Research ArticleParameter Estimation and Joint Confidence Regions forthe Parameters of the Generalized Lindley Distribution

Wenhao Gui and Man Chen

Department of Mathematics Beijing Jiaotong University Beijing 100044 China

Correspondence should be addressed to Wenhao Gui guiwenhaogmailcom

Received 23 October 2015 Accepted 28 December 2015

Academic Editor Vladimir Turetsky

Copyright copy 2016 W Gui and M ChenThis is an open access article distributed under the Creative CommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We deal with the problem of estimating the parameters of the generalized Lindley distribution Besides the classical estimatorinverse moment and modified inverse estimators are proposed and their properties are investigated A condition for the existenceand uniqueness of the inverse moment and modified inverse estimators of the parameters is established Monte Carlo simulationsare conducted to compare the estimatorsrsquo performances Two methods for constructing joint confidence regions for the twoparameters are also proposed and their performances are discussed A real example is presented to illustrate the proposedmethods

1 Introduction

Lindley [1] originally introduced the Lindley distribution toillustrate a difference between fiducial distribution and pos-terior distributionThis distribution is becoming increasinglypopular for modeling lifetime data and has a wide applicabil-ity in survival and reliability as closed forms for the survivaland hazard functions and good flexibility of fit Its densityfunction is given by

119891 (119905) =1205822

1 + 120582(1 + 119905) 119890

minus120582119905 119905 120582 gt 0 (1)

We denoted this by writing LD(120582) The Lindley distributionis a mixture of an exponential distribution with scale 120582 and agamma distribution with shape 2 and scale 120582 where themixing proportion is 119901 = 120582(1 + 120582)

Ghitany et al [2] provided a comprehensive treatmentof the statistical properties of the Lindley distributionMazucheli and Achcar [3] used the Lindley distribution asa good alternative to analyze lifetime data within the com-peting risks approach as compared with the use of standardexponential or even theWeibull distribution commonly usedin this area Krishna and Kumar [4] considered the reliabilityestimation in Lindley distribution with progressively type IIright censored sample Al-Mutairi et al [5] dealt with the esti-mation of the stress-strength parameter when the variables

are independent Lindley random variables with differentshape parameters

Some researchers have proposed and studied new classesof distributions based on the Lindley distribution See forexample Sankaran [6] Ghitany et al [7] Bakouch et al [8]Shanker et al [9] and Ghitany et al [10] In this paper wefocus on the generalized Lindley distribution (GLD) intro-duced by Nadarajah et al [11] It has the attractive featureof allowing for monotonically decreasing monotonicallyincreasing and bath tub shaped hazard rate functions whilenot allowing for constant hazard rate functions It has betterhazard rate properties than the gamma lognormal and theWeibull distributions

The cumulative distribution function and the probabilitydensity function are respectively given by

119865 (119909 120582 120572) = [1 minus1 + 120582 + 120582119909

1 + 120582119890minus120582119909]120572

119909 gt 0 (2)

119891 (119909 120582 120572)

=1205721205822 (119909 + 1) 119890minus120582119909

120582 + 1(1 minus

119890minus120582119909 (120582 + 120582119909 + 1)

120582 + 1)

120572minus1

119909 gt 0

(3)

where 120572 gt 0 and 120582 gt 0 are two parameters We denotethis distribution as GLD(120582 120572) When 120572 = 1 the generalized

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 7946828 13 pageshttpdxdoiorg10115520167946828

2 Mathematical Problems in Engineering

Lindley distribution reduces to the one parameter Lindleydistribution

Singh et al [12] developed the Bayesian estimation forthe generalized Lindley distribution under squared error andgeneral entropy loss functions in case of complete sampleof observations Singh et al [13] considered the generalizedLindley distribution and proposed the progressive type IIcensoring scheme which allows the removal of the live unitsfrom a life-test with beta-binomial probability law during theexecution of the experiment

Nadarajah et al [11] considered the classical maximum-likelihood estimation of the parameters of a generalized Lind-ley distribution The results showed that the bias is not satis-fied especially for a small or even moderate sample size Asfor the moment estimates two nonlinear equations need tobe solved simultaneously and the existence and uniqueness ofthe roots are not clear and guaranteed

In this paper we consider the problem of estimatingthe two parameters of the generalized Lindley distributionWe propose inverse moment and modified inverse momentestimators and study their properties The conditions of theexistence and uniqueness of the estimators are establishedMonte Carlo simulations are used to compare the perfor-mances of the estimatorsWe also investigate themethods forconstructing joint confidence regions for the two parametersand study their performances

The rest of this paper is organized as follows In Section 2we briefly review the classical maximum-likelihood estima-tion of the parameters of the generalized Lindley distributionIn Section 3 the moment estimator is discussed In Section 4we propose two new methods of estimating the parametersand study their properties Joint confidence regions for thetwo parameters are proposed in Section 5 Section 6 conductssimulations to assess the methods Finally in Section 7 a realexample is presented to illustrate the proposed methods

2 Maximum-Likelihood Estimation

In this section we briefly review the MLEs of the parametersof GLD distribution Let 119883

1 1198832 119883

119899be a random sample

from GLD(120582 120572) with pdf and cdf as (3) and (2) respectivelyThe log-likelihood function is given by

119871 (120582 120572) = (120572 minus 1)119899

sum119894=1

log[1 minus119890minus120582119909119894 (120582119909

119894+ 120582 + 1)

120582 + 1]

minus 120582119899

sum119894=1

119909119894+119899

sum119894=1

log (119909119894+ 1) + 119899 log (120572)

+ 2119899 log (120582) minus 119899 log (120582 + 1)

(4)

The score equations are thus as follows120597119871 (120582 120572)

120597120582

= minus (120572 minus 1)119899

sum119894=1

120582119909119894((120582 + 1) 119909

119894+ 120582 + 2)

(120582 + 1) ((120582 + 1) (1 minus 119890120582119909119894) + 120582119909119894)

minus119899

sum119894=1

119909119894+119899 (120582 + 2)

120582 (1 + 120582)

(5)

120597119871 (120582 120572)

120597120572=119899

sum119894=1

log[1 minus119890minus120582119909119894 (120582119909

119894+ 120582 + 1)

120582 + 1] +

119899

120572 (6)

From (6) we obtain the MLE of 120572 as a function of 120582

= minus119899

sum119899

119894=1log [1 minus 119890minus120582119909119894 (120582119909

119894+ 120582 + 1) (120582 + 1)]

(7)

The MLE of 120582 is the root of the following equation

119866 (120582) =119899

sum119899

119894=1log [1 minus 119890minus120582119909119894 (120582119909

119894+ 120582 + 1) (120582 + 1)]

sdot119899

sum119894=1

120582119909119894((120582 + 1) 119909

119894+ 120582 + 2)

(120582 + 1) ((120582 + 1) (1 minus 119890120582119909119894) + 120582119909119894)

+119899

sum119894=1

120582119909119894((120582 + 1) 119909

119894+ 120582 + 2)

(120582 + 1) ((120582 + 1) (1 minus 119890120582119909119894) + 120582119909119894)minus119899

sum119894=1

119909119894

+119899 (120582 + 2)

120582 (1 + 120582)= 0

(8)

Such nonlinear equation does not have closed formsolution We can apply numerical method such as Newton-Raphson method to compute 120582

3 Moment Estimation of Parameters

In this section we discuss the moment estimation (MOM)of the parameters of GLD distribution Let119883

1 1198832 119883

119899be

a random sample from GLD(120582 120572) with pdf and cdf as (3)and (2) respectively 119909

1 1199092 119909

119899are the observed values

Let 1198981

= (1119899)sum119899

119894=1119909119894 1198982

= (1119899)sum119899

119894=11199092119894 1198981 1198982are

sample moments For the population moments we need thefollowing lemma (Nadarajah et al [11])

Lemma 1 Let119870(120572 120582 119899) = intinfin

0119909119899(119909+1)119890minus120582119909(1minus119890minus120582119909(120582+120582119909+

1)(120582 + 1))120572minus1119889119909 One has

119870 (120572 120582 119899) =infin

sum119894=0

119894

sum119895=0

119895+1

sum119896=0

(120572 minus 1

119894)(

119894

119895)(

119895 + 1

119896)

sdot(minus1)119894 120582119895Γ (119899 + 1 + 119896)

(1 + 120582)119895 (120582119894 + 120582)119899+1+119896

(9)

Let119883 denote a GLD random variable It follows that

E119883119899 =1205721205822

1 + 120582119870 (120572 120582 119899) (10)

By equating the population moments with the samplemoments we obtain

1205721205822

1 + 120582119870 (120572 120582 1) = 119898

1 (11)

1205721205822

1 + 120582119870 (120572 120582 2) = 119898

2 (12)

The method of moments estimators is the roots of the twoequations Similar to the MLEs such nonlinear equationsdo not have closed form solutions We can apply numericalmethod such as Newton-Raphson method to determine theroots

Mathematical Problems in Engineering 3

4 Inverse Moment Estimation of Parameters

Unlike the regularmethod ofmoments the idea of the inversemoment estimation (IME) is as follows for a given randomsample119883

1 119883

119899from a distribution with unknown param-

eters first transform the original sample to a quasisample1198841 119884

119899 where 119884

119894contains the unknown parameters but its

distribution does not depend on the unknown parametersthat is 119884

119894is a pivot variable 119894 = 1 119899 The population

moments of the new sample do not depend on the unknownparameters while the sample moments do Let the populationmoments of the quasisample equal the sample moments andsolve for the unknown parameters

Let 1198831 119883

119899form a sample from GLD(120582 120572) with pdf

given in (3) it is known that 119865(119883119894) 119894 = 1 119899 follow

the uniform distribution 119880(0 1) and thus minus log119865(119883119894) 119894 =

1 119899 follow standard exponential distribution Exp(1) Bythe method of inverse moment estimation we let

1

119899

119899

sum119894=1

[minus log119865 (119883119894)] = 1 (13)

that is

minus120572

119899

119899

sum119894=1

log[1 minus119890minus120582119909119894 (120582119909

119894+ 120582 + 1)

120582 + 1] = 1 (14)

Thus the IME of 120572 is obtained as a function of 120582

= minus119899

sum119899

119894=1log [1 minus 119890minus120582119909119894 (120582119909

119894+ 120582 + 1) (120582 + 1)]

(15)

which is identical to the MLE of 120572 In the following wedetermine the IME of 120582

Lemma 2 Let 119885(1)

le 119885(2)

le sdot sdot sdot le 119885(119899)

be the order statisticsfrom the standard exponential distribution Then the randomvariables119882

11198822 119882

119899 where

119882119894= (119899 minus 119894 + 1) (119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899 (16)

with 119885(0)

equiv 0 are independent and follow standard exponen-tial distributions

Proof The proof can be found by Arnold et al [14]

Lemma 3 Let 11988211198822 119882

119899be iid standard exponential

variables 119878119894= 1198821+ sdot sdot sdot +119882

119894119880119894= (119878119894119878119894+1)119894 119894 = 1 2 119899minus1

119880119899= 1198821+ sdot sdot sdot + 119882

119899 then

(1) 1198801 1198802 119880

119899are independent

(2) 1198801 1198802 119880

119899minus1follow the uniform distribution

119880(0 1)

(3) 2119880119899follows 1205942(2119899)

Proof The proof can be found by Wang [15]

For the sample 1198831 119883

119899from GLD(120582 120572) considering

the order statistics119883(1)

le sdot sdot sdot le 119883(119899) we have that

minus log119865 (119883(119899)) le sdot sdot sdot le minus log119865 (119883

(1)) (17)

are 119899-order statistics from standard exponential distributionLet119885(119894)= minus120572 log[1minus119890minus120582119883(119899minus119894+1)(120582119883

(119899minus119894+1)+120582+1)(120582+1)] equiv

minus120572 log119866(119883(119899minus119894+1)

) 119894 = 1 119899 where 119866(119909) = 1 minus 119890minus120582119909(120582119909 +120582 + 1)(120582 + 1) Thus 119885

(1)le 119885(2)

le sdot sdot sdot le 119885(119899)

are the first119899-order statistics from the standard exponential distributionBy Lemma 2 119882

119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899form a sample from standard exponential distribution

Let 119878119894= 1198821+ sdot sdot sdot + 119882

119894 119880119894= (119878119894119878119894+1)119894 119894 = 1 2 119899 minus 1

and 119880119899= 1198821+ sdot sdot sdot + 119882

119899 by Lemma 3 we have

minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

) = 2119899minus1

sum119894=1

log(119878119899

119878119894

)

sim 1205942 (2119899 minus 2)

(18)

where

119878119899

119878119894

=1198821+ sdot sdot sdot + 119882

119899

1198821+ sdot sdot sdot + 119882

119894

=119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119899)

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119894minus1)

+ (119899 minus 119894 + 1) 119885(119894)

=log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

(19)

Note that the mean of 1205942(2119899 minus 2) is 2119899 minus 2 Thus we obtainan inverse moment equation for 120582 as follows

119899minus1

sum119894=1

log[log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] = 119899 minus 1 (20)

4 Mathematical Problems in Engineering

Solve the equation and we obtain the inverse estimate IMEof 120582 Plugging IME into (15) we obtain the inverse estimate

IME In addition considering that the mode of 1205942(2119899 minus 2) is2119899 minus 4 we can obtain a modified equation for 120582

119899minus1

sum119894=1

log[log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] = 119899 minus 2 (21)

Solve the equation and we obtain the modified inverseestimate MIME of 120582 Plugging MIME into (15) we obtainthe modified inverse estimate MIME Unlike the momentestimation herewe only need to solve one nonlinear equationinstead of two equations

In the following we prove the existence and uniquenessof the root in (20) and (21)

Lemma 4 The following limits hold

(1) One has

lim120582rarr0

log119866 (119886)

log119866 (119887)

= lim120582rarr0

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= 1

119891119900119903 119886 gt 0 119887 gt 0

(22)

(2) One has

lim120582rarrinfin

log119866 (119886)

log119866 (119887)

= lim120582rarrinfin

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= 0

119891119900119903 119886 gt 119887 gt 0

(23)

(3) One has

lim120582rarrinfin

log119866 (119886)

log119866 (119887)

= lim120582rarrinfin

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= +infin 119891119900119903 119887 gt 119886 gt 0

(24)

Lemma 5 For 119905 gt 0 119891(119905) = (120582 + 119905 + 2 minus 119890119905[120582 minus (120582 + 1)119905 +2])(120582 minus (120582 + 1)119890119905 + 119905 + 1) is a decreasing function of 119905

Theorem 6 Let119882119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899form a sample from standard exponential distribution 119878

119894=

1198821+ sdot sdot sdot + 119882

119894 then for 119905 gt 0 equation sum

119899minus1

119894=1log(119878119899119878119894) = 119905

has a unique positive solution

Proof By Lemma 4 we obtain

lim120582rarr0

119878119899

119878119894

= lim120582rarr0

1198821+ sdot sdot sdot + 119882

119899

1198821+ sdot sdot sdot + 119882

119894

= lim120582rarr0

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119899)

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119894minus1)

+ (119899 minus 119894 + 1) 119885(119894)

= lim120582rarr0

log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

= lim120582rarr0

[log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))] log119866 (119883

(119899))

[log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899))=119899

119899= 1

(25)

Thus lim120582rarr0

sum119899minus1

119894=1log(119878119899119878119894) = 0 On the other hand

lim120582rarrinfin

119878119899

119878119894

= lim120582rarrinfin

log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

= 1 + lim120582rarrinfin

log119866 (119883(119899minus119894)

) + sdot sdot sdot + log119866 (119883(1)) minus (119899 minus 119894) log119866 (119883

(119899minus119894+1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+1)) + (119899 minus 119894) log119866 (119883

(119899minus119894+1))

= 1 + lim120582rarrinfin

[log119866 (119883(119899minus119894)

) + sdot sdot sdot + log119866 (119883(1)) minus (119899 minus 119894) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899minus119894+1))

[log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+1)) + (119899 minus 119894) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899minus119894+1))= +infin

(26)

Mathematical Problems in Engineering 5

Thus lim120582rarrinfin

sum119899minus1

119894=1log(119878119899119878119894) = infin Therefore for 119905 gt 0

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has one positive solution For

the uniqueness of the solution we consider the derivative of119878119899119878119894with respect to 120582 Note that for 119894 = 1 119899

119882119894= (119899 minus 119894 + 1) 120572 [log119866 (119883

(119899minus119894+2)) minus log119866 (119883

(119899minus119894+1))]

119889119882119894

119889120582= (119899 minus 119894 + 1) 120572

120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+2)

)minus120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+1)

) = 119882

119894

sdot120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+2)

) minus 120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+1)

)

log119866 (119883(119899minus119894+2)

) minus log119866 (119883(119899minus119894+1)

)

(119878119899

119878119894

)1015840

= (1 +119882119894+1

+ sdot sdot sdot + 119882119899

1198821+ sdot sdot sdot + 119882

119894

)1015840

=1

(sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

[1198821015840119895119882119896minus1198821198951198821015840119896] =

1

120582 (sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

119882119895119882119896 [119860 (120582) minus 119861 (120582)]

(27)

where

119860 (120582)

=120582119883(119899minus119895+2)

119890minus120582119883(119899minus119895+2)120582 [(120582 + 1)119883(119899minus119895+2)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+2)

) minus 120582119883(119899minus119895+1)

119890minus120582119883(119899minus119895+1)120582 [(120582 + 1)119883(119899minus119895+1)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+1)

)

log119866(119883(119899minus119895+2)

) minus log119866(119883(119899minus119895+1)

)

119861 (120582)

=120582119883(119899minus119896+2)

119890minus120582119883(119899minus119896+2)120582 [(120582 + 1)119883(119899minus119896+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+2)

) minus 120582119883(119899minus119896+1)

