Testing Homogeneity in a Semiparametric Two-Sample Problem

Abstract

We study a two-sample homogeneity testing problem, in which one sample comes from a population with density 𝑓(𝑥) and the other is from a mixture population with mixture density (1−𝜆)𝑓(𝑥)+𝜆𝑔(𝑥). This problem arises naturally from many statistical applications such as test for partial differential gene expression in microarray study or genetic studies for gene mutation. Under the semiparametric assumption 𝑔(𝑥)=𝑓(𝑥)𝑒^{𝛼+𝛽𝑥}, a penalized empirical likelihood ratio test could be constructed, but its implementation is hindered by the fact that there is neither feasible algorithm for computing the test statistic nor available research results on its theoretical properties. To circumvent these difficulties, we propose an EM test based on the penalized empirical likelihood. We prove that the EM test has a simple chi-square limiting distribution, and we also demonstrate its competitive testing performances by simulations. A real-data example is used to illustrate the proposed methodology.

1. Introduction

Let 𝑥₁,…,𝑥_𝑛0 be a random sample from a population with distribution function 𝐹, and let 𝑦₁,…,𝑦_𝑛1 be a random sample from a population with distribution function 𝐻. Testing whether the two populations have the same distribution, that is, 𝐻₀∶𝐹=𝐻 versus 𝐻₁∶𝐹≠𝐻, with both 𝐹 and 𝐻 completely unspecified, will require a nonparametric test. Since 𝐻₁∶𝐹≠𝐻 is a very broad hypothesis, many times one may want to consider some more specified alternative, for example, the two populations only differ in location. In the present paper, we will consider a specified alternative in which one of the two samples has a mixture structure. More specifically, we have𝑥₁,…,𝑥∼_𝑛0^i.i.d.𝑓(𝑥),𝑦₁,…,𝑦∼_𝑛1^i.i.d.ℎ(𝑦)=(1−𝜆)𝑓(𝑦)+𝜆𝑔(𝑦),(1.1) where 𝑓(𝑥)=𝑑𝐹(𝑥)/𝑑𝑥, 𝑔(𝑦)=𝑑𝐺(𝑦)/𝑑𝑦, ℎ(𝑦)=𝑑𝐻(𝑦)/𝑑𝑦, and 𝜆∈(0,1) is an unknown parameter sometimes called contamination proportion. The problem of interest is to test 𝐻₀∶𝑓=ℎ or equivalently 𝜆=0. This particular two-sample problem arises naturally in a variety of statistical applications such as test for partial differential gene expression in microarray study, genetic studies for gene mutation, case-control studies with contaminated controls, or the test of a treatment effect in the presence of nonresponders in biological experiments (see Qin and Liang [1] for details).

If no auxiliary information is available, this is merely the usual two-sample goodness-of-fit problem. There has been extensive literature on it; see Zhang [2] and references therein. However, these tests are not suitable for the specific alternative with a mixture structure as they might be inferior comparing with methods that are designed for the specific alternative. In this paper, we will propose an empirical likelihood-based testing procedure for this specific mixture alternative under Anderson’s semiparametric assumption [3]. Motivated by the logistic regression model, the semiparametric assumption proposed by Anderson [3] links the two distribution functions 𝐹 and 𝐺 through the following equation:log𝑔(𝑥)𝑓(𝑥)=𝛼+𝛽𝑥,(1.2) where 𝛼 and 𝛽 are both unknown parameters. There are many examples where the logarithm of the density ratio is linear in the observations.

Example 1.1. Let 𝐹 and 𝐺 be the distribution functions of Binomial (𝑚,𝑝₁) and Binomial (𝑚,𝑝₂), respectively. We refer the densities 𝑓 and 𝑔 to the probability mass functions corresponding to 𝐹 and 𝐺, respectively. Then, log𝑔(𝑥)𝑓(𝑥)=𝑚log11−−𝑝𝑝²₁+log𝑝²1−𝑝₁𝑝₁1−𝑝₂𝑥.(1.3)

Example 1.2. Let 𝐹 be the distribution function of 𝑁(𝜇₁,𝜎²) and 𝐺 the distribution function of 𝑁(𝜇₂,𝜎²). Then, log𝑔(𝑥)𝑓(𝑥)=21𝜎²𝜇²¹−𝜇²²+1𝜎²𝜇₂−𝜇₁𝑥.(1.4)
In practice, one may need to apply some sort of transformation to the data (e.g., logarithm transformation) in order to justify the use of the semiparametric model assumption (1.2).

