Bayesian Estimation of Constant-Stress Life Test Model Using Type-I Censored Data from the Linear Failure Rate Distribution

This paper considers both likelihood and Bayesian estimations of a constant-stress partially accelerated life test model with type-I censored data from the linear failure rate distribution. The maximum likelihood estimates of the model parameters are obtained using Newton–Raphson technique. The posterior means and posterior variances are obtained under the squared error loss function using Lindley’s approximation procedure. The advantages of this approximation are exposed. Monte Carlo simulations are prepared under different sizes of samples and different parameter values for comparing and evaluating the proposed methods of estimation.

Appendix

The derivation of posterior means and posterior variances: here, there are three parameters in the model. That is, m = 3. Let the subscripts 1, 2, and 3 refer to β, α, and θ, respectively. It is not easy to obtain the posterior moments analytically. Therefore, using Lindley expansion, the posterior mean (i.e., Bayesian estimator under squared-error loss function) and the posterior variance of β are given, respectively, in the form:

$$ {\upbeta}^{\ast }=E\left(\left.\upbeta \right| data\right)\left[\upbeta -\left(\frac{\upsigma_{11}}{\upbeta}+\frac{\upsigma_{12}}{\uptheta}+\frac{\upsigma_{13}}{\upalpha}\right)+\frac{1}{2}\left({\upsigma}_{11}{E}_1+{\upsigma}_{12}{E}_2+{\upsigma}_{13}{E}_3\right)\right]\downarrow \Theta $$

(4)

and

$$ \operatorname{var}\left(\left.\upbeta \right| data\right)=E\left(\left.{\upbeta}^2\right|y\right)-{\left({\upbeta}^{\ast}\right)}^2={\upsigma}_{11}-{\left[\left(\frac{\upsigma_{11}}{\upbeta}+\frac{\upsigma_{12}}{\uptheta}+\frac{\upsigma_{13}}{\upalpha}\right)-\frac{1}{2}\left({\upsigma}_{11}{E}_1+{\upsigma}_{12}{E}_2+{\upsigma}_{13}{E}_3\right)\right]}^2\downarrow \Theta . $$

(5)

Similarly, for the shape parameter α, the posterior mean and the posterior variance are given by

$$ {\upalpha}^{\ast }=E\left(\left.\upalpha \right| data\right)\left[\upalpha -\left(\frac{\upsigma_{31}}{\upbeta}+\frac{\upsigma_{32}}{\uptheta}+\frac{\upsigma_{33}}{\upalpha}\right)+\frac{1}{2}\left({\upsigma}_{31}{E}_1+{\upsigma}_{32}{E}_2+{\upsigma}_{33}{E}_3\right)\right]\downarrow \Theta, $$

(6)

$$ \operatorname{var}\left(\left.\upalpha \right| data\right)={\upsigma}_{33}-{\left[\left(\frac{\upsigma_{31}}{\upbeta}+\frac{\upsigma_{32}}{\uptheta}+\frac{\upsigma_{33}}{\upalpha}\right)-\frac{1}{2}\left({\upsigma}_{31}{E}_1+{\upsigma}_{32}{E}_2+{\upsigma}_{33}{E}_3\right)\right]}^2\downarrow \Theta . $$

(7)

Applying the same technique, the posterior mean and posterior variance of the scale parameter θ take the following form:

$$ {\uptheta}^{\ast }=E\left(\left.\uptheta \right| data\right)\left[\uptheta -\left(\frac{\upsigma_{21}}{\upbeta}+\frac{\upsigma_{22}}{\uptheta}+\frac{\upsigma_{23}}{\upalpha}\right)+\frac{1}{2}\left({\upsigma}_{21}{E}_1+{\upsigma}_{22}{E}_2+{\upsigma}_{23}{E}_3\right)\right]\downarrow \Theta $$

(8)

and

$$ \operatorname{var}\left(\left.\uptheta \right| data\right)={\upsigma}_{22}-{\left[\left(\frac{\upsigma_{21}}{\upbeta}+\frac{\upsigma_{22}}{\uptheta}+\frac{\upsigma_{23}}{\upalpha}\right)-\frac{1}{2}\left({\upsigma}_{21}{E}_1+{\upsigma}_{22}{E}_2+{\upsigma}_{23}{E}_3\right)\right]}^2\downarrow \Theta, $$

(9)

where

$$ {E}_1=\sum \limits_{i,j}{\upsigma}_{ij}{L}_{ij1}^{(3)},\kern1em {E}_2=\sum \limits_{i,j}{\upsigma}_{ij}{L}_{ij2}^{(3)},\kern1em {E}_3=\sum \limits_{i,j}{\upsigma}_{ij}{L}_{ij3}^{(3)}, $$

for i, j = 1, 2, 3, σ_ij are the elements of the inverse of the asymptotic Fisher-information matrix of the ML estimators of β, θ, and α in the case of type-I censored data and i, j = 1, 2, 3, ${L}_{ijk}^{(3)}$ is the third derivatives of the natural logarithm of the likelihood function in type-I censoring.

