# Asymptotic Properties for Estimators in a Semiparametric EV Model with NA Errors and Missing Responses

Authors: Jing-jing Zhang; Cheng-Liang Liu
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

This article deals with the semiparametric errors-in-variables (EV) modelyi=ξiβ+gti+ϵi, xi=ξi+μi, where yi are the random missing response variables, ξi,ti are the design points, ξi are the potential variables observed with measurement errors μi, and random errors ϵi are negatively associated (NA) variables. In addition, we need to estimate the unknown slope parameter β and nonparametric component g·. In order to solve the missing responses, we introduce three different approaches to get the estimators of β and g·. Meanwhile, we study the asymptotic properties for all the estimators. The final results show that the strong consistent rates for all the estimators can achieve on−1/4. The asymptotic normality for all the estimators is considered. We compare the finite sample performance of the three estimators by simulation as well.


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## Body

## 1. Introduction

The following semiparametric EV model is considered:(1)yi=ξiβ+gti+ϵi,xi=ξi+μi,where yi are the response variables, ξi,ti are design points, ξi are the potential variables observed with measurement errors μi, and Eμi=0 are random errors with Eϵi=0. The unknown parameter β∈ℛ needs to be estimated. g· is an unknown function defined on a close interval 0,1, and h· is a known function defined on 0,1 satisfying(2)ξi=hti+vi,where vi are also design points.Model (1) has attracted more and more attention from scholars because of its practicality in recent years. When μi≡0, ξi are observed exactly, the model (1) transforms into the standard semiparametric model, which was initially mentioned by Engle et al. [1] to study the effect of weather on electricity. In reality, many applications in our daily lives involve covariate measurement errors, hence the model (1) is more relevant than a normal regression model in several ways. Furthermore, when g·≡0, the model (1) reverts to the standard linear EV model, which has been investigated and utilized by a number of researchers. For example, Liu and Chen [2] researched consistency for the estimators of the ordinary least-square estimators (LSE). Miao et al. [3] discussed the central limit theorem for the LSE. Research in other directions is included in Miao and Liu [4], Fan et al. [5], and so on.The model (1) for complete data has also been studied by many authors; Cui and Li [6] presented the asymptotic distributions for estimators of β, g· and error variance; Liang et al. [7] derived a consistent estimator of β and its asymptotic distribution; Zhou et al. [8] considered estimation and inference procedures for the fixed-effects model. Because of its applicability and variability, model (1) has been widely used in many aspects such as economy, biology, and so on. However, incomplete data is also common in practical applications of models, which often exist in opinion polls, social-economic investigations, market research surveys, and other scientific experiments. For example, Bakoyannis et al. [9] discussed semiparametric regression and risk prediction with competing risk data under missing causes of failure; Wang et al. [10] studied semiparametric regression analysis with missing responses; Xu et al. [11] considered hypothesis tests in partial linear EV models with missing responses. In recent years, the study of incomplete data has attracted the attention of many scholars.On the other hand, in the practical application of the model, the assumption of error independence is not always satisfied. For example, a series of continuously collected economic data, population, enterprise, education, science and technology, health, and other social survey data are independent. Common dependent structures include autoregressive (AR) sequences, moving average (MA) sequences,α-mixing sequences, ρ-mixing sequences, negative associated (NA) sequences, and so on. Among others, the NA sequence is very common. Models with NA errors have been researched by many authors. For example, Tang et al. [12] discussed the asymptotic normality of a wavelet estimator; Yang et al. [13] studied the complete consistency of estimators for regression models; Zhang et al. [14] considered the weak consistency of the M-estimator in a linear regression model.In the literature of incomplete data, scholars generally assume that the error is independent. But there are also a lot of dependency errors. In the common dependency structure, the NA sequence is common and easy to be satisfied. In conclusion, we decide to study the asymptotic properties of the semiparametric EV model under incomplete data and NA errors. In this article, it is assumed that we have obtained incomplete datayi,δi,xi,ti from the model (1), where δi=0, if yi is missing at random, otherwise δi=1. And ϵi are NA errors. We cite three approaches to estimate β and g· with missing responses and study the strong consistency rates and asymptotic normality for the estimators.


