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• •   CrossRef (0) Modified inverse moment estimation: its principle and applications  Wenhao Gui

aDepartment of Mathematics, Beijing Jiaotong University, China
Correspondence to: 1Department of Mathematics, Beijing Jiaotong University, Beijing 100044, China. E-mail: whgui@bjtu.edu.cn
Received October 30, 2016; Revised November 11, 2016; Accepted November 11, 2016.
Abstract

In this survey, we present a modified inverse moment estimation of parameters and its applications. We use a specific model to demonstrate its principle and how to apply this method in practice. The estimation of unknown parameters is considered. A necessary and sufficient condition for the existence and uniqueness of maximum-likelihood estimates of the parameters is obtained for the classical maximum likelihood estimation. Inverse moment and modified inverse moment estimators are proposed and their properties are studied. Monte Carlo simulations are conducted to compare the performances of these estimators. As far as the biases and mean squared errors are concerned, modified inverse moment estimator works the best in all cases considered for estimating the unknown parameters. Its performance is followed by inverse moment estimator and maximum likelihood estimator, especially for small sample sizes.

Keywords : inverse moment estimators, maximum likelihood estimates, existence and uniqueness, joint confidence regions, small sample size, Weibull distribution, inverted exponential Pareto distribution, Monte Carlo simulation
1. Introduction

The inverse estimation method was originally proposed by Wang (1992) and was applied to study parameter estimation for Weibull distribution. Different from the regular method of moments, the idea of the inverse moment estimation (IME) is as follows.

For a sample X1, …, Xn from a distribution with unknown parameters, first transform the original sample to a quasi-sample Y1, …, Yn, where Yi contains the unknown parameters but its distribution does not depend on unknown parameters, that is, Yi is a pivot variable, i = 1, …, n. The population moments of the new sample do not dependent on unknown parameters while the sample moments do. Let the population moments of the quasi-sample equal the sample moments and solve the unknown parameters.

Wang (2004) obtained the inverse moment estimators and the interval estimation based on type II progressively censored data under the Weibull distribution. The simulation results showed that the mean square errors of the inverse moment estimators are less than the maximum likelihood estimates (MLE)’s. Gu and Yue (2013) considered the problem of estimating parameters of the generalized exponential distribution based on a complete sample. They proposed the inverse moment estimators of the parameters of the generalized exponential distribution. The precisions of MLEs and IMEs are compared through numerical simulations. Gui (2015) studied the problem of estimating unknown shape and scale parameters of exponentiated half logistic distribution. Inverse moment and modified inverse moment estimators were derived. Monte Carlo simulations were conducted to compare the performances.

Wang et al. (2010) obtained some exponential inequalities for a negatively orthant dependent sequence and used the exponential inequalities to study the asymptotic approximation of an inverse moment for negatively orthant dependent random variables. Ye and Yang (2013) proposed a new method for dimension reduction in regression using the first two inverse moments. Yang et al. (2014) discussed the asymptotic approximation of an inverse moment for nonnegative random variables. Cheng et al. (2014) established inverse moment bounds for sample autocovariance matrices based on a detrended time series.

In this paper, we focus on the problem of parameter estimation for the inverted exponential Pareto distribution and demonstrate the inverse estimation and its modified version. We begin with the classical MLE and obtain a necessary and sufficient condition for the existence and uniqueness of MLE of the parameters. We propose inverse moment and modified inverse moment estimators and study their properties. Monte Carlo simulations are used to compare the performances. We also propose the methods for constructing joint confidence regions for the two parameters and study their performances.

Gupta et al. (1998) introduced the exponential Pareto distribution Y to model failure time data. The probability density function of Y is given by

$f(y;λ,α)=αλ[1-(1+y)-λ]α-1 (1+y)-(λ+1), y>0,$

where α > 0 and λ > 0 are two parameters. The corresponding cumulative distribution function is

$F(y;λ,α)=[1-(1+y)-λ]α, y>0.$

A random variable X = 1/Y is said to have the inverted exponential Pareto distribution. Its cumulative distribution function (cdf) and probability density function (pdf) are specified by

$F(x;λ,α)=1-[1-(1+1x)-λ]α, x>0,$

and

$f(x;λ,α)=αλ(1x+1)-λ-1 [1-(1x+1)-λ]α-1x2, x>0,$

respectively, where λ > 0 and α > 0 are the parameters. We denote this distribution as IEPD(λ, α). When α = 1, the inverted exponential Pareto distribution reduces to the inverted Pareto distribution. In this paper, we will show that the two-parameter inverted exponential Pareto distribution can be quite effective in modeling lifetime data.

Shawky and Abu-Zinadah (2009) considered the maximum likelihood estimation of the different parameters of an exponential Pareto distribution. Afify (2010) obtained Bayes and classical estimators for two parameters exponentiated Pareto distribution when a sample is available from complete, type I and type II censoring scheme. Ali et al. (2010) derived the distribution of the ratio of two independent exponentiated Pareto random variables and studied its properties. Singh et al. (2013) proposed maximum likelihood estimators and Bayes estimators of parameters of exponentiated Pareto distribution under general entropy loss function and squared error loss function for progressive type-II censored data with binomial removals.

The rest of this paper is organized as follows. In Section 2, we discuss the classical maximum likelihood estimation of the parameters of the inverted exponential Pareto distribution. In Section 3, we propose the inverse and modified inverse estimation methods to estimate the parameters and study their properties. Joint confidence regions for the two parameters are also proposed in Section 4. Section 5 conducts simulations to compare the estimators and the confidence regions. In Section 6, a numerical example is presented to illustrate the superiorities of the proposed methods. Finally, Section 7 concludes.