To compute the posterior means and the posterior variances of β, θ, and α derived before, both second and third derivatives of the natural logarithm of the likelihood function given in ((2) must be obtained.

The second derivatives can be given via the following equations:

$$ {\displaystyle \begin{array}{c}\frac{\partial^2\ln L}{\partial {\upalpha}^2}=-\sum \limits_{i=1}^{n\left(1-\uppi \right)}{\updelta}_{u_i}\left[\frac{1}{{\left(\upalpha +\uptheta {t}_i\right)}^2}\right]-{\upbeta}^2\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left[\frac{1}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^2}\right],\\ {}\frac{\partial^2\ln L}{\mathrm{\partial \upalpha \partial \uptheta }}=\frac{\partial^2\ln L}{\mathrm{\partial \uptheta \partial \upalpha }}=-\sum \limits_{i=1}^{n\left(1-\uppi \right)}{\updelta}_{u_i}\left[\frac{t_i}{{\left(\upalpha +\uptheta {t}_i\right)}^2}\right]-{\upbeta}^3\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left[\frac{x_j}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^2}\right],\\ {}\frac{\partial^2\ln L}{\mathrm{\partial \upalpha \partial \upbeta }}=\frac{\partial^2\ln L}{\mathrm{\partial \upbeta \partial \upalpha }}=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left[\frac{-\left({\uptheta \upbeta}^2{x}_j\right)}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^2}-{x}_j\right]+\upeta \left({n}_a-n\uppi \right),\\ {}\frac{\partial^2\ln L}{\partial {\uptheta}^2}=-\sum \limits_{i=1}^{n\left(1-\uppi \right)}{\updelta}_{u_i}\left[\frac{t_i^2}{{\left(\upalpha +\uptheta {t}_i\right)}^2}\right]-{\upbeta}^4\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left[\frac{x_j^2}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^2}\right],\\ {}\frac{\partial^2\ln L}{\mathrm{\partial \uptheta \partial \upbeta }}=\frac{\partial^2\ln L}{\mathrm{\partial \upbeta \partial \uptheta }}=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left[\frac{\left({\upalpha \upbeta}^2{x}_j\right)}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^2}-\upbeta {x}_j^2\right]+{\upbeta \upeta}^2\left({n}_a-n\uppi \right),\\ {}\frac{\partial^2\ln L}{\partial {\upbeta}^2}=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left[\frac{-\left({\upalpha}^2+2\upalpha \uptheta \upbeta {x}_j+2{\left(\uptheta \upbeta {x}_j\right)}^2\right)}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^2}-\uptheta {x}_j^2\right]+{\uptheta \upeta}^2\left({n}_a-n\uppi \right).\end{array}} $$

Third derivatives are calculated as follows:

