极限理论是概率论中备受关注的问题之一, 中心极限定理和大数律的研究, 可上溯至十八世纪初, 不过那时的结果主要针对的是单序列的部分和(求和的个数是确定的). Kolmogorov在20世纪三四十年代, 对行内独立的二重序列确定和的极限定理(包括中心极限定理和重对数律等)做了系统的研究, 并得出一系列奠基性的结果^[1-2].

在此之后, 由于实际的需要, 中心极限定理的研究对象开始从单纯的行内独立的随机变量序列的确定和转向随机个随机变量之和.如胡迪鹤^[3]通过单序列随机和的中心极限定理结果, 将Donsker不变原理推广到分枝过程的情形.

1970—1971年, Heyde^[4]得到关于分枝过程的一系列极限定理结果, 其中包括中心极限定理和重对数律.其后一些工作讨论了随机环境中分枝过程的极限理论^[5-6].

1 背景知识

本文讨论一致渐进可忽略(u.a.n.)二重随机变量序列的随机和的P-收敛性, 包括变化环境中的分枝过程的P-收敛性.对二重随机变量序列的确定和而言, P-收敛性的讨论通常是弱收敛研究的铺垫^[2].

令$(\mathit{\Omega} ,\mathscr{F},P)$为一个概率空间, 本文的所有随机变量均定义在此空间上.文中总记$\mathbb{N}$为全体非负整数构成的集合, ${\mathbb{N}^ + }$为全体正整数构成的集合.

定义1.1   令Z₀=1且对于任意的n≥0,

${Z_{n + 1}} = \left\{ \begin{array}{l} \sum\limits_{j = 1}^{{Z_n}} {{X_{n,j}}\left( x \right),\;{Z_n} \ne 0;} \\ 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;{Z_n} = 0, \end{array} \right.$

其中，$\left\{ {{X_{0,1}},{X_{n,j}}:n \ge 0,j \ge 1} \right\}$相互独立. X_{0, 1}的分布为$\left\{ {{p_0}\left( j \right),j \in \mathbb{N}} \right\}$, 对于任意的$n \ge 1,\left\{ {{X_{n,j}}:j \in \mathbb{N}} \right\}$有相同的分布$\left\{ {{p_n}\left( k \right),k \in \mathbb{N}} \right\},\left\{ {{Z_n}:n \ge 0} \right\}$称为变化环境中的分枝过程.记μ₀=E(X_{0, 1}), μ_n=E(X_{n, j}), n≥1, 又记m₀=μ₀, m_n=E(Z_n)=μ₀…μ_n-1, 当1 < μ_n < ∞时, 称该分枝过程为上临界的.在本文中总假定p_n(0)=0, 0 < p_n(1) < 1.特别地, 当p_n(k)=p(k), n≥0, 则{Z_n:n≥0}为经典的G-W分枝过程.

定义1.2   设$\left\{ {{\xi _n}:n \ge 1} \right\}$为一列定义在$\left( {\mathit{\Omega} ,\mathscr{F},P} \right)$上且取值于$\mathbb{N}$的随机变量序列, 又设ξ_n→∞a.s.当n→∞, 称{V_{n, k}:n≥1, k≥1}为关于{ξ_n:n≥1}的一致渐进可忽略序列, 简称u.a.n.序列, 如果它满足下面的条件

$\forall \epsilon > 0,\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} P\left( {\left| {{V_{n,k}}} \right| \ge \epsilon} \right) = 0\;\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$

注1.1   当ξ_n≡k_n时(k_n为正整数, $\forall n \in {\mathbb{N}^ + }$), 则{V_{n, k}:n≥1, k≥1}就是经典意义下的u.a.n.序列(见文献[7]).

引理1.1   沿用前面的符号, 设{Z_n, n≥0}为定义1.1中所定义的变化环境中的上临界的分枝过程, {X_{n, j}:j≥1}的期望为μ_n.假定Z_n→∞a.s.当$n \to \infty ,E\left| {{X_{n,j}} - {\mu _n}} \right| \le M < \infty ,\;\forall j,n \in {\mathbb{N}^ + }$, 则对任意单调上升函数f且limx→∞f(x)=+∞, 如下序列

${Y_{n,j}}: = \frac{{{X_{n,j}} - {\mu _n}}}{{f\left( {{Z_n}} \right)}},$

是关于{Z_n, n≥0}的u.a.n.序列.

证明   首先根据上临界分枝过程的性质和p_n(0)≡0的假设知: Z_n→∞a.s., 当n→∞.

现在只需要对于任意的$ \epsilon > 0$验证下式

$\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le j \le {Z_n}} P\left( {\frac{{\left| {{X_{n,j}} - {\mu _n}} \right|}}{{f\left( {{Z_n}} \right)}} \ge \epsilon} \right) = 0\;{\rm{a}}{\rm{.s}}{\rm{.}}$

给定$1 > \epsilon > 0$, 由于$\mathop {\lim }\limits_{n \to \infty } {Z_n} = \infty \;{\rm{a}}{\rm{.s}}{\rm{.}}$, 故对于任意的η>0, C>0, 总存在Ω₀Ω以及n₀>0使得P(Ω₀)>1-η且f(Z_n)>C, ω∈Ω₀, n≥n₀(此处Ω₀和n₀仅依赖于η和C).由Chebyshev不等式有:当n≥n₀时,

