Maximum precision estimation for a step-stress model using two-stage methodologies

Abstract

In this paper, we consider a two-stage sequential estimation procedure to estimate the parameters of a cumulative exposure model under an accelerated testing scenario. In particular, we focus on a step-stress model where the stress level changes after a pre-specified number of failures occur, which is also random. This is termed as a ‘random stress change time’ in the literature. We further aim to estimate these parameters using maximum precision and hence use a certain variance optimality criteria. Our proposed two-stage estimation procedures follow interesting efficiency properties and their applicability is seen through extensive simulation analyses and a pseudo-real data example from reliability studies.

1. Introduction

In reliability experiments, accelerated testing has been on the forefront in recent times due to its several advantages over the traditional Type-I or Type-II censoring schemes. Type-I censoring occurs when the subjects are right-censored after a predetermined time has elapsed, whereas Type-II censoring occurs when the subjects are right-censored after a fixed number of failures are observed. As it is almost impossible to formulate the lifetime distributions under both Type-I and Type-II censoring plans, one induces higher stress levels in accelerated testing to obtain early failures. One may refer to [18] for an expansive overview on accelerated testing.

In this paper, we will focus on a subset of accelerated testing called as step-stress testing. In a conventional step-stress testing model, several units are put on an initial stress level say $x_0$ and at pre-specified times say $t_1,t_2,\dots ,t_m$, the stress levels are changed to $x_1,x_2,\dots ,x_m$. If a Type-II censoring is in place, one would stop after a pre-specified number of items (say, r) fail. This kind of setup is studied extensively in the literature and one may refer to [8,15,27,28] etc.

However, a drawback of the above-mentioned procedure is that the parameters under the failure time distributions are not always estimable at different stress levels, even for some of the classical distributions , e.g. even under an exponential distribution setup, if no failures occur until time $t_1$, or all of the r failures happen before $t_1$, some of the parameter MLEs do not exist. Moreover, in general, the analytical forms of MLEs may not exist and obtaining confidence intervals may also not be very easy. One may refer to [4] for more details. To overcome this issue, a useful procedure will be where the stress level changes after a pre-specified number of failures occur, which is random. This is called as a ‘random stress change time’ procedure. This is clearly different from a conventional step-stress procedure, where the pre-specified number of failures are fixed. This procedure also ensures that all the underlying parameters are estimable at all stress levels. A short outline of a random stress change time procedure with m levels is: for a sample size n, let $n_1,n_2,\dots ,n_m$ be such that $n_1+n_2+\cdots +n_m<n$ and each $n_i\ge 1$, for $i=1,2,\dots ,m$. Further, let $t_{1:n},t_{2:n},\dots$ be the ordered failure times. Now, let n units be placed on an experiment at a stress level $x_1$. At the time of $n_1$th failure, the stress level is changed to $x_2$, at the time of $(n_1+n_2)$th failure, the stress level is changed to $x_3$ and so on until the stress level is changed to $x_m$ at the time of $(n_1+n_2+\cdots +n_{m-1})$th failure. The experiment is stopped when $(n_1+n_2+\cdots +n_m)$ failures are observed. Here, choosing a value of m is at the discretion of the experimenter. A higher value of m indicates more frequent failures, intuitively. One may refer to [12,27,28] or [29] for more details. We will especially focus on a setup as introduced by Kundu and Balakrishnan [12], where the above step-stress model only contains two stress levels $x_1$ and $x_2$, which we will denote as a simple step-stress model. The distributions at these stress levels are assumed to be exponential with means ${\theta }_1$ and ${\theta }_2,$ respectively. Further, we will assume that the data comes from a cumulative exposure model as introduced by Nelson [18]. A cumulative exposure model relates the lifetime distribution of units at a certain level to the lifetime distribution of units at the next stress level. Thus, the residual lifetime of the experimental units depends only on the cumulative exposure the units have experienced, with no memory of exposure accumulation.

The observed lifetime data will be used for estimating any underlying parameters of the simple step-stress model. Thus, the main question to answer for such a model is when to change the stress level in order to have good estimates of parameters at both the stress levels. In this paper, we try to connect a step-stress model with certain sequential methodologies to propose estimates of the underlying parameters under a certain optimality criterion. Clearly, it is important to choose an optimum value of $n_1$, given r and n, where r denotes the number of failures after which the experiment is stopped, which in this case happens to be $n_1+n_2$ and in general becomes $n_1+n_2+\cdots +n_m$. Moreover, one may adopt one of several optimality criteria some of which are ‘variance optimality’, ‘D optimality’, ‘A optimality’, etc. In the current context, we will focus on the variance optimality criterion. One may refer to works by Bai et al. [2], Alhadeed and Yang [1], Balakrishnan and Han [3], Basak and Balakrishnan [6] or Samanta et al. [21] among others for a more in-depth grasp on the various optimality procedures. On the other hand, there has been a vast literature on sequential and multi-stage procedures both on the estimation and testing front. Chow and Robbins [7] developed a purely sequential procedure to find a fixed-width confidence interval for an unknown parameter. Sen [23], Ghosh et al. [9], and Mukhopadhyay and de Silva [16] are in-depth books covering a large ground on sequential estimation and other problems. A few recent works involving reliability examples are by Pal et al. [19], who designed a multiple step-stress models when the data is Type-I censored, Khalifeh et al. [11], who constructed a fixed accuracy confidence interval for the stress-strength reliability parameter for an exponential distribution, Samanta et al. [22], who worked on the order restricted inference of multiple exponential parameters under a step-stress model, Bapat [5] who worked on obtaining confidence intervals under a stress-strength setup and Mukhopadhyay and Zhuang [17] who presented yet another confidence interval for Fisher's ‘Nile’ example. However, we are unaware of any existing works which connect an accelerated testing estimation problem with a sequential procedure.

In this paper, we mainly focus on providing effective sequential procedures to determine the number of failures to observe before changing the stress level. More specifically, we develop two-stage procedures to determine the value of $n_1$ out of $r=n_1+n_2$ total failures with n items being put on test. Our procedures are proved to enjoy appealing properties and they will provide very accurate estimations for the underlying parameters of the model, ${\theta }_1$ and ${\theta }_2$.

This article proceeds as follows: In Section 2, we describe a brief description of the problem along with an appropriate optimality constraint. Section 3 looks at obtaining a two-stage methodology to estimate the unknown parameters. We also propose some interesting properties for our stopping rule in question. In Section 4, we introduce a slight variation of the same problem by taking a different setting in terms of parameterization. Section 5 includes a discussion on other lifetime models and a similar two-stage sampling strategy under a random stress change time step-stress setup for Weibull distribution. Section 6 includes simulations and pseudo-real data analysis and Section 7 provides brief conclusions.

