On Bayesian Inference of $R=P(Y < X)$ for Weibull Distribution

Abstract

In this paper, we consider the Bayesian inference on the stress-strength parameter $R = P(Y < X)$, when X and Y follow independent Weibull distributions. We have considered different cases. It is assumed that the random variables X and Y have different scale parameters and (a) a common shape parameter or (b) different shape parameters. Moreover, both stress and strength may depend on some known covariates also. When the two distributions have a common shape parameter, Bayesian inference on R is obtained based on the assumption that the shape parameter has a log-concave prior, and given the shape parameter, the scale parameters have Dirichlet-Gamma prior. The Bayes estimate cannot be obtained in closed form, and we propose to use Gibbs sampling method to compute the Bayes estimate and also to compute the associated highest posterior density (HPD) credible interval. The results have been extended when the covariates are also present. We further consider the case when the two shape parameters are different. Simulation experiments have been performed to see the effectiveness of the proposed methods. One data set has been analyzed for illustrative purposes and finally, we conclude the paper.

Appendix

Proof of Theorem 3: Note that

$$\begin{aligned}\ln \pi _3(\alpha |\beta _1, \beta _2,data)= & {} const. + \ln \phi (\alpha ) + (m+n)\ln \alpha + \alpha \ln S - \\&(m+a_1) \ln (b_0 + \sum _{i=1}^m x_i^{\alpha } e^{{{\beta }_1}^T {{u}}_i}) - (n+a_2) \ln (b_0 + \sum _{j=1}^n y_j^{\alpha } e^{{{\beta }_2}^T {{v}}_j}). \end{aligned}$$

It has been shown in the proof of Theorem 2 of Kundu [17], that

$$ \frac{d^2}{d\alpha ^2} \ln (b_0 + \sum _{i=1}^m x_i^{\alpha } e^{{{\beta }_1}^T {{u}}_i}) \ge 0 \ \ \hbox {and} \ \ \frac{d^2}{d\alpha ^2} \ln (b_0 + \sum _{j=1}^n y_j^{\alpha } e^{{{\beta }_2}^T {{v}}_j}) \ge 0. $$

Since $\pi (\alpha )$ is log-concave, it immediately follows that

$$ \frac{d^2}{d\alpha ^2} \ln \pi _3(\alpha |\beta _1, \beta _2,data) \le 0. $$