Analysis of an Inverted Modified Lindley Distribution Using Dual Generalized Order Statistics

In this paper, the estimation of the unknown parameters of the inverted modified Lindley distribution using the concept of dual generalized order statistics is investigated from both classical and Bayesian viewpoints. Based on this considered distribution, first, the maximum likelihood estimator and approximate confidence interval of the model parameter are obtained using order statistics and lower record values. Next, we consider Bayes estimation under the symmetric loss function using a gamma prior. We also derived the highest posterior density credible interval based on the Metropolis–Hastings algorithm. Moreover, the results for single and product moments based on dual generalized order statistics using this distribution are derived. Based on order statistics and lower record values, Monte Carlo simulations are carried out to examine the enforcement of the proposed estimators, and a dataset is investigated for both order statistics and lower record values for illustrative purposes.

APPENDIX

Lemma 1. For (a₁, a₂, and a₃) > 0, let

$$ {\Delta }_1\left({a}_1,{a}_2,{a}_3\right)={\int}_0^{\infty }{y}^{\alpha_1-2}\left(\left(1+{a}_3\right){e}^{\frac{a_3}{y}}+\frac{2{a}_3}{y}-1\right)\exp \left(-\frac{a_2{a}_3}{y}\right) dy. $$

(16)

Then

$$ {\Delta }_1\left({a}_1,{a}_2,{a}_3\right)=\left(\frac{\left(1+{a}_3\right)\Gamma \left(1-{a}_1\right)}{{\left({a}_3\left({a}_2-1\right)\right)}^{1-{a}_1}}+\frac{2{a}_3\Gamma \left(2-{a}_1\right)}{{\left({a}_2{a}_3\right)}^{2-{a}_1}}-\frac{\Gamma \left(1-{a}_1\right)}{{\left({a}_2{a}_3\right)}^{1-{a}_1}}\right). $$

Proof. We can write Eq. (16) as

$$ {\displaystyle \begin{array}{c}{\Delta }_1\left({a}_1,{a}_2,{a}_3\right)=\left(1+{a}_3\right){\int}_0^{\infty }{y}^{\alpha_1-2}\exp \left(-\frac{\left({a}_2-1\right){a}_3}{y}\right) dy+2{a}_3{\int}_0^{\infty }{y}^{\alpha_1-3}\exp \left(-\frac{a_2{a}_3}{y}\right) dy\kern0.5em \\ {}-{\int}_0^{\infty }{y}^{\alpha_1-2}\exp \left(-\frac{a_2{a}_3}{y}\right) dy\kern0.5em \\ {}\begin{array}{c}=\frac{\left(1+{a}_3\right)}{{\left({a}_3\left({a}_2-1\right)\right)}^{1-{a}_1}}{\int}_0^{\infty }{u}^{-{a}_1}{e}^{-u} du+\frac{2{a}_3}{{\left({a}_2{a}_3\right)}^{2-{a}_1}}{\int}_0^{\infty }{v}^{1-{a}_1}{e}^{-v} dv\\ {}-\frac{1}{{\left({a}_2{a}_3\right)}^{1-{a}_1}}{\int}_0^{\infty }{v}^{-{a}_1}{e}^{-v} dv,\end{array}\end{array}} $$

where $u=\frac{a_3\left({a}_2-1\right)}{y}$ and$v=\frac{a_2{a}_3}{y}.$

Lemma 2. For positive real numbers a₁, a₂, a₃, a₄ and a₅,, let

$$ {\displaystyle \begin{array}{c}{\Delta }_2\left({a}_1,{a}_2,{a}_3,{a}_4,{a}_5\right)={\int}_0^{\infty }{\int}_y^{\infty }{y}^{\alpha_1-2}{z}^{\alpha_3-2}{e}^{-\frac{a_5{a}_2}{y}}\left(\left(1+{a}_5\right){e}^{\frac{a_5}{y}}+\frac{2{a}_5}{y}-1\right)\\ {}\times {e}^{-\frac{a_5{a}_4}{z}}\left(\left(1+{a}_5\right){e}^{\frac{a_5}{z}}+\frac{2{a}_5}{z}-1\right) dzdy.\end{array}} $$

(17)