2. Problem formulation

To recap, a general setup in particular for this problem is as follows: n items are put on a test at a certain stress level $x_1$, and after $n_1$ of those have failed, we change the stress level to $x_2$. We stop the experiment after $r=n_1+n_2$ have failed. Further, we assume that lifetimes at stress levels $x_1$ and $x_2$ are exponentially distributed with mean values being $1/{\theta }_1$ and $1/{\theta }_2,$ respectively. Here, our exponential distribution has a density function as follows:

$$f(x;{\theta }_i)={\theta }_i{e}^{-{\theta }_{i}x},i=1,2$$ (1)

with ${\theta }_i>0,i=1,2$.

In the light of [30], suppose the stress level will be changed at a random time point γ, the probability density function for the lifetime variable becomes

$$f(t;\theta 1,\theta 2)=\left\{\begin{array}{ll}{\theta }_1{e}^{-{\theta }_{1}t},& 0\le t\le \gamma ;\\ {\theta }_2{e}^{-\gamma {\theta }_1-(t-\gamma ){\theta }_2},& t>\gamma .\end{array}\right.$$ (2)

Back to our setup of the simple step-stress model, the experiment gets terminated at the time of the rth failure out of n units under test. The stress level is changed after seeing $n_1$ failures on the first stress level, and the experiment is terminated with $n_2$ failures on the second stress level. Further, we use $t_{i:n}$ to denote the lifetime of the i th item which fails. Hence clearly, $t_{n_1:n}$ stands for observing $n_1$ failures at time $t_{n_1:n}$. Thus, the maximum likelihood estimators (MLEs) for ${\theta }_1$ and ${\theta }_2$ are derived as

$${\hat{\theta }}_1=\frac{n_1}{T_1}\mathrm{a}\mathrm{n}\mathrm{d}{\hat{\theta }}_2=\frac{n_2}{T_2},$$ (3)

where $(T_1,T_2)$ happens to be a complete sufficient statistic and both of them take the following forms:

$$\begin{array}{rl}{T}_1& =\sum _{i=1}^{n_1}{t}_{i:n}+(n-n_1)t_{n_1:n},\\ T_2& =\sum _{i=n_1+1}^r(t_{i:n}-t_{n_1:n})+(n-r)(t_{r:n}-t_{n_1:n}).\end{array}$$

Obviously, the MLEs for ${\theta }_1$ and ${\theta }_2$ are highly dependent on how many failures one observes under each stress level when putting a certain number of items for the experiment. A genuine question here is that how to determine the random stress changing time point to change the stress level from $x_1$ to $x_2$? We propose a method for the same in the following sections.

2.1. Problem motivation

Determining the optimum value of $n_1$ is important as it amounts to changing the stress level from $x_1$ to $x_2$ accurately. Here, in order to achieve maximum precision for estimating ${\theta }_1$ and ${\theta }_2$, we will make use of a certain ‘variance optimality’ criterion as given below. Variance optimality is widely discussed in the literature. One may refer to [12,14,24] for more details. For our simple step-stress model, the overall variance can be written as

$${\tau }_1\left(n_1\right)=\frac{1}{n_1{\theta }_1^2}+\frac{1}{(r-n_1){{\theta }_2}^2}.$$ (4)

Hence, variance optimality signifies minimizing ${\tau }_1\left(n_1\right)$, for $1\le n_1\le (r-1)$, for given ${\theta }_1$ and ${\theta }_2$, which will help us determine the optimum number of testing items, $n_1$, to be included in the first stage, under the stress level $x_1$. The number of testing items in the second stage under a stress level $x_2$ will then be $n_2=r-n_1$.

By taking derivatives of ${\tau }_1\left(n_1\right),$ we get

$${\tau }_1^{\prime }\left(n_1\right)=\frac{-1}{{{\theta }_1^2{n}_1}^2}+\frac{1}{{{\theta }_2}^2(r-n_1{)}^2}$$

and one can easily get the following optimum value for $n_1$:

$$n_1^{\ast }=\frac{{\theta }_{2}r}{{\theta }_1+{\theta }_2}.$$ (5)

Since ${\tau }_1^{″}\left(n_1\right)>0$, ${\tau }_1\left(n_1\right)$ achieves its minimum at $n_1^{\ast }$, and the minimum variance becomes

$${\tau }_1\left(n_1^{\ast }\right)=\frac{({\theta }_1+{\theta }_2{)}^2}{{\theta }_1^2{\theta }_2^{2}r}.$$ (6)

Now, unfortunately, this minimization is only valid and can be performed if both ${\theta }_1,{\theta }_2$ are known or at least their ratio is known. If no information is available on the parameters, this minimization is not possible. One can also find a similar statement from [12]. In the following section, we propose a suitable sequential strategy that provide an appealing solution to this problem. That is, based on a pilot sample, we first get preliminary estimates for both of the parameters. And then, using the preliminary information, we develop a stopping rule to determine the optimum sample size, $n_1^{\ast }$.

3. Two stage methodologies

One can note that since $n_1^{\ast }$ from (5) remains unknown as both ${\theta }_1$ and ${\theta }_2$ are unknown, there is no way to find the value of the sample size. But the number of failures to observe under first stress level is crucial as to the time point of changing stress levels for the simple step-stress model. We hence propose a two-stage strategy to obtain the magnitude of $n_1^{\ast }$ with a two-stage ‘stopping’ variable defined as

$$N_1=max\left\{m,⌊\frac{{\hat{\theta }}_{2m}r}{{\hat{\theta }}_{1m}+{\hat{\theta }}_{2m}}⌋+1\right\}.$$ (7)

Here, m is a number between $(0,r)$, $\lfloor x\rfloor$ denotes the largest integer smaller than x. ${\hat{\theta }}_{1m}$ and ${\hat{\theta }}_{2m}$ are estimated values of ${\theta }_1$ and ${\theta }_2,$ respectively, with m failures at the first stress level $x_1$ and r−m failures at the second stress level $x_2$. In this step-stress life testing setting, the two-stage method can be conducted using two sets of items that are obtained through a random even split of the n items. One set is used for a pilot study to get preliminary estimates of ${\theta }_1$ and ${\theta }_2$. The value of $N_1$ can then be determined by applying the two-stage stopping rule (7), which denotes the number of failures at the first stress level of the test. The final determined value of $N_1$ should be a good estimate of $n_1^{\ast }$ (the optimal value of $n_1$) as per (5), to achieve variance optimality in estimating the parameters ${\theta }_1$ and ${\theta }_2$.

To be more specific, the following outline gives an implementation of this whole estimation procedure using the two-stage method (7):

1. Step 1: We take a sample of n items for a life-testing experiment and we randomly put n/2 items to be subjected to stress level $x_1$. At the time of mth failure, the stress level is changed to $x_2$ and the test continues until a total of r failures are observed where we fix $r\le n/2$. We will then obtain preliminary estimates of ${\theta }_1$ and ${\theta }_2$ according to (3).

