﻿ 指数分布下可靠性参数的推断

# 指数分布下可靠性参数的推断Inference of Reliability Parameters under Exponential Distribution

Abstract: In this paper, we study the interval estimation method of parameter   under single pa-rameter exponential distribution. Two kinds of generalized pivots of parameter   in single parameter exponential distribution, analytical solutions of generalized confidence interval of parameter and theoretical proof of frequency property are given. At the same time, the generalized p value of hypothesis testing problem is given. In addition, three existing methods, namely Bayes method, approximate estimation method of large sample and Bootstrap resampling method are given. Four methods are simulated by Monte Carlo method. The simulation results show that the coverage probability of generalized inference and Bayes method remains near the confidence level when sample size is small, and the average confidence length is smaller. In addition, this paper compares the error 1 probability and the power for hypothesis testing. The simulation results verify the good performance of the generalized inference method.

1. 引言

2. 变量为单参数指数分布情形

$T=P\left(X>Y\right)={\int }_{0}^{+\infty }\text{ }{\lambda }_{1}{\text{e}}^{-{\lambda }_{1}y}\text{d}x{\int }_{0}^{x}\text{ }{\lambda }_{2}{\text{e}}^{-{\lambda }_{2}y}\text{d}y=\frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}.$

2.1. 广义推断的方法

Tsui K. W.，Weerahandi [6] 和Weerahandi [7] 提出了广义推断的理论，并且给出广义推断方法来求检验的广义p值及参数的广义置信区间。

${X}_{1},\cdots ,{X}_{m}$${Y}_{1},\cdots ,{Y}_{n}$ 分别为从指数分布总体 $\mathrm{exp}\left({\lambda }_{1}\right)$$\mathrm{exp}\left({\lambda }_{2}\right)$ 中抽取的样本，由于 ${\sum }_{i=1}^{m}\text{ }{X}_{i}$${\sum }_{j=1}^{n}\text{ }{Y}_{j}$ 是独立的充分统计量，且有：

$U=2{\lambda }_{1}\underset{i=1}{\overset{m}{\sum }}\text{ }{X}_{i}~{\chi }^{2}\left(2m\right),\text{ }V=2{\lambda }_{2}\underset{j=1}{\overset{n}{\sum }}\text{ }{Y}_{j}~{\chi }^{2}\left(2n\right),$

${R}_{{\lambda }_{1}}=\frac{U}{2{\sum }_{i=1}^{m}{x}_{i}},\text{ }{R}_{{\lambda }_{2}}=\frac{V}{2{\sum }_{j=1}^{n}{y}_{j}},$ (1)

${R}_{T}=\frac{{R}_{{\lambda }_{2}}}{{R}_{{\lambda }_{1}}+{R}_{{\lambda }_{2}}},$ (2)

$\begin{array}{l}{H}_{0}:T\le {T}_{0},\text{ }{H}_{1}:T>{T}_{0}\hfill \end{array}$ (3)

$\begin{array}{c}p={P}_{r}\left({R}_{T}\left(X,Y,x,y,{\lambda }_{1},{\lambda }_{2}\right)\le r\left(x,y,x,y,{\lambda }_{1},{\lambda }_{2}\right)|T={T}_{0}\right)\\ ={P}_{r}\left(\frac{V/\left(2{\sum }_{j=1}^{n}{y}_{j}\right)}{U/\left(2{\sum }_{i=1}^{m}{x}_{i}\right)+V/\left(2{\sum }_{j=1}^{n}{y}_{i}\right)}\le {T}_{0}\right)\\ ={P}_{r}\left(V\le \frac{{T}_{0}\cdot U\cdot {\sum }_{j=1}^{n}{y}_{i}}{\left(1-{T}_{0}\right)\cdot {\sum }_{i=1}^{m}{x}_{i}}\right)\\ ={E}_{U}\left({F}_{V}\left(\frac{{T}_{0}\cdot U\cdot {\sum }_{j=1}^{n}{y}_{i}}{\left(1-{T}_{0}\right){\sum }_{i=1}^{m}{x}_{i}}\right)\right),\end{array}$

