Inference for generalized inverse Lindley distribution based on generalized order statistics

Abstract

This article addresses the problem of deriving the explicit expressions for single and product moments of order statistics, generalized order statistics and dual generalized order statistics from the generalized Lindley distribution. The means and variances of order statistics, lower record values and progressively type-II censored order statistics for various values of the parameters are tabulated. Furthermore, the maximum likelihood estimators for the parameters of the model using generalized order statistics are obtained. The Bayes estimators under squared error and LINEX (Linear exponential) loss functions using Markov Chain Monte Carlo (MCMC) technique are obtained. Finally, three real data examples to lower record values, type-II censored and progressively type-II censored order statistics have been analyzed for illustrative purposes.

Appendix

Lemma 1

Let f(x) and F(x) be given by (1) and (2), respectively. For $a_{1} > 0$ and $a_{2} > 0$, let

$$\begin{aligned} \displaystyle I (a_{1}, a_{2}) = \int _0^\infty x^{a_{1}} F^{a_{2}} (x) f(x) dx. \end{aligned}$$

Then,

$$\begin{aligned} \displaystyle I(a_{1}, a_{2}) = \sum _{i=0}^{a_{2}}{a_{2} \atopwithdelims ()i}\frac{\eta ^{i+2}}{(1+\eta )^{i+1}}\left[ \frac{\Gamma \left( \frac{\tau (i+1)-a_{1}}{\tau }+1\right) }{(\eta (a_{2}+1))^{\frac{\tau (i+1)-a_{1}}{\tau }+1}}+\frac{\Gamma \left( \frac{\tau (i+2)-a_{1}}{\tau }+1\right) }{(\eta (a_{2}+1))^{\frac{\tau (i+2)-a_{1}}{\tau }+1}}\right] , \end{aligned}$$

where $\Gamma (\lambda )=\int _{0}^{\infty }z^{\lambda -1}~e^{-z}dz.$

Proof

We can write

$$\begin{aligned} \displaystyle \int _0^\infty x^{a_{1}} F^{a_{2}} (x) f(x) dx = \frac{\tau \eta ^{2}}{1+\eta }\int _0^\infty x^{a_{1}-2\tau -1}~(1+x^{\tau })~e^{-\frac{\eta (a_{2}+1)}{ x^{\tau }}}\left( 1+\frac{\eta }{1+\eta }\frac{1}{x^{\tau }}\right) ^{a_{2}}dx. \end{aligned}$$

Using the binomial expression, the above can be rewritten as

$$\begin{aligned} \displaystyle \int _0^\infty x^{a_{1}} F^{a_{2}} f(x) dx= & {} \displaystyle \sum _{i=0}^{a_{2}}{a_{2} \atopwithdelims ()i}\frac{\eta ^{i+2}}{(1+\eta )^{i+1}}\int _0^\infty x^{a_{1}-\tau (i+2)-1}~(1+x^{\tau })~e^{-\frac{\eta (a_{2}+1)}{ x^{\tau }}}dx \\= & {} \displaystyle \sum _{i=0}^{a_{2}}{a_{2} \atopwithdelims ()i}\frac{\eta ^{i+2}}{(1+\eta )^{i+1}}\Bigg [\int _0^\infty x^{a_{1}-\tau (i+2)-1}~e^{-\frac{\eta (a_{2}+1)}{ x^{\tau }}}dx\\&+ \displaystyle \int _0^\infty x^{a_{1}-\tau (i+3)-1}~e^{-\frac{\eta (a_{2}+1)}{ x^{\tau }}}dx\Bigg ] \\= & {} \displaystyle \sum _{i=0}^{a_{2}}{a_{2} \atopwithdelims ()i}\frac{\eta ^{i+2}}{(1+\eta )^{i+1}}\Bigg [\frac{1}{\tau ~(\eta (a_{2}+1))^{\frac{\tau (i+1)-a_{1}}{\tau }+1}}\int _0^\infty z^{\frac{\tau (i+1)-a_{1}}{\tau }}~e^{-z}dz \\&+ \displaystyle \frac{1}{\tau ~(\eta (a_{2}+1))^{\frac{\tau (i+2)-a_{1}}{\tau }+1}}\int _0^\infty z^{\frac{\tau (i+2)-a_{1}}{\tau }}~e^{-z}dz\Bigg ]. \end{aligned}$$

$\square$

Lemma 2

For $a_{1} > 0$, $a_{2} > 0$, $a_{3} > 0$ and $a_{4} > 0$, let

$$\begin{aligned} \displaystyle K (a_{1}, a_{2}, a_{3}, a_{4}) = \int _0^\infty \int _x^\infty x^{a_{3}} y^{a_{4}} F^{a_{1}} (x) F^{a_{2}} (y) f(x) f(y) dy dx. \end{aligned}$$

