# On point estimators for Gamma and Beta distributions

Let X_1,…,X_n be a random sample from the Gamma distribution with density f(x)=λ^αx^α-1e^-λ x/Γ(α), x>0, where both α>0 (the shape parameter) and λ>0 (the reciprocal scale parameter) are unknown. The main result shows that the uniformly minimum variance unbiased estimator (UMVUE) of the shape parameter, α, exists if and only if n≥ 4; moreover, it has finite variance if and only if n≥ 6. More precisely, the form of the UMVUE is given for all parametric functions α, λ, 1/α and 1/λ. Furthermore, a highly efficient estimating procedure for the two-parameter Beta distribution is also given. This is based on a Stein-type covariance identity for the Beta distribution, followed by an application of the theory of U-statistics and the delta-method. MSC: Primary 62F10; 62F12; Secondary 62E15. Key words and phrases: unbiased estimation; Gamma distribution; Beta distribution; Ye-Chen-type closed-form estimators; asymptotic efficiency; U-statistics; Stein-type covariance identity; delta-method.

## Authors

• 5 publications
12/11/2020

11/30/2020

### The statistical properties of RCTs and a proposal for shrinkage

We abstract the concept of a randomized controlled trial (RCT) as a trip...
04/16/2021

### A Bivariate Beta Distribution with Arbitrary Beta Marginals and its Generalization to a Correlated Dirichlet Distribution

We discuss a bivariate beta distribution that can model arbitrary beta-d...
05/24/2021

### Asymptotic Moments Matching to Uniformly Minimum Variance Unbiased Estimation under Ewens Sampling Formula

The Ewens sampling formula is a distribution related to the random parti...
03/26/2019

### Predicting the scoring time in hockey

In this paper, we propose a Bayesian predictive density estimator to pre...
11/08/2019

### Relation between Blomqvist's beta and other measures of concordance of copulas

An investigation is presented of how a comprehensive choice of four most...
02/13/2018

### Maturation Trajectories of Cortical Resting-State Networks Depend on the Mediating Frequency Band

The functional significance of resting state networks and their abnormal...
##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

Let , , be the Gamma function. The Gamma distribution, , with density

 f(x)=λαΓ(α)xα−1e−λx,    x>0,

and parameters (the shape) and (the reciprocal scale), generates one of the most useful statistical models, especially when the data are nonnegative. This distribution belongs to the Pearson family, and in particular, to the Integrated Pearson family of distributions, see, e.g., Afendras and Papadatos (2015). Therefore, estimating procedures for the parameters , , and/or and , are of fundamental importance in applications. It is well-known that the maximum likelihood estimators (MLEs) are not tractable and, consequently, one has to apply numerical procedures for their evaluation. Moreover, to the best of our knowledge, it is not even known whether an unbiased estimator for (and ) exists, for some sample size . In the contrary, unbiased estimators for the reciprocal parameters and

do exist; recently, Ye and Chen (2017) obtained closed-form unbiased estimators for the reciprocal parameters with very high asymptotic efficiency. It should be noted that an augmented-sample (simulation) technique for obtaining exact confidence intervals for

is introduced by Iliopoulos (2016).

In the present work we show that an unbiased estimator for exists if and only if the sample size is at least , and the same is true for (provided that is not too small). More precisely, in Section 2 we obtain the uniformly minimum variance unbiased estimators (UMVUEs) of , , , . In particular, the UMVUE of , based on two observations, can be written as a positive symmetric kernel, whence, the corresponding -statistic generates the Ye-Chen (2017) estimator; see Remark 2.2.

Let , , , be the Beta function. The Beta distribution has density

 f(x)=1B(α,β)xα−1(1−x)β−1,    0

This distribution belongs to the Integrated Pearson family, and, at least to our knowledge, it is not known whether the parameters can be unbiasedly estimated. The MLEs do not have closed forms, and one has to overcome similar difficulties as for the Gamma case. In Section 3 we obtain closed-form, Ye-Chen-type, estimators for the parameters. Our method is different than that of Ye and Chen (2017), since it is based on a Stein-type covariance identity for the Beta distribution, followed by an application of the theory of -statistics. The covariance identity generates an unbiased bivariate symmetric kernel for the parameter . Then, an application of the classical theory of -statistics produces an unbiased estimator of , which, in turn, yields the desired estimators of and . The asymptotic efficiency of the proposed estimators is studied in some detail, and it is shown to be very high (see Table 1). The Beta model is useful not only for analyzing data with bounded support, but also when positive data are mapped to the interval by means of a transformation, e.g., .

