数学物理学报, 2026, 46(1): 305-317

研究论文

爆炸状态带线性漂移项 Ornstein-Uhlenbeck 过程的精细大偏差

李沁文,, 赵守江,*

三峡大学数理学院 湖北宜昌 443002

三峡大学三峡数学研究中心 湖北宜昌 443002

Sharp Large Deviations of the Non-Stationary Ornstein-Uhlenbeck Process with Linear Drift

Li Qinwen,, Zhao Shoujiang,*

School of Mathematics and Physics, China Three Gorges University, Hubei Yichang 443002

Three Gorges Mathematical Research Center, China Three Gorges University, Hubei Yichang 443002

通讯作者: *赵守江, Email: shjzhao@163.com

收稿日期: 2025-01-21   修回日期: 2025-04-21  

基金资助: 国家自然科学基金(11601267)

Received: 2025-01-21   Revised: 2025-04-21  

Fund supported: NSFC(11601267)

作者简介 About authors

李沁文,Email:2943769958@qq.com

摘要

Ornstein-Uhlenbeck (O-U) 过程作为一种重要的扩散过程, 在统计学、金融学、物理学等领域起重要作用. 该文利用测度变换技巧, 研究爆炸状态下带线性漂移项 O-U 过程的精细大偏差, 给出了极大似然估计尾概率的精细刻画. 作为应用得到了大偏差原理, 结果表明爆炸状态下有无线性漂移项 O-U 过程的极大似然估计具有相同的大偏差原理.

关键词: Ornstein-Uhlenbeck 过程; 极大似然估计; 精细大偏差

Abstract

The Ornstein-Uhlenbeck (O-U) process, as an important diffusion process, plays a significant role in fields such as statistics, finance, and physics. In this paper, the sharp large deviations of the maximum likelihood estimation for the O-U process with linear drift in the explosive cases are studied by change of measure, and a refined characterization of the tail probability is obtained. As an application, the large deviation principle is obtained. The results demonstrate that the maximum likelihood estimations of the O-U process with and without linear drift in the explosive cases have the same large deviations.

Keywords: Ornstein-Uhlenbeck process; maximum likelihood estimation; sharp large deviations

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本文引用格式

李沁文, 赵守江. 爆炸状态带线性漂移项 Ornstein-Uhlenbeck 过程的精细大偏差[J]. 数学物理学报, 2026, 46(1): 305-317

Li Qinwen, Zhao Shoujiang. Sharp Large Deviations of the Non-Stationary Ornstein-Uhlenbeck Process with Linear Drift[J]. Acta Mathematica Scientia, 2026, 46(1): 305-317

1 引言与主要结果

考虑下列带线性漂移项的 Ornstein-Uhlenbeck (O-U) 过程

${\rm d}X_{t} = \theta X_{t}{\rm d}t+\gamma {\rm d}t+{\rm d}B_{t}, X_{0}=0,$

其中 $\theta$, $\gamma $ 为未知参数, $\left \{ {B_{t}}, t \in \left [ 0,+\infty \right ) \right \} $ 是标准布朗运动. O-U 过程在物理学、生物学、统计学、金融学等诸多领域应用广泛, 常被用于模拟受随机干扰的动力系统的演变以及刻画生物和控制论中的随机现象, 具体详情可参考文献 [1,2]. 未知参数 $\theta $$\gamma$ 的极大似然估计量 (MLE) 为

$\widehat{\theta }_{T}=\frac{T \int_{0}^{T} X_{t}{\rm d}X_{t}-X_{T} \int_{0}^{T} X_{t}{\rm d}t}{T \int_{0}^{T} X_{t}^{2}{\rm d}t-\left(\int_{0}^{T} X_{t}{\rm d}t\right)^{2}}\quad \text{和}\quad \widehat{\gamma }_{T}=\frac{X_{T} \int_{0}^{T} X_{t}^{2}{\rm d}t-\int_{0}^{T} X_{t}{\rm d}X_{t} \int_{0}^{T} X_{t}{\rm d}t}{T \int_{0}^{T} X_{t}^{2}{\rm d}t-\left(\int_{0}^{T} X_{t}{\rm d}t\right)^{2}}.$

$\theta < 0$ 时, O-U 过程是平稳的. Kutoyants 在文献 [3] 中研究了 $\widehat{\theta }_{T}$$\widehat{\gamma }_{T}$ 的联合渐近分布

$\sqrt{T}\binom{\widehat{\theta }_{T}-\theta }{\widehat{\gamma }_{T}-\gamma } \xrightarrow{L} N(0, L),$

其中

$L=2\left(\begin{array}{ll} \theta & \quad \gamma \\ \gamma & \frac{2\gamma ^{2} + \theta }{2\theta} \end{array}\right).$

Jiang 在文献 [4] 中研究了带线性漂移项 O-U 过程 MLE 的重对数律和 Berry-Esseen 界, Jiang 和 Xie 在文献 [5] 中探究了该过程中轨道滤波估计量的渐近性质, 文献 [6,7] 分析了带线性漂移项 O-U 模型对数似然比过程的渐近性质. Gao 和 Jiang 在文献 [8] 中借助偏差不等式得到了 MLE 的中偏差, Bercu 和 Richou 在文献 [9] 中探讨了带线性漂移项 O-U 过程的极大似然估计的大偏差性质. 对于 $\gamma = 0$ 这一特殊情形, Florens-Landais 和 Pham 在文献 [10] 中通过 Gärtner-Ellis 定理获得了 MLE 大偏差原理, Bercu 和 Rouault 在文献 [11] 中给出了它的精细大偏差, Bercu 和 Richou 在文献 [12] 中利用测度变换技巧构建了 O-U 过程漂移系数的 MLE 大偏差, 有关中偏差结果可参见文献 [13,14].

$\theta > 0$ 时, O-U 过程是爆炸的. Jiang 和 Dong 在参考文献 [15] 中研究了非平稳情形下带线性漂移项 O-U 过程的参数估计问题, 得到了 $\widehat{\theta }_{T}$ 的渐近行为 (与平稳状态下完全不同)

$$ {\rm e}^{\theta T}(\widehat{\theta }_{T}-\theta)\xrightarrow{L}\frac{\sqrt{2\theta}\nu}{\eta-\frac{\gamma}{\theta}},$$

其中 $\nu$$\eta$ 是两个独立的高斯随机变量.

Bercu, Coutin 和 Savy 在文献 [16] 中论证了 $\gamma = 0$ 情形下的精细大偏差原理. 受文献 [16] 的启发, 本文将文献 [9] 中结果推广至爆炸状态, 研究带线性漂移项 O-U 过程极大似然估计的精细大偏差, 提供了比中心极限定理, 大偏差等极限理论更精细的数值计算逼近. 作为推论还得到了它的大偏差结果, 结果表明爆炸状态下无论 O-U 过程是否存在线性漂移项, 极大似然估计 $\widehat{\theta }_{T}$ 均具有相同的大偏差原理.

$I_{\theta} (x)=\left\{\begin{aligned}-\frac{(x-\theta )^{2}}{4x},& \, \text{ 当} x\le-\theta,\\\theta, & \text{ 当}\left | x \right|<\theta, \\0, & \text{ 当} x=\theta, \\2x-\theta,& \text{ 当} x>\theta.\end{aligned}\right.$

下面我们给出极大似然估计的精细大偏差结果.