$\begin{array}{l} P\left( {\frac{{\left| {{X_{n,j}} - {\mu _n}} \right|}}{{f\left( {{Z_n}} \right)}} \ge \epsilon} \right)\\ \le P\left( {\frac{{\left| {{X_{n,j}} - {\mu _n}} \right|}}{{f\left( {{Z_n}} \right)}} \ge \epsilon,{\mathit{\Omega} _0}} \right) + \eta \\ \le \frac{1}{{\epsilon f\left( C \right)}}E\left( {\left| {{X_{n,j}} - {\mu _n}} \right|} \right) + \eta \\ \le \frac{M}{{\epsilon f\left( C \right)}} + \eta , \end{array}$

令C→∞, 则有对a.s.ω

$\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le j \le {Z_n}} P\left( {\frac{{\left| {{X_{n,j}}{\mu _n}} \right|}}{{f\left( {{Z_n}} \right)}} \ge \epsilon} \right) \le \eta ,$

而η是任意的, 故$\left\{ {{Y_{n,j}}:n \ge 1,j \ge 1} \right\}$是关于$\left\{ {{Z_n},n \ge 0} \right\}$的u.a.n.序列.

注1.2   类似于上述证明可得, 在同样条件下

$\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le j \le {Z_n}} P\left( {\frac{{\left| {{X_{n,j}}{\mu _n}} \right|}}{{f\left( {{Z_n}} \right)}} \ge \epsilon\left| {{Z_n}} \right.} \right) = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$

此时称$\left\{ {\frac{{\left| {{X_{n,j}} - {\mu _n}} \right|}}{{f\left( {{Z_n}} \right)}}:n \ge 1,j \ge 1} \right\}$为关于{Z_n, n≥0}的条件u.a.n.序列.

注1.3   令Z_{n, k}^(j)为定义1.1中的分枝过程第n代第j个个体的第k代子孙的数量, 记$W_n^{\left( j \right)}: = \mathop {\min }\limits_{n \to \infty } Z_{n,k}^{\left( j \right)}{\left( {\prod\limits_{i = n + 1}^{n + k} {{\mu _i}} } \right)^{ - 1}}$(详见Athreya和Ney^[8]和高振龙^[5]), 仿照上述证明可知, 当$E\left| {1 - W_n^{\left( j \right)}} \right| \le K < \infty $时, ${U_{n,j}}: = \frac{{1 - W_n^{\left( j \right)}}}{{\sqrt {{Z_n}} }}$也为关于$\left\{ {{Z_n},n \ge 0} \right\}$的u.a.n.序列. Heyde^[4]讨论了$\left\{ {{U_{n,j}},n \ge 1,j \ge 1} \right\}$的中心极限定理.

2 主要结果

命题2.1   设$\left\{ {{V_{n,k}}:n \ge 1,k \ge 1} \right\}$为关于$\left\{ {{\xi _n}:n \ge 1} \right\}$的u.a.n.序列, 记F_{n, k}(t)为V_{n, k}的分布函数, f_{n, k}(t)为V_{n, k}的特征函数, 则下面6个条件等价:

1)$\forall \epsilon > 0,\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} P\left( {\left| {{V_{n,k}}} \right| \ge \epsilon} \right) = 0\;\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$;

2)在任意的有限区间[a, b]上对a.s.ω有, $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} {\mathop{\rm Re}\nolimits} \left( {1 - {f_{n,k}}\left( t \right)} \right)0\forall t \in \left[ {a,b} \right]$, 其中Re(1-f_{n, k}(t))为1-f_{n, k}(t)的实部, 上述收敛对于t∈[a, b]是一致的;

3)在任意的有限区间[a, b]上对a.s.ω有, $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} \left| {1 - {f_{n,k}}\left( t \right)} \right| = 0\forall t \in \left[ {a,b} \right]$, 上述收敛对于t∈[a, b]是一致的;

4)对任意的t, $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} \left| {1 - {f_{n,k}}\left( t \right)} \right| = 0\;{\rm{a}}{\rm{.s}}{\rm{.}}$;

5)对任意的t, $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} {\mathop{\rm Re}\nolimits} \left( {1 - {f_{n,k}}\left( t \right)} \right) = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$;

6)对任意的t, $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} \int_{ - \infty }^\infty {\frac{{{x^2}}}{{1 + {x^2}}}{\rm{d}}{F_{n,k}}\left( x \right) = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.}}} $.

证明   1)$ \Rightarrow $2).选取T>0使得$\left[ {a,b} \right] \subset \left[ { - T,T} \right]$, 给定$ \epsilon > 0$, 对于任意的$t \in \left[ { - T,T} \right]$，

$\begin{array}{c} {\mathop{\rm Re}\nolimits} \left( {1 - {f_{n,k}}\left( t \right)} \right)\\ = \int_{ - \infty }^\infty {\left( {1 - \cos xt} \right){\rm{d}}{F_{n,k}}\left( x \right)} \\ = \int_{\left| x \right| < \epsilon } {\left( {1 - \cos \;xt} \right){\rm{d}}{F_{n,k}}\left( x \right) + } \\ \int_{\left| x \right| \ge \epsilon} {\left( {1 - \cos xt} \right){\rm{d}}{F_{n,k}}\left( x \right)} \\ \le \frac{{{\epsilon^2}{T^2}}}{2} + \int_{\left| x \right| \ge \epsilon} {2{\rm{d}}{F_{n,k}}\left( x \right)} \\ = \frac{{{\epsilon^2}{T^2}}}{2} + 2P\left( {\left| {{X_{n,k}}} \right| \ge \epsilon} \right), \end{array}$

由1)知存在Ω₀与t无关且P(Ω₀)=1, 使得$\forall \omega \in {\mathit{\Omega} _0}$

$\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} P\left( {\left| {{V_{n,k}}} \right| \ge \epsilon} \right) = 0,$

故有

$\mathop {\lim \;\;\sup }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} {\mathop{\rm Re}\nolimits} \left( {1 - {f_{n,k}}\left( t \right)} \right) \le \frac{{{\epsilon^2}{T^2}}}{2},$

上述收敛性关于$t \in \left[ { - T,T} \right]\;$是一致的, 再让$ \epsilon \to 0$可以得到2).

