## E-Bayesian Estimation and E-MSE of Failure Probability and Its Applications

Han Ming,

School of Science, Ningbo University of Technology, Zhejiang Ningbo 315211

 基金资助: 宁波市自然科学基金.  2019A610041

 Fund supported: the NSF of Ningbo Municipality.  2019A610041

Abstract

In order to measure the estimated error, this paper based on the E-Bayesian estimation (expected Bayesian estimation) introduced the definition of E-MSE (expected mean square error), and derive the expressions of E-Bayesian estimation of failure probability and their the E-MSE under different loss functions (including: squared error loss function and LINEX loss function). By Monte Carlo simulations compared with the performances of the proposed the estimation method (the comparison of the results is based on the E-MSE). Finally, combined with the engine reliability problem, used respectively E-Bayesian estimation method and the MCMC method the calculation and analysis are performed. When considering evaluating the E-Bayesian estimations under different loss functions, this paper proposed the E-posterior risk as an evaluation standard.

Keywords： E-Bayesian estimation ; E-MSE ; Failure probability ; Monte Carlo simulation ; MCMC method

Han Ming. E-Bayesian Estimation and E-MSE of Failure Probability and Its Applications. Acta Mathematica Scientia[J], 2022, 42(6): 1790-1801 doi:

## 1 引言

Lindley & Smith[9]介绍了层次先验分布的思想. 层次贝叶斯方法需要两个阶段来完成先验分布的设置, 因此它比贝叶斯方法更具有稳健性. 近年来, 层次贝叶斯方法被应用到数据分析中, 更多细节见文献[4, 8, 1015]等. 然而层次贝叶斯估计往往涉及到复杂积分的计算, 此时一些计算方法如MCMC(Markov chain Monte Carlo)方法可用. 层次贝叶斯方法和E-Bayes方法都是处理先验分布中含有未知参数(超参数)的方法. 关于E-Bayes方法的相关研究, 详见文献[4, 8, 1314, 1626]等.

### 2.1 $p_{i}$的E-Bayes估计的定义

\begin{align} \pi(p_{i}|a, b)=\frac{{p_{i}}^{a-1}(1-p_{i})^{b-1}}{B(a, b)}, \end{align}

### 2.2 $\widehat{p_{i}}_{EB}$的E-MSE的定义

$\widehat{p_{i}}_{EB}$的E-MSE (expected mean square error). 其中$\int\!\!\!\int_D MSE[\widehat p_{iB}(a, b)]\pi(a, b){\rm d}a{\rm d}b$是存在的, $D=\{(a, b):0<a<1, 1<b<c \}$, $\pi(a, b) $$a$$ b$在区域$D$上的密度函数, MSE$[\widehat p_{iB}(a, b)] $$\widehat p_{iB}(a, b) 的MSE(用超参数 a$$ b$表示), $i=1, 2, \cdots, k$.