119890minus120582119883(119899minus119896+1)120582 [(120582 + 1)119883(119899minus119896+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+1)

)

log119866 (119883(119899minus119896+2)

) minus log119866 (119883(119899minus119896+1)

)

(28)

By Cauchyrsquos mean-value theorem for 119895 = 119894 + 1 119899 119896 =1 119894 there exist 120585

1isin (120582119883

(119899minus119895+1) 120582119883(119899minus119895+2)

) and 1205852

isin(120582119883(119899minus119896+1)

120582119883(119899minus119896+2)

) such that

119860 (120582) =120582 + 1205851+ 2 minus 119890120585

1[120582 minus (120582 + 1) 120585

1+ 2]

120582 minus (120582 + 1) 119890120585

1+ 1205851+ 1

119861 (120582) =120582 + 1205852+ 2 minus 119890120585

2[120582 minus (120582 + 1) 120585

2+ 2]

120582 minus (120582 + 1) 119890120585

2+ 1205852+ 1

(29)

Note that 1205851lt 1205852 by Lemma 5 119860(120582) minus 119861(120582) gt 0 (119878

119899119878119894)1015840 gt 0

thussum119899minus1119894=1

log(119878119899119878119894) is a strictly increasing function of 120582 and

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has a unique positive solution

5 Joint Confidence Regions for 120582 and 120572

Let 1198831 1198832 119883

119899form a sample from GLD(120582 120572) and

119883(1)

le 119883(2)

le sdot sdot sdot le 119883(119899)

are the order statistics Let119885(119894)

= minus120572 log[1 minus 119890minus120582119883(119899minus119894+1)(120582119883(119899minus119894+1)

+ 120582 + 1)(120582 + 1)] =minus120572 log119866(119883

(119899minus119894+1)) 119894 = 1 119899Thus 119885

(1)le 119885(2)

le sdot sdot sdot le 119885(119899)

are the first 119899-order statistics from the standard exponentialdistribution By Lemma 2 119882

119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

)119894 = 1 2 119899 form a sample from standard exponential

distribution Let 119878119894= 1198821+ sdot sdot sdot + 119882

119894 119880119894= (119878119894119878119894+1)119894 119894 =

1 2 119899 minus 1 and 119880119899= 1198821+ sdot sdot sdot + 119882

119899 Hence

119881 = 21198781= 21198821= 2119899119885

(1)

= minus2119899120572 log[1 minus119890minus120582119883(119899) (120582119883

(119899)+ 120582 + 1)

120582 + 1]

sim 1205942 (2)

119880 = 2 (119878119899minus 1198781) = 2

119899

sum119894=2

119882119894

= 2 [119885(1)

+ sdot sdot sdot + 119885(119899)

minus 119899119885(1)] sim 1205942 (2119899 minus 2)

(30)

It is obvious that 119880 and 119881 are independent Define

1198791=119880 (2119899 minus 2)

1198812=

119878119899minus 1198781

(119899 minus 1) 1198781

sim 119865 (2119899 minus 2 2)

1198792= 119880 + 119881 = 2119878

119899sim 1205942 (2119899)

(31)

We obtain that 1198791and 119879

2are independent using the known

bank-post office story in statisticsLet 119865120574(V1 V2) denote the percentile of 119865 distribution with

left-tail probability 120574 and V1and V

2degrees of freedom Let

1205942120574(V) denote the percentile of 1205942 distribution with left-tail

probability 120574 and V degrees of freedom

6 Mathematical Problems in Engineering

By using the pivotal variables1198791and1198792 a joint confidence

region for the two parameters 120582 and 120572 can be constructed asfollows

Theorem 7 (method 1) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198791and1198792

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582119871le 120582 le 120582

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(32)

where 120582119871is the root of 120582 for the equation 119879

1= 119865(1minusradic1minus120574)2

(2119899minus

2 2) and 120582119880

is the root of 120582 for the equation 1198791

=119865(1+radic1minus120574)2

(2119899 minus 2 2)

Proof 1198791= (1(119899 minus 1)) log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 +

1)] + sdot sdot sdot + log[1 minus 119890minus120582119883(1)(120582119883(1)

+ 120582 + 1)(120582 + 1)] minus 119899 log[1 minus119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 + 1)]119899 log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 +

1)(120582+1)] is a function of 120582 and does not depend on 120572 FromTheorem 6 we have lim

120582rarr01198791= (1(119899 minus 1))lim

120582rarr0(1198781198991198781minus

1) = 0 lim120582rarrinfin

1198791= (1(119899 minus 1))lim

120582rarrinfin(1198781198991198781minus 1) = infin

and 11987910158401= (1(119899 minus 1))(119878

119899119878119894)1015840 gt 0 Therefore for any 119905 gt 0

equation 1198791= 119905 has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2)

le 1198791le 119865(1+radic1minus120574)2

(2119899 minus 2 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899)

le 1198792le 1205942(1+radic1minus120574)2

(2119899))

= 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2) le 1198791

le 119865(1+radic1minus120574)2

(2119899 minus 2 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582119871le 120582

le 120582119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(33)

On the other hand by Lemma 3 we have

1198793= minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

)

= 2119899minus1

sum119894=1

log(119878119899

119878119894

) sim 1205942 (2119899 minus 2)

(34)

1198792and119879

3are also independent By using the pivotal variables

1198792and 119879

3 a joint confidence region for the two parameters 120582

and 120572 can be constructed as follows

Theorem 8 (method 2) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198792and1198793

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582lowast119871le 120582 le 120582lowast

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(35)

where 120582lowast119871is the root of 120582 for the equation 119879

3= 1205942(1minusradic1minus120574)2

(2119899minus

2) and 120582lowast119880is the root of 120582 for the equation119879

3= 1205942(1+radic1minus120574)2

(2119899minus

2)

Proof 1198793= 2sum

119899minus1

119894=1log(119878119899119878119894) is a function of 120582 and does not

depend on 120572 FromTheorem 6 for any 119904 gt 0 equation 1198793= 119904

has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582lowast119871le 120582

le 120582lowast119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(36)

6 Simulation Study

61 Comparison of the Four Estimation Methods In this sec-tion we conduct simulations to compare the performancesof the MIMEs IMEs MLEs and MOMs mainly with respectto their biases and mean squared errors (MSEs) for varioussample sizes and for various true parametric values

The random data 119883 from the GLD(120582 120572) distribution canbe generated as follows

119883 =minus120582 minus 1 minus119882(119890minus120582minus1 (120582 + 1) (1198801120572 minus 1))

120582 (37)

Mathematical Problems in Engineering 7

Table 1 Average relative estimates and MSEs of 120572

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 12705 (04298) 12350 (03510) 12965 (05115) 12770 (04460) 13522 (06256)MLE 11310 (01459) 11703 (02029) 11590 (02240) 11693 (02267) 11911 (02888)IME 10956 (01267) 11292 (01728) 11154 (01906) 11234 (01930) 11395 (02400)MIME 10482 (01063) 10755 (01420) 10590 (01570) 10635 (01572) 10754 (01939)

40

MOM 11791 (02212) 11954 (02262) 12046 (02693) 12299 (03163) 12536 (03961)MLE 10878 (00934) 11175 (01220) 11164 (01237) 11359 (01596) 11034 (01464)IME 10631 (00838) 10894 (01098) 10858 (01089) 11025 (01412) 10703 (01315)MIME 10294 (00739) 10516 (00951) 10456 (00937) 10591 (01207) 10267 (01139)

50

MOM 11363 (01557) 11396 (01767) 11570 (02069) 11747 (02296) 11842 (02652)MLE 10746 (00701) 10971 (01027) 11064 (01102) 10979 (01059) 11064 (01289)IME 10553 (00648) 10747 (00922) 10823 (00994) 10726 (00960) 10783 (01145)MIME 10288 (00584) 10451 (00822) 10505 (00879) 10394 (00849) 10432 (01009)

60

MOM 11233 (01227) 11281 (01249) 11480 (01679) 11312 (01608) 11511 (01892)MLE 10592 (00587) 10646 (00601) 10663 (00695) 10832 (00849) 10877 (00911)IME 10431 (00546) 10468 (00548) 10470 (00637) 10625 (00786) 10656 (00827)MIME 10214 (00502) 10232 (00499) 10218 (00580) 10354 (00709) 10371 (00743)

80

MOM 10954 (00843) 10836 (00856) 10984 (00974) 11136 (01105) 11220 (01224)MLE 10579 (00426) 10556 (00480) 10624 (00553) 10533 (00515) 10700 (00651)IME 10459 (00400) 10426 (00452) 10485 (00517) 10385 (00484) 10539 (00612)MIME 10297 (00372) 10251 (00421) 10296 (00478) 10189 (00449) 10329 (00563)

100

MOM 10704 (00659) 10704 (00736) 10835 (00859) 10763 (00888) 10780 (00851)MLE 10332 (00299) 10383 (00344) 10504 (00428) 10344 (00370) 10478 (00482)IME 10244 (00288) 10286 (00330) 10393 (00407) 10230 (00356) 10353 (00459)MIME 10119 (00274) 10149 (00313) 10245 (00382) 10077 (00337) 10189 (00432)

where 119880 follows uniform distribution over [0 1] and 119882(119886)giving the principal solution for 119908 in 119886 = 119908119890119908 is pronouncedas Lambert119882 function see Jodra [16]

We obtain MLE by solving (8) and MLE by (7) MOM andMOM can be obtained by solving (11) and (12) simultaneouslyIME and MIME can be obtained by solving (20) and (21)respectively IME and MIME can be obtained from (15)

We consider sample sizes 119899 = 30 40 50 60 80 100 and120572 = 20 25 30 35 40 We take 120582 = 2 in all our computa-tions For each combination of sample size 119899 and parameter120572 we generate a sample of size 119899 from GLD(120582 = 2 120572) andestimate the parameters120582 and120572 by theMLEMOM IME andMIMEmethodsThe average values of 120572 and 2 as well asthe correspondingMSEs over 1000 replications are computedand reported

Table 1 reports the average values of 120572 and the corre-sponding MSE is reported within parenthesis Figures 1(a)1(b) 1(c) and 1(d) show the relative biases and the MSEs ofthe four estimators of 120572 for sample sizes 119899 = 40 and 119899 = 80Figures 1(e) and 1(f) show the relative biases and the MSEsof the four estimators of 120572 for 120572 = 30 The other cases aresimilar

Table 2 reports the average values of 120582 = 2 and thecorresponding MSE is reported within parenthesis Figures2(a) 2(b) 2(c) and 2(d) show the relative biases and theMSEsof the four estimators of 120582 for sample sizes 119899 = 40 and 119899 = 80Figures 2(e) and 2(f) show the relative biases and the MSEs

of the four estimators of 120582 for 120572 = 30 The other cases aresimilar

From Tables 1 and 2 it is observed that for the fourmethods the average relative biases and the average relativeMSEs decrease as sample size 119899 increases as expected Theasymptotic unbiasedness of all the estimators is verified TheaverageMSEs of 120572 and 120582 = 2 depend on the parameter120572 For the four methods the average relative MSEs of 2decrease as 120572 goes up The average relative MSEs of 120572increase as120572 goes up Considering onlyMSEs we can observethat the estimation of 120572rsquos is more accurate for smaller valueswhile the estimation of 120582rsquos is more accurate for larger valuesof 120572 MOM MLE and IME overestimate both of the twoparameters 120572 and 120582 MIME overestimates only 120572

As far as the biases and MSEs are concerned it is clearthat MIME works the best in all the cases considered forestimating the two parameters Its performance is followedby IME MLE and MOM especially for small sample sizesThe four methods are close for larger sample sizes

Considering all the points MIME is recommended forestimating both parameters of the GLD(120582 120572) distributionMOM is not suggested

62 Comparison of the Two Joint Confidence Regions InSection 5 two methods to construct the confidence regionsof the two parameters 120582 and 120572 are proposed In this sectionwe conduct simulations to compare the two methods

8 Mathematical Problems in Engineering

Rela

tive b

ias

MLEIME

MIMEMOM

035

030

025

020

015

010

005

000

120572

20 25 30 35 40

(a) Relative biases (119899 = 40)

MLEIME

MIMEMOM

Rela

tive M

SEs

120572

20 25 30 35 40

05

04

03

02

01

00

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

015

010

005

000

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

025

020

015

010

005

000

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

Rela

tive M

SEs

035

030

025

020

015

010

005

000

n

MLEIME

MIMEMOM

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 1 Average relative biases and MSEs of 120572

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

Lindley distribution reduces to the one parameter Lindleydistribution

Singh et al [12] developed the Bayesian estimation forthe generalized Lindley distribution under squared error andgeneral entropy loss functions in case of complete sampleof observations Singh et al [13] considered the generalizedLindley distribution and proposed the progressive type IIcensoring scheme which allows the removal of the live unitsfrom a life-test with beta-binomial probability law during theexecution of the experiment

Nadarajah et al [11] considered the classical maximum-likelihood estimation of the parameters of a generalized Lind-ley distribution The results showed that the bias is not satis-fied especially for a small or even moderate sample size Asfor the moment estimates two nonlinear equations need tobe solved simultaneously and the existence and uniqueness ofthe roots are not clear and guaranteed

In this paper we consider the problem of estimatingthe two parameters of the generalized Lindley distributionWe propose inverse moment and modified inverse momentestimators and study their properties The conditions of theexistence and uniqueness of the estimators are establishedMonte Carlo simulations are used to compare the perfor-mances of the estimatorsWe also investigate themethods forconstructing joint confidence regions for the two parametersand study their performances

The rest of this paper is organized as follows In Section 2we briefly review the classical maximum-likelihood estima-tion of the parameters of the generalized Lindley distributionIn Section 3 the moment estimator is discussed In Section 4we propose two new methods of estimating the parametersand study their properties Joint confidence regions for thetwo parameters are proposed in Section 5 Section 6 conductssimulations to assess the methods Finally in Section 7 a realexample is presented to illustrate the proposed methods

2 Maximum-Likelihood Estimation

In this section we briefly review the MLEs of the parametersof GLD distribution Let 119883

1 1198832 119883

119899be a random sample

from GLD(120582 120572) with pdf and cdf as (3) and (2) respectivelyThe log-likelihood function is given by

119871 (120582 120572) = (120572 minus 1)119899

sum119894=1

log[1 minus119890minus120582119909119894 (120582119909

119894+ 120582 + 1)

120582 + 1]

minus 120582119899

sum119894=1

119909119894+119899

sum119894=1

log (119909119894+ 1) + 119899 log (120572)

+ 2119899 log (120582) minus 119899 log (120582 + 1)

(4)

The score equations are thus as follows120597119871 (120582 120572)

120597120582

= minus (120572 minus 1)119899

sum119894=1

120582119909119894((120582 + 1) 119909

119894+ 120582 + 2)

(120582 + 1) ((120582 + 1) (1 minus 119890120582119909119894) + 120582119909119894)

minus119899

sum119894=1

119909119894+119899 (120582 + 2)

120582 (1 + 120582)

(5)

120597119871 (120582 120572)

120597120572=119899

sum119894=1

log[1 minus119890minus120582119909119894 (120582119909

119894+ 120582 + 1)

120582 + 1] +

119899

120572 (6)

From (6) we obtain the MLE of 120572 as a function of 120582

= minus119899

sum119899

119894=1log [1 minus 119890minus120582119909119894 (120582119909

119894+ 120582 + 1) (120582 + 1)]

(7)

The MLE of 120582 is the root of the following equation

119866 (120582) =119899

sum119899

119894=1log [1 minus 119890minus120582119909119894 (120582119909

119894+ 120582 + 1) (120582 + 1)]

sdot119899

sum119894=1

120582119909119894((120582 + 1) 119909

119894+ 120582 + 2)

(120582 + 1) ((120582 + 1) (1 minus 119890120582119909119894) + 120582119909119894)

+119899

sum119894=1

120582119909119894((120582 + 1) 119909

119894+ 120582 + 2)

(120582 + 1) ((120582 + 1) (1 minus 119890120582119909119894) + 120582119909119894)minus119899

sum119894=1

119909119894

+119899 (120582 + 2)

120582 (1 + 120582)= 0

(8)

Such nonlinear equation does not have closed formsolution We can apply numerical method such as Newton-Raphson method to compute 120582

3 Moment Estimation of Parameters

In this section we discuss the moment estimation (MOM)of the parameters of GLD distribution Let119883

1 1198832 119883

119899be

a random sample from GLD(120582 120572) with pdf and cdf as (3)and (2) respectively 119909

1 1199092 119909

119899are the observed values

Let 1198981

= (1119899)sum119899

119894=1119909119894 1198982

= (1119899)sum119899

119894=11199092119894 1198981 1198982are

sample moments For the population moments we need thefollowing lemma (Nadarajah et al [11])

Lemma 1 Let119870(120572 120582 119899) = intinfin

0119909119899(119909+1)119890minus120582119909(1minus119890minus120582119909(120582+120582119909+

1)(120582 + 1))120572minus1119889119909 One has

119870 (120572 120582 119899) =infin

sum119894=0

119894

sum119895=0

119895+1

sum119896=0

(120572 minus 1

119894)(

119894

119895)(

119895 + 1

119896)

sdot(minus1)119894 120582119895Γ (119899 + 1 + 119896)

(1 + 120582)119895 (120582119894 + 120582)119899+1+119896

(9)

Let119883 denote a GLD random variable It follows that

E119883119899 =1205721205822

1 + 120582119870 (120572 120582 119899) (10)

By equating the population moments with the samplemoments we obtain

1205721205822

1 + 120582119870 (120572 120582 1) = 119898

1 (11)

1205721205822

1 + 120582119870 (120572 120582 2) = 119898

2 (12)

The method of moments estimators is the roots of the twoequations Similar to the MLEs such nonlinear equationsdo not have closed form solutions We can apply numericalmethod such as Newton-Raphson method to determine theroots