2. Maximum likelihood estimation

In this section, we discuss the MLEs of the parameters of inverted exponential Pareto distribution (IEPD) based on a complete sample. Let X1, X2, …, Xn be a random sample from IEPD(λ, α) with pdf and cdf as (1.4) and (1.3), respectively. The log-likelihood function is given by

$L(λ,α)=(α-1)∑i=1nlog [1-(1xi+1)-λ]-(λ+1)∑i=1nlog (1xi+1)+n log α+n log λ-2∑i=1nlog (xi).$

The score equations are as:

$∂L(λ,α)∂λ=(α-1)∑i=1n(1xi+1)-λ log (1xi+1)1-(1xi+1)-λ-∑i=1nlog (1xi+1)+nλ,$$∂L(λ,α)∂α=∑i=1nlog [1-(1xi+1)-λ]+nα.$

Consider the case when x1 = · · · = xn = x, the MLEs λ̂ and α̂ are

$α^=-1log [1-(1x+1)-λ]$

and λ̂ is the solution of the equation

$J(λ)=J1(λ)-J2(λ),$

where

$J1(λ)=-(1x+1)-λ log (1x+1)1-(1x+1)-λ-log (1x+1)+1λ,J2(λ)=1log [1-(1x+1)-λ](1x+1)-λ log (1x+1)1-(1x+1)-λ.$

Note that

$J1′(λ)=λ2 (1x+1)λ log2 (1x+1)-((1x+1)λ-1)2λ2 ((1x+1)λ-1)2<0,$

since (t − 1)2 > (log t)2t, for all t > 1.

$J2′(λ)=-log2 (1x+1)[(1x+1)λ log (1-(1x+1)-λ)+1][(1x+1)λ-1]2log2 (1-(1x+1)-λ)>0,$

since (1/t) log(1 − t) + 1 < 0, for all t > 0. J′(λ) < 0, J(λ) is a decreasing function of λ. Moreover, we have J(0) = ∞ and J(∞) = 0. Thus, J(λ) has no roots in the interval (0,∞).

In the following, we discuss the existence and uniqueness of MLEs in the case of at least two non-identical observed values of the sample.

### Theorem 1

Let X1, X2, …, Xnbe a random sample from IEPD(λ, α), if the observed values of the sample are not identical, that is xix(n) = max{x1, …, xn} for at least one i ∈ {1, …, n − 1}, then MLEs of λ and α exist and unique.

Proof

From (2.3) we obtain the MLE of α as a function of λ,

$α^=-n∑i=1nlog [1-(1xi+1)-λ].$

The MLE of λ is the root of the following equation

$G(λ)=nλ-∑i=1nlog (1xi+1)1-(1xi+1)-λ-n∑i=1nlog [1-(1xi+1)-λ]∑i=1n(1xi+1)-λ log (1xi+1)1-(1xi+1)-λ=0.$

Firstly, we prove that the equation G(λ) = 0 has a positive root. We calculate the limits G(0) and G(+∞) respectively.

$G(0)=nλ-∑i=1nlog (1xi+1)1-(1xi+1)-λ-n∑i=1nlog [1-(1xi+1)-λ]∑i=1n(1xi+1)-λ log (1xi+1)1-(1xi+1)-λ=∑i=1nlog (1xi+1)limλ→0[1λ log (1xi+1)-11-(1xi+1)-λ]-limλ→0n∑i=1nλ log [1-(1xi+1)-λ]∑i=1nλ(1xi+1)-λ log (1xi+1)1-(1xi+1)-λ=-12∑i=1nlog (1xi+1)-limλ→0n2∑i=1nλ log [1-(1xi+1)-λ]=-12∑i=1nlog(1xi+1)-n2∑i=1n(1/log (1xi+1)) limλ→0 λ log (1xi+1) log [1-(1xi+1)-λ]=+∞.$$G(+∞)=-∑i=1nlog (1xi+1)-limλ→∞n∑i=1n(1xi+1)-λ log (1xi+1)∑i=1nlog [1-(1xi+1)-λ]=-∑i=1nlog (1xi+1)+limλ→∞n∑i=1n(1xi+1)-λ log (1xi+1)∑i=1n(1xi+1)-λ=-∑i=1nlog (1xi+1)+limλ→∞n∑i=1n(1xi+1)-λ (1x(n)+1)λ log (1xi+1)∑i=1n(1xi+1)-λ (1x(n)+1)λ=-∑i=1nlog (1xi+1)+n log (1x(n)+1)=-∑i=1n[log (1xi+1)-log (1x(n)+1)]<0.$

It follows that the equation G(λ) = 0 has a positive real root.

Secondly, we show that the root is unique. We rewrite G(λ) as

$G(λ)=G1(λ)-G2(λ),$

where

$G1(λ)=nλ-∑i=1nlog (1xi+1)1-(1xi+1)-λ and G2(λ)=n∑i=1nlog [1-(1xi+1)-λ]∑i=1n(1xi+1)-λ log (1xi+1)1-(1xi+1)-λ.G1′(λ)=∑i=1n[(1xi+1)-λ log2 (1xi+1)(1-(1xi+1)-λ)2-1λ2]=∑i=1n-[(1xi+1)λ-1]2+λ2 (1xi+1)λ log2 (1xi+1)λ2 [(1xi+1)λ-1]2<0,$

since (t − 1)2 > t(log t)2 for all t > 1.

