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

Dorsaf Laribi; Afif Masmoudi; Imen Boutouria

doi:10.1515/ms-2017-0481

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

Dorsaf Laribi , Afif Masmoudi and Imen Boutouria

Published/Copyright: April 14, 2021

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From the journal Mathematica Slovaca Volume 71 Issue 2

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

Keywords: Gamma-Lindley distribution; characterization; survival function; order statistics; truncated moments

(Communicated by Gejza Wimmer )

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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 + b ∈ R+∗,,

(a+b)α=∑p=0∞αpbpaα−p,if|b|<|a|∑p=0∞αpbα−pap,if|b|>|a|

where

αp=α(α−1)(α−2)…(α−p+1)p!andp≤α.(6.1)

Lemma 6.1

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

If ∣x_k∣ < ∣y∣ andP_i ∈ ℕ, i = 1, …, n,
∏k=1n(xk+y)α−1=yn(α−1)∑Pn=0∞∑Pn−1=0Pn∑Pn−2=0Pn−1⋯∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−1−Pn−2×α−1Pn−Pn−1x1P1x2P2−P1⋯xnPn−Pn−1(1y)Pn.
If ∣x_k∣ > ∣y∣ andP_i ∈ ℕ, i = 1, …, n,
∏k=1n(xk+y)α−1=∑Pn=0∞∑Pn−1=0Pn∑Pn−2=0Pn−1⋯∑P1=0P2α−1P1⋯α−1Pn−1−Pn−2α−1Pn−Pn−1×x1α−P1−1x2α−P2+P1−1⋯xnα−Pn+Pn−1−1(y)Pn.

Mathematical induction is used to prove this lemma.

Proof

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

Qn:∏k=1n(xk+y)α−1=yn(α−1)∑Pn=0∞∑Pn−1=0Pn∑Pn−2=0Pn−1⋯∑P2=0P3∑P1=0P2α−1P1×α−1P2−P1⋯α−1Pn−1−Pn−2α−1Pn−Pn−1×x1P1x2P2−P1⋯xnPn−Pn−1(1y)Pn.

For n = 1,

(x1+y)α−1=y(α−1)∑P1=0∞α−1P1(1y)P1x1P1,

for n = 2,

(x1+y)α−1(x2+y)α−1=y2(α−1)∑p=0∞α−1p(1y)px1p∑q=0∞α−1q(1y)qx2q,=y2(α−1)∑P2=0∞∑P1=0P2α−1P1α−1P2−P1x1P1x2P2−P1(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)α−1=∏k=1n(xk+y)α−1(xn+1+y)α−1=yn(α−1)∑Pn=0∞∑Pn−1=0Pn∑Pn−2=0Pn−1⋯∑P2=0P3∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−1−Pn−2α−1Pn−Pn−1×x1P1x2P2−P1⋯xnPn−Pn−1(1y)Pny(α−1)×∑Pn+1=0∞α−1Pn+1(1y)Pn+1x1Pn+1,=y(n+1)(α−1)∑Pn+1=0∞∑Pn=0Pn+1∑Pn−1=0Pn⋯∑P2=0P3∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−1−Pn−2×α−1Pn−Pn−1α−1Pn+1−Pnx1P1x2P2−P1⋯xnPn−Pn−1xn+1Pn+1−Pn(1y)Pn+1.

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

Qn:∏k=1n(xk+y)α−1=∑Pn=0∞∑Pn−1=0Pn∑Pn−2=0Pn−1⋯∑P2=0P3∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−1−Pn−2×α−1Pn−Pn−1x1α−P1−1x2α−P2+P1−1⋯xnα−Pn+Pn−1−1×(y)Pn.

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

Let

τk(x)=∫x∞(θy)αke−θyk(Γ(α,θy))n−r−kdy,=∫x∞(θy)αke−θyk∫θy∞⋯∫θy∞z1α−1z2α−1⋯zn−r−kα−1×exp⁡(−∑j=1n−r−kzj)dz1⋯dzn−r−k,=∫x∞(θy)αke−θyk∫0∞⋯∫0∞(u1+θy)α−1(u2+θy)α−1⋯(un−r−k+θy)α−1×exp⁡(−∑j=1n−r−kuj)e−(n−r−k)θydu1⋯dun−r−k.

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

Lemma 6.2

For all ∣u₁∣, ∣u₂∣, …, ∣u_n−r−k∣ < ∣θy∣, andk = 0, 1, …, n − r, if (α − 1)(n − r − 1) + k > 0,

τk(x)=∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k⋯∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−r−k−Pn−r−k−1×Γ(P1+1)Γ(P2−P1+1)⋯Γ(Pn−r−k−Pn−r−k−1+1)×Γ(αk+(α−1)(n−r−k)−Pn−r−k+1,(n−r)θx)(θ(n−r))αk+(α−1)(n−r−k)−Pn−r−k+1.

