On point estimators for Gamma and Beta distributions

1 Introduction

Let $Γ (α) = \int_{0}^{\infty} x^{α - 1} e^{- x} d x$ , $α > 0$ , be the Gamma function. The Gamma distribution, $G (α, λ)$ , with density

f (x) = \frac{λ^{α}}{Γ (α)} x^{α - 1} e^{- λ x}, x > 0,

and parameters $α > 0$ (the shape) and $λ > 0$ (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 $α$ , $λ$ , and/or $1 / α$ and $1 / λ$ , 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 $α$ (and $λ$ ) exists, for some sample size $n$ . In the contrary, unbiased estimators for the reciprocal parameters $1 / α$ and $1 / λ$

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

α

is introduced by Iliopoulos (2016).

In the present work we show that an unbiased estimator for $α$ exists if and only if the sample size $n$ is at least $4$ , and the same is true for $λ$ (provided that $α$ is not too small). More precisely, in Section 2 we obtain the uniformly minimum variance unbiased estimators (UMVUEs) of $α$ , $λ$ , $1 / α$ , $1 / λ$ . In particular, the UMVUE of $1 / λ$ , based on two observations, can be written as a positive symmetric kernel, whence, the corresponding $U$ -statistic generates the Ye-Chen (2017) estimator; see Remark 2.2.

Let $B (α, β) = Γ (α) Γ (β) / Γ (α + β)$ , $α > 0$ , $β > 0$ , be the Beta function. The Beta distribution has density

f (x) = \frac{1}{B (α, β)} x^{α - 1} (1 - x)^{β - 1}, 0 < x < 1.

(1.1)

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 $U$ -statistics. The covariance identity generates an unbiased bivariate symmetric kernel for the parameter $θ = 1 / (α + β)$ . Then, an application of the classical theory of $U$ -statistics produces an unbiased estimator of $θ$ , which, in turn, yields the desired estimators of $α$ and $β$ . 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 $(0, 1)$ by means of a transformation, e.g., $x \mapsto x / (1 + x)$ .

2 Unbiased estimation of the parameters of Gamma distribution

Let $\boldmathX=(X1,…,Xn)$ ( $n \geq 2$ ) be a random sample from $Γ (α, λ)$ , where the unknown parameter $\boldmathθ=(α,λ)∈\boldmathΘ=(0,∞)2$ . It is well-known that the complete, sufficient statistic can be written in the form $\boldmathT(\boldmathX)=(X,Y)$ , where $X = \sum_{i = 1}^{n} X_{i}$ and $Y^{1 / n}$

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

X, Y

are independent, as follows, e.g., from Basu’s theorem, because the distribution of

Y

is free of

λ

, while

X

is complete for

λ

when

α > 0

is known. The random variable

X

follows a

G (n α, λ)

distribution, while

Y

is distributed like

\prod_{i = 1}^{n - 1} B_{α} (i / n)

, where the random variables

B_{α} (i / n)

i = 1, \dots, n - 1

, are independent, and

B_{α} (i / n)

has a Beta distribution (1.1) with parameters

α

and

β = i / n

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

f_{Y} (y; α) = K (α) y^{α - 1} (1 - y)^{(n - 3) / 2} G (y), 0 < y < 1,

(2.1)

where $K (α) = Γ (α)^{- n} \prod_{i = 1}^{n - 1} Γ (α + i / n) / Γ (i / n)$ , and the (power-series) function $G$ is given by

G (y) = \infty \sum m = 0 d_{m} (1 - y)^{m}, 0 < y < 1.

(2.2)

Notice that our notation suspends the dependence of $K (α)$ and $G (y)$ on $n$ .