$MSE[\widehat p_{iB}(a, b)]$对超参数$a $$b 的数学期望. ## 3 不同损失函数下的Bayes估计 损失函数在贝叶斯统计推断中很重要, 以下分别介绍平方损失函数、LINEX损失函数, 并介绍相应的Bayes估计. 根据文献[6], 有如下的引理3.1. 引理3.1 在平方损失函数 L_{1}(\theta, \delta)=(\theta-\delta)^2 下(这里 \delta 是参数 \theta 的一个估计), 对于 \theta 的任意先验分布 \pi(\theta) , 则参数 \theta 的贝叶斯估计为 \widehat{\theta}_{B_{1}}(x)=E(\theta|x). 根据文献[2], 有如下的引理3.2. 引理3.2 在LINEX损失函数 L_{2}(\theta, \delta)=\exp[k(\delta-\theta)]-k(\delta-\theta)-1 下, k\in {{\Bbb R}} , k\neq 0 (这里 \delta 是参数 \theta 的一个估计), 对于 \theta 的任意先验分布 \pi(\theta) , 则参数 \theta 的贝叶斯估计为 \widehat{\theta}_{B_{2}}(x)=-\frac{1}{k}\ln E[\exp(-k\theta)|x]. ## 4 不同损失函数下 p_{i} 的E-Bayes估计 本节以下分别在平方损失函数和LINEX损失函数下, 给出参数 p_{i} 的E-Bayes估计. 定理4.1 对某产品进行 k 次I型截尾试验, 获得的试验数据为 \{(n_{i}, r_{i}, t_{i}), i=1, 2, \cdots, k\} , 记 s_{i}=\sum\limits_{j=i}^k n_{j}, i=1, 2, \cdots, k.$$ p_{i}$的先验密度函数$\pi(p_{i}|a, b)$由(2.1)式给出, a b 的先验密度函数为 \begin{align} \pi(a, b)=\pi(a)\pi(b)=\frac{1}{c}, \ 0<a<1, 1<b<c, \end{align} 则有如下结论: (ⅰ) 在平方损失下, p_{i} 的Bayes估计为 \widehat p_{iB1}(a, b)=\frac{a+r_{i}}{a+b+s_{i}}, E-Bayes估计为 (ⅱ)在LINEX损失下, p_{i} 的Bayes估计为 E-Bayes估计为 对某产品进行 k 次I型截尾试验, 获得的试验数据为 \{(n_{i}, r_{i}, t_{i}), i=1, 2, \cdots, k\} , 似然函数为 L(r_{i}|p_{i})=C^{r_{i}}_{s_{i}}p_{i}^{r_{i}}(1-p_{i})^{s_{i}-r_{i}}, 其中 s_{i}=\sum\limits_{j=i}^k n_{j}, i=1, 2, \cdots, k. p_{i}的先验密度函数$\pi(p_{i}|a, b)$由(2.1)式给出, 根据Bayes定理, 则$p_{i}$的后验密度函数为

(ⅰ) 在平方损失下, 根据引理3.1, $p_{i}$的Bayes估计为

## 5 不同损失函数下$\widehat p_{iEBj}$的E-MSE的估计

(ⅰ) 在平方损失下, $\widehat p_{iB1}(a, b)$的MSE为

$\widehat{p}_{iEB1}$的E-MSE为

(ⅱ)在LINEX损失下, $\widehat p_{iB2}(a, b)$的MSE为

(ⅰ) 根据定理4.1, $p_{i}$的后验分布为Beta$(a+r_{i}, b+s_{i}-r_{i})$, 则有

## 6 Monte Carlo模拟计算

 $s_{i}$ $r_{i}$ $\widehat p_{iEB1}$ $\widehat p_{iEB2}$ $\widehat p_{iEB1}-\widehat p_{iEB2}$ 10 0 0.0364 0.0350 0.0014 1 0.1171 0.1123 0.0048 5 0.4586 0.4533 0.0053 10 0.8806 0.8732 0.0074 20 0 0.0193 0.0189 4.0000e-004 5 0.2500 0.2416 0.0084 10 0.4761 0.4670 0.0091 20 0.9298 0.9213 0.0085 50 0 0.0111 0.0110 1.0000e-004 10 0.2026 0.1991 0.0035 30 0.5865 0.5804 0.0061 50 0.9711 0.9706 5.0000e-004 100 0 0.0047 0.0046 1.0000e-004 20 0.2015 0.1989 0.0026 50 0.4952 0.4901 0.0051 100 0.9849 0.9830 0.0019

 si ri E-MSE$\left(\widehat{p}_{i E B 1}\right)$ E-MSE$\left(\widehat{p}_{i E B 2}\right)$ E-MSE$\left(\widehat{p}_{i E B 2}\right)-$E-MSE$\left(\widehat{p}_{i E B 1}\right)$ 10 0 0.0025 0.0027 2.0000e-004 1 0.0079 0.0081 2.0000e-004 5 0.0190 0.0194 4.0000e-004 10 0.0081 0.0086 5.0000e-004 20 0 8.1042e-004 8.1169e-004 1.2700e-006 5 0.0081 0.0084 3.0000e-004 10 0.0030 0.0034 4.0000e-004 20 0.0028 0.0033 5.0000e-004 50 0 2.0616e-004 2.1619e-004 1.0030e-005 10 0.2026 0.2033 7.0000e-004 30 0.0046 0.0051 5.0000e-004 50 5.2910e-004 5.3048e-004 1.3800e-006 100 0 4.4859e-005 4.5969e-005 1.1100e-006 20 0.0016 0.0018 2.0000e-004 50 0.0024 0.0028 4.0000e-004 100 1.4368e-004 1.4859e-004 4.9100e-006

