2. 上海师范大学 数理学院, 上海 200234
2. Mathematics and Science College, Shanghai Normal University, Shanghai 200234, China
Birnbaum-Saunders疲劳寿命分布是概率物理方法中的一个重要失效模型,该模型是BIRNBANUM和SAUDERS[1]于1969年在研究主因裂纹扩展导致材料失效过程中推导而来,主要应用于疲劳失效研究,它比常用寿命分布如Weibull分布等更适合描述某些由疲劳引起失效的产品寿命失效规律.
设T服从两参数Birnbaum-Saunders疲劳寿命分布BS(α, β),其分布函数与密度函数分别为:
$ {F\left( t \right) = \mathit{\Phi }\left[ {\frac{1}{\alpha }\left( {\sqrt {\frac{t}{\beta }} - \sqrt {\frac{\beta }{t}} } \right)} \right],\;\;\;\;t > 0,} $ |
$ {f\left( t \right) = \frac{1}{{2\alpha \sqrt \beta }}\left( {\frac{1}{{\sqrt t }} + \frac{\beta }{{t\sqrt t }}} \right)\varphi \left[ {\frac{1}{\alpha }\left( {\sqrt {\frac{t}{\beta }} - \sqrt {\frac{\beta }{t}} } \right)} \right],\;\;\;\;t > 0,} $ |
其中,α>0为形状参数,β>0为刻度参数(或者称为尺度参数),φ(x), Φ(x)分别为标准正态分布的密度函数与分布函数,即
$ \varphi \left( x \right) = \frac{1}{{\sqrt {2{\rm{ \mathsf{ π} }}} }}{{\rm{e}}^{ - \frac{{{x^2}}}{2}}},\;\;\;\;\mathit{\Phi }\left( x \right) = \int_{ - \infty }^x {\varphi \left( y \right){\rm{d}}y} . $ |
关于两参数Birnbaum-Saunders疲劳寿命分布BS(α, β)的统计分析方法已有众多研究[1-29].需要指出的是,2014年BALAKRISHNAN等[28]证明了在定数截尾和定时截尾下两参数BS分布参数的MLE存在且唯一,这一结果说明当形状参数α>2时,文献[4]关于似然函数有多个极值点这一看法是不正确的.关于两参数BS分布刻度参数的区间估计通常采用文献[19-20]所提出的2种方法,徐晓岭等[29]通过Monte-Carlo模拟说明这2种方法可能无法得到参数β的区间估计.
论文通过对数变换给出了求两参数BS疲劳寿命分布BS(α, β)在全样本场合下参数的对数矩估计.基于对数变换通过一阶泰勒展开,将两参数BS疲劳寿命分布BS(α, β)近似看作两参数对数正态分布,由此得到了2个参数α, β的近似区间估计,通过Monte-Carlo模拟发现,本文的近似方法比原有方法更精确.最后通过若干实例说明方法的可行性.
1 参数点估计的综合比较分析 1.1 方法1:参数点估计的新方法——对数矩估计设T1, T2, …, Tn为来自Birnbaum-Saunders疲劳寿命分布总体T~BS(α, β)的一个容量为n的简单随机样本,其样本观察值为t1, t2, …, tn.
记参数μ=lnβ,令Y=lnT, Yi=lnTi,i=1, 2, …, n,则Y1, Y2, …, Yn为来自分布函数为
令
$ {{F_Z}\left( z \right) = \mathit{\Phi }\left[ {\frac{1}{\alpha }\left( {{{\rm{e}}^z} - {{\rm{e}}^{ - z}}} \right)} \right],} $ |
$ {{f_Z}\left( z \right)\frac{1}{\alpha }\left( {{{\rm{e}}^z} - {{\rm{e}}^{ - z}}} \right)\varphi \left[ {\frac{1}{\alpha }\left( {{{\rm{e}}^z} - {{\rm{e}}^{ - z}}} \right)} \right], - \infty < z < + \infty .} $ |
若k为奇数时,E(Zk)=0;若k为偶数时,
$ E\left( {{Z^k}} \right) = 2\int_0^{ + \infty } {{{\left( {\ln \frac{{\alpha t + \sqrt {{\alpha ^2}{t^2} + 4} }}{2}} \right)}^k}\varphi \left( t \right){\rm{d}}t} . $ |
Y的分布函数FY(y)和密度函数fY(y)分别为:对-∞<y<+∞,
$ {F_Y}\left( y \right) = \mathit{\Phi }\left\{ {\frac{1}{\alpha }\left[ {\exp \left( {\frac{{y - \mu }}{2}} \right) - \exp \left( { - \frac{{y - \mu }}{2}} \right)} \right]} \right\}, $ |
$ \begin{array}{l} {f_Y}\left( y \right) = \frac{1}{{2\alpha }}\left[ {\exp \left( {\frac{{y - \mu }}{2}} \right) - \exp \left( { - \frac{{y - \mu }}{2}} \right)} \right] \times \\ \;\;\;\;\;\;\;\;\;\;\;\;\varphi \left\{ {\frac{1}{\alpha }\left[ {\exp \left( {\frac{{y - \mu }}{2}} \right) - \exp \left( { - \frac{{y - \mu }}{2}} \right)} \right]} \right\}. \end{array} $ |
记
$ E\left( Y \right) = \mu + 2E\left( Z \right) = \mu , $ |
