Bayesian inference for Maxwell Boltzmann distribution on step-stress partially accelerated life test under progressive type-II censoring with binomial removals

Abstract

This article deals with the estimation problem in step-stress partially accelerated life test of Maxwell Boltzmann distribution in presence of progressive type-II censoring with binomial removals. The maximum likelihood and Bayes estimators of the parameter are obtained under symmetric and asymmetric loss functions. Furthermore, the performances of the obtained estimators are compared in terms of risks. The proposed methodology is illustrated through the time to failure (in days) of Aluminium reduction cells and survival times (in weeks) for male rats that were exposed to a high level of radiation.

Appendix

Proposition 1

The second partial derivatives of the log-likelihood function are calculated as follows:

$$\begin{aligned} \xi _{\theta _1}=-\frac{3}{2\theta }\sum _{i=1}^{n_{1}} \frac{ R_{i}\{ \Gamma (\frac{x_{i}^2}{\theta },\frac{5}{2})-\Gamma (\frac{x_{i}^2}{\theta },\frac{3}{2}) \} }{1-\frac{1}{\Gamma (\frac{3}{2})}\Gamma (\frac{x_{i}^2}{\theta },\frac{3}{2})}, \end{aligned}$$

and

$$\begin{aligned} \xi _{\theta _2}=-\frac{3}{2\theta }\sum _{n_{1}+1}^{m}\frac{R_{i}\{ \Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{5}{2})-\Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}) \} }{1-\frac{1}{\Gamma (\frac{3}{2})}\Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2})}. \end{aligned}$$

Proof

We have,

$$ \begin{aligned} \xi _{{\theta _{1} }} = & \frac{\partial }{{\partial \theta }}\left\{ {\sum\limits_{{i = 1}}^{{n_{1} }} {R_{i} } \ln \left( {1 - \frac{1}{{\Gamma (\frac{3}{2})}}\Gamma \left( {\frac{{x_{i}^{2} }}{\theta }\frac{3}{2}} \right)} \right)} \right\} \\ = & - \sum\limits_{{i = 1}}^{{n_{i} }} {\frac{{R_{i} \{ \frac{\partial }{{\partial \theta }}\Gamma (\frac{{x_{i}^{2} }}{\theta },\frac{3}{2})\} }}{{1 - \frac{1}{{\Gamma (\frac{3}{2})}}\Gamma (\frac{{x_{i}^{2} }}{\theta },\frac{3}{2})}}} \\ \end{aligned} $$

(36)

Let $I=\frac{\partial }{\partial \theta }\left( \Gamma (\frac{x_{i}^2}{\theta },\frac{3}{2}) \right)$, we have $\Gamma (x,a)=\frac{1}{\Gamma (a)}\int _{0}^{x}e^{-w}w^{a-1}dw$, is an incomplete gamma ratio, so that

$$\begin{aligned} I_{1}=\frac{\partial }{\partial \theta }\int _{0}^{x_{i}^2/\theta }e^{-w}w^{a-1}dw \end{aligned}$$

(37)

After putting $w=\frac{v^2}{\theta }$, $dw=(\frac{2v}{\theta })dv$, at $w=0$ and $w=\frac{x_{i}^2}{\theta }$, $v=x$, above Eq. 37 becomes

