d}\Big) +r\Big(\frac{d}{16}+\frac{3}{8}+\frac{1}{2d}\Big), $

$ \chi_{r, d}= \frac{d}{48}-\frac{1}{32}+\frac{1}{24d}+r\frac{1}{64}(d+2)(d+4)\Big( \frac{r}{2}+ \frac{r-2}{3d}\Big ). $

注1.1  此推论中出现的随机变量$ {\cal V}_1, {\cal V}_2 , { {\cal V}_2^z}, { {\cal V}_3}, { {\cal V}_4} $是下面序列的极限

$ \begin{eqnarray} {\cal V}_j ={ \lim\limits_{n\rightarrow \infty} N_{ j, n}\ {\rm a.s.}} , \quad j=1, 2, 3, 4, \quad { {\cal V}_2^z}= \lim\limits_{n\rightarrow \infty} {N_{ 2, n}^z}, \ {{\rm a.s.}}, \end{eqnarray} $

其中

$ N_{1, n}= \frac{1}{m^n}\sum\limits_{u\in {\mathbb{T}}_n} S_u, N_{2, n}= \frac{1}{m^n}\sum\limits_{u\in {\mathbb{T}}_n} [ {\left\Vert {S_u} \right\Vert}^2 -n(1-r) ], $

$ {N_{2, n}^z}= \frac{1}{m^n}\sum\limits_{u\in {\mathbb{T}}_n}\left(\langle {{S_u}}, {{z}}\rangle^2 - \frac{n(1-r)}{d}{\left\Vert {z} \right\Vert}^2\right), $

$ {N_{3, n}}= \frac{1}{m^n}\sum\limits_{u\in {\mathbb{T}}_n} \left[ {\left\Vert {S_u} \right\Vert}^2 S_u -\Big(1+\frac{2}{d}\Big)n(1-r) S_u\right], $

$ {N_{4, n}} = \frac{1}{m^n}\sum\limits_{u\in {\mathbb{T}}_n} \left[ {\left\Vert {S_u} \right\Vert}^4- \Big(2+\frac{4}{d}\Big) n (1-r) {\left\Vert {S_u} \right\Vert}^2+ \Big(1+\frac{2}{d}\Big) \Big( (1-r)^2 n^2-r n \Big)+ \frac{2}{d} n \right]. $

最后对定理的证明做一些说明. 论证过程采用了文献[18, 19]中的基本思想, 即通过分枝过程的可加性质将$ Z_n(z) $分解为条件独立项的和. 但是证明的细节需要更细致的估计和更多的技巧性. 特别地, 此项工作特别之处在于前面定义的那些新的鞅和展开式的确切表达. 从技术层面上, 证明的计算比没有随机环境的情形复杂很多.

文章的结构安排如下: 在下节中, 证明了一个一般性地结论定理2.1, 作为其推论得到命题1.1. 在第3节中, 推导了$ {\mathbb{Z}}^d $上依时变化的随机环境中的随机游动的局部极限定理的二阶渐近展开. 在这些辅助结果的基础上, 在最后一节给出了定理1.1的证明.

这一节证明一个一般性的结果定理2.1, 命题1.1作为其推论来推出.

定理2.1  给定环境$ \xi $. 令$ {Q} \geq 0 $, $ C>0 $为给定常数. 假设$ f_\xi $是一个定义在$ {{\mathbb{R}}}^d\times {\mathbb{N}} $上的向量值(或实值) 函数, 满足$ {\left\Vert {f_\xi(x, n)} \right\Vert}\leq C({\left\Vert {x} \right\Vert}^{ {Q}} +n^{{Q}/2}) $及

$ \begin{eqnarray} {\mathbb{E}}_\xi f_\xi(x+X_{n}, n+1)= f_\xi(x, n), \forall (x, n)\in {{\mathbb{R}}}^d\times {\mathbb{N}}, \end{eqnarray} $

这里每个$ X_n (n\in {\mathbb{N}} ) $都是一个服从分布律$ G_{\xi_{n}} $的随机向量, 即

$ \begin{eqnarray} {\mathbb{P}}_\xi (X_{n}={\bf 0})=r_{n}, {\mathbb{P}}_\xi (X_{n}=\pm{\bf e}_s)=\frac{1}{2d}(1- r_{n}), 1\leq s\leq d. \end{eqnarray} $

那么由下式

(2.1) $ \begin{equation} {\cal M}_{n}= \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n}f_\xi(S_u, n) \end{equation} $

定义的序列$ \{ {\cal M}_{n} \} $是关于$ \{{\cal D}_n\}_{n\geq 0} $的鞅. 进而如果存在$ \lambda\geq {Q} $使得(1.3)式成立, 那么$ \{ {\cal M}_{n} \} $几乎必然收敛至一个随机变量$ {\cal V} $, 且

(2.2) $ \begin{equation} {\cal M}_{n}- {\cal V}= o(n^{-(\lambda- {Q}) }), \quad \mbox{当 }\ n\rightarrow \infty. \end{equation} $

注2.1  当$ f_\xi \equiv 1 $, $ {Q}=0 $时, $ {\cal M}_{n} $变成了$ W_n $, 于是定理化为下面已知的结论.

定理2.2^{[40, 定理1.2]}  设对某个$ \lambda>0 $, 条件(1.3)成立, 那么有

(2.3) $ \begin{equation} W_n-W=o(n^{-\lambda}). \end{equation} $

在Galton-Watson过程情形下, 这一结论首先是$ \rm Asmussen $在1976年文献[41] 中得到的.

为证明定理2.1, 此处借鉴了文献[41]中的思想. 该方法中的关键是利用恰当的截断来证明级数$ \sum\limits_{n} a_n ( {\cal M}_{n+1} - {\cal M}_{n}) $的收敛性, 从而推得$ {\cal M}_{n} $的收敛速度. 证明依赖于下面引理.

引理2.1^{[41, 引理2]}  设$ \{\alpha_n, \beta_n, n\geq 1 \} $是一列实数. 如果$ 0<\alpha_n\nearrow \infty, $且级数$ \sum\limits_{n=1}^\infty \alpha_n \beta_n $收敛, 那么

$ \sum\limits_{n=r}^\infty \beta_n= o\Big(\frac{1}{\alpha_r}\Big). $

证  先来证明$ \{ {\cal M}_{n} \} $是适应于$ \{ {\cal D}_n\} $的鞅.

定义条件概率和条件期望

$ {\mathbb{P}}_{\xi, n}(\cdot)= {\mathbb{P}}_\xi(\cdot|{\cal D}_n), \phantom{sss} {\mathbb{E}}_{\xi, n}(\cdot)= {\mathbb{E}}_\xi(\cdot|{\cal D}_n), $

则

$ \begin{eqnarray} {\mathbb{E}}_{\xi, n} {\cal M}_{n+1} &= & {\mathbb{E}}_{\xi, n} \frac{1}{\Pi_{n+1}}\sum\limits_{u\in {\mathbb{T}}_n}\sum\limits_{i=1}^{N_u} f_\xi(S_{ui}, n+1) \\ &=&\frac{1}{\Pi_{n+1}} \sum\limits_{u\in {\mathbb{T}}_n} {\mathbb{E}}_{\xi, n} \bigg[ \sum\limits_{i=1}^{N_u} {\mathbb{E}}_{\xi, n} f_\xi(S_{ui}, n+1) \bigg] \\ &= &\frac{1}{\Pi_{n+1}} \sum\limits_{u\in {\mathbb{T}}_n} f_\xi(S_{u}, n) {\mathbb{E}}_{\xi, n} N_u \\ &= & \frac{1}{\Pi_{n+1}} \sum\limits_{u\in {\mathbb{T}}_n} f_\xi(S_{u}, n) m_n= \frac{1}{\Pi_{n}} \sum\limits_{u\in {\mathbb{T}}_n} f_\xi(S_{u}, n)= {\cal M}_n. \end{eqnarray} $

由此可以推得$ \{ {\cal M}_n\} $是一个适应于$ \{ {\cal D}_n\} $的鞅.

下面来证明鞅的收敛速度.

首先引入下面记号

$ \lambda_Q= \lambda- {Q}, \quad X_u= f_{\xi}(S_u, n), \qquad \mbox{对}\ u\in {\mathbb{T}}_n, $

$ {I}_n= {\cal M}_{n+1}- {\cal M}_{n} = \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n}\Big ( \frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u\Big) , $

$ {I}_n' = \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n}\Big ( \frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u\Big) {\bf 1}_{\{N_u(\log N_u)^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}} \}} . $

如果能证明级数

(2.4) $ \begin{equation} \sum\limits_{n=1}^\infty n^{\lambda_{Q}} {I}_n \quad \mbox{几乎必然收敛}. \end{equation} $

那么令$ V_3= \sum\limits_{n=1}^\infty {I}_n + {\cal M}_1 $, 并应用引理2.1, 即可得所欲证.

为证断言(2.4), 只需证明下列级数几乎必然收敛

(2.5) $ \begin{equation} \sum\limits_{n=1}^\infty n^{\lambda_{Q}} ({I}_n-{I}_n' ), \quad \sum\limits_{n=1}^\infty n^{\lambda_{Q}} ({I}_n'- {\mathbb{E}}_{\xi, n}{I}_n' ), \quad \sum\limits_{n=1}^\infty n^{\lambda_{Q}} {\mathbb{E}}_{\xi, n}{I}_n'. \end{equation} $

而为证明(2.5)式, 需要以下辅助结果.

由零平均独立随机变量和的矩不等式^[42]可得如下结果.

引理2.2  对$ j\in {\mathbb{N}} $, 随机向量$ \widehat{L}_j $服从分布律$ G_{\xi_{j }} $. 令$ \widehat{S}_n=\sum\limits_{j=0}^{n-1}\widehat{L}_j $. 那么对$ r\geq 2 $, 有

(2.6) $ \begin{equation} {\mathbb{E}}_\xi {\left\Vert { \widehat{S}_n} \right\Vert}^r \leq K_r n^{\frac{r}{2}-1} \sum\limits_{j=0}^{n-1} {\mathbb{E}}_\xi {\left\Vert { \widehat{L}_j} \right\Vert} ^r \leq K_r n^{\frac{r}{2}}, \end{equation} $

这里$ K_r $是一个只依赖于$ r $的绝对常数.