$G2′(λ)=n [(∑i=1n-log (1-(1xi+1)-λ))∑i=1n(1xi+1)λ log2 (1xi+1)((1xi+1)λ-1)2-(∑i=1nlog (1xi+1)(1xi+1)λ-1)2](∑i=1nlog (1-(1xi+1)-λ))2≥n [(∑i=1n(1xi+1)-λ)∑i=1n(1xi+1)λ log2 (1xi+1)((1xi+1)λ-1)2-(∑i=1nlog (1xi+1)(1xi+1)λ-1)2](∑i=1nlog (1-(1xi+1)-λ))2≥0.$

Since − log(1 − t) ≥ t for all t > 0. The second inequality holds by Cauchy-Schwartz inequality. Therefore, G′(λ) < 0, G(λ) is decreasing and G(λ) = 0 has unique root over (0, +∞).

3. Inverse and modified inverse moment estimation

In Statistics, there are many methods available for estimating the parameter(s) of interest. One of the oldest methods is the method of moments. It is based on the assumption that sample moments should provide adequate estimates of the corresponding population moments. Suppose we want to estimate θ = (θ1, …, θk), the procedure is:

• Find k population moments, μi, i = 1, 2, …, k. μi will include one or more parameters θ1, …, θk.

• Determine the corresponding k sample moments, mi, i = 1, 2, …, k.

• Let μi = mi, i = 1, 2, …, k, solve for the parameters. The solution is a moment estimator.

The method of moments is simple and easy to compute. However, the estimator may not be unique or not exist. In this section, we propose an inverse moment estimation. The superiority of the new estimator is its existence and uniqueness.

### Definition 1

Suppose X ~ F(x, θ), where θ = (θ1, …, θk) is a parameter vector to be estimated. Transform X to a pivotal variable Y = g(X, θ) whose distribution does not depend on θ. The population moments$μi′ (i=1,…,k)$of Y will not contain θ. The sample Yi = g(Xi, θ) (i = 1, …, k) is called quasi-sample since it is a function of sample (X1, …, Xn) and parameter θ. The moments of the quasi-sample$mi′ (i=1,…,k)$is also a function of θ. Let$μi′=mi′$, i = 1, 2, …, k, solve for the parameters. The solution is an inverse moment estimator.

Let X1, …, Xn form a sample from IEPD(λ, α) with pdf given in (1.4), it is known that F(Xi), 1 − F(Xi), i = 1, …, n follow uniform distribution U(0, 1), and − log[1 − F(Xi)], i = 1, …, n follow standard exponential distribution Exp(1). By the method of inverse moment estimation, we set

$1n∑i=1n{-log[1-F(Xi)]}=1,$

that is,

$-αn∑i=1nlog [1-(1Xi+1)-λ]=1.$

Thus, the IME of α is obtained as a function of λ,

$α^=-n∑i=1nlog [1-(1Xi+1)-λ],$

which is identical with the MLE of α.

### Lemma 1

Let Z(1)Z(2) ≤ · · · ≤ Z(n)be the order statistics from the standard exponential distribution. Then, the random variables W1, W2, …, Wn, where

$Wi=(n-i+1)(Z(i)-Z(i-1)), i=1,2,…,n$

with Z(0) ≡ 0, are independent and follow standard exponential distributions.

Proof

The proof can be found in Arnold et al. (1992).

### Lemma 2

Let W1, W2, …, Wnbe iid standard exponential variables, Si = W1 + · · · + Wi, Ui = (Si/Si+1)i, i = 1, 2, …, n − 1, Un = W1 + · · · + Wn, then

• U1, U2, …, Unare independent;

• U1, U2, …, Un−1follow the uniform distribution U(0, 1);

• 2Unfollows χ2(2n).

Proof

The proof can be found in Wang (1992).

Now we consider the IME of λ. For the sample X1, …, Xn from IEPD(λ, α), for the order statistics X(1) ≤ · · · ≤ X(n), we have

$-log[1-F(X(1))]≤⋯≤-log[1-F(X(n))]$

are n order statistics from standard exponential distribution Exp(1).

Let Z(i) = −α log[1 − (1/X(i) + 1)λ], i = 1, …, n. Thus, Z(1)Z(2) ≤ · · · ≤ Z(n) are the first n order statistics from the standard exponential distribution. By Lemma 1, Wi = (ni + 1)(Z(i)Z(i−1)), i = 1, 2, …, n form a random sample from standard exponential distribution.

Let Si = W1 + · · · +Wi, Ui = (Si/Si+1)i, i = 1, 2, …, n − 1, Un = W1 + · · · +Wn, by Lemma 2, we have

$-2∑i=1n-1log Ui=-2∑i=1n-1i log (SiSi+1)=2∑i=1n-1log (SnSi)~χ2(2n-2),$

where

$SnSi=Z(1)+Z(2)+⋯+Z(n)Z(1)+Z(2)+⋯+Z(i-1)+(n-i+1)Z(i)=log [1-(1X(1)+1)-λ]+log [1-(1X(2)+1)-λ]+⋯+log [1-(1X(n)+1)-λ]log [1-(1X(1)+1)-λ]+⋯+log [1-(1X(i-1)+1)-λ]+(n-i+1) log [1-(1X(i)+1)-λ].$

Noting that the mean of χ2(2n − 2) is 2n − 2. Thus, we obtain an inverse moment equation for λ as follows:

$∑i=1n-1log [log [1-(1X(1)+1)-λ]+log [1-(1X(2)+1)-λ]+⋯+log [1-(1X(n)+1)-λ]log [1-(1X(1)+1)-λ]+⋯+log [1-(1X(i-1)+1)-λ]+(n-i+1) log [1-(1X(i)+1)-λ]]=n-1.$

Solve the equation and we obtain the inverse estimate λ̂IME of λ. Plugging λ̂IME into (3.3), we obtain the inverse estimate α̂IME. In addition, noting that the mode of χ2(2n − 2) is 2n − 4, we can obtain a modified equation for λ:

$∑i=1n-1log [log [1-(1X(1)+1)-λ]+log [1-(1X(2)+1)-λ]+⋯+log [1-(1X(n)+1)-λ]log[1-(1X(1)+1)-λ]+⋯+log [1-(1X(i-1)+1)-λ]+(n-i+1) log [1-(1X(i)+1)-λ]]=n-2.$

Solve the equation and we obtain the modified inverse estimate λ̂MIME of λ. Plugging λ̂MIME into (3.3), we obtain the modified inverse estimate α̂MIME.

In the following, we prove the existence and uniqueness of the root in the equation (3.7) and (3.8).

### Lemma 3

The following limits hold: (1) limλ→0 log[1 − (a + 1)λ]/log[1 − (b + 1)λ] = 1, for a > 0, b > 0. (2) limλ→∞ log[1 − (a + 1)λ]/log[1 − (b + 1)λ] = 0, for a > b > 0. (3) limλ→∞ log[1− (a + 1)λ]/log[1 − (b + 1)λ] = +∞, for b > a > 0.

### Lemma 4

For t > 0, f(t) = (1 + tetet)/(1 − et) is a decreasing function of t.

### Theorem 2

Let Wi = (ni+1)(Z(i)Z(i−1)), i = 1, 2, …, n form a sample from standard exponential distribution, Si = W1 + · · · + Wi, then for t > 0, equation$∑i=1n-1log(Sn/Si)=t$has a unique positive solution.

Proof

By Lemma 3, we obtain

$limλ→0SnSi=limλ→0log [1-(1X(1)+1)-λ]+log [1-(1X(2)+1)-λ]+⋯+log [1-(1X(n)+1)-λ]log [1-(1X(1)+1)-λ]+⋯+log [1-(1X(i-1)+1)-λ]+(n-i+1) log [1-(1X(i)+1)-λ]=limλ→0log[1-(1X(1)+1)-λ]+log[1-(1X(2)+1)-λ]+⋯+log[1-(1X(n)+1)-λ]log[1-(1X(n)+1)-λ]log[1-(1X(1)+1)-λ]+⋯+log[1-(1X(i-1)+1)-λ]+(n-i+1) log[1-(1X(i)+1)-λ]log[1-(1X(n)+1)-λ]=nn=1.$

Thus, $limλ→0∑i=1n-1log(Sn/Si)=0$. However,

$limλ→∞SnSi=limλ→∞log [1-(1X(1)+1)-λ]+log [1-(1X(2)+1)-λ]+⋯+log [1-(1X(n)+1)-λ]log [1-(1X(1)+1)-λ]+⋯+log [1-(1X(i-1)+1)-λ]+(n-i+1) log [1-(1X(i)+1)-λ]=1+limλ→∞log [1-(1X(i+1)+1)-λ]+⋯+log [1-(1X(n)+1)-λ]-(n-i) log [1-(1X(i)+1)-λ]log [1-(1X(1)+1)-λ]+⋯+log [1-(1X(i)+1)-λ]+(n-i) log [1-(1X(i)+1)-λ]=1+limλ→∞log[1-(1X(i+1)+1)-λ]+⋯+log[1-(1X(n)+1)-λ]-(n-i) log[1-(1X(i)+1)-λ]log[1-(1X(i)+1)-λ]log[1-(1X(1)+1)-λ]+⋯+log[1-(1X(i)+1)-λ]+(n-i) log[1-(1X(i)+1)-λ]log[1-(1X(i)+1)-λ]=+∞.$

Thus, $limλ→∞∑i=1n-1log(Sn/Si)=∞$. Therefore, for t > 0, equation $∑i=1n-1log(Sn/Si)=t$ exists a positive solution. For the uniqueness of the solution, we consider the derivative of Sn/Si with respect to λ.

Noting that, for i = 1, …, n,

$dWidλ=(n-i+1)α[(1X(i-1)+1)-λ log (1X(i-1)+1)1-(1X(i-1)+1)-λ-(1X(i)+1)-λ log (1X(i)+1)1-(1X(i)+1)-λ]=Wi(1X(i-1)+1)-λ log(1X(i-1)+1)1-(1X(i-1)+1)-λ-(1X(i)+1)-λ log(1X(i)+1)1-(1X(i)+1)-λlog (1-(1X(i-1)+1)-λ)-log (1-(1X(i)+1)-λ).(SnSi)′=(1+Wi+1+⋯+WnW1+⋯+Wi)′=1(∑k=1iWk)2∑j=i+1n∑k=1i[Wj′Wk-WjWk′]=1λ(∑k=1iWk)2∑j=i+1n∑k=1iWjWk [A(λ)-B(λ)],$

where

$A(λ)=(1X(j-1)+1)-λλ log(1X(j-1)+1)1-(1X(j-1)+1)-λ-(1X(j)+1)-λλ log(1X(j)+1)1-(1X(j)+1)-λlog (1-(1X(j-1)+1)-λ)-log (1-(1X(j)+1)-λ)$

and

$B(λ)=(1X(k-1)+1)-λλ log(1X(k-1)+1)1-(1X(k-1)+1)-λ-(1X(k)+1)-λλ log(1X(k)+1)1-(1X(k)+1)-λlog (1-(1X(k-1)+1)-λ)-log (1-(1X(k)+1)-λ).$