Mathematical Problems in Engineering 3

4 Inverse Moment Estimation of Parameters

Unlike the regularmethod ofmoments the idea of the inversemoment estimation (IME) is as follows for a given randomsample119883

1 119883

119899from a distribution with unknown param-

eters first transform the original sample to a quasisample1198841 119884

119899 where 119884

119894contains the unknown parameters but its

distribution does not depend on the unknown parametersthat is 119884

119894is a pivot variable 119894 = 1 119899 The population

moments of the new sample do not depend on the unknownparameters while the sample moments do Let the populationmoments of the quasisample equal the sample moments andsolve for the unknown parameters

Let 1198831 119883

119899form a sample from GLD(120582 120572) with pdf

given in (3) it is known that 119865(119883119894) 119894 = 1 119899 follow

the uniform distribution 119880(0 1) and thus minus log119865(119883119894) 119894 =

1 119899 follow standard exponential distribution Exp(1) Bythe method of inverse moment estimation we let

1

119899

119899

sum119894=1

[minus log119865 (119883119894)] = 1 (13)

that is

minus120572

119899

119899

sum119894=1

log[1 minus119890minus120582119909119894 (120582119909

119894+ 120582 + 1)

120582 + 1] = 1 (14)

Thus the IME of 120572 is obtained as a function of 120582

= minus119899

sum119899

119894=1log [1 minus 119890minus120582119909119894 (120582119909

119894+ 120582 + 1) (120582 + 1)]

(15)

which is identical to the MLE of 120572 In the following wedetermine the IME of 120582

Lemma 2 Let 119885(1)

le 119885(2)

le sdot sdot sdot le 119885(119899)

be the order statisticsfrom the standard exponential distribution Then the randomvariables119882

11198822 119882

119899 where

119882119894= (119899 minus 119894 + 1) (119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899 (16)

with 119885(0)

equiv 0 are independent and follow standard exponen-tial distributions

Proof The proof can be found by Arnold et al [14]

Lemma 3 Let 11988211198822 119882

119899be iid standard exponential

variables 119878119894= 1198821+ sdot sdot sdot +119882

119894119880119894= (119878119894119878119894+1)119894 119894 = 1 2 119899minus1

119880119899= 1198821+ sdot sdot sdot + 119882

119899 then

(1) 1198801 1198802 119880

119899are independent

(2) 1198801 1198802 119880

119899minus1follow the uniform distribution

119880(0 1)

(3) 2119880119899follows 1205942(2119899)

Proof The proof can be found by Wang [15]

For the sample 1198831 119883

119899from GLD(120582 120572) considering

the order statistics119883(1)

le sdot sdot sdot le 119883(119899) we have that

minus log119865 (119883(119899)) le sdot sdot sdot le minus log119865 (119883

(1)) (17)

are 119899-order statistics from standard exponential distributionLet119885(119894)= minus120572 log[1minus119890minus120582119883(119899minus119894+1)(120582119883

(119899minus119894+1)+120582+1)(120582+1)] equiv

minus120572 log119866(119883(119899minus119894+1)

) 119894 = 1 119899 where 119866(119909) = 1 minus 119890minus120582119909(120582119909 +120582 + 1)(120582 + 1) Thus 119885

(1)le 119885(2)

le sdot sdot sdot le 119885(119899)

are the first119899-order statistics from the standard exponential distributionBy Lemma 2 119882

119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899form a sample from standard exponential distribution

Let 119878119894= 1198821+ sdot sdot sdot + 119882

119894 119880119894= (119878119894119878119894+1)119894 119894 = 1 2 119899 minus 1

and 119880119899= 1198821+ sdot sdot sdot + 119882

119899 by Lemma 3 we have

minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

) = 2119899minus1

sum119894=1

log(119878119899

119878119894

)

sim 1205942 (2119899 minus 2)

(18)

where

119878119899

119878119894

=1198821+ sdot sdot sdot + 119882

119899

1198821+ sdot sdot sdot + 119882

119894

=119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119899)

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119894minus1)

+ (119899 minus 119894 + 1) 119885(119894)

=log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

(19)

Note that the mean of 1205942(2119899 minus 2) is 2119899 minus 2 Thus we obtainan inverse moment equation for 120582 as follows

119899minus1

sum119894=1

log[log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] = 119899 minus 1 (20)

4 Mathematical Problems in Engineering

Solve the equation and we obtain the inverse estimate IMEof 120582 Plugging IME into (15) we obtain the inverse estimate

IME In addition considering that the mode of 1205942(2119899 minus 2) is2119899 minus 4 we can obtain a modified equation for 120582

119899minus1

sum119894=1

log[log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] = 119899 minus 2 (21)

Solve the equation and we obtain the modified inverseestimate MIME of 120582 Plugging MIME into (15) we obtainthe modified inverse estimate MIME Unlike the momentestimation herewe only need to solve one nonlinear equationinstead of two equations

In the following we prove the existence and uniquenessof the root in (20) and (21)

Lemma 4 The following limits hold

(1) One has

lim120582rarr0

log119866 (119886)

log119866 (119887)

= lim120582rarr0

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= 1

119891119900119903 119886 gt 0 119887 gt 0

(22)

(2) One has

lim120582rarrinfin

log119866 (119886)

log119866 (119887)

= lim120582rarrinfin

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= 0

119891119900119903 119886 gt 119887 gt 0

(23)

(3) One has

lim120582rarrinfin

log119866 (119886)

log119866 (119887)

= lim120582rarrinfin

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= +infin 119891119900119903 119887 gt 119886 gt 0

(24)

Lemma 5 For 119905 gt 0 119891(119905) = (120582 + 119905 + 2 minus 119890119905[120582 minus (120582 + 1)119905 +2])(120582 minus (120582 + 1)119890119905 + 119905 + 1) is a decreasing function of 119905

Theorem 6 Let119882119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899form a sample from standard exponential distribution 119878

119894=

1198821+ sdot sdot sdot + 119882

119894 then for 119905 gt 0 equation sum

119899minus1

119894=1log(119878119899119878119894) = 119905

has a unique positive solution

Proof By Lemma 4 we obtain

lim120582rarr0

119878119899

119878119894

= lim120582rarr0

1198821+ sdot sdot sdot + 119882

119899

1198821+ sdot sdot sdot + 119882

119894

= lim120582rarr0

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119899)

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119894minus1)

+ (119899 minus 119894 + 1) 119885(119894)

= lim120582rarr0

log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

= lim120582rarr0

[log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))] log119866 (119883

(119899))

[log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899))=119899

119899= 1

(25)

Thus lim120582rarr0

sum119899minus1

119894=1log(119878119899119878119894) = 0 On the other hand

lim120582rarrinfin

119878119899

119878119894

= lim120582rarrinfin

log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

= 1 + lim120582rarrinfin

log119866 (119883(119899minus119894)

) + sdot sdot sdot + log119866 (119883(1)) minus (119899 minus 119894) log119866 (119883

(119899minus119894+1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+1)) + (119899 minus 119894) log119866 (119883

(119899minus119894+1))

= 1 + lim120582rarrinfin

[log119866 (119883(119899minus119894)

) + sdot sdot sdot + log119866 (119883(1)) minus (119899 minus 119894) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899minus119894+1))

[log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+1)) + (119899 minus 119894) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899minus119894+1))= +infin

(26)

Mathematical Problems in Engineering 5

Thus lim120582rarrinfin

sum119899minus1

119894=1log(119878119899119878119894) = infin Therefore for 119905 gt 0

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has one positive solution For

the uniqueness of the solution we consider the derivative of119878119899119878119894with respect to 120582 Note that for 119894 = 1 119899

119882119894= (119899 minus 119894 + 1) 120572 [log119866 (119883

(119899minus119894+2)) minus log119866 (119883

(119899minus119894+1))]

119889119882119894

119889120582= (119899 minus 119894 + 1) 120572

120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+2)

)minus120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+1)

) = 119882

119894

sdot120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+2)

) minus 120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+1)

)

log119866 (119883(119899minus119894+2)

) minus log119866 (119883(119899minus119894+1)

)

(119878119899

119878119894

)1015840

= (1 +119882119894+1

+ sdot sdot sdot + 119882119899

1198821+ sdot sdot sdot + 119882

119894

)1015840

=1

(sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

[1198821015840119895119882119896minus1198821198951198821015840119896] =

1

120582 (sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

119882119895119882119896 [119860 (120582) minus 119861 (120582)]

(27)

where

119860 (120582)

=120582119883(119899minus119895+2)

119890minus120582119883(119899minus119895+2)120582 [(120582 + 1)119883(119899minus119895+2)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+2)

) minus 120582119883(119899minus119895+1)

119890minus120582119883(119899minus119895+1)120582 [(120582 + 1)119883(119899minus119895+1)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+1)

)

log119866(119883(119899minus119895+2)

) minus log119866(119883(119899minus119895+1)

)

119861 (120582)

=120582119883(119899minus119896+2)

119890minus120582119883(119899minus119896+2)120582 [(120582 + 1)119883(119899minus119896+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+2)

) minus 120582119883(119899minus119896+1)

119890minus120582119883(119899minus119896+1)120582 [(120582 + 1)119883(119899minus119896+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+1)

)

log119866 (119883(119899minus119896+2)

) minus log119866 (119883(119899minus119896+1)

)

(28)

By Cauchyrsquos mean-value theorem for 119895 = 119894 + 1 119899 119896 =1 119894 there exist 120585

1isin (120582119883

(119899minus119895+1) 120582119883(119899minus119895+2)

) and 1205852

isin(120582119883(119899minus119896+1)

120582119883(119899minus119896+2)

) such that

119860 (120582) =120582 + 1205851+ 2 minus 119890120585

1[120582 minus (120582 + 1) 120585

1+ 2]

120582 minus (120582 + 1) 119890120585

1+ 1205851+ 1

119861 (120582) =120582 + 1205852+ 2 minus 119890120585

2[120582 minus (120582 + 1) 120585

2+ 2]

120582 minus (120582 + 1) 119890120585

2+ 1205852+ 1

(29)

Note that 1205851lt 1205852 by Lemma 5 119860(120582) minus 119861(120582) gt 0 (119878

119899119878119894)1015840 gt 0

thussum119899minus1119894=1

log(119878119899119878119894) is a strictly increasing function of 120582 and

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has a unique positive solution

5 Joint Confidence Regions for 120582 and 120572

Let 1198831 1198832 119883

119899form a sample from GLD(120582 120572) and

119883(1)

le 119883(2)

le sdot sdot sdot le 119883(119899)

are the order statistics Let119885(119894)

= minus120572 log[1 minus 119890minus120582119883(119899minus119894+1)(120582119883(119899minus119894+1)

+ 120582 + 1)(120582 + 1)] =minus120572 log119866(119883

(119899minus119894+1)) 119894 = 1 119899Thus 119885

(1)le 119885(2)

le sdot sdot sdot le 119885(119899)

are the first 119899-order statistics from the standard exponentialdistribution By Lemma 2 119882

119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

)119894 = 1 2 119899 form a sample from standard exponential

distribution Let 119878119894= 1198821+ sdot sdot sdot + 119882

119894 119880119894= (119878119894119878119894+1)119894 119894 =

1 2 119899 minus 1 and 119880119899= 1198821+ sdot sdot sdot + 119882

119899 Hence

119881 = 21198781= 21198821= 2119899119885

(1)

= minus2119899120572 log[1 minus119890minus120582119883(119899) (120582119883

(119899)+ 120582 + 1)

120582 + 1]

sim 1205942 (2)

119880 = 2 (119878119899minus 1198781) = 2

119899

sum119894=2

119882119894

= 2 [119885(1)

+ sdot sdot sdot + 119885(119899)

minus 119899119885(1)] sim 1205942 (2119899 minus 2)

(30)

It is obvious that 119880 and 119881 are independent Define

1198791=119880 (2119899 minus 2)

1198812=

119878119899minus 1198781

(119899 minus 1) 1198781

sim 119865 (2119899 minus 2 2)

1198792= 119880 + 119881 = 2119878

119899sim 1205942 (2119899)

(31)

We obtain that 1198791and 119879

2are independent using the known

bank-post office story in statisticsLet 119865120574(V1 V2) denote the percentile of 119865 distribution with

left-tail probability 120574 and V1and V

2degrees of freedom Let

1205942120574(V) denote the percentile of 1205942 distribution with left-tail

probability 120574 and V degrees of freedom

6 Mathematical Problems in Engineering

By using the pivotal variables1198791and1198792 a joint confidence

region for the two parameters 120582 and 120572 can be constructed asfollows

Theorem 7 (method 1) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198791and1198792

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582119871le 120582 le 120582

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(32)

where 120582119871is the root of 120582 for the equation 119879

1= 119865(1minusradic1minus120574)2

(2119899minus

2 2) and 120582119880

is the root of 120582 for the equation 1198791

=119865(1+radic1minus120574)2

(2119899 minus 2 2)

Proof 1198791= (1(119899 minus 1)) log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 +

1)] + sdot sdot sdot + log[1 minus 119890minus120582119883(1)(120582119883(1)

+ 120582 + 1)(120582 + 1)] minus 119899 log[1 minus119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 + 1)]119899 log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 +

1)(120582+1)] is a function of 120582 and does not depend on 120572 FromTheorem 6 we have lim

120582rarr01198791= (1(119899 minus 1))lim

120582rarr0(1198781198991198781minus

1) = 0 lim120582rarrinfin

1198791= (1(119899 minus 1))lim

120582rarrinfin(1198781198991198781minus 1) = infin

and 11987910158401= (1(119899 minus 1))(119878

119899119878119894)1015840 gt 0 Therefore for any 119905 gt 0

equation 1198791= 119905 has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2)

le 1198791le 119865(1+radic1minus120574)2

(2119899 minus 2 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899)

le 1198792le 1205942(1+radic1minus120574)2

(2119899))

= 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2) le 1198791

le 119865(1+radic1minus120574)2

(2119899 minus 2 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582119871le 120582

le 120582119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(33)

On the other hand by Lemma 3 we have

1198793= minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

)

= 2119899minus1

sum119894=1

log(119878119899

119878119894

) sim 1205942 (2119899 minus 2)

(34)

1198792and119879

3are also independent By using the pivotal variables

1198792and 119879

3 a joint confidence region for the two parameters 120582

and 120572 can be constructed as follows

Theorem 8 (method 2) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198792and1198793

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582lowast119871le 120582 le 120582lowast

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(35)

where 120582lowast119871is the root of 120582 for the equation 119879

3= 1205942(1minusradic1minus120574)2

(2119899minus

2) and 120582lowast119880is the root of 120582 for the equation119879

3= 1205942(1+radic1minus120574)2

(2119899minus

2)

Proof 1198793= 2sum

119899minus1

119894=1log(119878119899119878119894) is a function of 120582 and does not

depend on 120572 FromTheorem 6 for any 119904 gt 0 equation 1198793= 119904

has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582lowast119871le 120582

le 120582lowast119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(36)

6 Simulation Study

61 Comparison of the Four Estimation Methods In this sec-tion we conduct simulations to compare the performancesof the MIMEs IMEs MLEs and MOMs mainly with respectto their biases and mean squared errors (MSEs) for varioussample sizes and for various true parametric values

The random data 119883 from the GLD(120582 120572) distribution canbe generated as follows

119883 =minus120582 minus 1 minus119882(119890minus120582minus1 (120582 + 1) (1198801120572 minus 1))

120582 (37)

Mathematical Problems in Engineering 7

Table 1 Average relative estimates and MSEs of 120572

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 12705 (04298) 12350 (03510) 12965 (05115) 12770 (04460) 13522 (06256)MLE 11310 (01459) 11703 (02029) 11590 (02240) 11693 (02267) 11911 (02888)IME 10956 (01267) 11292 (01728) 11154 (01906) 11234 (01930) 11395 (02400)MIME 10482 (01063) 10755 (01420) 10590 (01570) 10635 (01572) 10754 (01939)

40

MOM 11791 (02212) 11954 (02262) 12046 (02693) 12299 (03163) 12536 (03961)MLE 10878 (00934) 11175 (01220) 11164 (01237) 11359 (01596) 11034 (01464)IME 10631 (00838) 10894 (01098) 10858 (01089) 11025 (01412) 10703 (01315)MIME 10294 (00739) 10516 (00951) 10456 (00937) 10591 (01207) 10267 (01139)

50

MOM 11363 (01557) 11396 (01767) 11570 (02069) 11747 (02296) 11842 (02652)MLE 10746 (00701) 10971 (01027) 11064 (01102) 10979 (01059) 11064 (01289)IME 10553 (00648) 10747 (00922) 10823 (00994) 10726 (00960) 10783 (01145)MIME 10288 (00584) 10451 (00822) 10505 (00879) 10394 (00849) 10432 (01009)

60

MOM 11233 (01227) 11281 (01249) 11480 (01679) 11312 (01608) 11511 (01892)MLE 10592 (00587) 10646 (00601) 10663 (00695) 10832 (00849) 10877 (00911)IME 10431 (00546) 10468 (00548) 10470 (00637) 10625 (00786) 10656 (00827)MIME 10214 (00502) 10232 (00499) 10218 (00580) 10354 (00709) 10371 (00743)

80

MOM 10954 (00843) 10836 (00856) 10984 (00974) 11136 (01105) 11220 (01224)MLE 10579 (00426) 10556 (00480) 10624 (00553) 10533 (00515) 10700 (00651)IME 10459 (00400) 10426 (00452) 10485 (00517) 10385 (00484) 10539 (00612)MIME 10297 (00372) 10251 (00421) 10296 (00478) 10189 (00449) 10329 (00563)

