近年来, 诸多领域的实际数据呈现出单峰、非对称等偏正态特征.为此, Azzalini^{[2, 3]}首次提出偏正态分布的概念, 并且给出其密度函数表达式.设随机变量$ Y \sim SN(\xi , {\eta ^2}, \lambda ) $, 其中位置参数为$ \xi \in R $, 尺度参数为$ {\eta ^2} \in {R^ + } $, 偏度参数为$ \lambda \in R $, 则$ Y $的密度函数可表示为

(1.1) $ \begin{equation} f(y;\xi , {\eta ^2}, \lambda ){\rm{ = }}2\phi (y;\xi , {\eta ^2})\Phi [\lambda {\eta ^{ - 1}}(y - \xi )], \end{equation} $

其中$ \phi ( \cdot ) $和$ \Phi ( \cdot ) $分别表示标准正态分布的密度函数和分布函数.若$ \lambda = 0 $, 则(1.1)式退化成均值为$ \xi $, 方差为$ {\eta ^2} $的正态分布.鉴于偏正态分布应用的广泛性, 学者们已经对其性质进行了深入的研究, 包括分布特征^[4]、特征函数^[5]、二次型分布^[6]、偏度度量^[7]、极值矩的近似表达式^[8]和独立变量和的精确密度^[9]等.

进一步, 针对单个偏正态总体, Wang等^[10]探讨了在变异系数和偏度参数已知时, 位置参数的区间估计问题. Gui和Guo^[11]基于近似似然方程, 给出位置参数和尺度参数的显式估计. Ma等^[12]在尺度参数和偏度参数已知时, 研究了位置参数的区间估计和假设检验问题.但是在实际应用中, 对多个偏正态总体的共同位置参数进行研究是无法回避的.例如, 在可靠性和寿命测试的层面中, 对多个偏正态总体的共同位置参数研究就等同于对共同保修期限的估计.然而, 目前对共同位置参数的研究主要集中在高斯分布、逆高斯分布、对数正态分布以及指数分布中^[13-22].但实际数据更符合偏正态分布特征.鉴于此, 本文针对多个尺度参数和偏度参数未知的偏正态总体, 基于Bootstrap方法构造其共同位置参数的置信区间.

本文安排如下.第2小节针对多个偏正态总体, 分别给出未知参数的矩估计和极大似然估计.第3小节将徐礼文^[1]对多个正态总体共同均值的探讨推广到多个偏正态总体, 进而构造共同位置参数的Bootstrap置信区间和Bootstrap检验统计量.第4小节给出上述方法的Monte Carlo模拟结果, 以验证上述方法的统计优良性.第5小节将本文所给方法应用于中国区域生产总值和生物利用度数据的案例分析.第6小节给出本文小结.第4小节中的模拟结果在附录7中.

首先, 本小节考虑单个偏正态总体参数的估计问题.设$ {Y_1}, \cdots , {Y_n} $是取自偏正态总体$ SN(\xi, \eta^2, \lambda) $的一组样本.则样本均值、样本二阶中心距和样本三阶中心距可分别表示为

(2.1) $ \begin{equation} \bar{Y} = \frac{1}{n}\sum\limits_{i = 1}^n Y_i, \;S_2 = \frac{1}{n}\sum\limits_{i = 1}^n (Y_i- \bar{Y})^2, \;S_3 = \frac{1}{n}\sum\limits_{i = 1}^n (Y_i-\bar{Y})^3. \end{equation} $

定理2.1  设$ \delta = \lambda /{(1 + {\lambda ^2})^{1/2}} $.若$ Y \sim SN(\xi , {\eta ^2}, \lambda ) $, 则参数$ (\xi , {\eta ^2}, \lambda ) $的矩估计可表示为

(2.2) $ \begin{equation} \hat{\xi} = \bar{Y} - c{S_3}^{1/3}, \;\hat{\eta}^2 = S_2 + {c^2}{S_3}^{2/3}, \;\hat{\lambda} = \hat{\delta}/(1-\hat{\delta}^2)^{1/2}, \end{equation} $

其中$ b = {{(2/\pi )}^{1/2}} $, $ c = {{[2/(4-\pi )]}^{1/3}} $, $ \hat{\delta } = \frac{cS_{3}^{1/3}}{b{{\left( {{S}_{2}}+{{c}^{2}}S_{3}^{2/3} \right)}^{1/2}}} $.

证  令$ (\bar y, {s_2}, {s_3}) $表示$ (\bar Y, {S_2}, {S_3}) $的观测值.令$ {X_i} = ({Y_i} - \bar y)/\sqrt {{s_2}} $, 则$ {X_1}, \cdots , {X_n} $是来自于$ X \sim SN({\xi _s}, \eta _s^2, \lambda ) $的一组样本.其中

(2.3) $ \begin{equation} {\xi _s}{\rm{ = (}}\xi - {\kern 1pt} {\kern 1pt} \bar y{\rm{)}}/\sqrt {{s_2}} , \; {\eta _s} = \eta /\sqrt {{s_2}}. \end{equation} $

易见, $ X $的矩生成函数为

(2.4) $ \begin{equation} {M_X}(t) = 2\exp \left( {t{\xi _s} + \frac{{{t^2}\eta _s^2}}{2}} \right)\Phi \left( {t{\eta _s}\delta } \right), \end{equation} $

则由(2.4)式可得

(2.5) $ \begin{eqnarray} M_X'(t)|_{(t = 0)}& = &\xi_s + b\eta_s\delta = 0, \\ M_X''(t)|_{(t = 0)}& = &\xi_s^2 + 2b\xi_s\eta_s\delta + \eta_s^2 = 1, \\ M_X'''(t)|_{(t = 0)}& = &\xi_s^3 + 3b\xi_s^2\eta_s\delta + 3\xi_s\eta_s^2 + 3b\eta_s^3\delta - b\eta_s^3\delta^3 = s_2^{-3/2}s_3. \end{eqnarray} $

根据(2.3)和(2.5)式, 可得$ (\xi, \eta^2, \lambda) $的矩估计值为

$ {\hat \xi ^{\rm{*}}}{\rm{ = }}\bar y - cs_3^{1/3}, \;{\hat \eta ^{{{\rm{*}}^2}}} = {s_2} + {c^2}s_3^{2/3}, \;{\hat \lambda ^*}{\rm{ = }}\frac{{{{\hat \delta }^*}}}{{\sqrt {1 - {{\hat \delta }^{{{\rm{*}}^2}}}} }}, $

其中$ {{\hat{\delta }}^{*}} = \frac{cs_{3}^{1/3}}{b{{\left( {{s}_{2}}+{{c}^{2}}s_{3}^{2/3} \right)}^{1/2}}} $.参数$ (\xi, \eta^2, \lambda) $的矩估计由(2.1)式给出.证毕.

