Estimation $\mathit{R}={\mathit{P}}_{\mathit{r}}(\mathit{Y}>\mathit{X})$ for a Family of Lifetime Distributions by Transformation Method

Surinder Kumar and Prem Lata Gautam^*

Department of Statistics, School for Physical and Decision Sciences, Babasaheb Bhimrao Ambedkar University, Lucknow, India

E-mail: surinderntls@gmail.com; premgautm61@gmail.com

${}^{*}$Corresponding Author

Received 15 March 2021; Accepted 15 July 2021; Publication 23 August 2021

Abstract

For a Family of lifetime distributions proposed by Chaturvedi and Singh (2008) [6]. The problem of estimating $R(t)=P(X>t)$, which is defined as the probability that a system survives until time t and $R=P(Y>X)$, which represents the stress-strength model are revisited. In order to obtain the maximum likelihood estimators (MLE’S), uniformly minimum variance unbiased estimators (UMVUS’S), interval estimators and the Bayes estimators for the considered model. The technique of transformation method is used.

Keywords: Family of lifetime distributions, uniformly minimum variance unbiased estimator, maximum likelihood estimator, confidence interval, bayes estimator.

1 Introduction

The reliability of an item or system can be defined as a function of time ‘t’ i.e, $R(t)=P(X>t)$, which defines the failure free operation of items/components until time ‘t’. One another important measure of reliability under the stress-strength model is $R=P_r(Y>X)$, which represents the reliability of an item or system for the random strength Y and random stress X.

A lot of work has been done in the literature on the point estiamtion of R. For a brief review literature one may refer to Pugh (1963) [12], Basu (1964) [3], Church and Harris (1970) [8], Enis and Geisser (1971) [10], Downton (1973) [9], Tong (1974) [19], Kelly et al. (1976) [11], Sinha and Kale (1980) [15], Sathe and Shah (1981) [14], Chao (1982) [4], Awad and Gharraf (1986) [2], Chaturvedi and Surinder (1999) [7], Rezaei et al. (2010) [13], Chaturvedi and Pathak (2012) [5], Surinder and Mayank(2014) [18], Surinder and Mukesh (2015) [16] and Surinder and Mukesh (2016) [17].

2 The Family of Lifetime Distributions

Chaturvedi and Singh (2008) [6] derived a family of lifetime distributions with the help of Weibull distribution. Let the random variable X follows a family of lifetime distributions, then the pdf is presented as

$f(x;a,\lambda ,\underset{¯}{\theta })={\displaystyle \frac{G^{\prime }(x;a,\underset{¯}{\theta })}{\lambda }}exp\left({\displaystyle \frac{-G(x;a,\underset{¯}{\theta })}{\lambda }}\right);x>a\ge 0,\lambda >0$

(1)

Here, $G(x;a,\underset{¯}{\theta })$ is a function of $x$ and may also depend on the parameters a and $\underset{¯}{\theta }$. $\underset{¯}{\theta }$ may be vector valued. $G^{\prime }(x;a,\underset{¯}{\theta })$ represents the derivative of $G(x;a,\underset{¯}{\theta })$ with respect to $x$.

The presented model (1) covers the following lifetime distributions as specific cases:

1. For $G(x;a,\underset{¯}{\theta })=x$ and a=0, we get the one-parameter exponential distribution.

2. For $G(x;a,\underset{¯}{\theta })=x^p,(p>0)$ and a=0, we get the Weibull distribution.

3. For $G(x;a,\underset{¯}{\theta })=x^2$ and a=0, we get the Rayleigh distribution.

4. For $G(x;a,\underset{¯}{\theta })=log(1+x^b),b>0$ and a=0, we get the Burr distribution.

5. For $G(x;a,\underset{¯}{\theta })=log(\frac{x}{a})$, we get the Pareto distribution.

6. For $G(x;a,\underset{¯}{\theta })=log(1+\frac{x}{\nu }),\nu >0$ and a=0, we get the Lomax distribution.

7. For $G(x;a,\underset{¯}{\theta })=log\left(1+\frac{x^b}{\nu }\right),b>0,\nu >0$ and a=0, we get the Burr distribution with scale parameter $\nu (>0)$.

8. For $G(x;a,\underset{¯}{\theta })=x^{\gamma }exp(\nu x),\gamma >0,\nu >0$ and a=0, we get the modified Weibull distribution.

9. For $G(x;a,\underset{¯}{\theta })=(x-a)+\frac{\nu }{\lambda }log\left(\frac{x+\nu }{a+\lambda }\right),\nu >0,\lambda >0$, we get the generalised Pareto distribution.

10. For $G(x;a,\underset{¯}{\theta })=bx+\frac{\theta }{2}{x}^2,\theta >0,b>0$ and a=0, we get the linear exponential distribution.

11. For $G(x;a,\underset{¯}{\theta })={(1+x^b)}^{\theta }-1,\theta >0,b>0$ and a=0, we get the generalised power Weibull distribution.

12. For $G(x;a,\underset{¯}{\theta })=\frac{\beta }{b}(e^{bx}-1),\beta >0,b>0$ and a=0, we get the Gompertz distribution.

13. For $G(x;a,\underset{¯}{\theta })=(e^{x^b}-1),b>0$ and a=0, we get the Chen distribution.

14. For $G(x;a,\underset{¯}{\theta })=(x-a)$, we get the two-parameter exponential distribution.

3 MLE of $R=P_r(Y>X)$

In the following theorem, MLE of R is derived through the transformation method

Theorem 1: The MLE of $R$ is

$$\ddot{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$$

(2)

where, $\overline{T}(y)={\displaystyle \frac{1}{n_2}}{\sum }_{j=1}^{n_2}H(y_j;a_2,{\theta }_2)$ and $\overline{T}(x)={\displaystyle \frac{1}{n_1}}{\sum }_{i=1}^{n_1}G(x_i;a_1,{\theta }_1)$

Proof: Let the random variable X follows a Family of lifetime distribution with pdf

$f(x;a_1,{\lambda }_1,{\theta }_1)={\displaystyle \frac{G^{\prime }(x;a_1,{\theta }_1)}{{\lambda }_1}}exp\left({\displaystyle \frac{-G(x;a_1,{\theta }_1)}{{\lambda }_1}}\right);$
$x>a_1\ge 0,{\lambda }_1>0$	(3)

For the given equation (3), let us consider the transformation $G(x;a_1,{\theta }_1)=t$. Then the distribution become

$$f(t;\alpha )=\frac{1}{\alpha }exp\left(\frac{-t}{\alpha }\right)$$

(4)

where, $\alpha ={\lambda }_1$.

