Bayesian Estimation and Optimal Censoring of Inverted Generalized Linear Exponential Distribution Using Progressive First Failure Censoring

Abstract

This article studies the problem of statistical estimation and optimal censoring from three-parameter inverted generalized linear exponential distribution under progressive first failure censored samples. The maximum likelihood estimation is presented to estimate the unknown parameters. Approximate confidence interval is constructed to compute the interval estimation for the parameters and the delta method is used to compute the interval estimation for survival, hazard rate, and reversed hazard rate functions. The Gibbs sampler with the Metropolis-Hastings algorithm is applied to generate the Markov chain Monte Carlo samples from the posterior functions to approximate the Bayes estimation using several loss functions and to establish the symmetric credible interval for the parameters. A two real data sets are used to study the suggested censoring schemes and the optimal censoring is used to show the performance of the censoring schemes using maximum likelihood estimator and Bayes estimator. Also, a new vision is studied to obtain the optimal censoring using Bayes estimator under varying loss functions. Finally, a simulation study is presented to compare the different estimation methods based on mean square error and average absolute bias.

Appendices

Appendix 1

To prove Lemma 1, we want to prove that $\frac{\partial ^2 \log {(\delta _{1}(\lambda |\mu ,\theta ,\mathbf{x} ))}}{\partial \lambda ^2}$ and $\frac{\partial ^2 \log {(\delta _{2}(\mu |\lambda ,\theta ,\mathbf{x} ))}}{\partial \mu ^2}$ are negative. The second derivative of $\log {(\delta _{1}(\lambda |\mu ,\theta ,\mathbf{x} ))}$ w.r.t. $\lambda$ can be written as

$$\begin{aligned}&\frac{\partial ^2 \log {(\delta _{1}(\lambda |\mu ,\theta ,\mathbf{x} ))}}{\partial \lambda ^2}=-\sum _{i=1}^m \frac{\theta \ (\theta -1)}{x_{i}^2} \left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta -2 }-\sum _{i=1}^m \frac{1}{x_i^4 \left( \frac{\lambda }{x_{i}^2}+\frac{\mu }{x_{i}^3}\right) ^2} \\&\quad - (\theta -1) \sum _{i=1}^m \frac{1}{x_{i}^2 \left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^2}+\sum _{i=1}^m\frac{\theta \ (k(R_{i}+1)-1) \ e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} \left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta -2 }}{x_{i}^2 \ \Biggr (\ 1-e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} \Biggr )^2 \ } \\&\quad \times \Biggr [\ - \Biggr (\ 1-e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} \Biggr )\ -\theta \Biggr (\ e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} +\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }-1 \Biggr )\ \Biggr ], \end{aligned}$$

and the second derivative of $\log {(\delta _{2}(\mu |\lambda ,\theta ,\mathbf{x} ))}$ w.r.t. $\mu$ can be written as

$$\begin{aligned}&\frac{\partial ^2 \log {(\delta _{2}(\mu |\lambda ,\theta ,\mathbf{x} ))}}{\partial \mu ^2}=-\sum _{i=1}^m \frac{\theta \ (\theta -1)}{4 \ x_{i}^4} \left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta -2 }-\sum _{i=1}^m \frac{1}{x_i^6 \left( \frac{\lambda }{x_{i}^2}+\frac{\mu }{x_{i}^3}\right) ^2} \\&\quad - (\theta -1) \sum _{i=1}^m \frac{1}{4 \ x_{i}^4 \left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^2}+\sum _{i=1}^m\frac{\theta \ (k(R_{i}+1)-1) \ e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} \left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta -2 }}{4 \ x_{i}^4 \ \Biggr (\ 1-e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} \Biggr )^2 \ } \\&\quad \times \Biggr [\ - \Biggr (\ 1-e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} \Biggr )\ -\theta \Biggr (\ e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} +\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }-1 \Biggr )\ \Biggr ]. \end{aligned}$$

Since $\Biggr (\ e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} +\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta } \Biggr )\ > 1$, for $\lambda >0$, $\mu >0$, $\theta >0$ and $x_{i}>0,i=1,,m$, then the result is satisfied.

Appendix 2

To prove Lemma 2, we want to prove that $\frac{\partial ^2 \log {(\delta _{3}(\theta |\mu ,\lambda ,\mathbf{x} ))}}{\partial \theta ^2}$ is negative. The second derivative of $\log {(\delta _{3}(\theta |\mu ,\lambda ,\mathbf{x} ))}$ w.r.t. $\theta$ can be written as

$$\begin{aligned}&\frac{\partial ^2 \log {(\delta _{3}(\theta |\mu ,\lambda ,\mathbf{x} ))}}{\partial \theta ^2}= \frac{-1}{\theta ^2} (m+a_{1}-1)-\sum _{i=1}^m \left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta } \log ^2\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) \\&\quad + \sum _{i=1}^m \frac{ (k(R_{i}+1)-1) \ e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} \left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta -2 } \log ^2\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) }{ \Biggr (\ 1-e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} \Biggr )^2 \ } \\&\quad \times \Biggr [\ -\Biggr (\ e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} +\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }-1 \Biggr )\ \Biggr ]\ \end{aligned}$$

Since $\Biggr (\ e^{-\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta }} +\left( \frac{\lambda }{x_{i}}+\frac{\mu }{2 \ x_{i}^2}\right) ^{\theta } \Biggr )\ > 1$, for $\lambda >0$, $\mu >0$, $\theta >0$ and $x_{i}>0,i=1,,m$, then the result is satisfied.