$$ \begin{aligned} I_{1} = & \frac{\partial }{{\partial \theta }}\int_{0}^{x} {e^{{ - \frac{{v^{2} }}{\theta }}} } \left( {\frac{{v^{2} }}{\theta }} \right)^{{\frac{3}{2} - 1}} dv = \int_{0}^{x} 2 v^{2} \frac{\partial }{{\partial \theta }}\left( {\frac{{e^{{ - \frac{{x^{2} }}{\theta }}} }}{{\theta ^{{3/2}} }}} \right) \\ I_{1} = & \frac{\partial }{{\partial \theta }}\int_{0}^{x} {e^{{ - \frac{{v^{2} }}{\theta }}} } \left( {\frac{{v^{2} }}{\theta }} \right)^{{\frac{3}{2} - 1}} dv = \int_{0}^{x} 2 v^{2} \frac{\partial }{{\partial \theta }}\left( {\frac{{e^{{ - \frac{{x^{2} }}{\theta }}} }}{{\theta ^{{3/2}} }}} \right) \\ = & \int_{0}^{x} 2 v^{2} \left( {\frac{{v^{2} e^{{ - \frac{{v^{2} }}{\theta }}} (1/\theta ^{{1/2}} ) - (3/2)\theta ^{{1/2}} e^{{ - \frac{{v^{2} }}{\theta }}} }}{{\theta ^{3} }}} \right)dv \\ = & \int_{0}^{x} {\left( {\frac{{2v^{4} e^{{ - \frac{{v^{2} }}{\theta }}} }}{{\theta ^{{7/2}} }} - \frac{{3v^{2} e^{{ - \frac{{v^{2} }}{\theta }}} }}{{\theta ^{{5/2}} }}} \right)} dv \\ \end{aligned} $$

(38)

On putting the $t=\frac{v^2}{\theta }$, $dt=\frac{2v}{\theta }dv$ at $v=0,~ t=0$ and $vx,~ t=\frac{x^2}{\theta }$, Eq. 38 becomes

$$\begin{aligned} I_{1}=\int _{0}^{x^2/\theta } \left( \frac{ 2t^{3/2} e^{-t}}{\theta } - \frac{3t^{1/2} e^{-t}}{2\theta } \right) dt = \frac{1}{\theta }\left[ \Gamma \left( \frac{5}{2}\right) \Gamma \left( \frac{x^2}{\theta }, \frac{5}{2}\right) - \frac{3}{2}\Gamma \left( \frac{3}{2}\right) \Gamma \left( \frac{x^2}{\theta }, \frac{3}{2}\right) \right] \end{aligned}$$

(39)

On putting the value from Eqs. 39 into 36, we get

$$\begin{aligned} \xi _{\theta _1}=-\frac{3}{2\theta }\sum _{i=1}^{n_{1}} \frac{ R_{i}\{ \Gamma (\frac{x_{i}^2}{\theta },\frac{5}{2})-\Gamma (\frac{x_{i}^2}{\theta },\frac{3}{2}) \} }{(1-\frac{1}{\Gamma (\frac{3}{2})}\Gamma (\frac{x_{i}^2}{\theta },\frac{3}{2}))}. \end{aligned}$$

Again differentiate above equation

$$ \begin{gathered} \xi ^{\prime}_{{\theta _{1} }} = \frac{\partial }{{\partial \theta }}(\xi _{{\theta _{1} }} ) = - \frac{\partial }{{\partial \theta }}\left\{ {\frac{3}{{2\theta }}\sum\limits_{{i = 1}}^{{n_{1} }} {\frac{{R_{i} \{ \Gamma (\frac{{x_{i}^{2} }}{\theta },\frac{5}{2}) - \Gamma (\frac{{x_{i}^{2} }}{\theta },\frac{3}{2})\} }}{{(1 - \frac{1}{{\Gamma (\frac{3}{2})}}\Gamma (\frac{{x_{i}^{2} }}{\theta },\frac{3}{2}))}}} } \right\} \hfill \\ = - \frac{3}{2}\frac{\partial }{{\partial \theta }}\left\{ {\sum\limits_{{i = 1}}^{{n_{1} }} {\frac{{R_{i} I_{2} }}{{I_{3} }}} } \right\} = \frac{3}{2}\left\{ {\sum\limits_{{i = 1}}^{{n_{1} }} {\frac{{R_{i} (I_{3} (\partial I_{2} /\partial \theta ) - I_{2} (\partial I_{3} /\partial \theta ))}}{{(I_{3} )^{2} }}} } \right\} \hfill \\ \end{gathered} $$

where, $I_{2}=\frac{1}{\theta } \left\{ \Gamma (\frac{x_{i}^2}{\theta },\frac{5}{2})-\Gamma (\frac{x_{i}^2}{\theta },\frac{3}{2}) \right\}$ and $I_{3}= \left\{ 1- \Gamma (\frac{x_{i}^2}{\theta },\frac{3}{2}) \right\}$; $i=1,2,\cdots ,n_{1}$.