$ \begin{array}{l} E\left( {{Y^2}} \right) = {\mu ^2} + 4E\left( {{Z^2}} \right) = \\ \;\;\;\;\;\;\;\;{\mu ^2} + 8\int_0^{ + \infty } {{{\left( {\ln \frac{{\alpha t + \sqrt {{\alpha ^2}{t^2} + 4} }}{2}} \right)}^2}\varphi \left( t \right){\rm{d}}t} . \end{array} $ |
由此,利用矩估计的思想可得参数μ的矩估计为
引理1[30] 设X1, X2, …, Xn是来自总体X的容量为n的一个简单随机样本,记E(X)=μ,D(X)=σ2<+∞,该总体X的4阶中心矩v4=E(X-EX)4有限,若函数h(x)的四阶导数存在且有界,则有
$ E\left( {h\left( {\bar X} \right)} \right) = h\left( u \right) + \frac{1}{{2n}}h''\left( \mu \right){\sigma ^2} + O\left( {{n^{ - 2}}} \right), $ |
$ \begin{array}{*{20}{c}} {D\left[ {h\left( {\bar X} \right)} \right] = \frac{1}{n}{{\left[ {h'\left( \mu \right)} \right]}^2}{\sigma ^2} + \frac{1}{{{n^2}}}\left\{ {h'\left( \mu \right)h''\left( \mu \right){v_3} + } \right.}\\ {\left. {\frac{1}{2}{{\left[ {h''\left( \mu \right)} \right]}^2}{\sigma ^4} + h'\left( \mu \right)h''\left( \mu \right){\sigma ^4}} \right\} + O\left( {{n^{ - 3}}} \right).} \end{array} $ |
定理1
证明 Y=2Z+μ, E(Y)=μ, Y-E(Y)=2Z,则Y的一至四阶中心矩为:
$ {v_1} = E\left[ {Y - E\left( Y \right)} \right] = 0, $ |
$ \begin{array}{l} {v_2} = E{\left[ {Y - E\left( Y \right)} \right]^2} = 4E\left( {{Z^2}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;8\int_0^{ + \infty } {{{\left( {\ln \frac{{\alpha t + \sqrt {{\alpha ^2}{t^2} + 4} }}{2}} \right)}^2}\varphi \left( t \right){\rm{d}}t} , \end{array} $ |
$ {v_3} = E{\left[ {Y - E\left( Y \right)} \right]^3} = 8E\left( {{Z^3}} \right) = 0, $ |
$ \begin{array}{l} {v_4} = E{\left[ {Y - E\left( Y \right)} \right]^4} = 16E\left( {{Z^4}} \right) = \\ \;\;\;\;\;\;\;\;\;32\int_0^{ + \infty } {{{\left( {\ln \frac{{\alpha t + \sqrt {{\alpha ^2}{t^2} + 4} }}{2}} \right)}^2}\varphi \left( t \right){\rm{d}}t} . \end{array} $ |
令函数h(x)=ex,h(x)的任意阶导数任为ex,则
$ \begin{array}{l} E\left( {{{\mathop \beta \limits^ \wedge }_1}} \right) = E\left[ {\exp \left( {\bar Y} \right)} \right] = {{\rm{e}}^\mu } + \frac{1}{{2n}}{{\rm{e}}^\mu } \cdot 4E\left( {{Z^2}} \right) + O\left( {{n^{ - 2}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\beta + \frac{2}{n}\beta E\left( {{Z^2}} \right) + O\left( {{n^{ - 2}}} \right), \end{array} $ |
$ \begin{array}{l} D\left( {{{\mathop \beta \limits^ \wedge }_1}} \right) = D\left[ {\exp \left( {\bar Y} \right)} \right] = \frac{1}{{2n}}{{\rm{e}}^{2\mu }} \cdot 4E\left( {{Z^2}} \right) + \\ \frac{1}{{{n^2}}}\left\{ {\frac{1}{2}{{\rm{e}}^{2\mu }} \cdot 16{{\left[ {E\left( {{Z^2}} \right)} \right]}^2} + {{\rm{e}}^{2\mu }} \cdot 16{{\left[ {E\left( {{Z^2}} \right)} \right]}^2}} \right\} + \\ O\left( {{n^{ - 3}}} \right) = \frac{4}{n}{\beta ^2}E\left( {{Z^2}} \right) + \frac{{24}}{{{n^2}}}{\beta ^2}{\left[ {E\left( {{Z^2}} \right)} \right]^2} + O\left( {{n^{ - 3}}} \right). \end{array} $ |
易见,
则
$ \int_0^{ + \infty } {{{\left( {\ln \frac{{\alpha t + \sqrt {{\alpha ^2}{t^2} + 4} }}{2}} \right)}^2}\varphi \left( t \right){\rm{d}}t} = \frac{{\overline {{Y^2}} - {{\bar Y}^2}}}{8}. $ |
易证上述方程有唯一正实根.