Now, let us consider Y be a random variable with pdf

$f(y;a_2,{\lambda }_2,{\theta }_2)={\displaystyle \frac{H^{\prime }(y;a_2,{\theta }_2)}{{\lambda }_2}}exp\left({\displaystyle \frac{-H(y;a_2,{\theta }_2)}{{\lambda }_2}}\right);$
$y>a_2\ge 0,{\lambda }_2>0$	(5)

Similarly, let us take the transformation $z=H(y;a_2,{\theta }_2)$ and $\beta ={\lambda }_2$, we get

$$f(z;\beta )=\frac{1}{\beta }exp\left(-\frac{z}{\beta }\right)$$

(6)

Let t and z be two independent random variable which follows exponential distribution (4) and (6) with parameters $\alpha$ and $\beta$, respectively, where $t=G(x;a_1,{\theta }_1)$ and $z=H(y;a_2,{\theta }_2)$. The relaibility model is

$$\begin{array}{cc}\hfill R=P_r(z>t)& ={\int }_{z=0}^{\mathrm{\infty }}{\int }_{t=0}^{\mathrm{\infty }}f(t;\alpha )f(z;\beta )\mathit{d}t\mathit{d}z\hfill \\ & ={\int }_{z=0}^{\mathrm{\infty }}\left[1-exp\left(-\frac{z}{\alpha }\right)\right]\frac{1}{\beta }exp\left(-\frac{z}{\beta }\right)\mathit{d}z\hfill \end{array}$$

After solving, we get

$$R=\frac{\beta }{\beta +\alpha }$$

(7)

On replacing the $\alpha$ and $\beta$ by their MLE’S i.e, $\ddot{\alpha }=\overline{t}$ and $\ddot{\beta }=\overline{z}$. The MLE of $R=P_r(z>t)$ is

$$\frac{\overline{z}}{\overline{z}+\overline{t}}$$

where, $\overline{t}=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{t}_i$ and $\overline{z}=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{z}_j$. Finally, MLE of $R$ is

$$\ddot{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$$

where, $\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}H(y_j;a_2,{\theta }_2)$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}G(x_i;a_1,{\theta }_1)$.

Hence, the theorem follows.

1. Implication
Here, we consider the different cases for the distributions to obtain the MLE of $R=P_r(Y>X)$ given in (2)

Values of parameters for The MLE of $R\mathrm{=}{P}_r\mathrm{(}Y\mathrm{>}X\mathrm{)}$
Distributions	Values of Parameter
The one-parameter exponential distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i$
Weibull distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j^p$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i^p$ for $p>0$
Rayleigh distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j^2$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i^2$
Burr distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}log(1+y_j^b)$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}log(1+x_i^b)$
	for $b>0$
Pareto distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}log\left(\frac{y_j}{a_2}\right)$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}log\left(\frac{x_i}{a_1}\right)$
Lomax distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}log\left(1+\frac{y_j}{\nu }\right)$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}log\left(1+\frac{x_i}{\nu }\right)$
	for $\nu >0$
Burr distribution with scale parameter $\nu (>0)$	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}log\left(1+\frac{y_j^b}{\nu }\right)$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}log\left(1+\frac{x_i^b}{\nu }\right)$
	for $b>0,\nu >0$
The modified Weibull distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j^{\gamma }exp(\nu y_j)$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i^{\gamma }exp(\nu x_i)$
	for $\gamma >0,\nu >0$
The generalised	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}$ $\left[(y_j-a_2)+\frac{\nu }{{\lambda }_2}log\left(\frac{y_j+\nu }{a_2+{\lambda }_2}\right)\right]$
Pareto distribution	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}$ $\left[(x_i-a_1)+\frac{\nu }{{\lambda }_1}log\left(\frac{x_i+\nu }{a_1+{\lambda }_1}\right)\right]$
	for ${\lambda }_1,{\lambda }_2>0$,  $\nu >0$
The linear exponential distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}\left[b{y}_j+\frac{{\theta }_2}{2}{y}_j^2\right]$ $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}\left[b{x}_i+\frac{{\theta }_1}{2}{x}_i^2\right]$
	for ${\theta }_1,{\theta }_2>0$ and $b>0$
The generalised power Weibull distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}\left[{(1+y_j^b)}^{{\theta }_2}\right]-1$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}\left[{(1+x_i^b)}^{{\theta }_1}\right]-1$
	${\theta }_1,{\theta }_2>0$ and $b>0$
The Gompertz distribution	$\overline{T}(y)=\frac{1}{n_2}\frac{\beta }{b}(e^{b{\mathrm{\Pi }}_{j=1}^{n_2}{y}_j}-1)$ and $\overline{T}(y)=\frac{1}{n_1}\frac{\beta }{b}(e^{b{\mathrm{\Pi }}_{i=1}^{n_1}{x}_i}-1)$
	$\beta ,b>0$
Chen distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}(e^{y_j^b}-1)$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}(e^{x_i^b}-1)$
	$b>0$
The two-parameter exponential distribution	$\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}(y_j-a_2)$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}(x_i-a_1)$

4 UMVUE of $R=P_r(Y>X)$

In the following theorem, UMVUE of R is derived through the transformation method

Theorem 2: The UMVUE of $R$ is

$\acute{R}=\{\begin{array}{cc}\sum _{i=0}^{n_2-1}{(-1)}^i{\displaystyle \frac{\mathrm{\Gamma }(n_1)\mathrm{\Gamma }(n_2)}{\mathrm{\Gamma }(n_2-i)\mathrm{\Gamma }(n_1+i)}}{\left({\displaystyle \frac{T(x)}{T(y)}}\right)}^i;\hfill & T(x)<T(y)\hfill \\ \sum _{i=0}^{n_1-2}{(-1)}^i{\displaystyle \frac{\mathrm{\Gamma }(n_1)\mathrm{\Gamma }(n_2)}{\mathrm{\Gamma }(n_2+i+1)\mathrm{\Gamma }(n_1-i-1)}}{\left({\displaystyle \frac{T(y)}{T(x)}}\right)}^{i+1};\hfill & T(x)\ge T(y)\hfill \end{array}$ (8)

where, $T(y)={\sum }_{i=1}^{n_2}H(y_j;a_2,{\theta }_2)$ and $T(x)={\sum }_{i=1}^{n_1}G(x_i;a_1,{\theta }_1)$.

