Applying the Laplace Transform Procedure, Testing Exponentiality against the NBRU_mgf Class

Abstract

This paper addresses a hypothesis testing problem for comparing exponentially distributed data against a new class termed “New Better than Renewal Used in Moment Generating Function” ($NBR{U}_{mgf}$). A measure of departure from exponentiality is constructed using the Laplace transform, followed by the development of a U-statistic-based test for the hypothesis. Additionally, a test based on the goodness of fit approach is examined as a special case. The asymptotic normality of the proposed statistic is introduced, and Pitman’s asymptotic efficiency of the two tests is computed and compared with other tests. Percentiles of the test statistics are computed for certain sample sizes in the case of complete data, and the powers of the tests are computed for popular reliability distributions. Finally, practical applications of the proposed tests are demonstrated in multiple cases.

1. Introduction

The investigation of several life distribution classes was introduced in reliability by Barlow and Proschan (1963) [1], Siddiqui and Bryson (1969) [2], Barlow and Proschan (1981) [3], Lee (1990) [4], and Serfling (1980) [5]. These kinds of life distributions are currently used in biometrics, engineering, maintenance, and the social and biological sciences. Consequently, there has been an increasing interest from statisticians and reliability analysts in modeling survival data with life distributions classified according to certain features of aging.

The NBRU_mgf class is particularly significant in this context because it generalizes and encompasses a variety of other well-known classes of life distributions, such as new better than used (NBU), New Better than Used in Expectation (NBUE), and Increasing Failure Rate (IFR). This makes the NBRU_mgf class a versatile and powerful tool for reliability modeling and life data analysis.

Everestus et al. (2022) [6] examined 91 distinct exponentiality tests. It was claimed that, while some tests were against specific kinds of alternative distributions, others were universally consistent. Their simulation study’s outcomes revealed that some tests that performed well under one set of shape alternatives would not do as well under another set of shape alternatives. Conversely, some tests that seemed less potent at small sample numbers also seemed more potent at large sample sizes. Ultimately, there was total concurrence between the outcomes of the simulated studies and the real-life implementations.

In more recent times, several classes of aging distributions of life have been further established within documented texts. For instance, using the goodness of fit and U-statistic approaches, respectively, Abu-Yousef et al. (2020, 2022) [7,8] introduced two novel test statistics for comparing the new is preferred over the used convex ordering function which generates moments (NBUC_mgf) life class distribution to the exponential distribution. In regard to a goodness of fit methodology, Abu Yousef and El-toony (2022) [9] suggested a new test statistic for comparing the better average increasing concave in Laplace transform order (UBAC (2)L) class of life distribution to exponentiality.

A new nonparametric class of life distribution known as new better than renewal used in moment-generating function (NBRU_mgf) was proposed by Hassan and Said (2021) [10,11]. The NBRU_mgf class extends the concepts of NBU and NBUE by incorporating moment-generating functions, thus providing a more comprehensive framework for analyzing life distributions with various aging properties. Subsequently, two new test statistics were developed based on moment inequalities and the U-statistic, respectively, for testing exponentiality versus (NBRU_mgf). El-Arishy et al. (2019) have identified several classes, among which are renewal new better than used in moment-generating function (RNBU_mgf) [12]; renewal new superior to that utilized in the Laplace transform order (RNBUL), as noted by Mahmoud et al. (2019) [13]; new superior to that utilized in the Laplace transform ($NBRUL$), as noted by Mahmoud et al. (2018) [14]; El-Arishy et al. (2017) discuss the application of “new better than renewal” ($NBR{U}_{rp}$) in the RP order [15]; Al-Zahrani and Stoyanov (2010) [16] discuss harmonic new better than renewal used in expectation ($HNBRUE$); Elbatal (2009) discusses renewal new better than renewal used ($RNBRU$) [17]; Mahmoud et al. (2008) [18] discuss the new better-than-average failure rate ($NBAFR$) and the new better-than-used failure rate ($NBUFR$); Abdel-Aziz (2007) [19] discusses renewal new better than renewal used in expectation ($RNBRUE$); Mahmoud et al. (2004) [20] discuss renewal new better than used; and Abouammoh et al. (1994) [21] discusses renewal new better than renewal used classes ($NBRU$).

The following is the paper’s composition: A novel category of life distributions known as new better than renewal is defined, and its relationships are used in the moment-generating function (NBRU_mgf) in Section 2. In Section 2, a test statistic computed using the Laplace transform technique is provided for evaluating exponentiality against NBRU_mgf. As a specific example, a test based on the goodness of fit approach is also suggested, and its properties are examined. Section 3 calculates PAE and PARE for a few widely used reliability distributions. In Section 4, we simulate the critical points of the Monte Carlo null distribution and tabulate the critical values of our two-test statistic.

For sample sizes $n=5,10,11,15\left(5\right) 40,43,45,50.$ The power of two tests is also computed for a few widely used reliability distributions in Section 5. In Section 6, applications for full data are covered. In Section 7, we finally provide a conclusion for our work.

2. Moment-Generating Function Class Uses New Better than Renewal

Consider a lifetime component $X$ with distribution function $F\left(x\right)$. A series of mutually identical components that are independent of the initial component will replace the component in the event of a failure. Over time, the remaining life distribution of the component operating at time $t$ is referred to as the stationary renewal distribution if the system is renewed indefinitely. This can be defined as follows:

$$W_F\left(t\right)={\mu }^{-1}{\int }_0^t\overline{F}\left(u\right)du,\hspace{1em}t\ge 0$$

(1)

The corresponding density function of the stationary renewal distribution $w_F\left(t\right)$ and the renewal survival function is given by

$${\overline{W}}_F\left(t\right)={\mu }^{-1}{\int }_t^{\infty }\overline{F}\left(u\right)du$$

(2)