Then,

$$\begin{aligned}&K(a_{1}, a_{2}, a_{3}, a_{4}) = \sum _{i=0}^{a_{2}}\sum _{j=0}^{i_1}{\left( {\begin{array}{c}a_{2}\\ i\end{array}}\right) } {\left( {\begin{array}{c}a_{1}\\ j\end{array}}\right) }\frac{\eta ^{i+j+4}}{(1+\eta )^{i+j+3}} \\&\left[ \frac{\left( \frac{a_{2}+1}{a_{1}+1}\right) ^{\frac{\tau (i+1)-a_{4}}{\tau }} \Gamma \left( \frac{\tau (i+1)-a_{4}}{\tau }+\frac{\tau (j+1)-a_{3}}{\tau }+1\right) }{\left( \frac{\tau (i+1)-a_{4}}{\tau }\right) [\eta (a_{1}+1)]^{\frac{\tau (j+1)-a_{3}}{\tau }+\frac{\tau (i+1)}{\tau }+2}\left( \frac{a_{2}+1}{a_{1}+1}+1 \right) ^{\frac{\tau (j+1)-a_{3}}{\tau }+\frac{\tau (i+1)-a_{4}}{\tau }+1}}\right. \\&\quad \times ~2F_{1}\left( 1,\frac{\tau (i+1)-a_{4}}{\tau }+\frac{\tau (j+1)- a_{3}}{\tau }+1;\frac{\tau (j+1)-a_{3}}{\tau }+2;\frac{a_{1}+1}{a_{1}+a_{2}+2}\right) \\&\quad +\frac{\left( \frac{a_{2}+1}{a_{1}+1}\right) ^{\frac{\tau (i+2)- a_{4}}{\tau }}\Gamma \left( \frac{\tau (i+2)-a_{4}}{\tau }+ \frac{\tau (j+2)-a_{3}}{\tau }+1\right) }{\left( \frac{\tau (i+2)- a_{4}}{\tau }\right) [\eta (a_{1}+1)]^{\frac{\tau (j+2)-a_{3}}{\tau } +\frac{\tau (i+2)}{\tau }+2}\left( \frac{a_{2}+1}{a_{1}+1}+1 \right) ^{\frac{\tau (j+2)-a_{3}}{\tau }+\frac{\tau (i+2)-a_{4}}{\tau }+1}}. \\&\quad \times ~2F_{1}\left( 1,\frac{\tau (i+2)-a_{4}}{\tau }+\frac{\tau (j+2)- a_{3}}{\tau }+1;\frac{\tau (j+2)-a_{3}}{\tau }+2;\frac{a_{1}+1}{a_{1} +a_{2}+2}\right) \\&\quad +\frac{\left( \frac{a_{2}+1}{a_{1}+1}\right) ^{\frac{\tau (i+2)- a_{2}}{\tau }}\Gamma \left( \frac{\tau (i+2)-a_{4}}{\tau }+ \frac{\tau (j+2)-a_{3}}{\tau }+1\right) }{\left( \frac{\tau (i+2)- a_{4}}{\tau }\right) [\eta (a_{1}+1)]^{\frac{\tau (j+2)-a_{3}}{\tau }+ \frac{\tau (i+2)}{\tau }+2}\left( \frac{a_{2}+1}{a_{1}+1}+1 \right) ^{\frac{\tau (j+2)-a_{3}}{\tau }+\frac{\tau (i+2)-a_{4}}{\tau }+1}}. \\&\quad \times ~2F_{1}\left( 1,\frac{\tau (i+2)-a_{4}}{\tau }+ \frac{\tau (j+2)-a_{3}}{\tau }+1;\frac{\tau (j+2)-a_{3}}{\tau }+2; \frac{a_{1}+1}{a_{1}+a_{2}+2}\right) \\&\quad +\frac{\left( \frac{a_{2}+1}{a_{1}+1}\right) ^{\frac{\tau (i+3)- a_{4}}{\tau }}\Gamma \left( \frac{\tau (i+3)-a_{4}}{\tau }+ \frac{\tau (j+3)-a_{3}}{\tau }+1\right) }{\left( \frac{\tau (i+3)- a_{4}}{\tau }\right) [\eta (a_{1}+1)]^{\frac{\tau (j+3)-a_{3}}{\tau } +\frac{\tau (i+3)}{\tau }+2}\left( \frac{a_{2}+1}{a_{1}+1}+1 \right) ^{\frac{\tau (j+3)-a_{3}}{\tau }+\frac{\tau (i+3)-a_{4}}{\tau }+1}}. \\&\quad \left. \times ~2F_{1}\left( 1,\frac{\tau (i+3)-a_{4}}{\tau } +\frac{\tau (j+3)-a_{3}}{\tau }+1;\frac{\tau (j+3)-a_{3}}{\tau }+2; \frac{a_{1}+1}{a_{1}+a_{2}+2}\right) \right] , \end{aligned}$$

where ${}_2 F_1(a,b;~c;~x)$ denotes the Gauss hypergeometric function defined by

$$\begin{aligned} {}_2 F_1(a_{1},a_{2};~a_{3};~x)=\sum _{p=0}^{\infty }\frac{(a_{1})_{p}~ (a_{2})_{p}}{(a_{3})_{p}}\frac{x^{p}}{p!}, \end{aligned}$$

where $(f)_{p}=f(f+1)\ldots (f+p-1)$ denotes the ascending factorial.