由此引理可得, 对于$ |u|=n $, 有

(2.7) $ \begin{equation} {\mathbb{E}}_\xi {\left\Vert {X_u} \right\Vert} \leq {K_Q} n^{{Q}/2}; \quad {\mathbb{E}}_\xi {\left\Vert {X_u} \right\Vert}^2 \leq {K_Q} n^{{Q}}. \end{equation} $

对于(2.5)式中的第一个级数, 注意到

$ \begin{eqnarray} {\mathbb{E}}_\xi {\left\Vert { I_n-I_n' } \right\Vert} &\leq &\frac{1}{\Pi_n} {\mathbb{E}}_\xi \sum\limits_{u\in {\mathbb{T}}_n} {\left\Vert {\frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u } \right\Vert} {\bf 1}_{\{ N_u(\log N_u)^{\nu }> \Pi_{n+1}/n^{\lambda_{Q}} \}} \\ &\leq& \frac{K_Q n^{{Q}/2}}{\Pi_n} {\mathbb{E}}_\xi \sum\limits_{u\in {\mathbb{T}}_n} (N_u/m_n +1) {\bf 1}_{\{N_u(\log N_u)^{\nu }> \Pi_{n+1}/n^{\lambda_{Q}}\}} \\ &=& K_Q n^{{Q}/2} {\mathbb{E}}_\xi (\widehat{N}_n /m_n+1) {\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }> \Pi_{n+1}/n^{\lambda_{Q}}\}}. \end{eqnarray} $

由大数定律得

(2.8) $ \begin{equation} \lim\limits_{n\rightarrow \infty} \frac{1}{n}\log\frac{\Pi_{n+1}}{n^{\lambda_{Q}}}= {\mathbb{E}} \log m_0 \mbox{ a.s. }. \end{equation} $

由此可推出存在$ \sigma_1>1 $, 对$ n $充分大, 有

$ \frac{\Pi_{n+1}}{n^{\lambda_{Q}}} \geq \sigma_1^n. $

那么可得

$ {\mathbb{E}}_\xi (\widehat{N}_n /m_n+1) {\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }> \Pi_{n+1}/n^{\lambda_{Q}}\}}\leq {\mathbb{E}}_\xi (\widehat{N}_n /m_n+1) {\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }>\sigma_1^{n}\}}. $

易见

$ \begin{eqnarray} && {\mathbb{E}}\Big(\sum\limits_{n=0}^\infty n^{\lambda_{Q}+{Q}/2} {\mathbb{E}}_\xi (\widehat{N}_n /m_n+1) {\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }>\sigma_1^{n}\}}\Big) \\ &= & \sum\limits_{n=0}^\infty n^{\lambda_{Q}+{Q}/2} {\mathbb{E}} (\widehat{N}_0 /m_0+1) {\bf 1}_{\{\widehat{N}_0 (\log \widehat{N}_0 )^{\nu }>\sigma_1^{n}\}} \\ &=& {\mathbb{E}}\Big[ (\widehat{N}_0 /m_0+1) \sum\limits_{n=0}^\infty n^{\lambda_{Q}+{Q}/2} {\bf 1}_{\{\widehat{N}_0 (\log \widehat{N}_0 )^{\nu }>\sigma_1^{n}\}} \Big] \\ &= & {\mathbb{E}} (\widehat{N}_0 /m_0+1) \sum\limits_{n=0}^{n(\widehat{N}_0)} n^{\lambda_{Q}+{Q}/2}\leq {\mathbb{E}} (\widehat{N}_0 /m_0+1) {n(\widehat{N}_0)}^{\lambda_{Q}+{Q}/2+1}, \end{eqnarray} $

其中

$ n(\widehat{N}_0)=\max\{ 0, \lceil(\log \sigma_1)^{-1} (\log\widehat{N}_0 +\nu \log\log\widehat{N}_0 ) -1\rceil\}, $

这里$ \lceil a\rceil =\min\{ k\in {\mathbb{Z}}:k\geq a\} $. 那么

$ \begin{eqnarray} && {\mathbb{E}}\Big(\sum\limits_{n=0}^\infty n^{\lambda_{Q}} n^{{Q}/2} {\mathbb{E}}_\xi (\widehat{N}_n /m_n+1) {\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }>\sigma_1^{n}\}}\Big) \\ &\leq & {\mathbb{E}} (\widehat{N}_0 /m_0+1) {n(\widehat{N}_0)}^{\lambda_{Q}+{Q}/2+1} \\ &\leq & K {\mathbb{E}} (\widehat{N}_0 /m_0+1) \Big( \log^+ \widehat{N}_0\Big) ^{\lambda_{Q} +{Q}/2 + 1 } \\ &\leq & K {\mathbb{E}} (\widehat{N}_0 /m_0+1) \Big( \log^+ \widehat{N}_0\Big) ^{1+\lambda } <\infty, \end{eqnarray} $

其中$ K $是一个充分大的常数. 从而可推出

$ \sum\limits_{n=0}^\infty n^{\lambda_{Q}+{Q}/2} {\mathbb{E}}_\xi (\widehat{N}_n /m_n+1) {\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }>\sigma_1^{n}\}} <\infty\qquad \mbox{a.s.}, $

于是

$ \sum\limits_{n=0}^\infty n^{\lambda_{Q}+{Q}/2} {\mathbb{E}}_\xi (\widehat{N}_n /m_n+1) {\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }> \Pi_{n+1}/n^{\lambda_{Q}}\}} <\infty\qquad \mbox{a.s.}, $

那么

$ {\mathbb{E}}_{\xi} {\left\Vert { \sum\limits_{n=1}^\infty n^{\lambda_{Q}} ({I}_{ n}-{I}_{n}')} \right\Vert} \leq \sum\limits_{n=1}^\infty n^{\lambda_{Q}} {\mathbb{E}}_{\xi} {\left\Vert {{I}_{n}-{I}_{n}'} \right\Vert} <\infty, $

$ {\mathbb{E}}_{\xi} {\left\Vert { \sum\limits_{n=1}^\infty n^{\lambda_{Q}} {\mathbb{E}}_{\xi, n} {I}_{n}' } \right\Vert} = {\mathbb{E}}_{\xi} {\left\Vert { \sum\limits_{n=1}^\infty n^{\lambda_{Q}} {\mathbb{E}}_{\xi, n} ({I}_{n}- {I}_{n}') } \right\Vert} \leq \sum\limits_{n=1}^\infty n^{\lambda_{Q}} {\mathbb{E}}_{\xi} {\left\Vert {{I}_{n}-{I}_{n}'} \right\Vert} <\infty. $

所以可得到级数$ \sum\limits_{n=1}^\infty n^{\lambda_{Q}} ({I}_{ n}-{I}_{ n}') $与$ \sum\limits_{n=1}^\infty n^{\lambda_{Q}} {\mathbb{E}}_{\xi, n} {I}_{ n}' $几乎必然收敛.

接下来还需证明$ \sum\limits_{n=1}^\infty n^{\lambda_{Q}} ({I}_n'- {\mathbb{E}}_{\xi, n}{I}_n' ) $几乎必然收敛. 引用事实$ \sum\limits_{k=1}^{n} k^{\lambda_{Q}} ({I}_k'- {\mathbb{E}}_{\xi, k}{I}_k' ) $是适应于$ \{ {\cal D}_{n+1}\} $的鞅及$ L^2 $有界鞅的几乎必然收敛性^{[43, p251, Ex 4.9]}, 只需要证明

$ \sum\limits_{n=1}^\infty n^{2\lambda_{Q}} {\mathbb{E}}_\xi({I}_n'- {\mathbb{E}}_{\xi, n}{I}_n' )^2 \quad \mbox { 几乎必然收敛}. $

为此, 注意到

$ \begin{eqnarray} && {\mathbb{E}}_{\xi}\bigg[\Big \| \frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u\Big\| ^2 {\bf 1}_{\{N_u (\log N_u )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}}\Big| {\cal F}_n\bigg] \\ &=& {\mathbb{E}}_\xi \bigg\{ {\mathbb{E}}_\xi \Big[ \Big \| \frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u\Big\| ^2 \Big|N_u\Big]{\bf 1}_{\{N_u (\log N_u)^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}} \bigg\} \\ &\leq& {\mathbb{E}}_\xi \bigg\{ 2 \Big( N_u \sum\limits_{i=1}^{N_u} \frac{ {\mathbb{E}}_\xi {\left\Vert { X_{ui}} \right\Vert}^2}{{m_n^2}} + {\mathbb{E}}_\xi {\left\Vert {X_u} \right\Vert}^2\Big) {\bf 1}_{\{N_u (\log N_u )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}} \bigg\} \\ &\leq&_{(1.7)} K_Q n^{{Q}}\Big( {\mathbb{E}}_\xi \frac{N_u^2}{m_n^2}{\bf 1}_{\{ N_u(\log N_u )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}} +1 \Big ) \\ &= &K_Q n^{{Q}}\Big( {\mathbb{E}}_\xi \frac{\widehat{N}_n^2}{m_n^2}{\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}} +1 \Big ). \end{eqnarray} $

观察到

$ \begin{eqnarray} && n^{2\lambda_{Q}} {\mathbb{E}}_\xi {\left\Vert {{I}_n'- {\mathbb{E}}_{\xi, n}{I}_n'} \right\Vert} ^2 \\ &= &\frac{n^{2\lambda_{Q}}}{\Pi_n^2} {\mathbb{E}}_\xi \sum\limits_{u\in {\mathbb{T}}_n} {\mathbb{E}}_{\xi, n}\Big\|\Big ( \frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u\Big) {\bf 1}_{\{ N_u(\log N_u )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}} \\ && - {\mathbb{E}}_{\xi, n} \Big ( \frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u\Big) {\bf 1}_{\{ N_u(\log N_u )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}}\Big\|^2 \\ &\leq& \frac{n^{2\lambda_{Q}}}{\Pi_n^2} {\mathbb{E}}_\xi \sum\limits_{u\in {\mathbb{T}}_n} {\mathbb{E}}_{\xi, n}\Big\|\Big ( \frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u\Big) {\bf 1}_{\{ N_u(\log N_u )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}} \Big\|^2 \\ &= & \frac{n^{2\lambda_{Q}}}{\Pi_n^2} {\mathbb{E}}_\xi \sum\limits_{u\in {\mathbb{T}}_n} {\mathbb{E}}_{\xi}\Big[\Big\|\frac{1}{m_n} \sum\limits_{i=1}^{N_u} X_{ui} -X_u\Big\| ^2{\bf 1}_{\{ N_u(\log N_u )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}}\Big| {\cal F}_n\Big] \\ &\leq& \frac{K_Q n^{ {Q}+2\lambda_{Q}}}{\Pi_{n} } {\mathbb{E}}_\xi \frac{\widehat{N}_n^2}{m_n^2}{\bf 1}_{\{\widehat{N}_n (\log \widehat{N}_n )^{\nu }\leq \Pi_{n+1}/n^{\lambda_{Q}}\}}+\frac{K_Qn^{ {Q}+2\lambda_{Q}}}{\Pi_{n} } \\ &=&\frac{K_Q n^{ {Q}+2\lambda_{Q}}}{\Pi_{n} } {\mathbb{E}}_\xi \frac{\widehat{N}_n^2}{m_n^2}\Big( {\bf 1}_{\{ \widehat{N}_n (\log \widehat{N}_n )^{\nu } \leq e^{1+\lambda+\nu} \}}+ {\bf 1}_{\{ e^{1+\lambda+\nu}< \widehat{N}_n (\log \widehat{N}_n )^{\nu } \leq n^{-\lambda_{Q}}\Pi_{n+1} \}} \Big) +\frac{K_Q n^{{Q}+2\lambda_{Q}}}{\Pi_{n} } \\ & \leq& \frac{K_Q n^{{Q}+2\lambda_{Q}}}{\Pi_{n} } {\mathbb{E}}_\xi \frac{\widehat{N}_n^2}{m_n^2} {\bf 1}_{\{ e^{1+\lambda+\nu}< \widehat{N}_n (\log \widehat{N}_n )^{\nu } \leq n^{-\lambda_{Q}}\Pi_{n+1} \}}+\frac{K_\xi n^{{Q}+2\lambda_{Q}}}{\Pi_{n} } \\ &\leq& \frac{K_\xi n^{{Q}+2\lambda_{Q}}}{\Pi_{n} }\frac{\Pi_{n+1}n^{-\lambda_{Q}} } {\big(\log \big( {\Pi_{n+1} }{n^{-\lambda_{Q}}} \big)\big)^{ 1+\lambda+\nu}} {\mathbb{E}}_\xi \frac{\widehat{N}_n^2}{m_n^2} \frac{\big( \log\big( 1 \vee \widehat{N}_n (\log\widehat{N}_n )^{\nu}\big)\big)^{1+\lambda+\nu} }{\widehat{N}_n (\log\widehat{N}_n )^{\nu} } +\frac{K_\xi n^{{Q}+2\lambda_{Q}}}{\Pi_{n} } \\ & \leq & K_\xi n^{{Q}+\lambda_{Q}-\lambda-1-\nu} {\mathbb{E}}_\xi \frac{\widehat{N}_n}{m_n}\big(\log^+\widehat{N}_n\big)^{1+\lambda} +\frac{K_\xi n^{{Q}+2\lambda_{Q}}}{\Pi_{n} } \\ &=& K_\xi n^{-1-\nu} {\mathbb{E}}_\xi \frac{\widehat{N}_n}{m_n}\big(\log^+\widehat{N}_n\big)^{1+\lambda} +K_\xi\frac{ n^{{Q}+2\lambda_{Q}}}{\Pi_{n} }, \end{eqnarray} $

