Bayesian Inference and Optimal Censoring Scheme Under Progressive Censoring

Abstract

In recent times an extensive work has been done related to the different aspects of the progressive censoring schemes. Here we deal with the statistical inference of the unknown parameters of a three-parameter Weibull distribution based on the assumption that the data are progressively Type-II censored. The maximum likelihood estimators of the unknown parameters do not exist due to the presence of the location parameter. Therefore, Bayesian approach seems to be a reasonable alternative. We assume here that the location parameter follows a uniform prior and the shape parameter follows a log-concave prior density function. We further assume that the scale parameter has a conjugate gamma prior given the shape and the location parameters. Based on these priors the Bayes estimate of any function of the unknown parameters under the squared error loss function and the associated highest posterior density credible interval are obtained using Gibbs sampling technique. We have also used one precision criterion to compare two different censoring schemes, and it can be used to find the optimal censoring scheme. Since finding the optimal censoring scheme is a challenging problem from the computational view point, we propose suboptimal censoring scheme, which can be obtained quite conveniently. We have carried out some Monte Carlo simulations to observe the performances of the proposed method, and for illustrative purposes, we presented the analysis of one data set.

Appendix

Proof of Theorem 2: To prove Theorem 2, it is enough to prove that

$$ \frac{{{\text{d}}^{2} }}{{{\text{d}}\alpha^{2} }}\ln \left[ {b + \sum\limits_{i = 1}^{m} \left( {1 + R_{i} } \right)\left( {t_{i} - \mu } \right)^{\alpha } } \right] > 0. $$

Now consider

$$ g(\alpha ) = b + \sum\limits_{i = 1}^{m} \left( {1 + R_{i} } \right)\left( {t_{i} - \mu } \right)^{\alpha } ; $$

then

$$ g^{\prime } (\alpha ) = \frac{\text{d}}{{{\text{d}}\alpha }}g(\alpha ) = \sum\limits_{i = 1}^{m} \left( {1 + R_{i} } \right)\left( {t_{i} - \mu } \right)^{\alpha } \ln \left( {t_{i} - \mu } \right) $$

and

$$ g^{\prime \prime } (\alpha ) = \frac{{{\text{d}}^{2} }}{{{\text{d}}\alpha^{2} }}g(\alpha ) = \sum\limits_{i = 1}^{m} \left( {1 + R_{i} } \right)\left( {t_{i} - \mu } \right)^{\alpha } \left( {\ln \left( {t_{i} - \mu } \right)} \right)^{2} . $$

Since

$$ \begin{aligned} & \left( {\sum\limits_{i = 1}^{m} \left( {1 + R_{i} } \right)\left( {t_{i} - \mu } \right)^{\alpha } \left( {\ln \left( {t_{i} - \mu } \right)} \right)^{2} } \right)\left( {\sum\limits_{i = 1}^{m} \left( {1 + R_{i} } \right)\left( {t_{i} - \mu } \right)^{\alpha } } \right) \\ & \quad - \left( {\sum\limits_{i = 1}^{m} \left( {1 + R_{i} } \right)\left( {t_{i} - \mu } \right)^{\alpha } \ln \left( {t_{i} - \mu } \right)} \right)^{2} \\ & = \sum\limits_{1 \le i < j \le m} \left( {R_{i} + 1} \right)\left( {R_{j} + 1} \right)\left( {\ln \left( {t_{i} - \mu } \right) - \ln \left( {t_{j} - \mu } \right)} \right)^{2} \ge 0, \\ \end{aligned} $$

the result follows.