接下来, 我们考虑总体参数的极大似然估计.研究表明, 偏正态总体中直接参数$ (\xi, \eta^2, \lambda) $的似然方程不存在唯一解.为此, 本文基于中心参数化思想, 给出未知参数的极大似然估计^[23-25].令

$ \begin{eqnarray*} W = \frac{{Y - \xi }}{\eta } \sim SN(\lambda ), \;{Y_C} = \mu + \sigma \Big( {\frac{{W - E(W)}}{{\sqrt {{\mathop{D}} (W)} }}} \Big) \sim S{N_C}(\mu , {\sigma ^2}, \gamma ), \end{eqnarray*} $

其中$ S{N_C}(\mu , {\sigma ^2}, \gamma ) $表示均值为$ \mu \in R $, 方差为$ {\sigma ^2} \in {R^ + } $, 偏度系数为$ \gamma $的偏正态分布. Pewsey^[23]给出直接参数$ (\xi, \eta^2, \lambda) $与中心化参数$ (\mu , {\sigma ^2}, \gamma ) $之间的关系如下

(2.6) $ \begin{equation} \xi = \mu - c{\gamma ^{1/3}}\sigma, \;{\eta ^2} = {\sigma ^2}(1 + {c^2}{\gamma ^{2/3}}), \;\lambda = \frac{{c{\gamma ^{1/3}}}}{{\sqrt {{b^2} + {c^2}({b^2} - 1){\gamma ^{2/3}}} }}{\kern 1pt} {\kern 1pt} {\kern 1pt}. \end{equation} $

令$ {Y_{C1}}, \cdots , {Y_{Cn}} $是取自于偏正态总体$ Y_C \sim SN_C (\mu, {\sigma^2}, \gamma) $的一组样本, 则其样本均值、样本二阶中心矩和样本三阶中心矩可分别表示为

$ \begin{eqnarray*} {\bar Y_C}{\rm{ = }}\frac{1}{n}\sum\limits_{i = 1}^n {{Y_{Ci}}, \;{\rm{ }}} {S_{C2}} = \frac{1}{n}\sum\limits_{i = 1}^n {({Y_{Ci}} - } {\bar Y_C}{)^2}, \;{\rm{ }}{S_{C3}} = \frac{1}{n}\sum\limits_{i = 1}^n {({Y_{Ci}} - } {\bar Y_C}{)^3}{\rm{ }}. \end{eqnarray*} $

定理2.2  设$ Y \sim SN(\xi, \eta^2, \lambda) $, 令$ W = \frac {Y - \xi}{\eta} \;{\kern 1pt}{\kern 1pt} $, $ {\kern 1pt}{\kern 1pt} \; {Y_C} = \mu + \sigma \big( {\frac{{W - E(W)}}{{\sqrt {{\mathop D}(W)} }}} \big), $则$ Y_c = Y $.

证  首先, 对$ Y $的矩生成函数$ {{M}_{Y}}(t) $求前三阶导数

$ \begin{array}{r@{\; }l} &{\rm{E(Y)}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = \xi + b\eta \delta, \\ &{\rm{E(}}{{\rm{Y}}^2}{\rm{)}} = {\xi ^2} + 2b\xi \eta \delta + {\eta ^2}, \\ &{\rm{E(}}{{\rm{Y}}^3}{\rm{)}} = {\xi ^3} + 3b{\xi ^2}\eta \delta + 3\xi {\eta ^2} + 3b{\eta ^3}\delta - b{\eta ^3}{\delta ^3}, \end{array} $

则$ Y $的偏度系数$ \gamma $可表示为

(2.7) $ \begin{equation} \gamma {\rm{ = }}\frac{{E{{\left[ {Y - E\left( Y \right)} \right]}^3}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} }}{{{{\left\{ {E{{\left[ {Y - E\left( Y \right)} \right]}^2}} \right\}}^{3/2}}}}{\rm{ = }}\frac{{{b^3}{\delta ^3}}}{{{c^3}{{(1 - {b^2}{\delta ^2})}^{3/2}}}}. \end{equation} $

由(2.6)和(2.7)式可得

(2.8) $ \begin{equation} \sigma = \frac{\eta }{{\sqrt {1 + {c^2}{\gamma ^{2/3}}} }}{\kern 1pt} {\rm{ = }}{\kern 1pt} \eta {\kern 1pt} {\kern 1pt} \sqrt {1 - {b^2}{\delta ^2}} , \;\mu = \xi + c{\gamma ^{1/3}}\sigma {\kern 1pt} {\kern 1pt} {\kern 1pt} = \xi + b\eta \delta. \end{equation} $

由于$ W \sim SN(\lambda) $, 易得$ E(W) = b\delta, \;D(W) = 1-b^2{\delta}^2. $于是有

$ {Y_C}{\rm{ = }}\mu + \sigma \left( {\frac{{W - E(W)}}{{\sqrt {{\mathop D}(W)} }}} \right) = \xi + b\eta \delta {\rm{ + }}\eta {\kern 1pt} {\kern 1pt} \sqrt {1 - {b^2}{\delta ^2}} \left( {\frac{{\frac{{Y - \xi }}{\eta } - b\delta }}{{\sqrt {1 - {b^2}{\delta ^2}} }}} \right) = Y. $

定理2.2证毕.

注2.1  由(2.2)式可知, 若偏度参数$ \left| \lambda \right| \to \infty $, 则$ \left| \delta \right| \to 1 $.进而, 由(2.7)式可知$ \gamma \in (-0.99527, 0.99527) $, 更多细节见Pewsey ^[25].

下面进一步考虑中心化参数$ (\mu , {\sigma ^2}, \gamma ) $的极大似然估计.令$ ({\bar y_{_C}}, {s_{_{C2}}}, {s_{_{C3}}}) $分别表示$ ({\bar Y_C}, $$ {S_{C2}}, {S_{C3}}) $的观测值.类似地, 对$ {Y_{Ci}} $进行标准化, 即$ {Y_{si}} = ({Y_{Ci}} - {\bar y_{_C}})/\sqrt {{s_{_{C2}}}} $, $ i = 1, \cdots , n $.易见, $ {Y_{s1}}, \cdots , {Y_{sn}} $是来自于$ {Y_s} \sim S{N_C}({\mu _s}, \sigma _s^2, \gamma ) $的一组样本, 其中, $ {\mu _s}{\rm{ = (}}\mu - {\kern 1pt} {\kern 1pt} {\bar y_{_C}}{\rm{)}}/\sqrt {{s_{_{C2}}}} {\rm{ }} $, $ {\sigma _s} = \sigma /\sqrt {{s_{_{C2}}}} $.这里$ {Y_s} $的密度函数为