Similarly, we can obtain the expression for $\xi _{\theta _2}$ given below

$$\begin{aligned} \xi _{\theta _2}=-\frac{3}{2\theta }\sum _{i=n_{1}+1}^{m}\frac{R_{i} \{ \Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{5}{2})-\Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}) \} }{(1-\frac{1}{\Gamma (\frac{3}{2})}\Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}))}. \end{aligned}$$

and also, we can obtain the expression for $\xi '_{\theta _2}$. $\square$

Proposition 2

$$\begin{aligned} \xi _{\beta } = -\frac{2}{\pi \theta }\sum _{n_{1}+1}^{m}\frac{R_{i} \{ \left( \frac{2(\varphi (\beta ))^2}{\theta ^{1/2}}\right) e^{-\frac{(\varphi (\beta ))^2}{\theta }}(x_{i}-\tau ) \} }{\left( 1-\frac{1}{\Gamma (\frac{3}{2})}\Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2})\right) }. \end{aligned}$$

Proof

$$ \begin{aligned} \xi _{\beta } = & \frac{\partial }{{\partial \beta }}\left\{ {\sum\limits_{{i = n_{1} + 1}}^{m} {R_{i} } \log \left( {1 - \frac{1}{{\Gamma (\frac{3}{2})}}\Gamma \left( {\frac{{(\varphi (\beta ))^{2} }}{\theta },\frac{3}{2}} \right)} \right)} \right\} \\ = & - \frac{{\sum\limits_{{i = n_{1} + 1}}^{m} {R_{i} } \{ \frac{\partial }{{\partial \beta }}\Gamma (\frac{{(\varphi (\beta ))^{2} }}{\theta },\frac{3}{2})\} }}{{1 - \frac{1}{{\Gamma (\frac{3}{2})}}\Gamma (\frac{{(\varphi (\beta ))^{2} }}{\theta },\frac{3}{2})}} \\ \end{aligned} $$

We know that $\Gamma (x,a)=\int _{0}^{x}t^{a-1}e^{-t}dt$ and differentiating with respect x i.e. $\frac{\partial }{\partial x}\Gamma (x,a)=x^{a-1}e^{-x}$. Now

$$ \begin{gathered} \frac{\partial }{{\partial \beta }}\Gamma \left( {\frac{{(\varphi (\beta ))^{2} }}{\theta },\frac{3}{2}} \right) = \frac{\partial }{{\partial ((\varphi (\beta ))^{2} /\theta )}}\Gamma \left( {\frac{{(\varphi (\beta ))^{2} }}{\theta },\frac{3}{2}} \right) \hfill \\ \frac{{\partial ((\varphi (\beta ))^{2} /\theta )}}{{\partial \beta }} = \left( {\frac{{(\varphi (\beta ))^{2} }}{\theta }} \right)^{{\frac{3}{2} - 1}} e^{{ - \frac{{(\varphi (\beta ))^{2} }}{\theta }}} \frac{{\partial ((\varphi (\beta ))^{2} /\theta )}}{{\partial \beta }} \hfill \\ = \left( {\frac{{(\varphi (\beta ))^{2} }}{\theta }} \right)^{{\frac{1}{2}}} e^{{ - \frac{{(\varphi (\beta ))^{2} }}{\theta }}} \frac{{2(\varphi (\beta ))(x_{i} - \tau )}}{\theta } \hfill \\ = \left( {\frac{{2(\varphi (\beta ))^{2} }}{{\theta ^{{3/2}} }}} \right)e^{{ - \frac{{(\varphi (\beta ))^{2} }}{\theta }}} (x_{i} - \tau ). \hfill \\ \end{gathered} $$