Example 1.3. Let 𝐹 and 𝐺 be the distribution functions of log𝑁(𝜇₁,𝜎²) and log𝑁(𝜇₂,𝜎²), respectively. It is clear that the density ratio is a linear function of the log-transformed data: log𝑔(𝑥)𝑓(𝑥)=21𝜎²𝜇²¹−𝜇²²+1𝜎²𝜇₂−𝜇₁log𝑥.(1.5)

Example 1.4. Let 𝐹 and 𝐺 be the distribution functions of Gamma(𝑚₁,𝜃) and Gamma(𝑚₂,𝜃), respectively. In this case, log𝑔(𝑥)𝑓(𝑥)=logΓ𝑚¹Γ𝑚₂+𝑚¹−𝑚₂log𝜃+𝑚₂−𝑚₁log𝑥.(1.6)

The semiparametric modeling assumption (1.2) is very flexible and has the advantage of not putting any specific restrictions on the functional form of 𝑓. Under this assumption, various approaches have been proposed to test homogeneity in the two-sample problem (see [1, 4, 5] and references therein). This paper adds to this literature by introducing a new type of test statistics which are based on the empirical likelihood [6, 7].

The empirical likelihood (EL) is a nonparametric likelihood method which has many nice properties paralleling to the likelihood methods, for example, it is range-preserving, transform-respect, Bartlett correctable, and a systematic approach to incorporating auxiliary information [8–11]. In general, if the parameters are identifiable, the empirical likelihood ratio (ELR) test has a chi-square limiting distribution under null hypothesis. However, for the aforementioned testing problem, the parameters under 𝐻₀ are not identifiable, which results in an intractable null limiting distribution for the ELR test. To circumvent this problem, we would add a penalty to the log EL to penalize 𝜆 being too close to zero. Working like a soft threshold, the penalty makes the parameters roughly identifiable. Intuitively, the penalized (or modified) ELR test should restore the usual chi-square limiting distribution. Unfortunately two things hinder the direct use of the penalized ELR test. One is that, to the best of our knowledge, there is no feasible algorithm to compute the penalized ELR test statistic. The other one is that there has been no research on the asymptotic properties of the penalized ELR test. Therefore, one cannot obtain critical values for the penalized ELR test regardless through simulations or an asymptotic reference distribution. We find that the EM test [12, 13] based on the penalized EL is a nice solution to the testing problem.

The remainder of this paper is organized as follows. In Section 2, we introduce the ELR and the penalized ELR. The penalized EL-based EM test is given in Section 3. A key computational issue of the EM test is discussed in Section 4. Sections 5 and 6 contain a simulation study and a real-data application, respectively. For clarity, all proofs are postponed to the appendix.

2. Empirical Likelihood

Let {𝑡₁,…,𝑡_𝑛0,𝑡_𝑛0+1,…,𝑡_𝑛}={𝑥₁,…,𝑥_𝑛0,𝑦₁,…,𝑦_𝑛1} denote the combined two-sample data, where 𝑛=𝑛₀+𝑛₁. Under Anderson’s semiparametric assumption (1.2), the likelihood of two-sample data (1.1) is𝐿=^𝑛0_𝑖=1𝑑𝐹𝑡_𝑖^𝑛_𝑗=_𝑛0+11−𝜆+𝜆𝑒^{𝛼+𝛽𝑡𝑗}𝑑𝐹𝑡_𝑗.(2.1) Let 𝑝_ℎ=𝑑𝐹(𝑡_ℎ), ℎ=1,…,𝑛. The EL is just the likelihood 𝐿 with constraints 𝑝_ℎ≥0, ∑^𝑛ℎ=1𝑝_ℎ=1 and ∑^𝑛ℎ=1𝑝_ℎ(𝑒^{𝛼+𝛽𝑡ℎ}−1)=0. The corresponding log-EL is𝑙=^𝑛_ℎ=1log𝑝_ℎ+^𝑛1_𝑗=1log1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}.(2.2) We are interested in testing𝐻₀∶𝜆=0or(𝛼,𝛽)=(0,0).(2.3) Under the null hypothesis, the constraint ∑^𝑛ℎ=1𝑝_ℎ(𝑒^{𝛼+𝛽𝑡ℎ}−1)=0 will always hold and sup_𝐻0𝑙=−𝑛log𝑛. Under alternative hypothesis, for any fixed (𝜆,𝛼,𝛽), maximizing 𝑙 with respect to 𝑝_ℎ’s leads to the log-EL function of (𝜆,𝛼,𝛽):𝑙(𝜆,𝛼,𝛽)=−^𝑛_ℎ=1log1+𝜉𝑒^{𝛼+𝛽𝑡ℎ}−1−𝑛log𝑛+^𝑛1_𝑗=1log1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗},(2.4) where 𝜉 is the solution to the following equation:^𝑛_ℎ=1𝑒^{𝛼+𝛽𝑡ℎ}−11+𝜉𝑒^{𝛼+𝛽𝑡ℎ}−1=0.(2.5) Hence, the EL ratio function 𝑅(𝜆,𝛼,𝛽)=2{𝑙(𝜆,𝛼,𝛽)+𝑛log𝑛} and the ELR is denoted as 𝑅=sup𝑅(𝜆,𝛼,𝛽).