$\begin{array}{l}P\left({R}_{T}\left(X,Y,x,y,{\lambda }_{2},{\lambda }_{2}\right)\ge c\right)\\ ={P}_{r}\left(U\le \frac{V{\sum }_{i=1}^{m}{x}_{i}}{\left(1/c-1\right){\sum }_{j=1}^{n}{y}_{i}}\right)\\ ={E}_{V}\left({F}_{U}\left(\frac{V{\sum }_{i=1}^{m}{x}_{i}}{\left(1/c-1\right){\sum }_{j=1}^{n}{y}_{i}}\right)\right)\\ =\gamma ,\end{array}$

${E}_{V}\left({F}_{U}\left(\frac{V{\sum }_{i=1}^{m}{x}_{i}}{\left(1/{c}_{\gamma }-1\right){\sum }_{j=1}^{n}{y}_{i}}\right)\right)=\gamma .$

$\frac{\frac{{\lambda }_{1}}{m}{\sum }_{i=1}^{m}{X}_{i}}{\frac{{\lambda }_{2}}{n}{\sum }_{j=1}^{n}{Y}_{i}}~F\left(2m,2n\right),$

$\frac{{\lambda }_{1}\stackrel{¯}{X}}{{\lambda }_{2}\stackrel{¯}{Y}}~F\left(2m,2n\right),$

${R}_{\frac{{\lambda }_{1}}{{\lambda }_{2}}}=W\frac{\stackrel{¯}{y}}{\stackrel{¯}{x}},$

${R}_{T}=\frac{1}{1+{R}_{{\lambda }_{1}/{\lambda }_{2}}}.$

i) 分别从两个指数分布总体中抽取样本量分别为m和n的样本，得到观测值 $\left\{\left({x}_{1},\cdots ,{x}_{m}\right),\left({y}_{1},\cdots ,{y}_{n}\right)\right\}$

ii) 计算 ${\sum }_{i=1}^{m}\text{ }{x}_{i};{\sum }_{j=1}^{n}\text{ }{y}_{j}$

iii) 产生 $U~{\chi }^{2}\left(2m\right)$$V~{\chi }^{2}\left(2n\right)$ 的实现值；

iv) 按(1)和(2)给出的公式计算 ${R}_{T}$

v) 重复步骤(iii)-(iv) M次，得到M个 ${R}_{T}$ 的值，将这一系列 ${R}_{T}$ 从小到大排列，取其 $\alpha /2$ 分位点与 $1-\alpha /2$ 分位点，分别记为 ${\stackrel{^}{g}}_{L},{\stackrel{^}{g}}_{U}$ ，得到参数T的一个双侧广义置信区间。计算M个 ${R}_{T}$ 中小于等于真值T的比率，作为假设检验问题(3)的广义p值；

vi) 重复步骤i)~v) L次，计算这L次得到的T的广义置信区间中包含真实值的个数，作为置信区间的覆盖概率，计算广义p值小于0.05的概率，作为检验犯第一类错误的概率。

${\stackrel{^}{g}}_{\alpha }\left(x\right)=inf\left\{g:{F}_{{\left(G\right)}_{x}}\left(g\right)\ge \alpha \right\}$

${P}_{\theta }\left(g\left(\theta \right)<{\stackrel{^}{g}}_{\alpha }\left(X\right)\right)=\alpha$

$\begin{array}{l}P\left({\stackrel{^}{g}}_{L}<\frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}<{\stackrel{^}{g}}_{U}\right)\\ =P\left(\frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}<{\stackrel{^}{g}}_{U}\right)-P\left(\frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}<{\stackrel{^}{g}}_{L}\right)\\ =P\left({F}_{{R}_{T}}\left(\frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}\right)<1-\frac{\alpha }{2}\right)-P\left({F}_{{R}_{T}}\left(\frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}\right)<\frac{\alpha }{2}\right),\end{array}$

$P\left({F}_{{R}_{T}}\left(\frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}\right)<1-\frac{\alpha }{2}\right)=P\left(P\left(\frac{U/2{\sum }_{i=1}^{m}{x}_{i}}{U/2{\sum }_{i=1}^{m}{x}_{i}+V/2{\sum }_{j=1}^{n}{y}_{i}}\le \frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}\right)<1-\frac{\alpha }{2}\right)$