In (2.2) the coefficients $(d_{m})_{m \geq 0}$ are defined recurrently as follows:

	$γ_{1} (m) = I (m = 0) = {\begin{matrix} 1, & if & m = 0, 0, & if & m = 1, 2, \dots, \end{matrix}$
	$γ_{i + 1} (m) = \frac{Γ (m + i (i + 1) / (2 n))}{Γ (m + (i + 1) (i + 2) / (2 n))} m \sum k = 0 \frac{Γ (m - k + (i + 1) / n)}{(m - k)!} γ_{i} (k), i = 1, \dots, n - 2,$
	$d_{m} := γ_{n - 1} (m), m = 0, 1, \dots .$		(2.3)

Most of the preceding results can be found in Glaser (1976), Nandi (1980), and Tang and Gupta (1984). Obviously, $G (y) \equiv 1$ when $n = 2$ . Moreover, for $n \geq 3$ , $G$ is strictly decreasing and positive; more precisely, $d_{m} > 0$ for all $m$ , $G^{'} (y) < 0$ and $G (y) > G (1 -) = d_{0} > 0$ , $0 < y < 1$ . It should be noted at this point that, when $n \geq 3$ , the auxiliary random variable $Y_{1} = \prod_{i = 1}^{n - 2} B_{α} (i / n)$ has density (cf. Tang and Gupta, 1984)

f_{Y_{1}} (y; α) = K_{1} (α) y^{α - 1} (1 - y)^{(n - 1) (n - 2) / (2 n) - 1} G_{1} (y), 0 < y < 1,

(2.4)

where $K_{1} (α) = Γ (α)^{- (n - 1)} \prod_{i = 1}^{n - 2} Γ (α + i / n) / Γ (i / n)$ and

G_{1} (y) = \infty \sum k = 0 γ_{n - 2} (k) (1 - y)^{k}, 0 < y < 1.

(2.5)

The random variable $Y_{1}$ will be used in the proof of Theorem 2.1, below.

Definition 2.1.

An estimator $T(\boldmathX)$ will be called uniformly minimum variance unbiased estimator (UMVUE) for the parametric function $h (α, λ)$ if $T(\boldmathX)=u(X,Y)$ is a (Borel) function of the complete, sufficient statistic $(X, Y)$ , and $\rm IE (α,λ)u(X,Y)=h(α,λ)$ , for all $α > 0$ , $λ > 0$ , where the subscript $(α, λ)$ denotes expectation w.r.t. the joint density of $(X_{1}, \dots, X_{n})$ ,

f (x_{1}, \dots, x_{n}) = \frac{λ^{n α}}{Γ (α)^{n}} {(n \prod i = 1 x_{i})}_{i}^{α - 1} exp (- λ n \sum i = 1 x_{i}), (x_{1}, \dots, x_{n}) \in (0, \infty)^{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 $\rm IE (α,λ)u(X,Y)=h(α,λ)$ is merely satisfied for all $(α, λ)$ in a subset of $(0, \infty)^{2}$ 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 $I = (λ_{1}, λ_{2})$ where $0 \leq λ_{1} < λ_{2} \leq \infty$ . Suppose that for the Borel functions $g_{i} : (0, \infty) \to R$ , $i = 1, 2$ , it is true that $\int_{0}^{\infty} e^{- λ x} | g_{i} (x) | d x < \infty$ for all $λ \in I$ . If

\int_{0}^{\infty} e^{- λ x} g_{1} (x) d x = \int_{0}^{\infty} e^{- λ x} g_{2} (x) d x, λ \in I,

then $g_{1} (x) = g_{2} (x)$ for almost all $x \in (0, \infty)$ .

Proof: For $i = 1, 2$ write $g_{i} = g_{i}^{+} - g_{i}^{-}$ where $g_{i}^{+} (x) = max {g_{i} (x), 0}$ , $g_{i}^{-} (x) = max {- g_{i} (x), 0}$ . By assumption,