By Cauchy’s mean-value theorem, for j = i + 1, …, n, k = 1, …, i, there exist ξ1 ∈ (λ log(1/X(j) + 1), λ log(1/X(j−1) + 1)) and ξ2 ∈ (λ log(1/X(k) + 1), λ log(1/X(k−1) + 1)) such that

$A(λ)=1+ξ1eξ1-eξ11-eξ1, B(λ)=1+ξ2eξ2-eξ21-eξ2.$

Note that ξ1 < ξ2, by Lemma 4, A(λ) − B(λ) > 0, (Sn/Si)′ > 0, thus $∑i=1n-1log(Sn/Si)$ is a strictly increasing function of λ, equation $∑i=1n-1log(Sn/Si)=t$ has a unique positive solution.

4. Joint confidence regions for λ and α

Let X1, X2, …, Xn form a sample from IEPD(λ, α), X(1)X(2) ≤ · · · ≤ X(n) are the order statistics from this sample. Let Z(i) = −α log[1−(1/X(i)+1)λ], i = 1, …, n. Thus, Z(1)Z(2) ≤ · · ·≤ Z(n) are the first n order statistics from the standard exponential distribution. By Lemma 1, Wi = (ni+1)(Z(i)Z(i−1)), i = 1, 2, …, n form a sample from standard exponential distribution. Let Si = W1 + · · · + Wi, Ui = (Si/Si+1)i, i = 1, 2, …, n − 1, Un = W1 + · · · + Wn. Hence,

$V=2S1=2W1=2nZ(1)=-2nα log [1-(1X(1)+1)-λ]~χ2(2),$

and

$U=2(Sn-S1)=2∑i=2nWi=2 [Z(1)+⋯+Z(n)-nZ(1)]~χ2(2n-2).$

It is obvious that U and V are independent. Define

$T1=U/(2n-2)V/2=Sn-S1(n-1)S1~F(2n-2,2),$

and

$T2=U+V=2Sn~χ2(2n).$

We obtain that T1 and T2 are independent using the known bank-post office story in statistics.

Let Fγ(v1, v2) denote the percentile of F distribution with left-tail probability γ and v1 and v2 degrees of freedom. Let $χγ2(v)$ denote the percentile of χ2 distribution with left-tail probability γ and v degrees of freedom.

By using the pivotal variables T1 and T2, a joint confidence region for the two parameters λ and α can be constructed as follows.

### Theorem 3

(Method 1) Let X1, X2, …, Xnform a sample from IEPD(λ, α), then, based on the pivotal variables T1and T2, a 100(1 − γ)% joint confidence region for the two parameters λ and α is determined by the following inequalities:

${λL≤λ≤λU,χ1-1-γ22(2n)-2∑i=1nlog [1-(1X(i)+1)-λ]≤α≤χ1+1-γ22(2n)-2∑i=1nlog [1-(1X(i)+1)-λ],$

where λLis the root of λ for the equation$T1=F(1-1-γ)/2(2n-2,2)$and λUis the root of λ for the equation$T1=F(1+1-γ)/2(2n-2,2)$.

Proof

The function of λ is

$T1=1n-1log [1-(1X(1)+1)-λ]+⋯log [1-(1X(n)+1)-λ]-n log [1-(1X(1)+1)-λ]n log [1-(1X(1)+1)-λ]$

and does not depend on α. From Theorem 2, we have limλ→0T1 = (1/(n − 1)) limλ→0(Sn/S1 −1) = 0, limλ→∞T1 = (1/(n − 1)) limλ→∞(Sn/S1 − 1) = ∞, $T1′=(1/(n-1))(Sn/Si)′>0$. Therefore, for any t > 0, equation T1 = t has a unique positive root of λ.

$1-γ=1-γ1-γ=P (F1-1-γ2(2n-2,2)≤T1≤F1+1-γ2(2n-2,2))×P (χ1-1-γ22(2n)≤T2≤χ1+1-γ22(2n))=P (F1-1-γ2(2n-2,2)≤T1≤F1+1-γ2(2n-2,2)),χ1-1-γ22(2n)≤T2≤χ1+1-γ22(2n))=P (λL≤λ≤λU,χ1-1-γ22(2n)-2∑i=1nlog [1-(1X(i)+1)-λ]≤α≤χ1+1-γ22(2n)-2∑i=1nlog [1-(1X(i)+1)-λ]).$

However, by Lemma 2, we have

$T3=-2∑i=1n-1log Ui=-2∑i=1n-1i log (SiSi+1)=2∑i=1n-1log (SnSi)~χ2(2n-2).$

T2 and T3 are also independent. By using the pivotal variables T2 and T3, a joint confidence region for the two parameters λ and α can be constructed as follows.

### Theorem 4

(Method 2) Let X1, X2, …, Xnform a sample from IEPD(λ, α), then, based on the pivotal variables T2and T3, a 100(1 − γ)% joint confidence region for the two parameters λ and α is determined by the following inequalities:

${λL*≤λ≤λU*,χ1-1-γ22(2n)-2∑i=1nlog [1-(1X(i)+1)-λ]≤α≤χ1+1-γ22(2n)-2∑i=1nlog [1-(1X(i)+1)-λ],$

where$λL*$is the root of λ for the equation$T3=χ(1-1-γ)/22(2n-2)$and$λU*$is the root of λ for the equation$T3=χ(1+1-γ)/22(2n-2)$.