需要指出的是,
${{\overset{\wedge }{\mathop{\alpha }}\,}_{1}}=\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( \sqrt{\frac{{{t}_{i}}}{{{\overset{\wedge }{\mathop{\beta }}\,}_{1}}}}-\sqrt{\frac{{{\overset{\wedge }{\mathop{\beta }}\,}_{1}}}{{{t}_{i}}}} \right)}^{2}}}}.$ |
似然函数为:
$ \begin{array}{l} L\left( {\alpha ,\beta } \right) = \\ \prod\limits_{i = 1}^n {\left\{ {\frac{1}{{2\alpha \sqrt \beta }}\left( {\frac{1}{{\sqrt {{t_i}} }} + \frac{\beta }{{{t_i}\sqrt {{t_i}} }}} \right)\varphi \left[ {\frac{1}{\alpha }\left( {\sqrt {\frac{{{t_i}}}{\beta }} - \sqrt {\frac{\beta }{{{t_i}}}} } \right)} \right]} \right\}} . \end{array} $ |
令
$ \left\{ \begin{array}{l} - \frac{n}{\alpha } + \frac{1}{{{\alpha ^3}}}\sum\limits_{i = 1}^n {{{\left( {\sqrt {\frac{{{t_i}}}{\beta }} - \sqrt {\frac{\beta }{{{t_i}}}} } \right)}^2}} = 0\\ - \frac{n}{{2\beta }} + \sum\limits_{i = 1}^n {\frac{1}{{{t_i} + \beta }} - \frac{1}{{2{\alpha ^2}}}} \sum\limits_{i = 1}^n {\left( { - \frac{{{t_i}}}{{{\beta ^2}}} + \frac{1}{{{t_i}}}} \right) = 0} , \end{array} \right. $ |
化简得仅含参数β的超越方程:
$ \frac{{\sum\limits_{i = 1}^n {\left( { - {t_i} + \frac{{{\beta ^2}}}{{{t_i}}}} \right)} }}{{\sum\limits_{i = 1}^n {\left( {{t_i} + \frac{{{\beta ^2}}}{{{t_i}}} - 2\beta } \right)} }} - \frac{2}{n}\sum\limits_{t = 1}^n {\frac{\beta }{{{t_i} + \beta }} + 1 = 0.} $ |
从上述超越方程中可以解出
${{\overset{\wedge }{\mathop{\alpha }}\,}_{2}}=\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( \sqrt{\frac{{{t}_{i}}}{{{\overset{\wedge }{\mathop{\beta }}\,}_{2}}}}-\sqrt{\frac{{{\overset{\wedge }{\mathop{\beta }}\,}_{2}}}{{{t}_{i}}}} \right)}^{2}}}}.$ |
需要指出的是,极大似然估计需要解非线性方程组,且计算较为复杂.
1.3 方法3:矩估计作简单运算,可得BS(α, β)分布的如下特征.
引理2[14] 设随机变量T~BS(α, β),则
(1) T-1~BS(α, β-1);
(2)
$ D\left( T \right) = \frac{{{\beta ^2}}}{4}\left( {5{\alpha ^4} + 4{\alpha ^2}} \right); $ |
(3)
(4) 变异系数为
记
$ \frac{\beta }{2}\left( {{\alpha ^2} + 2} \right) = \bar T,\frac{{{\beta ^2}}}{2}\left( {3{\alpha ^4} + 4{\alpha ^2} + 2} \right) = \overline {{T^2}} . $ |
化简得仅含参数α的超越方程:
$ 2\frac{{3{\alpha ^4} + 4{\alpha ^2} + 2}}{{{{\left( {{\alpha ^2} + 2} \right)}^2}}} = \frac{{\overline {{T^2}} }}{{{{\bar T}^2}}}. $ |
记
$ 6{\alpha ^4} + 8{\alpha ^2} + 4 = c{\alpha ^4} + 4c{\alpha ^2} + 4c, $ |
即
$ \left( {c - 6} \right){\alpha ^4} + 4\left( {c - 2} \right){\alpha ^2} + 4\left( {c - 1} \right) = 0, $ |
$ \begin{array}{l} \Delta = 16{\left( {c - 2} \right)^2} - 16\left( {c - 6} \right)\left( {c - 1} \right) = \\ \;\;\;\;\;\;16\left( {3c - 2} \right) > 0, \end{array} $ |
又
$ \begin{array}{l} {\left( {c - 2} \right)^2} - \left( {3c - 2} \right) = {c^2} - 7c + 6 = \\ \;\;\;\;\;\left( {c - 1} \right)\left( {c - 6} \right). \end{array} $ |
于是可知:当c=6时,方程无实根;当1<c<6时,方程存在唯一正实根:
由此,当1<c<6时,可得参数α的矩估计为
要使方法3的矩估计存在,则样本数据应满足:1<c<6,否则点估计不存在.下表 1给定样本容量n=10, 20, 30, 40, 50以及参数真值β=1,α=0.1, 0.5, 1, 2, 5, 7, 10, 15,通过10 000次模拟统计了其数据满足1<c<6的次数,从中可以看到,在α较小时,方法3的矩估计基本是存在的.
由于T-1~BS(α, β-1),记
$ \frac{\beta }{2}\left( {{\alpha ^2} + 2} \right) = \bar T,\;\;\;\frac{1}{{2\beta }}\left( {{\alpha ^2} + 2} \right) = \overline {{T^{ - 1}}} , $ |
从中可解得参数β, α的矩估计分别为:
${{\overset{\wedge }{\mathop{\beta }}\,}_{4}}=\sqrt{\frac{{\bar{T}}}{\overline{{{T}^{-1}}}}},\quad {{\overset{\wedge }{\mathop{\alpha }}\,}_{4}}=\sqrt{2\left( \sqrt{\bar{T}\overline{{{T}^{-1}}}}-1 \right)}.$ |
文献[15]给出了参数的逆矩估计如下:
${{\overset{\wedge }{\mathop{\beta }}\,}_{5}}=\frac{\sum\limits_{i=1}^{n}{\sqrt{{{T}_{i}}}}}{\sum\limits_{i=1}^{n}{\sqrt{\frac{1}{{{T}_{i}}}}}}=\frac{\overline{\sqrt{T}}}{\overline{\sqrt{{{T}^{-1}}}}},$ |
${{\overset{\wedge }{\mathop{\alpha }}\,}_{5}}=\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( \sqrt{\frac{{{T}_{i}}}{{{\overset{\wedge }{\mathop{\beta }}\,}_{5}}}}-\sqrt{\frac{{{\overset{\wedge }{\mathop{\beta }}\,}_{5}}}{{{T}_{i}}}} \right)}^{2}}}},$ |
其中,
设T1, T2, …, Tn为来自Birnbaum-Saunders疲劳寿命分布总体T~BS(α, β)的一个容量为n的简单随机样本,其次序统计量记为T(1)≤T(2)≤…≤T(n).