Proof: Considering the transfomation $G(x;a_1,{\theta }_1)=t$ and $z=H(y;a_2,{\theta }_2)$, we have the transform Equations (4) and (6). To obtain the measure of reliabilIty estimate $P_r(z>t)$, we required to obtain the UMVUE of $f(t;\alpha )$ and $f(z;\beta )$ i.e, $\acute{f}(t;\alpha )$ and $\acute{f}(z;\beta )$ respectively, which is given by

$\acute{f}(t;\alpha )={\displaystyle \frac{(n_1-1)G^{\prime }(t;a_1,{\theta }_1)}{n_1\overline{t}}}{\left[1-{\displaystyle \frac{G(t;a_1,{\theta }_1)}{n_1\overline{t}}}\right]}^{n_1-2};$
$G(t;a_1,{\theta }_1)<n_1\overline{t}$	(9)

and

$\acute{f}(z;\beta )={\displaystyle \frac{(n_2-1)H^{\prime }(z;a_2,{\theta }_2)}{n_2\overline{z}}}{\left[1-{\displaystyle \frac{H(z;a_2,{\theta }_1)}{n_2\overline{z}}}\right]}^{n_2-2};$
$H(z;a_2,{\theta }_1)<n_2\overline{z}$	(10)

Now to obtain UMVUE of R we have,

$$\begin{array}{cc}\hfill \acute{R}& =P_r(z>t)\hfill \\ & ={\int }_{t=0}^{\mathrm{\infty }}{\int }_{z=t}^{\mathrm{\infty }}\acute{f}(t;\alpha )\acute{f}(z;\beta )\mathit{d}z\mathit{d}t\hfill \end{array}$$

using (9) and (10)

$$\begin{array}{cc}\hfill \acute{R}=& {\int }_{t=0}^{n_1\overline{t}}{\int }_{z=t}^{n_2\overline{z}}\frac{(n_1-1)(n_2-1)H^{\prime }(z;a_2,{\theta }_2)G^{\prime }(t;a_1,{\theta }_1)}{n_1{n}_2\overline{t}\overline{z}}\hfill \\ & {\left[1-\frac{G(t;a_1,{\theta }_1)}{n_1\overline{t}}\right]}^{n_1-2}{\left[1-\frac{H(z;a_2,{\theta }_1)}{n_2\overline{z}}\right]}^{n_2-2}dzdt\hfill \end{array}$$

let $\left[1-\frac{H(z;a_2,{\theta }_1)}{n_2\overline{z}}\right]=w$

$={\displaystyle {\int }_{t=0}^{min(n_1\overline{t},n_2\overline{z})}}{\displaystyle \frac{(n_1-1)(n_2-1)G^{\prime }(t;a_1,{\theta }_1)}{n_1\overline{t}}}{\left[1-{\displaystyle \frac{G(t;a_1,{\theta }_1)}{n_1\overline{t}}}\right]}^{n_1-2}$

$\mathrm{\hspace{1em}}{\left[{\displaystyle \frac{w^{n_2-1}}{n_2-1}}\right]}_0^{1-\frac{H(t;a_2,{\theta }_1)}{n_2\overline{z}}}dt$

$={\displaystyle {\int }_{t=0}^{min(n_1\overline{t},n_2\overline{z})}}{\displaystyle \frac{(n_1-1)G^{\prime }(t;a_1,{\theta }_1)}{n_1\overline{t}}}{\left[1-{\displaystyle \frac{G(t;a_1,{\theta }_1)}{n_1\overline{t}}}\right]}^{n_1-2}$

$\mathrm{\hspace{1em}}{\left[1-{\displaystyle \frac{H(t;a_2,{\theta }_1)}{n_2\overline{z}}}\right]}^{n_2-1}dt$

$={\displaystyle {\int }_{t=0}^{min(n_1\overline{t},n_2\overline{z})}}{\displaystyle \frac{(n_1-1)G^{\prime }(t;a_1,{\theta }_1)}{n_1\overline{t}}}{\left[1-{\displaystyle \frac{G(t;a_1,{\theta }_1)}{n_1\overline{t}}}\right]}^{n_1-2}$

$\mathrm{\hspace{1em}}{\displaystyle \sum _{i=0}^{n_2-1}}{(-1)}^i\left({\displaystyle \genfrac{}{}{0pt}{}{n_2-1}{i}}\right){\left[{\displaystyle \frac{H(t;a_2,{\theta }_1)}{n_2\overline{z}}}\right]}^{i}dt$

Now consider the case $n_1\overline{t}<n_2\overline{z}$. Let $1-\frac{G(t;a_1,{\theta }_1)}{n_1\overline{t}}=u$, for solving the integral assuming $G(t;a_1,{\theta }_1)=H(t;a_2,{\theta }_2)$ i.e., $a_1=a_2$ and ${\theta }_1={\theta }_2$.

$$\begin{array}{cc}\hfill \acute{R}& ={\int }_0^1(n_1-1)\sum _{i=0}^{n_2-1}{(-1)}^i\left(\genfrac{}{}{0pt}{}{n_2-1}{i}\right){\left[\frac{n_1\overline{t}(1-u)}{n_2\overline{z}}\right]}^i{u}^{n_1-1}du\hfill \\ & =\sum _{i=0}^{n_2-1}{(-1)}^i\frac{\mathrm{\Gamma }(n_1)\mathrm{\Gamma }(n_2)}{\mathrm{\Gamma }(n_2-i)\mathrm{\Gamma }(n_1+i)}{\left(\frac{n_1\overline{t}}{n_2\overline{z}}\right)}^i\hfill \end{array}$$

In a same manner, we tackle the case when $n_1\overline{t}>n_2\overline{z}$:

$$\acute{R}=\sum _{i=0}^{n_1-2}{(-1)}^i\frac{\mathrm{\Gamma }(n_1)\mathrm{\Gamma }(n_2)}{\mathrm{\Gamma }(n_2+i+1)\mathrm{\Gamma }(n_1-i-1)}{\left(\frac{n_2\overline{z}}{n_1\overline{t}}\right)}^{i+1}$$

The UMVUE of $R=P_r(Y>X)$ is obtained by substituting $n_2\overline{z}=T(y)={\sum }_{j=1}^{n_2}H(y_j;a_2,{\theta }_2)$ and $n_1\overline{t}=T(x)={\sum }_{i=1}^{n_1}G(x_i;a_1,{\theta }_1)$.