(ⅰ)从表 1我们发现, 对于相同的$s_{i} $$r_{i} , \widehat p_{iEBj}\ (j= 1, 2) 的值非常接近. (ⅱ)从表 2我们发现, 对于相同的 s_{i}$$ r_{i}$, E-MSE$(\widehat p_{iEBj})\ (j= 1, 2)$的值具有以下顺序关系: E-MSE$(\widehat p_{iEB1})<$E-MSE$(\widehat p_{iEB2})$. 这也表明, 如果用E-MSE作为评价标准, 那么$\widehat p_{iEB1}$优于$\widehat p_{iEB2}$.

### 7 应用实例

 $i$ $t_{i}$ $n_{i}$ $r_{i}$ $s_{i}$ 1 450 10 0 70 2 650 10 0 60 3 850 10 0 50 4 1050 10 0 40 5 1250 10 1 30 6 1450 10 1 20 7 1650 10 2 10

### 7.1 E-Bayes估计及其E-MSE

 $i$ $t_{i}$ $\widehat p_{iEB1}$ $\widehat p_{iEB2}$ $\widehat p_{iEB1}-\widehat p_{iEB2}$ 1 450 0.0070 0.0069 1.0000e-004 2 650 0.0081 0.0080 1.0000e-004 3 850 0.0097 0.0096 1.0000e-004 4 1050 0.0120 0.0118 2.0000e-004 5 1250 0.0475 0.0471 4.0000e-004 6 1450 0.0694 0.0688 6.0000e-004 7 1650 0.2160 0.2151 9.0000e-004

 i ti E-MSE$\left(\widehat{p}_{i E B 1}\right)$ E-MSE$\left(\widehat{p}_{i E B 2}\right)$ E-MSE$\left(\widehat{p}_{i E B 2}\right)-$E-MSE$\left(\widehat{p}_{i E B 1}\right)$ 1 450 9.4992e-005 1.5319e-004 5.8198e-005 2 650 1.2779e-004 2.0635e-004 7.8560e-005 3 850 1.8106e-004 2.9287e-004 1.1181e-004 4 1050 2.7615e-004 3.8963e-004 1.1348e-004 5 1250 0.0014 0.0033 0.0019 6 1450 0.0029 0.0063 0.0034 7 1650 0.0047 0.0158 0.0111

 i ti $\widehat{p}_{i E B 1}$ $\widehat{p}_{i E B 2}$ $\widehat{p}_{i E B 1}-\widehat{p}_{i E B 2}$ 1 450 0.0069 0.0068 1.0000e-004 2 650 0.0080 0.0079 1.0000e-004 3 850 0.0096 0.0095 1.0000e-004 4 1050 0.0119 0.0117 2.0000e-004 5 1250 0.0468 0.0464 4.0000e-004 6 1450 0.0680 0.0672 8.0000e-004 7 1650 0.2080 0.2071 9.0000e-004

 i ti E-MSE$\left(\widehat{p}_{i E B 1}\right)$ E-MSE$\left(\widehat{p}_{i E B 2}\right)$ E-MSE$\left(\widehat{p}_{i E B 2}\right)-$E-MSE$\left(\widehat{p}_{i E B 1}\right)$ 1 450 9.3829e-005 9.3932e-005 1.0300e-007 2 650 1.2598e-004 1.2610e-004 1.2000e-007 3 850 1.7801e-004 1.7907e-004 1.0600e-006 4 1050 2.7044e-004 2.7349e-004 3.0500e-006 5 1250 0.0013 0.0019 6.0000e-004 6 1450 0.0027 0.0035 8.0000e-004 7 1650 0.0044 0.0107 0.0063