由于
当n为奇数时,
${{\overset{\wedge }{\mathop{\beta }}\,}_{6}}={{T}_{\left( \left[ \left( n+1 \right)/2 \right] \right)}};$ |
当n为偶数时,
${{\overset{\wedge }{\mathop{\beta }}\,}_{6}}=\frac{{{T}_{\left( n/2 \right)}}+{{T}_{\left( n/2+1 \right)}}}{2}.$ |
进而参数α的点估计可取为
${{\overset{\wedge }{\mathop{\alpha }}\,}_{6}}=\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( \sqrt{\frac{{{T}_{i}}}{{{\overset{\wedge }{\mathop{\beta }}\,}_{6}}}}-\sqrt{\frac{{{\overset{\wedge }{\mathop{\beta }}\,}_{6}}}{{{T}_{i}}}} \right)}^{2}}}}.$ |
文献[21]利用回归分析模型,给出了如下参数的最小二乘估计:
${{\overset{\wedge }{\mathop{\beta }}\,}_{7}}=\sqrt{\frac{\sum\limits_{i=1}^{n}{{{T}_{i}}}}{\sum\limits_{i=1}^{n}{\frac{1}{{{T}_{i}}}}}}=\sqrt{\frac{{\bar{T}}}{\overline{{{T}^{-1}}}}},$ |
${{\overset{\wedge }{\mathop{\alpha }}\,}_{7}}=\sqrt{\frac{2n}{n-1}\left( \sqrt{\bar{T}\overline{{{T}^{-1}}}}-1 \right)},$ |
注意到,
由于
$ \frac{1}{\alpha }\left( {\sqrt {\frac{T}{\beta }} - \sqrt {\frac{\beta }{T}} } \right) \sim N\left( {0,1} \right), $ |
$ \frac{1}{{\alpha \sqrt \beta }}\left( {\sqrt T - \frac{\beta }{{\sqrt T }}} \right) \sim N\left( {0,1} \right), $ |
则
$ Z = \sqrt T - \frac{\beta }{{\sqrt T }} \sim N\left( {0,{\alpha ^2}\beta } \right). $ |
令
$ Q = \sum\limits_{i = 1}^n {{{\left( {\sqrt {{T_i}} - \frac{\beta }{{\sqrt {{T_i}} }}} \right)}^2}} , $ |
$ \begin{array}{l} \frac{{{\rm{d}}Q}}{{{\rm{d}}\beta }} = - 2\sum\limits_{i = 1}^n {\left[ {\frac{1}{{\sqrt {{T_i}} }}\left( {\sqrt {{T_i}} - \frac{\beta }{{\sqrt {{T_i}} }}} \right)} \right]} = \\ \;\;\;\;\;\;\;\; - 2\sum\limits_{i = 1}^n {\left( {1 - \frac{\beta }{{{T_i}}}} \right)} . \end{array} $ |
令
解得
${{\hat \beta }_8} = \frac{n}{{\sum\limits_{i = 1}^n {\frac{1}{{{T_i}}}} }} = \frac{1}{{\overline {{T^{ - 1}}} }},$ |
进而,
$\begin{array}{*{35}{l}} \overset{\wedge }{\mathop{\alpha _{8}^{2}}}\,{{\overset{\wedge }{\mathop{\beta }}\,}_{8}}=\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{\left( \sqrt{{{T}_{i}}}-\frac{{{\overset{\wedge }{\mathop{\beta }}\,}_{8}}}{\sqrt{{{T}_{i}}}} \right)}^{2}}}= \\ \ \ \ \ \ \ \ \ \ \frac{1}{n-1}\sum\limits_{i=1}^{n}{\left( {{T}_{i}}+\frac{\overset{\wedge }{\mathop{\beta _{8}^{2}}}\,}{{{T}_{i}}}-2{{\overset{\wedge }{\mathop{\beta }}\,}_{8}} \right)=} \\ \ \ \ \ \ \ \ \ \ \frac{1}{n-1}\left( \sum\limits_{i=1}^{n}{{{T}_{i}}+\overset{\wedge }{\mathop{\beta _{8}^{2}}}\,\sum\limits_{i=1}^{n}{\frac{1}{{{T}_{i}}}-2n{{\overset{\wedge }{\mathop{\beta }}\,}_{8}}}} \right)= \\ \ \ \ \ \ \ \ \ \ \frac{n}{n-1}\left( \bar{T}-\frac{1}{\overline{{{T}^{-1}}}} \right), \\ \end{array}$ |
${{\overset{\wedge }{\mathop{\alpha }}\,}_{8}}=\sqrt{\frac{n}{n-1}\left( \bar{T}\overline{{{T}^{-1}}}-1 \right)}.$ |
由于