The failure rate of the renewal distribution $W_F\left(t\right)$ is given by

$${\lambda }_w\left(t\right)=\frac{w_F\left(t\right)}{{\overline{W}}_F\left(t\right)}=\frac{\overline{F}\left(t\right)}{{\int }_t^{\infty }\overline{F}\left(u\right)du}={( \mu \left(t\right) )}^{-1},\hspace{1em}0\le t<\infty ,$$

(3)

where the mean time to failure $\left(\mu \right)$ is defined by

$$\mu ={\int }_0^{\infty }\overline{F}\left(t\right)dt$$

(4)

The random residual at age t life is defined by $X_t=[T-t|T>t]$ with survival function ${\overline{F}}_{t }\left(x\right)=\frac{\overline{F}(x+t)}{\overline{F}\left(t\right)} , x, t\ge 0$. The average residual life of $X_t$ is given by

$$\mu \left(t\right)=\frac{{\int }_t^{\infty }\overline{F}\left(x\right)dx}{\overline{F}\left(t\right)}, t, x\ge 0 ; \overline{F}(t)>0$$

(5)

For more details, see Barlow and Proschan [3].

Definition 1.

If $X$ is a random variable with a non-negative survival function $\overline{F}\left(x\right),$ then X is said to have new better (worse) than renewal used distribution, denoted by $F\in NBRU$ ($NWRU$), if

$${\overline{W}}_F\left(x+t\right)\le \left(\ge \right){\overline{W}}_F\left(t\right)\overline{F}\left(x\right); x\ge 0,t\ge 0$$

(6)

Definition 2.

A non-negative random variable X is said to have new better than renewal used in moment-generating function (denoted by $F\in {NBRU}_{mgf}$) if, and only if,

$${\int }_0^{\infty }{e}^{sx}{\overline{W}}_F\left(x+t\right)dx \le {\overline{W}}_F\left(t\right){\int }_0^{\infty }{e}^{sx}\overline{F}\left(x\right)dx ; x, t , s\ge 0$$

(7)

Definition 3.

Put $s=0$ in Equation (7); a non-negative random variable X is said to have new better than renewal used in the convex ordering (denoted by $F\in NBRUC$) if, and only if,

$${\int }_x^{\infty }{\overline{W}}_F\left(x+t\right)dx \le {\overline{W}}_F\left(t\right){\int }_x^{\infty }\overline{F}\left(x\right)dx , x, t\ge 0$$

(8)

Then, we have

$$NBRU\subseteq NBRUC\subseteq {NBRU}_{mgf}$$

3. Testing Exponentiality Versus NBRU_mgf Class for Complete Data

This section tests the null hypothesis. Hypothesis $H_{0 }:F$ is exponential with mean ${\mu }_F$ against an alternative hypothesis that $H_1:F$ belongs to $NBR{U}_{mgf}$ and is not exponential. We draw a sample $X_1,X_2,X_3,\dots .X_{n }$ from a population with distribution F and use the symbol $\delta (s,\beta )$ as a measure of departure from exponentiality. According to Equation (7), the measure of departure from exponentiality based on the Laplace transform technique is defined by the following equation:

$$\delta \left(s,\beta \right)={\int }_0^{\infty }{e}^{-\beta t} [ {\overline{W}}_F\left(t\right){\int }_0^{\infty }{e}^{sx}\overline{F}\left(x\right)dx-{\int }_0^{\infty }{e}^{sx}{\overline{W}}_F\left(x+t\right)dx ] dt; s, \beta \ge 0$$

(9)

In the theorem below, the measure of departure from exponentiality for the NBRU_mgf class is derived and assumed that every moment is temporary and exists.

Theorem 1.

If X is a random variable with distribution function $F\left(x\right)$ and $F\left(x\right)$ relates to the NBRU_mgf class, then

$$\delta \left(s,\beta \right)=\frac{\phi \left(s\right)\varnothing \left(\beta \right)}{s{\beta }^2}+\left[\frac{\mu }{s\beta }-\frac{1}{s^2\left(\beta +s\right)}-\frac{1}{s{\beta }^2}\right]\phi \left(s\right)-\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2\left(\beta +s\right)}\right]\varnothing \left(\beta \right)+\frac{1}{s^2\beta }$$

(10)

where

$$\phi \left(s\right)={\int }_0^{\infty }{e}^{sx}\hspace{0.33em}dF\left(x\right) ; \varnothing \left(\beta \right)={\int }_0^{\infty }{e}^{-\beta x}\hspace{0.33em}dF\left(x\right)$$

Proof. 

Since $F$ is $NBR{U}_{mgf}$, then by multiplying Equation (7) by $e^{-\beta t}$and integrating both sides with respect to $t$ over [0, ∞), we find

$${\int }_0^{\infty }{e}^{-\beta t}{\int }_0^{\infty }{e}^{sx}{\overline{W}}_F\left(x+t\right)dxdt \le {\int }_0^{\infty }{e}^{-\beta t}{ \overline{W}}_F\left(t\right)dt{\int }_0^{\infty }{e}^{sx}\overline{F}\left(x\right)dx$$

(11)

Setting

$$\begin{array}{ll}{I}_1& ={\int }_0^{\infty }{e}^{sx}\overline{F}\left(x\right)dx=E\left[{\int }_0^{\infty }I\left(X>x\right)e^{sx}\hspace{0.33em}dx\right]\\ & =E[{\int }_0^x{e}^{sx}\hspace{0.33em}dx] =\frac{1}{s}\left(\phi \left(s\right)-1\right)\end{array}$$

(12)

$$\begin{array}{ll}{I}_2& ={\int }_0^{\infty }{e}^{-\beta t}{\overline{W}}_F\left(t\right)dt=E\hspace{0.33em}\left[{\int }_0^{\infty }{e}^{-\beta t}\left(X-t\right)I\left(X>t\right)dt\right]\\ & =E\left[{\int }_0^x{e}^{-\beta t}\left(X-t\right)dt \right]\\ & =\frac{1}{\mu }(\frac{\mu }{ \beta }+\frac{\varnothing \left(\beta \right)}{{\beta }^2}-\frac{1}{{\beta }^2} )\end{array}$$