0 \leq \int_{0}^{\infty} e^{- λ x} w_{1} (x) d x = \int_{0}^{\infty} e^{- λ x} w_{2} (x) d x < \infty, λ \in I,

where $w_{1} = g_{1}^{+} + g_{2}^{-} \geq 0$ , $w_{2} = g_{1}^{-} + g_{2}^{+} \geq 0$ . Fix $λ_{0} \in I$ and set $ν_{0} := \int_{0}^{\infty} e^{- λ_{0} x} w_{1} (x) d x = \int_{0}^{\infty} e^{- λ_{0} x} w_{2} (x) d x \geq 0$ . Since $w_{i} \geq 0$ , it is clear that $ν_{0} = 0$ implies that $w_{1} = 0$ and $w_{2} = 0$ a.e.; then, $w_{1} = w_{2}$ a.e. and, thus, $g_{1} = g_{2}$ a.e. If $ν_{0} > 0$

, we may define the probability densities

f_{i} (x) = e^{- λ_{0} x} w_{i} (x) / ν_{0}

i = 1, 2

; then, for

λ \in I

\int_{0}^{\infty} e^{(λ_{0} - λ) x} f_{1} (x) d x = \frac{1}{ν_{0}} \int_{0}^{\infty} e^{- λ x} w_{1} (x) d x = \frac{1}{ν_{0}} \int_{0}^{\infty} e^{- λ x} w_{2} (x) d x = \int_{0}^{\infty} e^{(λ_{0} - λ) x} f_{2} (x) d x .

If $M_{i}$

is the moment generating function of

f_{i}

, it follows that

M_{1} (t) = M_{2} (t) < \infty

for all

t \in J := (λ_{0} - λ_{2}, λ_{0} - λ_{1})

, and since

J

contains the origin,

f_{1} = f_{2}

a.e.; this implies that

w_{1} = w_{2}

a.e., and consequently,

g_{1} = g_{2}

a.e.

Proposition 2.1.

Let $s (x, y) : (0, \infty)^{2} \to R$ be a Borel measurable function satisfying

\int_{0}^{\infty} \int_{0}^{\infty} | s (x, y) | exp (- λ x - α y) x^{n α - 1} d y d x < \infty for all α > 0, λ > 0 .

Then, the function

H (α, λ) := \int_{0}^{\infty} \int_{0}^{\infty} s (x, y) exp (- λ x - α y) x^{n α - 1} d y d x

is continuous in $(0, \infty)^{2}$ .

Proof: Consider an arbitrary sequence $(α_{m}, λ_{m}) \to (α, λ) \in (0, \infty)^{2}$ , as $m \to \infty$ . Then, we can find $m_{0}$ such that $\frac{α}{2} < α_{m} < 2 α$ and $\frac{λ}{2} < λ_{m} < 2 λ$ for $m \geq m_{0}$ , The function $δ_{1} (λ) = e^{- λ x}$ is obviously decreasing (for each fixed $x > 0$ ), so that $e^{- λ_{m} x} < e^{- \frac{λ}{2} x}$ , $m \geq m_{0}$ . Consider also the function $δ_{2} (α) = e^{- α y} x^{n α - 1}$ (for fixed $x > 0$ , $y > 0$ ). Then, $log δ_{2} (α) = (n log x - y) α - log x$ is a linear function of $α$ , and thus, $δ_{2} (α_{m}) < max {δ_{2} (α / 2), δ_{2} (2 α)} < δ_{2} (α / 2) + δ_{2} (2 α)$ , $m \geq m_{0}$ . Combining the above we obtain the inequality

e^{- λ_{m} x - α_{m} y} x^{n α_{m} - 1} < e^{- \frac{λ}{2} x - \frac{α}{2} y} x^{n \frac{α}{2} - 1} + e^{- \frac{λ}{2} x - 2 α y} x^{2 n α - 1}, (x, y) \in (0, \infty)^{2}, m \geq m_{0} .

It follows that the sequence

w_{m} (x, y) := s (x, y) e^{- λ_{m} x - α_{m} y} x^{n α_{m} - 1}, (x, y) \in (0, \infty)^{2}, m = m_{0}, m_{0} + 1, \dots,

is dominated by $W (x, y) := | s (x, y) | e^{- \frac{λ}{2} x - \frac{α}{2} y} x^{n \frac{α}{2} - 1} + | s (x, y) | e^{- \frac{λ}{2} x - 2 α y} x^{2 n α - 1}$ , and, by assumption, $W$ is integrable on $(0, \infty)^{2}$ . Clearly, $w_{m} (x, y) \to w (x, y) := s (x, y) e^{- λ x - α y} x^{n α - 1}$ as $m \to \infty$ . Thus, by dominated convergence,

completing the proof.