$ \frac{1}{\alpha }\left( {\sqrt {\frac{T}{\beta }} - \sqrt {\frac{\beta }{T}} } \right) \sim N\left( {0,1} \right), $ |
$ \frac{{\sqrt \beta }}{\alpha }\left( {\frac{{\sqrt T }}{\beta } - \frac{1}{{\sqrt T }}} \right) \sim N\left( {0,1} \right), $ |
则
$Z = \frac{{\sqrt T }}{\beta } - \frac{1}{{\sqrt T }} \sim N\left( {0,\frac{{{\alpha ^2}}}{\beta }} \right).$ |
令
$ Q = \sum\limits_{i = 1}^n {{{\left( {\frac{{\sqrt {{T_i}} }}{\beta } - \frac{1}{{\sqrt {{T_i}} }}} \right)}^2}} , $ |
$ \begin{array}{l} \frac{{{\rm{d}}Q}}{{{\rm{d}}\beta }} = - \frac{2}{{{\beta ^2}}}\sum\limits_{i = 1}^n {\left[ {\sqrt {{T_i}} \left( {\frac{{\sqrt {{T_i}} }}{\beta } - \frac{1}{{\sqrt {{T_i}} }}} \right)} \right]} = \\ \;\;\;\;\;\;\;\; - \frac{2}{{{\beta ^2}}}\sum\limits_{i = 1}^n {\left( {\frac{{{T_i}}}{\beta } - 1} \right)} , \end{array} $ |
令
解得
$ \begin{array}{l} \frac{{\mathop {\alpha _9^2}\limits^ \wedge }}{{\mathop {{\beta _9}}\limits^ \wedge }} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {\frac{{\sqrt {{T_i}} }}{{\mathop {{\beta _9}}\limits^ \wedge }} - \frac{1}{{\sqrt {{T_i}} }}} \right)}^2}} = \\ \;\;\;\;\;\;\;\;\frac{1}{{n - 1}}\sum\limits_{i = 1}^n {\left( {\frac{{{T_i}}}{{\mathop {\beta _9^2}\limits^ \wedge }} - \frac{2}{{\mathop {{\beta _9}}\limits^ \wedge }} + \frac{1}{{{T_i}}}} \right)} , \end{array} $ |
$ \begin{array}{l} \mathop {\alpha _9^2}\limits^ \wedge = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {\left( {\frac{{{T_i}}}{{\mathop {{\beta _9}}\limits^ \wedge }} - 2 + \frac{{\mathop {{\beta _9}}\limits^ \wedge }}{{{T_i}}}} \right)} = \\ \;\;\;\;\;\;\;\;\frac{n}{{n - 1}}\left( {\bar T\overline {{T^{ - 1}}} - 1} \right), \end{array} $ |
$\overset{\wedge }{\mathop{{{\alpha }_{9}}}}\,=\sqrt{\frac{n}{n-1}\left( \bar{T}\overline{{{T}^{-1}}}-1 \right)}.$ |
比较上述9种点估计方法,方法2由于涉及复杂的超越方程求解,在此不推荐使用.
下面通过10 000次Monte-Carlo模拟比较方法1、方法3~方法9的β点估计的精度.给定样本容量n=10, 20, 30, 35,参数α=0.5, 1, 1.5,β=1,通过10 000次模拟得参数β的估计均值与均方差(其中方法3都满足1<c<6),结果列于表 2.
(1) 固定n,随着α的增大,参数β估计的均值与真值的偏差增大,均方误差也增大;
(2) 固定参数α的真值,参数β的估计随着n的增加变精确;
(3) 方法1,4,5中参数β估计的无偏性较为明显,同时其均方误差也较小,相对而言,方法4的均方误差更小,方法5与方法1均方误差很接近.
下面通过10 000次Monte-Carlo模拟,比较方法1、方法3~方法8(方法9与方法8同)的α点估计的精度.给定样本容量n=10, 20, 30, 35,参数α=0.5, 1, 1.5,β=1,通过10 000次模拟得参数α的估计均值与均方差(其中方法3都满足1<c<6),结果列于表 3.
(1) 固定α,随着n的增大,参数α估计的均方误差在减少,即估计变精确;
(2) 方法1、方法4~方法7的均方误差相差不大,考虑到其均值,使用方法6或方法7更合理.
2 参数区间估计的比较分析首先,给出利用对数正态分布求参数的区间估计方法(记为方法2),进而通过Monte-Carlo模拟与文献[20-21]的方法(记为方法1)比较分析(在方法1能得到刻度参数β的区间估计的前提下).
记参数μ=ln β,令Y=lnT, Yi=lnTi,i=1, 2, …, n,则Y1, Y2, …, Yn为来自分布函数为
${{F}_{Y}}\left( y \right)=\mathit{\Phi }\left\{ \frac{1}{\alpha }\exp \left( \frac{y-\mu }{2} \right)-\exp \left( -\frac{y-\mu }{2} \right) \right\}$ |
的一个容量为n的简单随机样本,其次序观察值记为y1, y2, …, yn.
