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Original Articles

A Consistent Method of Estimation For The Three-Parameter Gamma Distribution

, &
Pages 3905-3926 | Received 12 Jan 2012, Accepted 16 Jul 2012, Published online: 26 Aug 2014
 

Abstract

For the three-parameter gamma distribution, it is known that the method of moments as well as the maximum likelihood method have difficulties such as non-existence in some range of the parameters, convergence problems, and large variability. For this reason, in this article, we propose a method of estimation based on a transformation involving order statistics from the sample. In this method, the estimates always exist uniquely over the entire parameter space, and the estimators also have consistency over the entire parameter space. The bias and mean squared error of the estimators are also examined by means of a Monte Carlo simulation study, and the empirical results show the small-sample superiority in addition to the desirable large sample properties.

Mathematics Subject Clssification:

A. Proofs

A.1 Proof of Proposition 2.1.

Denote for convenience F( · ; α, 1, 0) by G( · ; α) and f( · ; α, 1, 0) by g( · ; α). Suppose Z1, ⋅⋅⋅, Zn are n independent random variables from the standard gamma distribution with shape parameter α. For i = 1, ⋅⋅⋅, n, let Z(i) be the i-th order statistic among Z1, ⋅⋅⋅, Zn.

For n − 2 real values 0 ⩽ w2 ⩽ ⋅⋅⋅ ⩽ wn − 1 ⩽ 1, let us consider (13)

For every u, v such that v > u > 0, n > 2 and α > 0, the integrand in (Equation13), i.e., n!g(u ; α)g(v ; α)∏n − 1i = 2G(u + (vu)wi ; α), has a partial derivative n!g(u ; α)g(v ; α)∏n − 1i = 2(vu)g(u + (vu)wi ; α), with respect to wi, i = 2, …, n − 1. Moreover, we note that (n − 2)!∏n − 1i = 2(vu)g(u + (vu)wi ; α)/{G(v ; α) − G(u ; α)}n − 2 is bounded above. From the boundedness of (n − 2)!∏n − 1i = 2(vu)g(u + (vu)wi ; α)/{G(v ; α) − G(u ; α)}n − 2, we have where M1 is a positive constant, and

Then, by applying Part (ii) of Theorem 16.8 of Billingsley (Citation1994), the partial derivative of with respect to wi, i = 2, …, n − 1, is given by (14) For w2, …, wn − 1 for which 0 ⩽ w2 ⩽ ⋅⋅⋅ ⩽ wn − 1 ⩽ 1 is not satisfied, the partial derivative of with respect to wi, i = 2, …, n − 1, is always equal to 0 since After suitable transformations of variables u and v, the proof of Proposition 2.1 gets completed.

A.2 Proof of Lemma 2.1

For α > 0, given w2, ⋅⋅⋅, wn − 1 such that 0 ⩽ w2 ⩽ ⋅⋅⋅ ⩽ wn − 1 ⩽ 1, L(α; w2, ⋅⋅⋅, wn − 1) is expressed as where η(α, u, v; w2, …, wn − 1)=− nlog Γ(α)+ (n − 2)log v + (α − 1)∑ni = 1log (u + wiv)− ∑ni = 1(u + wiv), and w1 = 0 and wn = 1. Without loss of generality, we denote η(α, u, v; w2, …, wn − 1) by η(α, u, v) in the remaining part of this appendix. For every u > 0, v > 0, n > 2, α > 0 and w2, …, wn − 1 such that 0 ⩽ w2 ⩽ … ⩽ wn − 1 ⩽ 1, the partial derivative of exp {η(α, u, v)} with respect to α is given by η′(α, u, v)exp {η(α, u, v)}, where η′(α, u, v) is the partial derivative of η(α, u, v) with respect to α, which is − nψ(α) + ∑ni = 1log (u + wiv), where ψ( · ) is the digamma function.

For every u > 0, v > 0, n > 2, α > 0 and w2, …, wn − 1 such that 0 ⩽ w2 ⩽ ⋅⋅⋅ ⩽ wn − 1 ⩽ 1, |η′(α, u, v) exp {η(α, u, v)}| is bounded above and there then exists a positive constant M2 such that |η′(α, u, v)exp {η(α, u, v)/2}| ⩽ M2. Thus, ∫00|η′(α, u, v)exp {η(α, u, v)}| dvduM200exp {η(α, u, v)/2} dvduM2(∫00exp {η(α, u, v)} dvdu)1/2 < ∞. The second last inequality is due to Lyapunov’s inequality while the last inequality holds from Proposition 2.1.

Then, by applying Part (ii) of Theorem 16.8 of Billingsley (Citation1994), the derivative of L(α; w2, …, wn − 1) is given by (15) The proof of Lemma 2.1 thus follows.

A.3 Proof of Theorem 2.1

First, we shall show that the likelihood equation has at least one solution.

