Skip to content
Licensed Unlicensed Requires Authentication Published by De Gruyter April 14, 2021

Characterization of generalized Gamma-Lindley distribution using truncated moments of order statistics

  • Dorsaf Laribi EMAIL logo , Afif Masmoudi and Imen Boutouria
From the journal Mathematica Slovaca

Abstract

Having only two parameters, the Gamma-Lindley distribution does not provide enough flexibility for analyzing different types of lifetime data. From this perspective, in order to further enhance its flexibility, we set forward in this paper a new class of distributions named Generalized Gamma-Lindley distribution with four parameters. Its construction is based on certain mixtures of Gamma and Lindley distributions. The truncated moment, as a characterization method, has drawn a little attention in the statistical literature over the great popularity of the classical methods. We attempt to prove that the Generalized Gamma-Lindley distribution is characterized by its truncated moment of the first order statistics. This method rests upon finding a survival function of a distribution, that is a solution of a first order differential equation. This characterization includes as special cases: Gamma, Lindley, Exponential, Gamma-Lindley and Weighted Lindley distributions. Finally, a simulation study is performed to help the reader check whether the available data follow the underlying distribution.

MSC 2010: Primary 62-XX
  1. (Communicated by Gejza Wimmer )

References

[1] Abd El-Monsef, M. M. E.—Hassanein, W. A.—Kilany, N. M.: Erlang-Lindley Distribution, Comm. Statist. Theory Methods 46 (2016), 9494–9506.10.1080/03610926.2016.1212069Search in Google Scholar

[2] Arnold, B. C.—Balakrishnan, N.—Nagaraja, H. N.: A First Course in Order Statistics, John Wiley and Sons, New York, 1992.Search in Google Scholar

[3] Afify, A.—Nofal, Z.—Ahmed, A.: Characterization of exponential and power function distributions using Sth truncated moments of order statistics, J. Adv. Math. 4(3) (2013), 486–494.Search in Google Scholar

[4] Ahsanullah, M.—Hamedani, G. G.: Certain characterizations of power function and beta distributions based on order statistics, J. Statist. Theor. Appl. 6 (2007), 220–226.Search in Google Scholar

[5] Ahsanullah, M.—Shakil, M.: Characterizations of Rayleigh distribution based on order statistics and record values, Bull. Malays. Math. Sci. Soc. 36(3) (2013), 625–635.Search in Google Scholar

[6] Ahsanullah, M.—Ghitany, M. E.—Al-Mutairi, D. K.: Characterizations of Lindley distributions by truncated moments, Comm. Statist. Theory Methods 46 (2016), 6222–6227.10.1080/03610926.2015.1124117Search in Google Scholar

[7] Boutouria, I.—Bouzida, I.—Masmoudi, A.: On characterizing the Exponential q-distribution, Bull. Malays. Math. Sci. Soc. 42(6) (2019), 3303–3322.10.1007/s40840-018-0670-5Search in Google Scholar

[8] Glaser, R. E..: Bathtube and related failure rate characterizations, J. Amer. Statist. Assoc. 75 (1980), 667–672.10.1080/01621459.1980.10477530Search in Google Scholar

[9] Galambos, J.—Kotz, S.: Characterizations of Probability Distributions. Lecture Notes in Math. 675, Springer-Verlag, Berlin, Germany, 1978.10.1007/BFb0069530Search in Google Scholar

[10] Gupta, P. L.—Gupta, R. C.: On the moments of residual life in reliability and some characterization results, Comm. Statist. Theory Methods 12(4) (1983), 449–461.10.1080/03610928308828471Search in Google Scholar

[11] Gupta, R. D.—Kundu, D.: Generalized exponential distributions, Aust. N. Z. J. Stat. 41(2) (1999), 173–188.10.1111/1467-842X.00072Search in Google Scholar

[12] Ghitany, M. E.—Atieh, B.—Nadarajah, S.: Lindley distribution and its application, Math. Comput. Simulation 78 (2008), 493–506.10.1016/j.matcom.2007.06.007Search in Google Scholar