Hence, the theorem follows.

2. Implication
Here, we consider the different cases for the distributions to obtain the UMVUE of $R=P_r(Y>X)$ given in (4)

Values of parameters for The UMVUE of $R\mathrm{=}{P}_r\mathrm{(}Y\mathrm{>}X\mathrm{)}$
Distributions	Values of Parameter
The one-parameter exponential distribution	$T(y)={\sum }_{j=1}^{n_2}{y}_j$ and $T(x)={\sum }_{i=1}^{n_1}{x}_i$
Weibull distribution	$T(y)={\sum }_{j=1}^{n_2}{y}_j^p$ and $T(x)={\sum }_{i=1}^{n_1}{x}_i^p$ for $p>0$
Rayleigh distribution	$T(y)={\sum }_{j=1}^{n_2}{y}_j^2$ and $T(x)={\sum }_{i=1}^{n_1}{x}_i^2$
Burr distribution	$T(y)={\sum }_{j=1}^{n_2}log(1+y_j^b)$ and $T(x)={\sum }_{i=1}^{n_1}log(1+x_i^b)$
	for $b>0$
Pareto distribution	$T(y)={\sum }_{j=1}^{n_2}log\left(\frac{y_j}{a_2}\right)$ and $T(x)={\sum }_{i=1}^{n_1}log\left(\frac{x_i}{a_1}\right)$
Lomax distribution	$T(y)={\sum }_{j=1}^{n_2}log\left(1+\frac{y_j}{\nu }\right)$ and $T(x)={\sum }_{i=1}^{n_1}log\left(1+\frac{x_i}{\nu }\right)$
	for $\nu >0$
Burr distribution with scale parameter $\nu (>0)$	$T(y)={\sum }_{j=1}^{n_2}log\left(1+\frac{y_j^b}{\nu }\right)$ and $T(x)={\sum }_{i=1}^{n_1}log\left(1+\frac{x_i^b}{\nu }\right)$
	for $b>0,\nu >0$
The modified Weibull distribution	$T(y)={\sum }_{j=1}^{n_2}{y}_j^{\gamma }exp(\nu y_j)$ and $T(x)={\sum }_{i=1}^{n_1}{x}_i^{\gamma }exp(\nu x_i)$
	for $\gamma >0,\nu >0$
The generalised Pareto distribution	$T(y)={\sum }_{j=1}^{n_2}\left[(y_j-a_2)+\frac{\nu }{{\lambda }_2}log\left(\frac{y_j+\nu }{a_2+{\lambda }_2}\right)\right]$
	$T(x)={\sum }_{i=1}^{n_1}\left[(x_i-a_1)+\frac{\nu }{{\lambda }_1}log\left(\frac{x_i+\nu }{a_1+{\lambda }_1}\right)\right]$
	for ${\lambda }_1,{\lambda }_2>0$,  $\nu >0$
The linear exponential distribution	$T(y)={\sum }_{j=1}^{n_2}\left[b{y}_j+\frac{{\theta }_2}{2}{y}_j^2\right]$ $T(x)={\sum }_{i=1}^{n_1}\left[b{x}_i+\frac{{\theta }_1}{2}{x}_i^2\right]$
	for ${\theta }_1,{\theta }_2>0$ and $b>0$
The generalised power	$T(y)={\sum }_{j=1}^{n_2}\left[{(1+y_j^b)}^{{\theta }_2}\right]-1$ and $T(x)={\sum }_{i=1}^{n_1}\left[{(1+x_i^b)}^{{\theta }_1}\right]-1$
Weibull distribution	${\theta }_1,{\theta }_2>0$ and $b>0$
The Gompertz distribution	$T(y)=\frac{\beta }{b}(e^{b{\mathrm{\Pi }}_{j=1}^{n_2}{y}_j}-1)$ and $T(x)=\frac{\beta }{b}(e^{b{\mathrm{\Pi }}_{i=1}^{n_1}{x}_i}-1)$
	$\beta ,b>0$
Chen distribution	$T(y)={\sum }_{j=1}^{n_2}(e^{y_j^b}-1)$ and $T(x)={\sum }_{i=1}^{n_1}(e^{x_i^b}-1)$
	$b>0$
The two-parameter exponential distribution	$T(y)={\sum }_{j=1}^{n_2}(y_j-a_2)$ and $T(x)={\sum }_{i=1}^{n_1}(x_i-a_1)$

5 Confidence Interval of $R=P_r(Y>X)$

In the following theorem, confidence interval of R is derived through the transformation method

Theorem 3: The confidence interval of $R=P_r(Y>X)$ is

$P({\displaystyle \frac{n_2\tilde{R}c}{n_1(1-\tilde{R})(1-c)+n_2\tilde{R}c}}<R<{\displaystyle \frac{n_2\tilde{R}d}{n_1(1-\tilde{R})(1-d)+n_2\tilde{R}d}})=1-\sigma$

(11)

where, $\ddot{R}={\displaystyle \frac{\overline{z}}{\overline{z}+\overline{t}}}$ and $0<c<d$.

