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Interval Estimation of the Unknown Exponential Parameter Based on Time Truncated Data

Arnab Koley1 & Debasis Kundu2
Abstract

In this paper we consider the statistical inference of the unknown parameter of an exponential distribution based on the time truncated data. The time truncated data occurs quite often in the reliability analysis for type-I or hybrid censoring cases. All the results available today are based on the conditional argument that at least one failure occurs during the experiment. In this paper we provide some inferential results based on the unconditional argument. We extend the results for some two-parameter distributions also.

Key Words and Phrases: Exponential distribution; type-I censoring; hybrid censoring; conjugate prior; confidence interval; credible interval.

AMS Subject Classifications: 62F10, 62F03, 62H12.

1 Department of Mathematics and Statistics, Indian Institute of Technology Kanpur, Pin 208016, India. E-mail: arnab@iitk.ac.in.

2 Department of Mathematics and Statistics, Indian Institute of Technology Kanpur, Pin 208016, India. Corresponding author, E-mail: kundu@iitk.ac.in, Phone no. 91-512-2597141, Fax no. 91-512-2597500.

1 Introduction

Let X1,,XnX_{1},\ldots,X_{n} be a random sample from an exponential distribution with parameter λ\lambda, then it has the following probability density function (PDF)

f(x;λ)={λeλx,if x0,0,if x<0.f(x;\lambda)=\begin{cases}\lambda e^{-\lambda x},&\text{if $x\geq 0$},\\ 0,&\text{if $x<0$.}\end{cases} (1)

Here λ(0,)\lambda\in(0,\infty) is the natural parameter space. Since for λ\lambda = 0, f(x;λ)f(x;\lambda) as defined in (1) is not a proper PDF, we are not including the point 0 in the parameter space. Suppose nn items are tested and their ordered failure times are denoted by x1:n<x2:n<<xn:nx_{1:n}<x_{2:n}<\ldots<x_{n:n}. If the experiment is stopped at a prefixed time TT then it results in a simple type-I censoring case. Let there be DD failures in [0,T][0,T]. Then x1:n<x2:n<<xD:nTx_{1:n}<x_{2:n}<\ldots<x_{D:n}\leq T and T<xD+1:n<xn:nT<x_{D+1:n}<\ldots x_{n:n}, although xD+1:n,,xn:nx_{D+1:n},\ldots,x_{n:n} are not observed. Here DD is a random variable that can take the values 0,1,,n0,1,\ldots,n. The main aim of this note is to draw inference on λ\lambda, based on DD observations.

This is an old problem and Bartholomew (1963) was the first to consider it. He considered the following form of the exponential PDF;

f(x;θ)={1θexθ,if x0,0,if x<0.f(x;\theta)=\begin{cases}\frac{1}{\theta}e^{-\frac{x}{\theta}},&\text{if $x\geq 0$},\\ 0,&\text{if $x<0$}.\end{cases} (2)

He considered [0,)[0,\infty) as the parameter space of θ\theta. In this case the maximum likelihood estimator (MLE) of θ\theta does not exist when D=0D=0. Hence all the inferences related to θ\theta are based on the conditional argument that D1D\geq 1. Later on a series of papers starting with Chen and Bhattacharyya (1988) and then continuing with Gupta and Kundu (1998) and Childs et al. (2003) considered type-I hybrid censoring case for the model (2), and obtained the exact inference on θ\theta based on the conditional argument that D1D\geq 1.

In case of type-I censoring, P(D=0)=exp(nλT)\displaystyle P(D=0)=exp(-n\lambda T) and this probability can be quite high for small value of nλTn\lambda T. The natural question here is whether it is possible to draw any inference on λ\lambda, when D=0D=0. The second aim of the study is to determine if there exists any significant difference between the conditional and unconditional inference. For example, in this paper it has been shown that it is possible to construct an exact 100(1α)%(1-\alpha)\% confidence interval of λ\lambda even when DD = 0. Following the approach of Chen and Bhattacharyya (1998), an exact 100(1α)%(1-\alpha)\% confidence interval of λ\lambda also can be obtained based on the conditional MLE of λ\lambda, conditioning on the event that D1D\geq 1. The question is whether the lengths of the confidence intervals based on the two different approaches are significantly different or not. We perform some simulation experiments to compare the confidence intervals based on the two different methods. We further obtain the Bayes estimate and the associated credible interval of the unknown parameter based on the non-informative prior. Finally the results are extended for the two-parameter exponential, Weibull and generalized exponential distributions also.