Proof

Lemma 6.1 implies

(u1+θy)α−1⋯(un−r−k+θy)α−1=(θy)(n−r−k)(α−1)∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k∑Pn−r−k−2=0Pn−r−k−1⋯∑P2=0P3∑P1=0P2α−1P1×α−1P2−P1⋯α−1Pn−r−k−1−Pn−r−k−2α−1Pn−r−k−Pn−r−k−1×u1P1u2P2−P1⋯un−r−kPn−r−k−Pn−r−k−1(1θy)Pn−r−k.

It follows that

τk(x)=∫x∞(θy)αke−θyk∫0∞⋯∫0∞(u1+θy)α−1(u2+θy)α−1⋯(un−r−k+θy)α−1×exp⁡(−∑j=1n−r−kuj)e−(n−r−k)θydu1⋯dun−r−k=∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k∑Pn−r−k−2=0Pn−r−k−1⋯∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−r−k−Pn−r−k−1×Γ(P1+1)⋯Γ(Pn−r−k−Pn−r−k−1+1)∫x∞(θy)αk+(n−r−k)(α−1)−Pn−r−ke−(n−r)θydy,=∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k∑Pn−r−k−2=0Pn−r−k−1⋯∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−r−k−Pn−r−k−1×Γ(P1+1)⋯Γ(Pn−r−k−Pn−r−k−1+1)∫x∞(θy)(n−r)(α−1)+k−Pn−r−ke−(n−r)θydy.

If (α − 1)(n − r − 1) + k > 0, then

τk(x)=∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k∑Pn−r−k−2=0Pn−r−k−1⋯∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−r−k−Pn−r−k−1×Γ(P1+1)⋯Γ(Pn−r−k−Pn−r−k−1+1)×Γ(αk+(α−1)(n−r−k)−Pn−r−k+1,(n−r)θx)(θ(n−r))αk+(α−1)(n−r−k)−Pn−r−k+1.

□

Lemma 6.3

For all ∣u₁∣, ∣u₂∣, …, ∣u_n−r−k∣> ∣θy∣,

τk(x)=∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k∑Pn−r−k−2=0Pn−r−k−1⋯∑P1=0P2α−1P1α−1P2−P1⋯α−1Pn−r−k−Pn−r−k−1×Γ(α−P1)⋯Γ(α−Pn−r−k+Pn−r−k−1)Γ(αk+Pn−r−k+1,(n−r)θx)(θ(n−r))αk+Pn−r−k+1.

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

We now assert that

H1=∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k⋯∑P1=0P2α−1P1⋯α−1Pn−r−k−Pn−r−k−1×Γ(P1+1)⋯Γ(Pn−r−k−Pn−r−k−1+1).

This expression is needed to prove what follows.

H1×(θx)−Pn−r−k=e(n−r−k)θx(θx)−(n−r−k)(α−1)(Γ(α,θx))n−r−k,

which implies

H1×(θx)−Pn−r−k=(θx)−Pn−r−k∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k⋯∑P2=0P3∑P1=0P2α−1P1⋯α−1Pn−r−k−Pn−r−k−1×∫0∞z1P1e−z1dz1∫0∞z2P2−P1e−z2dz2⋯∫0∞zn−r−kPn−r−k−Pn−r−k−1e−zn−r−kdzn−r−k,=∫0∞⋯∫0∞∑Pn−r−k=0∞∑Pn−r−k−1=0Pn−r−k⋯∑P2=0P3∑P1=0P2α−1P1⋯α−1Pn−r−k−Pn−r−k−1×z1P1×z2P2−P1×⋯×zn−r−kPn−r−k−Pn−r−k−1(θx)−Pn−r−k×e−∑i=1n−r−kzidz1⋯dzn−r−k,=(θx)−(n−r−k)(α−1)∫0∞⋯∫0∞(z1+θx)α−1(z2+θx)α−1⋯(zn−r−k+θx)α−1×e−∑i=1n−r−kzidz1⋯dzn−r−k,=e(n−r−k)θx(θx)−(n−r−k)(α−1)×(Γ(α,θx))n−r−k.

Received: 2020-03-12

Accepted: 2020-07-15

Published Online: 2021-04-14

Published in Print: 2021-04-27

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Articles in the same Issue

https://doi.org/10.1515/ms-2017-0481

Keywords for this article

Gamma-Lindley distribution; characterization; survival function; order statistics; truncated moments