Proof: From the Theorem 1, the MLE of R is $\frac{\beta }{\beta +\alpha }$ or $\frac{\overline{z}}{\overline{z}+\overline{t}}$. As we know $n_1\overline{t}$ and $n_2\overline{z}$ follows Gamma distribution with parameters $(\alpha ,n_1)$ and $(\beta ,n_2)$, respectively. For Confidence Interval of R, we must obtain the exact distribution of the variable

$$\delta =\frac{\alpha n_1\overline{t}}{\alpha n_1\overline{t}+\beta n_2\overline{z}}$$

(12)

Let $\rho =\alpha n_1\overline{t}$ and $\varrho =\beta n_2\overline{z}$ and observe that $\rho$ and $\varrho$ have gamma distribution with the parameters $(1,n_1)$ and $(1,n_2)$ respectively. New set of varible is $\delta =\frac{\rho }{\rho +\varrho }$.

On taking $\psi =\varrho$ and expressing the old variable in terms of new ones $\rho =\frac{\delta \psi }{(1-\delta )}$. The Jacobian of transformation is $J={(1-\delta )}^{-2}\psi$. The joint pdf of $\delta$ and $\psi$

$$P_r(\delta ,\psi )=\frac{e^{-\left(\frac{\psi }{1-\delta }\right)}{\psi }^{n_1+n_2-1}{\delta }^{n_1-1}}{\mathrm{\Gamma }(n_1)\mathrm{\Gamma }(n_2){(1-\delta )}^{n_1+1}}$$

(13)

Intergrating out $\psi$, we have the maginal distribution of $\delta$

$$P_r(\delta )={\left[B(n_1,n_2)\right]}^{-1}{\delta }^{n_1-1}{(1-\delta )}^{n_2-1};0<\delta <1$$

Here, $\delta$ has a beta distribution with the known parameters $n_1$ and $n_2$. So we have, for any $0<c<d$

$$P_r(c<\delta <d)=I_d(n_1,n_2)-I_c(n_1,n_2)$$

(14)

where, $I_x(n_1,n_2)={\left[B(n_1,n_2)\right]}^{-1}{\int }_0^x{z}^{n_1-1}{(1-z)}^{n_2-1}\mathit{d}z$ is the incomplete beta function. After calculation for the conection of $\delta$ and $\ddot{R}$, we have the pivotal quantity

$$\delta ={\left[1+\frac{n_2\ddot{R}(1-R)}{n_{2}R(1-\ddot{R})}\right]}^{-1}$$

where, $R=\frac{\beta }{\beta +\alpha }$ and $\ddot{R}=\frac{\overline{z}}{\overline{z}+\overline{t}}$.

If c and d in (14) are such that for a given $\sigma$

$$I_d(n_1,n_2)-I_c(n_1,n_2)=1-\sigma$$

then,

$$P(c<{[1+\frac{n_2\ddot{R}(1-R)}{n_{2}R(1-\ddot{R})}]}^{-1}<d)=1-\sigma$$

(15)

After solving the equation (15) for R.

$$P(\frac{n_2\tilde{R}c}{n_1(1-\tilde{R})(1-c)+n_2\tilde{R}c}<R<\frac{n_2\tilde{R}d}{n_1(1-\tilde{R})(1-d)+n_2\tilde{R}d})=1-\sigma$$

The above equation is valid for any values of $n_1$ and $n_2$, large or small.

Hence the theorem follows.

3. Implication
Here, we consider the different cases for the distributions to obtain the Confidence Interval of $R=P_r(Y>X)$ given in (11)

Values of parameters for The Confidence Integral of $R\mathrm{=}{P}_r\mathrm{(}Y\mathrm{>}X\mathrm{)}$
Distributions	Values of Parameter
The one-parameter exponential distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i$
Weibull distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j^p$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i^p,$
	$p>0$
Rayleigh distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j^2$ and $\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i^2$
Burr distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}log(1+y_j^b)$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}log(1+x_i^b),b>0$
Pareto distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}log\left(\frac{y_j}{a_2}\right)$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}log\left(\frac{x_i}{a_1}\right),b>0$
Lomax distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}log\left(1+\frac{y_j}{\nu }\right)$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}log\left(1+\frac{x_i}{\nu }\right)$, for $\nu >0$
Burr distribution with scale parameter $\nu (>0)$	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}log\left(1+\frac{y_j^b}{\nu }\right)$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}log\left(1+\frac{x_i^b}{\nu }\right),\nu >0$ and $b>0$
The modified Weibull distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j^{\gamma }exp(\nu y_j)$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i^{\gamma }exp(\nu x_i),$  $\nu >0$ and $\gamma >0$
The generalised Pareto distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(y)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}$ $\left[(y_j-a_2)+\frac{\gamma }{{\lambda }_2}log\left(\frac{y_j+\nu }{a_2+{\lambda }_2}\right)\right]$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}$ $\left[(x_i-a_1)+\frac{\gamma }{{\lambda }_1}log\left(\frac{x_i+\nu }{a_1+{\lambda }_1}\right)\right],\nu >0$ and $\gamma >0$
The linear exponential distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}\left[b{y}_j+\frac{{\theta }_2}{2}{y}_j^2\right]$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}\left[b{x}_i+\frac{{\theta }_1}{2}{x}_i^2\right],{\theta }_1,{\theta }_2>0$ and $b>0$
The generalised power	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}\left[{(1+y_j^b)}^{{\theta }_2}-1\right]$ and
Weibull distribution	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}\left[{(1+x_i^b)}^{{\theta }_1}-1\right],{\theta }_1,{\theta }_2>0$ and $b>0$
The Gompertz distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}\left[\frac{\beta }{b}(e^{b{y}_j}-1)\right]$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}\left[\frac{\beta }{b}(e^{b{x}_i}-1)\right],\beta >0$ and $b>0$
Chen distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}(e^{y_j^b}-1)$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}(e^{x_i^b}-1),b>0$
The two-parameter exponential distribution	$\tilde{R}=\frac{\overline{T}(y)}{\overline{T}(y)+\overline{T}(x)}$ $\forall \mathit{\hspace{1em}}\overline{T}(y)=\frac{1}{n_2}{\sum }_{j=1}^{n_2}(y_j-a_2)$ and
	$\overline{T}(x)=\frac{1}{n_1}{\sum }_{i=1}^{n_1}(x_i-a_1),a_1,a_2>0$

6 Bayes Estimator of $R=P_r(Y>X)$

In the following theorem, Bayes estimator of R is derived through the Transformation method

