1 Introduction

Properties of the Weibull cumulative exposure model

YOSHIO KOMORI

Department of Systems Innovation and Informatics,

Kyushu Institute of Technology, Iizuka 820-8502, Japan

This article is aimed at the investigation of some properties of the Weibull cumulative exposure model on multiple-step step-stress accelerated life test data. Although the model includes a probabilistic idea of Miner’s rule in order to express the effect of cumulative damage in fatigue, our result shows that the application of only this is not sufficient to express degradation of specimens and the shape parameter must be larger than 1. For a random variable obeying the model, its average and standard deviation are investigated on a various sets of parameter values. In addition, a way of checking the validity of the model is illustrated through an example of the maximum likelihood estimation on an actual data set, which is about time to breakdown of cross-linked polyethylene-insulated cables.

## 1 Introduction

In many industrial fields it is requested for lots of products to operate for a long period of time. Accompanied with that, it is very important to give reliability in relation to the lifetime of products. In such cases, however, life testing under a normal stress can lead to a lengthy procedure with expensive cost. As a means to cope with these problems, the study of accelerated life test (ALT) has been developed. The test makes it possible to quickly obtain information on the life distribution of products by inducing early failure with stronger stress than normal.

One important way in ALT is step-stress accelerated life test (SSALT). There are mainly two types of SSALTs, a simple SSALT and a multiple-step SSALT. In the simple SSALT there is a single change of stress during the test. Miller and Nelson (1983) have shown optimum simple SSALT plans in an exponential cumulative exposure (CE) model. Xiong (1999) has studied an exponential CE model with a threshold parameter in the simple SSALT. Park and Yum (1998) have shown optimum modified simple SSALT plans in an exponential CE model, under the consideration that it is desirable to increase the stress at some finite rate. Lu and Rudy (2002) have dealt with the Weibull CE model with the inverse power law in the simple SSALT.

On the other hand, in the multiple-step SSALT there are changes of stress more than once. Yeo and Tang (1999) have investigated a three-step SSALT in an exponential CE model. Khamis (1997) has proposed an exponential CE model with explanatory variables and investigated it on three-step SSALT data. McSorley, Lu and Li (2002) have shown the properties of the maximum likelihood (ML) estimators of parameters in the Weibull CE model with a log-linear function of stress on three-step SSALT data. Nelson (1980, 1990) has proposed an important idea, which gives the basic CE model for life as a function of constant stress from SSALT data. This is a probabilistic analog of Miner’s rule (Miner, 1945), which is stated on a deterministic situation, and gives the basis of all models mentioned above. He also performed the ML estimation in the Weibull CE model with the inverse power law on multiple-step SSALT data concerning time to breakdown of an electrical insulation. Hirose (1996) has proposed a generalized Weibull CE model, which has a threshold parameter.

As we have seen, there are many kinds of studies about SSALTs on the basis of the CE model and these provide significant understanding of the Weibull and exponential models and SSALTs. However, the validity of the models is not necessarily clear (Bagdanavicius, 1978; Nelson, 1980, 1990). In this article we devote ourselves to considering the following questions:

• Can the models really express degradation of products?

• If so, what condition on the parameters is necessary for it?

• When a random variable obeys the Weibull CE model with a threshold parameter, how do its average and standard deviation behave under a condition?

In Section 2 we introduce the CE model with a threshold parameter, which is a generalization of the CE model provided by Nelson. After giving the Weibull CE model with a threshold parameter in Section 3, we analyze it in Section 4. In Section 5 we give an ML estimation procedure. In Section 6 we illustrate an example of the ML estimation and a goodness of fit test on an actual data set and lastly give the conclusions.

## 2 Cumulative exposure model

We construct a generalized CE model with the help of the CE model proposed by Nelson (1980, 1990), whose model gives the distribution function of a random variable for failure time. Although the general model is obtained in a similar way to Nelson’s, it differs in having a threshold parameter that decides whether a specimen is influenced by stress or not.

The assumptions to obtain the CE model were given by Nelson as follows:

1. The remaining life of specimens depends only on the current cumulative fraction accumulated.

2. If held at the current stress, survivors will fail according to the distribution function for that stress but starting at the previously accumulated fraction failed.

Using a distribution function of a non-negative random variable with an explanatory variable and a threshold , we construct the distribution function of a random variable for failure time in a sequential way. Denote by a stress that a specimen is subjected to in an interval (i=1,2,…).