2) $ \Rightarrow $3).由于

$1 - {f_{n,k}}\left( t \right) = {\mathop{\rm Re}\nolimits} \left( {1 - {f_{n,k}}\left( t \right)} \right) - \int_{ - \infty }^\infty {{\rm{i}}\sin xt{\rm{d}}{F_{n,k}}\left( x \right)} ,$

又因为

$\begin{array}{l} {\left| {\int_{ - \infty }^\infty {{\rm{i}}\sin xt{\rm{d}}{F_{n,k}}\left( x \right)} } \right|^2}\\ \le \int_{ - \infty }^\infty {{{\sin }^2}tx{\rm{d}}{F_{n,k}}\left( x \right)} \\ = \frac{1}{2}\int_{ - \infty }^\infty {\left( {1 - \cos 2tx} \right){\rm{d}}{F_{n,k}}\left( x \right)} \\ = \frac{1}{2}{\mathop{\rm Re}\nolimits} \left( {1 - {f_{n,k}}\left( {2t} \right)} \right), \end{array}$

故由2)可得3).

3)$ \Rightarrow $4)$ \Rightarrow $5).显然.

5)$ \Rightarrow $6).由文献[7]47页(4.8)式知:对任意特征函数f(t)以及与之对应的概率分布函数F(x)均有

$\int_{ - \infty }^\infty {\frac{{{x^2}}}{{1 + {x^2}}}{\rm{d}}F\left( x \right) \le M\left( t \right)\int_0^t {\left( {1 - {\mathop{\rm Re}\nolimits} \left( {f\left( v \right)} \right)} \right){\rm{d}}v,\forall t > 0.} } $

其中, $M\left( t \right) = {\left[ {\mathop {\inf }\limits_{x \in \mathbb{R}} \left( {t\left( {1 - \frac{{\sin \;tx}}{{tx}}} \right)\frac{{1 + {x^2}}}{{{x^2}}}} \right)} \right]^{ - 1}},Re\left( {f\left( v \right)} \right)$为f(v)的实部.

于是

$\begin{array}{l} \mathop {\max }\limits_{1 \le k \le {\xi _n}} \int_{ - \infty }^\infty {\frac{{{x^2}}}{{1 + {x^2}}}{\rm{d}}{F_{n,k}}\left( x \right)} \\ \le \mathop {\max }\limits_{1 \le k \le {\xi _n}} M\left( t \right)\int_0^t {{\mathop{\rm Re}\nolimits} } \left( {1 - {f_{n,k}}\left( v \right)} \right){\rm{d}}v\\ \le M\left( t \right)\int_0^t {\mathop {\max }\limits_{1 \le k{\xi _n}} {\mathop{\rm Re}\nolimits} \left( {1 - {f_{n,k}}\left( v \right)} \right){\rm{d}}v,\forall t > 0,} \end{array}$

由Lebesgue控制收敛定理, 对上式两端取极限, 立得6).

6)$ \Rightarrow $1).任意固定$ \epsilon > 0$, 我们注意到$\frac{{1 + {x^2}}}{{{x^2}}} \le \frac{{1 + {\epsilon^2}}}{{{\epsilon^2}}}$, 当$\left| x \right| \ge \epsilon$, 故

$\begin{array}{l} \mathop {\max }\limits_{1 \le k \le {\xi _n}} P\left( {\left| {{v_{n,k}}} \right| \ge \epsilon} \right)\\ = \mathop {\max }\limits_{1 \le k \le {\xi _n}} \int_{\left| x \right| \ge \epsilon} {\frac{{1 + {x^2}}}{{{x^2}}}\frac{{{x^2}}}{{1 + {x^2}}}{\rm{d}}{F_{n,k}}\left( x \right)} \\ \le \frac{{1 + {\epsilon^2}}}{{{\epsilon^2}}}\mathop {\max }\limits_{1 \le k \le {\xi _n}} \int_{\left| x \right| \ge \epsilon} {\frac{{{x^2}}}{{1 + {x^2}}}{\rm{d}}{F_{n,k}}\left( x \right),} \end{array}$

由6)立得1).

注2.1   经典的u.a.n.序列的相应结果见文献[7].

命题2.2   设$\left\{ {{\xi _n}:n \ge 1} \right\}$为取值于自然数集的随机变量序列, $\left\{ {{V_{n,k}}:n \ge 1,k \ge 1} \right\}$为关于$\left\{ {{\xi _n}:n \ge 1} \right\}$的u.a.n.序列, 给定$\tau > 0$, 记m_{n, k}为V_{n, k}的中位数, F_{n, k}为V_{n, k}的分布函数, 则:

1) $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} \left| {{m_{n,k}}} \right| = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$;

2) $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} \int_{\left| x \right| xy \tau } {{{\left| x \right|}^r}{\rm{d}}{F_{n,k}}\left( x \right) = 0} \;\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$, 其中常数r>0;

3) $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} \left| {{a_{n,k}}\left( \tau \right)} \right| = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$., 其中${a_{n,k}}\left( \tau \right) = \int_{\left| x \right| \le \tau } {x{\rm{d}}\;{F_{n,k}}\left( x \right)} $, 从而$\left\{ {{V_{n,k}} - {a_{n,k}}\left( \tau \right):k,n \ge 1} \right\}$也为关于$\left\{ {{\xi _n}:n \ge 1} \right\}$的u.a.n.序列;

4) $\mathop {\lim }\limits_{n \to \infty } \mathop {\max }\limits_{1 \le k \le {\xi _n}} \left| {\exp \left[ { - i{a_{n,k}}\left( \tau \right)t} \right]{f_{n,k}}\left( t \right) - 1} \right| = 0\;\;{\mathbb{\rm{a}}}{\rm{.s}}{\rm{.}}\;{\rm{,}}\;\forall t \in {\rm{R}}$.

