research article parameter estimation and joint confidence
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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
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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OptimizationJournal of
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International Journal of
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Operations ResearchAdvances in
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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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Algebra
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
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
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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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
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
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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
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Operations ResearchAdvances in
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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
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Algebra
Discrete Dynamics in Nature and Society
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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
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Mathematical Problems in Engineering
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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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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
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International 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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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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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
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
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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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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
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
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OptimizationJournal of
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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
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