将
$ \bar Y = \frac{1}{n}\sum\limits_{i = 1}^n {{Y_i}} = \frac{1}{n}\sum\limits_{i = 1}^n {\ln {T_i}} , $ |
$ S_Y^2 = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {{Y_i} - \bar Y} \right)}^2}} = \frac{n}{{n - 1}}\left( {\overline {{Y^2}} - {{\bar Y}^2}} \right). $ |
由于
$ \left[ {\bar Y - \frac{{{S_Y}}}{{\sqrt n }}{t_{\gamma /2}}\left( {n - 1} \right),\bar Y + \frac{{{S_Y}}}{{\sqrt n }}{t_{\gamma /2}}\left( {n - 1} \right)} \right]. $ |
进而参数β的置信水平1-γ的近似区间估计为
$ \left[ {\exp \left[ {\bar Y - \frac{{{S_Y}}}{{\sqrt n }}{t_{\gamma /2}}\left( {n - 1} \right)} \right],\exp \left[ {\bar Y + \frac{{{S_Y}}}{{\sqrt n }}{t_{\gamma /2}}\left( {n - 1} \right)} \right]} \right], $ |
又
$ \frac{{\left( {n - 1} \right)S_Y^2}}{{{\alpha ^2}}}\dot \sim {\chi ^2}\left( {n - 1} \right). $ |
参数α的置信水平1-γ的近似区间估计为
$ \left[ {\sqrt {\frac{{\left( {n - 1} \right)S_Y^2}}{{\chi _{\gamma /2}^2\left( {n - 1} \right)}}} ,\sqrt {\frac{{\left( {n - 1} \right)S_Y^2}}{{\chi _{1 - \gamma /2}^2\left( {n - 1} \right)}}} } \right]. $ |
下面通过10 000次Monte-Carlo模拟,比较方法1、方法2关于参数α, β的区间估计的优劣.给定样本容量n=5, 10, 15,参数α=0.5, 1, 1.5,β=1,置信水平1-γ=0.90,通过10 000次模拟(方法1中的参数β的区间估计都存在)得参数α, β区间估计的平均下限、平均上限、平均长度,以及区间估计包含参数真值的次数,结果列于表 4.可知方法1、方法2所得的区间估计包含参数的真值次数都大于9 000;方法2所得的区间估计的平均长度较方法1要短.可知方法2比方法1更优.
例1[14] 数据来自BJERKEDAL(1960), 也被GUPTAETAL(1997)分析过.数据表示几内亚猪注射不同剂量结核杆菌的生存时间.几内亚猪对结核杆菌的敏感性比人类更高,首先关注在相同笼子里用相同养殖法养殖的猪.对于文中的养殖方法,有如下72个观测值:
12,15,22,24,24,32,32,33,34,38,38,43,44,48,52,53,54,54,55,56,57,58,58,59,60,60,60,60,61,62,63,65,65,67,68,70,70,72,73,75,76,76,81,83,84,85,87,91,95,96,98,99,109,110,121,127,129,131,143,146,146,175,175,211,233,258,258,263,297,341,341,376.
经计算得:c=1.651 2, d=0.063 1.
取置信水平为0.90,
$ {t_{0.05}}\left( {71} \right) = 1.6666,\;\;\;\;\chi _{0.05}^2\left( {71} \right) = 91.6702, $ |
$ \chi _{0.95}^2\left( {71} \right) = 52.6003. $ |
利用区间估计方法1、方法2得参数β, α的置信水平为0.90的区间估计,如表 6如示.
例2[7] 本实例中的数据集为6061-T6铝合金的疲劳寿命.铝合金的切取位置应平行于轧制方向,振荡频率为18Hz.该数据集包括101个观测值,最大应力为31 000Pa.数据如下:
70,90,96,97,99,100,103,104,104,105,107,108,108,108,109,109,112,112,113,114,114,114,116,119,120,120,120,121,121,123,124,124,124,124,124,128,128,129,129,130,130,130,131,131,131,131,131,132,132,132,133,134,134,134,134,134,136,136,137,138,138,138,139,139,141,141,142,142,142,142,142,142,144,144,145,146,148,148,149,151,151,152,155,156,157,157,157,157,158,159,162,163,163,164,166,166,168,170,174,196,212.
经计算得:c=1.027 7, d=0.003 6.
文献[7]还给出了其他几种估计:
取置信水平为0.90,
$ {t_{0.05}}\left( {100} \right) = 1.66023,\;\;\;\;\chi _{0.05}^2\left( {100} \right) = 124.342, $ |
$ \chi _{0.95}^2\left( {100} \right) = 77.9295. $ |
取置信水平为0.95,
$ {t_{0.025}}\left( {100} \right) = 1.98397,\;\;\;\;\chi _{0.025}^2\left( {100} \right) = 129.561, $ |
$ \chi _{0.975}^2\left( 9 \right) = 74.2219. $ |
利用区间估计方法1、方法2得参数β, α的置信水平为0.90, 0.95时的区间估计,如表 9如示.
例3[31] 对于空气污染物浓度,假定数据不相关且相互独立,因此,一昼夜或循环趋势分析是无必要的.这一假设得到了很多学者(包括GOKHALE和KHARE等)的支持.例如,环境数据有时以均值作为指标,因此空间-时间依赖性不复存在.以下数据对应的是1973年5~9月纽约每日的臭氧浓度(由纽约州环境保护部提供):41,36,12,18,28,23,19,8,7,16,11,14,18,14,34,6,30,1,11,4,32,23,45,115,37,29,71,39,23,21,37,20,12,13,49,32,64,40,77,97,97,85,10,27,7,48,35,61,79,63,16,108,20,52,82,50,64,59,39,9,16,78,35,66,122,89,110,44,65,22,59,23,31,44,21,9,45,168,73,76,118,84,85,96,78,91,47,32,20,23,21,24,44,21,28,9,13,46,18,13,24,16,23,36,7,14,30,14,18,20,11,135,80,28,73,13.
经计算得:c=1.607 8, d=0.092 8.
取置信水平为0.90,
$ {t_{0.05}}\left( {115} \right) = 1.65821,\;\;\;\;\chi _{0.05}^2\left( {115} \right) = 141.03, $ |
$ \chi _{0.95}^2\left( {115} \right) = 91.2422. $ |
利用区间估计方法1、方法2得参数β, α的置信水平为0.90时的区间估计,如表 11如示.