这里用到了函数$ x(\log x)^{-1-\lambda-\nu} $ 在 $ x > e^{1+\lambda+\nu} $时是递增的. 易见

$ \sum\limits_{n=1}^\infty \frac{ n^{{Q}+2\lambda_{Q}}}{\Pi_{n} }<\infty. $

注意到

$ {\mathbb{E}} \sum\limits_{n=0}^\infty n^{-1-\nu} {\mathbb{E}}_\xi \frac{\widehat{N}_n}{m_n}\Big(\log^+\widehat{N}_n\Big)^{1+\lambda} = \sum\limits_{n=0}^\infty n^{-1-\nu} {\mathbb{E}} \frac{\widehat{N}_0}{m_0}\Big(\log^+\widehat{N}_0\Big)^{1+\lambda}<\infty. $

于是级数

$ \sum\limits_{n=0}^\infty n^{-1-\nu} {\mathbb{E}}_\xi \frac{\widehat{N}_n}{m_n}\Big(\log^+\widehat{N}_n\Big)^{1+\lambda} $

几乎必然收敛, 由此可推得级数

$ \sum\limits_{n=1}^\infty n^{2\lambda_{{Q}}} {\mathbb{E}}_\xi({I}_n'- {\mathbb{E}}_{\xi, n}{I}_n' )^2 $

几乎必然收敛.

综上已证(2.5)式中三个级数几乎必然收敛, 从而(1.4)式成立.

现在令$ {\cal V}= \sum\limits_{n=1}^\infty {I}_n + { {\cal M}}_{1} $, 即可得$ {\cal M}_{n}- {\cal V}= \sum\limits_{j=n}^\infty {I}_j $, 那么定理2.1即由引理2.1推得.

命题1.1的证明  先来证明$ \{N_{4, n}\}_n $是适应于$ \{{\cal D}_n\}_n $的鞅. 回顾

$ \begin{eqnarray} N_{4, n} = \frac{1}{\Pi_n} \sum\limits_{u\in {\mathbb{T}}_n} \Big[{\left\Vert {S_u} \right\Vert}^4- (2+4/d) {b}_n{\left\Vert {S_u} \right\Vert}^2+ (1+2/d)( {b}_n^2+ {b}_{2, n}) - {b}_n\Big]. \end{eqnarray} $

对于$ u\in {\mathbb{T}}_n $, $ 1\leq i\leq N_u $, 有

$ \begin{eqnarray} {\left\Vert {S_{ui}} \right\Vert}^2&=& \langle {{S_u+L_i(u)}}, {{S_u+L_i(u)}}\rangle ={\left\Vert {S_u} \right\Vert}^2+2\langle {{S_u}}, {{L_i(u)}}\rangle+{\left\Vert {L_i(u)} \right\Vert}^2, \\ {\left\Vert {S_{ui}} \right\Vert}^4&=&{\left\Vert {S_u} \right\Vert}^4+{\left\Vert {L_i(u)} \right\Vert}^4+4\langle {{S_u}}, {{L_i(u)}}\rangle^2 \\ &&+4\langle {{S_u}}, {{L_i(u)}}\rangle{\left\Vert {S_u} \right\Vert}^2+4\langle {{S_u}}, {{L_i(u)}}\rangle{\left\Vert {Li(u)} \right\Vert}^2+2{\left\Vert {S_u} \right\Vert}^2{\left\Vert {L_i(u)} \right\Vert}^2. \end{eqnarray} $

于是

$ {\mathbb{E}}_{\xi, n} {\left\Vert {S_{ui}} \right\Vert}^2={\left\Vert {S_u} \right\Vert}^2 + (1-r_n), $

$ {\mathbb{E}}_{\xi, n} {\left\Vert {S_{ui}} \right\Vert}^4={\left\Vert {S_u} \right\Vert}^4+ (1-r_n)+\Big(\frac{4}{d} +2\Big)(1-r_n) {\left\Vert {S_u} \right\Vert}^2. $

于是对于下面定义的函数

$ \begin{eqnarray} f_\xi(x, n)= {\left\Vert {x} \right\Vert}^4- (2+4/d) {b}_n{\left\Vert {x} \right\Vert}^2+ (1+2/d)( {b}_n^2+ {b}_{2, n}) - {b}_n, \end{eqnarray} $

可以看到

$ \begin{eqnarray} & & {\mathbb{E}}_{\xi, n} f_\xi (S_{ui}, n+1) \\& =& {\mathbb{E}}_{\xi, n}\Big[ {\left\Vert {S_{ui}} \right\Vert}^4- (2+4/d) {b}_{n+1}{\left\Vert {S_{ui}} \right\Vert}^2+ (1+2/d)( {b}_{n+1}^2+ {b}_{2, n+1}) - {b}_{n+1}\Big] \\ & = &{\left\Vert {S_u} \right\Vert}^4+\Big( {4}/{d} +2\Big)(1-r_n) {\left\Vert {S_u} \right\Vert}^2+ (1-r_n) - \Big( 2+{4}/{d}\Big) {b}_{n+1}\Big ({\left\Vert {S_u} \right\Vert}^2 + (1-r_n)\Big) \\ & & + (1+2/d)( {b}_{n+1}^2+ {b}_{2, n+1}) - {b}_{n+1} \\ & = &{\left\Vert {S_u} \right\Vert}^4- (2+4/d) {b}_n{\left\Vert {S_u} \right\Vert}^2+ (1+2/d)( {b}_n^2+ {b}_{2, n}) - {b}_n=f_\xi(S_u, n) \end{eqnarray} $

及

$ |f_\xi(x, n)|\leq C({\left\Vert {x} \right\Vert}^4+ n^2). $

因此所需断言由定理1.2中$ Q=4 $可得.

同理, 可以证明对$ q=1, 2, 3 $, $ \{ N_{q, n}\} $收敛至$ {\cal V}_q $, 收敛速度为$ o(n^{-(\lambda-q) }) $, 及$ N_{2, n}^z $收敛至$ {\cal V}_2^z $, 且收敛速度为$ o(n^{-(\lambda-2) }) $. 为避免重复, 忽略细节.

令$ \xi $为前述环境. 考虑一列随机向量$ \{ X_n\}_{n=0}^\infty $. 当给定$ \xi=\{\xi_n:n=0, 1, \cdots\} $时, 在$ {\mathbb{P}}_\xi $下$ X_n $服从分布律$ G_{\xi_n}(\cdot) $, 即

$ {\mathbb{P}}_\xi (X_n={\bf 0})=r_n, \quad {\mathbb{P}}_\xi(X_n={\bf e}_s)= {\mathbb{P}}_\xi(X_n=-{\bf e}_s)=(1-r_n)/(2d), {\quad} 1\leq s\leq d. $

令

$ S_n=X_0+X_2+\cdots+X_{n-1}. $

那么$ \{S_n\} $是依时变化的随机环境中的随机游动.

设

$ M_n=\sum\limits_{i=0}^{n-1} \min\{ 2r_i, \delta_d(1-r_i) \}, \quad \mbox{} \delta_d=(1-\cos(d^{-1/2}))/d. $

则有下面结论.

定理3.1  对上面定义的随机游动$ \{S_n\} $, 对任意$ n $和$ x\in {\mathbb{Z}}^d $, 成立

(3.1) $ \begin{eqnarray} {\mathbb{P}}_{\xi}(S_n =x ) &=& \Big( \frac{d }{2\pi {b}_{n} }\Big)^{d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big\{ 1+ \frac{d}{ {b}_n} \Big (\frac{1}{8}d-\frac{1}{8} (d+2)\frac{ {b}_{2, n}}{ {b}_n}\Big ) {} \\ &&+ \frac{d^2}{ {b}_n^2}\Big[ Q_d( {b}_n, {b}_{2, n} , {b}_{3, n}) -\Big (\frac{1}{4}-\frac{1}{4d} (d+2)\frac{ {b}_{2, n}}{ {b}_n}\Big ){\left\Vert {x} \right\Vert}^2{}\\ && + {\cal P}\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n}, {b}_n, {b}_{2, n}, {b}_{3, n}\Big) \Big] \Big\}+ \frac{1}{n^{2+d/2}} \alpha_n(\xi, x) , \end{eqnarray} $

其中对每组$ (u, v, w\in {{\mathbb{R}}}) $, $ {\cal P}(t_1, \dots, t_d, u, v, w) $都表示一个关于$ (t_1, \cdots, t_d) $的4次的对称多项式且无常数项; $ {b}_{n} , {b}_{2, n}, {b}_{3, n} $及$ Q_d(u, v, w) $分别在(1.4)和(1.11)式中定义; $ \alpha_n(\xi, x) $是无穷小使得

(3.2) $ \begin{equation} \sup\limits_{x\in{{\mathbb{R}}}^d}|\alpha_n(\xi, x) | \leq n^{2+d/2}e ^{ -M_n}+ C_d \Big(\frac{n}{ {b}_n} \Big)^{4+d/2} \Big[n^{-1}+ n^{ 2+d/2}e^{-\frac{11 {b}_n}{24 \sqrt{n}d} }\Big]. \end{equation} $

进而, 对满足式$ {\left\Vert {x} \right\Vert}<n^{1/6} $的$ x\in {\mathbb{Z}}^d $, 有

(3.3) $ \begin{eqnarray} {\mathbb{P}}_\xi(S_n =x ) &=& \Big( \frac{d }{2\pi {b}_{n} }\Big)^{d/2}\Big\{ 1+ \frac{d}{ {b}_{n}} \Big[ -\frac{1}{2}{\left\Vert {x} \right\Vert}^2 +\frac{1}{8} d -\frac{1}{8} (d+2) \frac{ {b}_{2, n} }{ {b}_{n}} \Big] {} \\ && + \frac{ d^2}{ {b}_{n}^2} \Big[ \frac{1}{8}{\left\Vert {x} \right\Vert}^4- \frac{ (d+4)}{16}{\left\Vert {x} \right\Vert}^2 \Big( 1- \Big(1+\frac{2}{d}\Big)\frac{ {b}_{2, n} }{ {b}_{n}} \Big) {}\\ &&+ Q_d( {b}_n, {b}_{2, n}, {b}_{3, n}) \Big] \Big\}+ \frac{\widetilde{\alpha}_n(\xi, x) }{n^{2+d/2}} , \end{eqnarray} $

其中$ {b}_{n} , {b}_{2, n}, {b}_{3, n} $和$ Q_d(u, v, w) $分别如(1.4)和(1.11)式中所定义, 无穷小项$ \widetilde{\alpha}_n(x) $满足

(3.4) $ \begin{equation} |\widetilde{\alpha}_n(\xi, x) | \leq n^{2+d/2}e ^{ -M_n} + C_d \Big(\frac{n}{ {b}_n} \Big)^{ 5+d/2} \Big[n^{ 2+d/2}e^{-\frac{11 {b}_n}{24 \sqrt{n}d} } + n^{-1/3}+ n^{ -1} ({\left\Vert {x} \right\Vert} \vee 1 )^6 \Big]. \end{equation} $

注3.1  此结论推广完善了文献[44, 第2章]中的结论.