(2.9) $ \begin{eqnarray} f\left( {{y_s};{\mu _s}, \sigma _s^2, \gamma } \right)&{\rm{ = }}&\frac{2}{{{\sigma _s}\sqrt {{s_{_{C2}}}\left( {1 + {c^2}{\gamma ^{2/3}}} \right)} }}\phi \left[ {\left( {\frac{{{y_s} - {\mu _s}}}{{{\sigma _s}}} + c{\gamma ^{1/3}}} \right)\frac{1}{{\sqrt {1 + {c^2}{\gamma ^{2/3}}} }}} \right]{}\\ & &{\kern 1pt} \times \Phi \left\{ {\left( {\frac{{{y_s} - {\mu _s}}}{{{\sigma _s}}} + c{\gamma ^{1/3}}} \right)\frac{{c{\gamma ^{1/3}}}}{{\sqrt {\left( {1 + {c^2}{\gamma ^{2/3}}} \right)\left[ {{b^2} + {c^2}{\gamma ^{2/3}}\left( {{b^2} - 1} \right)} \right]} }}} \right\}. \end{eqnarray} $

由(2.9)式可得对数似然函数为

(2.10) $ \begin{eqnarray} l\left( {{y_{s1}}, \cdots , {y_{sn}};{\mu _s}, \sigma _s^2, \gamma } \right) & = &- n\log {\sigma _s} - \frac{n}{2}\log \left( {1 + {c^2}{\gamma ^{2/3}}} \right){}\\ & & + \sum\limits_{i = 1}^n {\log \phi \left[ {\left( {\frac{{{y_{si}} - {\mu _s}}}{{{\sigma _s}}} + c{\gamma ^{1/3}}} \right)\frac{1}{{{{\left( {1 + {c^2}{\gamma ^{2/3}}} \right)}^{{\rm{1/2}}}}}}} \right]}{}\\ & & + \sum\limits_{i = 1}^n {\log \Phi \left\{ {\frac{{\frac{{\left( {{y_{si}} - {\mu _s}} \right)c{\gamma ^{1/3}}}}{{{\sigma _s}}} + {c^2}{\gamma ^{2/3}}}}{{{{\left( {1 + {c^2}{\gamma ^{2/3}}} \right)}^{{\rm{1/2}}}}{{\left[ {{b^2} + {c^2}{\gamma ^{2/3}}\left( {{b^2} - 1} \right)} \right]}^{{\rm{1/2}}}}}}} \right\}}. \end{eqnarray} $

于是, 令$ (\tilde \mu _s^*, \tilde \sigma _s^{{*^2}}, {\tilde \gamma ^*}) $为来自(2.10)式的$ ({\mu _s}, \sigma _s^2, \gamma ) $极大似然估计值, 其初始值可取$ ({\mu _s}, \sigma _s^2, \gamma ) $的矩估计, 即

(2.11) $ \begin{equation} \hat \mu _s^* = - cs_{_{C2}}^{ - 1/2}s_{_{C3}}^{1/3}, \; \hat \sigma _s^{{*^2}} = 1 + cs_{_{C2}}^{ - 1/2} s_{_{C3}}^{2/3}{\kern 1pt} , {\kern 1pt} \;{\hat \gamma ^*} = \frac{{b{{\hat \delta }^{{{\rm{*}}^3}}}\big( {2{b^2} - 1} \big)}}{{{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {{\big( {1 - {b^2}{{\hat \delta }^{{{\rm{*}}^2}}}} \big)}^{3/2}}}}{\kern 1pt} {\kern 1pt} {\kern 1pt}. \end{equation} $

进一步, 可得$ (\mu , \sigma ^2) $的极大似然估计值为

(2.12) $ \begin{equation} {\tilde \mu ^*} = {\bar y_{_C}} + s_{_{{\rm{C2}}}}^{1/2}\tilde \mu _s^*, {\kern 1pt} \;{\tilde \sigma ^{{*^2}}} = {s_{_{C2}}}\tilde \sigma _s^{*^2}. \end{equation} $

由(2.6)式可得直接参数$ (\xi, \eta^2, \lambda) $的极大似然估计值分别为

$ \begin{eqnarray*} {\tilde \xi ^*} = {\tilde \mu ^*} - c{\tilde \gamma ^{{*^{1/3}}}}{\tilde \sigma ^*}, \;{\tilde \eta ^{{*^2}}} = {\tilde \sigma ^{{*^2}}}(1 + {c^2}{\tilde \gamma ^{{*^{2/3}}}}), \;{\tilde \lambda ^{\rm{*}}} = \frac{{c{{\tilde \gamma }^{{*^{1/3}}}}}}{{\sqrt {{b^2} + {c^2}\left( {{b^2} - 1} \right){{\tilde \gamma }^{{*^{2/3}}}}} }}. \end{eqnarray*} $

则$ \delta $的极大似然估计值为$ {\tilde \delta ^*} = {\tilde \lambda ^*}/{(1 + {\tilde \lambda ^{{{\rm{*}}^2}}})^{1/2}} $.于是, 未知参数$ (\xi, \eta^2, \lambda) $的极大似然估计由下述定理给出.

定理2.3  令$ ({\tilde \mu _s}, \tilde \sigma _s^2, \tilde \gamma ) $是对应于$ (\tilde \mu _s^*, \tilde \sigma _s^{{*^2}}, {\tilde \gamma ^*}) $的极大似然估计, 则直接参数$ (\xi, \eta^2, \lambda) $的极大似然估计分别为

(2.13) $ \begin{equation} \tilde \xi = \tilde \mu - c{\tilde \gamma ^{1/3}}\tilde \sigma {\kern 1pt} , \;{\tilde \eta ^2} = {\tilde \sigma ^2}(1 + {c^2}{\tilde \gamma ^{2/3}}), \; \tilde \lambda = \frac{{c{{\tilde \gamma }^{1/3}}}}{{\sqrt {{b^2} + {c^2}\left( {{b^2} - 1} \right){{\tilde \gamma }^{2/3}}} }}, \end{equation} $

其中, $ ({\tilde \mu}, \tilde \sigma^2) $是对应于$ ({\tilde \mu ^* }, \tilde \sigma ^{{*^2}}) $的极大似然估计.进而, $ \delta $的极大似然估计为

$ \tilde \delta = \tilde \lambda /{(1 + {\tilde \lambda ^2})^{1/2}}. $

现将一个偏正态总体拓展至多个具有共同位置参数且相互独立的偏正态总体.设$ {Y_{i1}}, $$ \cdots , {Y_{i{n_i}}} $是来自偏正态总体$ SN(\xi, \eta_i^2, \lambda_i) $的一组样本, $ i = 1, \cdots , k $.则样本均值、样本二阶中心矩和样本三阶中心矩可分别表示为

(2.14) $ \begin{equation} \bar{Y_i} = \frac{1}{n_i}\sum\limits_{j = 1}^{n_i} Y_{ij}, \;S_{2i} = \frac{1}{n_i}\sum\limits_{j = 1}^{n_i} (Y_{ij}- \bar{Y_i})^2, \;S_{3i} = \frac{1}{n_i}\sum\limits_{j = 1}^{n_i} (Y_{ij}-\bar{Y_i})^3, \; i = 1, \cdots , k. \end{equation} $