100

MOM 10704 (00659) 10704 (00736) 10835 (00859) 10763 (00888) 10780 (00851)MLE 10332 (00299) 10383 (00344) 10504 (00428) 10344 (00370) 10478 (00482)IME 10244 (00288) 10286 (00330) 10393 (00407) 10230 (00356) 10353 (00459)MIME 10119 (00274) 10149 (00313) 10245 (00382) 10077 (00337) 10189 (00432)

where 119880 follows uniform distribution over [0 1] and 119882(119886)giving the principal solution for 119908 in 119886 = 119908119890119908 is pronouncedas Lambert119882 function see Jodra [16]

We obtain MLE by solving (8) and MLE by (7) MOM andMOM can be obtained by solving (11) and (12) simultaneouslyIME and MIME can be obtained by solving (20) and (21)respectively IME and MIME can be obtained from (15)

We consider sample sizes 119899 = 30 40 50 60 80 100 and120572 = 20 25 30 35 40 We take 120582 = 2 in all our computa-tions For each combination of sample size 119899 and parameter120572 we generate a sample of size 119899 from GLD(120582 = 2 120572) andestimate the parameters120582 and120572 by theMLEMOM IME andMIMEmethodsThe average values of 120572 and 2 as well asthe correspondingMSEs over 1000 replications are computedand reported

Table 1 reports the average values of 120572 and the corre-sponding MSE is reported within parenthesis Figures 1(a)1(b) 1(c) and 1(d) show the relative biases and the MSEs ofthe four estimators of 120572 for sample sizes 119899 = 40 and 119899 = 80Figures 1(e) and 1(f) show the relative biases and the MSEsof the four estimators of 120572 for 120572 = 30 The other cases aresimilar

Table 2 reports the average values of 120582 = 2 and thecorresponding MSE is reported within parenthesis Figures2(a) 2(b) 2(c) and 2(d) show the relative biases and theMSEsof the four estimators of 120582 for sample sizes 119899 = 40 and 119899 = 80Figures 2(e) and 2(f) show the relative biases and the MSEs

of the four estimators of 120582 for 120572 = 30 The other cases aresimilar

From Tables 1 and 2 it is observed that for the fourmethods the average relative biases and the average relativeMSEs decrease as sample size 119899 increases as expected Theasymptotic unbiasedness of all the estimators is verified TheaverageMSEs of 120572 and 120582 = 2 depend on the parameter120572 For the four methods the average relative MSEs of 2decrease as 120572 goes up The average relative MSEs of 120572increase as120572 goes up Considering onlyMSEs we can observethat the estimation of 120572rsquos is more accurate for smaller valueswhile the estimation of 120582rsquos is more accurate for larger valuesof 120572 MOM MLE and IME overestimate both of the twoparameters 120572 and 120582 MIME overestimates only 120572

As far as the biases and MSEs are concerned it is clearthat MIME works the best in all the cases considered forestimating the two parameters Its performance is followedby IME MLE and MOM especially for small sample sizesThe four methods are close for larger sample sizes

Considering all the points MIME is recommended forestimating both parameters of the GLD(120582 120572) distributionMOM is not suggested

62 Comparison of the Two Joint Confidence Regions InSection 5 two methods to construct the confidence regionsof the two parameters 120582 and 120572 are proposed In this sectionwe conduct simulations to compare the two methods

8 Mathematical Problems in Engineering

Rela

tive b

ias

MLEIME

MIMEMOM

035

030

025

020

015

010

005

000

120572

20 25 30 35 40

(a) Relative biases (119899 = 40)

MLEIME

MIMEMOM

Rela

tive M

SEs

120572

20 25 30 35 40

05

04

03

02

01

00

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

015

010

005

000

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

025

020

015

010

005

000

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

Rela

tive M

SEs

035

030

025

020

015

010

005

000

n

MLEIME

MIMEMOM

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 1 Average relative biases and MSEs of 120572

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

4 Inverse Moment Estimation of Parameters

Unlike the regularmethod ofmoments the idea of the inversemoment estimation (IME) is as follows for a given randomsample119883

1 119883

119899from a distribution with unknown param-

eters first transform the original sample to a quasisample1198841 119884

119899 where 119884

119894contains the unknown parameters but its

distribution does not depend on the unknown parametersthat is 119884

119894is a pivot variable 119894 = 1 119899 The population

moments of the new sample do not depend on the unknownparameters while the sample moments do Let the populationmoments of the quasisample equal the sample moments andsolve for the unknown parameters

Let 1198831 119883

119899form a sample from GLD(120582 120572) with pdf

given in (3) it is known that 119865(119883119894) 119894 = 1 119899 follow

the uniform distribution 119880(0 1) and thus minus log119865(119883119894) 119894 =

1 119899 follow standard exponential distribution Exp(1) Bythe method of inverse moment estimation we let

1

119899

119899

sum119894=1

[minus log119865 (119883119894)] = 1 (13)

that is

minus120572

119899

119899

sum119894=1

log[1 minus119890minus120582119909119894 (120582119909

119894+ 120582 + 1)

120582 + 1] = 1 (14)

Thus the IME of 120572 is obtained as a function of 120582

= minus119899

sum119899

119894=1log [1 minus 119890minus120582119909119894 (120582119909

119894+ 120582 + 1) (120582 + 1)]

(15)

which is identical to the MLE of 120572 In the following wedetermine the IME of 120582

Lemma 2 Let 119885(1)

le 119885(2)

le sdot sdot sdot le 119885(119899)

be the order statisticsfrom the standard exponential distribution Then the randomvariables119882

11198822 119882

119899 where

119882119894= (119899 minus 119894 + 1) (119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899 (16)

with 119885(0)

equiv 0 are independent and follow standard exponen-tial distributions

Proof The proof can be found by Arnold et al [14]

Lemma 3 Let 11988211198822 119882

119899be iid standard exponential

variables 119878119894= 1198821+ sdot sdot sdot +119882

119894119880119894= (119878119894119878119894+1)119894 119894 = 1 2 119899minus1

119880119899= 1198821+ sdot sdot sdot + 119882

119899 then

(1) 1198801 1198802 119880

119899are independent

(2) 1198801 1198802 119880

119899minus1follow the uniform distribution

119880(0 1)

(3) 2119880119899follows 1205942(2119899)

Proof The proof can be found by Wang [15]

For the sample 1198831 119883

119899from GLD(120582 120572) considering

the order statistics119883(1)

le sdot sdot sdot le 119883(119899) we have that

minus log119865 (119883(119899)) le sdot sdot sdot le minus log119865 (119883

(1)) (17)

are 119899-order statistics from standard exponential distributionLet119885(119894)= minus120572 log[1minus119890minus120582119883(119899minus119894+1)(120582119883

(119899minus119894+1)+120582+1)(120582+1)] equiv

minus120572 log119866(119883(119899minus119894+1)

) 119894 = 1 119899 where 119866(119909) = 1 minus 119890minus120582119909(120582119909 +120582 + 1)(120582 + 1) Thus 119885

(1)le 119885(2)

le sdot sdot sdot le 119885(119899)

are the first119899-order statistics from the standard exponential distributionBy Lemma 2 119882

119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899form a sample from standard exponential distribution

Let 119878119894= 1198821+ sdot sdot sdot + 119882

119894 119880119894= (119878119894119878119894+1)119894 119894 = 1 2 119899 minus 1

and 119880119899= 1198821+ sdot sdot sdot + 119882

119899 by Lemma 3 we have

minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

) = 2119899minus1

sum119894=1

log(119878119899

119878119894

)

sim 1205942 (2119899 minus 2)

(18)

where

119878119899

119878119894

=1198821+ sdot sdot sdot + 119882

119899

1198821+ sdot sdot sdot + 119882

119894

=119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119899)

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119894minus1)

+ (119899 minus 119894 + 1) 119885(119894)

=log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

(19)

Note that the mean of 1205942(2119899 minus 2) is 2119899 minus 2 Thus we obtainan inverse moment equation for 120582 as follows

119899minus1

sum119894=1

log[log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] = 119899 minus 1 (20)

4 Mathematical Problems in Engineering

Solve the equation and we obtain the inverse estimate IMEof 120582 Plugging IME into (15) we obtain the inverse estimate

IME In addition considering that the mode of 1205942(2119899 minus 2) is2119899 minus 4 we can obtain a modified equation for 120582

119899minus1

sum119894=1

log[log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] = 119899 minus 2 (21)

Solve the equation and we obtain the modified inverseestimate MIME of 120582 Plugging MIME into (15) we obtainthe modified inverse estimate MIME Unlike the momentestimation herewe only need to solve one nonlinear equationinstead of two equations

In the following we prove the existence and uniquenessof the root in (20) and (21)

Lemma 4 The following limits hold

(1) One has

lim120582rarr0

log119866 (119886)

log119866 (119887)

= lim120582rarr0

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= 1

119891119900119903 119886 gt 0 119887 gt 0

(22)

(2) One has

lim120582rarrinfin

log119866 (119886)

log119866 (119887)

= lim120582rarrinfin

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= 0

119891119900119903 119886 gt 119887 gt 0

(23)

(3) One has

lim120582rarrinfin

log119866 (119886)

log119866 (119887)

= lim120582rarrinfin

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= +infin 119891119900119903 119887 gt 119886 gt 0

(24)

Lemma 5 For 119905 gt 0 119891(119905) = (120582 + 119905 + 2 minus 119890119905[120582 minus (120582 + 1)119905 +2])(120582 minus (120582 + 1)119890119905 + 119905 + 1) is a decreasing function of 119905

Theorem 6 Let119882119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899form a sample from standard exponential distribution 119878

119894=

1198821+ sdot sdot sdot + 119882

119894 then for 119905 gt 0 equation sum

119899minus1

119894=1log(119878119899119878119894) = 119905

has a unique positive solution

Proof By Lemma 4 we obtain

lim120582rarr0

119878119899

119878119894

= lim120582rarr0

1198821+ sdot sdot sdot + 119882

119899

1198821+ sdot sdot sdot + 119882

119894

= lim120582rarr0

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119899)

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119894minus1)

+ (119899 minus 119894 + 1) 119885(119894)

= lim120582rarr0

log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

= lim120582rarr0

[log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))] log119866 (119883

(119899))

[log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899))=119899

119899= 1

(25)

Thus lim120582rarr0

sum119899minus1

119894=1log(119878119899119878119894) = 0 On the other hand

lim120582rarrinfin

119878119899

119878119894

= lim120582rarrinfin

log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

= 1 + lim120582rarrinfin

log119866 (119883(119899minus119894)

) + sdot sdot sdot + log119866 (119883(1)) minus (119899 minus 119894) log119866 (119883

(119899minus119894+1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+1)) + (119899 minus 119894) log119866 (119883

(119899minus119894+1))

= 1 + lim120582rarrinfin

[log119866 (119883(119899minus119894)

) + sdot sdot sdot + log119866 (119883(1)) minus (119899 minus 119894) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899minus119894+1))

[log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+1)) + (119899 minus 119894) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899minus119894+1))= +infin

(26)

Mathematical Problems in Engineering 5

Thus lim120582rarrinfin

sum119899minus1

119894=1log(119878119899119878119894) = infin Therefore for 119905 gt 0

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has one positive solution For

the uniqueness of the solution we consider the derivative of119878119899119878119894with respect to 120582 Note that for 119894 = 1 119899

119882119894= (119899 minus 119894 + 1) 120572 [log119866 (119883

(119899minus119894+2)) minus log119866 (119883

(119899minus119894+1))]

119889119882119894

119889120582= (119899 minus 119894 + 1) 120572

120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+2)

)minus120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+1)

) = 119882

119894

sdot120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+2)

) minus 120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+1)

)

log119866 (119883(119899minus119894+2)

) minus log119866 (119883(119899minus119894+1)

)

(119878119899

119878119894

)1015840

= (1 +119882119894+1

+ sdot sdot sdot + 119882119899

1198821+ sdot sdot sdot + 119882

119894

)1015840

=1

(sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

[1198821015840119895119882119896minus1198821198951198821015840119896] =

1

120582 (sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

119882119895119882119896 [119860 (120582) minus 119861 (120582)]

(27)

where

119860 (120582)

=120582119883(119899minus119895+2)

119890minus120582119883(119899minus119895+2)120582 [(120582 + 1)119883(119899minus119895+2)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+2)

) minus 120582119883(119899minus119895+1)

119890minus120582119883(119899minus119895+1)120582 [(120582 + 1)119883(119899minus119895+1)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+1)

)

log119866(119883(119899minus119895+2)

) minus log119866(119883(119899minus119895+1)

)

119861 (120582)

=120582119883(119899minus119896+2)

119890minus120582119883(119899minus119896+2)120582 [(120582 + 1)119883(119899minus119896+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+2)

) minus 120582119883(119899minus119896+1)

119890minus120582119883(119899minus119896+1)120582 [(120582 + 1)119883(119899minus119896+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+1)

)

log119866 (119883(119899minus119896+2)

) minus log119866 (119883(119899minus119896+1)

)

(28)

By Cauchyrsquos mean-value theorem for 119895 = 119894 + 1 119899 119896 =1 119894 there exist 120585

1isin (120582119883

(119899minus119895+1) 120582119883(119899minus119895+2)

) and 1205852

isin(120582119883(119899minus119896+1)

120582119883(119899minus119896+2)

) such that

119860 (120582) =120582 + 1205851+ 2 minus 119890120585

1[120582 minus (120582 + 1) 120585

1+ 2]

120582 minus (120582 + 1) 119890120585

1+ 1205851+ 1

119861 (120582) =120582 + 1205852+ 2 minus 119890120585

2[120582 minus (120582 + 1) 120585

2+ 2]

120582 minus (120582 + 1) 119890120585

2+ 1205852+ 1

(29)

Note that 1205851lt 1205852 by Lemma 5 119860(120582) minus 119861(120582) gt 0 (119878

119899119878119894)1015840 gt 0

thussum119899minus1119894=1

log(119878119899119878119894) is a strictly increasing function of 120582 and

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has a unique positive solution

5 Joint Confidence Regions for 120582 and 120572

Let 1198831 1198832 119883

119899form a sample from GLD(120582 120572) and

119883(1)

le 119883(2)

le sdot sdot sdot le 119883(119899)

are the order statistics Let119885(119894)

= minus120572 log[1 minus 119890minus120582119883(119899minus119894+1)(120582119883(119899minus119894+1)

+ 120582 + 1)(120582 + 1)] =minus120572 log119866(119883

(119899minus119894+1)) 119894 = 1 119899Thus 119885

(1)le 119885(2)

le sdot sdot sdot le 119885(119899)

are the first 119899-order statistics from the standard exponentialdistribution By Lemma 2 119882

119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

)119894 = 1 2 119899 form a sample from standard exponential

distribution Let 119878119894= 1198821+ sdot sdot sdot + 119882

119894 119880119894= (119878119894119878119894+1)119894 119894 =

1 2 119899 minus 1 and 119880119899= 1198821+ sdot sdot sdot + 119882

119899 Hence

119881 = 21198781= 21198821= 2119899119885

(1)

= minus2119899120572 log[1 minus119890minus120582119883(119899) (120582119883

(119899)+ 120582 + 1)

120582 + 1]

sim 1205942 (2)

119880 = 2 (119878119899minus 1198781) = 2

119899

sum119894=2

119882119894

= 2 [119885(1)

+ sdot sdot sdot + 119885(119899)

minus 119899119885(1)] sim 1205942 (2119899 minus 2)

(30)

It is obvious that 119880 and 119881 are independent Define

1198791=119880 (2119899 minus 2)

1198812=

119878119899minus 1198781

(119899 minus 1) 1198781

sim 119865 (2119899 minus 2 2)

1198792= 119880 + 119881 = 2119878

119899sim 1205942 (2119899)

(31)

We obtain that 1198791and 119879

2are independent using the known

bank-post office story in statisticsLet 119865120574(V1 V2) denote the percentile of 119865 distribution with

left-tail probability 120574 and V1and V

2degrees of freedom Let

1205942120574(V) denote the percentile of 1205942 distribution with left-tail

probability 120574 and V degrees of freedom

6 Mathematical Problems in Engineering

By using the pivotal variables1198791and1198792 a joint confidence

region for the two parameters 120582 and 120572 can be constructed asfollows

Theorem 7 (method 1) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198791and1198792

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582119871le 120582 le 120582

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(32)

where 120582119871is the root of 120582 for the equation 119879

1= 119865(1minusradic1minus120574)2

(2119899minus

2 2) and 120582119880

is the root of 120582 for the equation 1198791

=119865(1+radic1minus120574)2

(2119899 minus 2 2)

Proof 1198791= (1(119899 minus 1)) log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 +

1)] + sdot sdot sdot + log[1 minus 119890minus120582119883(1)(120582119883(1)

+ 120582 + 1)(120582 + 1)] minus 119899 log[1 minus119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 + 1)]119899 log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 +

1)(120582+1)] is a function of 120582 and does not depend on 120572 FromTheorem 6 we have lim

120582rarr01198791= (1(119899 minus 1))lim

120582rarr0(1198781198991198781minus

1) = 0 lim120582rarrinfin

1198791= (1(119899 minus 1))lim

120582rarrinfin(1198781198991198781minus 1) = infin

and 11987910158401= (1(119899 minus 1))(119878

119899119878119894)1015840 gt 0 Therefore for any 119905 gt 0

equation 1198791= 119905 has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2)

le 1198791le 119865(1+radic1minus120574)2

(2119899 minus 2 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899)

le 1198792le 1205942(1+radic1minus120574)2

(2119899))

= 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2) le 1198791

le 119865(1+radic1minus120574)2

(2119899 minus 2 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582119871le 120582

le 120582119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(33)

On the other hand by Lemma 3 we have

1198793= minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

)

= 2119899minus1

sum119894=1

log(119878119899

119878119894

) sim 1205942 (2119899 minus 2)

(34)