Rest of the paper is organized as follows. In Section 2, we provide two different constructions of confidence intervals and the Bayesian credible intervals. The simulation results for confidence and credible intervals of the parameter λ\lambda are provided in Section 3. In Section 4, we extend the results for the two parameter exponential, Weibull and generalized exponential distributions, and finally we conclude the paper in Section 5.

2 Construction of Confidence and Credible Intervals

In this section we proceed to construct the confidence and credible intervals of the parameter of interest λ\lambda, based on a new estimator of λ\lambda and the posterior distribution of λ\lambda, respectively.

2.1 Confidence interval (CI)

Based on the observations x1:n<<xD:nx_{1:n}<\ldots<x_{D:n} from the model (1), the likelihood function becomes

L(λ|x1:n,,xD:n)={enλT,if D=0,n!(nD)!λDeλ[i=1Dxi:n+(nD)T],if D>0.L(\lambda|x_{1:n},\ldots,x_{D:n})=\begin{cases}e^{-n\lambda T},&\text{if $D=0$},\\ \frac{n!}{(n-D)!}\lambda^{D}e^{-\lambda\big{[}\sum_{i=1}^{D}x_{i:n}+(n-D)T\big{]}},&\text{if $D>0$}.\end{cases} (3)

Note that the MLE of λ\lambda does not exist when D=0D=0. It exists only if D>0D>0 and it is given by

λ^MLE=Di=1Dxi:n+(nD)T.\widehat{\lambda}_{MLE}=\frac{D}{\sum_{i=1}^{D}x_{i:n}+(n-D)T}.

Now based on (D,i=1Dxi:n)\displaystyle(D,\sum_{i=1}^{D}x_{i:n}), the joint sufficient statistic for λ\lambda, we define a new estimator of λ\lambda for all D0D\geq 0 as follows

λ^=Di=1Dxi:n+(nD)T.\widehat{\lambda}=\frac{D}{\sum_{i=1}^{D}x_{i:n}+(n-D)T}. (4)

Note that, λ^\widehat{\lambda} is equal to λ^MLE\widehat{\lambda}_{MLE} only when D>0D>0. Now we provide the exact distribution of λ^\widehat{\lambda}, which will be useful in constructing an exact confidence interval of λ\lambda.

Theorem 1

The distribution of λ^\widehat{\lambda} for x0x\geq 0, can be written as,

P(λ^x)\displaystyle P(\widehat{\lambda}\leq x) =\displaystyle= {enλT,if x=0,enλT+d=1nk=0dCk,dΓ(d,Ad(x,Tk,d)),if x>0,\displaystyle\begin{cases}e^{-n\lambda T},&\text{if $x=0$,}\\ e^{-n\lambda T}+\sum_{d=1}^{n}\sum_{k=0}^{d}C_{k,d}\Gamma(d,A_{d}(x,T_{k,d})),&\text{if $x>0$,}\end{cases} (5)

where

Ck,d=(1)k(nd)(dk)eλT(nd+k),Tk,d=(nd+k)T/d,C_{k,d}=(-1)^{k}{n\choose{d}}{d\choose{k}}e^{-\lambda T(n-d+k)},\ \ \ T_{k,d}=(n-d+k)T/d,
Ak(x,a)={λk(1xa),if x<1a,0,if x1a,A_{k}(x,a)=\begin{cases}\lambda k\left(\frac{1}{x}-a\right),&\text{if $x<\frac{1}{a}$},\\ 0,&\text{if $x\geq\frac{1}{a}$},\end{cases}

and Γ(a,z)=1Γ(a)zta1et𝑑t\displaystyle\Gamma(a,z)=\frac{1}{\Gamma(a)}\int_{z}^{\infty}t^{a-1}e^{-t}dt, is the incomplete gamma function.