2. Step 2: Based on the information from Step 1, we determine the value of $N_1$ as per (7), and consider it to be the final observed optimum value of $n_1$.

3. Step 3: Compare the value of $N_1$ with m. If $N_1>m$, we will then put the other sample of n/2 items on a life-testing experiment but we test $N_1$ items before we change the stress level from $x_1$ to $x_2$; If $N_1<m$, we will use the observations from Step 1 as the final observations.

Upon termination, we propose the following estimators for the parameters ${\theta }_1$ and ${\theta }_2$:

$$\begin{array}{rl}{\hat{\theta }}_{1N_1}& =\frac{N_1}{T_{1N_1}},\end{array}$$ (8)

$$\begin{array}{rl}{\hat{\theta }}_{2N_1}& =\frac{r-N_1}{T_{2N_1}}.\end{array}$$ (9)

The estimated variance as per (4) becomes

$${\hat{\tau }}_1\left(N_1\right)=\frac{1}{N_1{\hat{\theta }}_{1N_1}^2}+\frac{1}{(r-N_1){\hat{\theta }}_{2N_1}^2}.$$ (10)

For convenience, Listing 1 is included in the supplementary materials to show a summary of the R code for simulating the procedures given in Steps 1–3 using the two-stage stopping variable (7).

Further, let us define the risk efficiency to measure the closeness between the achieved variance optimality in (10) and the minimum value for variance optimality from (6), in the light of [20,25].

$$\mathrm{R}\mathrm{i}\mathrm{s}\mathrm{k}\mathrm{E}\mathrm{f}\mathrm{f}\mathrm{i}\mathrm{c}\mathrm{i}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}:{\xi }_{N_1}\left(c\right)\equiv {\hat{\tau }}_1\left(N_1\right)/{\tau }_1\left(n_1^{\ast }\right).$$

The two-stage stopping variable from (7) along with (8)–(10) follows a number of appealing properties, namely (1) The two-stage variable (7) is asymptotic first-order efficient for estimating $n_1^{\ast }$ as per (5); (2) The final estimated variance ${\tau }_1\left(N_1\right)$ is asymptotic first-order efficient for estimating ${\tau }_1\left(n_1^{\ast }\right)$ as per (6). We summarize these properties in Theorem 3.1, and have included its proofs in Appendix, for convenience.

Theorem 3.1

For the two-stage sampling methodology (7)–(10), with ${\theta }_1,{\theta }_2,n,r,n_1,n_2,m$ fixed, we have the following results:

1. $N_1/n_1^{\ast }\to 1$ in probability, as $r\to \mathrm{\infty }$;

2. $E\left[N_1/n_1^{\ast }\right]\to 1$ as $r\to \mathrm{\infty }$ [Asymptotic First-Order Efficiency];

3. $\widehat{{\tau }_1}\left(N_1\right)/{\tau }_1\left(n_1^{\ast }\right)\to 1$ as $r\to \mathrm{\infty }$ [Asymptotic First-Order Risk Efficiency];

where $\stackrel{\stackrel{}{}}{n_1^{\ast }}$ comes from (5), and $N_1$ comes from (7). Here, one may note that $r\to \mathrm{\infty }$ is possible when we put a large enough sample on test, that is $n\to \mathrm{\infty }$.

4. Step-stress model with increased stress level

As one can note, the mean values of the two stress levels following an exponential distribution as given in (1) are $1/{\theta }_1$ and $1/{\theta }_2,$ respectively. It is hence reasonable to assume that one would increase the stress level rather than decrease it at a certain point of the test in order to observe early failures. Thus in this section, we assume that the expected value of the stress level increases from $\frac{1}{{\theta }_1}$ to $\frac{1}{{\theta }_1-c}$ where c is a preassigned positive constant. We also clearly require $c<{\theta }_1$. In such life testing experiments, the full information after $n_1+n_2$ failures can be utilized for estimating ${\theta }_1$. That is, if one decides to change the stress level at a certain point, then they may get more information for estimating the single unknown parameter ${\theta }_1$.

To be more specific, let $r=n_1+n_2$ be the observed samples from n items, where $t_{\left[1\right]},\dots ,t_{\left[n_1\right]}$ are the first $n_1$ smallest order statistics from a sample of size n from $\mathrm{E}\mathrm{x}\mathrm{p}\left({\theta }_1\right)$, and then $t_{[n_1+1]},\dots ,t_{\left[r\right]}$ are the first $n_2$ smallest order statistics from a sample of size $n-n_1$ from $\mathrm{E}\mathrm{E}\mathrm{x}\mathrm{p}({\theta }_1-c)$ given $t_{\left[1\right]},\dots ,t_{\left[n_1\right]}$. Then, the likelihood function of the observed data of size r will be

$$L(t;{\theta }_1)=D{{\theta }_1}^{n_1}({\theta }_1-c{)}^{n_2}{e}^{-T_1{\theta }_1-T_2({\theta }_1-c)},$$

where $D=\frac{\left(n\right)!}{(n-r)!}$ and $T_1,T_2$ are as defined in Section 2. The log-likelihood function will hence be

$$\log L\left({\theta }_1\right)=\log \left(D\right)+n_1\log \left({\theta }_1\right)+n_2\log ({\theta }_1-c)-T_1{\theta }_1-T_2({\theta }_1-c).$$

By setting $\frac{\mathrm{\partial }\log L\left({\theta }_1\right)}{\mathrm{\partial }{\theta }_1}=0$, one would get the following candidate solutions for the MLE of ${\theta }_1$:

$$\begin{array}{rl}{\hat{\theta }}_{1a}& =\frac{r+(T_1+T_2)c+\sqrt{(r+(T_1+T_2)c{)}^2-4(T_1+T_2)n_{1}c}}{2(T_1+T_2)};\end{array}$$ (11)

$$\begin{array}{rl}{\hat{\theta }}_{1b}& =\frac{r+(T_1+T_2)c-\sqrt{(r+(T_1+T_2)c{)}^2-4(T_1+T_2)n_{1}c}}{2(T_1+T_2)}.\end{array}$$ (12)

Now, it is not difficult to show that ${\hat{\theta }}_{1b}$ happens to be $<c$ for any combination of $n_1,r$ and c and hence we can rule it out. Further, $\frac{{\mathrm{\partial }}^2\log L\left({\theta }_1\right)}{\mathrm{\partial }{\theta }_1^2}=-\frac{n_1}{{\theta }_1^2}-\frac{n_2}{({\theta }_1-c{)}^2}$ is negative at ${\theta }_1\equiv {\hat{\theta }}_{1a}$ and we claim that ${\hat{\theta }}_{1a}$ from (11) is the MLE for ${\theta }_1$.