Theorem 4: The Bayes estimator of R is

$$\check{R}=\{\begin{array}{cc}\frac{{\mu }^{*}}{{\xi }^{*}+{\mu }^{*}}{\left(\frac{{\eta }^{*}}{{\omega }^{*}}\right)}^{-{\mu }^{*}}{}_{2}F_1({\mu }^{*}+{\xi }^{*},{\mu }^{*}+1,{\mu }^{*}+{\xi }^{*}+1;B),\text{for}\hspace{1em}B<1\hfill & \\ \hfill & \\ \frac{{\mu }^{*}}{{\xi }^{*}+{\mu }^{*}}{\left(\frac{{\omega }^{*}}{{\eta }^{*}}\right)}^{-{\xi }^{*}}{}_{2}F_1({\mu }^{*}+{\xi }^{*},{\xi }^{*},{\mu }^{*}+{\xi }^{*}+1;\frac{B}{1-B}),\text{for}\hspace{1em}B<-1\hfill & \end{array}$$

(16)

where ${}_{2}F_1(a,b,c;z)$ is the hypergeometric series and $B=\frac{{\omega }^{*}-{\eta }^{*}}{{\omega }^{*}}<1$.

Proof: Let us consider $\underset{¯}{t}$ and $\underset{¯}{z}$ be the independent samples from the pdfs (4) and (6). Here considering the conjugate prior, inverse gamma distributions for $\alpha$ and $\beta$ with the parameters $\mu$, $\eta$, and $\xi$, $\omega$, respectively. Prior is

$$\pi (\alpha ,\beta )\propto {\alpha }^{-\mu -1}{e}^{\left(-\frac{\eta }{\alpha }\right)}{\beta }^{-\xi -1}{e}^{\left(-\frac{\omega }{\beta }\right)};\mu ,\eta ,\xi ,\beta >0$$

(17)

The likelihood is

$$L(\alpha ,\beta |\underset{¯}{t},\underset{¯}{z})={\alpha }^{-n_1}{\beta }^{-n_2}\mathit{\hspace{1em}}exp\left[-\left(\frac{{\sum }_{i=1}^{n_1}{t}_i}{\alpha }+\frac{{\sum }_{j=1}^{n_2}{z}_j}{\beta }\right)\right]$$

(18)

Applying Bayes formula and using (17) and (18). The posterior density of $(\alpha ,\beta )$ is

$$\pi (\alpha ,\beta |\underset{¯}{t},\underset{¯}{z})\propto {\alpha }^{-\mu -n_1-1}{e}^{-\frac{(\eta +n_1\overline{t})}{\alpha }}{\beta }^{-\xi -n_2-1}{e}^{-\frac{(\omega +n_2\overline{z})}{\beta }}$$

(19)

Evidently the posterior risk is also the product of gamma pdfs with the updated parameters

$${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +n_1\overline{t},{\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +n_2\overline{z}$$

where, $\overline{t}$ and $\overline{z}$ are the sample means.

For posterior pdf of R, we consider a one-to-one transformation $F:R=\frac{\beta }{\beta +\alpha },{\vartheta }_R=\alpha +\beta$ with the inverse $Q:\alpha =R{\vartheta }_R,\beta =R(1-{\vartheta }_R)$. The Jacobian of transformation is ${\vartheta }_R$. The joint posterior density of R and ${\vartheta }_R$ becomes

${\pi }^{*}(R,{\vartheta }_R|\underset{¯}{t},\underset{¯}{z})\propto R^{{\mu }^{*}-1}{(1-R)}^{{\xi }^{*}-1}\mathit{\hspace{1em}}\vartheta _R{}^{{\mu }^{*}+{\xi }^{*}-1}{e}^{-{\vartheta }_R{\omega }^{*}(1-BR)};$
$0<R<1,{\vartheta }_R>0$	(20)

where $B=\frac{{\omega }^{*}-{\eta }^{*}}{{\omega }^{*}}<1$.

Intergrating the (20) for ${\vartheta }_R$

${\pi }_R(R|\underset{¯}{t},\underset{¯}{z})=C_R{R}^{{\mu }^{*}-1}{(1-R)}^{{\xi }^{*}-1}{(1-BR)}^{-({\mu }^{*}+{\xi }^{*})};0<R<1$ (21)

where, $C_R$ is the normalizing coefficient. For the Baye estimator we have

$$\check{R}=\int R{\pi }_R(R|\underset{¯}{t},\underset{¯}{z})\mathit{d}R$$

(22)

Using the (21) and solving (22), we obtain the bayes estimator of R

$$\check{R}=\{\begin{array}{cc}\frac{{\mu }^{*}}{{\xi }^{*}+{\mu }^{*}}{\left(\frac{{\eta }^{*}}{{\omega }^{*}}\right)}^{-{\mu }^{*}}{}_{2}F_1({\mu }^{*}+{\xi }^{*},{\mu }^{*}+1,{\mu }^{*}+{\xi }^{*}+1;B),\text{for}\hspace{1em}B<1\hfill & \\ \hfill & \\ \frac{{\mu }^{*}}{{\xi }^{*}+{\mu }^{*}}{\left(\frac{{\omega }^{*}}{{\eta }^{*}}\right)}^{-{\xi }^{*}}{}_{2}F_1({\mu }^{*}+{\xi }^{*},{\xi }^{*},{\mu }^{*}+{\xi }^{*}+1;\frac{B}{1-B}),\text{for}\hspace{1em}B<-1\hfill & \end{array}$$

where, ${}_{2}F_1(a,b,c;z)={\sum }_{j=1}^{\mathrm{\infty }}\frac{a(a+1)\mathrm{\dots }(a+j-1)b(b+1)\mathrm{\dots }(b+j-1)}{c(c+1)\mathrm{\dots }(c+j-1)}\frac{z^j}{j!}$ is the hypergeometric series.

For the Bayes estimator $\ddot{R}$, replacing the parameters as

$${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +n_1\overline{T}(x),{\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +n_2\overline{T}(y)$$

Hence, the theorem follows.