类似于一个总体的情况, 我们考虑多个偏正态总体中未知参数$ (\eta_i^2, \lambda_i) $的估计问题, $ i = 1, $$ \cdots , k $.由定理2.1, 可得$ (\eta_i^2, \lambda_i) $的矩估计为

(2.15) $ \begin{equation} \hat{\eta}_{i}^2 = S_{2i} + {c^2}{S}_{{3i}}^{2/3}, \;\hat{\lambda}_{i} = \hat{\delta}_{i}/(1-\hat{\delta}_i^2)^{1/2}, \;i = 1, \cdots , k, \end{equation} $

其中$ {{\hat{\delta }}_{i}} = \frac{cS_{3i}^{1/3}}{b\sqrt{{{S}_{2i}}+{{c}^{2}}S_{3i}^{2/3}}} $.则$ {{\bar{Y}}_{i}} $的期望和方差分别为

(2.16) $ \begin{equation} E({{\bar{Y}}_{i}}) = \xi +b{{\eta }_{i}}{{\delta }_{i}}, \;D({{\bar{Y}}_{i}}) = \frac{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}{{{n}_{i}}}, \;i = 1, \cdots , k. \end{equation} $

若$ (\eta_i^2, \delta_i) $已知, 基于第$ i $个样本可得共同位置参数$ \xi $的估计

$ {{\hat{\xi }}^{i}}{{|}_{({{\eta }_{i}}, {{\delta }_{i}})}} = {{\bar{Y}}_{i}}-b{{\eta }_{i}}{{\delta }_{i}}, \;D({{\hat{\xi }}^{i}}{{|}_{({{\eta }_{i}}, {{\delta }_{i}})}}) = \frac{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}{{{n}_{i}}}, \;i = 1, \cdots , k. $

借鉴Graybill-Deal估计的思想^[26], 可得

(2.17) $ \begin{equation} \hat{\xi } = \frac{\sum\limits_{i = 1}^{k}{\frac{1}{D({{{\hat{\xi }}}^{i}}{{|}_{({{\eta }_{i}}, {{\delta }_{i}})}})}}{{{\hat{\xi }}}^{i}}{{|}_{({{\eta }_{i}}, {{\delta }_{i}})}}}{\sum\limits_{i = 1}^{k}{\frac{1}{D({{{\hat{\xi }}}^{i}}{{|}_{({{\eta }_{i}}, {{\delta }_{i}})}})}}} = {{\bar{Y}}_{G}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{\delta }_{i}}}{{{\eta }_{i}}(1-{{b}^{2}}\delta _{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}}, \end{equation} $

其中$ {{\bar{Y}}_{G}} = \frac{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}{{{\bar{Y}}}_{i}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}}. $令$ (\hat{\eta}_i^{*^{2}}, \hat{\delta}_i^{*}, \hat{\lambda}_i^{*}) $是对应于$ ({\hat{\eta}_{i}}^2, \hat{\delta}_{i}, \hat{\lambda}_i) $的矩估计值, $ i = 1, \cdots , k $, 将$ ({\eta}_{i}^2, {\delta}_{i}) $的矩估计值代入(2.17)式, 可得$ {\xi} $的估计值为

(2.18) $ \begin{equation} {{\hat{\xi }}^{*}} = {{\bar{y}}_{G1}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}\hat{\delta }_{i}^{*}}{\hat{\eta }_{i}^{*}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{*2}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}, \end{equation} $

其中$ {\bar{y}}_{G1} $是$ {\bar{Y}}_{G1} $的观测值, $ {\bar{Y}}_{G1} $是将$ ({\eta}_{i}^2, {\delta}_{i}) $的矩估计代入$ {\bar{Y}}_{G} $所得.由定理2.3可得, 参数$ ({\eta }_{i}^{2}, {{{\delta }}_{i}}, {{{\lambda }}_{i}}) $的极大似然估计为$ (\tilde{\eta }_{i}^{2}, {{\tilde{\delta }}_{i}}, {{\tilde{\lambda }}_{i}}) $, $ i = 1, \cdots , k $.令$ (\tilde{\eta }_{i}^{*^2}, {{\tilde{\delta }}^{*}}_{i}, {{\tilde{\lambda }}^{*}}_{i}) $是对应于$ ({\tilde{\eta }_{i}^{2}}, {{\tilde{\delta }}_{i}}, {{\tilde{\lambda }}_{i}}) $的极大似然估计值, $ i = 1, \cdots , k $, 可得

(2.19) $ \begin{equation} {{\tilde{\xi }}^{*}} = {{\bar{y}}_{G2}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}\tilde{\delta }_{i}^{*}}{\tilde{\eta }_{i}^{*}(1-{{b}^{2}}\tilde{\delta }_{i}^{*2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{i}^{*2}(1-{{b}^{2}}\tilde{\delta }_{i}^{*2})}}}, \end{equation} $

其中$ {\bar{y}}_{G2} $是$ {\bar{Y}}_{G2} $的观测值, $ {\bar{Y}}_{G2} $是将$ ({\eta}_{i}^2, {\delta}_{i}) $的极大似然估计代入$ {\bar{Y}}_{G} $所得.

本小节利用Bootstrap方法探讨共同位置参数$ {\xi} $的区间估计和假设检验问题.由中心极限定理可得

(3.1) $ \begin{equation} Z = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}}\left( {{{\bar{Y}}}_{G}}-\xi -\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{\delta }_{i}}}{{{\eta }_{i}}(1-{{b}^{2}}\delta _{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}} \right). \end{equation} $

众所周知, 当$ n_i \rightarrow \infty , i = 1, \cdots , k $时, $ Z\sim N(0, 1) $.令$ U(\beta) $表示标准正态分布的第$ 100\beta $分位点, 可得$ \xi $的一个置信水平近似为$ 100(1-\alpha)\% $的置信区间

$ \left[ {{{\bar{Y}}}_{G}}-\frac{U(1-\alpha /2)}{\sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{\delta }_{i}}}{{{\eta }_{i}}(1-{{b}^{2}}\delta _{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}}, {{{\bar{Y}}}_{G}}-\frac{U(\alpha /2)}{\sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{\delta }_{i}}}{{{\eta }_{i}}(1-{{b}^{2}}\delta _{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}} \right]. $

然而, 在实际问题中$ (\eta_i, \delta_i) $往往是未知的.因此, 可将其矩估计和极大似然估计分别代入(3.1)式, $ i = 1, \cdots , k $.则有

(3.2) $ \begin{equation} {{Z}_{1}} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{2}(1-{{b}^{2}}\hat{\delta }_{i}^{2})}}}\left( {{{\bar{Y}}}_{G1}}-\xi -\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{{\hat{\delta }}}_{i}}}{{{{\hat{\eta }}}_{i}}(1-{{b}^{2}}\hat{\delta }_{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{2}(1-{{b}^{2}}\hat{\delta }_{i}^{2})}}} \right), \end{equation} $