定理 1.1 考虑由 $(1.1)$ 式给出的带线性漂移项 O-U 过程, 其中 $\theta > 0$.

(a) 当 $c < -\theta$ 时, 对于足够大的 $T$, 有

$P(\widehat{\theta }_T \leq c) = \frac{-\exp\{-TI_{\theta }(c) + H(a_c)\}}{a_c \sigma_c \sqrt{2 \pi T}}\left[ 1+\frac{b_{c,1}}{T} + O \left ( \frac{1}{T^{2} }\right )\right],$

其中

$a_c=\frac{c^2-\theta ^2}{2c}, \quad \sigma_c^2=-\frac{1}{2c}, \quad H(a_c)=-\frac{1}{2} \log \left(\frac{\theta ^2(c+\theta )(3c-\theta )}{4c^4}\right)+\frac{\gamma ^2(c-\theta )^2}{4c\theta ^2},$

系数 $b_{c,1}$ 可由 $\Lambda$$H$ (见引理 $2.1$)$a_{c}$ 处的导数表示

$\begin{aligned}b_{c,1}&=\frac{1}{\sigma _{c}^{2} }\left(-\frac{H_{2}}{2}-\frac{H_{1}^{2}}{2}+\frac{\Lambda _{4}}{8\sigma _{c}^{2}}+\frac{\Lambda _{3}H_{1}}{2\sigma _{c}^{2}}-\frac{5\Lambda _{3}^{2}}{24\sigma _{c}^{4}}+\frac{H_{1}}{a_{c}}-\frac{\Lambda _{3}}{2a_{c}\sigma _{c}^{2}}-\frac{1}{a_{c}^{2}}\right)\\&=\frac{G(c)}{-4c(c^{2}-\theta ^{2})^{2}(3c-\theta )^{2}}+\frac{\gamma ^{2}}{2\theta ^{2}},\end{aligned}$

其中 $\Lambda _{n} =\Lambda^{(n)} (a_{c} )$, $H _{n} =H^{(n)} (a_{c} )$,

$\begin{aligned}G(c)&=8c^{2}\left ( -7c^{2} +3\theta ^{2} \right )\left ( 3c-\theta \right )^{2}+4c^{2}\left ( c-\theta \right )^{2} \left ( -5c^{2}+2\theta c-\theta ^{2} \right )-30\left ( 3c-\theta \right ) ^{2}\left ( c^{2}-\theta ^{2} \right )^{2} \\&\quad +\left ( c-\theta \right )\left ( 5c^{2}+6\theta c-3\theta ^{2} \right )\left ( 37c^{3}-15\theta c^{2}-9\theta ^{2}c+3\theta ^{3} \right ).\end{aligned}$

(b) 当 $c > \theta$ 时, 对于足够大的 $T$, 存在系数 $d_{c, 1}$ (其取值与 $\Lambda$, $H$$a_c$ 处的导数有关),

$P(\widehat{\theta }_T \geq c) =\frac{\exp\{-TI_{\theta }(c) + K(c)\}}{a_c \sigma_c\sqrt{2 \pi T}}\left[1+\frac{b_{c,1}}{T} +O \left ( \frac{1}{T^{2} } \right ) \right],$

其中

$$a_c=2(c-\theta ),\quad\sigma_c^2= \frac{c^2}{2(2c-\theta )^3}, \quad K(c)=-\frac{1}{2} \log \left(\frac{\theta ^2(c-\theta )(3c-\theta )}{4c^2(2c-\theta )^2}\right)-\frac{\gamma ^2(2c-\theta )}{\theta ^2}.$$

(c) 当 $\vert c\vert < \theta$$c\neq 0$ 时, 对于足够大的 $T$, 存在系数 $e_{c, 1}$ (其取值与 $\Lambda$, $H$$a_c$ 处的导数有关),

$P(\widehat{\theta }_T \leq c) =\frac{\exp\{-TI_{\theta }(c) + J(c)\}}{a_c \sigma_c\sqrt{2 \pi T}}\left[ 1 +\frac{e_{c, 1} }{T}+O\left ( \frac{1}{T^{2} } \right ) \right],$

其中

$$a_c=\frac{\theta }{c+\theta },\quad\sigma_c^2= \frac{c^2}{2\theta ^3}, \quad J(c)=-\frac{1}{2} \log \left(\frac{(\theta -c)(c+\theta )}{4c^2}\right)-\frac{\gamma ^2}{\theta }.$$

(d) 当 $c = -\theta$ 时, 对于足够大的 $T$, 存在系数 $g_{c, 1} $ (其取值与 $\Lambda$, $H$$a_c$ 处的导数有关),

$P(\widehat{\theta }_T \leq c) =\frac{\exp\{-TI_{\theta }(c)-\frac{\gamma ^2}{\theta } \}}{2 \pi T^{1/4}}\frac{\Gamma (1/4)}{a_{\theta }^{3/4}\sigma_{\theta }}\left[1+\frac{g_{c,1}}{\sqrt{T} } +O\left ( \frac{1}{T } \right ) \right],$

其中

$a_{\theta }=\sqrt{\theta },\quad\sigma_{\theta }^2=\frac{1}{2\theta }.$

推理 1.1$\theta >0$ 时, $\widehat{\theta }_{T}$ 满足速度为 $T$, 速率函数为 $I_{\theta} (x)$ 的大偏差原理.

$\textbf{注 1.1}$ 爆炸状态下, 带线性漂移项 O-U 模型极大似然估计 $\widehat{\theta }_{T}$ 的大偏差原理与 $\gamma =0$ 这一特殊情形相同[16].

2 引理

$\overline{X}_{T}=\frac{1}{T} \int_{0}^{T} X_{t}{\rm d}t, \quad S_{T} =\int_{0}^{T} (X_{t}-\overline{X}_{T})^{2}{\rm d}t,$

$\hat{\theta }_{T}=\frac{\int_{0}^{T}\left ( X_{t}-\overline{X}_{T} \right ){\rm d}X_{t}}{S_{T} }.$

对于任意 $c\in R$, 有

$P(\hat{\theta }_{T}\le c )=P\left (\int_{0}^{T}\left ( X_{t}-\overline{X}_{T} \right ){\rm d}X_{t}-cS_{T} \le 0 \right).$

$\hspace{-1cm} Z_{T}(c)= \int_{0}^{T} (X_{t}-\overline{X}_{T}){\rm d}X_{t}-cS_{T}, \quad \varphi (a)=-\sqrt{\theta ^{2}+2ac},$
$\frac{{\rm d}P_{\varphi,0}}{{\rm d}P_{\theta,\gamma}}=\exp\left \{ \left ( \varphi -\theta \right )\int_{0}^{T} X_{t}{\rm d}X_{t}-\frac{1}{2} \left ( \varphi ^{2} -\theta ^{2} \right )\int_{0}^{T}X_{t}^{2}{\rm d}t+\zeta _{T} \right \},$

其中 $\zeta _{T}=\theta \gamma T\overline{X}_{T}-\gamma X_{T} +\frac{1}{2} \gamma ^{2}T$, 我们考虑 $Z_{T}(c)$ 的对数矩母函数

$\begin{aligned}\Lambda _{T}(a) &=\frac{1}{T} \log{E\left ( \exp\left \{ a Z_{T}(c)\right \} \right ) }\\&=\frac{1}{T} \log_{}{E_{\varphi,0 } \left ( \exp \left \{ (a+\theta +\varphi )\int_{0}^{T}X_{t}{\rm d}X_{t}+\frac{1}{2} (-2ac-\theta ^{2}+\varphi ^{2} )\int_{0}^{T}X_{t}^{2}{\rm d}t-\xi _{T} \right \}\right ) },\end{aligned}$

其中 $E_{\varphi, 0}$ 表示关于概率测度 $P_{\varphi, 0}$ 的期望, $ \xi _{T}=aX_{T}\overline{X}_{T}+\zeta _{T}-acT(\overline{X}_{T})^{2}. $ 利用参考文献 [9,引理 $3.2$] 完全相同的方法我们得到如下引理.