定理3.1的证明  易见下列事实成立: 对$ i=0, 1, 2, \cdots, $有

$ \begin{eqnarray} \psi_{{X}_i}(\lambda)={\mathbb{E}}_\xi e^{{\bf i} \langle {{\lambda }}, {{{X}_i}}\rangle }=r_i+\frac{1-r_i}{d}\sum\limits_{j=1}^{d}\cos\lambda_j, \qquad\lambda=(\lambda_1, \ldots, \lambda_d) \in[-\pi, \pi]^d, \end{eqnarray} $

其中$ {\bf i}=\sqrt{-1} $表示虚数单位. 由于$ {\{X_n\}} $独立, 可得

$ \begin{eqnarray} \psi_{{S}_n}(\lambda)={\mathbb{E}}_\xi e^{{\bf i}\langle\lambda, \sum\limits_{i=0}^{n-1}{X}_i\rangle}= \prod\limits_{i=0}^{n-1}{\mathbb{E}}_\xi e^{{\bf i}\langle\lambda, {X}_i\rangle}=\prod\limits_{i=0}^{n-1}\psi_{{X}_i}(\lambda). \end{eqnarray} $

那么由文献[44, 命题2.2.2], 可见

(3.5) $ \begin{equation} {\mathbb{P}}_\xi({S}_n=x) =\frac{1}{(2\pi)^d}\int_{[-\pi, \pi]^d}e^{-{\bf i}\langle\lambda, x\rangle}\psi_{{S}_n}(\lambda){\rm d}\lambda =\frac{1}{(2\pi)^d}\int_{[-\pi, \pi]^d}e^{-{\bf i}\langle\lambda, x\rangle}\prod\limits_{i=0}^{n-1}\psi_{{X}_i}(\lambda){\rm d}\lambda. \end{equation} $

易证

$ 2r_i-1\leq\psi_{{X}_i}(\lambda)\leq1, \quad \mbox{for}\quad r_i\in(0, 1), \lambda\in[-\pi, \pi]^d $

及

$ \psi_{{X}_i}(\lambda)=1 \Longleftrightarrow \lambda={\bf 0}. $

若$ \|\lambda\|<1 $, 由泰勒公式得

$ \psi_{X_i}(\lambda)=r_i+\frac{1-r_i}{d}\sum\limits_{j=1}^d\left(1-\frac12\lambda_j^2+\frac{\cos\eta_j}{24}\lambda_j^4\right), $

从而

(3.6) $ \begin{equation} |\psi_{X_i}(\lambda) |\leq 1-\frac{11(1-r_i)}{24d}\|\lambda\|^2\leq e^{-\frac{11}{24d}(1-r_i) \|\lambda\|^2}. \end{equation} $

而当$ \lambda\in[-\pi, \pi]^d $且$ \|\lambda\|\geq1 $时

$ -d\leq\sum\limits_{j=1}^d{\cos{\lambda_j}}\leq d-1+\cos (d^{-1/2})<d. $

则有

(3.7) $ \begin{eqnarray} |\psi_{X_i}(\lambda)| &\leq & \max \{|2r_i-1|, r_i+(1-\delta_d)(1-r_i)\} \\ & \leq & \exp\{- ( 1-\max\{|2r_i-1|, r_i+(1-\delta_d)(1-r_i)\} )\} \\ & =&e^{- \min\big\{2 r_i, \delta_ d(1-r_i)\big\} }. \end{eqnarray} $

基于上述讨论, 可记

(3.8) $ \begin{equation} {\mathbb{P}}({S}_n=x)={I}_n(x)+{J}_n(x), \end{equation} $

其中

$ {J}_n(x) \triangleq \frac{1}{(2\pi)^d}\int_{\lambda \in [-\pi, \pi]^d, \|\lambda\|\geq1}e^{-{\bf i}\langle\lambda, x\rangle}\prod\limits_{i=0}^{n-1}\psi_{{X}_i}(\lambda){\rm d}\lambda, $

$ {I}_n(x)\triangleq \frac{1}{(2\pi)^d}\int_{\|\lambda\|<1}e^{-{\bf i}\langle\lambda, x\rangle}\prod\limits_{i=0}^{n-1}\psi_{{X}_i}(\lambda){\rm d}\lambda. $

利用(3.7)式和$ M_n $的定义, 可见

(3.9) $ \begin{equation} |{J}_n(x)|\leq \frac{1}{(2\pi)^d}\int_{\lambda \in [-\pi, \pi]^d, \|\lambda\|\geq1} \prod\limits_{i=0}^{n-1} |\psi_{X_i}(\lambda)| {\rm d}\lambda\leq e ^{ -M_n}. \end{equation} $

下面估计$ I_n(x) $. 令$ \alpha=\sqrt{n}\lambda $. 则

$ \begin{eqnarray} {I}_n(x)= \frac{1}{(2\pi\sqrt{n})^d}\int_{\|\alpha\|<\sqrt{n} }e^{-{\bf i} \frac{1}{\sqrt{n}}\langle x, \alpha \rangle}\prod\limits_{i=0}^{n-1}\psi_{{X}_i}\Big(\frac{\alpha}{\sqrt{n}}\Big){\rm d}\alpha. \end{eqnarray} $

为估计此积分, 应用分解式

(3.10) $ \begin{equation} (2\pi\sqrt{n})^d{I}_n={I}_{1, n}(x)+{I}_{2, n}(x), \end{equation} $

其中

$ {I}_{1, n}= \int_{n^{1/4}\leq\|\alpha\|<\sqrt{n} }e^{-{\bf i} \frac{1}{\sqrt{n}}\langle x, \alpha \rangle}\prod\limits_{i=0}^{n-1}\psi_{{X}_i}\Big(\frac{\alpha}{\sqrt{n}}\Big){\rm d}\alpha, $

$ {I}_{2, n}=\int_{\|\alpha\|<n^{1/4}}e^{-{\bf i} \frac{1}{\sqrt{n}}\langle x, \alpha \rangle}\prod\limits_{i=0}^{n-1}\psi_{{X}_i}\Big(\frac{\alpha}{\sqrt{n}}\Big){\rm d}\alpha. $

由不等式(3.6), 有

(3.11) $ \begin{equation} |{I}_{1, n}(x)|\leq 2^dn^{d/2} \exp\Big\{-\frac{11 {b}_n}{24\sqrt{n}d} \Big\}. \end{equation} $

接下来估计$ {I}_{2, n}(x) $.

由泰勒展开, 可得

$ \begin{eqnarray} & & \prod\limits_{i=0}^{n-1}\psi_{{X}_i}\Big(\frac{\alpha}{\sqrt{n}}\Big) \\ & =&\prod\limits_{i=0}^{n-1}\Big(r_i+\frac{1-r_i}{d}\sum\limits_{j=1}^{d}\cos\frac{\alpha_j}{\sqrt{n}}\Big) \\ & =&\exp\Big\{\sum\limits_{i=0}^{n-1}\log\Big(1-(1-r_i)\frac{\|\alpha\|^2}{2nd}+\frac{(1-r_i)}{24n^2d}\sum\limits_{j=1}^{d}\alpha_{j}^4 - \frac{1-r_i}{6! n^3d}\sum\limits_{j=1}^d \alpha_j^6 + \frac{ (1-r_i)}{n^4}M_{1, n}(\alpha)\Big)\Big\} \\ &=&\exp\Big\{-\frac{ {b}_n}{2nd}\|\alpha\|^2+\frac{ {b}_n}{24n^2d}\sum\limits_{j=1}^{d}\alpha_{j}^4-\frac{ {b}_{2, n}}{8n^2d^2}\|\alpha\|^4- \frac{ {b}_n}{6!n^3d} \sum\limits_{j=1}^{d}\alpha_{j}^6-\frac{ {b}_{3, n}}{24n^3d^3} {\left\Vert {\alpha} \right\Vert}^6 \\ & & + \frac{ {b}_{2, n}}{48n^3d^2} \Big(\sum\limits_{j=1}^{d}\alpha_{j}^4\Big) {\left\Vert {\alpha} \right\Vert}^2 +\frac{1}{n^3}M_{2, n}(r, \alpha) \Big\} \\ & =&e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2}\Big\{1+ \frac{1}{ {b}_n} \Big( \frac{ {b}^2_n}{24n^2d}\sum\limits_{j=1}^d\alpha_{j}^4- \frac{ {b}_{2, n} {b}_n}{8n^2d^2}\|\alpha\|^4\Big) +\frac{1}{ {b}_n^2} \Big[ \frac{ {b}_{2, n} {b}_n^2}{48n^3d^2} (\sum\limits_{j=1}^{d}\alpha_{j}^4) {\left\Vert {\alpha} \right\Vert}^2 \\ &&- \frac{ {b}^3_n}{6!n^3d} \sum\limits_{j=1}^{d}\alpha_{j}^6-\frac{ {b}_n^2 {b}_{3, n}}{24n^3d^3} {\left\Vert {\alpha} \right\Vert}^6 + \frac{1}{2} \Big( \frac{ {b}_n^2}{24n^2d}\sum\limits_{j=1}^d\alpha_{j}^4- \frac{ {b}_n {b}_{2, n}}{8n^2d^2}\|\alpha\|^4\Big)^2\Big] + \frac{1}{n^3}M_{3, n}(\xi, \alpha)\Big\}, \end{eqnarray} $

其中余项$ M_{s, n} (\alpha)(s=1, 2, 3) $被$ {\left\Vert { \alpha} \right\Vert}^8 $的有界倍控制.

利用微积分和复分析技巧, 可得

$ \begin{eqnarray} {(1)} &&\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2}{\rm d}\alpha = (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} + \epsilon_{1, n}(x); \\ {(2)} &&\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} \sum\limits_{j=1}^d \alpha_{j}^4 {\rm d}\alpha \\ & =& (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{2+d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big( 3d-6d \frac{ {\left\Vert {x} \right\Vert}^2}{ {b}_n}+ \frac{4d^2}{ {b}_n^2}\sum\limits_{j=1}^dx_j^4 \Big) + \epsilon_{2, n}(x); \\ {(3)}& &\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} {\left\Vert {\alpha} \right\Vert}^4 {\rm d}\alpha \\ & =& (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{2+d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big[(d^2+2d)\Big(1-2\frac{{\left\Vert {x} \right\Vert}^2}{ {b}_n} \Big) +\frac{d^2}{ {b}_n^2}{\left\Vert {x} \right\Vert}^4 \Big]+ \epsilon_{3, n}(x); \\ {(4)} & &\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} {\left\Vert {\alpha} \right\Vert}^2 \sum\limits_{j=1}^d \alpha_{j}^4 {\rm d}\alpha \\ & =& (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{3+d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big[3d(d+4)+ {\cal P} _1\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n} \Big) \Big ]+ \epsilon_{4, n}(x); \\ {(5)}& &\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} \sum\limits_{j=1}^d \alpha_{j}^6 {\rm d}\alpha \\ & =& (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{3+d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big[ 15d + {\cal P} _2\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n} \Big) \Big] + \epsilon_{5, n}(x); \\ {(6)}& &\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} {\left\Vert {\alpha} \right\Vert}^6 {\rm d}\alpha \\ & =& (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{3+d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big[d(d+2)(d+4) + {\cal P} _3\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n} \Big) \Big] + \epsilon_{6, n}(x); \\ {(7)} &&\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} \Big( \sum\limits_{j=1}^d \alpha_{j}^4 \Big)^2 {\rm d}\alpha \\ & =& (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{4+d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big[3d(3d+32 )+ {\cal P} _4\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n} \Big) \Big] + \epsilon_{7, n}(x);\\ {(8)}& &\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} {\left\Vert {\alpha} \right\Vert}^4 \sum\limits_{j=1}^d \alpha_{j}^4 {\rm d}\alpha \\ & =& (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{4+d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2}\Big[ 3 d(d+4)(d+6) + {\cal P} _5\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n} \Big) \Big] + \epsilon_{8, n}(x); \\ {(9)}& &\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} {\left\Vert {\alpha} \right\Vert}^8 {\rm d}\alpha \\ & =& (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{4+d/2} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big[d(d+2 )(d+4)(d+6)+{\cal P} _6\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n} \Big) \Big] + \epsilon_{9, n}(x); \\ {(10)} && \bigg |\int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} M_{3, n}(\xi, \alpha) {\rm d}\alpha \bigg| \\ & \leq& \int_{{{\mathbb{R}}}^d} e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} {\left\Vert {\alpha} \right\Vert}^8 {\rm d}\alpha = (2\pi )^{d/2} \Big(\frac{nd}{ {b}_n} \Big)^{4+d/2} d(d+2 )(d+4)(d+6)=C_d \Big(\frac{n}{ {b}_n} \Big)^{4+d/2}, \end{eqnarray} $