1198792and119879

3are also independent By using the pivotal variables

1198792and 119879

3 a joint confidence region for the two parameters 120582

and 120572 can be constructed as follows

Theorem 8 (method 2) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198792and1198793

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582lowast119871le 120582 le 120582lowast

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(35)

where 120582lowast119871is the root of 120582 for the equation 119879

3= 1205942(1minusradic1minus120574)2

(2119899minus

2) and 120582lowast119880is the root of 120582 for the equation119879

3= 1205942(1+radic1minus120574)2

(2119899minus

2)

Proof 1198793= 2sum

119899minus1

119894=1log(119878119899119878119894) is a function of 120582 and does not

depend on 120572 FromTheorem 6 for any 119904 gt 0 equation 1198793= 119904

has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582lowast119871le 120582

le 120582lowast119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(36)

6 Simulation Study

61 Comparison of the Four Estimation Methods In this sec-tion we conduct simulations to compare the performancesof the MIMEs IMEs MLEs and MOMs mainly with respectto their biases and mean squared errors (MSEs) for varioussample sizes and for various true parametric values

The random data 119883 from the GLD(120582 120572) distribution canbe generated as follows

119883 =minus120582 minus 1 minus119882(119890minus120582minus1 (120582 + 1) (1198801120572 minus 1))

120582 (37)

Mathematical Problems in Engineering 7

Table 1 Average relative estimates and MSEs of 120572

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 12705 (04298) 12350 (03510) 12965 (05115) 12770 (04460) 13522 (06256)MLE 11310 (01459) 11703 (02029) 11590 (02240) 11693 (02267) 11911 (02888)IME 10956 (01267) 11292 (01728) 11154 (01906) 11234 (01930) 11395 (02400)MIME 10482 (01063) 10755 (01420) 10590 (01570) 10635 (01572) 10754 (01939)

40

MOM 11791 (02212) 11954 (02262) 12046 (02693) 12299 (03163) 12536 (03961)MLE 10878 (00934) 11175 (01220) 11164 (01237) 11359 (01596) 11034 (01464)IME 10631 (00838) 10894 (01098) 10858 (01089) 11025 (01412) 10703 (01315)MIME 10294 (00739) 10516 (00951) 10456 (00937) 10591 (01207) 10267 (01139)

50

MOM 11363 (01557) 11396 (01767) 11570 (02069) 11747 (02296) 11842 (02652)MLE 10746 (00701) 10971 (01027) 11064 (01102) 10979 (01059) 11064 (01289)IME 10553 (00648) 10747 (00922) 10823 (00994) 10726 (00960) 10783 (01145)MIME 10288 (00584) 10451 (00822) 10505 (00879) 10394 (00849) 10432 (01009)

60

MOM 11233 (01227) 11281 (01249) 11480 (01679) 11312 (01608) 11511 (01892)MLE 10592 (00587) 10646 (00601) 10663 (00695) 10832 (00849) 10877 (00911)IME 10431 (00546) 10468 (00548) 10470 (00637) 10625 (00786) 10656 (00827)MIME 10214 (00502) 10232 (00499) 10218 (00580) 10354 (00709) 10371 (00743)

80

MOM 10954 (00843) 10836 (00856) 10984 (00974) 11136 (01105) 11220 (01224)MLE 10579 (00426) 10556 (00480) 10624 (00553) 10533 (00515) 10700 (00651)IME 10459 (00400) 10426 (00452) 10485 (00517) 10385 (00484) 10539 (00612)MIME 10297 (00372) 10251 (00421) 10296 (00478) 10189 (00449) 10329 (00563)

100

MOM 10704 (00659) 10704 (00736) 10835 (00859) 10763 (00888) 10780 (00851)MLE 10332 (00299) 10383 (00344) 10504 (00428) 10344 (00370) 10478 (00482)IME 10244 (00288) 10286 (00330) 10393 (00407) 10230 (00356) 10353 (00459)MIME 10119 (00274) 10149 (00313) 10245 (00382) 10077 (00337) 10189 (00432)

where 119880 follows uniform distribution over [0 1] and 119882(119886)giving the principal solution for 119908 in 119886 = 119908119890119908 is pronouncedas Lambert119882 function see Jodra [16]

We obtain MLE by solving (8) and MLE by (7) MOM andMOM can be obtained by solving (11) and (12) simultaneouslyIME and MIME can be obtained by solving (20) and (21)respectively IME and MIME can be obtained from (15)

We consider sample sizes 119899 = 30 40 50 60 80 100 and120572 = 20 25 30 35 40 We take 120582 = 2 in all our computa-tions For each combination of sample size 119899 and parameter120572 we generate a sample of size 119899 from GLD(120582 = 2 120572) andestimate the parameters120582 and120572 by theMLEMOM IME andMIMEmethodsThe average values of 120572 and 2 as well asthe correspondingMSEs over 1000 replications are computedand reported

Table 1 reports the average values of 120572 and the corre-sponding MSE is reported within parenthesis Figures 1(a)1(b) 1(c) and 1(d) show the relative biases and the MSEs ofthe four estimators of 120572 for sample sizes 119899 = 40 and 119899 = 80Figures 1(e) and 1(f) show the relative biases and the MSEsof the four estimators of 120572 for 120572 = 30 The other cases aresimilar

Table 2 reports the average values of 120582 = 2 and thecorresponding MSE is reported within parenthesis Figures2(a) 2(b) 2(c) and 2(d) show the relative biases and theMSEsof the four estimators of 120582 for sample sizes 119899 = 40 and 119899 = 80Figures 2(e) and 2(f) show the relative biases and the MSEs

of the four estimators of 120582 for 120572 = 30 The other cases aresimilar

From Tables 1 and 2 it is observed that for the fourmethods the average relative biases and the average relativeMSEs decrease as sample size 119899 increases as expected Theasymptotic unbiasedness of all the estimators is verified TheaverageMSEs of 120572 and 120582 = 2 depend on the parameter120572 For the four methods the average relative MSEs of 2decrease as 120572 goes up The average relative MSEs of 120572increase as120572 goes up Considering onlyMSEs we can observethat the estimation of 120572rsquos is more accurate for smaller valueswhile the estimation of 120582rsquos is more accurate for larger valuesof 120572 MOM MLE and IME overestimate both of the twoparameters 120572 and 120582 MIME overestimates only 120572

As far as the biases and MSEs are concerned it is clearthat MIME works the best in all the cases considered forestimating the two parameters Its performance is followedby IME MLE and MOM especially for small sample sizesThe four methods are close for larger sample sizes

Considering all the points MIME is recommended forestimating both parameters of the GLD(120582 120572) distributionMOM is not suggested

62 Comparison of the Two Joint Confidence Regions InSection 5 two methods to construct the confidence regionsof the two parameters 120582 and 120572 are proposed In this sectionwe conduct simulations to compare the two methods

8 Mathematical Problems in Engineering

Rela

tive b

ias

MLEIME

MIMEMOM

035

030

025

020

015

010

005

000

120572

20 25 30 35 40

(a) Relative biases (119899 = 40)

MLEIME

MIMEMOM

Rela

tive M

SEs

120572

20 25 30 35 40

05

04

03

02

01

00

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

015

010

005

000

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

025

020

015

010

005

000

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

Rela

tive M

SEs

035

030

025

020

015

010

005

000

n

MLEIME

MIMEMOM

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 1 Average relative biases and MSEs of 120572

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

Solve the equation and we obtain the inverse estimate IMEof 120582 Plugging IME into (15) we obtain the inverse estimate

IME In addition considering that the mode of 1205942(2119899 minus 2) is2119899 minus 4 we can obtain a modified equation for 120582

119899minus1

sum119894=1

log[log119866 (119883

(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] = 119899 minus 2 (21)

Solve the equation and we obtain the modified inverseestimate MIME of 120582 Plugging MIME into (15) we obtainthe modified inverse estimate MIME Unlike the momentestimation herewe only need to solve one nonlinear equationinstead of two equations

In the following we prove the existence and uniquenessof the root in (20) and (21)

Lemma 4 The following limits hold

(1) One has

lim120582rarr0

log119866 (119886)

log119866 (119887)

= lim120582rarr0

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= 1

119891119900119903 119886 gt 0 119887 gt 0

(22)

(2) One has

lim120582rarrinfin

log119866 (119886)

log119866 (119887)

= lim120582rarrinfin

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= 0

119891119900119903 119886 gt 119887 gt 0

(23)

(3) One has

lim120582rarrinfin

log119866 (119886)

log119866 (119887)

= lim120582rarrinfin

log [1 minus 119890minus120582119886 (120582119886 + 120582 + 1) (120582 + 1)]log [1 minus 119890minus120582119887 (120582119887 + 120582 + 1) (120582 + 1)]

= +infin 119891119900119903 119887 gt 119886 gt 0

(24)

Lemma 5 For 119905 gt 0 119891(119905) = (120582 + 119905 + 2 minus 119890119905[120582 minus (120582 + 1)119905 +2])(120582 minus (120582 + 1)119890119905 + 119905 + 1) is a decreasing function of 119905

Theorem 6 Let119882119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

) 119894 = 1 2 119899form a sample from standard exponential distribution 119878

119894=

1198821+ sdot sdot sdot + 119882

119894 then for 119905 gt 0 equation sum

119899minus1

119894=1log(119878119899119878119894) = 119905

has a unique positive solution

Proof By Lemma 4 we obtain

lim120582rarr0

119878119899

119878119894

= lim120582rarr0

1198821+ sdot sdot sdot + 119882

119899

1198821+ sdot sdot sdot + 119882

119894

= lim120582rarr0

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119899)

119885(1)

+ 119885(2)

+ sdot sdot sdot + 119885(119894minus1)

+ (119899 minus 119894 + 1) 119885(119894)

= lim120582rarr0

log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

= lim120582rarr0

[log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))] log119866 (119883

(119899))

[log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899))=119899

119899= 1

(25)

Thus lim120582rarr0

sum119899minus1

119894=1log(119878119899119878119894) = 0 On the other hand

lim120582rarrinfin

119878119899

119878119894

= lim120582rarrinfin

log119866 (119883(119899)) + log119866 (119883

(119899minus1)) + sdot sdot sdot + log119866 (119883

(1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+2)) + (119899 minus 119894 + 1) log119866 (119883

(119899minus119894+1))

= 1 + lim120582rarrinfin

log119866 (119883(119899minus119894)

) + sdot sdot sdot + log119866 (119883(1)) minus (119899 minus 119894) log119866 (119883

(119899minus119894+1))

log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+1)) + (119899 minus 119894) log119866 (119883

(119899minus119894+1))

= 1 + lim120582rarrinfin

[log119866 (119883(119899minus119894)

) + sdot sdot sdot + log119866 (119883(1)) minus (119899 minus 119894) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899minus119894+1))

[log119866 (119883(119899)) + sdot sdot sdot + log119866 (119883

(119899minus119894+1)) + (119899 minus 119894) log119866 (119883

(119899minus119894+1))] log119866 (119883

(119899minus119894+1))= +infin

(26)

Mathematical Problems in Engineering 5

Thus lim120582rarrinfin

sum119899minus1

119894=1log(119878119899119878119894) = infin Therefore for 119905 gt 0

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has one positive solution For

the uniqueness of the solution we consider the derivative of119878119899119878119894with respect to 120582 Note that for 119894 = 1 119899

119882119894= (119899 minus 119894 + 1) 120572 [log119866 (119883

(119899minus119894+2)) minus log119866 (119883

(119899minus119894+1))]

119889119882119894

119889120582= (119899 minus 119894 + 1) 120572

120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+2)

)minus120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+1)

) = 119882

119894

sdot120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+2)

) minus 120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+1)

)

log119866 (119883(119899minus119894+2)

) minus log119866 (119883(119899minus119894+1)

)

(119878119899

119878119894

)1015840

= (1 +119882119894+1

+ sdot sdot sdot + 119882119899

1198821+ sdot sdot sdot + 119882

119894

)1015840

=1

(sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

[1198821015840119895119882119896minus1198821198951198821015840119896] =

1

120582 (sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

119882119895119882119896 [119860 (120582) minus 119861 (120582)]

(27)

where

119860 (120582)

=120582119883(119899minus119895+2)

119890minus120582119883(119899minus119895+2)120582 [(120582 + 1)119883(119899minus119895+2)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+2)

) minus 120582119883(119899minus119895+1)

119890minus120582119883(119899minus119895+1)120582 [(120582 + 1)119883(119899minus119895+1)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+1)

)

log119866(119883(119899minus119895+2)

) minus log119866(119883(119899minus119895+1)

)

119861 (120582)

=120582119883(119899minus119896+2)

119890minus120582119883(119899minus119896+2)120582 [(120582 + 1)119883(119899minus119896+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+2)

) minus 120582119883(119899minus119896+1)

119890minus120582119883(119899minus119896+1)120582 [(120582 + 1)119883(119899minus119896+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+1)

)

log119866 (119883(119899minus119896+2)

) minus log119866 (119883(119899minus119896+1)

)

(28)

By Cauchyrsquos mean-value theorem for 119895 = 119894 + 1 119899 119896 =1 119894 there exist 120585

1isin (120582119883

(119899minus119895+1) 120582119883(119899minus119895+2)

) and 1205852

isin(120582119883(119899minus119896+1)

120582119883(119899minus119896+2)

) such that

119860 (120582) =120582 + 1205851+ 2 minus 119890120585

1[120582 minus (120582 + 1) 120585

1+ 2]

120582 minus (120582 + 1) 119890120585

1+ 1205851+ 1

119861 (120582) =120582 + 1205852+ 2 minus 119890120585

2[120582 minus (120582 + 1) 120585

2+ 2]

120582 minus (120582 + 1) 119890120585

2+ 1205852+ 1

(29)

Note that 1205851lt 1205852 by Lemma 5 119860(120582) minus 119861(120582) gt 0 (119878

119899119878119894)1015840 gt 0

thussum119899minus1119894=1

log(119878119899119878119894) is a strictly increasing function of 120582 and

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has a unique positive solution

5 Joint Confidence Regions for 120582 and 120572

Let 1198831 1198832 119883

119899form a sample from GLD(120582 120572) and

119883(1)

le 119883(2)

le sdot sdot sdot le 119883(119899)

are the order statistics Let119885(119894)

= minus120572 log[1 minus 119890minus120582119883(119899minus119894+1)(120582119883(119899minus119894+1)

+ 120582 + 1)(120582 + 1)] =minus120572 log119866(119883

(119899minus119894+1)) 119894 = 1 119899Thus 119885

(1)le 119885(2)

le sdot sdot sdot le 119885(119899)

are the first 119899-order statistics from the standard exponentialdistribution By Lemma 2 119882

119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

)119894 = 1 2 119899 form a sample from standard exponential

distribution Let 119878119894= 1198821+ sdot sdot sdot + 119882

119894 119880119894= (119878119894119878119894+1)119894 119894 =

1 2 119899 minus 1 and 119880119899= 1198821+ sdot sdot sdot + 119882

119899 Hence

119881 = 21198781= 21198821= 2119899119885

(1)

= minus2119899120572 log[1 minus119890minus120582119883(119899) (120582119883

(119899)+ 120582 + 1)

120582 + 1]

sim 1205942 (2)

119880 = 2 (119878119899minus 1198781) = 2

119899

sum119894=2

119882119894

= 2 [119885(1)

+ sdot sdot sdot + 119885(119899)

minus 119899119885(1)] sim 1205942 (2119899 minus 2)

(30)

It is obvious that 119880 and 119881 are independent Define

1198791=119880 (2119899 minus 2)

1198812=

119878119899minus 1198781

(119899 minus 1) 1198781

sim 119865 (2119899 minus 2 2)

1198792= 119880 + 119881 = 2119878

119899sim 1205942 (2119899)

(31)

We obtain that 1198791and 119879

2are independent using the known

bank-post office story in statisticsLet 119865120574(V1 V2) denote the percentile of 119865 distribution with

left-tail probability 120574 and V1and V

2degrees of freedom Let

1205942120574(V) denote the percentile of 1205942 distribution with left-tail

probability 120574 and V degrees of freedom

6 Mathematical Problems in Engineering

By using the pivotal variables1198791and1198792 a joint confidence

region for the two parameters 120582 and 120572 can be constructed asfollows

Theorem 7 (method 1) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198791and1198792

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582119871le 120582 le 120582

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(32)

where 120582119871is the root of 120582 for the equation 119879

1= 119865(1minusradic1minus120574)2

(2119899minus

2 2) and 120582119880

is the root of 120582 for the equation 1198791

=119865(1+radic1minus120574)2

(2119899 minus 2 2)

Proof 1198791= (1(119899 minus 1)) log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 +

1)] + sdot sdot sdot + log[1 minus 119890minus120582119883(1)(120582119883(1)

+ 120582 + 1)(120582 + 1)] minus 119899 log[1 minus119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 + 1)]119899 log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 +

1)(120582+1)] is a function of 120582 and does not depend on 120572 FromTheorem 6 we have lim

120582rarr01198791= (1(119899 minus 1))lim

120582rarr0(1198781198991198781minus

1) = 0 lim120582rarrinfin

1198791= (1(119899 minus 1))lim

120582rarrinfin(1198781198991198781minus 1) = infin

and 11987910158401= (1(119899 minus 1))(119878

119899119878119894)1015840 gt 0 Therefore for any 119905 gt 0

equation 1198791= 119905 has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2)

le 1198791le 119865(1+radic1minus120574)2

(2119899 minus 2 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899)

le 1198792le 1205942(1+radic1minus120574)2

(2119899))

= 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2) le 1198791

le 119865(1+radic1minus120574)2

(2119899 minus 2 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582119871le 120582

le 120582119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(33)

On the other hand by Lemma 3 we have

1198793= minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

)

= 2119899minus1

sum119894=1

log(119878119899

119878119894

) sim 1205942 (2119899 minus 2)