Proposition 2.2.

Let $h (α) : (0, \infty) \to R$ be a parametric function of the shape parameter. If there exists an unbiased estimator $U1=U1(\boldmathX)$ for $h$ , then the UMVUE of $h (α)$ is a function of $Y$ , alone.

Proof: By assumption, $\rm IE (α,λ)|U1(\boldmathX)|<∞$ for all $α > 0$ , $λ > 0$ . The function $U_{1} : (0, \infty)^{n} \to R$ can always be taken to be Borel measurable. From the RB/LS Theorem, the UMVUE is given by $U(X,Y)=\rm IE (U1|X,Y)$ , where $U : (0, \infty)^{2} \to R$ can also be taken to be Borel measurable with no loss of generality. Unbiasedness of the estimator $U (X, Y)$ means that (recall that $X, Y$ are independent)

\int_{0}^{\infty} \int_{0}^{1} {\frac{λ^{n α}}{Γ (n α)} x^{n α - 1} e^{- λ x}} {K (α) y^{α - 1} (1 - y)^{(n - 3) / 2} G (y)} U (x, y) d y d x = h (α),

for all $α > 0$ , $λ > 0$ . This relation can be rewritten in the form

Since $\rm IE (α,λ)|U(X,Y)|<∞$ for all $α > 0$ and $λ > 0$ , we may apply Proposition 2.1 with $s (x, y) = (1 - e^{- y})^{(n - 3) / 2} G (e^{- y}) U (x, e^{- y})$ to conclude that the function $H (α, λ)$ is continuous, and hence, $h (α)$ is a continuous function of $α > 0$ ; notice that all functions $K (α)$ , $Γ (n α)$ , and $λ^{n α}$ , are positive and continuous.

Write now the relation $h(α)=\rm IE (α,λ)U(X,Y)$ in the form

\int_{0}^{\infty} e^{- λ x} h (α) x^{n α - 1} d x = \int_{0}^{\infty} e^{- λ x} K (α) x^{n α - 1} {\int_{0}^{1} y^{α - 1} (1 - y)^{(n - 3) / 2} G (y) U (x, y) d y} d x .

It follows that the Laplace transforms of the functions $f_{1} (x) := h (α) x^{n α - 1}$ and $f_{2} (x) := K (α) x^{n α - 1} \int_{0}^{1} y^{α - 1} (1 - y)^{(n - 3) / 2} G (y) U (x, y) d y$ are identical, and Lemma 2.1 implies that for every $α > 0$ ,

\frac{h (α)}{K (α)} = \int_{0}^{1} y^{α - 1} (1 - y)^{(n - 3) / 2} G (y) U (x, y) d y, for almost all x \in (0, \infty) .

(2.6)

Write $Q_{+} = Q \cap (0, \infty)$ and set $A_{α} = {x > 0 : (???) holds}$ , $N_{α} = (0, \infty) ∖ A_{α} = {x > 0 : (???) does not % hold}$ , $A = \cap_{α \in Q_{+}} A_{α}$ , $N = \cup_{α \in Q_{+}} N_{α} = (0, \infty) ∖ A$ . By construction, the Lebesgue measure of $N$ is zero. We chose an arbitrary $x_{0} \in A$ and define $˜ u (y) := U (x_{0}, y)$ . Notice that $˜ u : (0, \infty) \to R$ is not necessarily Borel measurable; it is, however, Lebesgue measurable and, hence, it is equal a.e. to a Borel function $u : (0, \infty) \to R$ . Obviously, substituting $u$ in place of $˜ u$ , the values of the above integrals are not affected. From (2.6) we have

\frac{h (α)}{K (α)} = \int_{0}^{1} y^{α - 1} (1 - y)^{(n - 3) / 2} G (y) u (y) d y, α \in Q_{+} .