Then,

$$ {\displaystyle \begin{array}{c}{\Delta }_2\left({a}_1,{a}_2,{a}_3,{a}_4,{a}_5\right)=\frac{{\left(1+{a}_5\right)}^2\Gamma \left(2-{a}_1-{a}_3\right)}{{\left(\left({a}_4-1\right){a}_5\right)}^{2-{a}_1-{a}_3}\left(1-{a}_3\right){\left(1+\frac{a_2-1}{a_4-1}\right)}^{2-{a}_1-{a}_3}}\\ {}\times 2{F}_1\left(1,2-{a}_1-{a}_3;2-{a}_3;\frac{1}{1+\frac{a_2-1}{a_4-1}}\right)\\ {}\begin{array}{c}+\frac{2{a}_5\left(1+{a}_5\right)\Gamma \left(2-{a}_1-{a}_3\right)}{\left(1-{a}_3\right){\left({a}_5\left({a}_4-1\right)\right)}^{3-{a}_1-{a}_3}{\left(1+\frac{a_2}{a_4-1}\right)}^{3-{a}_1-{a}_3}}2{F}_1\left(1,3-{a}_1-{a}_3;2-{a}_3;\frac{1}{1+\frac{a_2}{a_4-1}}\right)\\ {}-\frac{\left(1+{a}_5\right)\Gamma \left(2-{a}_1-{a}_3\right)}{\left(1-{a}_3\right){\left({a}_5\left({a}_4-1\right)\right)}^{2-{a}_1-{a}_3}{\left(1+\frac{a_2}{a_4-1}\right)}^{2-{a}_1-{a}_3}}2{F}_1\left(1,2-{a}_1-{a}_3;2-{a}_3;\frac{1}{1+\frac{a_2}{a_4-1}}\right)\\ {}\begin{array}{c}+\frac{2{a}_5\left(1+{a}_5\right)\Gamma \left(2-{a}_1-{a}_3\right)}{\left(2-{a}_3\right){\left({a}_5{a}_4\right)}^{3-{a}_1-{a}_3}{\left(1+\frac{a_2-1}{a_4}\right)}^{3-{a}_1-{a}_3}}2{F}_1\left(1,3-{a}_1-{a}_3;3-{a}_3;\frac{1}{1+\frac{a_2-1}{a_4}}\right)\\ {}+\frac{{\left(2{a}_5\right)}^2\Gamma \left(4-{a}_1-{a}_3\right)}{\left(2-{a}_3\right){\left({a}_5{a}_4\right)}^{3-{a}_1-{a}_3}{\left(1+\frac{a_2}{a_4}\right)}^{4-{a}_1-{a}_3}}2{F}_1\left(1,4-{a}_1-{a}_3;3-{a}_3;\frac{1}{1+\frac{a_2}{a_4}}\right)\\ {}\begin{array}{c}+\frac{2{a}_5\Gamma \left(3-{a}_1-{a}_3\right)}{\left(2-{a}_3\right){\left({a}_5{a}_4\right)}^{3-{a}_1-{a}_3}{\left(1+\frac{a_2}{a_4}\right)}^{3-{a}_1-{a}_3}}2{F}_1\left(1,3-{a}_1-{a}_3;3-{a}_3;\frac{1}{1+\frac{a_2}{a_4}}\right)\\ {}-\frac{\left(1+{a}_5\right)\Gamma \left(2-{a}_1-{a}_3\right)}{\left(1-{a}_3\right){\left({a}_5{a}_4\right)}^{2-{a}_1-{a}_3}{\left(1+\frac{a_2-1}{a_4}\right)}^{2-{a}_1-{a}_3}}2{F}_1\left(1,2-{a}_1-{a}_3;2-{a}_3;\frac{1}{1+\frac{a_2-1}{a_4}}\right)\\ {}\begin{array}{c}-\frac{2{a}_5\Gamma \left(3-{a}_1-{a}_3\right)}{\left(1-{a}_3\right){\left({a}_5{a}_4\right)}^{3-{a}_1-{a}_3}{\left(1+\frac{a_2}{a_4}\right)}^{2-{a}_1-{a}_3}}2{F}_1\left(1,3-{a}_1-{a}_3;2-{a}_3;\frac{1}{1+\frac{a_2}{a_4}}\right)\\ {}+\frac{\Gamma \left(2-{a}_1-{a}_3\right)}{\left(1-{a}_3\right){\left({a}_5{a}_4\right)}^{2-{a}_1-{a}_3}{\left(1+\frac{a_2}{a_4}\right)}^{2-{a}_1-{a}_3}}2{F}_1\left(1,2-{a}_1-{a}_3;2-{a}_3;\frac{1}{1+\frac{a_2}{a_4}}\right),\end{array}\end{array}\end{array}\end{array}\end{array}} $$

where

$$ 2{F}_1\left({b}_1,{b}_2,{b}_3;y\right)=\sum_{k=0}^{\infty}\frac{{\left({b}_1\right)}_k{\left({b}_2\right)}_k{y}^k}{{\left({b}_3\right)}_kk!}. $$