## 2 Unbiased estimation of the parameters of Gamma distribution

Let () be a random sample from , where the unknown parameter . It is well-known that the complete, sufficient statistic can be written in the form , where and

is the ratio of the geometric to the arithmetic mean of the data. The random variables

are independent, as follows, e.g., from Basu’s theorem, because the distribution of is free of , while is complete for when is known. The random variable follows a distribution, while is distributed like , where the random variables , , are independent, and has a Beta distribution (1.1) with parameters and .

By using the Beta product representation, it follows inductively that the random variable has density

 fY(y;α)=K(α)yα−1(1−y)(n−3)/2G(y),   0

where , and the (power-series) function is given by

 G(y)=∞∑m=0dm(1−y)m,  0

Notice that our notation suspends the dependence of and on .

In (2.2) the coefficients are defined recurrently as follows:

 γ1(m)=I(m=0)={1,ifm=0,0,ifm=1,2,…, γi+1(m)=Γ(m+i(i+1)/(2n))Γ(m+(i+1)(i+2)/(2n))m∑k=0Γ(m−k+(i+1)/n)(m−k)!γi(k),    i=1,…,n−2, dm:=γn−1(m),    m=0,1,… . (2.3)

Most of the preceding results can be found in Glaser (1976), Nandi (1980), and Tang and Gupta (1984). Obviously, when . Moreover, for , is strictly decreasing and positive; more precisely, for all , and , . It should be noted at this point that, when , the auxiliary random variable has density (cf. Tang and Gupta, 1984)

 fY1(y;α)=K1(α)yα−1(1−y)(n−1)(n−2)/(2n)−1G1(y),   0

where and

 G1(y)=∞∑k=0γn−2(k)(1−y)k,   0

The random variable will be used in the proof of Theorem 2.1, below.

###### Definition 2.1.

An estimator will be called uniformly minimum variance unbiased estimator (UMVUE) for the parametric function if is a (Borel) function of the complete, sufficient statistic , and , for all , , where the subscript denotes expectation w.r.t. the joint density of ,

 f(x1,…,xn)=λnαΓ(α)n(n∏i=1xi)α−1exp(−λn∑i=1xi),    (x1,…,xn)∈(0,∞)n.

Accordingly, an UMVUE may, or may not have finite variance. For completeness of the presentation, the term UMVUE will be also used in a wider sense, namely, in the case where the equality is merely satisfied for all in a subset of with nonempty interior.

In the sequel we shall make use of the following simple, but useful, observation, saying that if two functions have identical Laplace transforms in an arbitrarily small interval then they coincide.

###### Lemma 2.1.

Let where . Suppose that for the Borel functions , , it is true that for all . If

 ∫∞0e−λxg1(x)dx=∫∞0e−λxg2(x)dx,   λ∈I,

then for almost all .

Proof: For write where , . By assumption,

 0≤∫∞0e−λxw1(x)dx=∫∞0e−λxw2(x)dx<∞,   λ∈I,

where , . Fix and set . Since , it is clear that implies that and a.e.; then, a.e. and, thus, a.e. If

, we may define the probability densities

, ; then, for ,

 ∫∞0e(λ0−λ)xf1(x)dx=1ν0∫∞0e−λxw1(x)dx=1ν0∫∞0e−λxw2(x)dx=∫∞0e(λ0−λ)xf2(x)dx.

If

is the moment generating function of

, it follows that for all , and since contains the origin, a.e.; this implies that a.e., and consequently, a.e.

###### Proposition 2.1.

Let be a Borel measurable function satisfying

 ∫∞0∫∞0|s(x,y)|exp(−λx−αy)xnα−1dydx<∞  for all α>0, λ>0.

Then, the function

 H(α,λ):=∫∞0∫∞0s(x,y)exp(−λx−αy)xnα−1dydx

is continuous in .