The null hypothesis 𝐻₀ holds for 𝜆=0 regardless of (𝛼,𝛽), or (𝛼,𝛽)=(0,0) regardless of 𝜆. This implies that the parameter (𝜆,𝛼,𝛽) is not identifiable under 𝐻₀, resulting in rather complicated asymptotic properties of the ELR. One may consider the modified or penalized likelihood method [14] and define the penalized log-EL function 𝑝𝑙(𝜆,𝛼,𝛽)=𝑙(𝜆,𝛼,𝛽)+log(𝜆). Accordingly the penalized EL ratio function is 𝑝𝑅(𝜆,𝛼,𝛽)==2−{2𝑝𝑙_𝑛(𝜆,𝛼,𝛽)−𝑝𝑙(1,0,0)}_ℎ=1+2_𝑛1log1+𝜉𝑒^{𝛼+𝛽𝑡ℎ}−1_𝑗=1log1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}+2log(𝜆),(2.6) where 𝜉 is the solution to (2.5). The penalty function log(𝜆) goes to −∞ as 𝜆 approaches 0. Therefore, 𝜆 is bounded away from 0, and the null hypothesis in (2.3) then reduces to (𝛼,𝛽)=(0,0). That is, the parameters in the penalized log-EL function is asymptotically identifiable. However, the asymptotic behavior of the penalized ELR test is still complicated. Meanwhile, the computation of the penalized ELR test statistic is another obstacle of the implementation of the penalized ELR method. No feasible and stable algorithm has been found for this purpose. An EL-based EM test proposed in this paper provides an efficient way to solve the problem.

3. EL-Based EM Test

Motivated by Chen and Li [12] and Li et al. [13], we propose an EM test based on the penalized EL to test the hypothesis (2.3). The EM test statistics are derived iteratively. We first choose a finite set of Λ={𝜆₁,…,𝜆_𝐿}⊂(0,1], for instance, Λ={0.1,0.2,…,0.9,1.0}, and a positive integer 𝐾 (2 or 3 in general). For each 𝑙=1,…,𝐿, we proceed the following steps.

Step 1. Let 𝑘=1 and 𝜆^𝑙(𝑘)=𝜆_𝑙. Calculate (𝛼^𝑙(𝑘),𝛽^𝑙(𝑘))=argmax_𝛼,𝛽𝑝𝑅(𝜆^𝑙(𝑘),𝛼,𝛽).

Step 2. Update (𝜆,𝛼,𝛽) by using the following algorithm for 𝐾−1 times.
Substep 2.1. Calculate the posterior distribution, 𝑤^{(𝑘)𝑗𝑙}=𝜆^𝑙(𝑘)exp𝛼^𝑙(𝑘)+𝛽^𝑙(𝑘)𝑦_𝑗1−𝜆^𝑙(𝑘)+𝜆^𝑙(𝑘)exp𝛼^𝑙(𝑘)+𝛽^𝑙(𝑘)𝑦_𝑗,𝑗=1,…,𝑛¹,(3.1) and update 𝜆 by 𝜆^𝑙(𝑘+1)=argmax_𝜆^𝑛1_𝑗=11−𝑤^{(𝑘)𝑗𝑙}log(1−𝜆)+^𝑛1_𝑗=1𝑤^{(𝑘)𝑗𝑙}log(𝜆)+log(𝜆).(3.2)
Substep 2.2. Update (𝛼,𝛽) by (𝛼^𝑙(𝑘+1),𝛽^𝑙(𝑘+1))=argmax_𝛼,𝛽𝑝𝑅(𝜆^𝑙(𝑘+1),𝛼,𝛽).
Substep 2.3. Let 𝑘=𝑘+1 and continue.