## 2. Assumptions

In this section, we list some assumptions, which will be used in the proof of main results as follows.Definition 1.
A finite family of random variablesXi,1≤i≤n is said to be NA random variable if for every pair of disjoint subsets, A and B of 1,2,…,n , we have(3)Covf1Xi,i∈A,f2Xj,j∈B≤0,whenever f1 and f2 are coordinatewise increasing functions, such that the covariance exists. An infinite family of random variables is NA if every finite subfamily is NA.(i)
(A0) Letεi,1≤i≤n be a NA sequence, μi,1≤i≤n be independent random variables satisfying(i)
Eϵi=0, Eϵi2=1, Eμi=0, andEμi2=Ξμ2>0;(ii)
supiEϵip<∞, supiEμip<∞ for some p>4;(iii)
εi,1≤i≤n, μi,1≤i≤n are independent of each other.(i)
(A1) Letvi,1≤i≤n in (2) be a sequence satisfying(i)
limn⟶∞n−1∑i=1nvi2=∑0, limn⟶∞n−1∑i=1nδivi2=∑10<∑0,∑1<∞.(ii)
limn⟶∞supnnlogn−1·max1⩽m⩽n∑i=1mvji<∞; where j1,j2,…,jn is a permutation of 1,2,…,n.(i)
(A2)g· and h· are continuous functions satisfying the first-order Lipschitz condition on the close interval 0,1.(ii)
(A3) LetWnjct1≤j≤n be weight functions defined on [0, 1] and satisfy(i)
max1≤j≤n∑i=1nδjWnjcti=O1;(ii)
supt∑j=1nδjWnjctIt−tj>a⋅n−1/4=on−1/4 for any a>0;(iii)
suptWnjct=on−1/2log−1n.(i)
(A4) The probability weight functionsWnjt1≤j≤n are defined on 0,1 and satisfy(i)
max1≤j≤n∑i=1nWnjti=O1;(ii)
supt∑j=1nWnjtIt−tj>a·n−1/4=on−1/4 , for any a>0;(iii)
supt∑j=1nWnjt=on−1/2log−1n.(i)
(A5)supt∑j=1nδjWnjctIt−tj>b⋅n−1/2=on−1/2 for existing b>0.(ii)
(A6)supt∑j=1nWnjtIt−tj>b⋅n−1/2=on−1/2 for existing b>0.(iii)
(A7) There exist positive integersϕ≔ϕn and φ≔φn such that(i)
ϕ+φ≤n, ϕ⟶∞, φ⟶∞, φϕ−1⟶0, ϕ2=Onφ;(ii)
φϕ−11/2max1≤i≤nvi=o1.Remark 1.
Assumptions (A0)–(A6) are general conditions and have been widely used in some literature (see Chen et al. [15] and Gao et al. [16]);Remark 2.
(A7) can be satisfied easily. Such asϕ=n1/3, φ=logn, and max1≤i≤nvi=on1/6.Definition 2.
LetXt,t=0,±1,±2,… be a strictly stationary time series. For n=1,2,…, it is defined as(4)ρn=supX∈L2F−∞0,Y∈L2Fn∞CorrX,Y,where Fij denotes the σ-algebra generated by Xt,i≤t≤j, and L2Fij consists of Fij-measurable random variables with the finite second moment.Definition 3.
SupposeXi:i∈Z is a real-valued random variable sequence on a probability space Ω,ℬ,P. Let ℱmn denote the σ-filed generated by Xi:m≤i≤n. Let(5)αn=supPAB−PAPB:A∈ℱ−∞m,B∈ℱm+n∞.
The sequenceXi is said to be α-mixing or strong mixing if αn⟶0, as n⟶∞.


## 3. Main Results

### 3.1. Construction of Estimators

For model (1) with responses missing randomly, the most direct way to construct the estimators of β and g· is deleting all the missing data. Therefore, we can obtain the processed model δiyi=δiξiβ+δigti+δiϵi. If ξi can be observed, the parameter β can be estimated by using the least-squares estimation (LSE) approach. If the parameter β is known, we can define the estimator of g· by using the complete data δiyi,δixi,δiti,1≤i≤n and β, which is,(6)gn∗t,β=∑j=1nWnjctδjyj−δjxjβ,where Wnjct are weight functions satisfying (A3). On the other hand, in the case of semiparametric EV models, Liang et al. [17] improved LSE on the basis of the usual partial linear model and adopted the following formula to minimize the parameter estimation β:(7)SSβ=∑i=1nδiyi−xiβ−gn∗ti,β2−Ξμ2β2=min!.Therefore, one can obtain the modified LSE ofβ(8)β^c=∑i=1nδixic˜2−δiΞμ2−1∑i=1nδixic˜yic˜,where x˜ic=xi−∑j=1nδjWnjctixj, y˜ic=yi−∑j=1nδjWnjctiyj. We substitute (8) into gn∗t,β to define the following estimator of g·:(9)g^nct=∑j=1nδjWnjctyj−xjβ^c.Obviously, the composition of the estimatorsβ^c and g^nct does not use all the sample information reasonably. Hence, in order to make up for the missing data, we imply an imputation method from Wang and Sun [18] and let,(10)UiI=δiyi+1−δixiβ^c+g^ncti.Therefore, similar to (8) and (9) with using complete data UiI,xi,ti,1≤i≤n, we can get other estimators for β and g·, and they are,(11)β^I=∑i=1nx˜i2−δiΞμ2−1∑i=1nx˜iU˜iI,(12)g^nIt=∑j=1nWnjtUjI−xjβ^I,where U˜iI=UiI−∑j=1nWnjtUjI, x˜i=xi−∑j=1nWnjtixj, Wnjt are weight functions satisfying (A4).Thirdly, Wang and Sun [18] developed a so-called semiparametric regression surrogate approach. This method uses estimated semiparametric regression values instead of the corresponding response values to define estimators, whether the responses are observed or not. Let,(13)UiR=xiβ^c+g^ncti.,Therefore, similar to (8) and (9) with using complete data UiR,xi,ti,1≤i≤n, one can get the third estimators for β and g·, and they are,(14)β^R=∑i=1nx˜i2−1∑i=1nx˜iU˜iR,g^nRt=∑j=1nWnjtUjR−xjβ^R,where U˜iR=UiR−∑j=1nWnjtiUjR. Wnjt are weight functions satisfying (A4).