(34)

1198792and119879

3are also independent By using the pivotal variables

1198792and 119879

3 a joint confidence region for the two parameters 120582

and 120572 can be constructed as follows

Theorem 8 (method 2) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198792and1198793

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582lowast119871le 120582 le 120582lowast

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(35)

where 120582lowast119871is the root of 120582 for the equation 119879

3= 1205942(1minusradic1minus120574)2

(2119899minus

2) and 120582lowast119880is the root of 120582 for the equation119879

3= 1205942(1+radic1minus120574)2

(2119899minus

2)

Proof 1198793= 2sum

119899minus1

119894=1log(119878119899119878119894) is a function of 120582 and does not

depend on 120572 FromTheorem 6 for any 119904 gt 0 equation 1198793= 119904

has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582lowast119871le 120582

le 120582lowast119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(36)

6 Simulation Study

61 Comparison of the Four Estimation Methods In this sec-tion we conduct simulations to compare the performancesof the MIMEs IMEs MLEs and MOMs mainly with respectto their biases and mean squared errors (MSEs) for varioussample sizes and for various true parametric values

The random data 119883 from the GLD(120582 120572) distribution canbe generated as follows

119883 =minus120582 minus 1 minus119882(119890minus120582minus1 (120582 + 1) (1198801120572 minus 1))

120582 (37)

Mathematical Problems in Engineering 7

Table 1 Average relative estimates and MSEs of 120572

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 12705 (04298) 12350 (03510) 12965 (05115) 12770 (04460) 13522 (06256)MLE 11310 (01459) 11703 (02029) 11590 (02240) 11693 (02267) 11911 (02888)IME 10956 (01267) 11292 (01728) 11154 (01906) 11234 (01930) 11395 (02400)MIME 10482 (01063) 10755 (01420) 10590 (01570) 10635 (01572) 10754 (01939)

40

MOM 11791 (02212) 11954 (02262) 12046 (02693) 12299 (03163) 12536 (03961)MLE 10878 (00934) 11175 (01220) 11164 (01237) 11359 (01596) 11034 (01464)IME 10631 (00838) 10894 (01098) 10858 (01089) 11025 (01412) 10703 (01315)MIME 10294 (00739) 10516 (00951) 10456 (00937) 10591 (01207) 10267 (01139)

50

MOM 11363 (01557) 11396 (01767) 11570 (02069) 11747 (02296) 11842 (02652)MLE 10746 (00701) 10971 (01027) 11064 (01102) 10979 (01059) 11064 (01289)IME 10553 (00648) 10747 (00922) 10823 (00994) 10726 (00960) 10783 (01145)MIME 10288 (00584) 10451 (00822) 10505 (00879) 10394 (00849) 10432 (01009)

60

MOM 11233 (01227) 11281 (01249) 11480 (01679) 11312 (01608) 11511 (01892)MLE 10592 (00587) 10646 (00601) 10663 (00695) 10832 (00849) 10877 (00911)IME 10431 (00546) 10468 (00548) 10470 (00637) 10625 (00786) 10656 (00827)MIME 10214 (00502) 10232 (00499) 10218 (00580) 10354 (00709) 10371 (00743)

80

MOM 10954 (00843) 10836 (00856) 10984 (00974) 11136 (01105) 11220 (01224)MLE 10579 (00426) 10556 (00480) 10624 (00553) 10533 (00515) 10700 (00651)IME 10459 (00400) 10426 (00452) 10485 (00517) 10385 (00484) 10539 (00612)MIME 10297 (00372) 10251 (00421) 10296 (00478) 10189 (00449) 10329 (00563)

100

MOM 10704 (00659) 10704 (00736) 10835 (00859) 10763 (00888) 10780 (00851)MLE 10332 (00299) 10383 (00344) 10504 (00428) 10344 (00370) 10478 (00482)IME 10244 (00288) 10286 (00330) 10393 (00407) 10230 (00356) 10353 (00459)MIME 10119 (00274) 10149 (00313) 10245 (00382) 10077 (00337) 10189 (00432)

where 119880 follows uniform distribution over [0 1] and 119882(119886)giving the principal solution for 119908 in 119886 = 119908119890119908 is pronouncedas Lambert119882 function see Jodra [16]

We obtain MLE by solving (8) and MLE by (7) MOM andMOM can be obtained by solving (11) and (12) simultaneouslyIME and MIME can be obtained by solving (20) and (21)respectively IME and MIME can be obtained from (15)

We consider sample sizes 119899 = 30 40 50 60 80 100 and120572 = 20 25 30 35 40 We take 120582 = 2 in all our computa-tions For each combination of sample size 119899 and parameter120572 we generate a sample of size 119899 from GLD(120582 = 2 120572) andestimate the parameters120582 and120572 by theMLEMOM IME andMIMEmethodsThe average values of 120572 and 2 as well asthe correspondingMSEs over 1000 replications are computedand reported

Table 1 reports the average values of 120572 and the corre-sponding MSE is reported within parenthesis Figures 1(a)1(b) 1(c) and 1(d) show the relative biases and the MSEs ofthe four estimators of 120572 for sample sizes 119899 = 40 and 119899 = 80Figures 1(e) and 1(f) show the relative biases and the MSEsof the four estimators of 120572 for 120572 = 30 The other cases aresimilar

Table 2 reports the average values of 120582 = 2 and thecorresponding MSE is reported within parenthesis Figures2(a) 2(b) 2(c) and 2(d) show the relative biases and theMSEsof the four estimators of 120582 for sample sizes 119899 = 40 and 119899 = 80Figures 2(e) and 2(f) show the relative biases and the MSEs

of the four estimators of 120582 for 120572 = 30 The other cases aresimilar

From Tables 1 and 2 it is observed that for the fourmethods the average relative biases and the average relativeMSEs decrease as sample size 119899 increases as expected Theasymptotic unbiasedness of all the estimators is verified TheaverageMSEs of 120572 and 120582 = 2 depend on the parameter120572 For the four methods the average relative MSEs of 2decrease as 120572 goes up The average relative MSEs of 120572increase as120572 goes up Considering onlyMSEs we can observethat the estimation of 120572rsquos is more accurate for smaller valueswhile the estimation of 120582rsquos is more accurate for larger valuesof 120572 MOM MLE and IME overestimate both of the twoparameters 120572 and 120582 MIME overestimates only 120572

As far as the biases and MSEs are concerned it is clearthat MIME works the best in all the cases considered forestimating the two parameters Its performance is followedby IME MLE and MOM especially for small sample sizesThe four methods are close for larger sample sizes

Considering all the points MIME is recommended forestimating both parameters of the GLD(120582 120572) distributionMOM is not suggested

62 Comparison of the Two Joint Confidence Regions InSection 5 two methods to construct the confidence regionsof the two parameters 120582 and 120572 are proposed In this sectionwe conduct simulations to compare the two methods

8 Mathematical Problems in Engineering

Rela

tive b

ias

MLEIME

MIMEMOM

035

030

025

020

015

010

005

000

120572

20 25 30 35 40

(a) Relative biases (119899 = 40)

MLEIME

MIMEMOM

Rela

tive M

SEs

120572

20 25 30 35 40

05

04

03

02

01

00

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

015

010

005

000

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

025

020

015

010

005

000

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

Rela

tive M

SEs

035

030

025

020

015

010

005

000

n

MLEIME

MIMEMOM

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 1 Average relative biases and MSEs of 120572

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

Thus lim120582rarrinfin

sum119899minus1

119894=1log(119878119899119878119894) = infin Therefore for 119905 gt 0

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has one positive solution For

the uniqueness of the solution we consider the derivative of119878119899119878119894with respect to 120582 Note that for 119894 = 1 119899

119882119894= (119899 minus 119894 + 1) 120572 [log119866 (119883

(119899minus119894+2)) minus log119866 (119883

(119899minus119894+1))]

119889119882119894

119889120582= (119899 minus 119894 + 1) 120572

120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+2)

)minus120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2]

(120582 + 1)2 119866 (119883(119899minus119894+1)

) = 119882

119894

sdot120582119883(119899minus119894+2)

119890minus120582119883(119899minus119894+2) [(120582 + 1)119883(119899minus119894+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+2)

) minus 120582119883(119899minus119894+1)

119890minus120582119883(119899minus119894+1) [(120582 + 1)119883(119899minus119894+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119894+1)

)

log119866 (119883(119899minus119894+2)

) minus log119866 (119883(119899minus119894+1)

)

(119878119899

119878119894

)1015840

= (1 +119882119894+1

+ sdot sdot sdot + 119882119899

1198821+ sdot sdot sdot + 119882

119894

)1015840

=1

(sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

[1198821015840119895119882119896minus1198821198951198821015840119896] =

1

120582 (sum119894

119896=1119882119896)2

119899

sum119895=119894+1

119894

sum119896=1

119882119895119882119896 [119860 (120582) minus 119861 (120582)]

(27)

where

119860 (120582)

=120582119883(119899minus119895+2)

119890minus120582119883(119899minus119895+2)120582 [(120582 + 1)119883(119899minus119895+2)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+2)

) minus 120582119883(119899minus119895+1)

119890minus120582119883(119899minus119895+1)120582 [(120582 + 1)119883(119899minus119895+1)

+ 120582 + 2] (120582 + 1)2 119866(119883(119899minus119895+1)

)

log119866(119883(119899minus119895+2)

) minus log119866(119883(119899minus119895+1)

)

119861 (120582)

=120582119883(119899minus119896+2)

119890minus120582119883(119899minus119896+2)120582 [(120582 + 1)119883(119899minus119896+2)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+2)

) minus 120582119883(119899minus119896+1)

119890minus120582119883(119899minus119896+1)120582 [(120582 + 1)119883(119899minus119896+1)

+ 120582 + 2] (120582 + 1)2 119866 (119883(119899minus119896+1)

)

log119866 (119883(119899minus119896+2)

) minus log119866 (119883(119899minus119896+1)

)

(28)

By Cauchyrsquos mean-value theorem for 119895 = 119894 + 1 119899 119896 =1 119894 there exist 120585

1isin (120582119883

(119899minus119895+1) 120582119883(119899minus119895+2)

) and 1205852

isin(120582119883(119899minus119896+1)

120582119883(119899minus119896+2)

) such that

119860 (120582) =120582 + 1205851+ 2 minus 119890120585

1[120582 minus (120582 + 1) 120585

1+ 2]

120582 minus (120582 + 1) 119890120585

1+ 1205851+ 1

119861 (120582) =120582 + 1205852+ 2 minus 119890120585

2[120582 minus (120582 + 1) 120585

2+ 2]

120582 minus (120582 + 1) 119890120585

2+ 1205852+ 1

(29)

Note that 1205851lt 1205852 by Lemma 5 119860(120582) minus 119861(120582) gt 0 (119878

119899119878119894)1015840 gt 0

thussum119899minus1119894=1

log(119878119899119878119894) is a strictly increasing function of 120582 and

equation sum119899minus1

119894=1log(119878119899119878119894) = 119905 has a unique positive solution

5 Joint Confidence Regions for 120582 and 120572

Let 1198831 1198832 119883

119899form a sample from GLD(120582 120572) and

119883(1)

le 119883(2)

le sdot sdot sdot le 119883(119899)

are the order statistics Let119885(119894)

= minus120572 log[1 minus 119890minus120582119883(119899minus119894+1)(120582119883(119899minus119894+1)

+ 120582 + 1)(120582 + 1)] =minus120572 log119866(119883

(119899minus119894+1)) 119894 = 1 119899Thus 119885

(1)le 119885(2)

le sdot sdot sdot le 119885(119899)

are the first 119899-order statistics from the standard exponentialdistribution By Lemma 2 119882

119894= (119899 minus 119894 + 1)(119885

(119894)minus 119885(119894minus1)

)119894 = 1 2 119899 form a sample from standard exponential

distribution Let 119878119894= 1198821+ sdot sdot sdot + 119882

119894 119880119894= (119878119894119878119894+1)119894 119894 =

1 2 119899 minus 1 and 119880119899= 1198821+ sdot sdot sdot + 119882

119899 Hence

119881 = 21198781= 21198821= 2119899119885

(1)

= minus2119899120572 log[1 minus119890minus120582119883(119899) (120582119883

(119899)+ 120582 + 1)

120582 + 1]

sim 1205942 (2)

119880 = 2 (119878119899minus 1198781) = 2

119899

sum119894=2

119882119894

= 2 [119885(1)

+ sdot sdot sdot + 119885(119899)

minus 119899119885(1)] sim 1205942 (2119899 minus 2)

(30)

It is obvious that 119880 and 119881 are independent Define

1198791=119880 (2119899 minus 2)

1198812=

119878119899minus 1198781

(119899 minus 1) 1198781

sim 119865 (2119899 minus 2 2)

1198792= 119880 + 119881 = 2119878

119899sim 1205942 (2119899)

(31)

We obtain that 1198791and 119879

2are independent using the known

bank-post office story in statisticsLet 119865120574(V1 V2) denote the percentile of 119865 distribution with

left-tail probability 120574 and V1and V

2degrees of freedom Let

1205942120574(V) denote the percentile of 1205942 distribution with left-tail

probability 120574 and V degrees of freedom

6 Mathematical Problems in Engineering

By using the pivotal variables1198791and1198792 a joint confidence

region for the two parameters 120582 and 120572 can be constructed asfollows

Theorem 7 (method 1) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198791and1198792

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582119871le 120582 le 120582

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(32)

where 120582119871is the root of 120582 for the equation 119879

1= 119865(1minusradic1minus120574)2

(2119899minus

2 2) and 120582119880

is the root of 120582 for the equation 1198791

=119865(1+radic1minus120574)2

(2119899 minus 2 2)

Proof 1198791= (1(119899 minus 1)) log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 +

1)] + sdot sdot sdot + log[1 minus 119890minus120582119883(1)(120582119883(1)

+ 120582 + 1)(120582 + 1)] minus 119899 log[1 minus119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 + 1)]119899 log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 +

1)(120582+1)] is a function of 120582 and does not depend on 120572 FromTheorem 6 we have lim

120582rarr01198791= (1(119899 minus 1))lim

120582rarr0(1198781198991198781minus

1) = 0 lim120582rarrinfin

1198791= (1(119899 minus 1))lim

120582rarrinfin(1198781198991198781minus 1) = infin

and 11987910158401= (1(119899 minus 1))(119878

119899119878119894)1015840 gt 0 Therefore for any 119905 gt 0

equation 1198791= 119905 has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2)

le 1198791le 119865(1+radic1minus120574)2

(2119899 minus 2 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899)

le 1198792le 1205942(1+radic1minus120574)2

(2119899))

= 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2) le 1198791

le 119865(1+radic1minus120574)2

(2119899 minus 2 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582119871le 120582

le 120582119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(33)

On the other hand by Lemma 3 we have

1198793= minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

)

= 2119899minus1

sum119894=1

log(119878119899

119878119894

) sim 1205942 (2119899 minus 2)

(34)

1198792and119879

3are also independent By using the pivotal variables

1198792and 119879

3 a joint confidence region for the two parameters 120582

and 120572 can be constructed as follows

Theorem 8 (method 2) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198792and1198793

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582lowast119871le 120582 le 120582lowast

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(35)

where 120582lowast119871is the root of 120582 for the equation 119879

3= 1205942(1minusradic1minus120574)2

(2119899minus

2) and 120582lowast119880is the root of 120582 for the equation119879

3= 1205942(1+radic1minus120574)2

(2119899minus

2)

Proof 1198793= 2sum

119899minus1

119894=1log(119878119899119878119894) is a function of 120582 and does not

depend on 120572 FromTheorem 6 for any 119904 gt 0 equation 1198793= 119904

has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582lowast119871le 120582

le 120582lowast119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(36)

6 Simulation Study

61 Comparison of the Four Estimation Methods In this sec-tion we conduct simulations to compare the performancesof the MIMEs IMEs MLEs and MOMs mainly with respectto their biases and mean squared errors (MSEs) for varioussample sizes and for various true parametric values

The random data 119883 from the GLD(120582 120572) distribution canbe generated as follows

119883 =minus120582 minus 1 minus119882(119890minus120582minus1 (120582 + 1) (1198801120572 minus 1))

120582 (37)

Mathematical Problems in Engineering 7

Table 1 Average relative estimates and MSEs of 120572

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 12705 (04298) 12350 (03510) 12965 (05115) 12770 (04460) 13522 (06256)MLE 11310 (01459) 11703 (02029) 11590 (02240) 11693 (02267) 11911 (02888)IME 10956 (01267) 11292 (01728) 11154 (01906) 11234 (01930) 11395 (02400)MIME 10482 (01063) 10755 (01420) 10590 (01570) 10635 (01572) 10754 (01939)

40

MOM 11791 (02212) 11954 (02262) 12046 (02693) 12299 (03163) 12536 (03961)MLE 10878 (00934) 11175 (01220) 11164 (01237) 11359 (01596) 11034 (01464)IME 10631 (00838) 10894 (01098) 10858 (01089) 11025 (01412) 10703 (01315)MIME 10294 (00739) 10516 (00951) 10456 (00937) 10591 (01207) 10267 (01139)

50

MOM 11363 (01557) 11396 (01767) 11570 (02069) 11747 (02296) 11842 (02652)MLE 10746 (00701) 10971 (01027) 11064 (01102) 10979 (01059) 11064 (01289)IME 10553 (00648) 10747 (00922) 10823 (00994) 10726 (00960) 10783 (01145)MIME 10288 (00584) 10451 (00822) 10505 (00879) 10394 (00849) 10432 (01009)