Step 3. Define the test statistics 𝑀^𝑛(𝐾)(𝜆_𝑙)=𝑝𝑅(𝜆^𝑙(𝐾),𝛼^𝑙(𝐾),𝛽^𝑙(𝐾)).

The EM test statistic is defined asEM^𝑛(𝐾)=max𝑀^𝑛(𝐾)𝜆_𝑙,𝑙=1,…,𝐿.(3.3) We reject the null hypothesis 𝐻₀ when the EM test statistic is greater than some critical value determined by the following limiting distribution.

Theorem 3.1. Suppose 𝜌=𝑛₁/𝑛∈(0,1) is a constant. Assume the null hypothesis 𝐻₀ holds and 𝐸(𝑡_ℎ)=0 and Var(𝑡_ℎ)=𝜎²∈(0,∞) for ℎ=1,…,𝑛. For 𝑙=1,…,𝐿 and any fixed 𝑘, it holds that 𝜆^𝑙(𝑘)−𝜆_𝑙=𝑜_𝑝(1),𝛼^𝑙(𝑘)=𝑂_𝑝𝑛⁻¹,𝛽^𝑙(𝑘)=𝑦−𝑥𝜆_𝑙𝜎²+𝑜_𝑝𝑛^−1/2,(3.4) where 𝑥=(1/𝑛₀)∑^𝑛0_𝑖=1𝑥_𝑖 and 𝑦=(1/𝑛₁)∑^𝑛1_𝑗=1𝑦_𝑗.

Remark 3.2. The assumption 𝐸𝑡_ℎ=0 is only for convenience purpose and unnecessary. Otherwise, we can replace 𝑡_ℎ and 𝛼 with 𝑡_ℎ−𝐸(𝑡_ℎ) and 𝛼+𝛽𝐸(𝑡_ℎ).

Theorem 3.3. Assume the conditions of Theorem 3.1 hold and 1∈Λ. Under the null hypothesis (2.3), EM^𝑛(𝐾)→𝜒²¹ in distribution, as 𝑛→∞.

We finish this section with an additional remark.

Remark 3.4. We point out that the idea of the EM-test can also be generalized to more general models such as log(𝑔(𝑥)/𝑓(𝑥))=𝛼+𝛽₁𝑥+⋯+𝛽_𝑘𝑥^𝑘 for some integer 𝑘 or log(𝑔(𝑥)/𝑓(𝑥))=𝛼+𝛽₁𝑡₁(𝑥)+⋯+𝛽_𝑘𝑡_𝑘(𝑥)with 𝑡_𝑖(⋅)’s being known functions.

4. Computation of the EM Test

A key step of the EM test procedure is to maximize 𝑝𝑅(𝜆,𝛼,𝛽) with respect to (𝛼,𝛽) for fixed 𝜆. In this section, we propose a computation strategy which provides stable solution to this optimization problem. Throughout this section, 𝜆 is suppressed to be fixed.