例4[32] (1)第1个实例是有关洪水峰值的超过数,含1958~1984年共72个超过数,四舍五入到小数点后1位,数据如下:
1.7,2.2,14.4,1.1,0.4,20.6,5.3,0.7,1.9,13.0,12.0,9.3,1.4,18.7,8.5,25.5,11.6,14.1,22.1,1.1,2.5,14.4,1.7,37.6,0.6,2.2,39.0,0.3,15.0,11.0,7.3,22.9,1.7,0.1,1.1,0.6,9.0,1.7,7.0,20.1,0.4,2.8,14.1,9.9,10.4,10.7,30.0,3.6,5.6,30.8,13.3,4.2,25.5,3.4,11.9,21.5,27.6,36.4,2.7,64.0,1.5,2.5,27.4,1.0,27.1,20.2,16.8,5.3,9.7,27.5,2.5,27.0.
经计算得:c=2.001 2, d=0.247 5.
取置信水平为0.90,
$ {t_{0.05}}\left( {71} \right) = 1.666,\;\;\;\;\chi _{0.05}^2\left( {71} \right) = 91.6702, $ |
$ \chi _{0.95}^2\left( {71} \right) = 52.6003. $ |
利用区间估计的方法1、方法2得参数β, α的置信水平为0.90时的区间估计,如表 13如示.
(2) 第2个实例是有关50个工业设备的使用期,时间为零时进行寿命测试,数据如下:
0.1,0.2,1,1,1,1,1,2,3,6,7,11,12,18,18,18,18,18,21,32,36,40,45,46,47,50,55,60,63,63,67,67,67,67,72,75,79,82,82,83,84,84,84,85,85,85,85,85,86,86.
经计算得:c=1.506 2, d=0.382.
取置信水平为0.90,
$ {t_{0.05}}\left( {49} \right) = 1.67655,\;\;\;\;\chi _{0.05}^2\left( {49} \right) = 66.3386, $ |
$ \chi _{0.95}^2\left( {49} \right) = 33.9303. $ |
利用区间估计方法1、方法2得参数β, α的置信水平为0.90时的区间估计,如表 15如示.
[1] | BIRNBAUM Z W, SAUNDERS S C. A new family of life distribution[J]. Journal of Applied Probability, 1969, 6(2): 319–327. DOI:10.1017/S0021900200032848 |
[2] | DESMOND A F. Stochastic models of failure in random environments[J]. Canadian Journal of Statistics, 1985, 13(3): 171–183. DOI:10.2307/3315148 |
[3] | DESMOND A F. On the relationship between two fatigue-life models[J]. IEEE Transactions on Reliability, 1986, 35(2): 167–169. DOI:10.1109/TR.1986.4335393 |
[4] | BIRNBAUM Z W, SAUNDERS S C. Estimation for a family of life distributions with applications to fatigue[J]. Journal of Applied Probability, 1969, 6(2): 328–347. DOI:10.1017/S002190020003285X |
[5] | ENGELHARDT M, BAIN L J, WRIGHT F T. Inferences on the parameters of the Birnbaum Saunders fatigue life distribution based on maximum likelihood estimation[J]. Technometrics, 1981, 23(3): 251–256. DOI:10.2307/1267788 |
[6] | RIECK J R, NEDELMAN J R. A log-linear model for the Birnbaum Saunders distribution[J]. Techanometrics, 1991, 33(1): 51–60. |
[7] | NG H K T, KUNDU D, BALAKRISHNAN N. Modified moment estimation for the two-parameter Birnbaum-Saunders distribution[J]. Computational Statistics and Data Analysis, 2003, 43(3): 283–298. DOI:10.1016/S0167-9473(02)00254-2 |
[8] | DUPUIS D J, MILLS J E. Robust estimation of the Birnbaum-Saunders distribution[J]. IEEE Transactions on Reliability, 1998, 47(1): 88–95. DOI:10.1109/24.690913 |
[9] | CHANG D S, TANG L C. Reliability bounds and critical time for the Birnbaum-Saunders distribution[J]. IEEE Transactions on Reliability, 1993, 42(3): 464–469. DOI:10.1109/24.257832 |
[10] | RIECK J R. Parametric estimation for the Birnbaum-Saunders distribution based on symmetrically censored samples[J]. Communication in Statistics-Theory and Methods, 1995, 24(7): 1721–1736. DOI:10.1080/03610929508831581 |
[11] | OWEN W J, PADGETT W J. A Birnbaum-Saunders accelerated life model[J]. IEEE Transactions on Reliability, 2000, 49(2): 224–229. DOI:10.1109/24.877342 |
[12] | OWEN W J, PADGETT W J. Acceleration models for system strength based on Birnbaum-Saunders distribution[J]. Lifetime Data Analysis, 1999, 5(2): 133–147. DOI:10.1023/A:1009649428243 |
[13] | OWEN W J, PADGETT W J. Power-law accelerated Birnbaum-Saunders life models[J]. International Journal of Reliability Quality and Safety Engineering, 2000, 7(7): 1–15. |
[14] | KUNDU D, KANNAN N, BALAKRISHNAN N. On the hazard function of Birnbaum-Saunders distribution and associated inference[J]. Computational Statistics & Data Analysis, 2008, 52(5): 2692–2702. |
[15] |
王炳兴, 王玲玲. Birnbaum-Saunders疲劳寿命分布的参数估计[J].
华东师范大学学报:自然科学版, 1996(4): 10–15.