(2.7)

The right-hand side of (2.7) defines a continuous function of $α > 0$ , because

is the Laplace transform of the function $y \mapsto (1 - e^{- y})^{(n - 3) / 2} G (e^{- y}) u (e^{- y})$ . Since $h (α)$ and $K (α)$ are both continuous, it follows from (2.7) that for all $α > 0$ . Hence, $u (Y)$ is an unbiased estimator of $h (α)$ , and also, it is a function of the complete, sufficient statistic $(X, Y)$ ; 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 $G (α, λ)$ is given by

J = J (α, λ) = \frac{1}{λ^{2}} (\begin{matrix} λ^{2} ψ_{1} (α) & - λ - λ & α \end{matrix}), with inverse J^{- 1} = \frac{1}{α ψ_{1} (α) - 1} (\begin{matrix} α & λ λ & λ^{2} ψ_{1} (α) \end{matrix}),

where $ψ_{1} (α) = (log Γ (α))^{''} > 1 / α$ . The Cramér-Rao (CR) bound says that for any unbiased estimator $T=T(\boldmathX)$ of the parametric function $h (α, λ)$ , $n\rm Var T≥(∇h(α,λ))TJ−1∇h(α,λ)$ . Hence, in the particular case where $h = h (α)$ is a parametric function of the shape parameter alone (i.e., $h$ does not depend on $λ$ ), 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 $h (α)$ is a Borel function of $Y$ , say $u = u (Y)$ , we can apply the CR bound, $\rm Var u(Y)≥h′(α)2/JY$ , where $J_{Y}$ is the Fisher information of $Y$ . From (2.1), the density of $Y$ is given by

f_{Y} (y; α) = exp (a log y - A (α)) k (y),

where $A (α) = - log K (α)$ and $k (y) = y^{- 1} (1 - y)^{(n - 3) / 2} G (y) I_{(0, 1)} (y)$ , with $I_{B}$ denoting the indicator function of $B$ . This shows that ${f_{Y} (y; α), α > 0}$ defines a natural exponential family of Lebesgue densities, and thus, the regularity conditions are fulfilled. We calculate $- \frac{\partial^{2}}{\partial α^{2}} log f_{Y} (y; α) = A^{''} (α)$ and, therefore,

	$J_{Y}$	$=$	$(- log K (α))^{''} = {(n log Γ (α) + n - 1 \sum i = 1 log Γ (i / n) - n - 1 \sum i = 1 log Γ (α + i / n))}^{''}$
		$=$	$n ψ_{1} (α) - n - 1 \sum i = 1 ψ_{1} (α + i / n) .$

It follows that for any unbiased estimator $T(\boldmathX)$ of $h (α)$ ,

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 $ψ_{1} (α)$ is positive (the $Γ$ function is log-convex) and decreasing, we have $n^{- 1} ψ_{1} (α + i / n) < \int_{(i - 1) / n}^{i / n} ψ_{1} (α + t) d t$ , $i = 1, \dots, n$ . Therefore,

	$\frac{1}{n} n - 1 \sum i = 1 ψ_{1} (α + i / n) < \frac{1}{n} n \sum i = 1 ψ_{1} (α + i / n) < \int_{0}^{1} ψ_{1} (α + t) d t = \int_{α}^{α + 1} ψ_{1} (t) d t$
	$= (log Γ (α + 1))^{'} - (log Γ (α))^{'} = {(log \frac{Γ (α + 1)}{Γ (α)})}^{'} = (log α)^{'} = \frac{1}{α} .$

Thus, excluding the trivial case where $h (α)$ 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 $h (α)$ , 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 $c_{1}, c_{2}$ , the parametric function , where $ψ (α) = Γ^{'} (α) / Γ (α)$ , is efficiently estimated by $T(\boldmathX)=u(Y)=(c1/n)logY+c2$ , in the sense that $\rm IE (α,λ)T(\boldmathX)=h(α)$ for all $α, λ$ , and $n\rm Var T(\boldmathX)=c21(ψ1(α)−n−1∑n−1i=1ψ1(α+i/n))$ , 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, ${lim}_{n \to \infty} n^{- 1} \sum_{i = 1}^{n - 1} ψ_{1} (α + i / n) = \int_{α}^{α + 1} ψ_{1} (t) d t = 1 / α$ .