其中$ {\cal P} _v(t_1, \cdots, t_d) (v=1, 2, \cdots, 6) $表示无常数项的(3次或4次)对称多项式, 无穷小项$ \epsilon_{s, n}(x) $$ (1\leq s\leq 9) $满足

$ |\epsilon_{1, n}(x) |\leq C_d \Big(\frac{n}{ {b}_n} \Big)^{ d/2} e^{-\frac{ {b}_n}{2\sqrt{n} d} } n^{ (d-2)/4}; $

$ |\epsilon_{s, n}(x) |\leq C_d \Big(\frac{n}{ {b}_n} \Big)^{(4+d)/2}e^{-\frac{ {b}_n}{2\sqrt{n} d} } n^{ (2+d)/4}, \quad s=2, 3; $

$ |\epsilon_{s, n}(x)|\leq C_d \Big(\frac{n}{ {b}_n} \Big)^{(6+d)/2}e^{-\frac{ {b}_n}{2\sqrt{n} d} } n^{ (4+d)/4}, \quad s=4, 5, 6; $

$ |\epsilon_{s, n}(x) |\leq C_d \Big(\frac{n}{ {b}_n} \Big)^{(8+d)/2}e^{-\frac{ {b}_n}{2\sqrt{n} d} } n^{ (6+d)/4}, \quad s=7, 8, 9. $

上述积分的计算比较复杂, 但方法相似. 为方便读者, 以(2)的计算为例给出细节.

(2) 的计算  首先注意到

$ \int_{{\left\Vert {\alpha} \right\Vert}<n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} \sum\limits_{j=1}^d \alpha_{j}^4 {\rm d}\alpha=\int_{{{\mathbb{R}}}^d} e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} \sum\limits_{j=1}^d \alpha_{j}^4 {\rm d}\alpha + \epsilon_{2, n}(x), $

其中

$ \epsilon_{2, n}(x)=- \int_{{\left\Vert {\alpha} \right\Vert}>n^{1/4} } e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} \sum\limits_{j=1}^d \alpha_{j}^4 {\rm d}\alpha $

满足

$ |\epsilon_{2, n}(x) |\leq \int_{{\left\Vert {\alpha} \right\Vert} >n^ {1/4}} e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} {\left\Vert {\alpha} \right\Vert}^4 {\rm d}\alpha \leq C_d \Big(\frac{n}{ {b}_n} \Big)^{(4+d)/2}e^{-\frac{ {b}_n}{2\sqrt{n} d} } n^{ (2+d)/4}. $

为计算此积分, 做变量替换$ \alpha=\sqrt{ \frac{nd}{ {b}_n}}\zeta $, 则

$ \begin{eqnarray} & &\int_{{{\mathbb{R}}}^d} e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} \sum\limits_{j=1}^d \alpha_{j}^4 {\rm d}\alpha \\ & = &\Big( \sqrt{\frac{nd}{ {b}_n}}\Big)^{4+d}\int_{{{\mathbb{R}}}^d} \exp\Big\{ -{\bf i} \langle {{ \sqrt{ \frac{d}{ {b}_n}}x}}, {{\zeta}}\rangle -\frac{1}{2} {\left\Vert {\zeta} \right\Vert}^2\Big\} \sum\limits_{j=1}^d \zeta_j^4 {\rm d}\zeta \\ & = &\Big( \sqrt{\frac{nd}{ {b}_n}}\Big)^{4+d}\sum\limits_{j=1}^d \int_{{{\mathbb{R}}}^d} \exp\Big\{ -{\bf i} \langle {{ \sqrt{ \frac{d}{ {b}_n}}x}}, {{\zeta}}\rangle -\frac{1}{2} {\left\Vert {\zeta} \right\Vert}^2\Big\} \zeta_j^4 {\rm d}\zeta \\ & = & \Big( \sqrt{\frac{nd}{ {b}_n}}\Big)^{4+d} \sum\limits_{j=1}^d \int_{{{\mathbb{R}}}} {e}^{ -{\bf i} \sqrt{ \frac{d}{ {b}_n}} x_j\zeta_j-\frac{1}{2} \zeta_1^2 } \zeta_j^4 {\rm d}\zeta_j \prod\limits_{1\leq t\leq d, t\neq j } \int_{{{\mathbb{R}}}} e^{ -{\bf i} \sqrt{ \frac{d}{ {b}_n}} x_t\zeta_t-\frac{1}{2} \zeta_t^2} {\rm d}\zeta_t. \end{eqnarray} $

由柯西积分定理, 可得

$ \begin{eqnarray} & &\int_{{{\mathbb{R}}}} \exp\Big\{ -{\bf i} \sqrt{ \frac{d}{ {b}_n}} x_j\zeta_j-\frac{1}{2} \zeta_j^2\Big\} \zeta_j^4 {\rm d}\zeta_j \\ & =& e^{-\frac{d}{2{b}_n} x_j^2} \int_{{{\mathbb{R}}}} e^{-\frac{ 1}{2} \eta_j^2} \Big(\eta_j- {\bf i}\sqrt{ \frac{d}{{b}_n}}x_j\Big)^4 {\rm d}\eta_j \\ & =& e^{-\frac{d}{2{b}_n} x_j^2} \Big[ \int_{{\mathbb{R}}} e^{-\frac{ 1}{2} \eta_j^2}\eta_j^4 {\rm d}\eta_j + 6 \Big( - {\bf i}\sqrt{ \frac{d}{{b}_n}}x_j\Big)^2\int_{{\mathbb{R}}} e^{-\frac{ 1}{2} \eta_j^2}\eta_j^2 {\rm d}\eta_j + \Big( - {\bf i}\sqrt{ \frac{d}{{b}_n}}x_j\Big)^4\int_{{\mathbb{R}}} e^{-\frac{ 1}{2} \eta_j^2} {\rm d}\eta_j \Big] \\ & =& (2\pi)^{1/2} e^{-\frac{d}{2{b}_n} x_j^2} \Big(3-6 \frac{d}{{b}_n} x_j^2+ \frac{d^2}{{b}_n^2} x_j^4\Big) \end{eqnarray} $

及

$ \begin{eqnarray} \int_{{{\mathbb{R}}}} \exp\Big\{ -{\bf i} \sqrt{ \frac{d}{ {b}_n}} x_t\zeta_t-\frac{1}{2} \zeta_t^2\Big\} {\rm d}\zeta_t = e^{-\frac{d}{2{b}_n} x_t^2} \int_{{{\mathbb{R}}}} e^{-\frac{ 1}{2} \eta_t^2} {\rm d}\eta_t =(2\pi)^{1/2}e^{-\frac{d}{2{b}_n} x_t^2}. \end{eqnarray} $

综合以上诸式, 可得

$ \begin{eqnarray} & &\int_{{{\mathbb{R}}}^d} e^{-{\bf i} \langle {{ x}}, {{ \frac{\alpha }{\sqrt{n}} }}\rangle } e^{-\frac{ {b}_n}{2nd}\|\alpha\|^2} \sum\limits_{j=1}^d \alpha_{j}^4 {\rm d}\alpha \\ & =&\Big( \sqrt{\frac{nd}{ {b}_n}}\Big)^{4+d} \sum\limits_{j=1}^d \Big[ (2\pi)^{1/2} e^{-\frac{d}{2{b}_n} x_j^2} \Big(3-6 \frac{d}{{b}_n} x_j^2+ \frac{d^2}{{b}_n^2} x_j^4\Big) \prod\limits_{1\leq t\leq d, t\neq j } (2\pi)^{1/2}e^{-\frac{d}{2{b}_n} x_t^2}\Big] \\ & =&(2\pi)^{d/2}\Big( \sqrt{\frac{nd}{ {b}_n}}\Big)^{4+d} e^{-\frac{d}{2{b}_n} {\left\Vert {x} \right\Vert}^2}\Big(3d -6 \frac{d}{{b}_n} {\left\Vert {x} \right\Vert} ^2+ \frac{d^2}{{b}_n^2}\sum\limits_{j=1}^d x_j^4\Big). \end{eqnarray} $

从而得出(2).

利用(1.5)式和关系$ {b}_n \leq n, {b}_{2, n}\leq n, {b}_{3, n}\leq n $, 将(1)–(10)式代入, 可得

(3.12) $ \begin{eqnarray} {I }_{2, n}(x) & =& (2\pi d)^{d/2} \Big( \frac{n}{ {b}_n}\Big)^{d/2}e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big\{ 1+ \frac{d}{ {b}_n} \Big (\frac{1}{8}d-\frac{1}{8} (d+2)\frac{ {b}_{2, n}}{ {b}_n}\Big ) {} \\ &&+\frac{d^2}{ {b}_n^2}\Big[ Q_d( {b}_n, {b}_{2, n} , {b}_{3, n}) -\Big (\frac{1}{4}-\frac{1}{4d} (d+2)\frac{ {b}_{2, n}}{ {b}_n}\Big ){\left\Vert {x} \right\Vert}^2{} \\ && +{\cal P}\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n}, {b}_n, {b}_{2, n}, {b}_{3, n}\Big) \Big]\Big\} + \frac{1}{n^{2}}\epsilon_n(x) , \end{eqnarray} $

其中对任意$ u, v, w\in {{\mathbb{R}}} $, $ {\cal P}(t_1, \dots, t_d, u, v, w) $是关于$ (t_1, \cdots, t_d) $的不含常数项的4次对称多项式

$ \begin{eqnarray} {\cal P}\Big(t_1, \cdots, t_d, u, v, w \Big) &=&\frac{d}{6}u\sum\limits_{j=1}^d t_j^2 - \frac{1}{8} v\Big(\sum\limits_{j=1}^d t_j \Big) ^2 +\Big[\frac{1}{48d}\frac{v}{u} {\cal P}_1-\frac{1}{6!} {\cal P}_2 -\frac{1}{24d^2}\frac{w}{u}{\cal P}_3 \\ &&+\frac{1}{1152} {\cal P}_4- \frac{1}{192d}\frac{v}{u} {\cal P}_5+\frac{1}{128d^2}\frac{v^2}{u^2} {\cal P}_6\Big]\Big(t_1, \cdots, t_d \Big), \end{eqnarray} $

无穷小$ \epsilon_n(x) $满足

(3.13) $ \begin{eqnarray} |\epsilon_n(x)|&\leq& n^2 |\epsilon_{1, n}(x)|+ \frac{n}{24d}|\epsilon_{2, n}(x)| + \frac{n}{8d^2} |\epsilon_{3, n}(x)|+ \frac{1}{48d^2}|\epsilon_{4, n}(x)|\\ & & +\frac{1}{6!d } |\epsilon_{5, n}(x)| + \frac{1}{24d^3 } |\epsilon_{6, n}(x)| + \frac{1}{2\cdot 24^2 d^2} |\epsilon_{7, n}(x)|\\ &&+ \frac{1}{24\cdot 8 d^3} |\epsilon_{8, n}(x)| +\frac{1}{2\cdot 8^2d^4} |\epsilon_{9, n}(x)| +C_d \frac{1}{n} \Big(\frac{n}{ {b}_n} \Big)^{(8+d)/2} \\ &\leq & C_d \Big(\frac{n}{ {b}_n} \Big)^{(8+d)/2} \Big( e^{-\frac{ {b}_n}{2\sqrt{n} d} } n^{ (6+d)/4} + \frac{1}{n} \Big) \\ & \leq &C_d \Big(\frac{n}{ {b}_n} \Big)^{4+d/2} \Big(\frac{1}{n}+ e^{-\frac{11 {b}_n}{24\sqrt{n} d} } n^{ (6+d)/4}\Big). \end{eqnarray} $

然后将(3.8), (3.10), (3.12) 式与估计式(3.9), (3.11), (3.13)结合, 可得到所需式(3.1)及余项$ \alpha_n(\xi, x) $满足(3.2)式.