WANG B X, WANG L L. Parameter estimation of Birnbaum-Saunders fatigue life distribution[J]. Journal of East China Normal University:Natural Sciences, 1996(4): 10–15. |
[16] |
王炳兴, 王玲玲. Birnbaum-Saunders疲劳寿命分布在截尾试验情形的统计分析[J].
应用概率统计, 1996, 12(4): 369–375.
WANG B X, WANG L L. Statistical analysis of Birnbaum-Saunders fatigue life distribution in the censored test case[J]. Applied Probability and Statistics, 1996, 12(4): 369–375. |
[17] |
王蓉华, 费鹤良. 双边截尾场合下BS疲劳寿命分布的参数估计[J].
上海师范大学学报:自然科学版, 1999, 28(2): 17–22.
WANG R H, FEI H L. Parameter estimation for the BS fatigue life distribution under bilateral censoring[J]. Journal of Shanghai Normal University:Natural Sciences, 1999, 28(2): 17–22. |
[18] | WANG R H, FEI H L. Statistical analysis for the Birnbaum-Saunders fatigue life distribution under multiply type Ⅱ censoring[J]. Chinese Quarterly Journal of Mathematics, 2006, 21(1): 15–27. |
[19] | WANG R H, FEI H L. Statistical analysis for the Birnbaum-Saunders fatigue life distribution under type Ⅱ bilateral censoring and multiply type Ⅱ censoring[J]. Chinese Quarterly Journal of Mathematics, 2004, 19(2): 126–132. |
[20] |
孙祝岭. Birnbaum-Saunders疲劳寿命分布尺度参数的区间估计[J].
兵工学报, 2009, 30(11): 1558–1561.
SUN Z L. Interval estimation of scale parameter for Birnbaum-Saunders fatigue life distribution[J]. Acta Armamentarii, 2009, 30(11): 1558–1561. DOI:10.3321/j.issn:1000-1093.2009.11.028 |
[21] |
孙祝岭. Birnbaum-Saunders疲劳寿命分布参数的回归估计方法[J].
兵工学报, 2010, 31(9): 1260–1262.
SUN Z L. Regression estimation method of parameters for Birnbaum-Saunders fatigue life distribution[J]. Acta Armamentarii, 2010, 31(9): 1260–1262. |
[22] |
孙祝岭. 疲劳寿命分布变异系数的统计推断[J].
质量与可靠性, 2013(1): 13–15.
SUN Z L. Statistical inference of variable coefficient for fatigue life distribution[J]. Quality and Reliability, 2013(1): 13–15. |
[23] |
孙祝岭. Birnbaum-Saunders分布环境因子的置信限[J].
强度与环境, 2012, 39(4): 51–55.
SUN Z L. Confidence limit of environmental factor for Birnbaum-Saunders distribution[J]. Structure & Environment Engineering, 2012, 39(4): 51–55. |
[24] | WANG B X. Generalized interval estimation for the Birnbaum-Saunders distribution[J]. Computational Statistics and Data Analysis, 2012, 56(12): 4320–4326. DOI:10.1016/j.csda.2012.03.023 |
[25] | NIU C Z, GUO X, XU W L, et al. Comparison of several Birnbaum-Saunders distributions[J]. Journal of Statistical Computation and Simulation, 2014, 84(12): 2721–2733. DOI:10.1080/00949655.2014.881814 |
[26] |
周磊, 孙玲玲. 一种基于概率解释的新型互连线时延Slew模型[J].
电路与系统学报, 2009, 14(2): 7–10.
ZHOU L, SUN L L. A new slew interconnect delay model based on probability interpretation[J]. Journal of Circuits and Systems, 2009, 14(2): 7–10. |
[27] |
赵建印, 孙权, 彭宝华, 等. 基于加速退化数据的BS分布的统计推断[J].
电子产品可靠性与环境试验, 2006, 24(1): 11–14.
ZHAO J Y, SUN Q, PENG B H, et al. Statistical inference for BS distribution based on the accelerated degradation data[J]. Electronic Product Reliability and Environmental Test, 2006, 24(1): 11–14. |
[28] | BALAKRISHNAN N, ZHU X J. On the existence and uniqueness of the maximum likelihood estimates of the parameters of Birnbaum-Saunders distribution based on type-Ⅰ, type-Ⅱ and hybrid censored samples[J]. Statistics, 2014, 48(5): 1013–1032. DOI:10.1080/02331888.2013.800069 |
[29] |
徐晓岭, 王蓉华, 顾蓓青. 关于两参数Birnbaum-Saunders疲劳寿命分布统计分析的2个注记[J].
浙江大学学报:理学版, 2016, 43(5): 539–544.
XU X L, WANG R H, GU B Q. Two notes of statistical analysis about two-parameter Birnbaum-Saunders fatigue life distribution[J]. Journal of Zhejiang University:Science Edition, 2016, 43(5): 539–544. |
[30] |
徐晓岭, 王蓉华.
概率论与数理统计[M]. 北京: 人民邮电出版社, 2014: 48-178.
XU X L, WANG R H. Probability and Mathematical Statistics[M]. Beijing: Posts and Telecom Press, 2014: 48-178. |
[31] | VILCA F, SANTANA L, VICTOR LEIVA V, et al. Estimation of extreme percentiles in Birnbaum-Saunders distributions[J]. Computational Statistics and Data Analysis, 2011, 55(4): 1665–1678. DOI:10.1016/j.csda.2010.10.023 |
[32] | CORDEIRO G M, LEMONTE A J. The exponentiated generalized Birnbaum-Saunders distribution[J]. Applied Mathematics and Computation, 2014, 247: 762–779. DOI:10.1016/j.amc.2014.09.054 |