The objective function is 𝑝𝑅(𝜆,𝛼,𝛽)=𝐺(𝜉_∗,𝛼,𝛽) where𝐺(𝜉,𝛼,𝛽)=−2^𝑛_ℎ=1log1+𝜉𝑒^{𝛼+𝛽𝑡ℎ}−1+2^𝑛1_𝑗=1log1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}+2log(𝜆)(4.1) and 𝜉_∗=𝜉_∗(𝛼,𝛽) is the solution to𝜕𝐺𝜕𝜉=−2^𝑛_ℎ=1𝑒^{𝛼+𝛽𝑡ℎ}−11+𝜉𝑒^{𝛼+𝛽𝑡ℎ}−1=0.(4.2) If (𝛼,𝛽) is the maximum point of 𝑝𝑅(𝜆,𝛼,𝛽), it should generally satisfy𝜕𝐺𝜕𝛼=−2^𝑛_ℎ=1𝜉𝑒^{𝛼+𝛽𝑡ℎ}1+𝜉𝑒^{𝛼+𝛽𝑡ℎ}−1+2^𝑛1_𝑗=1𝜆𝑒^{𝛼+𝛽𝑦𝑗}1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}=0.(4.3) Combining (4.2) and (4.3) leads to𝜉=1𝑛^𝑛1_𝑗=1𝜆𝑒^{𝛼+𝛽𝑦𝑗}1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}.(4.4) Putting this expression of 𝜉 back into (4.1), we have a new function𝐻(𝛼,𝛽)=−2^𝑛_ℎ=1log1+𝑒^{𝛼+𝛽𝑡ℎ}−11𝑛^𝑛1_𝑗=1𝜆𝑒^{𝛼+𝛽𝑦𝑗}+2_𝑛11−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}_𝑗=1log1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}.(4.5) It can be verified that 𝐻(𝛼,𝛽) is almost surely concave in a neighborhood of (0,0) given 𝜆, which means that maximizing 𝐻(𝛼,𝛽) with respect to (𝛼,𝛽) gives the maximum of 𝑝𝑅(𝜆,𝛼,𝛽) for fixed 𝜆. The stability of the method is illustrated by the following simulation study.

5. Simulation Study

We consider two models in Examples 1.3 and 1.4 with 𝜇₁=0, 𝜇₂=𝜇, and 𝜎²=1 for Example 1.3, and 𝑚₁=1, 𝑚₂=𝑚, and 𝜃=1 for Example 1.4. Nominal levels of 0.01, 0.05, and 0.10 are considered. The logarithm transformation is applied to the original data before using the EM test. The initial set Λ={0.1,0.2,…,1} and iteration number 𝐾=3 are used to calculate the EM test statistic.

One competitive method for testing homogeneity under the semiparametric two-sample model is the score test proposed by Qin and Liang [1]. This method is based on𝑆(𝛼,𝛽)=𝜕𝑙(𝜆,𝛼,𝛽)𝜕𝜆|||_𝜆=0=^𝑛1_𝑗=1𝑒^{𝛼+𝛽𝑦𝑗}−1,(5.1) where 𝑙(𝜆,𝛼,𝛽) is the log empirical likelihood function given in (2.4). Let (𝛼₁,𝛽̂₁)=argmax_𝛼,𝛽𝑙(1,𝛼,𝛽). The score test statistic was defined as 𝑇₁=𝑆(𝛼₁,𝛽̂₁)/(1+𝑛₁/𝑛₀), which has a 𝜒²¹ limiting distribution under the null hypothesis.

We compare the EM test and the score test in terms of type I error and power. We calculate the type I errors of each method under the null hypothesis based on 20,000 repetitions and the power under the alternative models based on 2,000 repetitions. For fair comparison, simulated critical values are used to calculate the power. We consider two sample sizes: 50 and 200 and 𝐾=1,2,3. Tables 1 and 2 contain the simulation results for the log-normal models and Tables 3 and 4 for the gamma models.

Table 1 

Type I error and power comparisons (%) of the EM test and the score test (SC test) for log-normal model: 𝑛₀=𝑛₁=50.

Table 2 

Type I error and power comparisons (%) of the EM test and the score test (SC test) for log-normal model: 𝑛₀=𝑛₁=200.

Table 3 

Type I error and power comparisons (%) of the EM test and the score test (SC test) for gamma model:𝑛₀=𝑛₁=50.

Table 4 

Type I error and power comparisons (%) of the EM test and the score test (SC test) for gamma model: 𝑛₀=𝑛₁=200.

The results show that the EM test and the score test have similar type I errors. For both methods, the type I errors are somehow larger than the nominal levels when the sample size is 𝑛=50; they are close to the nominal levels when the sample size is increased to 𝑛=200. For the log-normal models, two methods have almost the same power when the alternatives are close to each other such as 𝜇=1; the EM test becomes much more powerful when the alternatives are distant and the sample size increases. In the case of 𝑛=50, 𝜆=0.2, 𝜇=3, and nominal level 0.01, the EM test has a 10% gain in power compared with the score test; the gain rushes up to almost 30% when 𝜆=0.1, 𝜇=3, and the sample size increases to 𝑛=200. For the gamma models, the advantage of the EM test is more obvious. For both sample sizes 𝑛=50 and 200, the EM test is more powerful than the score test.