First of all, we define by

 G(t)\lx@stackreldef={F(t−t0;V1)(V1>Vth),0(V1≤Vth) (2. 1)

for .

Next, for we define

 G(t)\lx@stackreldef={F(t−t1+s1;V2)(V2>Vth),F(s1;V2)(V2≤Vth). (2. 2)

Here, according to Assumption ii), is a positive value satisfying .

Similarly, for we define

 G(t)\lx@stackreldef={F(t−ti−1+si−1;Vi)(Vi>Vth),F(si−1;Vi)(Vi≤Vth),

where is a positive value satisfying .

Also in Hirose (1996), a similar formulation was given, provided that the step stress at the present time is not lower than that at the past time, which means holds for any when . Actually, in his formulation

 G(t)\lx@stackreldef={F(t−ti−1+si−1;Vi)(Vi>Vth),0(Vi≤Vth)

for . Note that our formulation is more general.

## 3 Model for SSALT

We deal with a multiple-step SSALT under the condition that specimens were subjected to a normal level of stress and did not fail before the test. As we will see in the next section, this setting has a possibility of throwing light on new aspects concerning the Weibull and exponential CE models. In this section, first we introduce the multiple-step SSALT and second we give the Weibull CE model under the condition.

### 3.1 Multiple-step SSALT

During the multiple-step SSALT, specimens are subjected to successively higher levels of stress as follows. After a specimen was used at a normal level of stress, it is subjected to an initial level of stress for a predetermined time interval at the first stage in the test. If it does not fail, it is subjected to a higher level of stress for a predetermined time interval at the next stage. In analogy, it is repeatedly subjected to higher levels of stress until it fails. The other specimens are tested similarly. The pattern of stress levels and time intervals is the same for all specimens.

### 3.2 Weibull CE Model with a threshold parameter

We construct the Weibull CE model by combining the basic CE model in Section 2 and a Weibull law, in which the underlying distribution is the two parameter Weibull and the scale parameter is replaced with the function of an explanatory variable. In addition, we give the log likelihood function in the case when the step-stress data are given under the condition mentioned above.

Let be an explanatory variable and a function of it . When the scale parameter is replaced with in the Weibull distribution function, the distribution is given by

 W(t;V)=1−exp⎡⎣−(tϕ(V))β⎤⎦.

We use this as in Section 2.

Denote by and a normal level of stress and the length of the time interval during that a specimen is used before the test, respectively. In addition, denote by the stress that a specimen is subjected to at the -st stage in the test, and let be the start time of the stage . Since the level of stress becomes higher as the stage in the test advances, the relationship holds when .

Before we consider the Weibull CE model under the condition mentioned in the first two lines of Section 3, as preliminaries, let us consider the model without assuming the condition holds. When we denote by the start time of the test and set at , (2. 1) gives

 G(t1)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1−exp⎡⎣−(Tsϕ(Vs))β⎤⎦\makebox[20.0pt](Vs>Vth),0\makebox[20.0pt](Vs≤Vth).

This and

 W(s1;V2)=1−exp⎡⎣−(s1ϕ(V2))β⎤⎦

yield

 s1=⎧⎪⎨⎪⎩Tsϕ(Vs)ϕ(V2)\makebox[20.0pt](Vs>Vth),0\makebox[20.0pt](Vs≤Vth).

Hence, (2. 2) gives

 G(t2)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩1−exp⎡⎣−(Tsϕ(Vs)+t2−t1ϕ(V2))β⎤⎦\raisebox0.0pt[0.0pt][20.0pt]\makebox[20.0pt](Vs>Vth, V2>Vth),1−exp⎡⎣−(Tsϕ(Vs))β⎤⎦\makebox[20.0pt](Vs>Vth, V2≤Vth),1−exp⎡⎣−(t2−t1ϕ(V2))β⎤⎦\makebox[20.0pt](Vs≤Vth, V2>Vth),0\makebox[20.0pt](Vs≤Vth, V2≤Vth).

By repeating similar calculations, we can obtain the cumulative distribution function in the -st stage:

 G(t)=1−exp[−εβ(t)],ti−1

where

 ε(t)\lx@stackreldef=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩Tsϕ(Vs)+tk−tk−1ϕ(Vk)+⋯+ti−1−ti−2ϕ(Vi−1)+t−ti−1ϕ(Vi)\makebox[20.0pt](Vs>Vth)\raisebox0.0pt[0.0pt][20.0pt],tk−tk−1ϕ(Vk)+⋯+ti−1−ti−2ϕ(Vi−1)+t−ti−1ϕ(Vi)\makebox[20.0pt](Vs≤Vth) (3. 2)

when

 V2

Since we are interested in the case of degradation of products under a normal level of stress, we assume in the sequel.