(13)

$$\begin{array}{ll}{I}_3& ={\int }_0^{\infty }\hspace{0.33em}{e}^{sv}\hspace{0.33em}{\overline{W}}_F\left(v\right)dv=E\hspace{0.33em}\left[{\int }_0^x{e}^{sv}\left(X-v\right)I\left(X>v\right)dv\right]\\ & =\frac{1}{\mu } \left(-\frac{\mu }{s}+\frac{\phi \left(s\right)}{s^2}-\frac{1}{s^2}\right)\end{array}$$

(14)

$$R.H.S. of \mathit{Equation}\left(11\right)=I_1\ast I_2=\frac{1}{\mu }[ \frac{\phi \left(s\right)\mu }{s\beta }+\frac{\phi \left(s\right)\varnothing \left(\beta \right)}{s{\beta }^2}-\frac{\phi \left(s\right)}{s{\beta }^2}-\frac{\mu }{s\beta }-\frac{\varnothing \left(\beta \right)}{s{\beta }^2}+\frac{1}{s{\beta }^2} ],$$

(15)

where

$$I\left(X>x\right)=\{\begin{array}{cc}0& if x\ge X\\ 1& if x<X\end{array}$$

Let $u=x+t$, then $du=dx$, and $v=t$, and then $dv=dt;\left|J\right|=1$;

$$\begin{array}{ll}L.H.S of\mathit{Equation}\left(11\right)& ={\int }_0^{\mathrm{\infty }} e^{-\beta v}{\int }_v^{\mathrm{\infty }} e^{s(u-v)} {\overline{W}}_F\left(u\right)dudv\\ & ={\int }_0^{\mathrm{\infty }} e^{-\beta u}{\int }_0^v e^{s(v-u)} {\overline{W}}_F\left(v\right) dudv\\ & ={\int }_0^{\mathrm{\infty }} e^{sv} {\overline{W}}_F\left(v\right){\int }_0^v e^{-u(\beta +s)} dudv\\ & =\frac{1}{(\beta +s)}\left[{\int }_0^{\mathrm{\infty }} e^{sv} {\overline{W}}_F\left(v\right)dv-{\int }_0^{\mathrm{\infty }} e^{-\beta v} {\overline{W}}_F\left(v\right)dv\right]\\ & =\frac{1}{(\beta +s)}\left[I_3-I_2\right]\\ & =\frac{1}{\mu }\left[-\frac{\mu }{s\beta }+\frac{\phi \left(s\right)}{s^2(\beta +s)}-\frac{\mathrm{\varnothing }\left(\beta \right)}{{\beta }^2(\beta +s)}+\frac{(s-\beta )}{{\beta }^2{s}^2}\right]\end{array}$$

(16)

Substituting from Equations (15) and (16) into Equation (9), we obtain Equation (10).

Note that under $H_0:\delta \left(s,\beta \right)=0$, while under $H_1:\delta \left(s,\beta \right)>0$. □

Corollary 1.

If we set $\beta =1$ in Equation (10), then the following equation becomes the measure of departure from exponentiality based on the goodness of fit approach:

$$\delta \left(s,1\right)=\frac{\phi \left(s\right)\varnothing \left(1\right))}{s}+\left[\frac{\mu }{s}-\frac{1}{s^2\left(1+s\right)}-\frac{1}{s}\right]\phi \left(s\right)-\left[\frac{1}{s}-\frac{1}{\left(1+s\right)}\right]\varnothing \left(1\right)+\frac{1}{s^2},$$

(17)

and note that under $H_0:\delta \left(s,1\right)=0$, while under $H_1:\delta \left(s,1\right)>0$.

3.1. Test Statistic Empirical for NBRU_mgf

To estimate $\delta \left(s,\beta \right)$, let $X_1,X_2,X_3,\dots .X_{n }$ be a sample selected at random from a distributed population F. To make the test statistic $\widehat{\delta }\left(s,\beta \right)$ scale invariant, we set

$$∆\left(s,\beta \right)=\frac{\delta \left(s,\beta \right)}{\overline{\mu }}$$

Suppose that ${\overline{F}}_n\left(x\right),\mathrm{a}\mathrm{n}\mathrm{d} {\widehat{∆}}_n\left(s,\beta \right)$ denote the empirical distribution of the survival function $\overline{F}\left(x\right)$ and the empirical estimate of $∆\left(s,\beta \right)$, respectively. We find

$${\widehat{∆}}_n\left(s,\beta \right)=\frac{1}{n^2\overline{X}} {\sum }_{i=1}^n{\sum }_{j=1}^n\varnothing (X_i,X_j);$$

(18)

$$\varnothing \left(X_i,X_j\right)=\frac{e^{s{X}_{i \times }}{e}^{-\beta X_j}}{s{\beta }^2}+\left[\frac{X_i}{s\beta }-\frac{1}{s^2\left(\beta +s\right)}-\frac{1}{s{\beta }^2}\right]e^{s{X}_j}-\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2\left(\beta +s\right)}\right]e^{-\beta X_i}+\frac{1}{s^2\beta }$$

(19)

$${\overline{F}}_n\left(x\right)=\frac{1}{n}\sum _{i=1}^{n}I(X_i>x), dF\left(x\right)=\frac{1}{n}$$

Note that $\mathrm{\varnothing }\left(X_i,X_j\right)$ is not symmetric, and we define the symmetric kernel as

$$\mathsf{\Psi }\left(X_1,X_2\right)=\frac{1}{2!}{\sum }_R\mathrm{\varnothing }\left(X_i,X_j\right)$$

(20)

where the summation is over all arrangements of $X_i,X_j$; this leads to $\widehat{∆}\left(s,\beta \right)$ corresponding to the U-statistic provided by the next equation (see Lee [4]).

$${\mathrm{U}}_n=\frac{1}{\left(\begin{array}{c}n\\ 2\end{array}\right)}{\sum }_{i<j}\mathsf{\Psi }\left(X_i,X_j\right)$$

(21)

The asymptotic normality of ${\widehat{∆}}_n\left(s,\beta \right)$ may be summed up by the upcoming theorem.