引理 2.1

$\varphi (a)=-\sqrt{\theta ^{2}+2ac},\quad h(a)=\frac{a+\theta }{\varphi(a)}.$

对于任意的 $a\in D_{\Lambda }$, 当 $T$ 足够大时有,

$\Lambda_{T}(a)= \Lambda(a)+\frac{1}{T} H(a)+\frac{1}{T^{2}}R_{T}(a),$

其中

$\Lambda (a)=-\frac{1}{2}(a+\theta +\sqrt{\theta ^{2} +2ac}),$
$H(a)=-\frac{1}{2}\log\left ( \frac{1 }{2}\left ( 1+h(a) \right ) \right ) +\frac{1}{2}\log\left ( 1+\frac{2ac}{\theta ^{2} } \right ) -\frac{\gamma ^{2}(a+\theta -\varphi (a))}{2\theta ^{2} },$
$D_{\Lambda} =\left \{ a\in R \mid \theta ^{2}+2ac >0 \text{且} a+\theta <\sqrt{\theta ^{2}+2ac}\right \}.$

余项 $R_T(\cdot )$ 为可显式计算的有理函数, 将其拓展至二维复平面后, 当 $Re(a)\in D_{\Lambda}$ 时为有界解析函数.

$\Delta_{\Lambda_{T} }= \left \{ z\in C, Re(z)\in D_{\Lambda_{T} } \right \}$, 关于 $\Lambda_{T}(\cdot )$ 我们有如下估计.

引理 2.2$T$ 足够大时, 对任意的 $(a, u) \in R^2$, 使得 $a + iu \in \Delta_{\Lambda_{T} }$, 存在正常数 $C_0$, 满足

$\begin{aligned}\left | \exp\{T(\Lambda _{T}(a+iu)- \Lambda _{T}(a))\} \right | ^{2}\le& C _{0}\ell (a)\left ( 1+\frac{4c^{2}u^{2} }{\varphi ^{4}(a) } \right ) ^{\frac{3}{4} } \\& \times \exp\left \{ \left ( \frac{\gamma^{2} }{\theta ^{2} }+T\right ) \frac{c^{2}u^{2} }{4\varphi ^{3}(a) }\left ( 1+\frac{4c^{2}u^{2} }{\varphi ^{4}(a) } \right ) ^{-\frac{3}{4} } \right \},\end{aligned}$

其中

$\ell (a)=\max\left \{ 1, \frac{\left | \varphi (a)+\theta \right | }{\left | \varphi (a) \right | } \right \}\times \max\left \{ 1, \frac{\left | \varphi (a)+2c-\theta \right | }{\left | \varphi (a) \right | } \right \}.$

引理 2.2 的证明见附录.

3 主要结论的证明

情况 1 定理 1.1(a) 的证明

$c<-\theta $ 时, 由于 $\sup\limits_{a\in \mathbb{R}}\left \{ -\Lambda (a) \right \}$$a_c=\frac{c^2-\theta ^2}{2c}$ 处取得, 且 $a_c$ 属于定义域 $D_{\Lambda} =\left ( -\infty,0 \right )$. 考虑如下测度变换

$\frac{{\rm d}P_{T}}{{\rm d} P}=\exp \left\{a_{c} Z_{T}(c)-T \Lambda _{T}\left(a_{c}\right)\right\},$

$E_{T}$ 表示关于概率测度 $P_{T}$ 的期望, 我们得到

$\begin{aligned}P\left(\widehat{\theta }_{T} \leq c\right) &=P\left(Z_{T}(c) \leq 0\right)=\exp\left \{ T\Lambda _{T}(a_{c} ) \right \} E_{T}\left[\exp \left\{-a_{c} Z_{T}(c)\right\}I_{\{Z_{T}(c) \leq 0\}}\right]\\&=\exp\left \{ T\Lambda _{T}(a_{c} ) \right \} E_{T}\left[\exp \left\{-a_{c}\sigma _{c}U_{T}\sqrt{T}I_{\{U_T \leq 0\}} \right\}\right]=: A_{T}B_{T},\end{aligned}$

其中 $U_{T}=\frac{Z_{T}(c) }{\sigma _{c}\sqrt{T} } $, $\sigma_c^2=-\frac{1}{2c}$. 根据引理 $2.1$, 可得关于 $A_T$ 的如下引理.

引理 3.1$c < -\theta$ 时, 对于足够大的 $T$, 有

$\begin{aligned}A_{T} =\exp\{-TI_{\theta }(c)+ H(a_{c})\}\left (1+O \left ( \frac{1}{T^{2} } \right ) \right ).\end{aligned}$

$\Phi_T\left ( \cdot \right )$ 为测度 $P_T$$U_T$ 的特征函数, 则

引理 3.2$c < -\theta$$T\to\infty$ 时, $U_T$ 依分布收敛于 $N(0, 1)$. 对于足够大的 $T$, 有

$\Phi_{T}(u)=\mathrm{e}^{-\frac{u^{2}}{2}}\left[1+\frac{\varphi_{1}(u)}{\sqrt{T}}+\frac{\varphi_{2}(u)}{T}+O\left(\frac{\max \left\{1,|u|^{r(2)}\right\}}{(\sqrt{T})^{3}}\right)\right],$

其中

$\varphi_{1}(u)=\frac{iuH_{1} }{\sigma_{c}}-\frac{iu^{3}\Lambda _{3} }{6\sigma _{c}^{3} },\quad\varphi_{2}(u)=-\frac{u^{2}H _{1}^{2} }{2\sigma _{c}^{2} } -\frac{u^{2}H _{2} }{2\sigma _{c}^{2} }+\frac{u^{4}\Lambda _{4} }{24\sigma _{c}^{4}}+\frac{u^{4}\Lambda _{3}H _{1} }{6\sigma _{c}^{4}}- \frac{u^{6}\Lambda _{3}^{2} }{72\sigma _{c}^{6}},$

$\sigma_c^2=-\frac{1}{2c}$, $\Lambda _{n} =\Lambda^{(n)} (a_{c} )$, $H _{n} =H^{(n)} (a_{c} )$, $r(2)$ 是整数, 并且当 $|u|\le sT^{\frac{1}{6} }$($s>0$) 时, 余项 $O$ 是一致的.