证明   1)不妨设对于所有的$\omega \in \mathit{\Omega} ,{\xi _n}\left( \omega \right) \to \infty $当n→∞, 任给$ \epsilon > 0,\omega \in \mathit{\Omega} $, 存在正整数$N = N\left( {\epsilon,\omega } \right)$, 使得n>N时

$\mathop {\max }\limits_{1 \le k \le {\xi _n}\left( \omega \right)} P\left( {\left| {{V_{n,k}}} \right| \ge \epsilon} \right) < \frac{1}{2}$

而$P\left( {\left| {{V_{n,k}}} \right| \ge \epsilon} \right) < \frac{1}{2}$意味着$\left| {{m_{n,k}}} \right| \le \epsilon$, 故而

$\mathop {\max }\limits_{1 \le k \le {\xi _n}\left( \omega \right)} \left| {{m_{n,k}}} \right| < \epsilon,n \ge N,\omega \in \mathit{\Omega} ,$

由$\epsilon$的任意性得到1).

2)任给r>0, τ>0以及$0 < \epsilon < \tau $, 注意到

$\begin{array}{l} \mathop {\max }\limits_{1 \le k \le {\xi _n}} \int_{\left| x \right| < \tau } {{{\left| x \right|}^r}{\rm{d}}{F_{n,k}}\left( x \right)} \\ \le \mathop {\max }\limits_{1 \le K \le {\xi _n}} \int_{\left| x \right| < \epsilon} {{{\left| x \right|}^r}{\rm{d}}{F_{n,k}}\left( x \right) + } \\ \mathop {\max }\limits_{1 \le k \le {\xi _n}} \int_{\epsilon \le \left| x \right| < \tau } {{{\left| x \right|}^r}{\rm{d}}{F_{n,k}}\left( x \right)} \\ \le {\epsilon^r} + {\tau ^r}\mathop {\max }\limits_{1 \le k \le {\xi _n}} P\left( {\left| {{V_{n,k}}} \right| \ge \epsilon} \right), \end{array}$

再根据$\epsilon$的任意性和$\left\{ {{V_{n,k}}:n \ge 1,k \ge 1} \right\}$是关于$\left\{ {{\xi _n}:n \ge 1} \right\}$的u.a.n.序列这一假设得到结论.

3)在2)中取r=1即可.

4)由3)可知$\left\{ {{V_{n,k}} - {a_{n,k}}\left( \tau \right):k,n \ge 1} \right\}$也是一个关于$\left\{ {{\xi _n}:n \ge 1} \right\}$的u.a.n.序列, 其特征函数恰为$\exp \left\{ { - {\rm{i}}{a_{n,k}}\left( \tau \right)t} \right\}{f_{n,k}}\left( t \right)$, 故由命题2.1中的条件4)可知结论成立.

注2.2   该命题推广了经典u.a.n.序列的相应结论, 见文献[7].

定理2.1   令${\rm{\mathbb{V}}} = \left\{ {{V_{n,k}}:n \ge 1,k \ge 1} \right\}$为行内独立的二重随机变量序列, 取值于正整数的随机变量序列$\left\{ {{\xi _n}:n \ge 1} \right\}$满足$\mathop {\lim }\limits_{n \to \infty } {\xi _n} = \infty ,{\rm{a}}{\rm{.s}}{\rm{.}}$, 而且$\left\{ {{\xi _n}:n \ge 1} \right\}$与$\left\{ {{V_{n,k}}:n \ge 1,k \ge 1} \right\}$独立, 又设$\left\{ {{V_{n,k}}:n \ge 1,k \ge 1} \right\}$为关于$\left\{ {{\xi _n}:n \ge 1} \right\}$的u.a.n.序列, 记m_{n, k}为V_{n, k}的中位数, F_{n, k}为V_{n, k}的分布函数.若下面2个条件成立:

(a) $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^{{\xi _n}} {\int_{\left| x \right| > 1} {{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right) = 0\;{\rm{a}}{\rm{.s}}{\rm{.}}} } $;

(b) $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^{{\xi _n}} {\int_{\left| x \right| \le 1} {{x^2}{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right) = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.}}} } $;

则存在随机变量序列$\left\{ {{V_n},n \ge 1} \right\}$使得

$\sum\limits_{j=1}^{{{\xi }_{n}}}{{{V}_{n,k}}-{{V}_{n}}\xrightarrow{P}0,}\ 当n\to \infty .$

另一方面, 若存在形如h(ξ_n)的随机变量(h是Borel可测函数), 对任意的$\epsilon$>0均有

$\mathop {\lim }\limits_{n \to \infty } \left( {\left| {\sum\limits_{j = 1}^{{\xi _n}} {{V_{n,j}} - h\left( {{\xi _n}} \right)} } \right| \ge \epsilon\left| {{\xi _n}} \right.} \right) = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.,}}$

则(a)与(b)成立.