[13] Ghitany, M. E.—Alqallaf, F.—Al-Mutairi, D. K.—Husain, H. A.: A two-parameter weighted Lindley distribution and its applications to survival data, Math. Comput. Simulation 81 (2011), 1190–1201.10.1016/j.matcom.2010.11.005Search in Google Scholar

[14] Glänzel, W.: Some consequences of a characterization theorem based on truncated moments, Statistics 21 (1990), 613–618.10.1080/02331889008802273Search in Google Scholar

[15] Glänzel, W.: A characterization theorem based on truncated moments and its application to some distribution families. In: Mathematical Statistics and Probability Theory (Bauer P., Konecny F., Wertz W., (eds.)), Vol B, Springer Netherlands, 1987, 75–84.10.1007/978-94-009-3965-3_8Search in Google Scholar

[16] Hamedani, G. G.: Characterizations of continuous univariate distributions based on the truncated moments of functions of order statistics, Studia Sci. Math. Hungar. 47 (2010), 462–484.10.1556/sscmath.2009.1143Search in Google Scholar

[17] Hussain, E.: The Non-linear Functions of Order Statistics and their Properties in Selected Probability Models, Ph.D. thesis, Department of Statistics, University of Karachi, Pakistan, 2006.Search in Google Scholar

[18] Kilany, N. M.—Atallah, H. M.: Inverted beta Lindley distribution, J. Adv. Math. 13(1) (2017), 4074–4085.10.24297/jam.v13i1.5857Search in Google Scholar

[19] Kilany, N. M.: Characterization of Lindley distribution based on truncated moments of order statistics, J. Stat. Appl. Pro. 6(2) (2017), 1–6.10.18576/jsap/060210Search in Google Scholar

[20] Laurent, A. G.: On characterization of some distributions by truncation properties, J. Amer. Statist. Assoc. 59 (1974), 823–827.10.1080/01621459.1974.10480213Search in Google Scholar

[21] Lindley, D. V.: Fiducial distributions and Bayes theorem, J. R. Stat. Soc. Ser. B Stat. Methodol. 20 (1958), 102–107.10.1111/j.2517-6161.1958.tb00278.xSearch in Google Scholar

[22] Elbatal, I.—Merovci, F.—Elgarhy, M.: A new generalized Lindley distibution, Mathematical Theory and Modeling 3 (2013), 30–47.Search in Google Scholar

[23] Masmoudi, K.—Masmoudi, A.: A new class of continuous Bayesian networks, Internat. J. Approx. Reason 109 (2019), 125–138.10.1016/j.ijar.2019.03.010Search in Google Scholar

[24] Nanda, A. K.: Characterization of distributions through failure rate and mean residual life functions, Statist. Prob. Lett. 80(9–10) (2010), 752–755.10.1016/j.spl.2010.01.006Search in Google Scholar

[25] Nadarajah, S.—Kotz, S.: The exponentiated type distributions, Acta Appl. Math. 92(2) (2006), 97–111.10.1007/s10440-006-9055-0Search in Google Scholar

[26] Su, N. C.—Huang, W. J.: Characterizations based on conditional expectations, Statist. Papers 41(4) (2000), 423–435.10.1007/BF02925761Search in Google Scholar

[27] Ghitany, E. M.—Suja, M. A.—Baqer, A. A.—Gupta, C. R.: Gompertz-Lindley distribution and associated inference, Comm. Statist. Simulation Comput. (2019), https://doi.org/10.1080/03610918.2019.1699113.Search in Google Scholar

[28] Sankaran, M.: The discrete Poisson-Lindley distribution, Test 18 (2009), 497–515.10.1007/s11749-008-0110-1Search in Google Scholar

[29] Shaked, M.—Shanthikumar, G.: Stochastic Orders, Springer: New York, 1994.Search in Google Scholar

[30] Torabi, H.—Falahati-Naeini, M.—Montazeri, N. H.: An extended generalized Lindley distribution and its applications to lifetime data, Journal of Statistical Research of Iran 11 (2014), 203–222.10.18869/acadpub.jsri.11.2.203Search in Google Scholar