$$ {\displaystyle \begin{array}{c}{L}_{111}^{(3)}=\frac{\partial^3\ln L}{\partial {\upbeta}^3}=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{2\left({\upalpha}^2+2\upalpha \uptheta \upbeta {x}_j+2{\uptheta}^2{\upbeta}^2{x}_j^2\right)\left(\upalpha +2\uptheta \upbeta {x}_j\right)-\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)\left(2\upalpha \uptheta {x}_j+4{\uptheta}^2\upbeta {x}_j^2\right)}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right),\\ {}{L}_{222}^{(3)}=\frac{\partial^3\ln L}{\partial {\upalpha}^3}=\sum \limits_{i=1}^{n\left(1-\uppi \right)}{\updelta}_{u_i}\left(\frac{2}{{\left(\upalpha +\uptheta {t}_i\right)}^3}\right)+\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{2{\left(\upbeta \right)}^3}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right),\\ {}{L}_{333}^{(3)}=\frac{\partial^3\ln L}{\partial {\uptheta}^3}=\sum \limits_{i=1}^{n\left(1-\uppi \right)}{\updelta}_{u_i}\left(\frac{2{t}_i^3}{{\left(\upalpha +\uptheta {t}_i\right)}^3}\right)+\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{(2){\left({\upbeta}^2{x}_j\right)}^3}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right),\\ {}{L}_{112}^{(3)}={L}_{211}^{(3)}={L}_{121}^{(3)}=\frac{\partial^3\ln L}{\partial {\upbeta}^2\mathrm{\partial \upalpha }}=\frac{\partial }{\mathrm{\partial \upbeta }}\left(\frac{\partial^2\ln L}{\mathrm{\partial \upbeta \partial \upalpha }}\right)=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{2\left({\uptheta}^2{\upbeta}^3{x}_j^2\right)}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right),\\ {}{L}_{113}^{(3)}={L}_{311}^{(3)}={L}_{131}^{(3)}=\frac{\partial^3\ln L}{\partial {\upbeta}^2\mathrm{\partial \uptheta }}=\frac{\partial }{\mathrm{\partial \upbeta }}\left(\frac{\partial^2\ln L}{\mathrm{\partial \upbeta \partial \uptheta }}\right)=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{-2\left({\upalpha \uptheta \upbeta}^3{x}_j^2\right)}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}-{x}_j^2\right)+{\upeta}^2\left({n}_a-n\uppi \right),\\ {}{L}_{221}^{(3)}={L}_{122}^{(3)}={L}_{212}^{(3)}=\frac{\partial^3\ln L}{\partial {\upalpha}^2\mathrm{\partial \upbeta }}=\frac{\partial }{\mathrm{\partial \upalpha }}\left(\frac{\partial^2\ln L}{\mathrm{\partial \upalpha \partial \upbeta }}\right)=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{2{\uptheta \upbeta}^3{x}_j}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right),\\ {}{L}_{331}^{(3)}={L}_{133}^{(3)}={L}_{313}^{(3)}=\frac{\partial^3\ln L}{\partial {\uptheta}^2\mathrm{\partial \upbeta }}=\frac{\partial }{\mathrm{\partial \uptheta }}\left(\frac{\partial^2\ln L}{\mathrm{\partial \uptheta \partial \upbeta }}\right)=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{-2{\upalpha \upbeta}^4{x}_j^2}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right),\\ {}{L}_{223}^{(3)}={L}_{322}^{(3)}={L}_{232}^{(3)}=\frac{\partial^3\ln L}{\partial {\upalpha}^2\mathrm{\partial \uptheta }}=\frac{\partial }{\mathrm{\partial \upalpha }}\left(\frac{\partial^2\ln L}{\mathrm{\partial \upalpha \partial \uptheta }}\right)=\sum \limits_{i=1}^{n\left(1-\uppi \right)}{\updelta}_{u_i}\left(\frac{2{t}_i}{{\left(\upalpha +\uptheta {t}_i\right)}^3}\right)+\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{2{\upbeta}^4{x}_j}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right),\\ {}{L}_{332}^{(3)}={L}_{233}^{(3)}={L}_{323}^{(3)}=\frac{\partial^3\ln L}{\partial {\uptheta}^2\mathrm{\partial \upalpha }}=\frac{\partial }{\mathrm{\partial \uptheta }}\left(\frac{\partial^2\ln L}{\mathrm{\partial \uptheta \partial \upalpha }}\right)=\sum \limits_{i=1}^{n\left(1-\uppi \right)}{\updelta}_{u_i}\left(\frac{2{t}_i^2}{{\left(\upalpha +\uptheta {t}_i\right)}^3}\right)+\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{2{\upbeta}^5{x}_j^2}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right),\\ {}{L}_{123}^{(3)}={L}_{321}^{(3)}={L}_{231}^{(3)}={L}_{312}^{(3)}={L}_{213}^{(3)}={L}_{132}^{(3)}=\frac{\partial^3\ln L}{\mathrm{\partial \upalpha \partial \uptheta \partial \upbeta }}=\frac{\partial }{\mathrm{\partial \upbeta }}\left(\frac{\partial^2\ln L}{\mathrm{\partial \upalpha \partial \uptheta }}\right)=\sum \limits_{j=1}^{n\uppi}{\updelta}_{a_j}\left(\frac{{\uptheta \upbeta}^4{x}_j^2-{\upalpha \upbeta}^3{x}_j}{{\left(\upalpha \upbeta +{\uptheta \upbeta}^2{x}_j\right)}^3}\right).\end{array}} $$