60

MOM 11233 (01227) 11281 (01249) 11480 (01679) 11312 (01608) 11511 (01892)MLE 10592 (00587) 10646 (00601) 10663 (00695) 10832 (00849) 10877 (00911)IME 10431 (00546) 10468 (00548) 10470 (00637) 10625 (00786) 10656 (00827)MIME 10214 (00502) 10232 (00499) 10218 (00580) 10354 (00709) 10371 (00743)

80

MOM 10954 (00843) 10836 (00856) 10984 (00974) 11136 (01105) 11220 (01224)MLE 10579 (00426) 10556 (00480) 10624 (00553) 10533 (00515) 10700 (00651)IME 10459 (00400) 10426 (00452) 10485 (00517) 10385 (00484) 10539 (00612)MIME 10297 (00372) 10251 (00421) 10296 (00478) 10189 (00449) 10329 (00563)

100

MOM 10704 (00659) 10704 (00736) 10835 (00859) 10763 (00888) 10780 (00851)MLE 10332 (00299) 10383 (00344) 10504 (00428) 10344 (00370) 10478 (00482)IME 10244 (00288) 10286 (00330) 10393 (00407) 10230 (00356) 10353 (00459)MIME 10119 (00274) 10149 (00313) 10245 (00382) 10077 (00337) 10189 (00432)

where 119880 follows uniform distribution over [0 1] and 119882(119886)giving the principal solution for 119908 in 119886 = 119908119890119908 is pronouncedas Lambert119882 function see Jodra [16]

We obtain MLE by solving (8) and MLE by (7) MOM andMOM can be obtained by solving (11) and (12) simultaneouslyIME and MIME can be obtained by solving (20) and (21)respectively IME and MIME can be obtained from (15)

We consider sample sizes 119899 = 30 40 50 60 80 100 and120572 = 20 25 30 35 40 We take 120582 = 2 in all our computa-tions For each combination of sample size 119899 and parameter120572 we generate a sample of size 119899 from GLD(120582 = 2 120572) andestimate the parameters120582 and120572 by theMLEMOM IME andMIMEmethodsThe average values of 120572 and 2 as well asthe correspondingMSEs over 1000 replications are computedand reported

Table 1 reports the average values of 120572 and the corre-sponding MSE is reported within parenthesis Figures 1(a)1(b) 1(c) and 1(d) show the relative biases and the MSEs ofthe four estimators of 120572 for sample sizes 119899 = 40 and 119899 = 80Figures 1(e) and 1(f) show the relative biases and the MSEsof the four estimators of 120572 for 120572 = 30 The other cases aresimilar

Table 2 reports the average values of 120582 = 2 and thecorresponding MSE is reported within parenthesis Figures2(a) 2(b) 2(c) and 2(d) show the relative biases and theMSEsof the four estimators of 120582 for sample sizes 119899 = 40 and 119899 = 80Figures 2(e) and 2(f) show the relative biases and the MSEs

of the four estimators of 120582 for 120572 = 30 The other cases aresimilar

From Tables 1 and 2 it is observed that for the fourmethods the average relative biases and the average relativeMSEs decrease as sample size 119899 increases as expected Theasymptotic unbiasedness of all the estimators is verified TheaverageMSEs of 120572 and 120582 = 2 depend on the parameter120572 For the four methods the average relative MSEs of 2decrease as 120572 goes up The average relative MSEs of 120572increase as120572 goes up Considering onlyMSEs we can observethat the estimation of 120572rsquos is more accurate for smaller valueswhile the estimation of 120582rsquos is more accurate for larger valuesof 120572 MOM MLE and IME overestimate both of the twoparameters 120572 and 120582 MIME overestimates only 120572

As far as the biases and MSEs are concerned it is clearthat MIME works the best in all the cases considered forestimating the two parameters Its performance is followedby IME MLE and MOM especially for small sample sizesThe four methods are close for larger sample sizes

Considering all the points MIME is recommended forestimating both parameters of the GLD(120582 120572) distributionMOM is not suggested

62 Comparison of the Two Joint Confidence Regions InSection 5 two methods to construct the confidence regionsof the two parameters 120582 and 120572 are proposed In this sectionwe conduct simulations to compare the two methods

8 Mathematical Problems in Engineering

Rela

tive b

ias

MLEIME

MIMEMOM

035

030

025

020

015

010

005

000

120572

20 25 30 35 40

(a) Relative biases (119899 = 40)

MLEIME

MIMEMOM

Rela

tive M

SEs

120572

20 25 30 35 40

05

04

03

02

01

00

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

015

010

005

000

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

025

020

015

010

005

000

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

Rela

tive M

SEs

035

030

025

020

015

010

005

000

n

MLEIME

MIMEMOM

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 1 Average relative biases and MSEs of 120572

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

By using the pivotal variables1198791and1198792 a joint confidence

region for the two parameters 120582 and 120572 can be constructed asfollows

Theorem 7 (method 1) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198791and1198792

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582119871le 120582 le 120582

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(32)

where 120582119871is the root of 120582 for the equation 119879

1= 119865(1minusradic1minus120574)2

(2119899minus

2 2) and 120582119880

is the root of 120582 for the equation 1198791

=119865(1+radic1minus120574)2

(2119899 minus 2 2)

Proof 1198791= (1(119899 minus 1)) log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 +

1)] + sdot sdot sdot + log[1 minus 119890minus120582119883(1)(120582119883(1)

+ 120582 + 1)(120582 + 1)] minus 119899 log[1 minus119890minus120582119883(119899)(120582119883

(119899)+ 120582 + 1)(120582 + 1)]119899 log[1 minus 119890minus120582119883(119899)(120582119883

(119899)+ 120582 +

1)(120582+1)] is a function of 120582 and does not depend on 120572 FromTheorem 6 we have lim

120582rarr01198791= (1(119899 minus 1))lim

120582rarr0(1198781198991198781minus

1) = 0 lim120582rarrinfin

1198791= (1(119899 minus 1))lim

120582rarrinfin(1198781198991198781minus 1) = infin

and 11987910158401= (1(119899 minus 1))(119878

119899119878119894)1015840 gt 0 Therefore for any 119905 gt 0

equation 1198791= 119905 has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2)

le 1198791le 119865(1+radic1minus120574)2

(2119899 minus 2 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899)

le 1198792le 1205942(1+radic1minus120574)2

(2119899))

= 119875 (119865(1minusradic1minus120574)2

(2119899 minus 2 2) le 1198791

le 119865(1+radic1minus120574)2

(2119899 minus 2 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582119871le 120582

le 120582119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(33)

On the other hand by Lemma 3 we have

1198793= minus2119899minus1

sum119894=1

log119880119894= minus2119899minus1

sum119894=1

119894 log(119878119894

119878119894+1

)

= 2119899minus1

sum119894=1

log(119878119899

119878119894

) sim 1205942 (2119899 minus 2)

(34)

1198792and119879

3are also independent By using the pivotal variables

1198792and 119879

3 a joint confidence region for the two parameters 120582

and 120572 can be constructed as follows

Theorem 8 (method 2) Let 1198831 1198832 119883

119899form a sample

from119866119871119863(120582 120572) then based on the pivotal variables1198792and1198793

a 100(1 minus 120574) joint confidence region for the two parameters(120582 120572) is determined by the following inequalities

120582lowast119871le 120582 le 120582lowast

119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(35)

where 120582lowast119871is the root of 120582 for the equation 119879

3= 1205942(1minusradic1minus120574)2

(2119899minus

2) and 120582lowast119880is the root of 120582 for the equation119879

3= 1205942(1+radic1minus120574)2

(2119899minus

2)

Proof 1198793= 2sum

119899minus1

119894=1log(119878119899119878119894) is a function of 120582 and does not

depend on 120572 FromTheorem 6 for any 119904 gt 0 equation 1198793= 119904

has a unique positive root of 120582

1 minus 120574 = radic1 minus 120574radic1 minus 120574 = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2)) 119875 (1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875 (1205942(1minusradic1minus120574)2

(2119899 minus 2) le 1198793

le 1205942(1+radic1minus120574)2

(2119899 minus 2) 1205942

(1minusradic1minus120574)2(2119899) le 119879

2

le 1205942(1+radic1minus120574)2

(2119899)) = 119875(120582lowast119871le 120582

le 120582lowast119880

1205942(1minusradic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894))le 120572

le1205942(1+radic1minus120574)2

(2119899)

minus2sum119899

119894=1log119866 (119883

(119894)))

(36)

6 Simulation Study

61 Comparison of the Four Estimation Methods In this sec-tion we conduct simulations to compare the performancesof the MIMEs IMEs MLEs and MOMs mainly with respectto their biases and mean squared errors (MSEs) for varioussample sizes and for various true parametric values

The random data 119883 from the GLD(120582 120572) distribution canbe generated as follows

119883 =minus120582 minus 1 minus119882(119890minus120582minus1 (120582 + 1) (1198801120572 minus 1))

120582 (37)

Mathematical Problems in Engineering 7

Table 1 Average relative estimates and MSEs of 120572

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 12705 (04298) 12350 (03510) 12965 (05115) 12770 (04460) 13522 (06256)MLE 11310 (01459) 11703 (02029) 11590 (02240) 11693 (02267) 11911 (02888)IME 10956 (01267) 11292 (01728) 11154 (01906) 11234 (01930) 11395 (02400)MIME 10482 (01063) 10755 (01420) 10590 (01570) 10635 (01572) 10754 (01939)

40

MOM 11791 (02212) 11954 (02262) 12046 (02693) 12299 (03163) 12536 (03961)MLE 10878 (00934) 11175 (01220) 11164 (01237) 11359 (01596) 11034 (01464)IME 10631 (00838) 10894 (01098) 10858 (01089) 11025 (01412) 10703 (01315)MIME 10294 (00739) 10516 (00951) 10456 (00937) 10591 (01207) 10267 (01139)

50

MOM 11363 (01557) 11396 (01767) 11570 (02069) 11747 (02296) 11842 (02652)MLE 10746 (00701) 10971 (01027) 11064 (01102) 10979 (01059) 11064 (01289)IME 10553 (00648) 10747 (00922) 10823 (00994) 10726 (00960) 10783 (01145)MIME 10288 (00584) 10451 (00822) 10505 (00879) 10394 (00849) 10432 (01009)

60

MOM 11233 (01227) 11281 (01249) 11480 (01679) 11312 (01608) 11511 (01892)MLE 10592 (00587) 10646 (00601) 10663 (00695) 10832 (00849) 10877 (00911)IME 10431 (00546) 10468 (00548) 10470 (00637) 10625 (00786) 10656 (00827)MIME 10214 (00502) 10232 (00499) 10218 (00580) 10354 (00709) 10371 (00743)

80

MOM 10954 (00843) 10836 (00856) 10984 (00974) 11136 (01105) 11220 (01224)MLE 10579 (00426) 10556 (00480) 10624 (00553) 10533 (00515) 10700 (00651)IME 10459 (00400) 10426 (00452) 10485 (00517) 10385 (00484) 10539 (00612)MIME 10297 (00372) 10251 (00421) 10296 (00478) 10189 (00449) 10329 (00563)

100

MOM 10704 (00659) 10704 (00736) 10835 (00859) 10763 (00888) 10780 (00851)MLE 10332 (00299) 10383 (00344) 10504 (00428) 10344 (00370) 10478 (00482)IME 10244 (00288) 10286 (00330) 10393 (00407) 10230 (00356) 10353 (00459)MIME 10119 (00274) 10149 (00313) 10245 (00382) 10077 (00337) 10189 (00432)

where 119880 follows uniform distribution over [0 1] and 119882(119886)giving the principal solution for 119908 in 119886 = 119908119890119908 is pronouncedas Lambert119882 function see Jodra [16]

We obtain MLE by solving (8) and MLE by (7) MOM andMOM can be obtained by solving (11) and (12) simultaneouslyIME and MIME can be obtained by solving (20) and (21)respectively IME and MIME can be obtained from (15)

We consider sample sizes 119899 = 30 40 50 60 80 100 and120572 = 20 25 30 35 40 We take 120582 = 2 in all our computa-tions For each combination of sample size 119899 and parameter120572 we generate a sample of size 119899 from GLD(120582 = 2 120572) andestimate the parameters120582 and120572 by theMLEMOM IME andMIMEmethodsThe average values of 120572 and 2 as well asthe correspondingMSEs over 1000 replications are computedand reported

Table 1 reports the average values of 120572 and the corre-sponding MSE is reported within parenthesis Figures 1(a)1(b) 1(c) and 1(d) show the relative biases and the MSEs ofthe four estimators of 120572 for sample sizes 119899 = 40 and 119899 = 80Figures 1(e) and 1(f) show the relative biases and the MSEsof the four estimators of 120572 for 120572 = 30 The other cases aresimilar

Table 2 reports the average values of 120582 = 2 and thecorresponding MSE is reported within parenthesis Figures2(a) 2(b) 2(c) and 2(d) show the relative biases and theMSEsof the four estimators of 120582 for sample sizes 119899 = 40 and 119899 = 80Figures 2(e) and 2(f) show the relative biases and the MSEs

of the four estimators of 120582 for 120572 = 30 The other cases aresimilar

From Tables 1 and 2 it is observed that for the fourmethods the average relative biases and the average relativeMSEs decrease as sample size 119899 increases as expected Theasymptotic unbiasedness of all the estimators is verified TheaverageMSEs of 120572 and 120582 = 2 depend on the parameter120572 For the four methods the average relative MSEs of 2decrease as 120572 goes up The average relative MSEs of 120572increase as120572 goes up Considering onlyMSEs we can observethat the estimation of 120572rsquos is more accurate for smaller valueswhile the estimation of 120582rsquos is more accurate for larger valuesof 120572 MOM MLE and IME overestimate both of the twoparameters 120572 and 120582 MIME overestimates only 120572

As far as the biases and MSEs are concerned it is clearthat MIME works the best in all the cases considered forestimating the two parameters Its performance is followedby IME MLE and MOM especially for small sample sizesThe four methods are close for larger sample sizes

Considering all the points MIME is recommended forestimating both parameters of the GLD(120582 120572) distributionMOM is not suggested

62 Comparison of the Two Joint Confidence Regions InSection 5 two methods to construct the confidence regionsof the two parameters 120582 and 120572 are proposed In this sectionwe conduct simulations to compare the two methods

8 Mathematical Problems in Engineering

Rela

tive b

ias

MLEIME

MIMEMOM

035

030

025

020

015

010

005

000

120572

20 25 30 35 40

(a) Relative biases (119899 = 40)

MLEIME

MIMEMOM

Rela

tive M

SEs

120572

20 25 30 35 40

05

04

03

02

01

00

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

015

010

005

000

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

025

020

015

010

005

000

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

Rela

tive M

SEs

035

030

025

020

015

010

005

000

n

MLEIME

MIMEMOM

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 1 Average relative biases and MSEs of 120572

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Applied MathematicsJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

Table 1 Average relative estimates and MSEs of 120572

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 12705 (04298) 12350 (03510) 12965 (05115) 12770 (04460) 13522 (06256)MLE 11310 (01459) 11703 (02029) 11590 (02240) 11693 (02267) 11911 (02888)IME 10956 (01267) 11292 (01728) 11154 (01906) 11234 (01930) 11395 (02400)MIME 10482 (01063) 10755 (01420) 10590 (01570) 10635 (01572) 10754 (01939)

40

MOM 11791 (02212) 11954 (02262) 12046 (02693) 12299 (03163) 12536 (03961)MLE 10878 (00934) 11175 (01220) 11164 (01237) 11359 (01596) 11034 (01464)IME 10631 (00838) 10894 (01098) 10858 (01089) 11025 (01412) 10703 (01315)MIME 10294 (00739) 10516 (00951) 10456 (00937) 10591 (01207) 10267 (01139)

50

MOM 11363 (01557) 11396 (01767) 11570 (02069) 11747 (02296) 11842 (02652)MLE 10746 (00701) 10971 (01027) 11064 (01102) 10979 (01059) 11064 (01289)IME 10553 (00648) 10747 (00922) 10823 (00994) 10726 (00960) 10783 (01145)MIME 10288 (00584) 10451 (00822) 10505 (00879) 10394 (00849) 10432 (01009)

60

MOM 11233 (01227) 11281 (01249) 11480 (01679) 11312 (01608) 11511 (01892)MLE 10592 (00587) 10646 (00601) 10663 (00695) 10832 (00849) 10877 (00911)IME 10431 (00546) 10468 (00548) 10470 (00637) 10625 (00786) 10656 (00827)MIME 10214 (00502) 10232 (00499) 10218 (00580) 10354 (00709) 10371 (00743)

80

MOM 10954 (00843) 10836 (00856) 10984 (00974) 11136 (01105) 11220 (01224)MLE 10579 (00426) 10556 (00480) 10624 (00553) 10533 (00515) 10700 (00651)IME 10459 (00400) 10426 (00452) 10485 (00517) 10385 (00484) 10539 (00612)MIME 10297 (00372) 10251 (00421) 10296 (00478) 10189 (00449) 10329 (00563)

100

MOM 10704 (00659) 10704 (00736) 10835 (00859) 10763 (00888) 10780 (00851)MLE 10332 (00299) 10383 (00344) 10504 (00428) 10344 (00370) 10478 (00482)IME 10244 (00288) 10286 (00330) 10393 (00407) 10230 (00356) 10353 (00459)MIME 10119 (00274) 10149 (00313) 10245 (00382) 10077 (00337) 10189 (00432)

where 119880 follows uniform distribution over [0 1] and 119882(119886)giving the principal solution for 119908 in 119886 = 119908119890119908 is pronouncedas Lambert119882 function see Jodra [16]

We obtain MLE by solving (8) and MLE by (7) MOM andMOM can be obtained by solving (11) and (12) simultaneouslyIME and MIME can be obtained by solving (20) and (21)respectively IME and MIME can be obtained from (15)