Proof

$T3=2∑i=1n-1log(Sn/Si)$ is a function of λ and does not depend on α. From Theorem 2, for any s > 0, equation T3 = s has a unique positive root of λ.

$1-γ=1-γ1-γ=P (χ1-1-γ22(2n-2)≤T3≤χ1+1-γ22(2n-2))×P (χ1-1-γ22(2n)≤T2≤χ1+1-γ22(2n))=P (χ1-1-γ22(2n-2)≤T3≤χ1+1-γ22(2n-2),χ1-1-γ22(2n)≤T2≤χ1+1-γ22(2n))=P (λL*≤λ≤λU*,χ1-1-γ22(2n)-2∑i=1nlog [1-(1X(i)+1)-λ]≤α≤χ1+1-γ22(2n)-2∑i=1nlog [1-(1X(i)+1)-λ]).$
5. Simulation study

### 5.1. Comparison of the three estimation methods

In this section, we conduct simulations to compare the performances of the MLEs, IMEs and MIMEs mainly with respect to their biases and mean squared errors (MSE’s), for various sample sizes and for various true parametric values.

Suppose X ~ IEPD(λ, α), the random data can be generated as: X = 1/[(1 −U1/α)−1/λ − 1], where U follows uniform distribution over [0, 1]. We obtain λ̂MLE by solving Equation (2.5) and α̂MLE by (2.4). The λ̂IME and λ̂MIME can be obtained by solving (3.7) and (3.8) respectively. The α̂IME and α̂MIME can be obtained from (3.3).

We consider sample sizes n = 30, 40, 50, 80, 100 and α = 2.0, 2.5, 3.0, 3.5, 4.0. We take λ = 4 in all our computations. For each combination of sample size n and parameter α, we generate a sample of size n from IEPD(λ = 4, α), and estimate the parameters λ and α by the MLE, IME, MIME methods. The average values of α̂/α and λ̂/4 as well as the corresponding MSEs over 1,000 replications are computed and reported.

For different cases, Table 1 reports the average values of α̂/α and the corresponding MSE is reported within parenthesis. Figure 1(a), (b), (c), and (d) show the relative biases and the MSEs of the three estimators of α for sample sizes n = 40 and n = 80. Figure 1(e) and (f) show the relative biases and the MSEs of the three estimators of α for α = 3.0. The other cases are also similar.

For different cases, Table 2 reports the average values of λ̂/λ = λ̂/4 and the corresponding MSE is reported within parenthesis. Figure 2(a), (b), (c) and (d) show the relative biases and the MSEs of the three estimators of λ for sample sizes n = 40 and n = 80. Figure 2(e) and (f) show the relative biases and the MSEs of the three estimators of λ for α = 3.0. The other cases are similar.

From Tables 1 and 2, we observe that

• The average relative biases and the average relative MSEs for the three methods decrease as sample size n increases as expected. The asymptotic unbiasedness and consistency of all the estimators are verified.

• For the three methods, the average biases and relative MSEs of λ̂/4 decrease as α goes up. The average biases and relative MSEs of α̂/α increase as α goes up.

• Considering only MSE’s, the estimation of α′s are more accurate for smaller values while the estimation of λ′s are more accurate for larger values of α.

• MLE and IME overestimate both of the two parameters α and λ. MIME overestimates only α.

As far as the biases and MSEs are concerned, it is clear MIME works the best in all the cases considered to estimate the two parameters. Its performance is followed by IME and MLE, especially for small sample sizes. The three methods are close for larger sample sizes. Considering all the points, MIME is recommended for estimating both the parameters of the IEPD(λ, α) distribution.

### 5.2. Comparison of the two joint confidence regions

In Section 4, two methods to construct the confidence regions of the two parameters λ and α are proposed. In this section, we conduct simulations to compare the two methods.

First, we assess the precisions of the two methods of interval estimators for the parameter λ. We take sample sizes n = 30, 40, 50, 80, 100 and α = 2.0, 2.5, 3.0, 3.5, 4.0. We take λ = 4 in all our computations. For each combination of sample size n and parameter α, we generate a sample of size n from IEPD(λ = 4, α), and estimate the parameters λ by the two proposed methods (4.5) and (4.7).

The mean widths as well as the coverage rates over 1,000 replications are computed. Here the coverage rate is defined as the rate of the confidence intervals that contain the true value λ = 4 among these 1,000 confidence intervals. The results are reported in Table 3. It is observed that:

• The mean widths of the intervals decrease as sample sizes n increase as expected.

• The mean widths of the intervals decrease as the parameter α increases.

• The coverage rates of the two methods are close to the nominal level 0.95.

Considering the mean widths, the interval estimate of λ obtained in method 2 performs better than that obtained in method 1. Method 2 for constructing the interval estimate of λ is recommended.

We consider the two joint confidence regions and the empirical coverage rates and expected areas. The results of the methods for constructing joint confidence regions for (λ, α) with confidence level γ = 0.95 are reported in

It is observed that:

• The mean areas of the joint regions decrease as sample sizes n increase as expected.

• The mean areas of the joint regions increase as the parameter α increases.

• The coverage rates of the two methods are close to the nominal level 0.95.

Considering the mean areas, the joint region of (λ, α) obtained in method 2 performs better than that obtained in method 1. Method 2 is recommended.