Theorem 2.

(1) 

$n\to \infty , \surd n({\widehat{∆}}_n\left(s,\beta \right)-∆\left(s,\beta \right))$ is asymptotically normal with zero mean and variance ${\sigma }^2$ where ${\sigma }^2$ is given by

$$\begin{array}{ll}{\sigma }^2& =Var [ \frac{e^{sX}\varnothing \left(\beta \right)+e^{-\beta X} \phi \left(s\right)}{s{\beta }^2}-\left[\frac{1}{s^2\left(\beta +s\right)}+\frac{1}{s{\beta }^2}\right]\left(\phi \left(s\right)+e^{sX}\right)+\frac{2}{s^2\beta }\\ & -\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2\left(\beta +s\right)}\right]\left(e^{-\beta X}+\varnothing \left(\beta \right) \right)+\frac{X\phi \left(s\right)+\mu e^{sX}}{s\beta }]\end{array}$$

(22)

(2) 

Under $H_0$, the variance ${\sigma }_0^2$ is

$${\sigma }_0^2=\frac{a_8}{\left(1+\beta \right)}-\frac{a_9}{\left(1-s\right)}+\frac{a_{10}}{{\left(1-s\right)}^2}+\frac{2a_{5 }}{s\beta {\left(1-s\right)}^3}+\frac{a_5^2}{\left(1-2s\right)}+a_{11}+a_{12}$$

(23)

where

$$\begin{array}{l}{a}_1=\left[\frac{1}{s^2(\beta +s)}+\frac{1}{s{\beta }^2}\right]; a_2=\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2(\beta +s)}\right]; a_3=\frac{1}{(1-s)(1+\beta )};\\ a_4=\frac{1}{(1+\beta )(1-2s)}; a_5=\left[\frac{1}{s\beta }-a_1\right]; a_6=\frac{1}{(1+\beta -s)}; a_7=\frac{1}{(1+2\beta )};\\ a_8=\left[\frac{a_4}{s^2{\beta }^4}-\frac{2a_2\left(a_6+a_3\right)}{s{\beta }^2}-\frac{4a_2{a}_3}{s\beta }-\frac{8a_2}{s^2\beta }\right]; a_9=\left[\frac{2\left(2a_1{a}_3+a_2{a}_7-a_5{a}_6\right)}{s{\beta }^2}-\frac{8a_5}{s^2\beta }\right]\\ a_{10}=\left[\frac{2a_3}{s^2{\beta }^3}+\frac{2}{s^2{\beta }^2}-\frac{2a_1}{s\beta }++\frac{a_7}{s^2{\beta }^4}+3a_2^2+a_1^2-2a_1{a}_5\right];\\ a_{11}=\left(\frac{2a_3{a}_6}{s^2{\beta }^4}+\frac{2a_4{a}_5}{s{\beta }^2}+\frac{6a_3}{s^3{\beta }^3}+\frac{2a_3^2}{s^2{\beta }^3}+\frac{4}{s^4{\beta }^2}\right);\\ a_{12}=2a_1{a}_2{a}_3-3a_2{a}_3{a}_5+a_7{a}_2^2-2a_2{a}_5{a}_6\end{array}$$

Proof. 

First, as $n⟶\infty$, we have $\overline{X}\stackrel{P}{\to }\mu$ and ${\widehat{∆}}_n\left(s,\beta \right)\stackrel{P}{\to }∆\left(s,\beta \right).$ Using the results of Serfling [5], we note that as $n⟶\infty ; {\widehat{∆}}_n\left(s,\beta \right)$ has an asymptotically normal with a mean of zero and a variance of ${\sigma }^2$ which is given by

$${\sigma }^2=Var({\varphi }_1\left(X\right)+{\varphi }_2(X\left)\right)$$

(24)

$${\varphi }_1\left(X\right)=E\left(\mathrm{\varnothing }\left(X_1,X_2\right)|X_1\right); {\varphi }_2\left(X\right)=E\left(\mathrm{\varnothing }\left(X_1,X_2\right)|X_2\right)$$

(25)

Recall the definition of $\mathrm{\varnothing }\left(X_1,X_2\right)$ in Equation (19); then, it is easy shown that

$${\varphi }_1\left(X\right)=\frac{e^{sX }E⌈{ e}^{-\beta X}⌉}{s{\beta }^2}+\left[\frac{X}{s\beta }-\frac{1}{s^2\left(\beta +s\right)}-\frac{1}{s{\beta }^2}\right]E\left[e^{sX}\right]-\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2\left(\beta +s\right)}\right]e^{-\beta X}+\frac{1}{s^2\beta }$$

(26)

$${\varphi }_2\left(X\right)=\frac{E\left[e^{sX}\right] e^{-\beta X}}{s{\beta }^2}+\left[\frac{\mu }{s\beta }-\frac{1}{s^2\left(\beta +s\right)}-\frac{1}{s{\beta }^2}\right]e^{sX}-\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2\left(\beta +s\right)}\right]E\left[e^{-\beta X}\right]+\frac{1}{s^2\beta }$$

(27)

From Equations (26) and (27), Equation (22) can be proved.

Under $H_0$, where $F_0\left(x\right)=1-e^{-x},x\ge 0,$ it is easy to prove that

$$\begin{array}{ll}{\mu }_0& =E\left({\varphi }_1\left(X\right)+{\varphi }_2\left(X\right)\right)\\ & =\frac{2\beta +2s-2\beta +2s}{s{\beta }^2(1+\beta )(1-s)(\beta +s)}+\frac{2(s{\beta }^2+s^2\beta -{\beta }^2-s\beta -s^2+{\beta }^2-{\beta }^{2}s+\beta s-\beta s^2)}{s^2{\beta }^2\left(\beta +s\right)\left(1-s\right)}=0\end{array}$$

and the variance ${\sigma }_0^2=E\left[{\left({\varphi }_1\left(X\right)+{\varphi }_2\left(X\right)\right)}^2\right]$ reduces to Equation (23). □

Corollary 2.