当$ x $满足$ {\left\Vert {x} \right\Vert}<n^{1/6} $时, 有

(3.14) $ \begin{equation} e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2}= 1- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2 + \frac{d^2}{8 {b}_n^2} {\left\Vert {x} \right\Vert}^4+ \frac{1}{n^2}\zeta_n(x), \end{equation} $

这里$ \zeta_n(x) $是无穷小且满足

(3.15) $ \begin{equation} |\zeta_n(x)|\leq \frac{d^3}{48}\Big(\frac{n}{ {b}_n}\Big)^3 \frac{1}{n}{\left\Vert {x} \right\Vert}^6. \end{equation} $

易证当$ x $满足$ {\left\Vert {x} \right\Vert}<n^{1/6} $时, 有

(3.16) $ \begin{eqnarray} && e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \bigg|{\cal P}\Big(\frac{x_1^2}{ {b}_n}, \cdots, \frac{x_d^2}{ {b}_n}, {b}_n, {b}_{2, n}, {b}_{3, n} \Big) \bigg| {} \\ & \leq & K\Big(\frac{n}{ {b}_n}\frac{{\left\Vert {x} \right\Vert}^4}{ {b}_n} + \frac{{\left\Vert {x} \right\Vert}^2}{ {b}_n} \Big( 1\vee \frac{{\left\Vert {x} \right\Vert}^2}{ {b}_n} \Big)^3e^{- \frac{d}{2 {b}_n} {\left\Vert {x} \right\Vert}^2} \Big)\leq K \Big(\frac{n}{ {b}_n}\Big)^2 \frac{1}{ n^{1/3}}. \end{eqnarray} $

将这些估计与式(3.1)和(3.2) 综合起来, 可以推出所需展式(3.3) 及余项$ \widetilde{\alpha}_n(\xi, x) $满足(3.4)式. 定理3.1得证.

这一节证明定理1.1. 该定理的证明基于一个重要的分解式. 此分解式早在文献[12]中已经引入, 之后在许多文献中广泛使用, 参见文献[2, 5, 17, 20]. 在介绍此分解之前, 先来引入一些记号.

令$ {\mathbb{T}}(u) $表示$ {\mathbb{T}} $的以$ u $为端点的平移树, 组成元素为$ \{N_{uv}\} $满足: 1) $ \varnothing \in {\mathbb{T}}(u) $; 2) $ vi\in {\mathbb{T}}(u)\Rightarrow v\in {\mathbb{T}}(u) $; 3) 如果$ v\in {\mathbb{T}}(u) $, 那么$ vi\in {\mathbb{T}}(u) $当且仅当$ 1\leq i\leq N_{uv} $. 令$ {\mathbb{T}}_n(u)=\{v\in {\mathbb{T}}(u): |v|=n\} $. 用$ | {\mathbb{T}}_n(u)| $表示$ {\mathbb{T}}_n(u) $的基数(即$ u $的第$ n $代子孙的个数).

对$ u\in ( {\mathbb{N}}^*)^k (k\geq 0) $及$ n\geq 1 $, 令$ Z_n(u, x ) $表示$ u $的第$ n $代后代中位于$ x+S_u\in {\mathbb{Z}}^d $处的粒子个数. 这意味着

$ Z_n(u, x)=\sum\limits_{v\in{\mathbb{T}}_n(u)}\delta_x(S_{uv}-S_u), \quad\mbox{ 其中 }\ \delta_x(y)= \left\{\begin{array}{ll} 1, & \quad x=y;\\ 0, & \quad{x\neq y}. \end{array}\right. $

那么$ Z_n(u, x) $在$ { {\mathbb{P}}}_{\xi} $下的分布等同于$ Z_n(x) $在$ { {\mathbb{P}}}_{\theta^k\xi} $下的分布.

令$ \kappa $是一个实数且满足$ \frac{d/2+3}{\lambda} <\kappa<\frac{1}{6} $. 设$ k_n=\lfloor n^{\kappa}\rfloor $, 表示不大于$ n^{\kappa} $的最大整数. 根据分枝过程的可加性质, 有下面的分解式

(4.1) $ \begin{equation} Z_n(x)=\sum\limits_{u\in {\mathbb{T}}_{k_n}} Z_{n-k_n}(u, x-S_u). \end{equation} $

由定义, 对$ u\in{\mathbb{T}}_{k_n} $, 成立

$ Z_{n-k_n}(u, x-S_u)=\sum\limits_{v\in{\mathbb{T}}_{n-k_n}(u)}\delta_{x}(S_{uv}). $

此外将需要下面的$ \sigma $ -代数

$ {\cal I}_0=\{\emptyset, \Omega\}, \quad {\cal I}_n=\sigma(\xi_k, N_u, L_i(u): k<n, i\geq1, |u|<n)\phantom{ss}\mbox{对}\phantom{ss} n\geq1, $

及相关的条件概率和条件期望

$ \phantom{sss} {\mathbb{P}}_n(\cdot)= {\mathbb{P}}(\cdot|{\cal I}_n), \phantom{sss} {\mathbb{E}}_n(\cdot)= {\mathbb{E}}(\cdot|{\cal I}_n). $

记$ \widehat{S}_{n}= S_{1_n} $ ($ n\geq 0 $), 其中$ 1_n = \underbrace{1 \cdots 1}_{n} $, 当$ n>1 $, $ 1_0= \varnothing $. 那么$ \widehat{S}_{n} $是随机环境$ \xi $下的随机游动.

因为环境序列是独立同分布的, 故对$ u\in {\mathbb{T}}_k $, $ Z_n(u, x) $在$ { {\mathbb{P}}}_{\xi} $下的分布律等价于$ Z_n(x) $在$ { {\mathbb{P}}}_{\theta^k\xi} $下的分布律. 则对$ u\in {\mathbb{T}}_{k_n} $, 有

(4.2) $ \begin{equation} {\mathbb{E}}_{\xi}\left(\frac{1}{ \Pi_{n-k_n}( \theta^{k_n}\xi )} Z_{n-k_n}(u, x)\right) = {\mathbb{P}}_{\theta^{k_n}\xi}(\widehat{S}_{n-{k_n}}=x), \end{equation} $

其中$ \Pi_{n}(\theta^k\xi)=m_k\cdots m_{k+n-1} $.

则有下面的分解式成立

(4.3) $ \begin{eqnarray} \frac{1}{\Pi_n}Z_n(z) &=& \frac{1}{\Pi_{k_n}}\sum\limits_{u\in{\mathbb{T}}_{k_n}}\Big(\frac{Z_{n-k_n}(u, z-S_u)}{\Pi_{n-k_n}( \theta^{k_n}\xi )}-\Big[{\mathbb{P}}_{\theta^{k_n}\xi}(\widehat{S}_{n-{k_n}}=z-y)\Big]_{y=S_u}\Big) {}\\ &&+\frac{1}{\Pi_{k_n}}\sum\limits_{u\in{\mathbb{T}}_{k_n}}\Big[{\mathbb{P}}_{\theta^{k_n}\xi}(\widehat{S}_{n-{k_n}}=z-y) \Big]_{y=S_u}=: {\mathbb{A}}_n+{\mathbb{B}}_n. \end{eqnarray} $

基于(4.3)式, 可将定理1.1的证明归结为两个引理.

引理4.1  假设定理1.1的条件成立. 则

(4.4) $ \begin{equation} n^{2+d/2} {\mathbb{A}}_n \mathop{\longrightarrow}_{\rm a.s.}^{n\longrightarrow \infty}0. \end{equation} $

引理4.2  假设定理1.1的条件成立. 则

(4.5) $ \begin{eqnarray} {\mathbb{B}}_n&=&\Big (\frac{ d}{2\pi {b}_{ n} } \Big)^{d/2}\Big \{ W + \frac{d}{ {b}_n} \Big [ F_{1}(z)-\frac{1}{8} \frac{ {b}_{2, n} }{ {b}_n} (d+2) W\Big ] {}\\ && + \frac{d^2}{ {b}_n^2} \Big[F_{2}(z) + \frac{ {b}_{2, n}}{ {b}_n} F_3(z) + Q_d( {b}_n, {b}_{2, n}, {b}_{3, n}) W\Big]+ \frac{1}{n^2}o(1)\Big\}, \end{eqnarray} $

其中$ F_1(z) $, $ F_2(z) $及$ F_3(z) $分别由式(1.8), (1.9) 与(1.10)定义.

引理4.1的证明  先来引入一些记号. 对$ u\in {\mathbb{T}}_{k_n} $, 置

$ X_{n, u}= \frac{Z_{n-k_n}(u, x-S_u)}{\Pi_{n-k_n}(\theta^{k_n}\xi)} - \Big[{\mathbb{P}}_{\theta^{k_n}\xi}(\widehat{S}_{n-{k_n}}=x-y)\Big]_{y=S_u} , $

$ \overline{X}_{n, u}= X_{n, u} {\bf 1}_{\{ |X_{n, u}|\leq \Pi_{k_n}\}}, {\quad} \overline{{\mathbb{A}}}_n = \frac{1}{\Pi_{k_n}} \sum\limits_{u\in {\mathbb{T}}_{k_n}} \overline{X}_{n, u}. $

易见下列事实成立

(4.6) $ \begin{equation} | X_{n, u}|\leq W_{n-k_n}(u) +1, \quad \mbox{ 其中 }\ W_{n-k_n}(u) = | {\mathbb{T}}_{n-k_n}(u)|/\Pi_{(n-k_n)}. \end{equation} $

回顾在$ {\mathbb{P}}_{\xi} $下, $ \{W_{n-k_n}(u): u\in {\mathbb{T}}_{k_n}\} $是独立同分布的, 其分布等同于$ W_{n-k_n} $在$ {\mathbb{P}}_{\theta^{k_n}\xi} $下的分布.

如果下面各式成立, 则引理可得

(4.7) $ \begin{eqnarray} & & {\mathbb{P}}_{k_n} ( \overline{{\mathbb{A}}}_n \neq {\mathbb{A}} _n \mbox{ i.o.} ) =0, \end{eqnarray} $

(4.8) $ \begin{eqnarray} & &{n^{d/2+2}}\Big(\overline{ {\mathbb{A}}}_n - {\mathbb{E}}_{\xi, k_n } \overline{ {\mathbb{A}}}_n \Big) \mathop{\longrightarrow}^{n \rightarrow \infty } 0\quad\mbox{ a.s. }, \end{eqnarray} $

(4.9) $ \begin{eqnarray} & & {n^{d/2+2}} {\mathbb{E}}_{\xi, k_n } \overline{ {\mathbb{A}}}_n \mathop{\longrightarrow}^{n \rightarrow \infty } 0\quad\mbox{ a.s. }. \end{eqnarray} $

为证明这些结论需要下面引理.