(18)

Proof. We have

$$ {\displaystyle \begin{array}{c}{\Delta }_2\left({p}_1,{p}_2,{q}_1,{q}_2\right)={\int}_0^1{y}^{p_1}{\left(2-{y}^{\alpha}\right)}^{p_2}\left({\int}_0^y{z}^{q_1}\left(2-{z}^{\alpha}\right) dz\right) dy\\ {}=\frac{2^{\frac{q_1+1}{\alpha }+{q}_2}}{\alpha }{\int}_0^1{y}^{p_1}{\left(2-{y}^{\alpha}\right)}^{p_2}\left({\int}_0^{\frac{y}{2}}{t}^{\frac{q_1+1}{\alpha }-1}{\left(1-t\right)}^{q_2} dt\right) dy\\ {}=\frac{2^{\frac{q_1+1}{\alpha }+{q}_2}}{\alpha }{\int}_0^1{y}^{p_1}{\left(2-{y}^{\alpha}\right)}^{p_2} Bet\left(\frac{q_1+1}{\alpha },{q}_2+1,\frac{y}{2}\right) dy,\end{array}} $$

(19)

where z^α = 2t. Next, we use the following formula

$$ Bet\left(a,b,y\right)={\int}_0^y{t}^{a-1}{\left(1-t\right)}^{b-1} dt={a}^{-1}{y}^a2{F}_1\left(a,1-b,a+1:y\right). $$

Now by using Eq. (19), we obtain

$$ {\Delta }_2\left({p}_1,{p}_2,{q}_1,{q}_2\right)=\frac{2^{q_2}}{q_1+1}{\int}_0^1{y}^{p_1+\frac{q_1+1}{\alpha }}{\left(1-{y}^{\alpha}\right)}^{p_2}2{F}_1\left(\frac{q_1+1}{\alpha },-{q}_2;\frac{q_1+1}{\alpha }+1,\frac{y}{2}\right) dy. $$

In view of Eq. (18), ∆₂(p₁, p₂, q₁, q₂) can be rewritten as

$$ {\displaystyle \begin{array}{c}{\Delta }_2\left({p}_1,{p}_2,{q}_1,{q}_2\right)=\frac{2^{q_2}}{q_1+1}\sum_{i=0}^{\infty}\frac{{\left(\frac{q_1+1}{\alpha}\right)}_i{\left(-{q}_2\right)}_i}{{\left(\frac{q_1+1}{\alpha}\right)}_ii!}{2}^{-i}{\int}_0^1{y}^{p_1+\frac{q_1+1}{\alpha }+i}{\left(1-{y}^{\alpha}\right)}^{p_2} dy\\ {}=\sum_{i=0}^{\infty}\frac{{\left(\frac{q_1+1}{\alpha}\right)}_i{\left(-{q}_2\right)}_i}{{\left(\frac{q_1+1}{\alpha}\right)}_ii!}{2}^{-i}\frac{1}{\alpha }{2}^{p_2+{q}_2+\frac{\frac{q_1+1}{\alpha }+{p}_1+i+1}{\alpha }}\\ {}\begin{array}{c}\times {\int}_0^{\frac{1}{2}}{v}^{\frac{\frac{q_1+1}{\alpha }+{p}_1+i+1}{\alpha }-1}{\left(1-v\right)}^{p_2} dv\\ {}=\frac{2^{p_2+{q}_2}}{\left({q}_1+1\right)\left(\frac{q_1+1}{\alpha }+{p}_1+1\right)}\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{{\left(\frac{q_1+1}{\alpha }+{p}_1+1\right)}_{i+j}{\left(\frac{q_1+1}{\alpha}\right)}_i{\left(-{q}_2\right)}_i{\left(-{p}_2\right)}_j}{{\left(\frac{\frac{q_1+1}{\alpha }+{p}_1+1}{\alpha }+1\right)}_{i+j}{\left(\frac{q_1+1}{\alpha }+1\right)}_i}\\ {}\times \frac{2^{-i}{2}^{-j}}{\left(\frac{q_1+1}{\alpha }+{p}_1+i+1\right)}\frac{1}{i!j!},\end{array}\end{array}} $$

where y^α = 2v.