If we option to $\beta =1$ in Equation (23), we obtain the variance in the case of goodness of fit approach as

$${\sigma }_0^2(s,1)=\frac{a_8}{2}-\frac{a_9}{\left(1-s\right)}+\frac{a_{10}}{{\left(1-s\right)}^2}+\frac{2a_{5 }}{s{\left(1-s\right)}^3}+\frac{a_5^2}{\left(1-2s\right)}+a_{11}+a_{12}$$

3.2. Pitman’s Asymptotic Efficiency

In order to assess this procedure’s quality, Pitman asymptotic efficiencies (PAEs) of our test are computed for the alternative distributions below in the class (since they are in the ${NBRU}_{mgf}$). Then, we perform comparison testing of various famous classes of life distribution with our suggested test based on PAEs.

(i)

Linear failure rate distribution (LFR):

$${\overline{F}}_1 \left(x\right)=e^{-X-\frac{\theta }{2}{X}^2}, x\ge 0, \theta \ge 0$$

(ii)

Makeham distribution:

$${\overline{F}}_2 \left(x\right)=e^{-x-\theta (X+e^{-X}-1)}, x\ge 0, \theta \ge 0$$

(iii)

Weibull distribution:

$${\overline{F}}_3 \left(x\right)=e^{-X^{\theta }} , x\ge 0, \theta \ge 0$$

Note that, for $\theta =$ 1, ${\overline{F}}_3 \left(x\right)$decreases the exponential distribution, but in the case of $\theta =0, {\overline{F}}_1 \left(x\right)$ and ${\overline{F}}_2 \left(x\right)$ reduce to exponential distribution. The $PAE\left({\delta }_{\theta }\left(s,\beta \right)\right)$ is defined by

$$\mathrm{PAE}\left({\delta }_{\theta }\left(s,\beta \right)\right)=\frac{1}{{\mathsf{\sigma }}_0}{\left|\frac{\mathrm{d}{\delta }_{\theta }\left(s,\beta \right)}{\mathrm{d}\mathsf{\theta }}\right|}_{\mathsf{\theta }\to {\mathsf{\theta }}_0}$$

(28)

where

$${\delta }_{\theta }\left(s,\beta \right)=\frac{{\phi }_{\theta }\left(s\right){\varnothing }_{\theta }\left(\beta \right)}{s{\beta }^2}+\left[\frac{{\mu }_{\theta }}{s\beta }-\frac{1}{s^2\left(\beta +s\right)}-\frac{1}{s{\beta }^2}\right]{\phi }_{\theta }\left(s\right)-\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2\left(\beta +s\right)}\right]{\varnothing }_{\theta }\left(\beta \right)+\frac{1}{s^2\beta }$$

The $PAE\left({\delta }_{\theta }\left(s,\beta \right)\right)$ can be written as

$$\begin{gathered} \mathrm{PAE}\left({\delta }_{\theta }\left(s,\beta \right)\right)= \\ \frac{1}{{\mathsf{\sigma }}_0}{\left|\begin{array}{c}\frac{1}{s{\beta }^2}\left[{\acute{\phi }}_{\theta }\left(s\right){\int }_0^{\infty }{e}^{-\beta x}\hspace{0.33em}d{F}_{\theta }\left(x\right)+{\int }_0^{\infty }{e}^{sx}\hspace{0.33em}d{F}_{\theta }\left(x\right) {\acute{\varnothing }}_{\theta }\left(\beta \right)\right]+\\ \left[\frac{{\mu }_{\theta }}{s\beta }-\frac{1}{s^2\left(\beta +s\right)}-\frac{1}{s{\beta }^2}\right]\acute{{\phi }_{\theta }}\left(s\right)+\left[\frac{{\acute{\mu }}_{\theta }}{s\beta }\right]{\phi }_{\theta }\left(s\right)-\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2\left(\beta +s\right)}\right]{\acute{\varnothing }}_{\theta }\left(\beta \right)\end{array}\right|}_{\theta \to {\theta }_0} \end{gathered}$$

(29)

Upon using Equations (28) and (29), we obtain the $PAE\left({\delta }_{\theta }\left(s,\beta \right)\right)$ of these above three alternative distributions of our procedure, which are, respectively,

(i)

Linear failure rate distribution (${\theta }_0=0)$:

$$PAE\left({\delta }_{\theta }\left(s,\beta \right),{\overline{F}}_1 \left(x\right)\right)=\frac{1}{{\sigma }_0}\left|\frac{a_3(1-s)+s{\beta }^2{a}_5}{{{\beta }^2\left(1-s\right)}^3}-\frac{1}{s\beta \left(1-s\right)}+\frac{a_3(1+\beta )-s{\beta }^2{a}_2}{{sp\left(1+\beta \right)}^3}\right|$$

(30)

(ii)

Makeham distribution (${\theta }_0=0)$ :

$$PAE\left({\delta }_{\theta }\left(s,\beta \right),{\overline{F}}_2 \left(x\right) \right)=\frac{1}{{\sigma }_0}\left|\frac{a_3{(1+\beta )}^2-s{\beta }^2{a}_2}{{s\beta (2+\beta )\left(1+\beta \right)}^3}-\frac{(1-s)a_3+2{\beta }^2{a}_5}{{\beta }^2\left(2-s\right){\left(1-s\right)}^2}-\frac{1}{2s\beta \left(1-s\right)}\right|$$

(31)

(iii)

Weibull distribution (${\theta }_0=1)$:

$$PAE\left({\delta }_{\theta }\left(s,\beta \right),{\overline{F}}_3 \left(x\right)\right)=\frac{1}{{\sigma }_0} \left|\begin{array}{c} \frac{\left[\left(1-s\right)+slog\left(1-s\right)\right]\left[\left(1-s\right)a_3+s{\beta }^2{a}_5\right]}{s{\beta }^2{\left(1-s\right)}^2 }-\frac{0.422784}{s\beta \left(1-s\right)} \\ +\frac{\left[\left(1+\beta \right)+\beta log\left(1+\beta \right)\right]\left[\left(1+\beta \right)a_3-s{\beta }^2{a}_2\right]}{s{\beta }^2{\left(1+\beta \right)}^2 } \end{array}\right|$$

(32)

By using Equations (30)–(32) and the Mathematica program 12.3, we obtain

Table 1. The Pitman asymptotic efficiencies of ${\delta }_{\theta }\left(s,\beta \right)$ at different values of s, β.