(3.3) $ \begin{equation} {{Z}_{2}} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{i}^{2}(1-{{b}^{2}}\tilde{\delta }_{i}^{2})}}}\left( {{{\bar{Y}}}_{G2}}-\xi -\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{{\tilde{\delta }}}_{i}}}{{{{\tilde{\eta }}}_{i}}(1-{{b}^{2}}\tilde{\delta }_{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{i}^{2}(1-{{b}^{2}}\tilde{\delta }_{i}^{2})}}} \right). \end{equation} $

显然, 枢轴量$ Z_1 $和$ Z_2 $的精确分布是未知的, 故无法找到其分位点的精确值.于是, 本文将基于Bootstrap方法构造共同位置参数$ \xi $的置信区间.令$ Y_{BMij}\sim SN({{\hat{\xi }}^{*}}, \hat{\eta }{{_{i}^{*}}^{^{2}}}, \hat{\lambda }_{i}^{*}), $$ i = 1, \cdots , k $, $ j = 1, \cdots , n_i $.则其样本均值、样本二阶中心矩和样本三阶中心矩分别为$ ({{\bar{Y}}_{BMi}}, $$ {{S}_{BM2i}}, $$ {{S}_{BM3i}}) $.由定理2.1可得, $ (\eta_{i}^2, \delta_{i}) $的矩估计为

(3.4) $ \begin{equation} \hat{\eta }_{BMi}^{2} = {{S}_{BM2i}}+{{c}^{2}}S_{BM3i}^{2/3}, \;{{\hat{\delta }}_{BMi}} = \frac{cS_{BM3i}^{1/3}}{b\sqrt{{{S}_{BM2i}}+{{c}^{2}}S_{BM3i}^{2/3}}}, \;i = 1, \cdots , k. \end{equation} $

令$ (\hat{\eta }_{BMi}^{*^2}, \hat{\delta }_{BMi}^{*}) $是对应于$ (\hat{\eta }_{BMi}^{2}, \hat{\delta }_{BMi}) $的矩估计值.同理, 令$ Y_{BLij}\sim SN({{\tilde{\xi }}^{*}}, \tilde{\eta }{{_{i}^{*}}^{^{2}}}, \tilde{\lambda }_{i}^{*}) $. $ ({{\bar{Y}}_{BLi}}, {{S}_{BL2i}}, {{S}_{BL3i}}) $分别表示$ Y_{BLij} $的样本均值、样本二阶中心矩和样本三阶中心矩.由定理2.3, 则$ (\eta_{i}^2, \delta_{i}) $的极大似然估计可表示为$ (\tilde{\eta }_{BLi}^{2}, {{\tilde{\delta }}_{BLi}}) $, $ i = 1, \cdots , k $.基于$ (\hat{\eta }_{BMi}^{2}, \hat{\delta }_{BMi}) $和$ (\tilde{\eta }_{BLi}^{2}, {{\tilde{\delta }}_{BLi}}) $, 分别定义

(3.5) $ \begin{equation} {{\bar{Y}}_{GB1}} = \frac{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{BMi}^{2}(1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}{{{\bar{Y}}}_{BMi}}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{BMi}^{2}(1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}}}, \end{equation} $

(3.6) $ \begin{equation} {{\bar{Y}}_{GB2}} = \frac{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{BLi}^{2}(1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}{{{\bar{Y}}}_{BLi}}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{BLi}^{2}(1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}}}. \end{equation} $

进而, 类似于(3.2)和(3.3)式, 分别构造Bootstrap枢轴量

(3.7) $ \begin{equation} {{Z}_{B1}} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{BMi}^{2} (1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}}} \left( {{{\bar{Y}}}_{GB1}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{{\hat{\delta }}}_{BMi}}}{{{{\hat{\eta }}}_{BMi}}(1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{BMi}^{2}(1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}}}-\big( {{{\bar{y}}}_{G1}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}\hat{\delta }_{i}^{*}}{\hat{\eta }_{i}^{*}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{*2}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}} \big) \right), \end{equation} $

(3.8) $ \begin{equation} {{Z}_{B2}} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{BLi}^{2} (1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}}} \left( {{{\bar{Y}}}_{GB2}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{{\tilde{\delta }}}_{BLi}}}{{{{\tilde{\eta }}}_{BLi}}(1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{BLi}^{2}(1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}}}-\big( {{{\bar{y}}}_{G2}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}\tilde{\delta }_{i}^{*}}{\tilde{\eta }_{i}^{*}(1-{{b}^{2}}\tilde{\delta }_{i}^{*2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{i}^{*2}(1-{{b}^{2}}\tilde{\delta }_{i}^{*2})}}} \big) \right). \end{equation} $

令$ Z_{B1}(\beta) $表示$ Z_{B1} $的第$ 100\beta $分位点, 可得$ \xi $的一个置信水平近似为$ 100(1-\alpha)\% $的Bootstrap置信区间

$ \left[ {{{\bar{y}}}_{G1}}-\frac{{{Z}_{B1}}(1-\alpha /2)}{\sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{*2}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}\hat{\delta }_{i}^{*}}{\hat{\eta }_{i}^{*}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{*2}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}, {{{\bar{y}}}_{G1}}-\frac{{{Z}_{B1}}(\alpha /2)}{\sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{*2}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}\hat{\delta }_{i}^{*}}{\hat{\eta }_{i}^{*}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{*2}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}} \right]. $

同理可得基于$ Z_{B2} $的一个置信水平近似为$ 100(1-\alpha)\% $的Bootstrap置信区间.

注3.1  对于$ i = 1, \cdots , k $, 当$ {{\lambda }_{i}} = 0 $时, 则$ \eta _{i}^{2} $的矩估计和极大似然估计相等, 即为$ S_{2i} $.于是, (2.17)式中$ \hat{\xi} $为$ \xi $的Graybill-Deal估计.此时, $ \xi $的Bootstrap置信区间退化为徐礼文^[1]的结果.综上所述, 本文将徐礼文^[1]结果从多个正态总体推广到多个偏正态总体.