### 3.2. Strong Consistency of Estimators

In recent years, statisticians at home and abroad have shown great interest in the field of incomplete data analysis. However, there are few theoretical results for the semiparametric EV model whose error is a NA sequence and responses are missing. For a good estimator, strong consistency is the essential property that must be possessed. And we can judge whether estimators are good or bad by their strong consistency rates. So in this section, we gain the estimators ofβ and g· when yi are missing at random and discuss the strong consistency of the estimators in the following process.Theorem 1.
Suppose that (A0)-(A3) are satisfied. For everyt∈0,1, we have(a)
β^c−β=on−1/4a.s.(b)
g^nct−gt=on−1/4a.s.Theorem 2.
Suppose that (A0)–(A4) are satisfied. For everyt∈0,1, we have(a)
β^I−β=on−1/4a.s.(b)
g^nIt−gt=on−1/4a.s.Theorem 3.
Suppose that (A0)–(A4) are satisfied. For everyt∈0,1, we have(a)
β^R−β=on−1/4a.s.(b)
g^nRt−gt=on−1/4a.s.


### 3.3. Asymptotic Normality of Estimators

Under the condition of independent errors, many scholars have extended the semiparametric model in many aspects and studied the properties of the estimators. In recent years, more and more attention has been paid to the estimation of semiparametric regression models under dependent conditions, and a lot of achievements have been made. However, there are few studies on the semiparametric EV model with NA errors. Because many models need to satisfy the normal distribution, asymptotic normality is significant and helps to construct confidence intervals for estimators. Based on this, we research the asymptotic normality of estimators forβ and g⋅.Firstly, let us go through several phrases that will be utilized in the conclusions and proofs that follow:(15)ξ˜ic=ξi−∑j=1nδjWnjctiξj,μ˜ic=μi−∑j=1nδjWnjctiμj,g˜ic=gti−∑j=1nδjWnjctigtj,ϵ˜ic=ϵi−∑j=1nδjWnjctiϵj,ξ˜i=ξi−∑j=1nWnjtiξj,μ˜i=μi−∑j=1nWnjtiμj,ηi=ϵi−μiβ,g˜i=gti−∑j=1nWnjtigtj,ϵ˜i=ϵi−∑j=1nWnjtiϵj,Sn2=∑i=1nξ˜i2,S1n2=∑i=1nδix˜i2−δiΞμ2,S2n2=∑i=1nx˜i2−Ξμ2,S3n2=∑i=1nx˜i2,Σ1n2=Var∑i=1nδiξ˜ic+μiϵi−μiβ,Γ1n2t=Var∑j=1nWnjctδjϵj−μjβ,V=∑i=1n1−δiΞμ2S1n2,U=∑i=1nξ˜i21−δiS1n2,Σ2n2=Var∑i=1nδiξ˜i+U+Vξ˜ic+1+U+Vμiϵi−μiβ,Γ2n2t=Var∑j=1nWnjtδjϵj−μjβ,Γ3n2t=Var∑j=1nWnjt∑i=1nWnictjδiϵi−μiβ.Theorem 4.
Suppose that (A0)–(A3), (A5), and (A7) are satisfied. For everyt∈0,1,(a)
ifΣ1n2≥Cn, we have S1n2β^c−β/Σ1n⟶DN0,1;(b)
ifnΓ1n2t⟶∞ as n⟶∞ is satisfied, we have g^nct−gt/Γ1nt⟶DN0,1.Theorem 5.
Suppose that (A0)–(A7) are satisfied. For everyt∈0,1,(a)
ifΣ2n2≥Cn, we have S2n2β^L−β/Σ2n⟶DN0,1;(b)
ifnΓ2n2t⟶∞ as n⟶∞ is satisfied, we have g^nIt−gt/Γ2nt⟶DN0,1.Theorem 6.
Suppose that (A0)–(A7) are satisfied. For everyt∈0,1,(a)
ifΣ1n2≥Cn, we have S3n2β^R−β/Σ1n⟶DN0,1;(b)
ifnΓ3n2t⟶∞ as n⟶∞ is satisfied, we have g^nRt−gt/Γ3nt⟶DN0,1.