[31] Zeghdoudi, H.—Nedjar, S.: On Gamma Lindley distribution: Properties and simulations, J. Comput. Appl. Math. 298 (2016), 167–174.10.1016/j.cam.2015.11.047Search in Google Scholar

[32] Zakerzadeh, H.—Dolati, A.: Generalized Lindley distribution, J. Math. Ext. 3(2) (2009), 13–25.Search in Google Scholar

6 Appendix A

To prove the characterization theorem, we need to use the following propositions which are based on the coming Newten′s generalized Binomial theorem for all α ∈ ℝ, a + bR+,,

(a+b)α=p=0αpbpaαp,if|b|<|a|p=0αpbαpap,if|b|>|a|

where

αp=α(α1)(α2)(αp+1)p!andpα.(6.1)

Lemma 6.1

For all n ≥ 1 andk = 1, 2, …, n, wherexk ∈ ℝ, y ∈ ℝ*andxk + yR+,we obtain according to Newten’s generalized Binomial theorem, the following expressions:

  1. Ifxk∣ < ∣yandPi ∈ ℕ, i = 1, …, n,

    k=1n(xk+y)α1=yn(α1)Pn=0Pn1=0PnPn2=0Pn1P1=0P2α1P1α1P2P1α1Pn1Pn2×α1PnPn1x1P1x2P2P1xnPnPn1(1y)Pn.
  2. Ifxk∣ > ∣yandPi ∈ ℕ, i = 1, …, n,

    k=1n(xk+y)α1=Pn=0Pn1=0PnPn2=0Pn1P1=0P2α1P1α1Pn1Pn2α1PnPn1×x1αP11x2αP2+P11xnαPn+Pn11(y)Pn.

Mathematical induction is used to prove this lemma.

Proof

For all n ≥ 1 and ∣xk∣ < ∣y∣, we denote

Qn:k=1n(xk+y)α1=yn(α1)Pn=0Pn1=0PnPn2=0Pn1P2=0P3P1=0P2α1P1×α1P2P1α1Pn1Pn2α1PnPn1×x1P1x2P2P1xnPnPn1(1y)Pn.

For n = 1,

(x1+y)α1=y(α1)P1=0α1P1(1y)P1x1P1,

for n = 2,

(x1+y)α1(x2+y)α1=y2(α1)p=0α1p(1y)px1pq=0α1q(1y)qx2q,=y2(α1)P2=0P1=0P2α1P1α1P2P1x1P1x2P2P1(1y)P2.

For n ≥ 2, we assume that the formula is true for n and we need now to show that it is also true for n + 1,

k=1n+1(xk+y)α1=k=1n(xk+y)α1(xn+1+y)α1=yn(α1)Pn=0Pn1=0PnPn2=0Pn1P2=0P3P1=0P2α1P1α1P2P1α1Pn1Pn2α1PnPn1×x1P1x2P2P1xnPnPn1(1y)Pny(α1)×Pn+1=0α1Pn+1(1y)Pn+1x1Pn+1,=y(n+1)(α1)Pn+1=0Pn=0Pn+1Pn1=0PnP2=0P3P1=0P2α1P1α1P2P1α1Pn1Pn2×α1PnPn1α1Pn+1Pnx1P1x2P2P1xnPnPn1xn+1Pn+1Pn(1y)Pn+1.

For all k = 1, 2, …, n, where ∣xk∣ > ∣y∣, we denote

Qn:k=1n(xk+y)α1=Pn=0Pn1=0PnPn2=0Pn1P2=0P3P1=0P2α1P1α1P2P1α1Pn1Pn2×α1PnPn1x1αP11x2αP2+P11xnαPn+Pn11×(y)Pn.

The same steps are applied to prove that Qn is true.□

Let

τk(x)=x(θy)αkeθyk(Γ(α,θy))nrkdy,=x(θy)αkeθykθyθyz1α1z2α1znrkα1×exp(j=1nrkzj)dz1dznrk,=x(θy)αkeθyk00(u1+θy)α1(u2+θy)α1(unrk+θy)α1×exp(j=1nrkuj)e(nrk)θydu1dunrk.

We need the following two lemmas to calculate τk(x) in two different ways.