进一步, 我们考虑共同位置参数$ \xi $的假设检验问题

(3.9) $ \begin{equation} {H_0}:\xi = {\xi _0}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} vs{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {H_1}:\xi \ne {\xi _0}{\kern 1pt} {\kern 1pt}, {\kern 1pt} \end{equation} $

其中$ \xi_0 $为预先给定值.在原假设成立的情况下, 由中心极限定理可得

(3.10) $ \begin{equation} {{Z}^{*}} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}}\left( {{{\bar{Y}}}_{G}}-{{\xi }_{0}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{\delta }_{i}}}{{{\eta }_{i}}(1-{{b}^{2}}\delta _{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}}} \right). \end{equation} $

若$ (\eta_i, \delta_i) $已知, 则$ {Z}^{*} $是假设检验问题(3.9)的近似检验统计量.然而, 在实际问题中, $ (\eta_i, \delta_i) $往往是未知的, 故可分别将$ (\eta_i, \delta_i) $的矩估计和极大似然估计代入(3.10)式, 则有

$ Z_{1}^{*} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{2}(1-{{b}^{2}}\hat{\delta }_{i}^{2})}}}\left( {{{\bar{Y}}}_{G1}}-{{\xi }_{0}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{{\hat{\delta }}}_{i}}}{{{{\hat{\eta }}}_{i}}(1-{{b}^{2}}\hat{\delta }_{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{i}^{2}(1-{{b}^{2}}\hat{\delta }_{i}^{2})}}} \right), $

$ Z_{2}^{*} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{i}^{2}(1-{{b}^{2}}\tilde{\delta }_{i}^{2})}}}\left( {{{\bar{Y}}}_{G2}}-{{\xi }_{0}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{{\tilde{\delta }}}_{i}}}{{{{\tilde{\eta }}}_{i}}(1-{{b}^{2}}\tilde{\delta }_{i}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{i}^{2}(1-{{b}^{2}}\tilde{\delta }_{i}^{2})}}} \right). $

类似于$ Z_1 $和$ Z_2 $, $ Z_{1}^{*} $和$ Z_{2}^{*} $的精确分布亦是未知的, 故无法建立其检验方法.因此, 本文利用Bootstrap方法构造假设检验问题(3.9)的检验统计量

(3.11) $ \begin{equation} \quad Z_{B1}^{*} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{BMi}^{2}(1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}}}\left( {{{\bar{Y}}}_{GB1}}-{{\xi }_{0}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{{\hat{\delta }}}_{BMi}}}{{{{\hat{\eta }}}_{BMi}}(1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\hat{\eta }_{BMi}^{2}(1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}}} \right), \end{equation} $

(3.12) $ \begin{equation} \quad Z_{B2}^{*} = \sqrt{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{BLi}^{2}(1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}}}\left( {{{\bar{Y}}}_{GB2}}-{{\xi }_{0}}-\frac{\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}{{{\tilde{\delta }}}_{BLi}}}{{{{\tilde{\eta }}}_{BLi}}(1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}}}{\sum\limits_{i = 1}^{k}{\frac{{{n}_{i}}}{\tilde{\eta }_{BLi}^{2}(1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}}} \right). \end{equation} $

分别基于$ Z_{B1}^{*} $和$ Z_{B2}^{*} $可得

(3.13) $ \begin{equation} {{p}_{i}} = 2\min \{P(Z_{Bi}^{*}>z_{i}^{*}), P(Z_{Bi}^{*}<z_{i}^{*})\}, \:i = 1, 2, \end{equation} $

其中, $ z_{1}^{*} $和$ z_{2}^{*} $分别表示$ Z_{1}^{*} $和$ Z_{2}^{*} $的观测值.若$ {{p}_{i}}<\alpha $, 则在名义显著性水平$ \alpha $下拒绝假设检验问题(3.9)中的原假设$ H_0 $, 即认为$ \xi $和$ \xi_0 $具有显著性差异.

注3.2  当$ \lambda_1 = \lambda_2 = \cdots = \lambda_k = 0 $时, 可得$ p_1 = p_2 $.此时假设检验问题(3.9)的Bootstrap检验$ p $值退化成徐礼文^[1]的结果.

此外, 本文基于不同权重给出第二种Bootstrap置信区间.令$ \bar{Y} = \sum\limits_{i = 1}^{k}{{{n}_{i}}{{{\bar{Y}}}_{i}}}/\tilde{n} $, 其中$ \tilde{n} = \sum\limits_{i = 1}^{k}{{{n}_{i}}} $.将$ {{\bar{Y}}} $进行标准化, 可得

(3.14) $ \begin{equation} T = \frac{\bar{Y}-\sum\limits_{i = 1}^{k}{{{{n}_{i}}(\xi +b{{\eta }_{i}}{{\delta }_{i}})}/{{\tilde{n}}}\;}}{\sqrt{\sum\limits_{i = 1}^{k}{{{{n}_{i}}\eta _{i}^{2}(1-{{b}^{2}}\delta _{i}^{2})}/{{{{\tilde{n}}}^{2}}}\;}}}. \end{equation} $

分别将$ ({\eta_i^2}, {\delta_i}) $的矩估计和极大似然估计代入(3.14)式, 可得

$ {{T}_{1}} = \frac{{{{\bar{Y}}}_{1}}-\sum\limits_{i = 1}^{k}{{{{n}_{i}}(\xi +b{{{\hat{\eta }}}_{i}}{{{\hat{\delta }}}_{i}})}/{{\tilde{n}}}\;}}{\sqrt{\sum\limits_{i = 1}^{k}{{{{n}_{i}}\hat{\eta }_{i}^{2}(1-{{b}^{2}}\hat{\delta }_{i}^{2})}/{{{{\tilde{n}}}^{2}}}\;}}}, \; \begin{matrix} {} \\ \end{matrix}{{T}_{2}} = \frac{{{{\bar{Y}}}_{2}}-\sum\limits_{i = 1}^{k}{{{{n}_{i}}(\xi +b{{{\tilde{\eta }}}_{i}}{{{\tilde{\delta }}}_{i}})}/{{\tilde{n}}}\;}}{\sqrt{\sum\limits_{i = 1}^{k}{{{{n}_{i}}\tilde{\eta }_{i}^{2}(1-{{b}^{2}}\tilde{\delta }_{i}^{2})}/{{{{\tilde{n}}}^{2}}}\;}}}. $

类似于$ Z_{B1} $和$ Z_{B2} $, 分别构造Bootstrap枢轴量

(3.15) $ \begin{equation} {{T}_{B1}} = \frac{{{{\bar{Y}}}_{B1}}-\sum\limits_{i = 1}^{k}{{{{n}_{i}}\left( \left( \sum\limits_{i = 1}^{k}{{{{n}_{i}}{{{\bar{y}}}_{i}}}/{{\tilde{n}}}\;-\sum\limits_{i = 1}^{k}{{b{{n}_{i}}\hat{\eta }_{i}^{*}\hat{\delta }_{i}^{*}}/{{\tilde{n}}}\;}} \right)+b{{{\hat{\eta }}}_{BMi}}{{{\hat{\delta }}}_{BMi}} \right)}/{{\tilde{n}}}\;}}{\sqrt{\sum\limits_{i = 1}^{k}{{{{n}_{i}}\hat{\eta }_{BMi}^{2}(1-{{b}^{2}}\hat{\delta }_{BMi}^{2})}/{{{{\tilde{n}}}^{2}}}\;}}}, \end{equation} $