4. Implication
Here, we consider the different cases for the distributions to obtain the Bayes estimators of $R=P_r(Y>X)$ given in (16)

Values of parameters for The Bayes estimators of $R\mathrm{=}{P}_r\mathrm{(}Y\mathrm{>}X\mathrm{)}$
Distributions	Values of Parameter
The one-parameter exponential	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +n_1\overline{x},$
distribution	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +n_2\overline{y}$
Weibull distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}{x}_i^p,$
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}{y}_j^p,$  $p>0$
Rayleigh distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}{x}_i^2,$
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}{y}_j^2,p>0$
Burr distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}log(1+x_i^b),$
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}log(1+y_j^b),b>0$
Pareto distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}log\left(\frac{x_i}{a_1}\right),$
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}log\left(\frac{y_j}{a_2}\right),a_1,a_2>0$
Lomax distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}log\left(1+\frac{x_i}{\nu }\right)$ ,
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}log\left(1+\frac{y_j}{\nu }\right),\nu ,b>0$
Burr distribution with scale parameter $\nu (>0)$	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}log\left(1+\frac{x_i^b}{\nu }\right)$ ,
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}log\left(1+\frac{y_j^b}{\nu }\right),\nu ,b>0$
The modified Weibull distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}{x}_i^{\gamma }exp(\nu x_i)$ ,
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}{y}_j^{\gamma }exp(\nu y_j),\gamma ,\nu >0$
The generalised Pareto distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}\left[(x_i-a_1)+\frac{\nu }{{\lambda }_1}log\left(\frac{x_i+\nu }{a_1+{\lambda }_1}\right)\right],$
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}\left[(y_j-a_2)+\frac{\nu }{{\lambda }_2}log\left(\frac{y_j+\nu }{a_2+{\lambda }_2}\right)\right],$
	$\gamma ,\nu >0$
The linear exponential distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}\left[b{x}_i+\frac{{\theta }_1}{2}{x}_i^2\right],$
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}\left[b{y}_j+\frac{{\theta }_2}{2}{y}_j^2\right],b>0$
The generalised power	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}\left[{(1+x_i^b)}^{{\theta }_1}-1\right],$
Weibull distribution	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}\left[{(1+y_j^b)}^{{\theta }_2}-1\right],b>0$
The Gompertz distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}\left[\frac{\beta }{b}(e^{b{x}_i}-1)\right]$,
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}\left[\frac{\beta }{b}(e^{b{y}_j}-1)\right],b>0$
Chen distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}\left[e^{x_i^b}-1\right],$
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}\left[e^{y_j^b}-1\right],b>0$
The two-parameter exponential distribution	${\mu }^{*}=-(n_1+\mu ),{\eta }^{*}=\eta +{\sum }_{i=1}^{n_1}(x_i-a_1),$
	${\xi }^{*}=-(\xi +n_2),{\omega }^{*}=\omega +{\sum }_{i=1}^{n_2}(y_j-a_2),a_1,a_2>0$

7 Discussion

The Family of lifetime distribution is used in order to obtained the MLES, UMVUES, Confidence intervals and Bayes estimators of R for the various distributions. Initially, the generalized expressions for obtaining the MLES, UMVUES, Confidence intervals and Bayes estimators of R are obtained, then the estimator of the corresponding distributions are simply obtained by just replacing their respective parameters. For example, consider the following examples:-

Example 1 – Consider the Weibull distribution
Let $X_1,X_2,\mathrm{\dots }{X}_n$ be a random sample from WE($\alpha ,{\lambda }_1$) and $Y_1,Y_2,\mathrm{\dots }{Y}_m$ be a random sample from WE($\alpha ,{\lambda }_2$). Amiri et al. (2013) [1] obtained the MLE and UMVUE of R for Weibull distribution, which is given as

$$\ddot{R}=\frac{\frac{m}{{\sum }_{j=1}^{m}y_j{}^{\alpha }}}{\frac{n}{{\sum }_{i=1}^{n}x_i{}^{\alpha }}+\frac{m}{{\sum }_{j=1}^{m}y_j{}^{\alpha }}}$$

and the UMVUE of $R$ is

$$\acute{R}=\{\begin{array}{cc}1-\sum _{i=0}^{m-1}{(-1)}^i\frac{\mathrm{\Gamma }(n)\mathrm{\Gamma }(m)}{\mathrm{\Gamma }(n+i)\mathrm{\Gamma }(m-i)}{\left(\frac{t_1}{t_2}\right)}^i;t_1<t_2\hfill & \\ \sum _{j=0}^{n-1}{(-1)}^j\frac{\mathrm{\Gamma }(n)\mathrm{\Gamma }(m)}{\mathrm{\Gamma }(n-j)\mathrm{\Gamma }(m+j)}{\left(\frac{t_2}{t_1}\right)}^j;t_1\ge t_2\hfill & \end{array}$$

where, $t_1={\sum }_{i=1}^{n}x_i{}^{\alpha }$ and $t_2={\sum }_{j=1}^{m}y_j{}^{\alpha }$ are the sufficient statistics for the ${\lambda }_1$ and ${\lambda }_2$.

Example 2 – Consider the Burr distribution
Let $X$ be a Burr random variable with parameters (p, b) and $Y$ is another Burr random variable with parameters (a, b). Awad and Gharraf (1986) [2] obtained the MLE and UMVUE of R for Burr distribution, which is given as

$$\ddot{R}=\frac{1}{1+\frac{n}{m}\frac{{\sum }_{j=1}^{m}log\left(1+y_j{}^b\right)}{{\sum }_{j=1}^{n}log\left(1+x_j{}^b\right)}}$$

and the UMVUE of $R$ is

$$\acute{R}=\{\begin{array}{cc}\sum _{j=0}^{m-1}{(-1)}^j\frac{(m-1)!(n-1)!}{(m-1+j)!(n-1-j)!}\hfill & \sum _{i=1}^m{v}_i\le \sum _{i=1}^n{w}_i\hfill \\ {\left(\frac{{\sum }_{i=1}^m{v}_i}{{\sum }_{i=1}^n{w}_i}\right)}^j;\hfill & \hfill \\ 1-\sum _{j=0}^{m-1}{(-1)}^j\frac{(m-1)!(n-1)!}{(m-1-j)!(n-1+j)!}\hfill & \sum _{i=1}^m{v}_i>\sum _{i=1}^n{w}_i\hfill \\ {\left(\frac{{\sum }_{i=1}^n{w}_i}{{\sum }_{i=1}^m{v}_i}\right)}^j;\hfill & \hfill \end{array}$$

where, ${\sum }_{i=1}^n{w}_i={\sum }_{j=1}^{n}log\left(1+x_j{}^b\right)$ and ${\sum }_{i=1}^m{v}_i={\sum }_{j=1}^{m}log\left(1+y_j{}^b\right)$