Next, we seek our target, that is, the cumulative distribution function under the condition that a specimen was subjected to a normal level of stress and did not fail before the test. From the statements above, the function is as follows:

 G(t|t>t1) = {G(t)−G(t1)}I{t>t1}(t)1−G(t1) (3. 3) = {1−exp[εβ(t1)−εβ(t)]}I{t>t1}(t),

where

 I{t>t1}(t)\lx@stackreldef={[]cc1(t>t1),0(t≤t1).

Thus, the log likelihood function under the condition is expressed by the following: if we denote by and the sample size and the level of the stage at which a specimen fails, and use superscript to show that a variable is related to the -th specimen,

 (3. 4)

where we express by in order to show clearly that each specimen has each .

## 4 Statistical properties

We consider the statistical properties of the model under the condition mentioned in the previous section. First we state the role of the shape parameter in the distribution function (3. 3) and second we investigate the relationship between statistical quantities and the values of parameters after we simplify the model without loss of generality. In the sequel we express by when it is necessary to show clearly the length of the time interval during that a specimen is used before the test. In analogy, we express by when it is necessary to show clearly the normal stress that a specimen is subjected to before the test. Since we are interested in elapsed time from the start time of test, in the sequel we suppose that is the base point in time. That is, we may consider equal to .

Depending on the magnitude of , the distribution function has a different aspect as follows.

###### Lemma 4.1

Assume that . Then, the following holds for such that .

1. If ,

 G(t|t>t1;Ta)>G(t|t>t1;Tb).
2. If ,

 G(t|t>t1;Ta)=G(t|t>t1;Tb).
3. If ,

 G(t|t>t1;Ta)t1;Tb).

Proof. The substitutions of and (3. 2) into (3. 3) yield

 G(t|t>t1;Ts)=1−exp⎡⎣{Tsϕ(Vs)}β−{Tsϕ(Vs)+i−1∑m=ktm−tm−1ϕ(Vm)+t−ti−1ϕ(Vi)}β⎤⎦.

By differentiating this with respect to and arranging it, we find

 ∂G(t|t>t1;Ts)∂Ts=βϕ(Vs)[εβ−1(t)−εβ−1(t1)]exp[εβ(t1)−εβ(t)].

Noting , we can see

1. if ,
is a strictly decreasing function of since ,

2. if ,
does not depend on since ,

3. if ,
is a strictly increasing function of since .

This completes the proof.

The statement i) in the lemma means that specimens become more durable as they are used longer before the test. This is clearly irrational. Thus, in this sense any value in is inadmissible for . The statement ii) deals with a situation when the underlying distribution is exponential. It indicates that the CE model inherits the memoryless property from the exponential distribution. The statement iii) expresses the most realistic situation, in which the durability of specimens decreases as the the duration of their use becomes longer before the test.

In a similar fashion, we can obtain the following lemma.

###### Lemma 4.2

Assume that . Then, the following holds for such that .

1. If ,

 G(t|t>t1;Va)>G(t|t>t1;Vb).
2. If ,

 G(t|t>t1;Va)=G(t|t>t1;Vb).
3. If ,

 G(t|t>t1;Va)t1;Vb).

From this lemma, we can know a similar fact to Lemma 4.1. Especially, note that the statement iii) expresses the most realistic situation, in which the durability of specimens decreases as the normal stress imposed before the test becomes higher.

In the sequel we assume the inverse power law in for :

 ϕ(V)=K(V−Vth)n, (4. 1)

where and are positive parameters and is a non-negative parameter. In addition, we assume that the length of the time interval and the breadth of upsurge of stress are constant in the test. That is, we set

 Δt\lx@stackreldef=ti−ti−1,ΔV\lx@stackreldef=Vi+1−Vi(i=2,3,…)andV2=ΔV.

Let us simplify (3. 2) and seek the expectation and second moment of a random variable obeying (3. 3).