6. Real Example

We apply our EM test procedure to the drug abuse data [15] in a study of addiction to morphine in rats. In this study, rats got morphine by pressing a lever and the frequency of lever presses (self-injection rates) after six-day treatment with morphine was recorded as response variable. The data consist of the number of lever presses for five groups of rats: four treatment groups with different dose levels and one saline group (control group).

We analyzed the response variables (the number of lever presses by rats) of the treatment group at the first dose level and the control group. The data is tabulated in Table  3 of Fu et al. [5]. Following Boos and Browine [16] and Fu et al. [5], we analyze the transformed data, log₁₀(𝑅+1) with 𝑅 being the number of lever presses by rats. Instead of using the parametric models as Boos and Browine [16] and Fu et al. [5], we adopt Anderson’s semiparametric approach. That is, we assume that the response variables in control group comes from 𝑓(𝑥), while the response variables in treatment group comes from ℎ(𝑥)=(1−𝜆)𝑓(𝑥)+𝜆𝑔(𝑥) with 𝑔(𝑥)/𝑓(𝑥)=exp(𝛼+𝛽𝑥). The EM test statistics for testing homogeneity under the semiparametric two-sample model are found to be EM^𝑛(1)=14.090, EM^𝑛(2)=14.150, and EM^𝑛(3)=14.167. Calibrated by the 𝜒²¹ limiting distribution, the 𝑃values are all around 0.02%. We also applied the score test of Qin and Liang [1]. The score test statistic is 9.417 with the 𝑃 value equal to 0.2% calibrated by the 𝜒²¹ limiting distribution. We also used the permutation methods to get the 𝑃values of the two types of tests. Based on 50,000 permutations, the 𝑃 values of the three EM test statistics are all around 0.03%, and the 𝑃 value of the score test is around 0.5%. In accordance with Fu et al. [5], both methods suggest a significant treatment effect, while the proposed EM test has much stronger evidence than the score test.

Appendix

Proofs

The proofs of Theorems 3.1 and 3.3 are based on the three lemmas given below. Lemma A.1 assesses the order of the maximum empirical likelihood estimators of 𝛼 and 𝛽 with 𝜆 bounded away from 0 under the null hypothesis. Lemma A.2 shows that the EM iteration updates the value of 𝜆 by the amount of order 𝑜_𝑝(1). Theorem 3.1 is then proved by iteratively using Lemmas A.1 and A.2. Lemma A.3 gives an approximation of the penalized ELR for any 𝜆 bounded away from 0, based on which we prove Theorem 3.3.

Lemma A.1. Assume the conditions of Theorem 3.1. Let 𝜆∈[𝜖,1] for some constant 𝜖>0 and (𝛼,𝛽)=argmax_𝛼,𝛽𝑝𝑅(𝜆,𝛼,𝛽). Then, we have 𝛼=𝑂_𝑝𝑛⁻¹,𝛽=𝑦−𝑥𝜆𝜎²+𝑜_𝑝𝑛^−1/2(A.1) with 𝑥=1/𝑛₀∑^𝑛0_𝑖=1𝑥_𝑖 and 𝑦=1/𝑛₁∑^𝑛1_𝑗=1𝑦_𝑗.