证明   首先证明第1部分.令

$\begin{align} & {{U}_{n}}=\sum\limits_{j=1}^{{{\xi }_{n}}}{{{V}_{n,j}},V_{n,k}^{\left( 1 \right)}={{V}_{n,k}}-{{m}_{n,k}},} \\ & F_{n,k}^{\left( 1 \right)}\left( x \right)=P\left. \left( V_{n,k}^{\left( 1 \right)}\le x \right) \right),V_{n,k}^{\left( 2 \right)}=V_{n,k}^{\left( 1 \right)}{{I}_{\left\{ \left| V_{n,k}^{\left( 1 \right)} \right|\le 1 \right\}}}, \\ & {{V}_{n}}=\sum\limits_{j=1}^{{{\xi }_{n}}}{\left( {{m}_{n,j}}+EV_{n,j}^{\left( 2 \right)} \right),}U_{n}^{\left( 1 \right)}=\sum\limits_{j=1}^{{{\xi }_{n}}}{V_{n,j}^{\left( 1 \right)},} \\ & U_{n}^{\left( 2 \right)}=\sum\limits_{j=1}^{{{\xi }_{n}}}{V_{n,j}^{\left( 2 \right)},{{B}_{n}}=\left\{ U_{n}^{\left( 1 \right)}=U_{n}^{\left( 2 \right)} \right\}.} \\ \end{align}$

由于$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^{{\xi _n}} {\int_{\left| x \right| > 1} {{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right) = 0} \;\;{\rm{a}}{\rm{.s}}{\rm{.}}} $且$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^{{\xi _n}} {\int_{\left| x \right| \le 1} {{x^2}{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right) = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.}}} } $, 故对任意的η>0, 存在${n_0} = {n_0}\left( \eta \right) \in {\rm{\mathbb{N}}}$, 当n≥n₀时, 令${\mathit{\Omega} _0}: = \left\{ {\left( {\int_{\left| x \right| > 1} {{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right)} } \right) \vee \left( {\int_{\left| x \right| \le 1} {{x^2}{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right)} } \right) \le 1} \right\}$满足P(Ω₀)≥1-η, 注意到Ω₀∈σ(ξ_n, n≥1), 故而Ω₀与{V_{n, k}, n≥1, k≥1}独立, 因此

$\begin{array}{l} P\left( {B_n^c \cap {\mathit{\Omega} _0}} \right)\\ = P\left( {\sum\limits_{j = 1}^{{\xi _n}} {V_{n,j}^{\left( 1 \right)} \ne \sum\limits_{j = 1}^{{\xi _n}} {V_{n,k}^{\left( 2 \right)},{\xi _n} = k,{\mathit{\Omega} _0}} } } \right)\\ = \sum\limits_{k = 1}^\infty {P\left( {{\xi _n} = k,{\mathit{\Omega} _0}} \right)P\left( {\sum\limits_{j = 1}^k {V_{n,j}^{\left( 1 \right)} \ne \sum\limits_{j = 1}^k {V_{n,j}^{\left( 2 \right)}} } } \right)} \\ \le \sum\limits_{k = 1}^\infty {P\left( {{\xi _n} = k,{\mathit{\Omega} _0}} \right)\sum\limits_{j = 1}^k {P\left( {V_{n,j}^{\left( 1 \right)} \ne V_{n,j}^{\left( 2 \right)}} \right)} } \\ = \sum\limits_{k = 1}^\infty {P\left( {{\xi _n} = k,{\mathit{\Omega} _0}} \right)\sum\limits_{j = 1}^k {P\left( {\left| {V_{n,j}^{\left( 1 \right)}} \right| > 1} \right)} } \\ = \sum\limits_{k = 1}^\infty {P\left( {{\xi _n} = k,{\mathit{\Omega} _0}} \right)\sum\limits_{j = 1}^k {\int_{\left| x \right| > 1} {{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right)} } }\\ = E\left( {\sum\limits_{j = 1}^{{\xi _n}} {\int_{\left| x \right| > 1} {{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right),{\mathit{\Omega} _0}} } } \right), \end{array}$

(1)

再利用控制收敛定理可得, $\mathop {\lim }\limits_{n \to \infty } P\left( {B_n^c \cap {\mathit{\Omega} _0}} \right) = 0.$注意:

$\begin{array}{c} P\left( {\left| {{U_n} - {V_n}} \right| \ge \epsilon,{\mathit{\Omega} _0}} \right)\\ = P\left[ {\left\{ {\left| {{U_n} - {V_n}} \right| \ge \epsilon} \right\} \cap {B_n} \cap {\mathit{\Omega} _0}} \right] + \\ P\left[ {\left\{ {\left| {{U_n} - {V_n}} \right| \ge \epsilon} \right\} \cap B_n^c \cap {\mathit{\Omega} _0}} \right], \end{array}$

(2)