(3.16) $ \begin{equation} {{T}_{B2}} = \frac{{{{\bar{Y}}}_{B2}}-\sum\limits_{i = 1}^{k}{{{{n}_{i}}\left( \left( \sum\limits_{i = 1}^{k}{{{{n}_{i}}{{{\bar{y}}}_{i}}}/{{\tilde{n}}}\;-\sum\limits_{i = 1}^{k}{{b{{n}_{i}}\tilde{\eta }_{i}^{*}\tilde{\delta }_{i}^{*}}/{{\tilde{n}}}\;}} \right)+b{{{\tilde{\eta }}}_{BLi}}{{{\tilde{\delta }}}_{BLi}} \right)}/{{\tilde{n}}}\;}}{\sqrt{\sum\limits_{i = 1}^{k}{{{{n}_{i}}\tilde{\eta }_{BLi}^{2}(1-{{b}^{2}}\tilde{\delta }_{BLi}^{2})}/{{{{\tilde{n}}}^{2}}}\;}}}. \end{equation} $

令$ {{T}_{B1}}(\beta ) $表示$ T_{B1} $的第$ 100\beta $分位点, 可得$ \xi $的一个置信水平近似为$ 100(1-\alpha)\% $的Bootstrap置信区间

$ \begin{eqnarray*} && \left[ {{{\bar{y}}}_{1}}-\frac{{{T}_{B1}}(1-\alpha /2)\sqrt{\sum\limits_{i = 1}^{k}{{{n}_{i}}\hat{\eta }_{i}^{*2}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}{{\tilde{n}}}-\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}\hat{\eta }_{i}^{*}\hat{\delta }_{i}^{*}}{{\tilde{n}}}}, \right. \\ &&\left. {{{\bar{y}}}_{1}}-\frac{{{T}_{B1}}(\alpha /2)\sqrt{\sum\limits_{i = 1}^{k}{{{n}_{i}}\hat{\eta }_{i}^{*2}(1-{{b}^{2}}\hat{\delta }_{i}^{*2})}}}{{\tilde{n}}}-\sum\limits_{i = 1}^{k}{\frac{b{{n}_{i}}\hat{\eta }_{i}^{*}\hat{\delta }_{i}^{*}}{{\tilde{n}}}} \right]. \end{eqnarray*} $

同理可得, 基于$ {{T}_{B2}} $的一个置信水平近似为$ 100(1-\alpha)\% $的Bootstrap置信区间.

在假设检验问题(3.9)成立的条件下, 类似于(3.14)式构造检验统计量为

$ {{T}_{1}}^{*} = \frac{{{{\bar{Y}}}_{1}}-\sum\limits_{i = 1}^{k}{{{{n}_{i}}({{\xi }_{0}}+b{{{\hat{\eta }}}_{i}}{{{\hat{\delta }}}_{i}})}/{{\tilde{n}}}\;}}{\sqrt{\sum\limits_{i = 1}^{k}{{{{n}_{i}}\hat{\eta }_{i}^{2}(1-{{b}^{2}}\hat{\delta }_{i}^{2})}/{{{{\tilde{n}}}^{2}}}\;}}}, \;\begin{matrix} {} \\ \end{matrix}{{T}_{2}}^{*} = \frac{{{{\bar{Y}}}_{2}}-\sum\limits_{i = 1}^{k}{{{{n}_{i}}({{\xi }_{0}}+b{{{\tilde{\eta }}}_{i}}{{{\tilde{\delta }}}_{i}})}/{{\tilde{n}}}\;}}{\sqrt{\sum\limits_{i = 1}^{k}{{{{n}_{i}}\tilde{\eta }_{i}^{2}(1-{{b}^{2}}\tilde{\delta }_{i}^{2})}/{{{{\tilde{n}}}^{2}}}\;}}}. $

类似于$ Z_{B1}^{*} $和$ Z_{B2}^{*} $, 构造假设检验问题(3.9)的Bootstrap检验统计量, 分别记为$ T_{B1}^{*} $和$ T_{B2}^{*} $.基于$ T_{B1}^{*} $和$ T_{B2}^{*} $可得

(3.17) $ \begin{equation} p_{i}^{*} = 2\min \{P(T_{_{Bi}}^{*}>{{t}_{i}}), P(T_{_{Bi}}^{*}<{{t}_{i}})\}, \: i = 1, 2, \end{equation} $

同样地, $ t_1 $和$ t_2 $分别表示$ T_{1}^{*} $和$ T_{2}^{*} $的观测值.若$ p_{i}^{*}<\alpha , i = 1, 2 $, 则在名义显著性水平$ \alpha $下拒绝假设检验问题(3.9)中的原假设$ H_0 $, 即认为$ \xi $和$ \xi_0 $具有显著性差异.

本小节通过Monte Carlo模拟, 从数值上研究上述置信区间的覆盖概率和区间长度的统计性质.为方便起见, 本小节针对共同位置参数$ \xi $的区间估计问题, 仅给出基于矩估计的Bootstrap置信区间的覆盖概率和区间长度的算法.

Step 1  对于给定的$ ({{\xi }}, {{n}_{i}}, \eta _{i}^{2}, {{\lambda }_{i}}) $, 生成一组样本$ {{Y}_{ij}}\sim SN({{\xi }}, \eta _{i}^{2}, {{\lambda }_{i}}) $, 由(2.14)式计算$ ({{\bar{Y}}_{i}}, {{S}_{2i}}, {{S}_{3i}}) $, $ i = 1, \cdots , k $, $ j = 1, \cdots , {{n}_{i}} $.

Step 2  由(2.15)和(2.17)式, 可得$ (\xi , \eta _{i}^{2}, {{\lambda }_{i}}) $, $ i = 1, \cdots , k $的矩估计值$ ({{\hat{\xi }}^{*}}, \hat{\eta }_{i}^{*^2}, \hat{\lambda }_{i}^{*}) $, $ i = 1, \cdots , k $.

Step 3  生成Bootstrap样本$ {{Y}_{BMij}}\sim SN({{\hat{\xi }}^{*}}, \hat{\eta }{{_{i}^{*}}^{^{2}}}, \hat{\lambda }_{i}^{*}) $, 并计算$ ({{\bar{Y}}_{BMi}}, {{S}_{2BMi}}, {{S}_{3BMi}}) $, $ i = 1, \cdots , k $, $ j = 1, \cdots , {{n}_{i}} $.

Step 4  由(2.15)式, 利用Bootstrap样本计算矩估计值$ (\hat{\eta }_{BMi}^{*^2}, \hat{\lambda }_{BMi}^{*}) $, $ i = 1, \cdots , k $.进一步, 分别由(3.7)和(3.15)式计算$ {{Z}_{B1}} $和$ {{T}_{B1}} $.

Step 5  将步骤3和步骤4重复$ {{n}_{1}} $次, 可得$ n_1 $个$ {{Z}_{B1}} $和$ {{T}_{B1}} $.进而, 分别给出$ \xi $的置信水平近似为$ 100(1-\alpha )\% $的Bootstrap置信区间.