对于 $\forall n\in N $, 由 (3.1) 式有

$\begin{aligned}\Lambda^{(n)}_{T}(a_{c} )&= \Lambda^{(n)}(a_{c})+\frac{1}{T} H^{(n)}(a_{c})+\frac{1}{T^{2}}R^{(n)}_{T}(a_{c}).\end{aligned}$

观察到 $\Lambda'(a_{c} ) =0, \Lambda''(a_{c} ) =\sigma_c^2=-\frac{1}{2c}$, 利用泰勒展开, 由 (3.2) 式可得

$\begin{aligned}\log \Phi_{T}(u)=& T\Lambda_{T} \left (a_c+\frac{iu}{ \sigma_{c}\sqrt{T} }\right )-T \Lambda _{T} (a_c)\\=&-\frac{u^{2}}{2}+T \sum_{n=3}^{4}\left(\frac{i u}{\sigma_{c} \sqrt{T}}\right)^{n} \frac{\Lambda ^{(n)}\left(a_{c}\right)}{n!} \\ &\quad+\sum_{n=1}^{2}\left(\frac{i u}{\sigma_{c} \sqrt{T}}\right)^{n} \frac{H^{(n)}\left(a_{c}\right)}{n!}+O\left(\frac{\max \left\{1, u^{5}\right\}}{T^{\frac{3}{2}}}\right),\end{aligned}$

最终, 对上述式子两边取指数可得到 (3.1) 式.

借助引理 $3.2$ 可得到 $B_{T}$ 的展开.

引理 3.3$c< - \theta$ 时, 对于足够大的 $T$, 有

${B}_{T}=-\frac{1}{a_c\sigma_c\sqrt{2\pi T}}\left [ 1+ \frac{\psi _{1} }{T}+O\left ( \frac{1}{T^{2} } \right ) \right ],$

其中 $\psi_{1}$ 的值与 $\Lambda$$H$$a_c$ 点处的导数有关.

在引理 $4.1$ (见附录) 中令 $s_{T} =sT^{\frac{1}{6}}$($s>0$), $\beta _{T} =\sigma _{c}\sqrt{T}$, 则有

$\min\left \{ \frac{Ts_{T}^{2}}{\beta _{T}^{2} }, \frac{T\sqrt{s_{T}} }{\sqrt{\left | \beta _{T} \right | } } \right \}\ge CT^{\frac{1}{3} },$

且存在正常数 $d$$D$ 使得

$\left |D_{T}\right | \le dT\exp\{-DT^{\frac{1}{3} }\}.$

最终, 借助 (3.1) 式以及关于 $N(0,1)$ 分布的标准计算得到上述 ${B}_{T}$ 的展开.

情况 2 定理 1.1(b) 的证明

$c > \theta$ 时, 其定义域 $D_{\Lambda } =\left ( 0, 2(c - \theta) \right )$, 函数 $\Lambda(\cdot)$$D_{\Lambda } $ 上单调递减, 且于 $a_{c}=2(c-\theta)$ 处取得最小值, 因此在 $D_{\Lambda }$ 内, 存在序列 $a_{T}$, 当 $T\to+\infty$ 时收敛于 $a_{c}=2(c-\theta)$, 且满足如下方程

$\Lambda ^{'} (a_{T})+\frac{1}{T} H^{'} (a_{T})=0.$

经计算, 上式可改写为

$\left ( \theta ^{2} T+\gamma ^{2} \right ) \tau (a_{T} )=\frac{\theta ^{2}[c(2\varphi (a_{T})+a_{T})+\theta \left ( 3c-\theta \right ) ] }{\varphi (a_{T})(\varphi (a_{T})-c)},$

其中 $\tau (a)=\varphi (a)+a+\theta$, $\varphi (a)=-\sqrt{\theta ^{2}+2ac }$, 令

$\frac{{\rm d}P_{T}}{{\rm d} P}=\exp \left\{a_{T} Z_{T}(c)-T \Lambda _{T}\left(a_{T}\right)\right\},$

$E_{T}$ 表示关于概率测度 $P_{T}$ 的期望, 我们得到

$\begin{aligned}P\left(\widehat{\theta }_{T}\ge c\right)=P\left(Z_{T}(c) \ge 0\right)=\exp\left \{ T\Lambda _{T}(a_{T} ) \right\}E_{T}\left[\exp \left\{a_{T}U_{T}TI_{\{U_T \leq 0\}} \right\}\right]=: A_{T}B_{T},\end{aligned}$

其中 $U_{T} =-\frac{Z_{T}(c) }{T } $. 根据引理 $2.1$, 可得关于 $A_T$ 的如下引理.

引理 3.4$c>\theta$ 时, 对于足够大的 $T$, 有

$\begin{aligned}A_{T} &=\exp \left\{-T I_{\theta }(c)-\frac{\gamma ^{2}}{\theta ^{2}}(2 c-\theta )\right\}\times \left(\frac{2 e T(2 c-\theta )^{3}(3 c-\theta )}{\theta ^{2}(c-\theta )}\right)^{1 / 2}\\&\quad \times \left ( 1+\frac{c(c^{2}-3\theta c+\theta ^{2})}{2T(c-\theta )(\theta -2c)(3c-\theta )^{2} } +O \left ( \frac{1}{T^{2} } \right ) \right ).\end{aligned}$

已知 $c>\theta$, $a_{c} =2(c-\theta )$$\varphi (a_{c} )=\theta -2c$, 故 $\varphi (a_{c} )-c=\theta -3c\ne 0$, 而 $\varphi(a_c) + a_c$$ + \theta = 0$. 由 (3.3) 式可得 $a_{T}$$\varphi (a_{T} )$ 的展开

$$a_{T}=a_{0}+\frac{a_{1} }{T }+\frac{a_{2} }{T^{2} } +O\left ( \frac{1}{T^{3} }\right ), \quad \varphi (a_{T} )=\varphi _{0}+\frac{\varphi _{1} }{T } +\frac{\varphi _{2} }{T^{2} } +O\left ( \frac{1}{T^{3} }\right ),$$

其中

$\begin{aligned}&a_{0} =2(c-\theta ),\quad a_{1} =\frac{\theta -2c}{3c-\theta },\quad a_{2}=\frac{13c^{3}-21\theta c^{2}+6\theta ^{2}c }{2(\theta -c)(3c-\theta )^{3} } +\frac{\gamma ^{2} }{\theta ^{2} }\frac{2c-\theta }{3c-\theta }; \\&\varphi _{0} =\theta -2c,\quad\varphi _{1} =\frac{c}{3c-\theta },\quad \, \varphi _{2}=\frac{16c^{4}-25\theta c^{3}+7\theta ^{2}c^{2}}{2(\theta -c)(\theta -2c)(3c-\theta )^{3} } -\frac{\gamma ^{2} }{\theta ^{2} }\frac{c}{3c-\theta }.\end{aligned}$

利用泰勒展开, 当 $T$ 足够大时有

$\exp\left \{ T\Lambda (a_{T} ) \right \} =\exp\left \{ -TI_{\theta } (c) +\frac{1}{2} \right \}\left [ 1+\frac{\alpha _{1}}{T} +O\bigg(\frac{1}{T^{2} } \bigg) \right ]$

