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Objective Bayesian analysis for generalized exponential stress–strength model

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Abstract

In reliability studies, a stress–strength model is often used to analyze a system that fails whenever the applied stress is greater than the strength. Statistical inference of reliability is widely used in a number of areas, such as engineering, clinical trials, and quality control. In addition to the common stress–strength model with one stress and one strength, the reliability of more complex systems has also been studied. In this study, we consider the reliability of a generalized stress–strength model that consists of a serial system with one stress and multiple strengths. We then develop the probability matching priors and reference priors for a generalized exponential stress–strength model. We demonstrate that the two-group reference prior and Jeffreys prior are not a matching prior. Through a simulation study and real data example, we also demonstrate that the proposed probability matching priors match the target coverage probabilities in a frequentist sense even for a small sample size.

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Acknowledgements

The research of Yongku Kim was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (No. 2018R1D1A1B07043352).

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Correspondence to Yongku Kim.

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Appendix: Derivation of first-order matching priors

Appendix: Derivation of first-order matching priors

Let \(\nabla g(\theta )=({\partial g(\theta )\over \partial \theta _1},\ldots ,{\partial g(\theta )\over \partial \theta _p})^T\) be the gradient vector of the parametric function \(g(\theta )\). Define vector \(\eta =(\eta _1,\ldots ,\eta _p)^T\) as

$$\begin{aligned} \eta ={I^{-1}\nabla g(\theta ) \over \sqrt{\nabla ^T g(\theta ) I^{-1} \nabla g(\theta )}}, \end{aligned}$$
(28)

where \(I^{-1}\) is the inverse of the Fisher information matrix for \(\theta \). The first-order cumulative distribution function matching prior for \(g(\theta )\) can then be derived (see Datta and Ghosh 1995). That is,

$$\begin{aligned} P_{\theta }\left[ {n^{1/2}(g(\theta )-g({\hat{\theta }}))\over b}\le z\right] =P_{\pi }\left[ {n^{1/2}(g(\theta )-g({\hat{\theta }}))\over b}\le z \vert X\right] =1-\alpha +o(n^{-{1\over 2}})\nonumber \\ \end{aligned}$$
(29)

for all z and \(\theta \) if and only if \(\pi (\cdot )\) satisfies the partial differential equation

$$\begin{aligned} \sum _{j=1}^p {\partial \over \partial \theta _j}\eta _j\pi (\theta ) = 0. \end{aligned}$$
(30)

Because \(g(\lambda )=\lambda _1/(\lambda _1+\cdots +\lambda _k)\) and \(I^{-1}=diag\{{\lambda _1^2/n_1},\ldots ,{\lambda _k^2/n_k}\}\), we obtain the following:

$$\begin{aligned} \nabla g(\lambda )= & {} \left( {\sum _{i=2}^k\lambda _i \over (\sum _{i=1}\lambda _i)^2}, -{\lambda _1 \over (\sum _{i=1}\lambda _i)^2},\ldots , -{\lambda _1 \over (\sum _{i=1}\lambda _i)^2}\right) ^T, \end{aligned}$$
(31)
$$\begin{aligned} \nabla ^T g(\lambda ) I^{-1} \nabla g(\lambda )= & {} {\lambda _1^2[(\sum _{i=2}^k\lambda _i)^2/n_1+\sum _{i=2}^k \lambda _i^2/n_i] \over (\sum _{i=1}\lambda _i)^4}, \end{aligned}$$
(32)
$$\begin{aligned} I^{-1} \nabla g(\lambda )= & {} \left( {\lambda _1^2\sum _{i=2}^k\lambda _i \over n_1(\sum _{i=1}\lambda _i)^2}, -{\lambda _1\lambda _2^2 \over n_2(\sum _{i=1}\lambda _i)^2},\ldots , -{\lambda _1\lambda _k^2 \over n_k(\sum _{i=1}\lambda _i)^2} \right) ^T. \end{aligned}$$
(33)

From (31) and (32), the differential equation (30) simplifies to

$$\begin{aligned}&{\partial \over \partial \lambda _1} \left\{ {\lambda _1\sum _{i=2}^k\lambda _i \over n_1 [(\sum _{i=2}^k\lambda _i)^2/n_1+\sum _{i=2}^k \lambda _i^2/n_i]^{1\over 2}} \pi (\lambda )\right\} \nonumber \\&\quad + \sum _{i=2}^k {\partial \over \partial \lambda _i} \left\{ {\lambda _i^2 \over n_i [(\sum _{i=2}^k\lambda _i)^2/n_1+\sum _{i=2}^k \lambda _i^2/n_i]^{1\over 2}} \pi (\lambda )\right\} =0. \end{aligned}$$
(34)

Thus, the solution of (34) is of the form

$$\begin{aligned} \pi (\lambda _1,\ldots ,\lambda _k)=\left( \prod _{i=2}^k\lambda _i^{-2+{n_i\over n_1}}\right) \left[ {(\sum _{i=2}^k\lambda _i)^2 \over n_1}+\sum _{i=2}^k {\lambda _i^2 \over n_i}\right] ^{1\over 2} d\left( \prod _{i=1}^k \lambda _i^{n_i}\right) , \end{aligned}$$
(35)

where \(d(\cdot )>0\) is an arbitrary function differentiable in its argument. Therefore, setting \(d=(\prod _{i=1}^k \lambda _i^{n_i})^{-{1\over n_1}}\) leads the matching prior of (35) to be

$$\begin{aligned} \pi (\lambda _1,\ldots ,\lambda _k)=\lambda _1{-1}\prod _{i=2}^k\lambda _i^{-2} \left[ {(\sum _{i=2}^k\lambda _i)^2 \over n_1}+\sum _{i=2}^k {\lambda _i^2 \over n_i}\right] ^{1\over 2}. \end{aligned}$$

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Kang, S.G., Lee, W.D. & Kim, Y. Objective Bayesian analysis for generalized exponential stress–strength model. Comput Stat 36, 2079–2109 (2021). https://doi.org/10.1007/s00180-021-01083-6

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