Without loss of generality, given w2, …, wn − 1 such that 0 ⩽ w2 ⩽ ⋅⋅⋅ ⩽ wn − 1 ⩽ 1, we denote L(α; w2, …, wn − 1) by L(α), and L′(α; w2, …, wn − 1) by L′(α) in the remaining part of this appendix. Since exp {η(α, u, v)} > 0, limα↓0η′(α, u, v)= ∞ and limα → ∞η′(α, u, v)= −∞ for every α > 0 and u and v > 0, there exists a positive real value δ1 such that L′(α) > 0 for all α in (0, δ1) and a positive real value δ2 such that L′(α) < 0 for all α > δ2. Also, for α > 0, L′(α) is continuous with respect to α. Thus, L′(α) = 0 always has at least one solution.

Next, we will show that the number of solutions is exactly one.

Let Tu, v= ∑ni = 1log (u + vwi) (− ∞ < Tu, v < ∞) for simplicity. Then, η(α, u, v) and η′(α, u, v) are rewritten as and

Note that Tu, v takes on values over the entire real line. Thus, there exists a unique value of Tu, v such that η′(α, Tu, v) = 0 for any fixed α > 0. We denote that value of Tu, v by T0(α). We see that η′(α, Tu, v) < 0 for Tu, v< T0(α) and η′(α, Tu, v) > 0 for Tu, v> T0(α) for every α > 0.

Let, for Δα, (16) then, L′(α + Δα) can be rewritten as (17) where Cn = n!.

From now on, let us focus on the case when Δα ⩾ 0.

We note that (18) while (19) for any α.

Let α* be one of the solutions of L′(α) = 0. Then, (20)

From (Equation18) and (Equation19), it is easy to see that φ(α*, Δα, Tu, v), for any (u, v) ∈ {Tu, vT0(α*)}, approaches 1 faster than for any (u′, v′) ∈ approaching 1 when Δα decreases. Hence, φ(α*, Δα, Tu, v) η′(α*, Tu, v) exp {η(α*, Tu, v)} dvdu approaches η′(α*, Tu, v)exp {η(α*, Tu, v)} dvdu faster than φ(α*,Δα,Tu, v)η′(α*, Tu, v)exp {η(α*, Tu, v)} dvdu approaching η′(α*,Tu, v)exp {η(α*,Tu, v)} dvdu when Δα decreases. Note further that η′(α*, Tu, v)exp {η(α*, Tu, v)} dvdu=η′(α*, Tu, v)exp {η(α*, Tu, v)} dvdu=0. Therefore, by the fundamental theory of differential calculus, the sign of the last equality of (Equation20) agrees with the sign of φ(α*, Δα, Tu, v) η′(α*, Tu, v) exp {η(α*, Tu, v)} dvdu for sufficiently small Δα > 0, which implies (21) since, for any Δα > 0, η′(α* + Δα, T0(α*))<0 and exp {η(α* + Δα, Tu, v)} > 0, and thus φ(α*, Δα, Tu, v) η′(α*, Tu, v) exp {η(α*, Tu, v)} dvdu= exp {η(α* + Δα, Tu, v)} dvdu<0.

Analogously, we obtain the following inequality: (22) The proof is very similar to the proof of (Equation21) and is therefore omitted for the sake of brevity.

It follows from (Equation21), (Equation22), and the fact that L′(α) is differentiable with respect to α (the proof is very similar to the proof of the differentiability of L(σ; w2, ⋅⋅⋅, wn − 1) in Appendix 6.2 and is therefore omitted for the sake of brevity) that holds. This clearly implies that L′(α) changes sign only once with respect to α.

From the facts established above, L′(α*)= 0 always has a unique solution with respect to α. The proof of Theorem 2.1 is thus completed.

A.4 Proof of Lemma 2.2

Let Wi, i = 2, …, n − 1, be the random variables whose order statistics are W(i), i = 2, …, n − 1. For every u and v such that 0 < u < v, under the conditions Z(1) = u, Z(n) = v, where Z(1) = (X(1) − γ0)/β0 and Z(n) = (X(n) − γ0)/β0, and β0 and γ0 are the true values of β and γ, respectively, since the conditional joint pdf of the order statistics of Wi’s, given Z(1) = u, Z(n) = v, is given by Wi, i = 2, …, n − 1, are distributed with common conditional pdf, given Z(1) = u, Z(n) = v, which is given by (23) and these are conditionally independent, given Z(1) = u, Z(n) = v.

Denote (n − 2)!∏n − 1i = 2(vu)g(u + (vu)W(i); α)/{G(v ; α) − G(u ; α)} by

Lu, v(α; W(2), ⋅⋅⋅, W(n − 1)).

For any fixed u and v such that 0 < u < v, under the conditions Z(1) = u, Z(n) = v, let us consider, for every α ≠ α0, n > 2, (24) By the law of large numbers, (Equation24) tends in probability to (25) where W is a random variable which is distributed with the conditional pdf in (Equation23), given Z(1) = u, Z(n) = v. By Jensen’s inequality, we have (26) It then follows that or (27) By the positivity and the integrability of Lu, v(α; W(2), ⋅⋅⋅, W(n − 1)) and

Lu, v0; W(2), ⋅⋅⋅, W(n − 1)), (Equation27) implies (28) Moreover, (29) and (30) since .|Z(1) = u, Z(n) = v) is bounded by 1.

Then, by applying the dominated convergence theorem, from (Equation29) and (Equation30), it follows that The proof of Lemma 2.2 is thus complete.

B. Tables of Bias and Rmse

See Tables and .

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