Proof: Consider an arbitrary sequence , as . Then, we can find such that and for , The function is obviously decreasing (for each fixed ), so that , . Consider also the function (for fixed , ). Then, is a linear function of , and thus, , . Combining the above we obtain the inequality

 e−λmx−αmyxnαm−1

It follows that the sequence

 wm(x,y):=s(x,y)e−λmx−αmyxnαm−1,   (x,y)∈(0,∞)2,  m=m0,m0+1,…,

is dominated by , and, by assumption, is integrable on . Clearly, as . Thus, by dominated convergence,

completing the proof.

###### Proposition 2.2.

Let be a parametric function of the shape parameter. If there exists an unbiased estimator for , then the UMVUE of is a function of , alone.

Proof: By assumption, for all , . The function can always be taken to be Borel measurable. From the RB/LS Theorem, the UMVUE is given by , where can also be taken to be Borel measurable with no loss of generality. Unbiasedness of the estimator means that (recall that are independent)

 ∫∞0∫10{λnαΓ(nα)xnα−1e−λx}{K(α)yα−1(1−y)(n−3)/2G(y)}U(x,y)dydx=h(α),

for all , . This relation can be rewritten in the form

Since for all and , we may apply Proposition 2.1 with to conclude that the function is continuous, and hence, is a continuous function of ; notice that all functions , , and , are positive and continuous.

Write now the relation in the form

 ∫∞0e−λxh(α)xnα−1dx=∫∞0e−λxK(α)xnα−1{∫10yα−1(1−y)(n−3)/2G(y)U(x,y)dy}dx.

It follows that the Laplace transforms of the functions and are identical, and Lemma 2.1 implies that for every ,

 h(α)K(α)=∫10yα−1(1−y)(n−3)/2G(y)U(x,y)dy,   for almost all x∈(0,∞). (2.6)

Write and set , , , . By construction, the Lebesgue measure of is zero. We chose an arbitrary and define . Notice that is not necessarily Borel measurable; it is, however, Lebesgue measurable and, hence, it is equal a.e. to a Borel function . Obviously, substituting in place of , the values of the above integrals are not affected. From (2.6) we have

 h(α)K(α)=∫10yα−1(1−y)(n−3)/2G(y)u(y)dy,    α∈Q+. (2.7)

The right-hand side of (2.7) defines a continuous function of , because

is the Laplace transform of the function . Since and are both continuous, it follows from (2.7) that for all . Hence, is an unbiased estimator of , and also, it is a function of the complete, sufficient statistic ; thus, it is the (unique) UMVUE.

###### Remark 2.1.

(A consequence to the Cramér-Rao bound). The Fisher information matrix based on a single observation from is given by

 J=J(α,λ)=1λ2(λ2ψ1(α)−λ−λα), with inverse J−1=1αψ1(α)−1(αλλλ2ψ1(α)),

where . The Cramér-Rao (CR) bound says that for any unbiased estimator of the parametric function , . Hence, in the particular case where is a parametric function of the shape parameter alone (i.e., does not depend on ), the CR bound takes the form

 n\rm Var T(\boldmathX)≥h′(α)2ψ1(α)−1/α,   whenever  \rm IE (α,λ)T(\boldmathX)=h(α)  for all α>0, λ>0. (2.8)

We shall now show that Proposition 2.2 yields a better lower bound. Indeed, since the UMVUE of is a Borel function of , say , we can apply the CR bound, , where is the Fisher information of . From (2.1), the density of is given by

 fY(y;α)=exp(alogy−A(α))k(y),

where and , with denoting the indicator function of . This shows that defines a natural exponential family of Lebesgue densities, and thus, the regularity conditions are fulfilled. We calculate and, therefore,

 JY = (−logK(α))′′=(nlogΓ(α)+n−1∑i=1logΓ(i/n)−n−1∑i=1logΓ(α+i/n))′′ = nψ1(α)−n−1∑i=1ψ1(α+i/n).

It follows that for any unbiased estimator of ,

 n\rm Var T(\boldmathX)≥n\rm Var u(Y)≥h′(α)2ψ1(α)−n−1∑n−1i=1ψ1(α+i/n). (2.9)

Since the function is positive (the function is log-convex) and decreasing, we have , . Therefore,

 1nn−1∑i=1ψ1(α+i/n)<1nn∑i=1ψ1(α+i/n)<∫10ψ1(α+t)dt=∫α+1αψ1(t)dt =(logΓ(α+1))′−(logΓ(α))′=(logΓ(α+1)Γ(α))′=(logα)′=1α.