We will now find the optimal values of $n_1$ and $n_2$ for a given sample of size n and fixed r. Choosing an optimum value for $n_1$ is beneficial since it gives the experimenter an idea as to when to change the stress level. We will again use the ‘Variance optimality’ condition to achieve maximum precision for the estimation. That is, we need to minimize

$${\tau }_1\left(n_1\right)=\frac{1}{n_1{\theta }_1^2}+\frac{1}{(r-n_1)({\theta }_1-c{)}^2}.$$ (13)

On differentiating with respect to $n_1,$ we get

$${\tau }_1^{\prime }\left(n_1\right)=\frac{-1}{{\theta }_1^2{n_1}^2}+\frac{1}{(r-n_1{)}^2({\theta }_1-c{)}^2},$$

and the optimum value of $n_1$ will be hence,

$$n_1^{\ast }\left(c\right)=\frac{r({\theta }_1-c)}{2{\theta }_1-c}.$$ (14)

Since ${\tau }_1^{″}\left(n_1\right)>0$, ${\tau }_1\left(n_1\right)$ achieves its minimum at $n_1^{\ast }\left(c\right)$, and the minimum variance becomes

$${\tau }_1\left(n_1^{\ast }\right(c\left)\right)=\frac{(2{\theta }_1-c{)}^2}{{\theta }_1^2({\theta }_1-c{)}^{2}r}.$$ (15)

Here, clearly, $n_1$ will be smaller than r, the total number of observed failures, since we assume ${\theta }_1-c>0$.

One can note that $n_1^{\ast }\left(c\right)$ remains unknown since ${\theta }_1$ is unknown and we hence propose the following two-stage methodology, which is seen to be similar to (7).

$$N_1=max\left\{m,⌊\frac{r({\hat{\theta }}_1-c)}{2{\hat{\theta }}_1-c}⌋+1\right\}.$$ (16)

The stopping rule $N_1$ given above also follows similar efficiency properties as Theorem 3.1 and we omit further details for brevity. Listing 2 is included in the supplementary materials to show the R code for simulating the two-stage procedures according to the stopping variable defined in (16).

Remark 4.1

Some comments on why we provide the two varying methods as given in Sections 3 and 4 are, (i) for experiments where people have no control of both the stress levels, they may go directly with the general method which is provided in Section 3; (ii) for experiments where people have some prior information about by how much the stress level should be increased, they are recommended to use the method from Section 4; (iii) both the methods are necessary as people acquire different information at different experimental scenarios, and in fact, the method in Section 4 provides better estimation for ${\theta }_1$, since the entire observed lifetime data is used for the estimation of a single parameter ${\theta }_1$; (iv) the general method provided in Section 3 does not translate into the method in Section 4 as they are implementable under different settings and parameter estimations are derived independently from their own unique likelihood functions.

5. Outlines for more advanced models

In literature, a step-stress procedure has been discussed extensively under other more flexible lifetime distributions as well. These include Weibull, generalized exponential, two parameter exponential, gamma, log-normal, log-logistic, Pareto, Birnbaum-Saunders among others. We would like to point out that depending on the distribution, one can construct a similar two-stage sampling strategy under a random stress change time step-stress setup. One needs to write down the corresponding likelihood function diligently and come up with maximum likelihood estimators, before proposing a two-stage rule. We will now briefly outline a strategy for a two-parameter Weibull distribution, under a similar setup as seen in Sections 2 and 3. We now assume that the lifetimes at stress levels $x_1$ and $x_2$ follow a Weibull distribution with parameters $(\alpha ,{\theta }_1)$ and $(\alpha ,{\theta }_2)$. For an easy comparison with the pdf given in Section 2, we have taken a re-parameterized common density asfollows:

$$f(x;\alpha ,{\theta }_i)=\left(\alpha {\theta }_i\right)(x{\theta }_i{)}^{\alpha -1}{e}^{-(x{\theta }_i{)}^{\alpha }},i=1,2$$ (17)

with $\alpha >0$ and ${\theta }_i>0$, i = 1, 2. One may refer to [10] or [13] for more details and references therein. Now suppose the stress level will be changed at a random time point γ, the probability density function for the lifetime variable becomes

$$f(t;\alpha ,\theta 1,\theta 2)=\left\{\begin{array}{ll}\alpha {\theta }_1^{\alpha }(t{\theta }_1{)}^{\alpha -1}{e}^{-(t{\theta }_1{)}^{\alpha }},& 0\le t\le \gamma ;\\ \alpha {\theta }_2^{\alpha }{\left(\frac{{\theta }_1}{{\theta }_2}\gamma +t-\gamma \right)}^{\alpha -1}{e}^{-(\gamma {\theta }_1+t{\theta }_2-\gamma {\theta }_2{)}^{\alpha }},& t>\gamma .\end{array}\right.$$ (18)

In this case, however, the maximum likelihood estimators of ${\theta }_1,{\theta }_2$ cannot be found out explicitly and require some numerical procedure like Newton–Raphson, etc. We will denote the MLEs as ${\theta }_1^{\ast }$ and ${\theta }_2^{\ast }$. Now similar to Section 2, the goal is to determine the optimum value of $n_1$, which gives a cut-off for changing the stress level. We will again deploy the variance optimality condition as before. Now, the overall variance can bewritten as

$${\tau }_1\left(n_1\right)=\frac{k^{\ast }}{n_1{\theta }_1^2}+\frac{k^{\ast }}{(r-n_1){\theta }_2^2},$$ (19)

where $k^{\ast }$ is a constant given by

$$k^{\ast }=\mathrm{\Gamma }\left(1+\frac{2}{\alpha }\right)-{\left(\mathrm{\Gamma }\left(1+\frac{1}{\alpha }\right)\right)}^2.$$ (20)

One can easily compare (19) with (4). Here again, variance optimality denotes minimizing the variance in (19) in terms of $n_1$ and the optimum value of $n_1$ hence becomes

$$n_1^{\ast \ast }=\frac{{\theta }_{2}r}{{\theta }_1+{\theta }_2}.$$ (21)

Now, clearly as both ${\theta }_1$ and ${\theta }_2$ are unknown, one may use a two-stage sampling scheme which is given as

$$N_1^{\ast }=max\left\{m^{\ast },⌊\frac{{\theta }_{1m^{\ast }}^{\ast }r}{{\theta }_{1m^{\ast }}^{\ast }+{\theta }_{2m^{\ast }}^{\ast }}⌋+1\right\},$$ (22)

where $m^{\ast }$ is a fixed number between $(0,r)$.

6. Simulation results and real data analysis

In this section, we include some selected simulation results to validate our methodologies from Sections 3 and 4. We also present pseudo-real data analysis for a specific scenario, which illustrates on how such a life testing procedure can be conducted.

6.1. Selected simulations results

To check the performances of our two-stage sequential procedures described in Sections 3 and 4, we have carried out simulations under different combinations of n, m, r, ${\theta }_1$ and ${\theta }_2$. These different values of n and r will be good representations of the sampling scenarios with small, moderate and large sample sizes for lifetime testings.