Proof: For x0x\geq 0, P(λ^x)P(\widehat{\lambda}\leq x) can be written as,

P(λ^x)\displaystyle P(\widehat{\lambda}\leq x) =\displaystyle= P(λ^x,D=0)+P(λ^x,D>0)\displaystyle P(\widehat{\lambda}\leq x,D=0)+P(\widehat{\lambda}\leq x,D>0) (6)
=\displaystyle= P(λ^x|D=0)P(D=0)+P(λ^x|D>0)P(D>0)\displaystyle P(\widehat{\lambda}\leq x|D=0)P(D=0)+P(\widehat{\lambda}\leq x|D>0)P(D>0)
=\displaystyle= {enλT,if x=0,enλT+d=1nk=0dCk,dΓ(d,Ad(x,Tk,d)),if x>0.\displaystyle\begin{cases}e^{-n\lambda T},&\text{if $x=0$},\\ e^{-n\lambda T}+\sum_{d=1}^{n}\sum_{k=0}^{d}C_{k,d}\Gamma(d,A_{d}(x,T_{k,d})),&\text{if $x>0$}.\end{cases}

Note that the expression of P(λ^x|D>0)P(\widehat{\lambda}\leq x|D>0) can be obtained by using the moment generating function approach. Refer to Corollary 2.2 of Childs et al. (2003) or equation (8) of Gupta and Kundu (1998) for it. It is possible to obtain (5) in terms of the chi-square integral as in Bartholomew (1963). The distribution of λ^\widehat{\lambda} is a mixture of discrete and continuous distributions. The following corollary comes from Theorem 1.

Corollary 1
P(λ^=0)=enλTP(\widehat{\lambda}=0)=e^{-n\lambda T}

and for x>0x>0, the PDF of λ^\widehat{\lambda} is given by

fλ^(x)=1x2d=1nk=0dCk,dg(1xTk,d;λd,d),x1nT,f_{\widehat{\lambda}}(x)=\frac{1}{x^{2}}\sum_{d=1}^{n}\sum_{k=0}^{d}C_{k,d}\ \ g\left(\frac{1}{x}-T_{k,d};\lambda d,d\right),\ \ \ \ x\geq\frac{1}{nT}, (7)

where

g(x;α,p)={αpΓ(p)xp1eαx,if x>0,0,otherwise.g(x;\alpha,p)=\begin{cases}\frac{\alpha^{p}}{\Gamma(p)}x^{p-1}e^{-\alpha x},&\text{if $x>0$},\\ 0,&\text{otherwise.}\end{cases} (8)

Now we consider the construction of an exact 100(1α)%(1-\alpha)\% confidence interval of λ\lambda. We need the following lemma for further development.

Lemma 1

For a fixed b0b\geq 0, Pλ(λ^b)P_{\lambda}(\widehat{\lambda}\leq b) is a monotonically decreasing function of λ\lambda.

Proof: The proof can be obtained along the same line as the proof of the three monotonic lemmas by Balakrishnan and Iliopoulos (2009).

A graphical plot (Figure 1) of Pλ(λ^b)P_{\lambda}(\widehat{\lambda}\leq b) as a function of λ\lambda for a fixed value of b0b\geq 0 is provided for a visual illustration. Here we have taken nn = 5, TT = 0.5 and bb = 1.

Refer to caption
Figure 1: The plot of Pλ(λ^b)P_{\lambda}(\widehat{\lambda}\leq b) as a function of λ\lambda, when bb is fixed.

We now provide an exact 100(1α)%(1-\alpha)\% confidence interval of λ\lambda for different values of DD.

Case-I: Construction of CI when D=0D=0:
Since Pλ(D=0)=exp(nλT)P_{\lambda}(D=0)=exp(-n\lambda T) and it is a decreasing function of λ\lambda, a one sided 100(1α)%(1-\alpha)\% confidence interval of λ\lambda can be obtained as A={λ:Pλ(D=0)1α}A=\{\lambda:P_{\lambda}(D=0)\geq 1-\alpha\}. Hence,

A=[(0,log(1α)/(nT)].A=[(0,-\log(1-\alpha)/(nT)].