$\begin{aligned}\exp \{H (a_{T} )\}&=-\frac{\varphi (a_{T})}{\theta } \exp\left \{ \frac{-\gamma ^{2} (a_{T} +\theta -\varphi (a_{T}))}{2\theta ^{2} } \right \}\sqrt{\frac{2\varphi (a_{T})T}{T(\varphi (a_{T} )+a_{T} +\theta )} } \\&=-\frac{\varphi (a_{T})}{\theta }\exp\left \{ \frac{-\gamma ^{2} (a_{T} +\theta -\varphi (a_{T}))}{2\theta ^{2} } \right \}\sqrt{\frac{2\varphi _{0}T}{\varphi _{1}+a_{1} } }\left [ 1+\frac{\beta _{1}}{T} +O\bigg(\frac{1}{T^{2} } \bigg) \right ],\end{aligned}$

其中

$$\alpha _{1} =\frac{7c^{3}-11\theta c^{2}+3\theta^{2} c }{2(\theta -2c)(\theta -c)(3c-\theta )^{2} } -\frac{\gamma ^{2} }{2\theta ^{2} }, \quad \beta _{1} =\frac{-4c^{3}+7\theta c^{2} -2\theta ^{2} c }{(\theta -c)(\theta -2c)(3c-\theta )^{2} } +\frac{\gamma ^{2 } }{2\theta ^{2} }.$$

注意到, $R (a_{T} )=O\left (\frac{1}{T^{2} } \right )$, 我们可得引理 $3.4$ 的证明.

$\Phi_T\left ( \cdot \right ) $ 为测度 $P_T$$U_T$ 的特征函数, 则有

引理 3.5$c>\theta$$T\to \infty$ 时, $U_{T}$ 依分布收敛到 $K(N^{2}-1 )$, 对于足够大的 $T$, 有

$\Phi_{T}(u)=\Phi(u)\left[1+\frac{\varphi_{1}(u)}{T}+\frac{\varphi_{2}(u)}{T^{2}}+O\left(\frac{\max \left\{1,|u|^{s(2)}\right\}}{T^{3}}\right)\right],$

其中

$\Phi (u)=\frac{\exp\{-iKu \}}{\sqrt{1-2iKu} },\quad K =\Lambda' (2(c-\theta ))=\frac{(3c-\theta )}{2(\theta -2c)},$

$N$ 为标准正态分布, $s(2)$ 是整数, $\varphi_{1}$, $\varphi_{2}$ 是关于 $u$ 的多项式, 并且当 $|u|\le sT^{\frac{2}{3} } $($s>0$) 时, 余项 $O$ 是一致的.

结合 (2.1) 式可知,

$\begin{aligned}\Phi_{T}(u)=\,&\exp\left \{ T\left ( \Lambda \left ( a_{T} -\frac{iu}{T} \right ) -\Lambda(a_{T} ) \right )\right. + H\left ( a_{T} -\frac{iu}{T} \right ) -H(a_{T} )\\&\left.+\frac{1}{T} \left ( R\left ( a_{T} -\frac{iu}{T} \right ) -R(a_{T} ) \right )\right \},\end{aligned}$

$\begin{aligned}\exp\left \{T\left ( \Lambda \left ( a_{T} -\frac{iu}{T} \right ) -\Lambda(a_{T} ) \right ) \right \} &=\exp\left \{ -\frac{T}{2}\left ( -\frac{iu}{T }-\varphi _{T}\left ( \left ( 1-\frac{ium_{T} }{T } \right ) ^{\frac{1}{2} } -1 \right ) \right ) \right \} \\&=\exp\left \{ -iuc_{T}-\frac{b_{T} u^{2} }{2T} +\frac{iu^{3} m_{T}^{3}\varphi _{T}T}{32T^{3} } +O\left (\frac{\left | u \right |^{4}}{T^{4} } \right ) \right \},\end{aligned}$

其中 $\varphi _{T} =-\sqrt{\theta ^{2}+2a_{T}c} $, $m_{T} =2c/\varphi _{T}^{2}$, $b_{T} =- \frac{\varphi _{T}m_{T}^{2}}{8}$$c_{T} =\frac{c-\varphi _{T} }{2\varphi _{T}} $.

依据 (2.3) 式可将 (3.4) 式第二项写为

$\begin{aligned}\exp\left \{ H\left ( a_{T}-\frac{iu}{T } \right ) -H(a_{T} ) \right \}=&\left ( \frac{\varphi (a_{T}-iuT^{-1})}{\varphi_{T}} \right )\exp\left \{ \frac{\gamma ^{2} }{2\theta ^{2} } \left (\varphi \left (a_{T} -\frac{iu}{T } \right )-\varphi _{T} +\frac{iu}{T } \right ) \right \}\\&\times \left ( \frac{\varphi_{T}+a_{T}+\theta }{\varphi_{T}+(a_{T}-iuT^{-1}+\theta )(1-ium_{T}T^{-1})^{-\frac{1}{2} }} \right ) ^{\frac{1}{2} }.\end{aligned}$

因此, 第一步

$\begin{aligned}\frac{\varphi (a_{T}-iuT^{-1})}{\varphi_{T}}= 1- \frac{ium_{T} }{2T} +\frac{u^{2} m^{2} _{T} }{8T^{2} }+\frac{iu^{3} m^{3} _{T} }{16T^{3} }+ O\left (\frac{\left | u \right |^{4}}{T^{4}}\right ).\end{aligned}$

第二步

$\begin{aligned}&\exp\left \{ \frac{\gamma ^{2} }{2\theta ^{2} } \left ( \varphi \left ( a_{T} -\frac{iu}{T }\right ) -\varphi (a_{T} )+\frac{iu}{T } \right ) \right \}\\&=\exp\left \{ \frac{\gamma ^{2} }{\theta ^{2}} \left ( -\frac{1}{2} \left (-\frac{iu}{T}-\varphi _{T}\left ( \left ( 1-\frac{ium_{T} }{T} \right )^\frac{1}{2} -1 \right ) \right ) \right ) \right \}\\&=\exp\left \{\frac{\gamma ^{2} }{\theta ^{2}}\left ( \frac{2iu-ium_{T}\varphi _{T}}{4T }+\frac{u^{2} m_{T} ^{2} \varphi _{T}}{16T^{2}}+\frac{iu^{3} m_{T}^{3}\varphi _{T}}{32T^{3} } \right ) +O\left (\frac{\left | u \right |^{4}}{T^{4} }\right )\right \}.\end{aligned}$

第三步

$\begin{aligned}& \left ( \frac{\varphi_{T}+a_{T}+\theta }{\varphi_{T}+(a_{T}-iuT^{-1}+\theta )(1-ium_{T}T^{-1})^{-\frac{1}{2} }} \right ) ^{\frac{1}{2} }\\&=\frac{1}{\sqrt{f_{T}(u)}}\left(1 +g_{T}(u) u^{2}+h_{T}(u)\left(-\frac{3u^{2}m_{T}^{2} }{8T^{2} }+\frac{5iu^{3}m_{T}^{3} }{16T^{3} } +O\left(\frac{|u|^{4}}{T^{4}}\right)\right)\right)^{-1 / 2},\end{aligned}$