By using the above constants and rewriting (3. 2) and (4. 1), we can obtain

 ε(t) = Δtϕ(Vs)~Ts+Δtϕ(Vk)+⋯+Δtϕ(Vi−1)+Δtϕ(Vi)t−ti−1Δt, (4. 2) Δtϕ(Vs) = (1−~Vth)n~K,Δtϕ(Vm)=((m−1)Δ~V−~Vth)n~K(m=k,k+1,…,i), (4. 3)

where

 ~Ts\lx@stackreldef=TsΔt,~K\lx@stackreldef=K(ΔtVs)n,Δ~V\lx@stackreldef=ΔVVs,~Vth\lx@stackreldef=VthVs

and holds. Here, remark that and are parameters to be estimated while and are quantities to be prespecified in order to decide a concrete model. These expressions indicate that we can take and as a unit of time and a unit of stress, respectively. Besides, we can suppose that when we deal with the case that .

By means of a similar procedure and the arrangement of expressions, we can obtain another for a different stress , say , in the following form:

 ε′(t)=Δtϕ(V′s)~Ts+Δtϕ(Vk)+⋯+Δtϕ(Vi−1)+Δtϕ(Vi)t−ti−1Δt,

where

 Δtϕ(V′s)=((V′s/Vs)−~Vth)n~K.

Note that only the first terms in the right-hand sides differ in the expressions of and . Thus, once we obtain the values of parameters, we can decide the distribution function in the case of another stress () by replacing only the first term in the right-hand side of (4. 2).

Let us seek the expectation of a random variable obeying (3. 3). We first seek the following conditional expectation as preliminaries: for ,

 E[T|T≤tm] = m∑i=k∫titi−1t∂∂tG(t|t>t1)dt = m∑i=k[[tG(t|t>t1)]titi−1−∫titi−1G(t|t>t1)dt] = −tmexp[εβ(t1)−εβ(tm)]+tk−1+m∑i=k∫titi−1exp[εβ(t1)−εβ(t)]dt.

In the last line of this equation the relationship is used, which holds by (3. 2).

When we denote by a positive integer such that holds for any , we can see that

 tmexp[−εβ(tm)] = (m−1)Δtexp⎡⎣−{Δtϕ(Vs)~Ts+m∑i=kΔtϕ(Vi)}β⎤⎦ <

and the right-hand side converges to as .

From the things above and , we obtain

 E[T]Δt = tk−1Δt+1Δt∞∑i=k∫titi−1exp[εβ(t1)−εβ(t)]dt (4. 4) = tk−1Δt+1β∞∑i=kϕ(Vi)Δt{−A(ti)+A(ti−1)}

as the expectation in the case that is used as a unit of time. Here,

The expression in the right-hand side of (4. 4) is useful for stable numerical calculations when takes a large value.

In a similar fashion we obtain

 E[T2](Δt)2 = (tk−1Δt)2+2(Δt)2∞∑i=k∫titi−1texp[εβ(t1)−εβ(t)]dt (4. 5) = (tk−1Δt)2+2β∞∑i=k(ϕ(Vi)Δt)2{−Bi(ti)+Bi(ti−1)}

as the second moment in the case that is used as a unit of time. Here,

 Bi(t) \lx@stackreldef= exp[εβ(t1)−εβ(t)] \makebox[20.0pt]×∫∞0{(u+εβ(t))1/β−(ε(ti−1)−ti−1−t1ϕ(Vi))}(u+εβ(t))1/β−1e−udu.

Using (4. 4) and (4. 5), we can calculate the mean and standard deviation of for the parameter values in Table 1. The results are shown on Figs 1 and 2. In these figures we show the difference in the pair of the values of and by the combination of the sort of line and the thickness or the color of line. That is, the solid, dash or dotted line means that , or , respectively. On the other hand, the thick, normal or gray one means , or , respectively.

## 5 Estimation procedure

We state the way of seeking the ML estimates of the parameters in (3. 3), (4. 2) and (4. 3). Some techniques below help us to obtain the estimates numerically stably.

We use a new parameter defined by for a constant instead of . The reason is because only the parameter possibly has an estimate that is much larger than those of the other parameters.