We now state the main result of this section. To the best of our knowledge, this result, as well as the existence of $u_{0}$ , below, is new.

Theorem 2.1.

Let $X_{1}, \dots, X_{n}$ , $n \geq 2$ , be a random sample from $G (α, λ)$ . For every $n \geq 4$ , the UMVUE of the shape parameter $α$ is given by

u_{0} (Y) := \frac{1}{2} (n - 3) \frac{Y}{1 - Y} - \frac{Y G^{'} (Y)}{G (Y)},

(2.10)

where $G$ is as in (2.2) and $Y^{1 / n}$ is the ratio of the geometric to the arithmetic mean of $X_{1}, \dots, X_{n}$ . The estimator $u_{0} (Y)$ has finite variance if and only if $n \geq 6$ . Moreover, for $n = 2, 3$ , no unbiased estimator of $α$ exists.

Proof: According to Proposition 2.2, if an unbiased estimator (of $α$ ) exists, the UMVUE must be a function of $Y$ , say $u (Y)$ . Then, the relation $\rm IE αu(Y)=α$ can be written as

\frac{1}{K (α)} = \frac{1}{α} \int_{0}^{1} y^{α - 1} (1 - y)^{(n - 3) / 2} G (y) u (y) d y, α > 0.

Observing that $1 / K (α) = \int_{0}^{1} y^{α - 1} (1 - y)^{(n - 3) / 2} G (y) d y$ and $1 / α = \int_{0}^{\infty} e^{- α x} d x$ , the substitution $y = e^{- x}$ yields

\int_{0}^{\infty} e^{- α x} (1 - e^{- x})^{(n - 3) / 2} G (e^{- x}) d x = {\int_{0}^{\infty} e^{- α x} d x} {\int_{0}^{\infty} e^{- α x} (1 - e^{- x})^{(n - 3) / 2} G (e^{- x}) u (e^{- x}) d x},

for all $α > 0$ . Since we have assumed that $\rm IE α|u(Y)|<∞$ for all $α > 0$ , we may apply Fubini’s theorem to the right-hand side of the above equation, obtaining

This relation shows that the Laplace transforms of $f_{1} (x) := (1 - e^{- x})^{(n - 3) / 2} G (e^{- x})$ and $f_{2} (x) := \int_{0}^{x} (1 - e^{- t})^{(n - 3) / 2} G (e^{- t}) u (e^{- t}) d t$ are identical. Hence, $f_{1} \equiv f_{2}$ (notice that both functions are continuous). Setting $x = - log y$ we get

(1 - y)^{(n - 3) / 2} G (y) = \int_{0}^{- log y} (1 - e^{- t})^{(n - 3) / 2} G (e^{- t}) u (e^{- t}) d t = \int_{y}^{1} \frac{1}{x} (1 - x)^{(n - 3) / 2} G (x) u (x) d x,

(2.11)

$0 < y < 1$ , and a differentiation of (2.11) leads to

- \frac{n - 3}{2} (1 - y)^{(n - 5) / 2} G (y) + (1 - y)^{(n - 3) / 2} G^{'} (y) = \frac{- u (y)}{y} (1 - y)^{(n - 3) / 2} G (y), for almost all y \in (0, 1) .