## 3.1. Construction of Estimators

For model (1) with responses missing randomly, the most direct way to construct the estimators of β and g· is deleting all the missing data. Therefore, we can obtain the processed model δiyi=δiξiβ+δigti+δiϵi. If ξi can be observed, the parameter β can be estimated by using the least-squares estimation (LSE) approach. If the parameter β is known, we can define the estimator of g· by using the complete data δiyi,δixi,δiti,1≤i≤n and β, which is,(6)gn∗t,β=∑j=1nWnjctδjyj−δjxjβ,where Wnjct are weight functions satisfying (A3). On the other hand, in the case of semiparametric EV models, Liang et al. [17] improved LSE on the basis of the usual partial linear model and adopted the following formula to minimize the parameter estimation β:(7)SSβ=∑i=1nδiyi−xiβ−gn∗ti,β2−Ξμ2β2=min!.Therefore, one can obtain the modified LSE ofβ(8)β^c=∑i=1nδixic˜2−δiΞμ2−1∑i=1nδixic˜yic˜,where x˜ic=xi−∑j=1nδjWnjctixj, y˜ic=yi−∑j=1nδjWnjctiyj. We substitute (8) into gn∗t,β to define the following estimator of g·:(9)g^nct=∑j=1nδjWnjctyj−xjβ^c.Obviously, the composition of the estimatorsβ^c and g^nct does not use all the sample information reasonably. Hence, in order to make up for the missing data, we imply an imputation method from Wang and Sun [18] and let,(10)UiI=δiyi+1−δixiβ^c+g^ncti.Therefore, similar to (8) and (9) with using complete data UiI,xi,ti,1≤i≤n, we can get other estimators for β and g·, and they are,(11)β^I=∑i=1nx˜i2−δiΞμ2−1∑i=1nx˜iU˜iI,(12)g^nIt=∑j=1nWnjtUjI−xjβ^I,where U˜iI=UiI−∑j=1nWnjtUjI, x˜i=xi−∑j=1nWnjtixj, Wnjt are weight functions satisfying (A4).Thirdly, Wang and Sun [18] developed a so-called semiparametric regression surrogate approach. This method uses estimated semiparametric regression values instead of the corresponding response values to define estimators, whether the responses are observed or not. Let,(13)UiR=xiβ^c+g^ncti.,Therefore, similar to (8) and (9) with using complete data UiR,xi,ti,1≤i≤n, one can get the third estimators for β and g·, and they are,(14)β^R=∑i=1nx˜i2−1∑i=1nx˜iU˜iR,g^nRt=∑j=1nWnjtUjR−xjβ^R,where U˜iR=UiR−∑j=1nWnjtiUjR. Wnjt are weight functions satisfying (A4).


## 3.2. Strong Consistency of Estimators

In recent years, statisticians at home and abroad have shown great interest in the field of incomplete data analysis. However, there are few theoretical results for the semiparametric EV model whose error is a NA sequence and responses are missing. For a good estimator, strong consistency is the essential property that must be possessed. And we can judge whether estimators are good or bad by their strong consistency rates. So in this section, we gain the estimators ofβ and g· when yi are missing at random and discuss the strong consistency of the estimators in the following process.Theorem 1.
Suppose that (A0)-(A3) are satisfied. For everyt∈0,1, we have(a)
β^c−β=on−1/4a.s.(b)
g^nct−gt=on−1/4a.s.Theorem 2.
Suppose that (A0)–(A4) are satisfied. For everyt∈0,1, we have(a)
β^I−β=on−1/4a.s.(b)
g^nIt−gt=on−1/4a.s.Theorem 3.
Suppose that (A0)–(A4) are satisfied. For everyt∈0,1, we have(a)
β^R−β=on−1/4a.s.(b)
g^nRt−gt=on−1/4a.s.