By differentiating (3. 4) with respect to each parameter and arranging each equation, we can obtain the likelihood equations in the following simplified form:

 N∑j=1λ(j)θdj+N∑j=1δθ(t(j)1;~T(j)s)=0,θ∈{β,n,ζ,~Vth}, (5. 1)

where

 dj \lx@stackreldef= exp(−εβ(t(j)l−1;~T(j)s))−exp(−εβ(t(j)l;~T(j)s)), λ(j)θ \lx@stackreldef= −δθ(t(j)l−1;~T(j)s)exp(−εβ(t(j)l−1;~T(j)s))+δθ(t(j)l;~T(j)s)exp(−εβ(t(j)l;~T(j)s)), δβ(ti;~Ts) \lx@stackreldef= εβ(ti;~Ts)lnε(ti;~Ts), δn(ti;~Ts) \lx@stackreldef= 1~K0εβ−1(ti;~Ts){i∑m=k((m−1)Δ~V−~Vth)nln((m−1)Δ~V−~Vth)}, δζ(ti;~Ts) \lx@stackreldef= εβ(ti;~Ts),δ~Vth(ti;~Ts)\lx@stackreldef=εβ−1(ti;~Ts)1~K0i∑m=k((m−1)Δ~V−~Vth)n−1.

In the expressions above, note that in (4. 2) is expressed by as usual. We seek the zero of (5. 1) by means of the damped Newton method (Bank and Rose, 1981) as follows.

In this model, the calculation for ML estimates is so sensitive that, depending on a vector of initial guesses, a sequence of approximate vectors by the damped Newton iteration can converge to a vector of estimates in which the estimate of is less 1 even if the value of the likelihood function is not a maximum value. According to our observation, when this phenomenon occurs, there is often a tendency that the estimate of tends to 0. Thus, we adopt the strategy below.

1. We seek the profile of (3. 4) with respect to . That is, while changing the value of from a value to another value in incremental steps, we seek the estimates of the other parameters in each step.

2. Among the points on the part of the profile, we select the point at that the profile achieves its maximum, and then seek the ML estimates of all parameters simultaneously by using the point as a vector of initial values and performing the damped Newton iteration without fixing .

The derivatives of the expressions in the left-hand side of (5. 1) are given in Appendix.

## 6 Example

On the basis of the results obtained in the previous two sections, we show an example of the ML estimation in (3. 3) on an actual data set. The data set is part of the step-stress data on time to breakdown of cross-linked polyethylene-insulated cables in Hirose (1997). We chose only data whose insulation class is 22kV (in 3-phase). The reason is because data whose insulation class is 33kV in the literature indicate that their durability is higher than those for 22kV. Thus, we concluded that we can not mix both of them. Also, note Lemma 4.2. Moreover, we did not incorporate into our sample data set the data coming from the cables that passed 26 years because their values are abnormal, compared with the others, and they push down the value of the likelihood function extremely. Finally, we perform a goodness of fit test by utilizing the Monte Carlo method.

### 6.1 Data set

We show the data set in Table 2. For each specimen , the first column indicates the length of the time interval during that the specimen is used before the test, and the second column indicates elapsed time from the start time of test by the start time of the stage on which a specimen fails. The unit of time is ten minutes. The third column indicates the number of data in each row, which is denoted by . The fourth and fifth columns indicate the average and standard deviation of data, respectively, in the case that an outrageous datum is not taken into account.

### 6.2 Maximum likelihood estimation

On the data set, we can see that because that kV (in 3-phase) kV (in single phase) and kV (in single phase). As a constant for , we set and used as a vector of initial guesses . These were roughly guessed from the comparison between Fig. 1 or 2 and the averages or standard deviations in Table 2. In addition, for seeking the profile we set , and the increment of at in each step.

Ultimately, we can obtain the following ML estimates:

 β=5.016812,n=1.603875,ζ=0.548237,~Vth=0.944054. (6. 1)

Then, the value of the likelihood function (3. 4) is . The mean and some statistical quantities of and the scatter plot of data are given on Fig. 3. In the figure the solid line indicates the mean and the upper or lower dotted line indicates the mean plus or minus the standard deviation, respectively. Each dot indicates for each sample .

### 6.3 Goodness of fit test

After the values of the parameters were estimated and a model was decided, we are often concerned with testing its validity. In the example, however, we can not perform a chi-square goodness of fit test on the grouped data because the number of samples is too small (Rao, 2002, p. 396). Hence, in almost the same way as that Ross (2002, p. 206) adopted in such a situation, we test the hypothesis that the model is consistent with the data set when the parameters are of the values in (6. 1).

#### Simulation conditions

We performed Monte Carlo simulation under the simulation conditions below.