Hence, solving for $u$ we see that $u (y) = ((n - 3) / 2) y / (1 - y) - y G^{'} (y) / G (y) = u_{0} (y)$ , a.e., with $u_{0}$ as in (2.10). The preceding argument shows that if an unbiased estimator $T (X_{1}, \dots, X_{n})$ exists then $u_{0} (Y)$ must be unbiased, but it does not prove that an unbiased estimator exists. To see this, let $n = 2$ . Then, since $G (y) \equiv 1$ when $n = 2$ , (2.11) reads as

\frac{1}{\sqrt{1 - y}} = \int_{y}^{1} \frac{u (x)}{x \sqrt{1 - x}} % d x .

(2.12)

The assumption $\rm IE α|u(Y)|<∞$ is equivalent to $\int_{0}^{1} x^{α - 1} (1 - x)^{- 1 / 2} | u (x) | d x < \infty$ for all $α > 0$ . Hence,

as $y \to 1 -$ , and this contradicts (2.12). When $n = 3$ , it is easy to see that the function $G$ of (2.2) is given by

G (y) = \infty \sum m = 0 \frac{(3 m)!}{(3^{m} m!)^{3}} (1 - y)^{m}, 0 < y < 1.

Hence, $G (1 -) = 1$ . Since $(1 - y)^{(n - 3) / 2} \equiv 1$ , assuming $\rm IE α|u(Y)|<∞$ we get $\int_{0}^{1} x^{α - 1} G (x) | u (x) | d x < \infty$ , $α > 0$ . Therefore,

∣ ∣ ∣ \int_{y}^{1} \frac{G (x)}{x} u (x) d x ∣ ∣ ∣ \leq \int_{y}^{1} \frac{G (x)}{x} | u (x) | d x = \int_{y}^{1} \frac{x^{α - 1} G (x) | u (x) |}{x^{α}} d x \leq \frac{1}{y^{α}} \int_{y}^{1} x^{α - 1} G (x) | u (x) | d x \to 0,

as $y \to 1 -$ . This contradicts (2.11), i.e.,

G (y) = \int_{y}^{1} \frac{1}{x} G (x) u (x) d x,

since the left-hand side of the above equation approaches $1$ as $y \to 1 -$ .

The preceding argument shows that there is no unbiased estimator when $n \leq 3$ . For $n \geq 4$ , however, the situation is completely different: The (positive) estimator $u_{0} (Y)$ of (2.10) has finite expectation for all $α > 0$ and, indeed, $\rm IE αu0(Y)=α$ . We now proceed to verify this claim. First observe that $- G^{'} (y) > 0$ so that $u_{0} (y) > 0$ for $0 < y < 1$ . Noting that $G^{'} (y) = - \sum_{m = 0}^{\infty} m d_{m} (1 - y)^{m - 1}$ , we calculate

$\rm IE αu0(Y)$	$=$	$K (α) \int_{0}^{1} y^{α} (1 - y)^{(n - 5) / 2} (\infty \sum m = 0 (\frac{n - 3}{2} + m) d_{m} (1 - y)^{m}) d y$
	$=$	$K (α) \infty \sum m = 0 (\frac{n - 3}{2} + m) d_{m} \int_{0}^{1} y^{α} (1 - y)^{m + (n - 5) / 2} d y$
	$=$	$α Γ (α) K (α) \infty \sum m = 0 \frac{Γ (m + (n - 1) / 2)}{Γ (α + m + (n - 1) / 2)} d_{m},$

where, since $d_{m} > 0$ , the interchanging of summation and integration is justified by Beppo Levi’s theorem. It remains to verify that

S := \infty \sum m = 0 \frac{Γ (m + (n - 1) / 2)}{Γ (α + m + (n - 1) / 2)} d_{m} = \frac{1}{Γ (α) K (α)}, α > 0.

On point estimators for Gamma and Beta distributions

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1 Introduction

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Definition 2.1.

Lemma 2.1.

Proposition 2.1.

Proposition 2.2.

Remark 2.1.

Theorem 2.1.

On point estimators for Gamma and Beta distributions

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This week in AI

1 Introduction

2 Unbiased estimation of the parameters of Gamma distribution

Definition 2.1.

Lemma 2.1.

Proposition 2.1.

Proposition 2.2.

Remark 2.1.

Theorem 2.1.