## 3.3. Asymptotic Normality of Estimators

Under the condition of independent errors, many scholars have extended the semiparametric model in many aspects and studied the properties of the estimators. In recent years, more and more attention has been paid to the estimation of semiparametric regression models under dependent conditions, and a lot of achievements have been made. However, there are few studies on the semiparametric EV model with NA errors. Because many models need to satisfy the normal distribution, asymptotic normality is significant and helps to construct confidence intervals for estimators. Based on this, we research the asymptotic normality of estimators forβ and g⋅.Firstly, let us go through several phrases that will be utilized in the conclusions and proofs that follow:(15)ξ˜ic=ξi−∑j=1nδjWnjctiξj,μ˜ic=μi−∑j=1nδjWnjctiμj,g˜ic=gti−∑j=1nδjWnjctigtj,ϵ˜ic=ϵi−∑j=1nδjWnjctiϵj,ξ˜i=ξi−∑j=1nWnjtiξj,μ˜i=μi−∑j=1nWnjtiμj,ηi=ϵi−μiβ,g˜i=gti−∑j=1nWnjtigtj,ϵ˜i=ϵi−∑j=1nWnjtiϵj,Sn2=∑i=1nξ˜i2,S1n2=∑i=1nδix˜i2−δiΞμ2,S2n2=∑i=1nx˜i2−Ξμ2,S3n2=∑i=1nx˜i2,Σ1n2=Var∑i=1nδiξ˜ic+μiϵi−μiβ,Γ1n2t=Var∑j=1nWnjctδjϵj−μjβ,V=∑i=1n1−δiΞμ2S1n2,U=∑i=1nξ˜i21−δiS1n2,Σ2n2=Var∑i=1nδiξ˜i+U+Vξ˜ic+1+U+Vμiϵi−μiβ,Γ2n2t=Var∑j=1nWnjtδjϵj−μjβ,Γ3n2t=Var∑j=1nWnjt∑i=1nWnictjδiϵi−μiβ.Theorem 4.
Suppose that (A0)–(A3), (A5), and (A7) are satisfied. For everyt∈0,1,(a)
ifΣ1n2≥Cn, we have S1n2β^c−β/Σ1n⟶DN0,1;(b)
ifnΓ1n2t⟶∞ as n⟶∞ is satisfied, we have g^nct−gt/Γ1nt⟶DN0,1.Theorem 5.
Suppose that (A0)–(A7) are satisfied. For everyt∈0,1,(a)
ifΣ2n2≥Cn, we have S2n2β^L−β/Σ2n⟶DN0,1;(b)
ifnΓ2n2t⟶∞ as n⟶∞ is satisfied, we have g^nIt−gt/Γ2nt⟶DN0,1.Theorem 6.
Suppose that (A0)–(A7) are satisfied. For everyt∈0,1,(a)
ifΣ1n2≥Cn, we have S3n2β^R−β/Σ1n⟶DN0,1;(b)
ifnΓ3n2t⟶∞ as n⟶∞ is satisfied, we have g^nRt−gt/Γ3nt⟶DN0,1.


## 4. Simulation Study

In this section, we carry out a simulation to study the finite sample performance of the proposed estimators. In particular,(i)
we compare the performance of the estimators amongβ^c, β^I, and β^R by their mean square error(MSE), the estimators among g^nct, g^nIt, and g^nRt by their generalized mean square error (GMSE).(ii)
we give the Q-Q plots for the three estimators ofg·.Observations are generated from,(16)yi=ξi+gti+εi,xi=ξi+μi,i=1,2,…,n,where β=1, gt=sin2πt, ti=i−0.5/n, and ξi=ti2+vi with vi=sini/n−1/3 for 1≤i≤n. μi,1≤i≤n is an i.i.d N0,0.22 sequence and εi,1≤i≤n is a NA sequence. Following Joag-Dev and Proschan (1983), we choose that εi,1≤i≤n is a multivariate normal distribution with Eϵ1,…,ϵn=0,…,0, Covϵi,ϵj=−6−j−i−1 for i≠j, and Varϵi=0.32 for 1≤i≤n. The weight functions for the proposed estimators are as follows:(17)Wnjct=Kt−ti/hn∑j=1nδjKt−tj/hn,Wnjt=Mt−ti/bn∑j=1nMt−tj/bn,where K· and M· are Gaussian kernel functions, hn and bn are two bandwidth sequences.


### 4.1. The MSE or GMSE for the Estimators ofβ and g·

In this subsection, we generate the observed data with sample sizesn=100,300,500 and the missing probability of the response variables is p=0.1,0.25,0.5. The MSE or GMSE of the estimators for β and g⋅ based on M=500 replications are defined as,(18)MSEβ^=1M∑l=1Mβ^l−β2,GMSEg^⋅=1Mn∑l=1M∑k=1ng^tk,l−gtk2.We compute the MSE or GMSE for each estimator based onM=500 replications and a grid of bandwidths from 0 to 1. We choose the optimal bandwidths to minimize the MSE or GMSE. The minimum MSE or GMSE and the corresponding optimal bandwidths for the estimators are reported in Table 1 and 2.Table 1
The MSE and corresponding optimal bandwidths for three estimators ofβ.
npβ^ch1β^Ih1b1β^Rh1b11000.11.0000∗10−30.3758.9316∗10−40.2500.3758.8058∗10−40.3750.0253000.13.1740∗10−40.3503.0293∗10−40.3750.3252.9280∗10−40.3250.2005000.12.1041∗10−40.3252.0127∗10−40.1750.3251.8867∗10−40.3000.2501000.251.5000∗10−30.4751.2000∗10−30.5000.4751.0000∗10−30.5000.4753000.254.9747∗10−40.4504.3934∗10−40.1750.4004.3723∗10−40.4500.1255000.252.6421∗10−40.4002.1610∗10−40.3750.4002.2831∗10−40.4250.1001000.51.7000∗10−30.4501.6000∗10−30.3750.3501.5000∗10−30.3750.3503000.57.6833∗10−40.4756.6091∗10−40.5000.3756.2785∗10−40.4500.1755000.53.5225∗10−40.3253.2059∗10−40.4250.2503.2181∗10−40.3000.075Table 2
The GMSE and corresponding optimal bandwidths for three estimators ofgt.
npg^nch1g^nIh1b1g^nRh1b11000.16.800∗10−30.0256.300∗10−30.1000.0255.700∗10−30.0250.0253000.12.100∗10−30.0252.100∗10−30.0250.0252.600∗10−30.0250.0255000.11.500∗10−30.0251.600∗10−30.0250.0252.200∗10−30.0250.0251000.251.000∗10−20.0509.400∗10−30.0250.0258.400∗10−30.0250.0253000.253.400∗10−30.0253.400∗10−30.0250.0253.600∗10−30.0250.0255000.252.500∗10−30.0252.500∗10−30.0250.0253.000∗10−30.0250.0251000.51.860∗10−20.0501.740∗10−20.0500.0251.820∗10−20.0500.0253000.58.100∗10−30.0257.300∗10−30.0200.0257.300∗10−30.0250.0255000.54.000∗10−30.0253.800∗10−30.0250.0253.900∗10−30.0250.025From Table1 and Table 2, it can be seen that,(i)
the strong consistency of all estimators is significant.(ii)
for every fixedn, the MSE or GMSE for all estimators increases as the missing probability increases.(iii)
for every fixed missing probability, the MSE or GMSE for all estimators decreases as the sample sizen increases.(iv)
compared toβ^c, the estimated value β^I and β^R are closer to the true value. Therefore, the compensation for missing data is meaningful. For g⋅, the results are similar.