The whole life-testing procedure is conducted according to Steps 1–3 as in Section 3, and we replicate the process $\mathrm{10,000}$ ( $=R$, say) times. In the tth replication, suppose that $n_{1t}$ is the observed value of $N_1$ from (16), with $t=1,\dots ,R$. Then ${\overline{n}}_t=R^{-1}{\mathrm{\Sigma }}_{t=1}^R{n}_{1t}$ should estimate $E\left(N_1\right)$, the expected number of $N_1$. Similarly, other important quantities are listed in (23).

$$\begin{array}{r}\overline{)\begin{array}{rl}{n}_1^{\ast }& :\mathrm{o}\mathrm{p}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{u}\mathrm{m}\mathrm{s}\mathrm{a}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{z}\mathrm{e}{n}_1,\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\left(5\right)\\ {\overline{n}}_t=R^{-1}{\mathrm{\Sigma }}_{t=1}^R{n}_{1t}& :\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{f}\stackrel{}{E\left[N_1\right];}\\ s.e\left({\overline{n}}_t\right)={\left\{\frac{1}{(R^2-R)}{\mathrm{\Sigma }}_{t=1}^R(n_{1t}-{\overline{n}}_t{)}^2\right\}}^{1/2}& :\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{f}\stackrel{}{{\overline{n}}_t;}\\ {\overline{\hat{\theta }}}_1=R^{-1}{\mathrm{\Sigma }}_{t=1}^R{\hat{\theta }}_{1t}& :\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{f}\stackrel{}{E\left[{\hat{\theta }}_1\right];}\\ s.e\left({\overline{\hat{\theta }}}_1\right)={\left\{\frac{1}{(R^2-R)}{\mathrm{\Sigma }}_{i=1}^R({\hat{\theta }}_1-{\overline{\hat{\theta }}}_1{)}^2\right\}}^{1/2}& :\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{f}{\hat{\theta }}_1.\\ {\overline{\hat{\theta }}}_2=R^{-1}{\mathrm{\Sigma }}_{t=1}^R{\hat{\theta }}_{2t}& :\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{f}\stackrel{}{E\left[{\hat{\theta }}_2\right];}\\ s.e\left({\overline{\hat{\theta }}}_2\right)={\left\{\frac{1}{(R^2-R)}{\mathrm{\Sigma }}_{i=1}^R({\hat{\theta }}_2-{\overline{\hat{\theta }}}_2{)}^2\right\}}^{1/2}& :\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{f}{\hat{\theta }}_2.\\ {\tau }_1\left(n_1^{\ast }\right)& :\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{z}\mathrm{e}\mathrm{d}\mathrm{v}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{p}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y},\mathrm{f}\mathrm{r}\mathrm{o}\mathrm{m}\left(6\right)\\ {\overline{\hat{\tau }}}_1\left(N_1\right)& :\mathrm{e}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{f}\stackrel{}{E\left[{\hat{\tau }}_1\right(N_1\left)\right].}\end{array}}\end{array}$$ (23)

In Table 1, we present some selected simulation results by fixing $n=50,100,300,600,1000$, $r=25,50,80,100,300$, ${\theta }_1=0.2$ and ${\theta }_2=0.1$ for the two-stage methodology as per (7). Columns four to seven clearly show superior first and second-order efficiencies as ${\overline{n}}_t$ is seen to be close to the values of $n_1^{\ast }$. Also, the estimated parameters are very close to the true values of ${\theta }_1$ and ${\theta }_2$, and all standard errors are seen to be very small.

Table 1.

Simulation results for the two-stage methodology (7) with ${\theta }_1=0.2$ and ${\theta }_2=0.1$ from 10,000 replicates.

n	m	r	$n_1^{\ast }$	${\overline{n}}_t$	${\overline{n}}_t/n_1^{\ast }$	${\overline{n}}_t-n_1^{\ast }$	$se\left({\overline{n}}_t\right)$	${\overline{\hat{\theta }}}_1$	$se\left({\overline{\hat{\theta }}}_1\right)$	${\overline{\hat{\theta }}}_2$	$se\left({\overline{\hat{\theta }}}_2\right)$
50	5	25	8.3333	8.6913	1.0430	0.3580	0.0255	0.2633	0.0014	0.1053	0.0003
50	10	25	8.3333	10.4611	1.2553	2.1278	0.0102	0.2359	0.0008	0.1009	0.0003
100	5	50	16.6667	16.6632	0.9998	−0.0035	0.0501	0.2183	0.0009	0.1034	0.0001
100	10	50	16.6667	17.0391	1.0223	0.3724	0.0385	0.2227	0.0007	0.1026	0.0002
100	15	50	16.6667	17.6564	1.0594	0.9897	0.0274	0.2304	0.0006	0.0997	0.0002
300	5	100	33.3333	32.5798	0.9774	−0.7535	0.0973	0.2078	0.0005	0.1014	0.0001
300	10	100	33.3333	33.3867	1.0016	0.0534	0.0732	0.2064	0.0004	0.1017	0.0001
300	15	100	33.3333	33.5997	1.0080	0.2664	0.0621	0.2076	0.0004	0.1013	0.0001
500	5	200	66.6667	64.2939	0.9644	−2.3728	0.1902	0.2033	0.0003	0.1007	0.0001
500	10	200	66.6667	65.8044	0.9871	−0.8623	0.1409	0.2031	0.0003	0.1009	0.0001
500	15	200	66.6667	66.2859	0.9943	−0.3808	0.1182	0.2030	0.0003	0.1007	0.0001
1000	15	300	100.0000	99.5010	0.9950	−0.4990	0.1753	0.2019	0.0002	0.1006	0.0001
1000	20	300	100.0000	99.5162	0.9952	−0.4838	0.1519	0.2021	0.0002	0.1005	0.0001
1000	30	300	100.0000	100.1395	1.0014	0.1395	0.1278	0.2018	0.0002	0.1004	0.0001

Note: The row in bold has relevance given in Section 6.3.

Table 2 displays results to appreciate the ‘variance optimality’ condition and the theoretical variance optimality, as defined in (10) is shown in Column 4. Column 5 contains values on average of the estimated ${\tau }_1$ associated with observed $N_1$ values. It is obvious from Column 7 that as one puts more items to test, that is, with more allowed failures, the differences between ${\tau }_1\left(n_1^{\ast }\right)$ and ${\overline{\hat{\tau }}}_1\left(N_1\right)$ are getting smaller, and the estimated ${\tau }_1$ values are estimated robustly by ${\hat{\tau }}_1\left(N_1\right)$ in general.

Table 2.