Case-II: Construction of CI when D>0D>0:
A symmetric 100(1α)%(1-\alpha)\% confidence interval, (λL,λU)(\lambda_{L},\lambda_{U}), of λ\lambda can be obtained by solving the following two non-linear equations

1α2=PλL(λ^λ^obs)andα2=PλU(λ^λ^obs).1-\frac{\alpha}{2}=P_{\lambda_{L}}(\widehat{\lambda}\leq\widehat{\lambda}_{obs})\ \ \ \hbox{and}\ \ \ \ \frac{\alpha}{2}=P_{\lambda_{U}}(\widehat{\lambda}\leq\widehat{\lambda}_{obs}). (9)

These non-linear equations are to be solved using standard non-linear solver viz. Newton Rapshon, bisection method etc.

2.2 Credible interval (CRI)

In this subsection we will discuss constructing a 100(1α)%(1-\alpha)\% credible interval of λ\lambda based on the conjugate prior on λ\lambda. It is assumed that λ\lambda has a natural gamma prior with the shape and scale parameters as a>0a>0 and b>0b>0, respectively with the following PDF

π(λ)={baΓaλa1eλb,if λ>0,0,otherwise.\pi(\lambda)=\begin{cases}\frac{b^{a}}{\Gamma a}\lambda^{a-1}e^{-\lambda b},&\text{if $\lambda>0$},\\ 0,&\text{otherwise}.\end{cases} (10)

Therefore the posterior density function of λ\lambda becomes

π(λ|data)λa+D1eλ[b+i=1Dxi:n+(nD)T],λ>0.\pi(\lambda|data)\propto\lambda^{a+D-1}e^{-\lambda\big{[}b+\sum_{i=1}^{D}x_{i:n}+(n-D)T\big{]}},\ \ \ \lambda>0. (11)

Hence the Bayes estimate of λ\lambda under the squared error loss function becomes

λ^Bayes=a+Db+i=1Dxi:n+(nD)T).\widehat{\lambda}_{Bayes}=\frac{a+D}{b+\sum_{i=1}^{D}x_{i:n}+(n-D)T)}. (12)

The associated 100(1α)%(1-\alpha)\% symmetric credible interval of λ\lambda can be obtained as (λLB,λUB)(\lambda_{LB},\lambda_{UB}), where λLB\lambda_{LB} and λUB\lambda_{UB} can be obtained as the solutions of

Γ(a+D,λLB(b+S))=(1α2)Γ(a+D)andΓ(a+D,λUB(b+S))=Γ(a+D)α2,\Gamma(a+D,\lambda_{LB}(b+S))=\left(1-\frac{\alpha}{2}\right)\Gamma(a+D)\ \ \ \ \hbox{and}\ \ \ \ \Gamma(a+D,\lambda_{UB}(b+S))=\Gamma(a+D)\ \frac{\alpha}{2}, (13)

respectively. Here Γ(a+D,x)\Gamma(a+D,x) is the incomplete gamma function and S=i=1Dxi:n+(nD)T\displaystyle S=\sum_{i=1}^{D}x_{i:n}+(n-D)T. Observe that, when a=b=0a=b=0, the Bayes estimate of λ\lambda matches with λ^\widehat{\lambda}, although it is an improper prior. Therefore, the comparison of the confidence intervals based on (9) and the credible interval based on (13) makes sense, although their interpretations are different. When DD = 0, the posterior density function of λ\lambda becomes improper for a=ba=b = 0. Due to this reason we propose to use a proper prior with a=ba=b\approx 0 as suggested by Congdon (2014).