其中

$$e_{T} =T(\varphi _{T} +a_{T}+\theta ), \quad f_{T}(u) =1-\frac{i u}{e_{T}}+\frac{\left(a_{T}+\theta \right) i u m_{T}}{2 e_{T}}, $$

$$\hspace{-1.5cm} g_{T}(u) =\frac{m_{T}}{2 T e_{T} f_{T}(u)}, \quad h_{T}(u) =\frac{T\left(a_{T}+\theta \right)-i u}{e_{T} f_{T}(u)}.$$

$T\to \infty $ 时, $m_{T} \to 2c/(\theta -2c)^{2}$, $ c_{T}\to K$, $b_{T} \to \sigma _{c}^{2}$, $e_{T}\to (\theta -c)/(3c-\theta )$, 由此可推断 $f_{T}(u)$ 收敛于 $ 1-2iK u$. 最后由 (3.5) 式, (3.6) 式, (3.7) 式和 (3.8) 式可推出

$\lim_{T \to \infty } \Phi (u)=\Phi (u)=\frac{\exp\{-iKu\}}{\sqrt{1-2iKu} },$

进而得到引理证明.

借助引理 $3.5$, 类似于引理 3.3 的证明可得到 $B_{T}$ 的展开.

引理 3.6$c>\theta$ 时, 对于足够大的 $T$, 有

$B_{T} =\frac{1}{a_{c}\delta \sqrt{2\pi {\rm e}} T}\left [ 1+ \frac{\delta _{1} }{T}+O\left ( \frac{1}{T^{2} } \right ) \right ],$

其中 $ \delta =\frac{3c-\theta }{2(2c-\theta)},$$\delta _{1}$ 的值与 $\Lambda$$H$$a_c$ 点处的导数有关.

情况 3 定理 1.1(c) 的证明

$|c|<\theta $$c \neq 0$ 时,

$D_{\Lambda } = \begin{cases} (-\infty,0), & \text{ 当} -\theta <c<0, \\ \left ( -\frac{\theta ^{2} }{2c}, 0 \right ), & \text{ 当} 0<c\le\frac{\theta }{2}, \\ \left ( 2(c-\theta ), 0 \right ), & \text{ 当} \frac{\theta }{2} \le c< \theta.\end{cases}$

函数 $\Lambda(\cdot)$ 在定义域 $D_{\Lambda }$ 上恒递减且在原点处取最小值, 因此在 $D_{\Lambda }$ 内, 存在序列 $a_{T}$, 当 $T\to+\infty$ 时收敛于 $a_{c}=0$, 且满足如下方程

${\Lambda}' (a_{T})+\frac{1}{T} {H }'(a_{T})=0.$

类似地, 我们得到

$\begin{aligned}P\left(\widehat{\theta }_{T} \leq c\right) =\exp\left \{ T\Lambda _{T}(a_{T} ) \right\} E_{T}\left[\exp \left\{-a_{T}U_{T}T I_{\{U_T \leq 0\}} \right\}\right]=: A_{T}B_{T},\end{aligned}$

其中 $U_{T} =\frac{Z_{T}(c) }{T } $. 类似于引理 $3.4$$3.6$ 的证明, 我们有

引理 3.7$|c|<\theta $$c \neq 0$ 时, 对于足够大的 $T$, 有

$\begin{aligned} A_{T} =\exp \left\{-TI_{\theta }(c)-\frac{\gamma ^{2}}{\theta } \right \} \left ( \frac{2{\rm e}\theta T(c+\theta )}{\theta -c} \right ) ^{\frac{1}{2} }\left ( 1 -\frac{c(c^{2} +\theta c -\theta ^{2} )}{2T\theta (c-\theta )(c+\theta )^{2} }+O\left ( \frac{1}{T^{2} } \right ) \right ).\end{aligned}$

引理 3.8$|c|<\theta $$c \neq 0$ 时, 对于足够大的 $T$, 有

$B_{T} =\frac{1}{a_{c}\delta \sqrt{2\pi {\rm e}} T}\left [ 1+ \frac{\delta _{1} }{T}+O\left ( \frac{1}{T^{2} } \right ) \right ],$

其中 $ \delta =-\frac{\theta +c }{2\theta }, $$\delta _{1}$ 的值与 $\Lambda$$H$$0$ 处的导数有关.

情况 4 定理 1.1(d) 的证明

$c = -\theta$ 时, 其定义域为 $D_{\Lambda }=\left ( -\infty,0 \right )$. 存在序列 $a_{T}$ 收敛于原点 ($T \to \infty$), 满足如下方程

${\Lambda}' (a_{T})+\frac{1}{T} {H }'(a_{T})=0.$

类似地, 我们有

$\begin{aligned}P\left(\widehat{\theta }_{T} \leq c\right) =\exp\left \{ T\Lambda _{T}(a_{T} ) \right \} E_{T}\left[\exp \left\{-a_{T}U_{T}\sqrt{T} I_{\{U_T \leq 0\}} \right\}\right]=: A_{T}B_{T},\end{aligned}$

其中 $U_{T} =\frac{Z_{T}(c) }{\sqrt{T} } $. 类似于引理 $3.4$$3.6$ 的证明, 我们有

引理 3.9$c = -\theta$ 时, 对于足够大的 $T$, 有

$\begin{aligned}A_{T}=\exp\left \{-TI_{\theta }(c)-\frac{\gamma ^{2}}{\theta } \right \} \left ({\rm e}T\theta \right ) ^{\frac{1}{4} }\left (1+\frac{3}{8\sqrt{T\theta } } +O\left ( \frac{1}{T } \right ) \right ).\end{aligned}$

引理 3.10$c = -\theta$ 时, 对于足够大的 $T$, 有

$\begin{aligned}B_{T}=\frac{1}{2\pi \sqrt{T} } {\rm e}^{-\frac{1}{4}} \Gamma \left ( \frac{1}{4} \right )\left [ 1+ \frac{ \delta _{1} }{\sqrt{T} }+O\left ( \frac{1}{T } \right ) \right ],\end{aligned}$

其中 $\delta _{1}$ 的值与 $\Lambda$$H$$0$ 处的导数有关.