引理4.3^{[45, 推论1.1]}  记$ W^*= \sup_n W_n $. 设存在$ \delta>0 $$ \lambda>0 $, 使得$ {\mathbb{E}} m_0^{-\delta}<\infty $与(1.3)式成立. 则

(4.10) $ \begin{equation} { {\mathbb{E}}}(W^*+1)(\log (W^*+1))^{\lambda} <\infty. \end{equation} $

下面证明分三步来证明(4.7), (4.8) 及(4.9)式.

第1步  先来证明(4.7)式. 为此只需证明

(4.11) $ \begin{equation} \sum\limits_{n=1}^\infty {\mathbb{P}}_{k_n}( \overline{{\mathbb{A}}}_n \neq {\mathbb{A}} _n ) <\infty. \end{equation} $

观察到

$ \begin{eqnarray} {\mathbb{P}}_{k_n} ( {\mathbb{A}}_n\neq \overline{ {\mathbb{A}}}_n ) & \leq& {\mathbb{E}} _{k_n}\sum\limits_{u\in {\mathbb{T}}_{k_n} } {\mathbb{P}}_{\xi, {k_n}}(X_{n, u}\neq \overline{X}_{n, u}) = \sum\limits_{u\in {\mathbb{T}}_{k_n} } {\mathbb{P}}_{k_n}(|X_{n, u}|\geq \Pi_{k_n}) \\&\leq&_{(4.6)} \sum\limits_{u\in {\mathbb{T}}_{k_n} } {\mathbb{P}} _{k_n}( W_{n-k_n} (u)+1 \geq \Pi_{k_n})\\ &=& | {\mathbb{T}}_{k_n}|\Big[ {\mathbb{P}} ( W_{n-k_n} +1 \geq t_n)\Big]_{t_n=\Pi_{k_n} } \\& \leq& W_{k_n} \Big[ {\mathbb{E}}\big((W_{n-k_n}+1 ) {\bf 1}_{ \{W_{n-k_n}+1 \geq t_n\}} \big)\Big]_{t_n=\Pi_{k_n} } \\ & \leq &W^* \Big[ {\mathbb{E}}\big((W^*+1 ) {\bf 1}_{ \{W^*+1 \geq t_n\}} \big) \Big]_{t_n=\Pi_{k_n} } \\& \leq &W^* (\log \Pi_{k_n}) ^{-\lambda} {\mathbb{E}} (W^*+1) \log ^{\lambda} (W^*+1) . \end{eqnarray} $

由于$ (\Pi_{n})^{\frac{1}{n}}\rightarrow \exp\{ {\mathbb{E}}\log m_0\} $ a.s. 及$ k_n\sim n^{\kappa} $, 可知$ \liminf\limits_{n\rightarrow \infty}\frac{\log \Pi_{k_n}}{n^{\kappa}}>0 $. 因此可由$ \lambda \kappa>1 $及(4.10)式得出(4.11)式.

第2步  证明(4.8)式. 为此需要下面不等式(参见文献[46] 中(5.3)式后的段落): 对$ 1<\alpha<2 $, 有

$ \begin{eqnarray} && {\mathbb{P}}_{\xi, k_n} \Big( \Big| \sum\limits_{u\in {\mathbb{T}}_{k_n}} \frac{1}{\Pi_{k_n} }(\bar{X}_{n, u}- {\mathbb{E}}_{\xi, {k_n} } \bar{X}_{n, u}) \Big| > \epsilon n^{-d/2-2} \Big) \\ & \leq & K_a \frac{n^{(d/2+2)\alpha}}{\epsilon^{\alpha}} \sum\limits_{u\in {\mathbb{T}}_{k_n}} \frac{1}{\Pi_{k_n}^\alpha} {\mathbb{E}}_{\xi, {k_n} } |X_{n, u}|^{\alpha} {{\bf 1}_{\left\{{ |X_{n, u}|\leq \Pi_{k_n}}\right\}}}, \end{eqnarray} $

其中$ K_a $是一个绝对常数. 在上式两边取条件期望可得

$ \begin{eqnarray} && {\mathbb{P}}_{k_n} \Big( \Big| \sum\limits_{u\in {\mathbb{T}}_{k_n}} \frac{1}{\Pi_{k_n} }(\bar{X}_{n, u}- {\mathbb{E}}_{\xi, {k_n} } \bar{X}_{n, u}) \Big| > \epsilon n^{-d/2-2} \Big) \\ & \leq & K_a \frac{n^{(d/2+2)\alpha}}{\epsilon^{\alpha}} \sum\limits_{u\in {\mathbb{T}}_{k_n}} \frac{1}{\Pi_{k_n}^\alpha} {\mathbb{E}}_{k_n} \Big( |X_{n, u}|^{\alpha} {{\bf 1}_{\left\{{ |X_{n, u}|\leq \Pi_{k_n}}\right\}}}\Big), \end{eqnarray} $

取常数$ b\in (1, e^{ {\mathbb{E}} \log m_0}) $. 观察到

$ \begin{eqnarray} && {\mathbb{E}}_{k_n}|X_{n, u}|^{\alpha} {{\bf 1}_{\left\{{ |X_{n, u}|\leq \Pi_{k_n}}\right\}}} \\ & =& \int_{0}^\infty \alpha x^{\alpha-1} {\mathbb{P}}_{k_n} ({{\bf 1}_{\left\{{ |X_{n, u}|\leq \Pi_{k_n}}\right\}}}|X_{n, u}|>x ) {\rm d}x \\ & \leq & \alpha \int_0^{\Pi_{k_n}} x^{\alpha-1} {\mathbb{P}}_{k_n} ( | W_{n-k_n}(u)+1 | >x) {\rm d}x\\ &=& \alpha \int_0^{\Pi_{k_n}} x^{\alpha-1} {\mathbb{P}} ( | W_{n-k_n}+1 | >x) {\rm d}x \\ & \leq& \alpha \int_0^{\Pi_{k_n}} x^{\alpha-1} {\mathbb{P}}( W^*+1 >x) {\rm d}x \\ & \leq& \alpha \int_e^{\Pi_{k_n} } x^{\alpha-2} (\log x)^{-\lambda} {\mathbb{E}} (W^*+1)(\log(W^*+1) )^{\lambda} {\rm d}x + e^\alpha \\ &\leq & \alpha {\mathbb{E}} (W^*+1)(\log(W^*+1) )^{\lambda} \left(\int_e^{b^{k_n}} x^{\alpha-2} (\log x)^{-\lambda} {\rm d}x + \int_{b^{k_n}}^{\Pi_{k_n}} x^{\alpha-2}(\log x)^{-\lambda} {\rm d}x \right)+ e^\alpha\\ & \leq & \alpha {\mathbb{E}} (W^*+1)(\log(W^*+1) )^{\lambda}\Big( \int_e^{b^{k_n}} x^{\alpha-2} {\rm d}x +(k_n\log b)^{-\lambda}\int_{b^{k_n}}^{\Pi_{k_n}} x^{\alpha-2} {\rm d}x \Big)+e^\alpha \\ & \leq & \frac{\alpha}{\alpha-1} {\mathbb{E}} (W^*+1)(\log(W^*+1) )^{\lambda} \Big( \big( b^{k_n}\big)^{\alpha-1} + (k_n\log b)^{-\lambda} \Pi_{k_n} ^{\alpha-1} \Big)+e^\alpha \\ &= & K(\lambda, \alpha) \Big( \big( b^{k_n}\big)^{\alpha-1} + (k_n\log b)^{-\lambda} \Pi_{k_n} ^{\alpha-1} \Big)+e^\alpha, \end{eqnarray} $

其中$ K(\lambda, \alpha) $是一个与$ \lambda $和$ \alpha $相关的常数. 则有

$ \begin{eqnarray} & & {\mathbb{P}}_{k_n} \Big( \Big| \sum\limits_{u\in {\mathbb{T}}_{k_n}} \frac{1}{\Pi_{k_n} }(\bar{X}_{n, u}- {\mathbb{E}}_{\xi, {k_n} } \bar{X}_{n, u}) \Big| > \epsilon n^{-d/2-2} \Big) \\ &\leq & K_a \frac{n^{(d/2+2)\alpha}}{\epsilon^{\alpha}} \frac{1}{\Pi_{k_n}^\alpha} \sum\limits_{u\in {\mathbb{T}}_{k_n}} \Big[ K(\lambda, \alpha) \Big( \big( b^{k_n}\big)^{\alpha-1} + (k_n\log b)^{-\lambda} \Pi_{k_n} ^{\alpha-1} \Big)+e^\alpha\Big] \\ & =& K_a \frac{n^{(d/2+2)\alpha}}{\epsilon^{\alpha}} W_{k_n} \Big[ K(\lambda, \alpha) \Big( \big( b^{k_n} /\Pi_{k_n}\big)^{\alpha-1} + (k_n\log b)^{-\lambda} \Big)+ e^\alpha\Pi_{k_n}^{1-\alpha}\Big]. \end{eqnarray} $

那么

$ \begin{eqnarray} & &\sum\limits_{n=1}^\infty {\mathbb{P}}_{k_n} \Big( n^{ d/2+2}\Big| \bar{ {\mathbb{A}}_n}- {\mathbb{E}}_{\xi, {k_n} } \bar{ {\mathbb{A}}_n}\Big| > \epsilon \Big) \\ & = &\sum\limits_{n=1}^\infty {\mathbb{P}}_{k_n} \Big(n^{ d/2+2} \Big| \sum\limits_{u\in {\mathbb{T}}_{k_n}} \frac{1}{\Pi_{k_n} }(\bar{X}_{n, u}- {\mathbb{E}}_{\xi, {k_n} } \bar{X}_{n, u}) \Big| > \epsilon \Big) \\ & \leq& K_a \frac{W^* }{\epsilon^{\alpha}} \Big[ K(\lambda, \alpha) \Big( \sum\limits_{n=1}^\infty n^{(d/2+2)\alpha} \big( b^{k_n} /\Pi_{k_n}\big)^{\alpha-1} + \sum\limits_{n=1}^\infty n^{(d/2+2) \alpha}(k_n\log b)^{-\lambda} \Big) \\ & & + e^\alpha \sum\limits_{n=1}^\infty n^{(d/2+2) \alpha}\Pi_{k_n}^{1-\alpha}\Big] , \end{eqnarray} $

取$ \alpha $充分接近1, 由$ \lambda\kappa>d/2+3 $可推出$ \lambda\kappa-(d/2+2)\alpha>1 $, 从而上式最后一项有限.