Proof

We can write

$$\begin{aligned}&\displaystyle \int _x^\infty y^{a_{4}} F^{a_{2}} (y) f(y) dy \\&\quad = \displaystyle \left( \frac{\tau \eta ^{2}}{1+\eta }\right) \int _x^\infty y^{a_{4}-2\tau -1}(1+y^{\tau })e^{-\frac{\eta (a_{2}+1)}{y^{\tau }}} \left( 1+\frac{\eta }{1+\eta }\frac{1}{y^{\tau }}\right) ^{a_{2}} dy \\&\quad = \displaystyle \left( \frac{\tau \eta ^{2}}{1+\eta }\right) \sum _{i = 0}^{a_{2}}{a_{2} \atopwithdelims ()i}\left( \frac{\eta }{1+\eta }\right) ^{i}\int _x^\infty y^{a_{4}-\tau (i+2)-1}(1+y^{\tau })e^{-\frac{\eta (a_{2}+1)}{y^{\tau }}}dy \\&\quad = \displaystyle \left( \frac{\tau \eta ^{2}}{1+\eta }\right) \sum _{i = 0}^{a_{2}}{a_{2} \atopwithdelims ()i}\left( \frac{\eta }{1+\eta }\right) ^{i}\Bigg [\int _x^\infty y^{a_{4}-\tau (i+2)-1}(1+y^{\tau })e^{-\frac{\eta (a_{2}+1)}{y^{\tau }}}dy \\&\qquad + \displaystyle \int _x^\infty y^{a_{4}-\tau (i+3)-1}(1+y^{\tau })e^{-\frac{\eta (a_{2}+1)}{y^{\tau }}}dy\Bigg ] \\&\quad = \displaystyle \left( \frac{\tau \eta ^{2}}{1+\eta }\right) \sum _{i = 0}^{a_{2}}{a_{2} \atopwithdelims ()i}\left( \frac{\eta }{1+\eta }\right) ^{i} \Bigg [\frac{1}{\tau (\eta (a_{2}+1))^{\frac{\tau (i+1)- a_{4}}{\tau }+1}}\int _{0}^{\frac{\eta (a_{2}+1)}{x^{\tau }}} z^{\frac{\tau (i+1)-a_{4}}{\tau }}~e^{-z}dz \\&\qquad + \displaystyle \frac{1}{\tau (\eta (a_{2}+1))^{\frac{\tau (i+2)-a_{4}}{\tau }+1}}\int _{0}^{\frac{\eta (a_{2}+1)}{x^{\tau }}} z^{\frac{\tau (i+2)-a_{4}}{\tau }}~e^{-z}dz\Bigg ] \\&\quad = \sum _{i= 0}^{a_{2}} {a_{2} \atopwithdelims ()i}\frac{\eta ^{i+2}}{(1+\eta )^{i+1}} \left[ \frac{\gamma \left( \frac{\tau (i+1)-a_{4}}{\tau }+ 1,~\frac{\eta (i+1)}{x^{\tau }}\right) }{(\eta (a_{2}+1))^{\frac{\tau (i+1) -a_{4}}{\tau }+1}}+\frac{\gamma \left( \frac{\tau (i+2)-a_{4}}{\tau }+1, ~\frac{\eta (a_{2}+1)}{x^{\tau }}\right) }{(\eta (a_{2}+1) )^{\frac{\tau ~(i+2)-a_{4}}{\tau }+1}}\right] , \end{aligned}$$

where we have set $z=\frac{\eta (a_{2}+1)}{y^{\tau }}$ and $\gamma (a_{1},x)=\int _{0}^{x}t^{a_{1}-1}exp(-t)dt$ denote the complementary incomplete gamma function. Hence,