4 附录

$\textbf{ 引理 2.2 的证明 }$

第一对于所有的 $a\in D_{\Lambda } $, $u\in R$, 由 (2.2) 式可得

$\Lambda (a+iu)-\Lambda (a)=-\frac{1}{2} \left ( \varphi (a)-\varphi (a+iu)+iu\right ),$

从而

$\left | \exp\{T(\Lambda(a+iu)-\Lambda(a) )\} \right | \le \exp\left \{ \frac{T}{2} ({\rm Re}(\varphi (a+iu)-\varphi (a))) \right \}.$

因为

${\rm Re}\left (\varphi (a+iu)-\varphi (a) \right )\le \frac{c^{2} u^{2} }{2\varphi ^{3}(a) }\left ( 1+\frac{4c^{2}u^{2} }{\varphi ^{4}(a) } \right )^{-\frac{3}{4}},$

所以

$\left | \exp\left \{ T(\Lambda (a+iu)-\Lambda (a) ) \right \} \right | ^{2}\le \exp\left \{ T\frac{c^{2}u^{2} }{4\varphi ^{3}(a) }\left ( 1+\frac{4c^{2}u^{2} }{\varphi ^{4}(a) } \right )^{-\frac{3}{4} } \right \}.$

第二对于所有 $a\in D_{\Lambda }$, $u, \alpha \in R$ 由 (2.3) 式可知

$\begin{aligned} \exp\left \{ H (a+iu)-H(a)\right \}& = \exp\left \{\frac{1}{2}\log\left ( \frac{1+h(a)}{1+h(a+iu)} \right ) \right \} \times \exp\left \{ \frac{1}{2}\log\left ( 1+\frac{2iuc}{\varphi ^{2}(a)} \right ) \right \}\\ &\times \exp\left \{ -\frac{\gamma ^{2} }{2\theta ^{2} } \left ( \varphi (a)-\varphi (a+iu)+iu \right ) \right \}.\end{aligned}$

首先, 由于

$1+h(a)=\frac{(\varphi (a)+\theta )(\varphi (a)+2c-\theta )}{2c\varphi (a)},\quad\frac{1}{\left | \varphi \left ( a+iu \right ) +\alpha \right | }\le \max\left \{ \frac{1}{\left | \varphi \left ( a \right ) \right |}, \frac{1}{\left | \varphi \left ( a \right )+\alpha \right | } \right \},$

因此

$\begin{aligned}\left | \frac{1+h(a)}{1+h(a+iu)} \right |&=\left | \frac{\varphi \left ( a+iu \right ) }{\varphi \left ( a\right )} \right |\times \left | \frac{\left ( \varphi \left ( a \right )+\theta \right )\left ( \varphi \left ( a \right )+2c-\theta \right ) }{\left ( \varphi \left ( a+iu \right )+\theta \right )\left ( \varphi \left ( a+iu \right )+2c-\theta \right ) } \right | \\&\le \left | \frac{\varphi \left ( a+iu \right ) }{\varphi \left ( a\right )} \right | \times \max\left \{ 1,\frac{\left | \varphi (a)+\theta \right | }{\left | \varphi (a) \right | } \right \}\times \max\left \{ 1,\frac{\left | \varphi (a)+2c-\theta \right | }{\left | \varphi (a) \right | } \right \}\\& \le \ell (a)\left ( 1+\frac{4c^{2}u ^{2} }{\varphi ^{4}(a) } \right ) ^{\frac{1}{4} },\end{aligned}$

所以

$\left | \exp\left \{ \frac{1}{2} \log\left ( \frac{1+h(a)}{1+h(a+iu)} \right ) \right \} \right |^{2}= \left | \frac{1+h(a)}{1+h(a+iu)}\right |\le \ell (a)\left ( 1+\frac{4c^{2}u ^{2} }{\varphi ^{4}(a) } \right ) ^{\frac{1}{4} }.$

其次

$\left | \exp\left \{ \frac{1}{2}\log\left ( 1+\frac{2iuc}{\varphi ^{2}(a) } \right ) \right \} \right | ^{2} =\left | \sqrt{1+\frac{2iuc}{\varphi ^{2} (a)} } \right | ^{2}=\left ( 1+\frac{4u^{2}c^{2} }{\varphi ^{4} (a)} \right )^{\frac{1}{2} }.$

最后由 (4.1) 式可以得出

$\left | \exp\left \{ -\frac{\gamma ^{2} }{2\theta ^{2} } \left ( \varphi (a)-\varphi (a+iu)+iu \right ) \right\} \right |^{2}\le \exp\left \{ \frac{\gamma ^{2} }{\theta ^{2} }\frac{c^{2}u^{2} }{4\varphi ^{3}(a) }\left ( 1+\frac{4c^{2}u^{2} }{\varphi ^{4}(a) } \right )^{-\frac{3}{4} } \right \}.$

因此, 由 (4.2) 式, (4.3) 式和 (4.4) 式可得

$\left |\exp\left \{ H (a+iu)-H(a) \right \}\right |^{2}\le \ell (a)\left ( 1+\frac{4c^{2}u ^{2} }{\varphi ^{4}(a) } \right ) ^{\frac{3}{4} }\exp\left \{ \frac{\gamma ^{2} }{\theta ^{2} }\frac{c^{2}u^{2} }{4\varphi ^{3}(a) }\left ( 1+\frac{4c^{2}u^{2} }{\varphi ^{4}(a) } \right )^{-\frac{3}{4} }\right \}.$

第三 $R(\cdot)$ 始终为有界函数, 存在正常数 $C_0$, 使得

$\left | \exp\left \{ R (a+iu)-R(a) \right \}\right |^{2}\le C _{0},$

综上引理 $2.2$ 得证.

$\alpha_{T}=\left\{\begin{array}{l} a_{c}, \text { 当 } c<-\theta, \\ a_{T}, \text { 其它, } \end{array} \quad \beta_{T}=\left\{\begin{array}{ll} \sigma_{c} \sqrt{T}, & \text { 当 } c<-\theta, \\ \sqrt{T}, & \text { 当 } c=-\theta, \\ T, & \text { 当 }|c|<\theta, \\ -T, & \text { 当 } c>\theta, \end{array}\right.\right.$

$$ \frac{{\rm d}P_{T}}{{\rm d} P}=\exp \left\{\alpha _{T} Z_{T}(c)-T \Lambda _{T}\left(\alpha _{T}\right)\right\},$$

$\Phi_T\left ( \cdot \right ) $ 为测度 $P_T$$U_T$ 的特征函数, 则

$\Phi_{T}(u)=\exp\left \{ T\Lambda_{T} \left (\alpha_{T}+\frac{iu}{\beta _{T} }\right )-T \Lambda _{T} (\alpha_{T}) \right \},$

由引理 $2.2$ 知, 当 $T$ 足够大时, $\Phi _{T}\left ( \cdot \right ) \in L^{2} (R)$.

引理 4.1 对于

$B_{T}=E_{T}\left(\exp \left\{-\alpha _{T}\beta _{T} {U}_{T}\right\} I_{\left\{{U}_{T} \le 0\right\}}\right),$

其中 $U_{T} =\frac{Z_{T}(c) }{\beta_{T}} $, 我们有如下分解$B_{T} =:C_{T} +D_{T},$ 其中

$C_{T}=-\frac{1}{2\pi\alpha _{T}\beta_{T}}\int_{\left | u \right |\le s_{T} }^{}\left ( 1+\frac{iu}{\alpha _{T}\beta_{T}} \right )^{-1}\Phi _{T} (u){\rm d}u,$
$D_{T}=-\frac{1}{2\pi\alpha _{T}\beta_{T}}\int_{\left | u \right |> s_{T} }^{}\left ( 1+\frac{iu}{\alpha _{T}\beta_{T}} \right )^{-1}\Phi _{T} (u){\rm d}u.$

且存在正常数 $d$, $D$ 满足

$\left | D_{T} \right | \le dT\exp\{-DT^{v}\},$

其中 $s_T$ 满足: 存在正常数 $C$$0 < v < 1$, 使得

$\min\left \{ \frac{Ts_{T}^{2}}{\beta _{T}^{2} },\frac{T\sqrt{s_{T}} }{\sqrt{\left | \beta _{T} \right | } } \right \}\ge CT^{v}.$

根据 Parseval 公式,

$\begin{aligned} B_{T} =-\frac{1}{2\pi\alpha _{T}\beta_{T}}\int_{R}\left ( 1+\frac{iu}{\alpha _{T}\beta_{T}} \right )^{-1}\Phi _{T} (u){\rm d}u=:C_{T} +D_{T}\end{aligned}$

其中 $C_T$, $D_T$ 的表达式见 (4.6) 式和 (4.7) 式, $s_T$ 的选取满足 (4.8) 式.