Simulation results for the estimation variance of the two-stage methodology (7) with ${\theta }_1=0.2$ and ${\theta }_2=0.1$ from 10,000 replicates.

n	m	r	${\tau }_1\left(n_1^{\ast }\right)$	${\overline{\hat{\tau }}}_1\left(N_1\right)$	${\overline{\hat{\tau }}}_1\left(N_1\right)/{\tau }_1\left(n_1^{\ast }\right)$	${\overline{\hat{\tau }}}_1\left(N_1\right)-{\tau }_1\left(n_1^{\ast }\right)$	s.e.( ${\overline{\hat{\tau }}}_1\left(N_1\right)$)
50	5	25	9.0000	9.7979	1.0887	0.7979	0.0445
50	10	25	9.0000	10.3344	1.1483	1.3344	0.0452
100	5	50	4.5000	4.9325	1.0961	0.4325	0.0167
100	10	50	4.5000	4.7693	1.0598	0.2693	0.0146
100	15	50	4.5000	4.7167	1.0482	0.2167	0.0139
300	5	100	2.2500	2.4311	1.0805	0.1811	0.0065
300	10	100	2.2500	2.3590	1.0485	0.1090	0.0052
300	15	100	2.2500	2.3385	1.0393	0.0885	0.0049
500	5	200	1.1250	1.2028	1.0692	0.0778	0.0024
500	10	200	1.1250	1.1638	1.0345	0.0388	0.0018
500	15	200	1.1250	1.1584	1.0297	0.0334	0.0017
1000	15	300	0.7500	0.7673	1.0230	0.0173	0.0009
1000	20	300	0.7500	0.7639	1.0186	0.0139	0.0009
1000	30	300	0.7500	0.7636	1.0182	0.0136	0.0009

Further, Tables 3 and 4 are organized in a similar manner as that of Tables 1 and 2, and they summarize the simulation results using procedures from Section 4 with the two-stage variable as defined in (10). The majority of the results turn out to be as expected, and they are similar to what we have in Tables 1 and 2. However, one can point out that within each block of fixed values of n, the differences ${\overline{\hat{\tau }}}_1\left(N_1\right)-{\tau }_1\left(n_1^{\ast }\right)$ seem to oscillate without a pattern. This would indicate ${\tau }_1$ being less robustly estimated by ${\hat{\tau }}_1\left(N_1\right)$, as compared to Section 3. However, as one puts more items on test, that is, as r gets larger, the estimation of $n_1^{\ast }$ gets more accurate and ${\overline{\hat{\tau }}}_1\left(N_1\right)$ gets closer to ${\tau }_1\left(n_1^{\ast }\right)$, in general. We omit further comments for brevity.

Table 3.

Simulation results for the two-stage methodology related to increased stress level (16) with ${\theta }_1=0.2$ and c = 0.1 from 10,000 replicates.

n	m	r	$n_1^{\ast }$	${\overline{n}}_t$	${\overline{n}}_t/n_1^{\ast }$	${\overline{n}}_t-n_1^{\ast }$	$se\left({\overline{n}}_t\right)$	${\overline{\hat{\theta }}}_1$	$se\left({\overline{\hat{\theta }}}_1\right)$
50	5	25	8.3333	8.8658	1.0639	0.5325	0.0066	0.2054	0.0003
50	10	25	8.3333	10.0040	1.2005	1.6707	0.0006	0.2048	0.0003
100	5	50	16.6667	17.2119	1.0327	0.5452	0.0086	0.2026	0.0002
100	10	50	16.6667	17.2058	1.0323	0.5391	0.0090	0.2026	0.0002
100	15	50	16.6667	17.2221	1.0333	0.5554	0.0093	0.2017	0.0002
300	5	100	33.3333	33.8729	1.0162	0.5396	0.0117	0.2013	0.0001
300	10	100	33.3333	33.8618	1.0159	0.5285	0.0118	0.2012	0.0001
300	15	100	33.3333	33.8621	1.0159	0.5288	0.0122	0.2013	0.0001
500	5	200	66.6667	67.1879	1.0078	0.5212	0.0160	0.2007	0.0001
500	10	200	66.6667	67.2204	1.0083	0.5537	0.0164	0.2007	0.0001
500	15	200	66.6667	67.2044	1.0081	0.5377	0.0164	0.2007	0.0001
1000	15	300	100.0000	100.5416	1.0054	0.5416	0.0199	0.2004	0.0001
1000	20	300	100.0000	100.5826	1.0058	0.5826	0.0198	0.2003	0.0001
1000	30	300	100.0000	100.5489	1.0055	0.5489	0.0201	0.2005	0.0001

Table 4.

Simulation results for the estimated variance of the two-stage methodology related to increased stress level (16) with ${\theta }_1=0.2$ and c = 0.1 from 10,000 replicates.

n	m	r	${\tau }_1\left(n_1^{\ast }\right)$	${\overline{\hat{\tau }}}_1\left(N_1\right)$	${\overline{\hat{\tau }}}_1\left(N_1\right)/{\tau }_1\left(n_1^{\ast }\right)$	${\overline{\hat{\tau }}}_1\left(N_1\right)-{\tau }_1\left(n_1^{\ast }\right)$	s.e.( ${\overline{\hat{\tau }}}_1\left(N_1\right)$)
50	5	25	9.0000	9.3825	1.0425	0.3825	0.0375
50	10	25	9.0000	9.6439	1.0715	0.6439	0.0402
100	5	50	4.5000	4.5898	1.0199	0.0898	0.0127
100	10	50	4.5000	4.5979	1.0218	0.0979	0.0128
100	15	50	4.5000	4.6889	1.0420	0.1889	0.0137
300	5	100	2.2500	2.2731	1.0103	0.0231	0.0045
300	10	100	2.2500	2.2756	1.0114	0.0256	0.0044
300	15	100	2.2500	2.2704	1.0091	0.0204	0.0044
500	5	200	1.1250	1.1296	1.0041	0.0046	0.0015
500	10	200	1.1250	1.1288	1.0034	0.0038	0.0016
500	15	200	1.1250	1.1297	1.0042	0.0047	0.0015
1000	15	300	0.7500	0.7531	1.0041	0.0031	0.0008
1000	20	300	0.7500	0.7535	1.0047	0.0035	0.0008
1000	30	300	0.7500	0.7508	1.0011	0.0008	0.0008

6.2. Specific pseudo-real data example

In this section, we will discuss a pseudo-real data example from a simple step-stress experiment, with detailed explanations, for illustrative purposes.