3 Numerical Comparisons

In this section we present some simulation results to compare the performances of the two confidence intervals and the corresponding credible intervals. We compare the performances in terms of the average lengths and the coverage percentages. Different values of nn, λ\lambda and TT are taken. We consider nn = 5, 10, 15, 20, λ\lambda = 0.5, 1.0, 2.0 and TT = 1, 2. For Bayesian inference we have taken a=ba=b = 0.001. All the results are based on 5,000 replications. The results are reported in Tables 1 to 3.

The following points have been revealed from these simulation experiments. In all these cases it is observed that biases and mean squared errors (MSEs) decrease as sample size increases, TT increases or λ\lambda increases as expected. Now comparing the confidence intervals and credible intervals in Tables 1 to 3 it is clear that for small sample sizes, small λ\lambda and small TT values, the confidence intervals based on unconditional distribution performs better that the credible intervals based on non-informative priors in terms of the coverage percentages. The coverage percentages of the confidence intervals based on unconditional distribution are very close to the nominal value (95%) in all cases. Although, as expected for large sample sizes they are almost equal. The confidence intervals based on the conditional distribution do not perform very well when the sample size is very small and λ\lambda is also small, although when the sample size is not very small it performs well. Performances of the two confidence intervals match as expected when sample size is large.

4 Some Related Models

In this section we consider some of the related two-parameter models, and construct an exact 100(1α)%(1-\alpha)\% confidence set of the two parameters, when nn items are tested and there is no observed failure during [0,T][0,T].

Model 1: A two-parameter exponential distribution with the location parameter μ\mu and scale parameter λ\lambda having the following PDF

f(x;λ,μ)={λeλ(xμ),if xμ,0,if x<μ.f(x;\lambda,\mu)=\begin{cases}\lambda e^{-\lambda(x-\mu)},&\text{if $x\geq\mu$,}\\ 0,&\text{if $x<\mu$}.\end{cases} (14)

Here λ>0\lambda>0 and <μ<-\infty<\mu<\infty. In this case P{λ,μ}(D=0)=enλ(Tμ)P_{\{\lambda,\mu\}}(D=0)=e^{-n\lambda(T-\mu)}. Hence, a 100(1α)%(1-\alpha)\% confidence set of (μ,λ)(\mu,\lambda) can be obtained as A={(μ,λ);λ>0,<μ<,P{λ,μ}(D=0)(1α)}A=\{(\mu,\lambda);\lambda>0,-\infty<\mu<\infty,P_{\{\lambda,\mu\}}(D=0)\geq(1-\alpha)\}. Therefore,

A={(μ,λ);λ>0,<μ<T,λ(Tμ)ln(1α)/n}.A=\{(\mu,\lambda);\lambda>0,-\infty<\mu<T,\lambda(T-\mu)\leq-\ln(1-\alpha)/n\}.

Model 2: Let us consider now a two-parameter Weibull distribution with the shape parameter β\beta and scale parameter λ\lambda which has the following PDF

f(x;λ,β)={βλxβ1eλxβ,if x0,0,if x<0.f(x;\lambda,\beta)=\begin{cases}\beta\lambda x^{\beta-1}e^{-\lambda x^{\beta}},&\text{if $x\geq 0$},\\ 0,&\text{if $x<0$}.\end{cases} (15)

Here β>0\beta>0 and λ>0\lambda>0. In this case P{λ,β}(D=0)=enλTβP_{\{\lambda,\beta\}}(D=0)=e^{-n\lambda T^{\beta}}. A 100(1α)%(1-\alpha)\% confidence set of (λ,β)(\lambda,\beta), can be obtained as A={(λ,β);λ>0,β>0,P{λ,β}(D=0)(1α)}A=\{(\lambda,\beta);\lambda>0,\beta>0,P_{\{\lambda,\beta\}}(D=0)\geq(1-\alpha)\}. Hence,

A={(λ,β);λ>0,β>0,λTβln(1α)/n}.A=\{(\lambda,\beta);\lambda>0,\beta>0,\lambda T^{\beta}\leq-\ln(1-\alpha)/n\}.