Proof. Since 𝜆≥𝜖>0, the parameters (𝛼,𝛽) in the empirical likelihood ratio are identifiable. Therefore, (𝛼,𝛽) are √𝑛-consistent to the true value (0,0), that is, 𝛼=𝑂_𝑝(𝑛^−1/2) and 𝛽=𝑂_𝑝(𝑛^−1/2) [10].
Following the arguments in Section 4, the maximum empirical likelihood estimate (𝛼,𝛽) should satisfy (here 𝜆 is suppressed to 𝜆) 𝜕𝐺𝜉,𝛼,𝛽𝜕𝛼=−2^𝑛_ℎ=1𝜉𝑒^{𝛼+𝛽𝑡ℎ}1+𝜉𝑒^{𝛼+𝛽𝑡ℎ}−1+2^𝑛1_𝑗=1𝜆𝑒^{𝛼+𝛽𝑦𝑗}1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}𝜕𝐺=0,(A.2)𝜉,𝛼,𝛽𝜕𝛽=−2^𝑛_ℎ=1𝜉𝑒^{𝛼+𝛽𝑡ℎ}𝑡_ℎ1+𝜉𝑒^{𝛼+𝛽𝑡ℎ}−1+2^𝑛1_𝑗=1𝜆𝑒^{𝛼+𝛽𝑦𝑗}𝑦_𝑗1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}=0(A.3) with 𝜉=1𝑛^𝑛1_𝑗=1𝜆𝑒^{𝛼+𝛽𝑦𝑗}1−𝜆+𝜆𝑒^{𝛼+𝛽𝑦𝑗}.(A.4)
Applying Taylor expansion on the right-hand side of (A.4), we get 𝜉=𝑛¹𝑛𝜆+𝑜_𝑝(1).(A.5) Further applying first-order Taylor expansion to (A.2) and using (A.4), we get 𝑛𝜉1−𝜉𝛼+𝜉1−𝜉^𝑛_ℎ=1𝑡_ℎ𝛽−𝑛₁𝜆1−𝜆𝛼−𝜆1−𝜆^𝑛1_𝑗=1𝑦_𝑗𝛽=𝑂_𝑝(𝑛)𝛼²+𝛽².(A.6) Note that both 𝛼 and 𝛽 are of order 𝑂_𝑝(𝑛^−1/2) and that both ∑^𝑛ℎ=1𝑡_ℎ and ∑^𝑛1_𝑗=1𝑦_𝑗 have order 𝑂_𝑝(𝑛^1/2). Combining (A.5) and (A.6) yields 𝑛₁𝜆(𝜆−𝑛₁/𝑛𝜆)𝛼=𝑂_𝑝(1). Therefore, 𝛼=𝑂_𝑝(𝑛⁻¹).
Similarly, first-order Taylor expansion of (A.3) results in 0=−𝜉^𝑛_ℎ=1𝑡_ℎ−𝜉1−𝜉^𝑛_ℎ=1𝑡_ℎ𝛼−𝜉1−𝜉^𝑛_ℎ=1+𝑡^2ℎ𝛽𝜆^𝑛1_𝑗=1𝑦_𝑗+𝜆1−𝜆^𝑛1_𝑗=1𝑦_𝑗𝛼+𝜆1−𝜆^𝑛1_𝑗=1𝑦^2𝑗𝛽+𝑂_𝑝(𝑛)𝛼²+𝛽².(A.7) With the same reasoning as for 𝛼, it follows from (A.7) that 𝑛¹𝜆1−𝑛¹𝑛𝜆𝜎²−𝑛₁𝜆1−𝜆𝜎²𝛽=𝜆^𝑛1_𝑗=1𝑦_𝑗−𝑛¹𝑛𝜆^𝑛_ℎ=1𝑡_ℎ+𝑜_𝑝𝑛^1/2.(A.8) After some algebra, we have 𝛽=(𝑦−𝑥)/(𝜆𝜎²)+𝑜_𝑝(𝑛^−1/2), which completes the proof.

Suppose that 𝜆, 𝛼, and 𝛽 have the properties given in Lemma A.1. For 𝑗=1,…,𝑛₁, let 𝑤_𝑗=𝜆exp(𝛼+𝛽𝑦_𝑗)/(1−𝜆+𝜆exp(𝛼+𝛽𝑦_𝑗)). The updated value of 𝜆 is𝜆^∗=argmax_𝜆^𝑛1_𝑗=11−𝑤_𝑗log(1−𝜆)+^𝑛1_𝑗=1𝑤_𝑗log(𝜆)+log(𝜆).(A.9) It can be verified that the close form of 𝜆^∗ is given by 𝜆^∗=(1/(𝑛₁+1))(∑^𝑛1_𝑗=1𝑤_𝑗+1). We now show that the above iteration only changes the value of 𝜆 by an 𝑜_𝑝(1) term.

Lemma A.2. Assume the conditions of Lemma A.1 hold. Then, 𝜆^∗=𝜆+𝑜_𝑝(1).