下面证明 $D_{T}$ 以指数速度趋近于 $0$. 根据 Cauchy-Schwarz 不等式, 有

$\left | D_{T} \right |^{2} \le \frac{1}{4\pi^{2}\alpha _{T}^{2}\beta _{T}^{2}} \int_{\left | u \right |> s_{T} }\left ( 1+\frac{u^2}{\left ( \alpha _{T}\beta_{T} \right ) ^{2} } \right )^{-1}{\rm d}u\int_{\left | u \right |>s_{T} } \left | \Phi _{T} (u) \right |^{2}{\rm d}u,$

经换元可得到

$\int_{\left | u \right |> s_{T} }^{}\left ( 1+\frac{u^2}{\left ( \alpha _{T}\beta_{T} \right ) ^{2} } \right )^{-1}{\rm d}u\le \left |\alpha _{T}\beta_{T}\right | \int_{R}^{} \frac{1}{1+v^{2} }{\rm d}v \le \left |\alpha _{T}\beta_{T} \right |\pi.$

$\varphi _{T} =\varphi (\alpha _{T} )$, $K_{T} = \frac{2\left | c \right |}{\left | \beta _{T} \right |\varphi ^{2} (\alpha _{T} ) } $.$T$ 足够大时, 根据引理 $2.2$ 及 (4.5) 式可得

$\left | \Phi _{T} (u) \right |^{2} \le C_{0} \ell (\alpha_T)(1+K_{T}^{2}u^{2})^{\frac{3}{4} } \exp\left \{\left (\frac{T\varphi _{T} }{16}+\frac{\gamma ^{2}\varphi _{T}}{16\theta ^{2}} \right )K_{T}^{2}u^{2}\left (1+K_{T}^{2}u^{2} \right )^{-\frac{3}{4} } \right \}.$

$T$ 足够大时, 存在一个正常数 $C_{\ell}$, 有 $\ell (\alpha_T)\le C_\ell T$, 使得 $\delta _{T} =K_{T} s_{T}$, 则有

$\begin{aligned} \int_{\left | u \right |>s_{T}} \left | \Phi_{T} (u) \right |^{2} {\rm d}u \le \frac{2C_{0}C_{\ell} T}{K_{T} } \int_{\delta_{T}}^{+\infty}\left ( 1+v^{2} \right )^{\frac{3}{4}} \exp\left \{\left ( \frac{T\varphi _{T} }{16}+\frac{\gamma ^{2}\varphi _{T}}{16\theta ^{2}}\right ) v^{2} \left ( 1+v^{2} \right ) ^{-\frac{3}{4} }\right \} {\rm d}v.\end{aligned}$

$g(v)=\frac{v^{2}}{(1+v^{2})^{\frac{3}{4} } },\quad h(v)=\frac{v^{\frac{3}{2} } }{(1+v^{2})^{\frac{3}{4} } }.$ 于是, 当 $v>\delta _{T}$ 时, $g(v)>\sqrt{v}h(\delta_{T} )$. 此外, 对于 $\forall v\in R^+$, 均有 $2^{\frac{3}{4}}g(v)\ge \min\{v^{2},\sqrt{v}\}$. 故 (4.11) 式可改写为

$\begin{aligned} \int_{\left | u \right |>s_{T}}^{} \left | \Phi_{T} (u) \right |^{2}{\rm d}u\!\le\! \frac{2C_{0}C_{\ell}T}{K_{T}}\exp\left \{\! \left (\frac{T\varphi _{T} }{32}+\frac{\gamma ^{2}\varphi _{T}}{32\theta ^{2} }\right ) g(\delta _{T}) \! \right \} \int_{\delta _{T}}^{\infty }2^{\frac{3}{4}} \max\{1,\sqrt{v}\}\exp\{e_{T}\sqrt{v}\}{\rm d}v,\end{aligned}$

其中 $ e_{T}=\left (\frac{T\varphi _{T} }{32}+\frac{\gamma ^{2}\varphi _{T}}{32\theta ^{2}}\right )h\left ( \delta _{T} \right ).$ 因为 $\varphi _{T}<0$, 所以

$\begin{aligned} \left ( \frac{T\varphi_{T} }{16}+ \frac{\gamma ^{2}\varphi_{T}}{16\theta ^{2} } \right ) g(\delta _{T})&\le \left ( \frac{T\varphi_{T} }{32}+ \frac{\gamma ^{2}\varphi _{T}}{32\theta ^{2} } \right ) \min\left \{ \delta _{T}^{2},\sqrt{\delta _{T}} \right \}\\&\le \max\left\{\frac{c^{2} }{8\varphi _{T}^{3} }, -\frac{\sqrt{2\left | c \right | } }{32} \right \}\min\left\{T\frac{s_{T}^{2} }{\beta _{T}^{2}},T\sqrt{\frac{s_{T} }{\left | \beta _{T} \right | } } \right \}\\& \quad +\max\left\{\frac{\gamma^{2}c^{2}}{8\theta ^{2}\varphi _{T}^{3}}, -\frac{\gamma ^{2}\sqrt{2\left | c \right | } }{32\theta ^{2} }\right\}\min\left\{ \frac{s_{T}^{2} }{\beta _{T}^{2}}, \sqrt{\frac{s_{T} }{\left | \beta _{T} \right | } } \right \} \\ &\le - C_1T^{v} - C_{2},\end{aligned}$

其中 $C_1,C_2$ 为正常数. 此外,

$\int_{\delta _{T} }^{+\infty } \max\{1,v^{\frac{1}{2}}\}\exp\{e_{T} \sqrt{v}\}{\rm d}v\le \int_{\delta _{T} }^{+\infty }\exp\{(e_{T}-1)\sqrt{v} \}{\rm d}v\le \frac{2}{(1-e_{T})^{2} }.$

$T\to+\infty$ 时, $e_{T} \to -\infty$, 因此, 上式趋近于 $0$, 这表明这个积分可任意小. 由 (4.12) 式可知存在正常数 $C_3$, 使得

$\begin{aligned}\int_{\left | u \right |>s_{T}}^{} \left | \Phi_{T} (u) \right |^{2}{\rm d}u\le C_3T\left | \beta _{T} \right |\varphi _{T}^{2} \exp\{- C_3T^{v}\}.\end{aligned}$

最后由 (4.9) 式, (4.10) 式和 (4.13) 式可得

$$\left | D_{T} \right | \le dT\exp\{-DT^{v}\},$$

其中 $d$, $D$ 为正常数.

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