Model 3: Similarly, if we consider two-parameter generalized exponential distribution which has the following PDF

f(x;λ,β)={βλeλx(1eλx)β1,if x0,0,if x<0.f(x;\lambda,\beta)=\begin{cases}\beta\lambda e^{-\lambda x}(1-e^{-\lambda x})^{\beta-1},&\text{if $x\geq 0$},\\ 0,&\text{if $x<0$}.\end{cases} (16)

Here β>0\beta>0 is the shape parameter and λ>0\lambda>0 is the scale parameter. Here, P{λ,β}(D=0)=(1(1eλT)β)nP_{\{\lambda,\beta\}}(D=0)=(1-(1-e^{-\lambda T})^{\beta})^{n}. A 100(1α)%(1-\alpha)\% confidence set of (λ,β)(\lambda,\beta), can be obtained as A={(λ,β);λ>0,β>0,P{λ,β}(D=0)(1α)}A=\{(\lambda,\beta);\lambda>0,\beta>0,P_{\{\lambda,\beta\}}(D=0)\geq(1-\alpha)\}. Therefore,

A={(λ,β);λ>0,β>0,(1eλT)β1(1α)1/n}.A=\{(\lambda,\beta);\lambda>0,\beta>0,(1-e^{-\lambda T})^{\beta}\leq 1-(1-\alpha)^{1/n}\}.

Joint Confidence Set

Just for illustrative purposes, taking nn = 5 and TT = 0.5, we provide the joint 100(1α)%(1-\alpha)\% confidence set as the shaded region of (i) (λ,μ)(\lambda,\mu) for two-parameter exponential distribution, (ii) (β,λ)(\beta,\lambda) for two-parameter Weibull distribution and (iii) (β,λ)(\beta,\lambda) for two-parameter generalized exponential distribution, in Figure 2 when DD = 0.

Refer to caption
(a) Joint confidence set of (λ,μ)(\lambda,\mu) for the two-parameter exponential distribution.
Refer to caption
(b) Joint confidence set of (β,λ)(\beta,\lambda) for the two-parameter Weibull distribution.
Refer to caption
(c) Joint confidence set of (β,λ)(\beta,\lambda) for the two-parameter generalized exponential distribution.
Figure 2: Joint confidence sets of the parameters for different distributions. Here ‘A’ denotes the required set.

5 Conclusions

In this paper we have considered the time truncated exponential distribution and provide the exact confidence interval of the unknown scale parameter based on the unconditional argument. All the existing results are based on the conditional argument and it does not provide any statistical inference of the unknown parameter when there is no observation. In this paper we have provided the inference on the scale parameter even when there is no observation. Simulation experiments are performed and it is observed that the proposed method works quite well even when the sample size is very small. The joint confidence sets have been provided for two-parameter exponential, Weibull and generalized exponential distributions also for time truncated case when there is no observation during the time period of the experiment. The results can be extended to the competing risks model also, as considered in Kundu, Kannan and Balakrishnan (2004).

Table 1: CI of λ\lambda based on unconditional distribution of λ^\widehat{\lambda} when TT=1, 2

T=1 T=2
nn Bias MSE CI CP Bias MSE CI CP
5 0.074 0.207 1.577 97.36 0.091 0.166 1.253 94.42
10 0.033 0.080 1.041 94.10 0.038 0.055 0.825 94.62
λ\lambda=0.5 15 0.020 0.049 0.833 94.86 0.023 0.033 0.659 94.58
20 0.015 0.035 0.716 94.86 0.016 0.023 0.565 94.96
5 0.182 0.662 2.505 94.42 0.226 0.619 2.240 94.50
10 0.076 0.221 1.649 94.62 0.092 0.181 1.434 94.60
λ\lambda=1.0 15 0.046 0.132 1.318 94.58 0.056 0.104 1.139 94.56
20 0.032 0.092 1.130 94.98 0.040 0.071 0.974 94.86
5 0.452 2.477 4.481 94.50 0.510 2.428 4.351 94.66
10 0.183 0.724 2.869 94.60 0.214 0.688 2.742 94.42
λ\lambda=2.0 15 0.112 0.416 2.278 94.56 0.131 0.390 2.162 94.92
20 0.081 0.284 1.948 94.86 0.094 0.260 1.843 94.92
Table 2: CI of λ\lambda based on the conditional distribution of λ^\widehat{\lambda} when TT=1, 2