Lemma 6.2

For allu1∣, ∣u2∣, …, ∣unrk∣ < ∣θy∣, andk = 0, 1, …, nr, if (α − 1)(nr − 1) + k > 0,

τk(x)=Pnrk=0Pnrk1=0PnrkP1=0P2α1P1α1P2P1α1PnrkPnrk1×Γ(P1+1)Γ(P2P1+1)Γ(PnrkPnrk1+1)×Γ(αk+(α1)(nrk)Pnrk+1,(nr)θx)(θ(nr))αk+(α1)(nrk)Pnrk+1.

Proof

Lemma 6.1 implies

(u1+θy)α1(unrk+θy)α1=(θy)(nrk)(α1)Pnrk=0Pnrk1=0PnrkPnrk2=0Pnrk1P2=0P3P1=0P2α1P1×α1P2P1α1Pnrk1Pnrk2α1PnrkPnrk1×u1P1u2P2P1unrkPnrkPnrk1(1θy)Pnrk.

It follows that

τk(x)=x(θy)αkeθyk00(u1+θy)α1(u2+θy)α1(unrk+θy)α1×exp(j=1nrkuj)e(nrk)θydu1dunrk=Pnrk=0Pnrk1=0PnrkPnrk2=0Pnrk1P1=0P2α1P1α1P2P1α1PnrkPnrk1×Γ(P1+1)Γ(PnrkPnrk1+1)x(θy)αk+(nrk)(α1)Pnrke(nr)θydy,=Pnrk=0Pnrk1=0PnrkPnrk2=0Pnrk1P1=0P2α1P1α1P2P1α1PnrkPnrk1×Γ(P1+1)Γ(PnrkPnrk1+1)x(θy)(nr)(α1)+kPnrke(nr)θydy.

If (α − 1)(nr − 1) + k > 0, then

τk(x)=Pnrk=0Pnrk1=0PnrkPnrk2=0Pnrk1P1=0P2α1P1α1P2P1α1PnrkPnrk1×Γ(P1+1)Γ(PnrkPnrk1+1)×Γ(αk+(α1)(nrk)Pnrk+1,(nr)θx)(θ(nr))αk+(α1)(nrk)Pnrk+1.

Lemma 6.3

For allu1∣, ∣u2∣, …, ∣unrk∣> ∣θy∣,

τk(x)=Pnrk=0Pnrk1=0PnrkPnrk2=0Pnrk1P1=0P2α1P1α1P2P1α1PnrkPnrk1×Γ(αP1)Γ(αPnrk+Pnrk1)Γ(αk+Pnrk+1,(nr)θx)(θ(nr))αk+Pnrk+1.

To prove this lemma, the same approach is applied as in the previous proposal.

We now assert that

H1=Pnrk=0Pnrk1=0PnrkP1=0P2α1P1α1PnrkPnrk1×Γ(P1+1)Γ(PnrkPnrk1+1).

This expression is needed to prove what follows.

H1×(θx)Pnrk=e(nrk)θx(θx)(nrk)(α1)(Γ(α,θx))nrk,

which implies

H1×(θx)Pnrk=(θx)PnrkPnrk=0Pnrk1=0PnrkP2=0P3P1=0P2α1P1α1PnrkPnrk1×0z1P1ez1dz10z2P2P1ez2dz20znrkPnrkPnrk1eznrkdznrk,=00Pnrk=0Pnrk1=0PnrkP2=0P3P1=0P2α1P1α1PnrkPnrk1×z1P1×z2P2P1××znrkPnrkPnrk1(θx)Pnrk×ei=1nrkzidz1dznrk,=(θx)(nrk)(α1)00(z1+θx)α1(z2+θx)α1(znrk+θx)α1×ei=1nrkzidz1dznrk,=e(nrk)θx(θx)(nrk)(α1)×(Γ(α,θx))nrk.
Received: 2020-03-12
Accepted: 2020-07-15
Published Online: 2021-04-14
Published in Print: 2021-04-27

© 2021 Mathematical Institute Slovak Academy of Sciences

Downloaded on 29.3.2024 from https://www.degruyter.com/document/doi/10.1515/ms-2017-0481/html
Scroll to top button