Distribution	β	s = 0.58	s = 0.59	s = 0.6	s = 0.7	s = 0.8	s = 0.9
LFR	1	1.7817	1.4408	1.2897	1.0789	1.2268	1.6276
	2	1.1115	1.1730	1.2425	2.5188	6.6169	30.933
	3	0.9666	1.0273	1.0947	2.3120	6.2829	30.698
	4	0.8793	0.9374	1.0017	2.1595	5.9807	29.925
	5	0.8156	0.8712	0.9326	2.0386	5.7253	29.189
Makeham	1	4.3495	3.4915	3.1033	2.4676	2.6535	3.3851
	2	1.2162	1.2416	1.2718	1.7986	2.9998	6.8103
	3	1.0070	1.0344	1.0652	1.5593	2.6799	6.3496
	4	0.9055	0.9330	0.9632	1.4374	2.5160	6.1036
	5	0.8364	0.8634	0.8929	1.3508	2.3971	5.9251
Weibull	1	1.5097	1.2191	1.0887	0.8530	0.7948	0.6836
	2	0.1815	0.2075	0.2356	0.6892	1.8969	6.8722
	3	0.1242	0.1479	0.1734	0.5945	1.7671	6.7673
	4	0.1028	0.1247	0.1484	0.5441	1.6603	6.5830
	5	0.0902	0.1107	0.1329	0.5081	1.5838	6.4152

below, which shows our suggested test’s Pitman asymptotic efficiencies ${\delta }_{\theta }\left(s,\beta \right)$ for LFR, Makeham, and Weibull distributions at different values of $s, \beta .$

From Table 1, it is shown that these alternative distributions have maximum values of ${\delta }_{\theta }\left(s,\beta \right)$ at $s=0.9 , \beta =2$ and minimum values at $s=0.58, \beta =5.$

In

Table 2. The (PAEs) for LFR, Makeham, and Weibull distributions.

Test	LFR	Makeham	Weibull
Abu-Yousef et al. (2022) [8]	12.8986	0.7462	3.532
Abu-Yousef et al. (2022) [9]	1.3000	0.5800	---------
Hassan et al. (2021) [10]	8.30159	2.90418	2.2092
Abu-Yousef et al. (2020) [7]	5.665	0.8765	2.1679
Mahmoud et al. (2019) [13]	0.9150	0.1720	0.6180
El. Arishy et al. (2019) [12]	0.95658	0.20395	0.73343
$\mathrm{Our} \mathrm{test} {\delta }_{\theta }\left(s=0.9,\beta =2\right)$	30.9330	6.8103	6.8722

, we display Pittman’s asymptotic efficiencies for the proposed test statistics ${\delta }_{\theta }\left(s,\beta \right)$ and other test statistics.

Using Equation (33), we obtain

Table 3. The Pitman’s asymptotic relative efficiencies (PAREs) of our test.

PARE	LFR	Makeham	Weibull
e ( ${\delta }_{\theta }\left(s=0.58,\beta =1\right)), T \mathrm{i}\mathrm{n}$ [8])	2.398	9.127	3.532
e ( ${\delta }_{\theta }\left(s=0.58,\beta =1\right)), T \mathrm{i}\mathrm{n}$ [9])	23.795	11.742	-------
e ( ${\delta }_{\theta }\left(s=0.9,\beta =2\right), T \mathrm{i}\mathrm{n}$ [10])	3.726	2.345	3.111
e ( ${\delta }_{\theta }\left(s=0.9,\beta =2\right), T \mathrm{i}\mathrm{n}$ [7])	5.460	7.770	3.170
e ( ${\delta }_{\theta }\left(s=0.9,\beta =2\right), T \mathrm{i}\mathrm{n}$ [13])	33.807	39.595	11.120
e ( ${\delta }_{\theta }\left(s=0.9,\beta =2\right), T \mathrm{i}\mathrm{n}$ [12])	32.337	33.392	9.370

below, which illustrates our test statistics’ Pitman’s asymptotic efficiency (${\widehat{\delta }}_n$) with respect to other test statistics (T).

$$PARE=e\left( {\delta }_{\theta }\left(s,\beta \right)), T\right)= \frac{PAE\left({\delta }_{\theta }\right)}{PAE\left(T\right)}$$

(33)

From Table 2 and Table 3, it is obvious that our test statistic is more efficient than other tests because $e\left({\delta }_{\theta }, T \right)>1$.

4. Critical Points of the Monte Carlo Null Distribution

The present part presents the Monte Carlo critical points for the null distribution for ${\widehat{∆}}_n\left(s,\beta \right)$ in Equation (18), which are simulated based on 10,000 samples of sizes $n=\mathrm{5,10,11,15}\left(5\right) \mathrm{40,43,46,50}$ from the standard exponential distributions using the Mathematica software 12.3.