Step 6  将步骤1和步骤5重复$ {{n}_{2}} $次, 可得$ {{n}_{2}} $个$ \xi $的置信水平近似为$ 1-\alpha $的Bootstrap置信区间.由上述$ {{n}_{2}} $个置信区间中包含$ \xi $的比例, 可得其覆盖概率的Monte Carlo模拟值; 而这$ {{n}_{2}} $个置信区间的平均长度即为Bootstrap区间长度的Monte Carlo模拟值.

在模拟研究中, 令名义置信水平为95%, 内循环数$ {{n}_{1} = 2500} $, 外循环数$ {{n}_{2} = 2500} $.对于两个总体, 设尺度参数$ \eta _{1}^{2} = ({{0.1}^{2}}, {{1}^{2}}), \eta _{2}^{2} = ({{0.3}^{2}}, {{2.5}^{2}}), \eta _{3}^{2} = ({{0.5}^{2}}, {{4}^{2}}), \eta _{4}^{2} = ({{0.7}^{2}}, {{7}^{2}}), $$ \eta _{5}^{2} = ({{0.9}^{2}}, {{10}^{2}}) $; 偏度参数$ ({{\lambda }_{1}}, {{\lambda }_{2}}) = (3, 4), (5, 6), (8, 9) $; 样本量$ ({{n}_{1}}, {{n}_{2}}) = (30, 40), (40, 60), $$ (60, 80), (90, 120), (120, 150), (150, 200) $.

对于三个总体, 设尺度参数$ \eta _{1}^{2} = ({{0.3}^{2}}, {{0.9}^{2}}, {{3}^{2}}), \eta _{2}^{2} = ({{0.4}^{2}}, {{1.1}^{2}}, {{4}^{2}}), \eta _{3}^{2} = ({{0.5}^{2}}, {{1.2}^{2}}, {{5}^{2}}) $, $ \eta _{4}^{2} = ({{0.7}^{2}}, {{1.5}^{2}}, {{7}^{2}}), \eta _{5}^{2} = ({{0.9}^{2}}, {{2.1}^{2}}, {{9}^{2}}) $; 偏度参数$ ({{\lambda }_{1}}, {{\lambda }_{2}}, {{\lambda }_{3}}) = (3, 3, 4), (5, 5, 6), (6, 8, 9) $; 样本量$ ({{n}_{1}}, {{n}_{2}}, {{n}_{3}}) = (30, 40, 40), (40, 50, 60), (60, 80, 90), (90, 100, 120), (120, 150, 180), (150, 200, $$ 240). $

对于五个总体, 设尺度参数$ \eta _{1}^{2} = ({{0.3}^{2}}, {{0.9}^{2}}{{2}^{2}}, {{3}^{2}}, {{3}^{2}}), $$ \eta _{2}^{2} = ({{0.4}^{2}}, {{1.2}^{2}}, {{4}^{2}}, {{4}^{2}}, {{5}^{2}}), $$ \eta _{3}^{2} = ({{0.5}^{2}}, $$ {{1.8}^{2}}, {{6}^{2}}, {{5}^{2}}, {{7}^{2}}), $$ \eta _{4}^{2} = ({{0.7}^{2}}, {{3}^{2}}, {{8}^{2}}, {{7}^{2}}, {{9}^{2}}), $$ \eta _{5}^{2} = ({{0.9}^{2}}, {{5.4}^{2}}, {{10}^{2}}, {{9}^{2}}, {{11}^{2}}) $; 偏度参数$ ({{\lambda }_{1}}, {{\lambda }_{2}}, $$ {{\lambda }_{3}}, {{\lambda }_{4}}, $$ {{\lambda }_{5}}) = (3, 3, 4, 4, 5), (4, 4, 5, 5, 6), (7, 7, 8, 8, 9) $; 样本量$ ({{n}_{1}}, {{n}_{2}}, {{n}_{3}}, {{n}_{4}}, {{n}_{5}}) = (30, 40, 40, 50, $$ 50), $$ (50, 50, 60, 60, 80), $$ (70, 70, 90, 90, 100), $$ (90, 90, 90, 120, 120), $$ (100, 120, 120, 150, 150), $$ (150, $$ 150, 200, 200, 200) $.

类似于第3节中给出的方法, 我们在模拟中增加两种基于惩罚极大似然方法的Bootstrap置信区间^[27], 分别记作$ Z_{mple} $和$ T_{mple} $.

对于两个总体, 表 1和表 4分别给出六种Bootstrap置信区间的模拟覆盖概率和区间长度.就覆盖概率而言, 当偏度参数较小时, $ Z_{mple} $、$ Z_{mle} $和$ Z_{mm} $表现较为自由.随着样本量和偏度参数的增加, 这种现象得到明显改善. $ T_{mple} $、$ T_{mle} $和$ T_{mm} $的实际置信水平接近于95%名义置信水平, 表现较好, 但随着偏度参数的增加, 它们略显保守.就区间长度而言, 六种Bootstrap置信区间的区间长度随着尺度参数的增加而增加, 而随着样本量的增加而减少.然而, 无论是基于矩估计还是极大似然估计, $ Z_{mple} $、$ Z_{mle} $和$ Z_{mm} $在区间长度方面均明显优于$ T_{mple} $、$ T_{mle} $和$ T_{mm} $.

表 2和表 3分别给出六种Bootstrap置信区间在三个和五个总体情况下的模拟覆盖概率. 表 5和表 6分别给出六种Bootstrap置信区间在三个和五个总体情况下的模拟区间长度.就覆盖概率而言, 当偏度参数和样本量较小时, $ Z_{mple} $、$ Z_{mle} $、$ Z_{mm} $和$ T_{mle} $四种方法均呈现出自由的现象.随着样本量和偏度参数的增加, 无论是三个总体还是五个总体, $ Z_{mple} $和$ Z_{mm} $的实际置信水平得到明显改善, 逐渐接近于95%名义置信水平.对于$ T_{mle} $, 仅在三个总体时呈现出$ Z_{mm} $的类似特征. $ T_{mple} $和$ T_{mm} $整体表现较好, 然而, 随着偏度参数的增加略显保守.此外, 四种Bootstrap置信区间长度在三个和五个总体下的模拟结果类似于两个总体.

注4.1  根据$ Z_{mple} $、$ Z_{mle} $、$ Z_{mm} $、$ T_{mple} $、$ T_{mle} $和$ T_{mm} $六种Bootstrap置信区间, 可以建立假设检验问题(3.9)的检验方法.由表 1-表 3可知, $ Z_{mple} $、$ Z_{mle} $、$ Z_{mm} $和$ T_{mle} $四种Bootstrap置信区间在某些参数设置下覆盖概率较低, 故而无法有效控制犯第一类错误的概率.而$ T_{mple} $和$ T_{mm} $均能有效控制犯第一类错误的概率, 且整体表现较好; 但随着偏度参数的增加略显保守.