前面已证等式右边第2项收敛到0.对第1项利用Chebyshev不等式, 可以得到

$\begin{array}{l} P\left[ {\left\{ {\left| {{U_n} - {V_n}} \right| \ge \epsilon} \right\} \cap {B_n} \cap {\mathit{\Omega} _0}} \right]\\ = P\left[ {\left\{ {\left| {\sum\limits_{j = 1}^{{\xi _n}} {\left( {V_{n,k}^{\left( 2 \right)} - EV_{n,k}^{\left( 2 \right)}} \right)} } \right| \ge \epsilon} \right\} \cap {B_n} \cap {\mathit{\Omega} _0}} \right]\\ \le \sum\limits_{k = 1}^\infty {P\left( {{\xi _n} = k,{\mathit{\Omega} _0}} \right)P\left( {\left| {\sum\limits_{j = 1}^k {\left( {V_{n,k}^{\left( 2 \right)} - EV_{n,j}^{\left( 2 \right)}} \right)} } \right| \ge \epsilon} \right)} \\ \le \sum\limits_{k = 1}^\infty {P\left( {{\xi _n} = k,{\mathit{\Omega} _0}} \right)\frac{1}{{{\epsilon^2}}}\sum\limits_{j = 1}^k {{\rm{Var}}\left( {V_{n,k}^{\left( 2 \right)}} \right)} } \\ = \frac{1}{{{\epsilon^2}}}E\left( {\sum\limits_{k = 1}^{{\xi _n}} {{\rm{Var}}\left( {V_{n,j}^{\left( 2 \right)}} \right),{\mathit{\Omega} _0}} } \right)\\ = \frac{1}{{{\epsilon^2}}}E\left( {\sum\limits_{j = 1}^{{\xi _n}} {\int_{\left| x \right| \le 1} {{x^2}{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right),{\mathit{\Omega} _0}} } } \right), \end{array}$

因为在Ω₀上$\left| {\sum\limits_{j = 1}^{{\xi _n}} {\int_{\left| x \right| \le 1} {{x^2}{\rm{d}}{F_{n,j}}\left( {x + {m_{n,j}}} \right)} } } \right| \le 1$, 所以由控制收敛定理知(2)式右端第1项亦收敛于0.总之

$\mathop {\lim }\limits_{n \to \infty } P\left( {\left| {{U_n} - {V_n}} \right| \ge \epsilon} \right) = 0.$

因此

$\sum\limits_{j=1}^{{{\xi }_{n}}}{{{V}_{n,j}}-{{V}_{n}}}\xrightarrow{P}0,n\to \infty .$

另一方面, 对于每个$\epsilon$>0, 由条件

$P\left( \left| \sum\limits_{j=1}^{{{\xi }_{n}}}{{{V}_{n,j}}-h\left( {{\xi }_{n}} \right)} \right|\ge \epsilon \left| {{\xi }_{n}} \right. \right)\to 0\ \ \text{a}\text{.s}\text{.}$

可得

$P\left( \sum\limits_{j=1}^{{{\xi }_{n}}}{{{V}_{n,j}}-h\left( {{\xi }_{n}} \right)\le x\left| {{\xi }_{n}} \right.} \right)\text{a}\text{.s}\text{.}\left\{ \begin{align} & 1,x>0, \\ & 0, < 0. \\ \end{align} \right.$

故任取T>0, 下式对u∈I=[-T, T]一致成立

$\mathop {\lim }\limits_{n \to \infty ]} E{\mkern 1mu} \left( {\exp \left( {{\rm{i}}u\left( {\sum\limits_{j = 1}^{{\xi _n}} {{V_{n,j}} - h\left( {{\xi _n}} \right)} } \right)} \right)\left| {{\xi _n}} \right.} \right)\left( \omega \right) = 1\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$

而

$\begin{align} & E\left( \exp \left( \text{i}u\left( \sum\limits_{j=1}^{{{\xi }_{n}}}{Vn,j-h\left( {{\xi }_{n}} \right)} \right) \right)\left| {{\xi }_{n}} \right. \right) \\ & \ \ \ =\exp \left\{ -\text{i}uh\left( {{\xi }_{n}} \right) \right\}\prod\limits_{j=1}^{{{\xi }_{n}}}{{{f}_{n,j}}\left( u \right)}, \\ \end{align}$

其中，f_{n, j}(u)为V_{n, j}的特征函数, 于是对于a.s.ω, $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^{{\xi _n}} {1 - {{\left| {{f_{n,j}}\left( u \right)} \right|}^2} = 1} $在I上一致成立.对上式两端取对数可得: $\forall u \in I$,

$\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{j=1}^{{{\xi }_{n}}}{\log }\left( 1-\left( 1-{{\left| {{f}_{n,j}}\left( u \right) \right|}^{2}} \right) \right)=0\ \ \text{a}\text{.s}\text{.}$

(3)

根据命题2.1, 对于a.s.ω, $\mathop {\max }\limits_{1 \le j \le {\xi _n}} \left| {1 - {{\left| {{f_{n,j}}\left( u \right)} \right|}^2}} \right|$在I上一致地收敛到零(当n→∞), 利用不等式$ - \log \left( {1 - x} \right) \ge x\left( {0 < x < 1} \right)$可得:对a.s.ω,

$\begin{align} & \text{-}\sum\limits_{j=1}^{{{\xi }_{n}}}{\log \left( 1-\left( 1-{{\left| {{f}_{n,j}}\left( u \right) \right|}^{2}} \right) \right)\ge \sum\limits_{j=1}^{{{\xi }_{n}}}{\left( 1-{{\left| {{f}_{n,j}}\left( u \right) \right|}^{2}} \right)}} \\ & \ \ \ge 0,\forall u\in I,n\ge {{n}_{0}}={{n}_{0}}\left( \omega \right), \\ \end{align}$

利用式(3)有:对于a.s.ω, $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^{{\xi _n}} {\left( {1 - {{\left| {{f_{n,j}}\left( u \right)} \right|}^2}} \right) = 0\left( {对u \in I一致} \right)} $.取I=[-1, 1], 由一致收敛性有

$\underset{n\to \infty }{\mathop{\lim }}\,\int_{-1}^{1}{\sum\limits_{j=1}^{{{\xi }_{n}}}{\left( 1-{{\left| {{f}_{n,j}}\left( u \right) \right|}^{2}} \right)\text{d}u=0\ \ \ \text{a}\text{.s}\text{.}}}$