$P\left(P\left(\frac{{U}^{*}}{{V}^{*}}\le \frac{U}{V}\right)<1-\alpha \right)={P}_{{U}^{*}/{V}^{*}}\left({F}_{U/V}\left(\frac{{U}^{*}}{{V}_{*}}\right)>\alpha /2\right)=1-\alpha /2,$

$P\left({F}_{T}\left(\frac{{\lambda }_{2}}{{\lambda }_{1}+{\lambda }_{2}}\right)<\alpha /2\right)=\alpha /2$

2.2. 基于渐近正态的大样本方法

${X}_{1},\cdots ,{X}_{m}$${Y}_{1},\cdots ,{Y}_{n}$ 分别为从指数分布总体 $\mathrm{exp}\left({\lambda }_{1}\right)$$\mathrm{exp}\left({\lambda }_{2}\right)$ 中抽取的样本，由于 ${\lambda }_{1}$${\lambda }_{2}$ 的极大似然估计分别为：

${\stackrel{^}{\lambda }}_{1}=\frac{1}{\stackrel{¯}{X}},\text{ }{\stackrel{^}{\lambda }}_{2}=\frac{1}{\stackrel{¯}{Y}},$

$\stackrel{^}{T}=\frac{\stackrel{¯}{X}}{\stackrel{¯}{X}+\stackrel{¯}{Y}}.$ (4)

$\left[\stackrel{^}{T}-\frac{{U}_{1-\alpha /2}\cdot \stackrel{^}{T}\left(1-\stackrel{^}{T}\right)}{\sqrt{nb\left(1-b\right)}},\stackrel{^}{T}+\frac{{U}_{1-\alpha /2}\cdot \stackrel{^}{T}\left(1-\stackrel{^}{T}\right)}{\sqrt{nb\left(1-b\right)}}\right],$

$p=1-\Phi \left(\frac{\sqrt{nb\left(1-b\right)}}{{T}_{0}\left(1-{T}_{0}\right)}\left(\stackrel{^}{T}-{T}_{0}\right)\right),$

2.3. Bootstrap-t方法

Bootstrap方法最早是由斯坦福大学教授Efron于1977年提出的，该方法认为经验分布函数能够较好地拟合总体分布，下面给出基于bootstrap方法的指数分布可靠性参数的区间估计中较常用的一种方法，Bootstrap-t区间估计 [9] 。

$\stackrel{^}{T}$ 是T的极大似然估计， $\stackrel{^}{V}\left(\stackrel{^}{T}\right)$$\stackrel{^}{T}$ 的方差估计， ${\stackrel{^}{T}}^{*}$ 是通过Boostrap样本得到的T的极大似然估计， $\stackrel{^}{V}\left({\stackrel{^}{T}}^{*}\right)$$\stackrel{^}{T}$ 的方差的Bootstrap估计。

i) 分别从两个指数分布总体抽取样本量为 ${n}_{1}$${n}_{2}$ 的样本集合，记为 ${W}_{1}$${W}_{2}$

ii) 通过样本 ${W}_{1}$${W}_{2}$ 利用公式(4)求出 $\stackrel{^}{T}$

iii) 分别从 ${W}_{1}$${W}_{2}$ 中再抽取样本量为 ${n}_{1}$${n}_{2}$ 的Bootstrap样本，记为 ${Q}_{1}$${Q}_{2}$

iv) 通过样本 ${Q}_{1}$${Q}_{2}$ 求出 ${Z}^{*}=\frac{{\stackrel{^}{T}}^{*}-\stackrel{^}{T}}{\stackrel{^}{V}{\left({\stackrel{^}{T}}^{*}\right)}^{1/2}}$