### 4.2. Simulation Study of Asymptotic Normality

In this subsection, we give the Q-Q plots for the estimators ofβ and g⋅. We consider all estimators of β and g⋅ with p=0.25, n=100,500. In Figures 1, 2, 3, we give the Q-Q plots for β^c, β^I, and β^R with n=100 and 500, respectively. In Figures 4, 5, 6, we provide the Q-Q plots for g^nc0.5, g^nI0.5, and g^nR0.5 with n=100 and 500, respectively.Figure 1
The Q-Q plot forβ^c with M = 500, n = 100, and n = 500, respectively.Figure 2
The Q-Q plot forβ^I with M = 500, n = 100, and n = 500, respectively.Figure 3
The Q-Q plot forβ^R with M = 500, n = 100, and n = 500, respectively.Figure 4
The Q-Q plot forg^nc0.5 with M = 500, n = 100, and n = 500, respectively.Figure 5
The Q-Q plot forg^nI0.5 with M = 500, n = 100, and n = 500, respectively.Figure 6
The Q-Q plot forg^nR0.5 with M = 500, n = = 100, and n = 500, respectively.From Figure1–6, one can see that:(i)
for 500 samples, distributions for the estimators ofβ and g0.5 are very close to normal distribution. So the asymptotic normality for all the estimators is obvious.(ii)
the sample variances of all the estimators decrease whenn goes from 100 to 500.


## 4.1. The MSE or GMSE for the Estimators ofβ and g·

In this subsection, we generate the observed data with sample sizesn=100,300,500 and the missing probability of the response variables is p=0.1,0.25,0.5. The MSE or GMSE of the estimators for β and g⋅ based on M=500 replications are defined as,(18)MSEβ^=1M∑l=1Mβ^l−β2,GMSEg^⋅=1Mn∑l=1M∑k=1ng^tk,l−gtk2.We compute the MSE or GMSE for each estimator based onM=500 replications and a grid of bandwidths from 0 to 1. We choose the optimal bandwidths to minimize the MSE or GMSE. The minimum MSE or GMSE and the corresponding optimal bandwidths for the estimators are reported in Table 1 and 2.Table 1
The MSE and corresponding optimal bandwidths for three estimators ofβ.
npβ^ch1β^Ih1b1β^Rh1b11000.11.0000∗10−30.3758.9316∗10−40.2500.3758.8058∗10−40.3750.0253000.13.1740∗10−40.3503.0293∗10−40.3750.3252.9280∗10−40.3250.2005000.12.1041∗10−40.3252.0127∗10−40.1750.3251.8867∗10−40.3000.2501000.251.5000∗10−30.4751.2000∗10−30.5000.4751.0000∗10−30.5000.4753000.254.9747∗10−40.4504.3934∗10−40.1750.4004.3723∗10−40.4500.1255000.252.6421∗10−40.4002.1610∗10−40.3750.4002.2831∗10−40.4250.1001000.51.7000∗10−30.4501.6000∗10−30.3750.3501.5000∗10−30.3750.3503000.57.6833∗10−40.4756.6091∗10−40.5000.3756.2785∗10−40.4500.1755000.53.5225∗10−40.3253.2059∗10−40.4250.2503.2181∗10−40.3000.075Table 2
The GMSE and corresponding optimal bandwidths for three estimators ofgt.
npg^nch1g^nIh1b1g^nRh1b11000.16.800∗10−30.0256.300∗10−30.1000.0255.700∗10−30.0250.0253000.12.100∗10−30.0252.100∗10−30.0250.0252.600∗10−30.0250.0255000.11.500∗10−30.0251.600∗10−30.0250.0252.200∗10−30.0250.0251000.251.000∗10−20.0509.400∗10−30.0250.0258.400∗10−30.0250.0253000.253.400∗10−30.0253.400∗10−30.0250.0253.600∗10−30.0250.0255000.252.500∗10−30.0252.500∗10−30.0250.0253.000∗10−30.0250.0251000.51.860∗10−20.0501.740∗10−20.0500.0251.820∗10−20.0500.0253000.58.100∗10−30.0257.300∗10−30.0200.0257.300∗10−30.0250.0255000.54.000∗10−30.0253.800∗10−30.0250.0253.900∗10−30.0250.025From Table1 and Table 2, it can be seen that,(i)
the strong consistency of all estimators is significant.(ii)
for every fixedn, the MSE or GMSE for all estimators increases as the missing probability increases.(iii)
for every fixed missing probability, the MSE or GMSE for all estimators decreases as the sample sizen increases.(iv)
compared toβ^c, the estimated value β^I and β^R are closer to the true value. Therefore, the compensation for missing data is meaningful. For g⋅, the results are similar.