T=1 T=2
nn Bias MSE CI CP Bias MSE CI CP
5 0.067 0.185 1.398 69.28 0.088 0.156 1.241 91.28
10 0.034 0.072 1.037 92.74 0.038 0.055 0.826 94.62
λ\lambda=0.5 15 0.021 0.048 0.837 94.96 0.023 0.033 0.659 94.58
20 0.016 0.035 0.718 94.88 0.016 0.023 0.565 94.96
5 0.176 0.623 2.482 91.28 0.226 0.618 2.245 94.48
10 0.076 0.219 1.652 94.62 0.092 0.181 1.435 94.60
λ\lambda=1.0 15 0.046 0.132 1.318 94.58 0.056 0.104 1.139 94.56
20 0.032 0.092 1.130 94.98 0.040 0.071 0.974 94.86
5 0.452 2.471 4.490 94.48 0.510 2.428 4.352 94.66
10 0.184 0.724 2.869 94.60 0.214 0.688 2.742 94.42
λ\lambda=2.0 15 0.112 0.416 2.278 94.56 0.131 0.390 2.162 94.92
20 0.081 0.284 1.948 94.86 0.094 0.260 1.843 94.92
Table 3: CRI of λ\lambda based on non-informative priors when TT=1, 2

T=1 T=2
nn Bias MSE CRI CP Bias MSE CRI CP
5 0.074 0.207 1.390 89.36 0.091 0.166 1.202 90.42
10 0.033 0.080 0.985 92.36 0.038 0.055 0.807 93.78
λ\lambda=0.5 15 0.020 0.049 0.804 94.94 0.023 0.033 0.649 93.76
20 0.015 0.035 0.697 93.28 0.016 0.023 0.559 94.28
5 0.182 0.662 2.404 90.42 0.226 0.619 2.212 93.54
10 0.076 0.221 1.614 93.78 0.092 0.181 1.424 94.04
λ\lambda=1.0 15 0.046 0.132 1.299 93.76 0.056 0.104 1.134 94.04
20 0.032 0.092 1.117 94.28 0.040 0.071 0.970 94.58
5 0.452 2.477 4.422 93.54 0.510 2.428 4.343 94.46
10 0.183 0.724 2.849 94.04 0.214 0.688 2.738 94.26
λ\lambda=2.0 15 0.112 0.416 2.267 94.06 0.131 0.390 2.161 94.76
20 0.081 0.284 1.941 94.58 0.094 0.260 1.842 94.80

References

  • [1] Bartholomew, D.J. (1963), ”The sampling distribution of an estimate arising in life testing”, Technometrics, vol. 5, 361-372.
  • [2] Balakrishnan, N. and Iliopoulos, G. (2009), “Stochastic monotonicity of the MLE of exponential mean under different censoring scheme”, Annals of the Institute of Statistical Mathematics, vol. 61, 753 - 772.
  • [3] Chen, S.M., Bhattayacharya, G.K. (1998), ”Exact confidence bound for an exponential parameter under hybrid censoring”, Communication in Statistics Theory and Methodology, 2429-2442.
  • [4] Childs, A., Chandrasekhar, B., Balakrishnan, N. and Kundu, D. (2003), “Exact likelihood inference based on type-I and type-II hybrid censored samples from the exponential distribution”, Annals of the Institute of Statistical Mathematics, vol. 55, 319 - 330.
  • [5] Congdon, P. (2014), Applied Bayesian Modelling, 2nd edition, Wiley, New York.
  • [6] Gupta, R.D. and Kundu, D. (1998), “Hybrid censoring schemes with exponential failure distribution”, Communications in Statistics - Theory and Methods, vol. 27, 3065 - 3083.
  • [7] Kundu, D., Kannan, N. and Balakrishnan, N. (2004), “Analysis of progressively censored competing risks data”, Handbook of Statistics, vol. 23, Eds. N. Balakrishnan and C.R. Rao, 331 - 348.