v) 重复步骤(iii)-(iv) M次，得到M个 $\stackrel{^}{T}$ ，从小到大排序，取其 $\alpha$ 分位点 ${z}_{\alpha /2}$$1-\alpha$ 分位点 ${z}_{1-\alpha /2}$ ，得到参数T的一个双侧Bootstrap置信区间 $\left(\stackrel{^}{T}-{z}_{1-\alpha /2}^{*}\stackrel{^}{V}{\left(\stackrel{^}{T}\right)}^{1/2},\stackrel{^}{T}-{z}_{\alpha /2}^{*}\stackrel{^}{V}{\left(\stackrel{^}{T}\right)}^{1/2}\right)$

vi) 重复步骤i)~v) L次，计算这L次得到的T的置信区间中包含真实值的概率，作为置信区间的覆盖概率。

2.4. Bayes方法

$L\left({\lambda }_{1},{\lambda }_{2}|x,y\right)={\lambda }_{1}^{m}{\lambda }_{2}^{n}{\text{e}}^{-{\lambda }_{1}{\sum }_{i=1}^{m}{x}_{i}-{\lambda }_{2}{\sum }_{j=1}^{n}{y}_{i}},{\lambda }_{1}>0,{\lambda }_{2}>0,$

${\lambda }_{1}$${\lambda }_{2}$ 的先验分布为非信息先验分布： $\pi \left({\lambda }_{i}\right)=\frac{1}{{\lambda }_{i}},{\lambda }_{i}>0,i=1,2$

${\Pi }_{1}\left({\lambda }_{1}|x\right)=\frac{{\lambda }_{1}^{m-1}\mathrm{exp}\left(-{\lambda }_{1}{\sum }_{i=1}^{m}{x}_{i}\right)}{{\int }_{0}^{+\infty }{\lambda }_{1}^{m-1}\mathrm{exp}\left(-{\lambda }_{1}{\sum }_{i=1}^{m}{x}_{i}\right)\text{d}{\lambda }_{1}}=\frac{{\lambda }_{1}^{m-1}{\left({\sum }_{i=1}^{m}{x}_{i}\right)}^{m}\mathrm{exp}\left(-{\lambda }_{1}{\sum }_{i=1}^{m}{x}_{i}\right)}{\Gamma \left(m\right)},$ (5)

${\Pi }_{2}\left({\lambda }_{2}|y\right)=\frac{{\lambda }_{2}^{n-1}\mathrm{exp}\left(-{\lambda }_{2}{\sum }_{j=1}^{n}{y}_{i}\right)}{{\int }_{0}^{+\infty }{\lambda }_{2}^{n-1}\mathrm{exp}\left(-{\lambda }_{2}{\sum }_{j=1}^{n}{y}_{i}\right)\text{d}{\lambda }_{1}}=\frac{{\lambda }_{2}^{n-1}{\left({\sum }_{j=1}^{n}{y}_{i}\right)}^{n}\mathrm{exp}\left(-{\lambda }_{2}{\sum }_{j=1}^{n}{y}_{i}\right)}{\Gamma \left(n\right)},$ (6)

i) 通过公式(5)和(6)产生 ${\lambda }_{1}$${\lambda }_{2}$

ii) 通过得到的 $\left({\lambda }_{1},{\lambda }_{2}\right)$ ，计算得到T；

iv) 重复步骤(i)-(ii) M次，得到 ${T}_{1},\cdots ,{T}_{M}$ ，并将它们从小到大排序；

v) 取其 $\alpha /2$ 分位点与 $1-\alpha /2$ 分位点，得到参数T的一个置信区间；

vi) 重复步骤i)~v) L次，计算这L次得到的置信区间中包含真实值T的概率作为置信区间的覆盖概率。

3. 模拟与结论

Table 1. The main coverage probability and average confidence length of the two-sided confidence interval of the single-parameter exponential distribution parameter T 95%

Table 2. The main coverage probability and average confidence length of the two-sided confidence interval of the single-parameter exponential distribution parameter T 95%

Table 3. The main coverage probability and average confidence length of the two-sided confidence interval of the single-parameter exponential distribution parameter T 95%

Table 4. The probability of making the first type of error in hypothesis test

Table 5. The probability of making the first type of error in hypothesis test

Table 6. The power of the hypothesis test

Table 7. The power of the hypothesis test

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