Table 4. The upper percentile points of the statistic ${\widehat{∆}}_n\left(s,\beta \right)$ with 10,000 replications at s = 0.9, β = 2.

n	90%	95%	99%
5	6.832471	9.111409	20.881602
10	3.767876	5.492639	16.612772
11	3.487769	5.054081	15.462071
15	2.767289	4.163040	12.836489
20	2.164377	3.087190	9.591994
25	1.841200	2.661309	8.859419
30	1.576625	2.301766	7.199750
35	1.366794	1.959073	5.227894
40	1.220647	1.811924	6.043173
43	1.127370	1.700669	5.076861
46	1.057042	1.515008	5.439105
50	0.997464	1.479134	4.638035

and

Table 5. The upper percentile points of the statistic ${\widehat{∆}}_n\left(s,\beta \right)$ with 10,000 replications at s = 0.58, β = 1.

n	90%	95%	99%
5	1.425526	1.654881	2.512848
10	0.688724	0.800892	1.272633
11	0.622419	0.708224	1.075039
15	0.463981	0.550187	1.034439
20	0.352117	0.415252	0.704349
25	0.282936	0.337393	0.573153
30	0.235479	0.276930	0.457279
35	0.204216	0.244069	0.409616
40	0.178988	0.213249	0.340378
43	0.165415	0.195653	0.328344
46	0.154761	0.181204	0.276341
50	0.145253	0.169742	0.279583

provide the statistic’s upper percentile points. ${\widehat{∆}}_n\left(s,\beta \right)$ with 10,000 replications at $s=0.9, \beta =2$ and $s=0.58,\beta =1,$respectively.

Observations reveal that the critical values in Table 4 and Table 5 increase as the confidence level rises and falls with the growing sample size, with the exception of n = 40, 46 for a confidence level of 99% at $s=0.9, \beta =2$ in Table 4.

It is clear that the critical values in Figure 1 and Figure 2 rise with increasing confidence levels and fall with increasing sample sizes, with the exception of n = 40, 46 for a confidence level of 99% at $s=0.9, \beta =2$ in Figure 1.

5. The Estimated Power

The power of a test is defined as the probability of rejecting $H_0$ when it is false. In the calculation of the power from simulated samples, we calculate the proportion of times $H_0$ is rejected when it is false. The power of ${\widehat{∆}}_n\left(s,\beta \right)$ is estimated at a confidence level of $\left(1-\alpha \right)\%, \alpha =0.05$ with suitable parameters of $\theta =1, 2, 3, 4, 5$ at $n=10, 20, 30$. This is in regard to the three selections: failure rate in a linear approach and Makeham and Weibull distributions based on 10,000 replications.

From

Table 6. The power estimates of ${\widehat{∆}}_n\left(s=0.9,\beta =2\right)$ at α = 0.05.

n	θ	LFR	Makeham	Weibull
10	1	0.1027	0.3316	0.4630
	2	0.4202	0.9797	0.6214
	3	0.6839	1.000	0.7641
	4	0.8395	1.000	0.9673
	5	0.9728	1.000	0.9754
20	1	0.1056	0.3687	0.5571
	2	0.4590	0.9985	0.6431
	3	0.7679	1.000	0.7883
	4	0.8893	1.000	0.9780
	5	0.9750	1.000	0.9832
30	1	0.1077	0.4772	0.5597
	2	0.5616	0.9992	0.6782
	3	0.8066	1.000	0.8542
	4	0.8905	1.000	0.9812
	5	0.9834	1.000	0.9931

and

Table 7. The power estimates of ${\widehat{∆}}_n\left(s=0.58,\beta =1\right)$ at α = 0.05.

n	θ	LFR	Makeham	Weibull
10	1	0.1965	0.2621	0.5335
	2	0.4099	0.9977	0.7452
	3	0.6220	1.000	0.8682
	4	0.7775	1.000	0.9786
	5	0.9902	1.000	0.9860
20	1	0.3105	0.4932	0.6900
	2	0.7065	1.000	0.7743
	3	0.9212	1.000	0.9123
	4	0.9827	1.000	0.9892
	5	0.9970	1.000	0.9976
30	1	04193	0.6597	0.7342
	2	0.8653	1.000	0.8547
	3	0.9870	1.000	0.9678
	4	0.9996	1.000	0.9901
	5	1.000	1.000	0.9987

, it is clear that the power estimate of our test is good, and it increases when the value of the parameter $\theta$ increases and the sample size n increases for LFR, Makeham, and Weibull distributions. Note that the critical values in Table 4 and Table 5 increase as the sample size decreases. This means that the critical values increase when the value of the parameter $\theta$ decreases. Therefore, the power shown in Table 6 and Table 7 decreases as the $NBR{U}_{mgf}$ approaches the exponential distribution, because the proportion of times $H_0$ is rejected when it is false decreases with the decrease in sample size and the value of the parameter $\theta$.

6. Applications for Complete Data

In this section, we apply the test on some real data sets to elucidate the applications of the NBRU_mgf at a 95% confidence level.

Data Set 1

Consider the data set below in Abouammoh et al. [21], these statistics pertain to a group of forty Saudi Arabian Ministry of Health hospital patients with leukemia, a kind of blood cancer. In this case, we find ${\widehat{∆}}_n\left(s=0.9,\beta =2\right)=2.645587$ and ${\widehat{∆}}_n\left(s=0.58,\beta =1\right)=1.160749$which is more than the critical values of 1.811924 of Table 4 and 0.213249 of Table 5, respectively, at the 95% upper percentile. Then, we accept $H_1$, the alternative hypotheses, which states that the data set has NBRU_mgfproperty and that it is not exponential.

0.315	0.496	0.616	1.135	1.208	1.263	1.414	2.025	2.036
2.162	2.211	2.370	2.532	2.693	2.805	2.910	2.912	3.192
3.263	3.348	3.348	3.427	3.499	3.534	3.767	3.751	3.858
3.986	4.049	4.244	4.323	4.381	4.392	4.397	4.647	4.753
4.929	4.973	5.074	4.381

Data Set 2

Consider the following data set given in Grubbs [22]. This data set provides the times between the arrivals of 25 customers at a facility. In this case, we find ${\widehat{∆}}_n\left(s=0.9,\beta =2\right)=3.755778$ and ${\widehat{∆}}_n\left(s=0.58,\beta =1\right)=5.131664$ which are greater than the critical values of 2.661309 from Table 4 and 0.337393 from Table 5, respectively, at the 95% upper percentile. Hence, we accept $H_{1 },$which states that the data set has NBRU_mgf property, and it is not exponential.