This example is an extension of the data example which was first simulated by Xiong [26] and then discussed by Zhu et al. [30] and many others. In [26], the first test had four failure times with a true expected stress level being 2.5, which is further reduced to 1.5 with 12 failure times. In this section, we use this same setup of the simple step-stress test, assuming we have 100 items to be put on test. That is, we fix n = 100. We then further fix r = 30 and m = 5, and assume that the true values of ${\theta }_1$ and ${\theta }_2$ are $1/1.5=0.67$ and $1/2.5=0.4,$ respectively. Now similar to Section 6.1, we simulate lifetime data of size 100, from an exponential distribution as given in (1) with ${\theta }_1=0.67$ and then randomly split them into two parts, say $P_1$ and $P_2$. For the first part, choose the first m order statistics and denote them as $t_1,\dots ,t_m$. Then generate $n/2-m$ lifetime data from an exponential distribution as given in (1) with ${\theta }_2=0.4$ and consider the first r−m order statistics, $y_1,\dots ,y_{r-m}$. By now, the required sample for the first part will be: $t_1,\dots ,t_m,y_1+t_m,\dots ,y_{r-m}+t_m$. Following the steps as outlined in Section 3, we determine the value of $N_1$ using the pilot study from the first part of the data. Then, similarly, we would consider the first $N_1$ order statistics from $P_2$, and take a new set of lifetime data, of size $r-N_1$ as the required sample for the second stage.

For the pilot study, the lifetime data for this simple step-stress experiment is as given in Table 5.

Table 5.

Lifetime data from initial simple step-stress experiment with m = 5, r = 30, $n/2=50$.

Stress	Lifetimes
$x_1$	0.056, 0.089, 0.199, 0.219, 0.221
$x_2$	0.304, 0.357, 0.729, 0.787, 0.823, 0.844, 1.044, 1.089, 1.233, 1.320,
	1.494, 1.531, 1.576. 1.658, 1.788, 1.927, 2.002, 2.025, 2.097, 2.126,
	2.132, 2.233, 2.578, 2.604, 2.625

Using the lifetime data from Table 5, one will obtain ${\hat{\theta }}_{1m}=0.467$ and ${\hat{\theta }}_{2m}=0.307$, and hence the value of $N_1$ is seen to be 19. Thus, for the second batch of the step-stress experiment, we will observe 19 failures before we change the stress level. Table 6 shows the lifetime data from the second stage of tests and the final estimates can be found out to be ${\hat{\theta }}_{1N_1}=0.489$ and ${\hat{\theta }}_{2N_2}=0.430$. Further, the estimated variance becomes ${\hat{\tau }}_1\left(N_1\right)=0.029$.

Table 6.

Lifetime data from the simple step-stress experiment with $N_1=19$, r = 30, $n/2=50$.

Stress	Lifetimes
$x_1$	0.078, 0.089, 0.135, 0.159, 0.210, 0.315, 0.316, 0.371, 0.428, 0.441,
	0.464, 0.486, 0.579, 0.670, 0.771, 0.832, 0.837, 0.892, 0.963
x2	1.105, 1.152, 1.231, 1.269, 1.270, 1.347, 1.365, 1.398, 1.844, 1.957, 1.977

6.3. Sensitivity analysis

In this section, we explore the relationships between terminated variance ${\hat{\tau }}_1\left(N_1\right)$ as defined in (10), and the total number of failures of the step-stress life testing experiment, r, as well as the terminated variance, ${\hat{\tau }}_1\left(N_1\right)$, with the number of failures at the first stress level of the life test, $n_1$.

We set the parameter values as ${\theta }_1=0.2$, ${\theta }_2=0.1$, and the number of items put on the life test being n = 300. We randomly pick 150 items and once m = 10 items have failed at stress level $x_1$, change the stress level to $x_2$ and further wait until we see r failures. We then follow Steps 1–3 as given in Section 3 to complete the experiment. In order to see the effect of r on ${\hat{\tau }}_1\left(N_1\right)$, we run a number of experiments for various values of $r=50,55,60,\dots ,150$ and we replicate the whole process 10, 000 times. The left hand plot in Figure 1 is a plot of ${\hat{\tau }}_1\left(N_1\right)$ against r. Intuitively enough one can see that as r increases, the estimated variance decreases.

Figure 1.

Plots of estimated variance against r and $N_1,$ respectively.

Using a similar setting as above, we now discuss the behaviour of ${\hat{\tau }}_1\left(N_1\right)$ values against various choices of $n_1$, which is the number of failures that one observes before the stress level is changed. We again let ${\theta }_1=0.2$, ${\theta }_2=0.1$, n = 300 and m = 10. Out of the 300 items we randomly pick 150 items and put them on test. We wait for m = 10 items to fail at stress level $x_1$, then change the stress level to $x_2$ and complete the experiment when we observe a total of 100 failures. Then, based on this pilot study, we get the preliminary estimates for ${\theta }_1$ and ${\theta }_2$ as per Step 1 from Section 3. Obviously, we will obtain the value of $N_1$ by further following Step 2 but under this second set of 150 items, we run a number of tests with varying values of $n_1=11,13,15,\dots ,99$, and stopping each time after observing 100 failures. The right-hand plot in Figure 1 shows the plot of ${\hat{\tau }}_1\left(N_1\right)$ against $n_1$. The ‘star marked’ data point represents the value of $n_1$ at the lowest estimated variance value. As one can note, this corresponds exactly to the value of ${\overline{n}}_t$ as well as $n_1^{\ast }$ given in the row in bold, in Table 1. This shows a sense of accuracy of our proposed two-stage method in minimizing the variance while controlling the sample size. One may note that these tests with various combinations of $(n_1,r)$ is only doable via simulations, and it is used to demonstrate that our procedure does help achieve the best time point of changing stress level by determining the number of failures to observe before the stress level is changed. In real life, the whole experiment only requires at most two sets of tests according to Steps 1–3 in Section 3.

7. Some concluding thoughts

In this paper, we developed two-stage sequential methodologies for estimating parameters of an exponential simple step-stress model with random change points under a certain optimality criterion called as ‘variance optimality’. We first randomly split our total items under test into two parts and use the first part as a preliminary study to get the idea on when to change the stress level. That is, the developed two-stage methodologies on the preliminary study help us determine the number of failures, $N_1$, to observe at the first stress level lifetime tests. The rest of the tests, $r-N_1$, are then conducted at the second stress level.

Our two-stage methodologies follow appealing properties such as asymptotic first-order efficiency and asymptotic first-order risk efficiency, which are proved in Theorem 3.1. All properties are double validated in Section 5 with simulation analyses and one specific pseudo-data example. Moreover, the sensitivity analysis shows: (1) as we take more items on the test, that is, as r gets larger, the variance optimality will be smaller. (2) For a fixed r, the minimum value of variance optimality is achieved at our derived changing time, which means that the two-stage stopping variable serves the purpose in determining the number of lifetime test failures to see before the stress level is changed.

In terms of the actual implementation of the proposed testing plans, one may have questions on how to set up the size of the pilot sample, m, and the number of failures to observe out of n/2 items put on test, r. It makes sense that practitioners start with a small value of m as it is called as the ‘pilot sample’ in sequential studies. We have used values of m as 5, 10 and 15, and the choice of m did not change the estimations much, which was also observed through our summarized simulation results. From the left-hand plot in Figure 1, it is obvious that the estimated variance goes down as the r value getting larger. However, we can never observe more than n/2 failures as we only put n/2 items on test. Thus, we would recommend people to use the full n/2 items before they stop the experiment if the time and cost allows them to do so.