6. Real illustrative example

In this section, we consider a real dataset. This data set represents the total seasonal annual rainfall (in inches) recorded at Los Angeles Civic Center during the last 25 years, from 1985 to 2009 (season 1 July–30 June). The observations are

12.82, 17.86, 7.66, 12.48, 8.08, 7.35, 11.99, 21.00, 27.36, 8.11, 24.35, 12.44, 12.40, 31.01, 9.09, 11.57, 17.94, 4.42, 16.42, 9.25, 37.96, 13.19, 3.21, 13.53, 9.08.

The dataset has been previously analyzed by Raqab (2006, 2013) and Ahmadi and Balakrishnan (2009). Here we fit the data with IEPD.

The MLEs of the parameters are λ̂MLE = 25.6869 and α̂MLE = 4.9629 with log-likelihood value −84.1822. The Kolmogorov-Smirnov distance and its corresponding p-value are D = 0.12 and p = 0.9955, respectively. The inverted exponential Pareto distribution can be effective in modeling the rainfall data.

Using the methods proposed in Section 3, we obtain the following estimates:

$λ^IME=24.7618, α^IME=4.6793, λ^MIME=23.7404, α^MIME=4.3801.$

Based on method 1, the 95% joint confidence region for the parameters (λ, α) is given by the following inequalities:

${7.4044≤λ≤35.9168,-15.17118∑i=125log (1-(1+1Xi)-λ)≤α≤-37.48729∑i=125log (1-(1+1Xi)-λ).$

Based on method 2, the 95% joint confidence region for the parameters (λ, α) is given by the following inequalities:

${14.3964≤λ≤36.4891,-15.17118∑i=125log (1-(1+1Xi)-λ)≤α≤-37.48729∑i=125log (1-(1+1Xi)-λ).$

Figure 3(a) and (b) show the 95% joint confidence regions of (λ, α).

7. Conclusions and remarks

In this article, we present the modified inverse moment estimation of parameters and its applications. We use the inverted exponential Pareto distribution as a specific model to demonstrate its principle and how to apply this method in practice. The estimation of unknown parameters is investigated. For the classical maximum likelihood estimation, a necessary and sufficient condition for the existence and uniqueness of MLEs of the parameters is obtained. Inverse moment and modified inverse moment estimators are proposed and their properties are studied.

Monte Carlo simulations are conducted to compare the performances of the three estimators. The simulation results show that the modified inverse moment estimator works the best in all the cases considered for estimating the unknown parameters in terms of biases and mean squared errors. Its performance is followed by inverse moment estimator and maximum likelihood estimator, especially for small sample sizes. We also discuss joint confidence regions for unknown parameters. Real rainfall dataset is analyzed and used to illustrate the proposed method.

The method discussed in this paper can be easily extended to other common distributions (such as generalized exponential, and inverse Weibull distribution), which are frequently used in practice. Future research topics should include a comparison of the proposed modified inverse moment estimator with Bayesian estimator, what is the relation between the proposed estimator and sufficient statistics.

Acknowledgements

The author’s work was partially supported by the program for the Fundamental Research Funds for the Central Universities (2014RC042; 2015JBM109).

Figures Fig. 1. Average relative biases and MSEs of ╬▒. Fig. 2. Average relative biases and MSEs of ╬╗. Fig. 3. The 95% joint confidence region of (╬╗, ╬▒).
TABLES

### Table 1

Average relative estimates and MSEs of α

nMethodsα = 2.0α = 2.5α = 3.0α = 3.5α = 4.0
30MLE1.1229 (0.1600)1.1518 (0.1753)1.1790 (0.2353)1.1735 (0.2472)1.1736 (0.2482)
IME1.0899 (0.1376)1.1149 (0.1508)1.1376 (0.2019)1.1287 (0.2061)1.1286 (0.2127)
MIME 1.0457 (0.1166) 1.0656 (0.1254) 1.0835 (0.1662) 1.0721 (0.1688) 1.0696 (0.1734)

40MLE1.0760 (0.0793)1.0948 (0.0914)1.1212 (0.1262)1.1195 (0.1320)1.1356 (0.1617)
IME1.0531 (0.0717)1.0690 (0.0820)1.0922 (0.1116)1.0897 (0.1179)1.1014 (0.1416)
MIME1.0220 (0.0638)1.0345 (0.0717)1.0543 (0.0964)1.0499 (0.1017)1.0590 (0.1212)

50MLE1.0720 (0.0661)1.0663 (0.0658)1.0941 (0.0831)1.0866 (0.0877)1.1038 (0.1028)
IME1.0537 (0.0609)1.0464 (0.0607)1.0711 (0.0745)1.0635 (0.0799)1.0798 (0.0943)
MIME1.0290 (0.0551)1.0199 (0.0549)1.0418 (0.0660)1.0329 (0.0711)1.0472 (0.0831)

80MLE1.0385 (0.0343)1.0555 (0.0442)1.0515 (0.0455)1.0530 (0.0452)1.0485 (0.0470)
IME1.0276 (0.0325)1.0438 (0.0419)1.0387 (0.0431)1.0390 (0.0422)1.0331 (0.0437)
MIME1.0129 (0.0306)1.0275 (0.0390)1.0214 (0.0401)1.0207 (0.0392)1.0141 (0.0407)

100MLE1.0345 (0.0311)1.0431 (0.0318)1.0382 (0.0322)1.0393 (0.0353)1.0419 (0.0374)
IME1.0260 (0.0301)1.0339 (0.0304)1.0281 (0.0307)1.0288 (0.0338)1.0298 (0.0353)
MIME1.0143 (0.0287)1.0210 (0.0287)1.0144 (0.0290)1.0144 (0.0319)1.0147 (0.0332)

MLE = maximum likelihood estimate; IME = inverse moment estimation; MIME = modified inverse moment estimation; MSE = mean square error.