1.80	2.89	2.93	3.03	3.15	3.43	3.48	3.57
3.85	3.92	3.98	4.06	4.11	4.13	4.16	4.23
4.34	4.37	4.53	4.62	4.65	4.84	4.91	4.99
5.17

Data Set 3

Consider the following data set which has been given in Lawless [23]. This data set shows the failure times, expressed in hours, for a particular kind of electrical insulation under continually rising voltage stress in an experiment. Here, we find ${\widehat{∆}}_n\left(s=0.9,\beta =2\right)=0.52761$ and ${\widehat{∆}}_n\left(s=0.58,\beta =1\right)=0.038167$, which are less than the critical values of 5.054081 from Table 4 and 0.708224 from Table 5, respectively, at the 95% upper percentile. Hence, we accept the null hypothesis, which states that the data set has exponential property.

0.205	0.363	0.407	0.770	0.720	0.782	1.178	1.255	1.592
1.635	2.310

Data Set 4

Consider the data set which has been given by Kots and Johnson [24], which represents the survival times (in years) after the diagnosis of 43 patients with a certain kind of leukemia. In this case, we find ${\widehat{∆}}_n\left(s=0.9,\beta =2\right)=4.251812$ and ${\widehat{∆}}_n\left(s=0.58,\beta =1\right)=1.239598,$ which are greater than the critical values of 1.700669 from Table 4 and 0.195653 from Table 5, respectively, at the 95% upper percentile. Hence, we accept $H_1,$ which states that the data set has NBRU_mgf property and is not exponential.

0.019	0.129	0.159	0.203	0.485	0.636	0.748	0.781
0.869	1.175	1.206	1.219	1.219	1.282	1.356	1.362
1.458	1.564	1.586	1.592	1.781	1.923	1.959	2.134
2.413	2.466	2.548	2.652	2.951	3.038	3.600	3.655
3.745	4.203	4.690	4.888	5.143	5.167	5.603	5.633
6.192	6.655	6.874

The present paper deals with a hypothesis testing problem for testing exponentiality against the $NBR{U}_{mgf}$ class. Two empirical measures of departure from exponentiality

$${\widehat{∆}}_n\left(s,\beta \right)=\frac{1}{n^2\overline{X}} {\sum }_{i=1}^n{\sum }_{j=1}^n\varnothing (X_i,X_j) ;\hspace{1em}\mathrm{and} {\widehat{∆}}_n\left(s,1\right)=\frac{1}{n^2\overline{X}} {\sum }_{i=1}^n{\sum }_{j=1}^n\varnothing \left(X_i,X_j\right);$$

where

$$\varnothing \left(X_i,X_j\right)=\frac{e^{s{X}_{i * }}{e}^{-\beta X_j}}{s{\beta }^2}+\left[\frac{X_i}{s\beta }-\frac{1}{s^2\left(\beta +s\right)}-\frac{1}{s{\beta }^2}\right]e^{s{X}_j}-\left[\frac{1}{s{\beta }^2}-\frac{1}{{\beta }^2\left(\beta +s\right)}\right]e^{-\beta X_i}+\frac{1}{s^2\beta }$$

has been obtained by applying the Laplace transform technique and goodness of fit approach, respectively.

Authors in [10,11] dealt with a hypothesis testing problem for testing exponentiality against the $NBR{U}_{mgf}$ class. Two empirical measures of departure from exponentiality

$${\widehat{\delta }}_n=\frac{1}{n^2{s}^3}{\sum }_i^n {\sum }_j^n \left[6\mathrm{s}{X}_i+3{\mathrm{s}}^2{X}_i^2+6+\left(s^3{X}_i^3-6\right)e^{s{X}_j}\right],$$

and

$${\widehat{\delta }}_n=\hspace{0.33em}\frac{1}{n^2}\hspace{0.33em}{\sum }_j^n[\left(s^2{X}_i^2{e}^{s{X}_j}\right)-2e^{s{X}_i}+2\mathrm{s}{X}_i+2]$$

were obtained by applying moment inequalities and the U-statistic, respectively.

All authors in the present paper [10,11] obtained the same result (they reject the null hypothesis) which states that data sets 1, 2, and 4 have the NBRU_mgf property and are not exponential (Large Data).

All authors in the present papers, refs. [8,10,11,13,25], obtained the same result (they accept the null hypothesis) which states that data set 3 has the exponential property (Small Data). Also, all authors in [7,8,9,13,25,26,27] obtained the same result (they reject the null hypothesis), which states that at least one of data set 1, data set 2, or data set 4 has the UBAC(2)L, NBUC_mgf, RNBUL, NBUL, ODL, and NBURL properties, and are not exponential (Large Data), respectively.

All authors in [14] obtained a different result (they accept the null hypothesis), which states that data set 2 has exponential property (Large Data), and also in [28], they obtained a different result (they accept the null hypothesis), which states that data set 1 has the exponential property (Large Data.)

So, our test does not perform equally well in small and large samples.

7. Conclusions

Based on the Laplace transform technique and goodness of fit approach, new two-test statistics for assessing exponentiality versus a new class of life distributions termed new better than renewal used in the moment-generating function ($NBR{U}_{mgf}$) is examined. The quality requirements of the two tests are validated using the widely used Pitman asymptotic efficiency criterion.

Pitman’s asymptotic efficiencies for the suggested tests are computed for a number of well-known alternative reliability distributions, including the Makeham, Weibull, and LFR distributions. It has been demonstrated that the test statistics we suggested are more effective than the others. The performance of the suggested tests is evaluated by computing and tabulating the upper percentiles and power. Lastly, the utility of our tests is demonstrated by applying them to some actual data.