 i ti $\widehat{p}_{i E B 1}$ $\widehat{p}_{i E B 2}$ $\widehat{p}_{i E B 1}-\widehat{p}_{i E B 2}$ 1 450 0.0069 0.0068 1.0000e-004 2 650 0.0080 0.0079 1.0000e-004 3 850 0.0095 0.0094 1.0000e-004 4 1050 0.0117 0.0116 1.0000e-004 5 1250 0.0461 0.0457 4.0000e-004 6 1450 0.0666 0.0658 8.0000e-004 7 1650 0.2000 0.1991 9.0000e-004

 i ti E-MSE$\left(\widehat{p}_{i E B 1}\right)$ E-MSE$\left(\widehat{p}_{i E B 2}\right)$ E-MSE$\left(\widehat{p}_{i E B 2}\right)-$E-MSE$\left(\widehat{p}_{i E B 1}\right)$ 1 450 9.2570e-005 9.3109e-005 5.3900e-007 2 650 1.2403e-004 1.2513e-004 1.1000e-006 3 850 1.7474e-004 1.7589e-004 1.1500e-006 4 1050 2.6436e-004 2.6789e-004 3.5300e-006 5 1250 0.0013 0.0017 4.0000e-004 6 1450 0.0026 0.0032 6.0000e-004 7 1650 0.0042 0.0095 0.0053

(ⅰ)从表 4, 表 6表 8我们发现, 对于相同的$t_{i} $$c , \widehat p_{iEBj}\ (j= 1, 2) 的值非常接近. (ⅱ)从表 5, 表 7表 9我们发现, 对于相同的 t_{i}$$ c$, E-MSE$(\widehat p_{iEBj})\ (j= 1, 2)$的值具有以下顺序关系: E-MSE$(\widehat p_{iEB1})<$E-MSE$(\widehat p_{iEB2})$. 这也表明, 如果用E-MSE作为评价标准, 那么$\widehat p_{iEB1}$优于$\widehat p_{iEB2}$.

(ⅲ)对于相同的$t_{i}$和不同的$c(c=1.1, 2, 3)$, $\widehat p_{iEBj} $$(j=1, 2) 和E-MSE (\widehat p_{iEBj})$$ (j=1, 2)$都是稳健的. 在应用中作者建议$c$在区间(1, 3]居中取值, 即$c=2$.

\begin{align} F(t)=1-\exp\left\{-\left(\frac{t}{\eta}\right)^m\right\}, \ \eta>0, \ m>0, \ t>0. \end{align}

\begin{align} \widehat{R}(t)=\exp\left\{-\left(\frac{t}{\widehat{\eta}}\right)^{\widehat{m}}\right\}, \end{align}

 $t$ 100 300 500 800 1000 1200 1500 1700 $\widehat{R}_{EB1}(t)$ 0.9999 0.9995 0.9971 0.987 0.973 0.9511 0.9003 0.8529 $\widehat{R}_{EB2}(t)$ 0.9999 0.9995 0.9972 0.9872 0.9735 0.9524 0.9037 0.8584

 $i$ $t_{i}$ mean sd MC-error val2.5pc val97.5pc start sample 1 450 0.0069 0.0075 9.710e-5 1.1374e-5 0.0245 1001 9000 2 650 0.0079 0.0085 1.208e-4 4.9915e-5 0.0287 1001 9000 3 850 0.0096 0.0104 1.563e-4 7.0336e-5 0.0366 1001 9000 4 1050 0.0118 0.0136 1.856e-4 9.0227e-5 0.0469 1001 9000 5 1250 0.0456 0.0370 3.923e-4 0.00252 0.1386 1001 9000 6 1450 0.0675 0.0535 5.319e-4 0.00409 0.2015 1001 9000 7 1650 0.2021 0.1145 0.001368 0.04232 0.4748 1001 9000

 估计方法 $\widehat{m}$ $\widehat{\eta}$ 定理1的(ⅰ) 3.3174965 2958.5512 定理1的(ⅱ) 3.2787667 3015.4834 MCMC方法 3.2942417 2993.9772 以上三个估计值之间的最大差值 0.0387298 56.9322

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