## 4.2. Simulation Study of Asymptotic Normality

In this subsection, we give the Q-Q plots for the estimators ofβ and g⋅. We consider all estimators of β and g⋅ with p=0.25, n=100,500. In Figures 1, 2, 3, we give the Q-Q plots for β^c, β^I, and β^R with n=100 and 500, respectively. In Figures 4, 5, 6, we provide the Q-Q plots for g^nc0.5, g^nI0.5, and g^nR0.5 with n=100 and 500, respectively.Figure 1
The Q-Q plot forβ^c with M = 500, n = 100, and n = 500, respectively.Figure 2
The Q-Q plot forβ^I with M = 500, n = 100, and n = 500, respectively.Figure 3
The Q-Q plot forβ^R with M = 500, n = 100, and n = 500, respectively.Figure 4
The Q-Q plot forg^nc0.5 with M = 500, n = 100, and n = 500, respectively.Figure 5
The Q-Q plot forg^nI0.5 with M = 500, n = 100, and n = 500, respectively.Figure 6
The Q-Q plot forg^nR0.5 with M = 500, n = = 100, and n = 500, respectively.From Figure1–6, one can see that:(i)
for 500 samples, distributions for the estimators ofβ and g0.5 are very close to normal distribution. So the asymptotic normality for all the estimators is obvious.(ii)
the sample variances of all the estimators decrease whenn goes from 100 to 500.


## 5. Preliminary Lemmas

In this section, we assume thatC,C1,⋯ are some finite positive constants, whose values are different in different places. Now, we introduce several lemmas, which will be used in the proof of the main results.Lemma 1 (see Baek and Liang [19], Lemma 3.1).
Letei,i≥1 be a sequence of NA random variables with Eei=0 and α>2. Assuming that ani,1≤i≤n is a triangular array of numbers with max1≤i≤nani=On−1/2 and ∑i=1nani2=on−2/αlog−1n. If supiEeip<∞ for some p>2α/α−1. Then(19)∑i=1naniei=on−1/αa.s..Remark 3.
In Lemma1, it is easy to see that ∑i=1naniei=o1a.s. when α⟶∞; ∑i=1naniei=on−1/4a.s. when α=4. In addition, if all of the ''o′′ is changed into ''O′′, the conclusion is also right.Lemma 2 (see Härdle et al. [20], Lemma A.3).
LetV1,…,Vn be independent random variables with EVi=0, finite variances and sup1≤j≤nEVjr≤C<∞r>2. Assuming that aki,k,i=1,…,n is a sequence of numbers such that sup1≤i,k≤naki=On−p1 for some 0<p1<1 and ∑j=1naji=Onp2 for p2≥max0,2/r−p1. Then(20)max1≤i≤n∑k=1nakiVk=On−slogna.s.fors=p1−p22.Lemma 3 (see Liu et al. [21]).
Letei,1≤i≤n be a stationary NA sequence with zero mean. Assuming an is an array of positive real numbers, and ∑n=1∞σn2/an2<∞, where σn2=Varen. If 0<ai↑∞. Then(21)∑i=1∞eiai=o1a.s..Lemma 4 (see Xu Bing [22]).
Letϵi be a sequence of strong mixing, and Eϵi=0, when p>2, supi≥1Eεip<∞. And suppose that ∑n=1∞∑i=1nani2lognp/2<∞ and ∑n=1∞αnp−2/p<∞. Then(22)∑i=1∞aniϵi=o1a.s.,where αn is the mixing coefficient and ani,i=1,2,… is a real sequence.Lemma 5.
(see Zhang et al. [23]).(a)
LetR˜i=Rti−∑j=1nWnjtiRtj, where R⋅=g⋅ or h⋅. Let R˜ic=Rti−∑j=1nδjWnjctiRtj, where R⋅=g⋅ or h⋅. Then, we can deduce that max1≤i≤nR˜i=on−1/4 and max1⩽i⩽n|R˜ic|=on−1/4 by (A0)-(A4).(b)
From (A0)-(A4), we can get thatn−1∑i=1nξ˜i2⟶Σ0, ∑i=1nξ˜i⩽C1n, n−1∑i=1nδiξ˜ic2⟶Σ1 and ∑i=1nδiξ˜ic≤C2n.Lemma 6 (see Liang and Baek [17]).
Letani,1≤i≤n,n≥1 be a sequence of a real number. ei,1≤i≤n is a stationary NA sequence with zero mean. For any k≥1 and when μ⟶∞, there are Σj:k−j≥μCovek,ej⟶0a.s and max1≤i≤nani=oΔn, ∑i=1nani2=OΔn2. If ∑i=1nani=O1. Then(23)∑i=1nanieiΔn⟶DN0,1.Lemma 7.
(see Yang [24]).(a)
Letr>2, δ>0. ei,1≤i≤n is a strong mixing sequence with Eei=0, Eeir+δ<∞. Suppose that λ>rr+δ/2δ and αn=On−λ. Then, for any ε>0, there exists a positive constant C=Cε,r,δ,λ, such that(24)Emax1≤m≤n∑i=1meir≤Cnε∑i=1nEeir+∑i=1n‖ei‖r+δ2r/2.(b)
IfEei=0 and Eei2+δ<∞ for some δ>0, then(25)E∑i=1nei2≤1+16∑l=1nαδ/2+δl∑i=1n‖ei‖r+δ2.Lemma 8 (see Volkonskii and Rozanov [25]).
LetV1,…,Vm be α-mixing random variables measurable with respect to the σ-algebra ℱi1j1,…,ℱimjm, with 1≤i1<j1<⋯<jn≤n, il+1−jl≥ω≥1 and Vj≤1 for l,j=1,2,…,m,respectively. Then, E∏j=1mVj−∏j=1mEVj≤16m−1αω, where ℱab=σVi,a≤i≤b and αω is the mixing coefficient.