Thus, excluding the trivial case where is the constant function, the lower bound in (2.9) is strictly larger than the CR bound of (2.8). This means that, for any given non-constant parametric function , no efficient estimator exists, when efficiency is defined in the traditional way, i.e., in terms of attainability of the bound in (2.8). In the contrary, there are functions of the shape parameter that can be efficiently estimated in the sense of (2.9). More precisely, it can be checked that for any constants , the parametric function , where , is efficiently estimated by , in the sense that for all , and , which is equal to the lower bound given by (2.9). Finally, it can be easily verified that all efficiently estimated functions are of the above form. Notice, however, that asymptotically the lower bounds do coincide, since, by the Riemann integral, .

We now state the main result of this section. To the best of our knowledge, this result, as well as the existence of , below, is new.

###### Theorem 2.1.

Let , , be a random sample from . For every , the UMVUE of the shape parameter is given by

 u0(Y):=12(n−3)Y1−Y−YG′(Y)G(Y), (2.10)

where is as in (2.2) and is the ratio of the geometric to the arithmetic mean of . The estimator has finite variance if and only if . Moreover, for , no unbiased estimator of exists.

Proof: According to Proposition 2.2, if an unbiased estimator (of ) exists, the UMVUE must be a function of , say . Then, the relation can be written as

 1K(α)=1α∫10yα−1(1−y)(n−3)/2G(y)u(y)dy,  α>0.

Observing that and , the substitution yields

 ∫∞0e−αx(1−e−x)(n−3)/2G(e−x)dx={∫∞0e−αxdx}{∫∞0e−αx(1−e−x)(n−3)/2G(e−x)u(e−x)dx},

for all . Since we have assumed that for all , we may apply Fubini’s theorem to the right-hand side of the above equation, obtaining

This relation shows that the Laplace transforms of and are identical. Hence, (notice that both functions are continuous). Setting we get

 (1−y)(n−3)/2G(y)=∫−logy0(1−e−t)(n−3)/2G(e−t)u(e−t)dt=∫1y1x(1−x)(n−3)/2G(x)u(x)dx, (2.11)

, and a differentiation of (2.11) leads to

 −n−32(1−y)(n−5)/2G(y)+(1−y)(n−3)/2G′(y)=−u(y)y(1−y)(n−3)/2G(y),   for almost all y∈(0,1).

Hence, solving for we see that , a.e., with as in (2.10). The preceding argument shows that if an unbiased estimator exists then must be unbiased, but it does not prove that an unbiased estimator exists. To see this, let . Then, since when , (2.11) reads as

 1√1−y=∫1yu(x)x√1−x% dx. (2.12)

The assumption is equivalent to for all . Hence,

as , and this contradicts (2.12). When , it is easy to see that the function of (2.2) is given by

 G(y)=∞∑m=0(3m)!(3mm!)3(1−y)m,   0

Hence, . Since , assuming we get , . Therefore,

 ∣∣∣∫1yG(x)xu(x)dx∣∣∣≤∫1yG(x)x|u(x)|dx=∫1yxα−1G(x)|u(x)|xαdx≤1yα∫1yxα−1G(x)|u(x)|dx→0,

as . This contradicts (2.11), i.e.,

 G(y)=∫1y1xG(x)u(x)dx,

since the left-hand side of the above equation approaches as .

The preceding argument shows that there is no unbiased estimator when . For , however, the situation is completely different: The (positive) estimator of (2.10) has finite expectation for all and, indeed, . We now proceed to verify this claim. First observe that so that for . Noting that , we calculate

 \rm IE αu0(Y) = K(α)∫10yα(1−y)(n−5)/2(∞∑m=0(n−32+m)dm(1−y)m)dy = K(α)∞∑m=0(n−32+m)dm∫10yα(1−y)m+(n−5)/2dy = αΓ(α)K(α)∞∑m=0Γ(m+(n−1)/2)Γ(α+m+(n−1)/2)dm,

where, since , the interchanging of summation and integration is justified by Beppo Levi’s theorem. It remains to verify that

 S:=∞∑m=0Γ(m+(n−1)/2)Γ(α+m+(n−1)/2)dm=1Γ(α)K(α),   α>0.