• Setting for generating simulated data
We used as prespecified values

 Δ~V=0.39,~K0=104,~Ts=157680,473040,…,1156320,

which are the same values as those for the data set in Table 2, and as the true values of the parameters

 β=5.016812,n=1.603875,ζ=0.548237,~Vth=0.944054, (6. 2)

which come from (6. 1).

For each value of , the number of simulated data is the same as that in Table 2. The outrageous datum in the table is, however, counted out. That is, for , the number of simulated data is not . Hence, the total number of data is in a sample set.

• Sample sets
Except sample sets where the ML estimates could not be obtained, 1000 sets of independent pseudo-random samples were considered.

• Setting for estimation
On the stage to seek the profile we set , and the increment of at 0.001 in each step.

#### Procedure for generating random sets

From (3. 2), (4. 2) and (4. 3)

 ε(t1)=Tsϕ(Vs)=(1−~Vth)n~K0ζ~Ts.

When we set for () , this and (3. 3) give

 ε(t)=⎡⎣−ln(1−u)+((1−~Vth)n~K0ζ~Ts)β⎤⎦1/β. (6. 3)

Let us denote by the expression in the right-hand side. Then, (4. 2) and (4. 3) yield

 i−1∑m=k((m−1)Δ~V−~Vth)n~K0ζ

Note that and since . Consequently, the procedure for generating a pseudo-random set is as follows:

1. generate a uniform random number and seek by (6. 3),

2. find that satisfies (6. 4), where the is the stress level in which failure occurs,

3. calculate ,

4. repeat the three steps above 74 times.

#### Procedure for a goodness of fit test

In Table 2 we choose some ’s and calculate the value of the test statistic

 T=κ∑i=1(mi−Ndpi)2Ndpi (6. 5)

for each of the ’s. Here, stands for the number of data in the th subinterval when the interval where failure time lies is divided into nonoverlapping subintervals, and is the probability that failure time lies in the th subinterval.

In fact, we chose , and and for each of them divided the interval into subintervals shown in Table 3. The values of the test statistic and other variables in (6. 5) are, for example, as in Table 4 for the data in Table 2 and (6. 2).

In the simulation process we sought parameter estimates, calculated on simulated data, and checked whether the value, say , was at least as large as the value of in Table 4.

#### Simulation result

The result is shown in Table 5. The second row indicates the number of the data sets on that held. The last column indicates the number of the data sets on that the inequality held simultaneously in the three ’s. From this, we can reject the hypothesis at any level because the p-value is less than .

The unsuccessful number was in finding a vector of ML estimates. Both bias and variance of were relatively large as shown in Table 6.

## 7 Conclusions

Under the condition that specimens were subjected to a normal level of stress and did not fail before the test, we considered the two-parameter Weibull CE model with the threshold parameter in the multiple-step SSALT. This consideration revealed that the shape parameter must be larger than for the model to fit the realistic situation in which the durability of specimens decreases as they are used longer or with higher stress.

After simplifying the model without loss of generality, for a various sets of parameter values we showed the average and standard deviation of failure time versus the duration of the specimen’s use before the test. As we have seen in Section 6, we can utilize them in calculations for ML estimates.

In the section we performed a goodness of fit test by means of Monte Carlo simulation. In general it is not clear whether the inverse power law holds for every stress appearing in the step-stress test, and it is not clear even whether the basic idea of the CE model is available. Including these things, the example we showed gives a way of checking the validity of the CE model.

Acknowledgments

The author would like to thank the referees for their helpful comments to improve this paper.

Appendix

: Derivatives needed for the ML estimation

For , the derivatives of the expressions in the left-hand side of (5. 1) are given as follows:

 ∂∂θ2⎧⎪⎨⎪⎩N∑j=1λ(j)θ1dj+N∑j=1δθ1(t(j)1;~T(j)s)⎫⎪⎬⎪⎭ =N∑j=1⎧⎪⎨⎪⎩−(1dj∂dj∂θ2)λ(j)θ1dj+1dj∂λ(j)θ1∂θ2+∂∂θ2δθ1(t(j)1;~T(j)s)⎫⎪⎬⎪⎭ =N∑j=1⎧⎪⎨⎪⎩−Cθ2λ(j)θ1djλ(j)θ2dj+1dj∂λ(j)θ1∂θ2+∂∂θ2δθ1(t(j)1;~T(j)s)⎫⎪⎬⎪⎭,

where

 Cθ \lx@stackreldef= ⎧