We consider sample sizes 119899 = 30 40 50 60 80 100 and120572 = 20 25 30 35 40 We take 120582 = 2 in all our computa-tions For each combination of sample size 119899 and parameter120572 we generate a sample of size 119899 from GLD(120582 = 2 120572) andestimate the parameters120582 and120572 by theMLEMOM IME andMIMEmethodsThe average values of 120572 and 2 as well asthe correspondingMSEs over 1000 replications are computedand reported

Table 1 reports the average values of 120572 and the corre-sponding MSE is reported within parenthesis Figures 1(a)1(b) 1(c) and 1(d) show the relative biases and the MSEs ofthe four estimators of 120572 for sample sizes 119899 = 40 and 119899 = 80Figures 1(e) and 1(f) show the relative biases and the MSEsof the four estimators of 120572 for 120572 = 30 The other cases aresimilar

Table 2 reports the average values of 120582 = 2 and thecorresponding MSE is reported within parenthesis Figures2(a) 2(b) 2(c) and 2(d) show the relative biases and theMSEsof the four estimators of 120582 for sample sizes 119899 = 40 and 119899 = 80Figures 2(e) and 2(f) show the relative biases and the MSEs

of the four estimators of 120582 for 120572 = 30 The other cases aresimilar

From Tables 1 and 2 it is observed that for the fourmethods the average relative biases and the average relativeMSEs decrease as sample size 119899 increases as expected Theasymptotic unbiasedness of all the estimators is verified TheaverageMSEs of 120572 and 120582 = 2 depend on the parameter120572 For the four methods the average relative MSEs of 2decrease as 120572 goes up The average relative MSEs of 120572increase as120572 goes up Considering onlyMSEs we can observethat the estimation of 120572rsquos is more accurate for smaller valueswhile the estimation of 120582rsquos is more accurate for larger valuesof 120572 MOM MLE and IME overestimate both of the twoparameters 120572 and 120582 MIME overestimates only 120572

As far as the biases and MSEs are concerned it is clearthat MIME works the best in all the cases considered forestimating the two parameters Its performance is followedby IME MLE and MOM especially for small sample sizesThe four methods are close for larger sample sizes

Considering all the points MIME is recommended forestimating both parameters of the GLD(120582 120572) distributionMOM is not suggested

62 Comparison of the Two Joint Confidence Regions InSection 5 two methods to construct the confidence regionsof the two parameters 120582 and 120572 are proposed In this sectionwe conduct simulations to compare the two methods

8 Mathematical Problems in Engineering

Rela

tive b

ias

MLEIME

MIMEMOM

035

030

025

020

015

010

005

000

120572

20 25 30 35 40

(a) Relative biases (119899 = 40)

MLEIME

MIMEMOM

Rela

tive M

SEs

120572

20 25 30 35 40

05

04

03

02

01

00

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

015

010

005

000

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

025

020

015

010

005

000

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

Rela

tive M

SEs

035

030

025

020

015

010

005

000

n

MLEIME

MIMEMOM

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 1 Average relative biases and MSEs of 120572

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

Rela

tive b

ias

MLEIME

MIMEMOM

035

030

025

020

015

010

005

000

120572

20 25 30 35 40

(a) Relative biases (119899 = 40)

MLEIME

MIMEMOM

Rela

tive M

SEs

120572

20 25 30 35 40

05

04

03

02

01

00

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

015

010

005

000

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

025

020

015

010

005

000

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

Rela

tive M

SEs

035

030

025

020

015

010

005

000

n

MLEIME

MIMEMOM

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 1 Average relative biases and MSEs of 120572

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

020

015

010

005

000

minus005

minus010

(a) Relative biases (119899 = 40)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0045

0040

0035

0030

0025

0020

0015

0010

(b) Relative MSEs (119899 = 40)

Rela

tive b

ias

MLEIME

MIMEMOM

120572

20 25 30 35 40

010

005

000

minus005

(c) Relative biases (119899 = 80)

Rela

tive M

SEs

MLEIME

MIMEMOM

120572

20 25 30 35 40

0025

0020

0015

0010

0005

(d) Relative MSEs (119899 = 80)

n

Rela

tive b

ias

MLEIME

MIMEMOM

006

004

002

000

minus002

30 40 50 60 70 80 90 100

(e) Relative biases (120572 = 30)

n

Rela

tive M

SEs

MLEIME

MIMEMOM

0035

0030

0025

0020

0015

0010

0005

0000

30 40 50 60 70 80 90 100

(f) Relative MSEs (120572 = 30)

Figure 2 Average relative biases and MSEs of 120582

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

Table 2 Average relative estimates and MSEs of 120582

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

MOM 10778 (00551) 10795 (00524) 10760 (00483) 10794 (00495) 10699 (00454)MLE 10596 (00381) 10516 (00357) 10449 (00315) 10529 (00335) 10460 (00315)IME 10356 (00343) 10276 (00323) 10218 (00286) 10302 (00308) 10233 (00288)MIME 10024 (00314) 09957 (00299) 09909 (00269) 09997 (00285) 09934 (00270)

40

MOM 10507 (00395) 10721 (00413) 10561 (00380) 10580 (00358) 10522 (00290)MLE 10294 (00234) 10345 (00209) 10447 (00256) 10317 (00228) 10315 (00195)IME 10122 (00221) 10174 (00195) 10273 (00238) 10152 (00214) 10154 (00184)MIME 09880 (00213) 09939 (00185) 10042 (00222) 09928 (00205) 09934 (00176)

50

MOM 10546 (00319) 10427 (00288) 10400 (00266) 10412 (00249) 10424 (00254)MLE 10338 (00202) 10307 (00198) 10195 (00169) 10292 (00179) 10327 (00162)IME 10197 (00190) 10165 (00187) 10061 (00160) 10161 (00169) 10193 (00154)MIME 10003 (00181) 09978 (00178) 09880 (00156) 09983 (00162) 10017 (00146)

60

MOM 10402 (00280) 10432 (00244) 10332 (00224) 10417 (00225) 10339 (00217)MLE 10230 (00159) 10181 (00151) 10304 (00146) 10226 (00140) 10247 (00129)IME 10109 (00152) 10067 (00145) 10190 (00139) 10113 (00134) 10132 (00122)MIME 09948 (00147) 09913 (00142) 10039 (00132) 09965 (00129) 09987 (00118)

80

MOM 10342 (00219) 10288 (00167) 10272 (00180) 10305 (00155) 10327 (00161)MLE 10130 (00116) 10238 (00121) 10201 (00106) 10229 (00111) 10199 (00095)IME 10041 (00113) 10149 (00116) 10119 (00103) 10148 (00107) 10116 (00092)MIME 09922 (00111) 10034 (00112) 10007 (00100) 10037 (00103) 10008 (00089)

100

MOM 10269 (00154) 10255 (00153) 10210 (00125) 10144 (00120) 10236 (00117)MLE 10129 (00087) 10135 (00095) 10149 (00080) 10150 (00080) 10133 (00075)IME 10059 (00085) 10067 (00092) 10083 (00077) 10084 (00077) 10069 (00073)MIME 09964 (00083) 09975 (00090) 09994 (00076) 09996 (00076) 09983 (00072)

First we assess the precisions of the two methods ofinterval estimators for the parameter 120582 We take sample sizes119899 = 30 40 50 60 80 100 and 120572 = 20 25 30 35 40 Wetake 120582 = 2 in all our computations For each combination ofsample size 119899 and parameter 120572 we generate a sample of size 119899from GLD(120582 = 2 120572) and estimate the parameter 120582 by the twoproposed methods (32) and (35)

The mean widths as well as the coverage rates over 1000replications are computed and reported Here the coveragerate is defined as the rate of the confidence intervals thatcontain the true value 120582 = 2 among these 1000 confidenceintervals The results are reported in Table 3

It is observed that the mean widths of the intervalsdecrease as sample sizes 119899 increase as expected The meanwidths of the intervals decrease as the parameter 120572 increasesThe coverage rates of the two methods are close to thenominal level 095 Considering themeanwidths the intervalestimate of 120582 obtained in method 2 performs better than thatobtained in method 1 Method 2 for constructing the intervalestimate of 120582 is recommended

Next we consider the two joint confidence regions and theempirical coverage rates and expected areasThe results of themethods for constructing joint confidence regions for (120582 120572)with confidence level 120574 = 095 are reported in Table 4

We can find that the mean areas of the joint regionsdecrease as sample sizes 119899 increase as expected The mean

areas of the joint regions increase as the parameter 120572increases The coverage rates of the two methods are close tothe nominal level 095 Considering the mean areas the jointregion of (120582 120572) obtained in method 2 performs better thanthat obtained in method 1 Method 2 is recommended

7 Real Illustrative Example

In this section we consider a real lifetime data set (Gross andClark [17]) and it shows the relief times of twenty patientsreceiving an analgesic The dataset has been previously ana-lyzed by Bain and Engelhardt [18] Kumar andDharmaja [19]Nadarajah et al [11] and so forth The relief times in hoursare shown as follows

11 14 13 17 19 18 16 22 17 27 41 18 15 12 14 3

17 23 16 2(38)

Nadarajah et al [11] fit the data with generalized Lindleydistribution and showed that it can be a better model thanthose based on the gamma lognormal and the WeibulldistributionsTheMLEs of the parameters are MLE = 25395and MLE = 278766 with log-likelihood value minus164044 TheKolmogorov-Smirnov distance and its corresponding119901 value

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 11

Table 3 Results of the methods for constructing intervals for 120582 with confidence level 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean width 24282 24056 23482 23093 22885Coverage rate 0947 0947 0941 0944 0951

(2)Mean width 13872 13424 12958 12705 12539Coverage rate 0948 0942 0955 0943 0952

40(1)

Mean width 22463 22405 22048 21733 21644Coverage rate 0945 095 0947 0957 0958

(2)Mean width 11937 11545 11209 10939 10828Coverage rate 0947 0938 0938 0958 0948

50(1)

Mean width 21506 2107 20635 20356 20293Coverage rate 0953 0945 0936 0954 0949

(2)Mean width 10566 10174 09953 09732 09563Coverage rate 0953 0945 0943 0947 095

60(1)

Mean width 20555 20197 198 19502 19397Coverage rate 0953 0952 0955 0948 0951

(2)Mean width 09627 09331 09021 08856 08692Coverage rate 094 0957 0952 0942 0952

80

(1)Mean width 19277 18859 18485 18418 18412Coverage rate 0958 0959 0956 0961 0941

(2)Mean width 0831 07995 0777 07599 07521Coverage rate 0952 0947 0942 0951 0951

100

(1)Mean width 18251 17883 17849 17729 17633Coverage rate 0954 0948 0951 0946 0956

(2)Mean width 07423 07119 06947 06804 06684Coverage rate 0937 0961 0946 0948 0947

120572

100

80

60

40

20

0

120582

10 15 20 25 30

(a) Method 1

120572

150

100

50

0

120582

15 20 25 30

(b) Method 2

Figure 3 The 95 joint confidence region of (120582 120572)

are 119863 = 01377 and 119901 = 07941 respectively The MOMs ofthe parameters are MOM = 21042 and MOM = 131280

Using the methods proposed in Section 3 we obtain thefollowing estimates

IME = 23687

IME = 217428

MIME = 22671

MIME = 187269

(39)

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Mathematical Problems in Engineering

Table 4 Results of the methods for constructing joint confidence regions for (120582 120572) with confidence level 120574 = 095

119899 Methods 120572 = 20 120572 = 25 120572 = 30 120572 = 35 120572 = 40

30

(1)Mean area 80701 105512 155331 176929 218188

Coverage rate 096 0949 0949 0952 0959

(2)Mean area 30543 37328 47094 52673 61423

Coverage rate 0964 0956 095 0949 0952

40(1)

Mean area 62578 80063 101501 128546 15722Coverage rate 0942 0963 0962 0942 0942

(2)Mean area 2211 26583 31734 36342 4174

Coverage rate 0949 0951 0956 0948 0953

50

(1)Mean area 47221 671 8016 100739 12373

Coverage rate 0951 0935 0958 0944 0955

(2)Mean area 16762 20329 2404 2798 3201

Coverage rate 095 0931 0956 0959 0947

60

(1)Mean area 43159 54371 71443 83226 102957

Coverage rate 0954 0939 0953 094 0951

(2)Mean area 14017 16443 19985 22489 25635

Coverage rate 0954 0943 0952 0944 096

80

(1)Mean area 31997 42806 53322 6392 77788

Coverage rate 0951 0956 0951 0957 0948

(2)Mean area 10114 12325 14396 16863 18538

Coverage rate 0947 0956 0955 0951 0942

100

(1)Mean area 26803 34994 4418 54509 63339

Coverage rate 0951 0947 0965 0947 0951

(2)Mean area 07938 09673 11306 13265 14548

Coverage rate 0947 0945 0962 0943 0964

In addition based on method 1 the 95 joint confidenceregion for the parameters (120582 120572) is given by the followinginequalities

07736 le 120582 le 31505

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(40)

Based on method 2 the 95 joint confidence region forthe parameters (120582 120572) is given by the following inequalities

14503 le 120582 le 34078

minus113532

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

le 120572

leminus313028

sum119899

119894=1log [1 minus 119890minus120582119883(119894) (120582119883

(119894)+ 120582 + 1) (120582 + 1)]

(41)

Figures 3(a) and 3(b) show the 95 joint confidence regionsof (120582 120572)

Considering the widths of 120582 method 2 is suggested

8 Conclusion

In this paper we study the problem of estimating the twoparameters of the generalized Lindley distribution intro-duced by Nadarajah et al [11] We propose the inversemoment estimator and modified inverse moment estimatorand study their statistical properties The existence anduniqueness of inversemoment andmodified inversemomentestimates of the parameters are proved Monte Carlo simula-tions are used to compare their performances We also inves-tigate the methods for constructing joint confidence regionsfor the two parameters and study their performances

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

Wenhao Guirsquos work was partially supported by the programfor the Fundamental Research Funds for the Central Univer-sities (nos 2014RC042 and 2015JBM109)

References

[1] D V Lindley ldquoFiducial distributions and Bayesrsquo theoremrdquoJournal of the Royal Statistical Society Series B Methodologicalvol 20 pp 102ndash107 1958

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 13

[2] M E Ghitany B Atieh and S Nadarajah ldquoLindley distributionand its applicationrdquoMathematics and Computers in Simulationvol 78 no 4 pp 493ndash506 2008

[3] J Mazucheli and J A Achcar ldquoThe lindley distribution appliedto competing risks lifetime datardquo Computer Methods andPrograms in Biomedicine vol 104 no 2 pp 188ndash192 2011

[4] H Krishna and K Kumar ldquoReliability estimation in Lindleydistribution with progressively type II right censored samplerdquoMathematics amp Computers in Simulation vol 82 no 2 pp 281ndash294 2011

[5] D K Al-Mutairi M E Ghitany and D Kundu ldquoInferences onstress-strength reliability from Lindley distributionsrdquo Commu-nications in StatisticsmdashTheory and Methods vol 42 no 8 pp1443ndash1463 2013

[6] M Sankaran ldquoThe discrete poisson-lindley distributionrdquo Bio-metrics vol 26 no 1 pp 145ndash149 1970

[7] M E Ghitany D K Al-Mutairi and S Nadarajah ldquoZero-truncated Poisson-Lindley distribution and its applicationrdquoMathematics and Computers in Simulation vol 79 no 3 pp279ndash287 2008

[8] H S Bakouch B M Al-Zahrani A A Al-Shomrani V AMarchi and F Louzada ldquoAn extended lindley distributionrdquoJournal of the Korean Statistical Society vol 41 no 1 pp 75ndash852012

[9] R Shanker S Sharma and R Shanker ldquoA two-parameter lind-ley distribution for modeling waiting and survival times datardquoApplied Mathematics vol 4 no 2 pp 363ndash368 2013

[10] M E Ghitany D K Al-Mutairi N Balakrishnan and L J Al-Enezi ldquoPower Lindley distribution and associated inferencerdquoComputational Statistics amp Data Analysis vol 64 pp 20ndash332013

[11] S Nadarajah H S Bakouch and R Tahmasbi ldquoA generalizedLindley distributionrdquo Sankhya B vol 73 no 2 pp 331ndash359 2011

[12] S K Singh U Singh and V K Sharma ldquoBayesian estimationand prediction for the generalized lindley distribution underassymetric loss functionrdquoHacettepe Journal of Mathematics andStatistics vol 43 no 4 pp 661ndash678 2014

[13] S K SinghU Singh andVK Sharma ldquoExpected total test timeand Bayesian estimation for generalized Lindley distributionunder progressively Type-II censored sample where removalsfollow the beta-binomial probability lawrdquo Applied Mathematicsand Computation vol 222 pp 402ndash419 2013

[14] B Arnold N Balakrishnan and H Nagaraja A First Course inOrder Statistics Classics in Applied Mathematics Society forIndustrial and Applied Mathematics (SIAM) Philadelphia PaUSA 1992

[15] B X Wang ldquoStatistical inference of Weibull distributionrdquoChinese Journal of Applied Probability amp Statistics vol 8 no 4pp 357ndash364 1992

[16] P Jodra ldquoComputer generation of random variables with Lind-ley or PoissonndashLindley distribution via the Lambert W func-tionrdquo Mathematics and Computers in Simulation vol 81 no 4pp 851ndash859 2010

[17] A J Gross and V A Clark Survival Distributions ReliabilityApplications in the Biomedical Sciences vol 11Wiley New YorkNY USA 1975

[18] L J Bain and M Engelhardt ldquoProbability of correct selectionof weibull versus gamma based on livelihood ratiordquo Communi-cations in StatisticsmdashTheory and Methods vol 9 no 4 pp 375ndash381 2007

[19] C S Kumar and S H Dharmaja ldquoOn some properties of KiesdistributionrdquoMetron vol 72 no 1 pp 97ndash122 2014

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of