$$\begin{aligned}&\displaystyle \int _0^\infty \int _x^\infty x^{a_{3}} y^{a_{4}} F^{a_{1}} (x) F^{a_{2}} (y) f(x) f(y) dy dx \\&\quad = \sum _{i= 0}^{a_{2}} {b_{2} \atopwithdelims ()i}\frac{\tau \eta ^{i+4}}{(1+\eta )^{i+1}} \int _{0}^{\infty }x^{p_{1}-2\tau -1}~(1+x^{\tau })~e^{-\frac{\eta (a_{1}+1)}{ x^{\tau }}} \left( 1+\frac{\eta }{(1+\eta )}\frac{1}{x^{\tau }}\right) ^{a_{1}} \\&\qquad \times \displaystyle \left[ \frac{\gamma \left( \frac{\tau (i+1) -a_{4}}{\tau }+1,~ \frac{\eta (i+1)}{x^{\tau }}\right) }{(\eta (a_{2}+1))^{\frac{\tau (i+1)-a_{4}}{\tau }+1}}+ \frac{\gamma \left( \frac{\tau (i+2)-a_{4}}{\tau }+1,~ \frac{\eta (a_{3}+1)}{x^{\tau }}\right) }{(\eta (a_{2}+ 1))^{\frac{\tau ~(i+2)-a_{4}}{\tau }+1}}\right] dx \\&\quad = \sum _{i = 0}^{a_{2}} \sum _{j = 0}^{a_{1}} {a_{2} \atopwithdelims ()i}{a_{1} \atopwithdelims ()j}\frac{\tau \eta ^{i+j+4}}{(1+\eta )^{i+j+2}} \\&\qquad \times \displaystyle \Bigg [\frac{1}{(\eta (a_{2}+1))^{\frac{\tau (i+1)-a_{4}}{\tau }+1}}\int _{0}^{\infty }x^{a_{3}-\tau (j+2)-1}~(1+x^{\tau }) e^{-\frac{\eta (a_{1}+1)}{x^{\tau }}} \\&\qquad \times \displaystyle \gamma \left( \frac{\tau (i+1)-a_{4}}{\tau }+1,~\frac{\eta (a_{2}+1)}{x^{\tau }}\right) dx \\&\qquad + \displaystyle \frac{1}{(\eta (a_{2}+1))^{\frac{\tau (i+2)-a_{4}}{\tau }+1}} \int _{0}^{\infty }x^{a_{3}-\tau (j+2)-1}~(1+x^{\tau })~ e^{-\frac{\eta (a_{1}+1)}{x^{\tau }}} \\&\qquad \times \displaystyle \gamma \left( \frac{\tau (i+2)-a_{4}}{\tau }+1, ~\frac{\eta (a_{2}+1)}{x^{\tau }}\right) dx\Bigg ] \\&\quad = \displaystyle \sum _{i = 0}^{a_{2}} \sum _{j = 0}^{a_{1}} {a_{2} \atopwithdelims ()i}{a_{1} \atopwithdelims ()j}\frac{\tau \eta ^{i+j+4}}{(1+\eta )^{i+j+2}} \\&\qquad \times \displaystyle \Bigg [\frac{1}{(\eta (a_{2}+1))^{\frac{\tau (i+1)- a_{4}}{\tau }+1}}\int _{0}^{\infty }x^{a_{3}-\tau (j+2)-1} ~e^{-\frac{\eta (a_{1}+1)}{x^{\tau }}} \\&\qquad \times \displaystyle \gamma \left( \frac{\tau (i+1)-a_{4}}{\tau }+1,~ \frac{\eta (a_{2}+1)}{x^{\tau }}\right) dx \\&\qquad + \displaystyle \frac{1}{(\eta (a_{2}+1))^{\frac{\tau (i+1)-a_{4}}{\tau }+1}} \int _{0}^{\infty }x^{a_{3}-\tau (j+3)-1}~ e^{-\frac{\eta (a_{1}+1)}{x^{\tau }}} \\&\qquad \times \displaystyle \gamma \left( \frac{\tau (i+1)-a_{4}}{\tau }+1,~\frac{\eta (a_{2}+1)}{x^{\tau }}\right) dx \\&\qquad + \displaystyle \frac{1}{(\eta (a_{2}+1))^{\frac{\tau (i+1)-a_{4}}{\tau }+1}} \int _{0}^{\infty }x^{a_{3}-\tau (j+2)-1}~e^{-\frac{\eta (a_{1}+1)}{x^{\tau }}} \\&\qquad \times \displaystyle \gamma \left( \frac{\tau (i+2)-a_{4}}{\tau }+1,~\frac{\eta (a_{2}+1)}{x^{\tau }}\right) dx \\&\qquad + \displaystyle \frac{1}{(\eta (a_{2}+1))^{\frac{\tau (i+1)-a_{4}}{\tau }+1}}\int _{0}^{\infty }x^{a_{3}-\tau (j+3)-1}~e^{-\frac{\eta (a_{1}+1)}{x^{\tau }}} \\ \end{aligned}$$

Using the transformation $t= \frac{\eta (a_{2}+1)}{x^{\tau }}$. The result follows by using equation (6.455.1) in Gradshteyn and Ryzhik (2014) to calculate the above integral. $\square$