Finally, we provided a general outline to come up with a similar sampling strategy under other competing lifetime distributions, and gave a thorough procedure in particular for a Weibull distribution.

Appendix. Proofs for Theorem 3.1.

Proof of part (i): From the two stage stopping rule (7), we have

$$\frac{{\hat{\theta }}_{2m}r}{{\hat{\theta }}_{1m}+{\hat{\theta }}_{2m}}<N_1<m+\frac{{\hat{\theta }}_{2m}r}{{\hat{\theta }}_{1m}+{\hat{\theta }}_{2m}}.$$ (A1)

Now dividing throughout by $n_1^{\ast }$, we have

$$\frac{{\hat{\theta }}_{2m}}{{\theta }_2}\frac{{\theta }_1+{\theta }_2}{{\hat{\theta }}_{1m}+{\hat{\theta }}_{2m}}<\frac{N_1}{n_1^{\ast }}<\frac{m}{n_1^{\ast }}+\frac{{\hat{\theta }}_{2m}}{{\theta }_2}\frac{{\theta }_1+{\theta }_2}{{\hat{\theta }}_{1m}+{\hat{\theta }}_{2m}}.$$ (A2)

As $r\to \mathrm{\infty }$, $n_1^{\ast }\to \mathrm{\infty }$. Also, we can claim that ${\hat{\theta }}_1\to {\theta }_1$ and ${\hat{\theta }}_2\to {\theta }_2$. Hence, ${\hat{\theta }}_1+{\hat{\theta }}_2\to {\theta }_1+{\theta }_2$ and Part (i) follows immediately.

Proof of part (ii): From the right-hand side of (A2), we have

$$\frac{N_1}{n_1^{\ast }}\le \left(\frac{{\theta }_1+{\theta }_2}{{\theta }_2}\right)\frac{{\hat{\theta }}_{2_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}+\frac{m}{n_1^{\ast }}.$$ (A3)

Taking expectations throughout (A3), we get

$$E\left(\frac{N_1}{n_1^{\ast }}\right)\le \left(\frac{{\theta }_1+{\theta }_2}{{\theta }_2}\right)E\left(\frac{{\hat{\theta }}_{2_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}\right)+\frac{m}{n_1^{\ast }}.$$ (A4)

Now clearly, $\frac{{\hat{\theta }}_{2_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}\le 1$ and as $r\to \mathrm{\infty }$, $\frac{{\hat{\theta }}_{2_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}\to \frac{{\theta }_2}{{\theta }_1+{\theta }_2}$. Hence on applying the Dominated Convergence Theorem, we have

$$E\left(\frac{{\hat{\theta }}_{2_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}\right)\to \frac{{\theta }_2}{{\theta }_1+{\theta }_2}$$

from which part (ii) follows.

Proof of part (iii): From (10), we have

$${\hat{\tau }}_1\left(N_1\right)=\frac{1}{N_1{\hat{\theta }}_{1N_1}^2}+\frac{1}{(r-N_1){\hat{\theta }}_{2N_1}^2},$$ (A5)

where $N_1$ is as per (7). Now let the left-hand side of (A1) be denoted as ‘a’ and right-hand side of (A1) be denoted as ‘b’. We hence have the following conclusions:

$$a<N_1\Rightarrow \frac{1}{a{\hat{\theta }}_{1N_1}^2}>\frac{1}{N_1{\hat{\theta }}_{1N_1}^2}\mathrm{a}\mathrm{n}\mathrm{d}a<N_1\Rightarrow \frac{1}{(r-a){\hat{\theta }}_{2N_1}^2}<\frac{1}{(r-N_1){\hat{\theta }}_{2N_1}^2}$$

also

$$N_1<b\Rightarrow \frac{1}{b{\hat{\theta }}_{1N_1}^2}<\frac{1}{N_1{\hat{\theta }}_{1N_1}^2}\mathrm{a}\mathrm{n}\mathrm{d}{N}_1<b\Rightarrow \frac{1}{(r-b){\hat{\theta }}_{2N_1}^2}>\frac{1}{(r-N_1){\hat{\theta }}_{2N_1}^2}.$$

Combining these results, we get

$$\begin{array}{rl}& \frac{1}{b{\hat{\theta }}_{1N_1}^2}+\frac{1}{(r-a){\hat{\theta }}_{2N_1}^2}<{\hat{\tau }}_1\left(N_1\right)<\frac{1}{a{\hat{\theta }}_{1N_1}^2}+\frac{1}{(r-b){\hat{\theta }}_{2N_1}^2}\end{array}$$ (A6)

$$\begin{array}{rl}\Rightarrow & \frac{1/{\hat{\theta }}_{1N_1}^2}{m+\frac{{\hat{\theta }}_{2_m}r}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}}+\frac{1/{\hat{\theta }}_{2N_1}^2}{\frac{{\hat{\theta }}_{1_m}r}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}}<{\hat{\tau }}_1\left(N_1\right)<\frac{1/{\hat{\theta }}_{1N_1}^2}{\frac{{\hat{\theta }}_{2_m}r}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}}+\frac{1/{\hat{\theta }}_{2N_1}^2}{\frac{{\hat{\theta }}_{1_m}r}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}-m}.\end{array}$$ (A7)

Now on dividing (A7) throughout by (6), we have

$$\frac{{\theta }_1^2{\theta }_2^2}{({\theta }_1+{\theta }_2{)}^2}\left[\frac{1/{\hat{\theta }}_{1N_1}^2}{\frac{m}{r}+\frac{{\hat{\theta }}_{2_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}}+\frac{1/{\hat{\theta }}_{2N_1}^2}{\frac{{\hat{\theta }}_{1_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}}\right]<\frac{{\hat{\tau }}_1\left(N_1\right)}{{\tau }_1\left(n_1^{\ast }\right)}<\frac{{\theta }_1^2{\theta }_2^2}{({\theta }_1+{\theta }_2{)}^2}\left[\frac{1/{\hat{\theta }}_{1N_1}^2}{\frac{{\hat{\theta }}_{2_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}}+\frac{1/{\hat{\theta }}_{2N_1}^2}{\frac{{\hat{\theta }}_{1_m}}{{\hat{\theta }}_{1_m}+{\hat{\theta }}_{2_m}}-\frac{m}{r}}\right].$$ (A8)

Now, as $r\to \mathrm{\infty }$ and again noting that ${\hat{\theta }}_{1N_1}\to {\theta }_1$ and ${\hat{\theta }}_{2N_1}\to {\theta }_2$, one can easily get part (iii).

Disclosure statement

No potential conflict of interest was reported by the author(s).