第3步  观察发现

$ \begin{eqnarray} && {\mathbb{P}}_{k_n} \Big(| {\mathbb{E}}_{\xi, k_n} \bar{ {\mathbb{A}}}_n | > \epsilon n^{-d/2-2} \Big ) \\ & \leq & \frac{ n^{d/2+2}}{\varepsilon} {\mathbb{E}}_{k_n} | {\mathbb{E}}_{\xi, k_n} \bar{ {\mathbb{A}}}_n | = \frac{ n^{d/2+2}}{\varepsilon} {\mathbb{E}}_{k_n} \Big| \frac{1}{\Pi_{k_n} } \sum\limits_{u\in {\mathbb{T}}_{k_n}} {\mathbb{E}}_{\xi, k_n} \bar{X}_{n, u} \Big| \\ & = &\frac{ n^{d/2+2}}{\varepsilon} {\mathbb{E}}_{k_n} \Big| \frac{1}{\Pi_{k_n} } \sum\limits_{u\in {\mathbb{T}}_{k_n}} (- {\mathbb{E}}_{\xi, k_n } X_{n, u} {\bf 1}_{ \{ |X_{n, u}| \geq \Pi_{k_n}\}} ) \Big| \\ & \leq & \frac{ n^{d/2+2}}{\varepsilon} \frac{1}{\Pi_{k_n} } \sum\limits_{u\in {\mathbb{T}}_{k_n}} {\mathbb{E}}_{k_n } ( W_{n-k_n} (u) +1) {\bf 1}_{ \{ W_{n-k_n}(u)+1 \geq \Pi_{k_n}\}} \\ & = & \frac{ n^{d/2+2}W_{k_n}}{\varepsilon} \Big[ {\mathbb{E}} ( W_{n-k_n} +1) {\bf 1}_{ \{ W_{n-k_n}+1 \geq t_n\}} \Big]_{t_n =\Pi_{k_n}} \\ & \leq& \frac{W^*}{\varepsilon} n^{d/2+2} \Big[ {\mathbb{E}} ( W^* +1) {\bf 1}_{ \{ W^*+1 \geq t_n\}} \Big]_{t_n =\Pi_{k_n}} \\ & \leq &\frac{W^*}{\varepsilon} \frac{ n^{d/2+2}}{(\log\Pi_{k_n})^{\lambda} } {\mathbb{E}} (W^*+1) \log ^{\lambda } ( W^*+1 ) \\ & \leq & \frac{W^*}{\varepsilon} K_\xi n^{d/2+2 -\lambda\kappa} {\mathbb{E}} (W^*+1) \log ^{\lambda } ( W^*+1 ). \end{eqnarray} $

那么由$ \lambda \kappa > d/2+3 $可推得

$ \sum\limits_{n=1}^\infty {\mathbb{P}}_{k_n} \Big(\Big| n^{d/2+2} {\mathbb{E}}_{\xi, k_n} \bar{ {\mathbb{A}}}_n \Big| > \epsilon \Big )<\infty. $

于是(4.9)式成立.

综上, 引理4.1得证.

引理4.2的证明  容易发现对$ u \in {\mathbb{T}}_{k_n} $, $ {\left\Vert {S_u} \right\Vert} \leq k_n. $回顾$ k_n<n^{1/6}. $故$ {\left\Vert {z+S_u} \right\Vert}<n^{1/6} $对充分大的$ n $成立. 由定理3.1中的(3.3)式可得

$ \begin{eqnarray} & &{\mathbb{P}}_{\theta^{k_n}\xi}(\widehat{S}_{n-k_n}=z-y)|_{y=S_u}\\ & =&\left(\frac{d}{2\pi ( {b}_n- {b}_{k_n})}\right)^{d/2} \Big\{ 1+ \frac{d}{ {b}_n- {b}_{k_n}} \Big[ -\frac{1}{2}{\left\Vert {z-S_u} \right\Vert}^2 +\frac{1}{8}\Big( d - (d+2) \frac{ {b}_{2, n} - {b}_{2, k_n} }{ {b}_n- {b}_{k_n}} \Big)\Big] \\ &&+ \frac{ d^2}{( {b}_n- {b}_{k_n})^2} \Big[ \frac{1}{8}{\left\Vert {z-S_u} \right\Vert}^4- \frac{(d+4)}{16} {\left\Vert {z-S_u} \right\Vert}^2 \Big( 1- (1+\frac{2}{d})\frac{ {b}_{2, n} - {b}_{2, k_n} }{ {b}_n- {b}_{k_n}} \Big) \\ & & + Q_d( {b}_n- {b}_{k_n}, {b}_{2, n}- {b}_{2, k_n}, {b}_{3, n}- {b}_{3, k_n}) \Big] +n^{-2}\widetilde{\alpha}_{n}(\theta^{k_n}\xi, z-S_u)\Big\} \\ & = &\Big (\frac{ d}{2\pi {b}_{ n} } \Big)^{d/2}\Big\{ 1+ \frac{d}{ {b}_n} \Big[ -\frac{1}{2}{\left\Vert {z} \right\Vert}^2+\frac{1}{8}\Big( d - (d+2) \frac{ {b}_{2, n} }{ {b}_n} \Big)+\langle {{z}}, {{S_u}}\rangle-\frac{1}{2}({\left\Vert {S_u} \right\Vert}^2 -b_{k_n})\Big] \\ & & + \frac{ d^2}{8 {b}_n^2} \Big[\Big({\left\Vert {S_u} \right\Vert}^4-\Big(2+\frac{4}{d}\Big) {b}_{k_n}{\left\Vert {S_u} \right\Vert}^2 + \Big(1+\frac{2}{d}\Big)( {b}_{k_n}^2 + {b}_{2, k_n})- {b}_{k_n} \Big) \\ & & -4 \langle {{z}}, {{{\left\Vert {S_u} \right\Vert}^2S_u -\Big(1+\frac{2}{d}\Big) {b}_{k_n}S_u }}\rangle+ 4\Big(\langle {{z}}, {{S_u}}\rangle^2 - \frac{1} {d} {b}_{k_n}{\left\Vert {z} \right\Vert}^2 \Big) \\ && + \Big(2{\left\Vert {z} \right\Vert}^2- \frac{1}{2} (d+4)\Big(1- \Big( 1+\frac{2}{d} \Big)\frac{ {b}_{2, n}}{ {b}_n}\Big) \Big)({\left\Vert {S_u} \right\Vert}^2 - {b}_{k_n}) \\ && +\Big[(d+4) \Big(1- \Big( 1+\frac{2}{d} \Big)\frac{ {b}_{2, n}}{ {b}_n}\Big)- 4{\left\Vert {z} \right\Vert}^2\Big ] \langle {{z}}, {{S_u}}\rangle \\ && +{\left\Vert {z} \right\Vert}^4 - \frac{1}{2} (d+4) {\left\Vert {z} \right\Vert}^2\Big(1- \Big( 1+\frac{2}{d} \Big)\frac{ {b}_{2, n}}{ {b}_n}\Big) +Q_d( {b}_n, {b}_{2, n}, {b}_{3, n})\Big]+ n^{-2} \epsilon_{n, u}\Big\}, \end{eqnarray} $

其中无穷小项$ \epsilon_{n, u}(u\in {\mathbb{T}}_{k_n}) $满足

$ \begin{eqnarray} \sup\limits_{u\in {\mathbb{T}}_{k_n} }\{ |\epsilon_{n, u} | \} & \leq & C_d (n-k_n)^{2+d/2}e ^{ -(M_n-M_{k_n})} + C_d \Big(\frac{n-k_n}{ {b}_n- {b}_{k_n}} \Big)^{ 5+d/2}\\ &&\times \Big[(n-k_n)^{ (6+d)/4}e^{-\frac{11( {b}_n- {b}_{k_n})}{24 \sqrt{n-k_n}d} } + (n-k_n)^{-1/3}+ (n-k_n)^{ -1} ({\left\Vert {z} \right\Vert}+k_n )^6 \Big], \end{eqnarray} $

则

(4.12) $ \begin{equation} \epsilon_n = \sup\limits_{u\in {\mathbb{T}}_{k_n} }\{ | \epsilon_{n, u} | \} \mathop{\longrightarrow}^{n\rightarrow \infty} 0. \end{equation} $

从而

(4.13) $ \begin{eqnarray} {\mathbb{B}}_{n}&=& \frac{1}{\Pi_{k_n}}\sum\limits_{u\in {\mathbb{T}}_{k_n}} {\mathbb{P}}_{\theta^{k_n}\xi} \big(\widehat{S}_{n-k_n} =z-y\big)|_{y=S_u} \\ & =& \Big (\frac{ d}{2\pi {b}_{ n} } \Big)^{d/2}\Big \{ W_{k_n} + \frac{d}{ {b}_n} \Big [ F_{1, k_n}(z)-\frac{1}{8} \frac{ {b}_{2, n} }{ {b}_n} (d+2) W_{k_n} \Big ] \\ & &+\frac{d^2}{ {b}_n^2} \Big[F_{2, k_n}(z) + \frac{ {b}_{2, n}}{ {b}_n} F_{3, k_n}(z) + Q_d( {b}_n, {b}_{2, n}, {b}_{3, n}) W_{k_n}\Big]+ \frac{1}{n^2}\frac{1}{\Pi^{k_n}}\sum\limits_{u\in {\mathbb{T}}_{k_n}}\epsilon_{n, u}\Big\}, \end{eqnarray} $

其中

$ \begin{eqnarray} F_{1, k_n}(z)&=& \Big[\frac{1}{8}d-\frac{1}{2}{\left\Vert {z} \right\Vert}^2\Big]W_{k_n}+\langle {{z}}, {{N_{1, k_n}}}\rangle-\frac{1}{2} N_{2, k_n}, \\ F_{2, k_n}(z)&= &\frac{1}{8} N_{4, k_n} - \frac{1}{2}\langle {{z}}, {{ N_{3, k_n}}}\rangle+\frac{1}{2} N^z_{2, k_n} + \Big( \frac{1}{4} {\left\Vert {z} \right\Vert}^2-\frac{d}{16}-\frac{1}{4}\Big) N_{2, k_n} \\ && + \Big( \frac{d}{8}+\frac{1}{2} -\frac{1}{2}{\left\Vert {z} \right\Vert}^2\Big)\langle {{z}}, {{ N_{1, k_n} }}\rangle +\Big[ \frac{1}{8}{\left\Vert {z} \right\Vert}^4 - \frac{1}{16} (d+4) {\left\Vert {z} \right\Vert}^2 \Big]W_{k_n}, \\ F_{3, k_n}(z)&=& (d+4)\Big(1+\frac{2}{d}\Big)\Big(\frac{1}{16} N_{2, k_n}-\frac{1}{8}\langle {{z}}, {{ N_{1, k_n}}}\rangle + \frac{1}{16}{\left\Vert {z} \right\Vert}^2 W_{k_n} \Big). \end{eqnarray} $

当条件(1.3)对$ \lambda>0 $成立, 由(2.3)式, 可得

$ W-W_n=o(n^{-\lambda}), \quad\mbox{a.s. 当 }\ n\rightarrow \infty. $

根据$ \kappa $的选取及$ k_n=n^{\kappa} $, 可见

(4.14) $ \begin{equation} W-W_{k_n}=o(n^{-\lambda\kappa})=o(n^{-2}) \quad\mbox{a.s. 当 }\ n\rightarrow \infty. \end{equation} $

类似地, 由命题1.1可得当$ n\rightarrow \infty $时, a.s.

$ \begin{eqnarray} &&N_{1, k_n}- {\cal V}_1= o(n^{-1}), N_{2, k_n}- {\cal V}_2= o(n^{-1}) , N^z_{2, k_n}- {\cal V}_2^z=o(1), \\ && N_{3, k_n} - {\cal V}_3 =o(1) , N_{4, k_n} - {\cal V}_4 =o(1) . \end{eqnarray} $

于是可得当$ n\rightarrow \infty $时,

(4.15) $ \begin{equation} F_{1, k_n}(z)-F_1(z)= o(n^{-1}), \ F_{2, k_n}(z)- F_2(z)=o(1), \ F_{3, k_n}(z)- F_3(z)=o(1) \ \mbox{a.s.}. \end{equation} $

而由(4.12)式, 有

(4.16) $ \begin{equation} \bigg|\frac{1}{\Pi^{k_n}} \sum\limits_{u\in {\mathbb{T}}_{k_n}}\epsilon_{n, u}\bigg|\leq \epsilon_n W_{k_n} \rightarrow 0. \end{equation} $

综合(4.14), (4.15), (4.16)与(4.13)式, 即可导出所需结论(4.5)式.