(4)

令$\left\{ {{{\bar V}_{n,k}},n \ge 1,k \ge 1} \right\}$为{V_{n, k}}的一个独立复制.令G_{n, j}(x)为$\left( {{{\bar V}_{n,j}} - {V_{n,j}}} \right)$的分布, 则${\left| {{f_{n,j}}\left( u \right)} \right|^2}为{G_{n,j}}\left( x \right)$对应的特征函数, 由Fubini定理得到

$\int_{-1}^{1}{\left( 1-{{\left| {{f}_{n,j}}\left( u \right) \right|}^{2}} \right)}\text{d}u=2\int_{-\infty }^{\infty }{\left( 1-\frac{\sin \ x}{x} \right)\text{d}\ {{G}_{n,j}}\left( x \right).}$

接下来分析一些简单的估计, 首先存在正常数K₁, 使得|x|>1时, $1 - \frac{{\sin \;x}}{x} \ge {K_1} > 0$.对于|x|≤1存在正常数K₂使得

$1-\frac{\sin x}{x}=1-\frac{1}{x}\left( x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-+\cdots \right)\ge {{K}_{2}}{{x}^{2}}.$

令K₃=min{K₁, K₂}>0, 令$\psi \left( x \right) = {x^2}{I_{\left\{ {\left| x \right| \le 1} \right\}}} + {I_{\left\{ {\left| x \right| > 1} \right\}}}$, 由文献[2]中169页引理1, $E\psi \left( {{{\bar V}_{n,j}} - {V_{n,j}}} \right) \ge \frac{1}{2}E\psi \left( {{{\bar V}_{n,j}} - {m_{n,j}}} \right)$则

$\begin{align} & \int\limits_{-1}^{1}{\sum\limits_{j=1}^{{{\xi }_{n}}}{\left( 1-{{\left| {{f}_{n,j}}\left( u \right) \right|}^{2}} \right)\text{d}u}} \\ & \ge {{K}_{3}}\left\{ \sum\limits_{j=1}^{{{\xi }_{n}}}{\int_{\left| x \right|\le 1}{{{x}^{2}}\text{d}{{G}_{n,j}}\left( x \right)+\sum\limits_{j=1}^{{{\xi }_{n}}}{\int_{\left| x \right|>1}{\text{d}{{G}_{n,j}}\left( x \right)}}}} \right\} \\ & ={{K}_{3}}\sum\limits_{j=1}^{{{\xi }_{n}}}{E\psi \left( {{{\bar{V}}}_{n,j}}-{{V}_{n,j}} \right)\ge \frac{{{K}_{3}}}{2}\sum\limits_{j=1}^{{{\xi }_{n}}}{E\psi \left( {{{\bar{V}}}_{n,j}}-{{m}_{n,j}} \right)}} \\ & =\frac{{{K}_{3}}}{2}\sum\limits_{j=1}^{{{\xi }_{n}}}{\left\{ \int_{\left| x \right|>1}{\text{d}{{F}_{n,j}}\left( x+{{m}_{n,j}} \right)}+ \right.} \\ & \left. \int_{\left| x \right|\le 1}{{{x}^{2}}\text{d}{{F}_{n,j}}\left( x+{{m}_{n,j}} \right)} \right\}\ge 0. \\ \end{align}$

由于括号内两式均非负, 对两边取极限得到(a)和(b).

下面讨论另一种情形.仍然沿用前面的符号.设{Z_n:n≥0}为定义1.1所定义的变化环境中的上临界的分枝过程, 记${Y_{n,k}} = \frac{{{X_{n,k}} - {\mu _n}}}{{f\left( {{Z_n}} \right)}}:n \ge 1,k \ge 1.$. f是单调上升函数且$\mathop {\lim }\limits_{x \to \infty } \left( x \right) = \infty $.记ν_{n, k}为Y_{n, k}关于Z_n的条件中位数(即$P\left( {{Y_{n,k}} \ge {v_{n,k}}\left| {{Z_n}} \right.} \right) \ge \frac{1}{2},P\left( {{Y_{n,k}} \le {v_{n,k}}\left| {{Z_n}} \right.} \right) \ge \frac{1}{2}$), F_{n, k}为Y_{n, k}关于Z_n的条件分布函数(即${F_{n,k}}\left( x \right) = P\left( {{Y_{n,k}} \le x\left| {{Z_n}} \right.} \right)$).

定理2.2   假设Z_n→∞a.s., 当n→∞, 若下面两个条件成立:

(a′)$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^{{Z_n}} {\int_{\left| x \right| > 1} {{\rm{d}}{F_{n,j}}\left( {x + {v_{n,j}}} \right) = 0} \;{\rm{a}}{\rm{.s}}{\rm{.}}} $;

(b′) $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^{{Z_n}} {\int_{\left| x \right| \le 1} {{x^2}{\rm{d}}{F_{n,j}}\left( {x + {v_{n,j}}} \right) = 0} \;{\rm{a}}{\rm{.s}}{\rm{.}}} $;

则存在随机变量序列$\left\{ {{H_n}:n \ge 1} \right\}$使得:对于任意的$\epsilon > 0$,

$\mathop {\lim }\limits_{n \to \infty } P\left( {\left| {\sum\limits_{j = 1}^{{Z_n}} {{Y_{n,j}} - {H_n}} } \right| \ge \epsilon\left| {{Z_n}} \right.} \right) = 0\;\;\;{\rm{a}}.{\rm{s}}.,$