Proof. Let 𝜆̂=∑^𝑛1_𝑗=1𝑤_𝑗/𝑛₁. According to Lemma A.1, 𝛼=𝑜_𝑝(1) and 𝛽=𝑜_𝑝(1). Applying the first-order Taylor expansion, we have 𝜆̂=1𝑛_1𝑛¹_𝑗=1𝜆exp𝛼+𝛽𝑦_𝑗1−𝜆+𝜆exp𝛼+𝛽𝑦_𝑗=𝜆+𝑂_𝑝(1)𝛼+𝛽=𝜆+𝑜_𝑝(1).(A.10) Some simple algebra work shows that 𝜆^∗−𝜆̂=1−𝜆𝑛₁+1=𝑜^𝑝(1).(A.11) Therefore, 𝜆^∗=𝜆+𝑜_𝑝(1), and this finishes the proof.

Proof of Theorem 3.1. With the above two technical lemmas, the proof is the same as that of Theorem  1 in Li et al. [13] and therefore is omitted.

The next lemma is a technical preparation for proving Theorem 3.3. It investigates the asymptotic approximation of the penalized ELR for any 𝜆 bounded away from 0.

Lemma A.3. Assume the conditions of Theorem 3.1 and 𝜆∈[𝜖,1] for some 𝜖>0. Then, 𝑝𝑅𝜆,𝛼,𝛽=𝑛𝜌(1−𝜌)𝜎⁻²𝑦−𝑥²+2log𝜆+𝑜_𝑝(1).(A.12)

Proof. With Lemma A.1, we have 𝛼=𝑂_𝑝(𝑛⁻¹) and 𝛽=𝑂_𝑝(𝑛^−1/2). Applying second-order Taylor expansion on 𝑝𝑅(𝜆,𝛼,𝛽) and noting that 𝜕𝑝𝑅/𝜕𝛼|_{(𝛼,𝛽)=(0,0)}=0, we have 𝑝𝑅𝜆,𝛼,𝛽=2−𝜉^𝑛_ℎ=1𝑡_ℎ+𝜆^𝑛1_𝑗=1𝑦_𝑗𝛽−𝜉1−𝜉^𝑛_ℎ=1𝑡^2ℎ−𝜆1−𝜆^𝑛1_𝑗=1𝑦^2𝑗𝛽²+2log𝜆+𝑜_𝑝(1).(A.13) Using (A.5) and the facts that both ∑^𝑛ℎ=1𝑡^2ℎ/𝑛 and ∑^𝑛1_𝑗=1𝑦^2𝑗/𝑛₁ converge to 𝜎² in probability, the above expression can be simplified to 𝑝𝑅𝜆,𝛼,𝛽=2𝑛¹𝑛₀𝑛𝜆𝑦−𝑥𝛽−𝑛¹𝑛₀𝑛𝜆²𝜎²𝛽²+2log𝜆+𝑜_𝑝(1).(A.14) Plugging in the approximation 𝛽=(𝑦−𝑥)/(𝜆𝜎²)+𝑜_𝑝(𝑛^−1/2), we get 𝑝𝑅𝜆,𝛼,𝛽=𝑛¹𝑛₀𝑛𝑦−𝑥²𝜎²=𝑛𝜌(1−𝜌)𝜎₋₂+2log𝜆+𝑜_𝑝(1)𝑦−𝑥²+2log𝜆+𝑜_𝑝(1).(A.15) This completes the proof.

Proof of Theorem 3.3. Without loss of generality, we assume 0<𝜆₁<𝜆₂<⋯<𝜆_𝐿=1. According to Theorem 3.1 and Lemma A.3, for 𝑙=1,…,𝐿, we have 𝑝𝑅𝜆^𝑙(𝐾),𝛼^𝑙(𝐾),𝛽^𝑙(𝐾)=𝑛𝜌(1−𝜌)𝜎⁻²𝑦−𝑥²+2log𝜆_𝑙+𝑜_𝑝(1).(A.16) This leads to EM^𝑛(𝐾)=max_{1≤𝑙≤𝐿}𝑝𝑅𝜆^𝑙(𝐾),𝛼^𝑙(𝐾),𝛽^𝑙(𝐾)=𝑛𝜌(1−𝜌)𝜎⁻²𝑦−𝑥²+𝑜_𝑝(1),(A.17) where the remainder is still 𝑜_𝑝(1) since the maximum is taken over a finite set.
Note that when 𝑛 tends to infinity, √𝑛(𝑦−𝑥)→𝑁(0,𝜎²/[𝜌(1−𝜌)])in distribution. Therefore, EM^𝑛(𝐾)⟶𝜒²¹(A.18) in distribution as 𝑛 goes to infinity. This completes the proof.