### Table 2

Average relative estimates and MSEs of λ

nMethodsα = 2.0α = 2.5α = 3.0α = 3.5α = 4.0
30MLE1.0742 (0.0576)1.0590 (0.0491)1.0678 (0.0472)1.0524 (0.0414)1.0557 (0.0459)
IME1.0451 (0.0516)1.0316 (0.0445)1.0406 (0.0423)1.0257 (0.0376)1.0294 (0.0419)
MIME1.0054 (0.0466)0.9940 (0.0410)1.0038 (0.0385)0.9901 (0.0350)0.9944 (0.0389)

40MLE 1.0628 (0.0396) 1.0569 (0.0395) 1.0530 (0.0364) 1.0480 (0.0310) 1.0385 (0.0299)
IME1.0416 (0.0363)1.0363 (0.0363)1.0336 (0.0339)1.0286 (0.0285)1.0200 (0.0280)
MIME1.0122 (0.0331)1.0081 (0.0336)1.0064 (0.0315)1.0022 (0.0265)0.9942 (0.0265)

50MLE1.0407 (0.0298)1.0370 (0.0251)1.0360 (0.0261)1.0346 (0.0237)1.0366 (0.0245)
IME1.0243 (0.0280)1.0207 (0.0235)1.0203 (0.0246)1.0198 (0.0225)1.0209 (0.0231)
MIME1.0013 (0.0264)0.9986 (0.0223)0.9988 (0.0234)0.9989 (0.0214)1.0004 (0.0220)

80MLE1.0234 (0.0179)1.0200 (0.0147)1.0162 (0.0145)1.0255 (0.0136)1.0161 (0.0139)
IME1.0130 (0.0172)1.0101 (0.0142)1.0064 (0.0140)1.0156 (0.0130)1.0067 (0.0135)
MIME0.9988 (0.0167)0.9964 (0.0138)0.9933 (0.0137)1.0026 (0.0125)0.9941 (0.0132)

100MLE1.0193 (0.0123)1.0178 (0.0130)1.0185 (0.0115)1.0189 (0.0105)1.0177 (0.0104)
IME1.0113 (0.0119)1.0101 (0.0127)1.0109 (0.0112)1.0116 (0.0102)1.0101 (0.0102)
MIME0.9999 (0.0116)0.9992 (0.0123)1.0004 (0.0109)1.0013 (0.0099)1.0000 (0.0099)

MLE = maximum likelihood estimate; IME = inverse moment estimation; MIME = modified inverse moment estimation; MSE = mean square error.

### Table 3

Results of the methods for constructing intervals for λ with confidence level 0.95

nMethodsα = 2.0α = 2.5α = 3.0α = 3.5α = 4.0
30(I)Mean width5.60435.54895.50525.44405.2401
Coverage rate0.9550.9530.9530.9420.954

(II)Mean width3.32483.17783.05463.02912.9178
Coverage rate0.9610.9500.9570.9470.958

40(I)Mean width5.28265.14105.05694.99304.8931
Coverage rate 0.9440.9500.9430.9590.949

(II)Mean width2.84762.71872.62282.55562.5152
Coverage rate0.9510.9400.9570.9580.951

50(I)Mean width4.97594.86864.72844.68794.6513
Coverage rate0.9380.9590.9680.9660.950

(II)Mean width2.51792.41052.33872.28442.2472
Coverage rate0.9480.9430.9510.9520.960

80(I)Mean width4.45674.33954.26834.19834.2200
Coverage rate0.9540.9500.9520.9490.941

(II)Mean width1.98241.90101.83081.78571.7608
Coverage rate0.9380.9520.9560.9490.961

100(I)Mean width4.29554.16994.12044.01313.9981
Coverage rate0.9520.9590.9530.9450.957

(II)Mean width1.75641.69381.64321.59511.5721
Coverage rate0.9570.9390.9510.9310.950

### Table 4

Results of the methods for constructing joint confidence regions for (λ, α) with confidence level γ = 0.95

nMethodsα = 2.0α = 2.5α = 3.0α = 3.5α = 4.0
30(I)Mean area18.144724.104029.322937.472345.1162
Coverage rate0.9430.9530.9480.9490.942

(II)Mean area7.26119.047310.434712.261614.0958
Coverage rate0.9400.9510.9570.9540.949

40(I)Mean area13.929817.286422.633728.343432.9806
Coverage rate0.9500.9530.9450.9380.956

(II)Mean area5.23596.17517.34368.69589.7041
Coverage rate 0.9520.9420.9440.9460.948

50(I)Mean area11.158614.136517.700521.105825.5788
Coverage rate0.9510.9540.9370.9510.948

(II)Mean area4.07314.85945.68456.49217.3824
Coverage rate0.9550.9530.9420.9530.949

80(I)Mean area7.21429.598711.888113.886417.4234
Coverage rate0.9460.9440.9420.9430.935

(II)Mean area4.03704.98866.10207.16018.2498
Coverage rate0.9410.9510.9380.9400.936

100(I)Mean area5.88197.79369.376311.353413.5202
Coverage rate0.9540.9490.9510.9500.950

(II)Mean area1.87882.25962.67273.01933.3800
Coverage rate0.9500.9520.9500.9420.953

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