进而

$\sum\limits_{j=1}^{{{Z}_{n}}}{{{Y}_{n,j}}-{{H}_{n}}\xrightarrow{P}0,}为n\to \infty .$

证明   首先令

$\begin{align} & Y_{n,k}^{\left( 1 \right)}={{Y}_{n,k}}-{{v}_{n,k}},F_{n,k}^{\left( 1 \right)}\left( x \right)=P\left( Y_{n,k}^{(1)}\le x\left| {{Z}_{n}} \right. \right), \\ & Y_{n,k}^{\left( 2 \right)}=Y_{n,k}^{\left( 1 \right)}{{I}_{\left\{ \left| Y_{n,k}^{\left( 1 \right)} \right|\le 1 \right\}}},{{H}_{n}}=\sum\limits_{j=1}^{{{Z}_{n}}}{\left( {{v}_{n,j}}+EY_{n,j}^{\left( 2 \right)} \right),} \\ & {{Y}_{n}}=\sum\limits_{j=1}^{{{Z}_{n}}}{{{Y}_{n,j}},Y_{n}^{\left( 1 \right)}=\sum\limits_{j=1}^{{{Z}_{n}}}{Y_{n,j}^{\left( 1 \right)}},} \\ & Y_{n}^{\left( 2 \right)}=\sum\limits_{j=1}^{{{Z}_{n}}}{Y_{n,j}^{\left( 2 \right)},{{C}_{n}}=\left\{ Y_{n}^{\left( 1 \right)}=Y_{n}^{\left( 2 \right)} \right\}.} \\ \end{align}$

注意到

$\begin{align} & P\left( C_{n}^{c}\left| {{Z}_{n}} \right. \right)=P\left( Y_{n}^{\left( 1 \right)}\ne Y_{n}^{\left( 2 \right)}\left| {{Z}_{n}} \right. \right), \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =P\left( \sum\limits_{j=1}^{{{Z}_{n}}}{Y_{n,j}^{\left( 1 \right)}\ne \sum\limits_{j=1}^{{{Z}_{n}}}{Y_{n,j}^{\left( 2 \right)}\left| {{Z}_{n}} \right.}} \right), \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \le \sum\limits_{k=1}^{{{Z}_{n}}}{P\left( Y_{n,j}^{\left( 1 \right)}\ne Y_{n,j}^{\left( 2 \right)}\left| {{Z}_{n}} \right. \right),} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{j=1}^{{{Z}_{n}}}{\int_{\left| x \right|>1}{\text{d}{{F}_{n,j}}\left( x+{{v}_{n,j}} \right),}} \\ \end{align}$

根据(a′)知, $\mathop {\lim }\limits_{n \to \infty } P\left( {C_n^c\left| {{Z_n}} \right.} \right) = 0\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$.又有

$\begin{align} & P\left( \left\{ \left| \sum\limits_{k=1}^{{{Z}_{n}}}{{{Y}_{n,j}}-{{H}_{n}}} \right|\ge \epsilon \right\},{{C}_{n}}\left| {{Z}_{n}} \right. \right) \\ & =P\left( \left| \sum\limits_{j=1}^{{{Z}_{n}}}{\left( Y_{n,j}^{\left( 2 \right)}-EY_{n,j}^{\left( 2 \right)} \right)\left| \ge \epsilon,{{C}_{n}} \right.} \right|{{Z}_{n}} \right) \\ & \le P\left( \left| \sum\limits_{j=1}^{{{Z}_{n}}}{\left( Y_{n,j}^{\left( 2 \right)}-EY_{n,j}^{\left( 2 \right)} \right)} \right|\ge \epsilon\left| {{Z}_{n}} \right. \right) \\ & \le \sum\limits_{k=1}^{{{Z}_{n}}}{P\left( \left| Y_{n,j}^{\left( 2 \right)}-EY_{n,j}^{\left( 2 \right)} \right|\ge \epsilon\left| {{Z}_{n}} \right. \right)} \\ & =\sum\limits_{k=1}^{\infty }{\frac{1}{{{\epsilon}^{2}}}\text{Var}\left( Y_{n,j}^{\left( 2 \right)}\left| {{Z}_{n}} \right. \right)} \\ & \le \sum\limits_{k=1}^{\infty }{\frac{1}{{{\epsilon}^{2}}}E\left( {{\left| Y_{n,j}^{\left( 2 \right)} \right|}^{2}}\left| {{Z}_{n}} \right. \right)} \\ & =\frac{1}{{{\epsilon}^{2}}}\sum\limits_{k=1}^{\infty }{\int_{\left| x \right|\le 1}{{{x}^{2}}\text{d}{{F}_{n,j}}\left( x+{{v}_{n,j}} \right),}} \\ \end{align}$

由(b′), $\mathop {\lim }\limits_{n \to \infty } P\left( {\left| {\sum\limits_{j = 1}^{{Z_n}} {{Y_{n,j}} - {H_n}} } \right| \ge \epsilon,{C_n}\left| {{Z_n}} \right.} \right) = 0\;\;{\rm{a}}{\rm{.s}}$总之

$\mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} P\left( {\left| {\sum\limits_{j = 1}^{{Z_n}} {{Y_{n,j}} - {H_n}} } \right| \ge \epsilon\left| {{Z_n}} \right.} \right) = 0\;\;\;{\rm{a}}{\rm{.s}}{\rm{.}}$

利用控制收敛定理得

$\sum\limits_{j=1}^{{{Z}_{n}}}{{{Y}_{n,j}}-{{H}_{n}}\xrightarrow{P}0,当n\to \infty .}$