Bayesian inference for Poisson-inverse exponential distribution under progressive type-II censoring with binomial removal

Abstract

This paper proposes a Poisson-inverse exponential distribution (PIED) as a lifetime model with initially increasing then decreasing failure model. Its statistical characteristics and important distributional properties are discussed along with its reliability and failure rate function. The maximum likelihood estimators (MLEs) and Bayes estimators of parameters of PIED under symmetric and asymmetric loss functions for progressive type-II censored data with binomial removals have been obtained. The MLEs and corresponding Bayes estimators are compared in terms of their simulated risks. The proposed methodology is illustrated through a real data set of bladder cancer.

Appendices

Appendix 1: Proof of Theorem

The conditional density of the random variable $X = max(Y_{1, 2, \ldots ,} Y_{N})$ is given by

$$\begin{aligned} f(x|N=n,\lambda )&=n f(x)\left[ F(x)\right] ^{n-1};\quad x> 0,\quad \lambda > 0\nonumber \\&= \frac{n\lambda }{ x^2} e^{-\frac{n\lambda }{x}}. \end{aligned}$$

Then the joint distribution of X and N is given by:

$$\begin{aligned} f(X, N)&= P(N = n)f(x|N=n,\lambda ) \nonumber \\&= \frac{n\lambda }{ x^2} e^{-\frac{n\lambda }{x}} \frac{e^{-\theta }\theta ^{n}}{n!\left[ 1-e^{-\theta }\right] };\quad x> 0,\quad \lambda , \theta > 0. \end{aligned}$$

The marginal pdf of X can be easily obtained as follows:

$$\begin{aligned} f(x)&= \sum _{n=1}^{\infty }f(X, N)\nonumber \\&= \sum _{n=1}^{\infty }\frac{n\lambda }{ x^2} e^{-\frac{n\lambda }{x}} \frac{e^{-\theta }\theta ^{n}}{n!\left[ 1-e^{-\theta }\right] } \nonumber \\&= \frac{\theta \lambda e^{-\theta -\frac{\lambda }{x}}}{(1-e^{-\theta })x^{2}}\sum _{n=1}^{\infty } \frac{\left[ \theta e^{-\frac{\lambda }{x}})\right] ^{n-1}}{(n-1)!}\nonumber \\&= \frac{\theta \lambda e^{-\theta -\frac{\lambda }{x}+\theta e^{-\frac{\lambda }{x}}}}{(1-e^{-\theta })x^2}. \end{aligned}$$

Appendix 2: Proof for h(0) and $h(\infty ) = 0$

Since the failure rate function with $\lim _{x\rightarrow 0} h(x; \lambda , \theta )$ and $\lim _{x\rightarrow \infty } h(x; \lambda , \theta )$ are given as

$$\begin{aligned} \lim _{x\rightarrow 0} h(x; \lambda , \theta )&=\lim _{x\rightarrow 0} {\frac{\theta e^{\theta \left( e^{-\frac{\lambda }{x}}-1\right) } \lambda e^{-\frac{\lambda }{x}}}{x^2 \left( 1-e^{\theta \left( e^{-\frac{\lambda }{x}}-1\right) }\right) }}\\&= \lim _{x\rightarrow 0}{\frac{\theta e^{\theta \left( e^{-\frac{\lambda }{x}}-1\right) }}{\left( 1-e^{\theta \left( e^{-\frac{\lambda }{x}}-1\right) }\right) }} \lim _{x\rightarrow 0} {\frac{ \lambda e^{-\frac{\lambda }{x}}}{x^2}}\\&=\left( \frac{\theta e^{-\theta }}{1-e^{-\theta }}\right) \lim _{x\rightarrow 0} {\frac{ \lambda e^{-\frac{\lambda }{x}}}{x^2}} = 0, \end{aligned}$$

and

$$\begin{aligned} \lim _{x\rightarrow \infty } h(x; \lambda , \theta )&=\lim _{x\rightarrow \infty } {\frac{\theta e^{\theta \left( e^{-\frac{\lambda }{x}}-1\right) } \lambda e^{-\frac{\lambda }{x}}}{x^2 \left( 1-e^{\theta \left( e^{-\frac{\lambda }{x}}-1\right) }\right) }}\\&= \lim _{x\rightarrow \infty }{\frac{\theta e^{\theta \left( e^{-\frac{\lambda }{x}}-1\right) }}{\left( 1-e^{\theta \left( e^{-\frac{\lambda }{x}}-1\right) }\right) }} \lim _{x\rightarrow \infty } {\frac{ \lambda e^{-\frac{\lambda }{x}}}{x^2}}\\&=\left( \frac{\theta e^{-\theta }}{1-e^{-\theta }}\right) \lim _{x\rightarrow \infty } {\frac{ \lambda e^{-\frac{\lambda }{x}}}{x^2}} = 0. \end{aligned}$$