We get,

$$\begin{aligned} \xi _{\beta } = -\frac{2}{\pi \theta }\sum _{n_{1}+1}^{m}\frac{R_{i} \{ \left( \frac{2(\varphi (\beta ))^2}{\theta ^{1/2}}\right) e^{-\frac{(\varphi (\beta ))^2}{\theta }}(x_{i}-\tau ) \} }{\left( 1-\frac{1}{\Gamma (\frac{3}{2})}\Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2})\right) }. \end{aligned}$$

Again differentiate above equation w. r. t. $\beta$ and $\theta$ i.e.

$$ \begin{gathered} \frac{{\partial ^{2} }}{{\partial \beta ^{2} }}\Gamma \left( {\frac{{(\varphi (\beta ))^{2} }}{\theta },\frac{3}{2}} \right) = \left( {\frac{{2(\varphi (\beta ))(x_{i} - \tau )}}{{\theta ^{{5/2}} }}} \right) \hfill \\ e^{{ - \frac{{(\varphi (\beta ))^{2} }}{\theta }}} \left\{ {2\theta (x_{i} - \tau ) + \varphi (\beta )} \right\}\frac{{\partial ^{2} }}{{\partial \beta \partial \theta }} \hfill \\ \Gamma \left( {\frac{{(\varphi (\beta ))^{2} }}{\theta },\frac{3}{2}} \right) = - \left( {\frac{{(\varphi (\beta ))^{2} (x_{i} - \tau )}}{{\theta ^{{5/2}} }}} \right) \hfill \\ e^{{ - \frac{{(\varphi (\beta ))^{2} }}{\theta }}} \left\{ {2\theta ^{{1/2}} (\varphi (\beta ))^{2} + 3} \right\} \hfill \\ \end{gathered} $$

respectively, and we obtain the second derivative given below,

$$\begin{aligned} \xi '_{\beta }=\sum _{n_{1}+1}^{m}\frac{R_{i}\left( \left( -\frac{1}{\Gamma (3/2)}\frac{\partial }{\partial \beta }\Gamma \left( \frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}\right) \right) \left( -\frac{1}{\Gamma (3/2)}\frac{\partial }{\partial \beta }\Gamma \left( \frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}\right) \right) -\left( 1-\frac{1}{\Gamma (3/2)}\Gamma \left( \frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}\right) \right) \left( -\frac{1}{\Gamma (3/2)}\frac{\partial ^2}{\partial \beta ^2}\Gamma \left( \frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}\right) \right) \right) }{\left( 1-\frac{1}{\Gamma (\frac{3}{2})}\Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2})\right) ^2}. \end{aligned}$$

and using the above differentials to obtain given below

$$\begin{aligned} \xi '_{\beta \theta }=\sum _{n_{1}+1}^{m}\frac{R_{i}\left( \left( -\frac{1}{\Gamma (3/2)}\frac{\partial }{\partial \beta }\Gamma \left( \frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}\right) \right) \left( -\frac{1}{\Gamma (3/2)}\frac{\partial }{\partial \theta }\Gamma \left( \frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}\right) \right) -\left( 1-\frac{1}{\Gamma (3/2)}\Gamma \left( \frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}\right) \right) \left( -\frac{1}{\Gamma (3/2)}\frac{\partial ^2}{\partial \beta \partial \theta }\Gamma \left( \frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2}\right) \right) \right) }{\left( 1-\frac{1}{\Gamma (\frac{3}{2})}\Gamma (\frac{(\varphi (\beta ))^2}{\theta },\frac{3}{2})\right) ^2}. \end{aligned}$$

respectively. $\square$