Example 3 – Consider the generalized Pareto distribution
Suppose $X_1,X_2,\mathrm{\dots }{X}_n$ be a random sample from GP($\alpha ,\lambda$) and $Y_1,Y_2,\mathrm{\dots }{Y}_n$ be a random sample from GP($\beta ,\lambda$). Rezaei et al. (2010) [13] obtained the MLE and UMVUE of R for generalized Pareto distribution, which is given as

$$\ddot{R}=\frac{\frac{m}{{\sum }_{j=1}^{m}ln(1+\lambda y_j)}}{\frac{n}{{\sum }_{i=1}^{n}ln(1+\lambda x_i)}+\frac{m}{{\sum }_{j=1}^m(1+\lambda y_j)}}$$

and the UMVUE of $R$ is

$$\acute{R}=\{\begin{array}{cc}1-{\sum }_{i=0}^{m-1}{(-1)}^i\frac{(m-1)!(n-1)!}{(m-i-1)!(n+i-1)!}{\left(\frac{T_1}{T_2}\right)}^i;T_1\le T_2\hfill & \\ {\sum }_{i=0}^{n-1}{(-1)}^i\frac{(m-1)!(n-1)!}{(m+i-1)!(n-i-1)!}{\left(\frac{T_2}{T_1}\right)}^i;T_2\le T_1\hfill & \end{array}$$

where, $T_1={\sum }_{i=1}^{n}ln(1+X_i)$ and $T_2={\sum }_{i=1}^{m}ln(1+Y_i)$

Remarks: All the above Example 1–3 are the specific cases of our generalized expressions. Thus, in this study we have suggested a very simple and approved method i.e, transformation method for obtaining the MLES, UMVUES, Confidence intervals and Bayes estimators of R for the different distributions.

References

[1] Amiri, N., Azimi, R., Yaghmaei, F. and Babanezhad, M. 2013: Estimation of stress-strength parameter for two-parameter weibull distribution. Int. J. of Adanced Stat. and prob., 1(1):4–8.

[2] Awad, A. M. and Gharraf, M. K. 1986: Estimation of $P(Y<X)$ in the Burr case, A Comparative Study. Commun. Statist. – Simul., 15(2):389–403.

[3] Basu, D. 1964: Estimates of reliability for some distributions useful in life testing. Technometrics, 6:215–219.

[4] Chao, A. 1982: On comparing estimators of $P(X>Y)$ in the exponential case. IEEE transactions on reliability, 31:389–392.

[5] Chaturvedi, A. and Pathak, A. 2012: Estimation of the reliability functions for exponentiated Weibull distribution. J. Stat. Appl., 7:1–8.

[6] Chaturvedi, A. and Singh, K. G. 2008: A family of lifetime distributions and related estimation and testing procedures for the reliability function. J. Appl. Stat. Sci., 16(2):35–50.

[7] Chaturvedi, A. and Surinder, K. 1999: Further remarks on estimating the reliability function of exponential distribution under Type-I and Type-II censorings. Brazilian Journal of Probability and Statistics, 13:29–39.

[8] Church, J. D. and Harries, B. 1970: The estimation of reliability from stress-strength relationships. Technometrics, 12:49–54.

[9] Downton, F. 1973: The estimation of $Pr(Y<X)$ in the normal case. Technometrics, 15:551–558.

[10] Enis, P. and Geisser, S. 1971: Estimation of the probability that $(Y>X)$. J. Amer. Statist. Asso., 66:162–168.

[11] Kelly, G. D., Kelly., J. A. and Schucany, W. R. 1976: Efficient estimation of $P(Y<X)$ in the exponential case. Technometrics, 18:359–360.

[12] Pugh, E. L. 1963: The best estimate of reliability in the exponential case. Operations Research, 11:57–61.

[13] Rezaei, S., Tahmasbi, R. and Mahmoodi, M. 2010: Estimation of $P(Y<X)$ for generalized Pareto distribution. J. Stat. Plan Inference, 140:480–494.

[14] Sathe, Y. S. and Shah, S. P. 1981: On estimating $P(X<Y)$ for the exponential distribution. Commun. Statist. Theor. Meth., A10:39–47.

[15] Sinha , S. K. and Kale, B. K. 1980: Life testing and Reliability Estimation. Wiley Eastern Ltd., New Delhi.

[16] Surinder, K. and Kumar, M. 2015: Study of the Stress-Strength Reliability among the Parameters of Generalized Inverse Weibull Distribution. Intern. Journal of Science, Technology and Management, 4:751–757.

[17] Surinder, K. and Kumar, M. 2016: Point and Interval Estimation of $R=P(Y>X)$ for Generalized Inverse Weibull Distribution by Transformation Method. J. Stat. Appl. Pro. Lett., 3:1–6.

[18] Surinder, K. and Mayank, V. 2014: On the estimation of $R=P(Y>X)$ for a class of Lifetime Distributions by Transformation Method. J. Stat. Appl. Pro., 3(3):369–378.

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Biographies

Surinder Kumar, Head, Dept. of Statistics, BBAU (A central University), Lucknow – India. He is having 26 years research experience in various research fields of Statistics such as Sequential Analysis, Reliability Theory, Business Statistics and Bayesian Inference. Prof. Kumar has published more than 60 research publications in various journals of national and international repute.

Prem Lata Gautam, Dept. of Statistics, BBAU (A Central University) Lucknow, India. She has research experiences of 6 years and has also published 6 research articles in various reputed journals in the field of Sequential analysis, Bayesian estimation and Reliability theory and wholesome knowledge of many softwares and language like R Software, Mathematica and Fortron.

Journal of Reliability and Statistical Studies, Vol. 14, Issue 2 (2021), 393–414.
doi: 10.13052/jrss0974-8024.1422
© 2021 River Publishers