## 6. Proof of Main Results

### 6.1. Proof of Strong Consistency

Proof ofTheorem 1(a). According to (8), one can write that,(26)β^c−β=S1n−2∑i=1nδiξ˜ic+μ˜icϵ˜ic−μ˜icβ+δiΞμ2β+∑i=1nδiξ˜icg˜ic+∑i=1nδiμ˜icg˜ic=S1n−2∑i=1nδiξ˜icϵi−μiβ+∑i=1nδiμiϵi−∑i=1nδiμi2−Ξμ2β+∑i=1nδiξ˜icg˜ic+∑i=1nδiμ˜icg˜ic−∑i=1n∑j=1nδiδjWnjctiξ˜icϵj−∑i=1n∑j=1nδiδjWnjctiϵiμj−∑i=1n∑j=1nδiδjWnjctiμiϵj+∑i=1n∑j=1nδiδjWnjctiξ˜icμjβ+2∑i=1n∑j=1nδiδjWnjctiμiμjβ+∑i=1n∑j=1n∑k=1nδiδjδkWnjctiWnkctiμjϵk−∑i=1n∑j=1n∑k=1nδiδjδkWnjctiWnkctiμjμkβ:=S1n−2∑k=112Akn.Hence, in order to prove Theorem1 (a), according to (26), we just deduce that S1n−2≤Cn−1a.s. and n−1Akn=on−1/4a.s. for k=1,2,…,12.Step 1. We first prove S1n−2≤Cn−1a.s.. One can write S1n2 as follows:(27)S1n2=∑i=1nδiξ˜ic2+∑i=1nδiμi2−Ξμ2+∑i=1nδi∑j=1nδjWnjctiμj2+2∑i=1nδiξ˜icμi−2∑i=1nδiξ˜ic∑j=1nδjWnjctiμj−2∑i=1nδiμi∑j=1nδjWnjctiμj:=B1n+B2n+B3n+B4n+B5n+B6n.By Lemma5 (b), we have n−1B1n⟶Σ1. Hence, it suffices to verify that Bkn=on a.s. for k=2,3,…,6. Let r=p/2>2, p1=1/2, p2=1/2, and δ=0 in Lemma 2 and by (A0), we can verify that(28)∑i=1nϖi−Eϖi=n1/2⋅∑i=1nn−1/2ϖi−Eϖi=On1/2logna.s.,where ϖi are independent random variables satisfying Eϖi=0 and sup1≤i≤nEϖip/2<∞. Therefore, we obtain B2n=On1/2logn=on a.s. from (A0) and (28). On the other hand, taking α=4, p>4 in Lemma 1, we have(29)max1≤i≤n∑j=1nδjWnjctiϖj=on−1/4a.s.,max1≤i≤n∑j=1nWnjtiϖj=on−1/4a.s.,where ϖi are independent random variables satisfying Eϖi=0 and sup1≤i,k≤nEϖip<∞. By (A0) and Lemma 5, taking r=4, Unω=∑m=1kEΠnm2⋅IΠnm>ω≤C∑m=1kEΠnmr=C∑m=1kE∑i=kmkm+ϕ−1Znir≤Cn−r/2∑m=1kϕε∑i=kmkm+ϕ−1Eωir+∑i=kmkm+ϕ−1ωip/22r/2≤Cn−1+τϕε∑m=1k∑i=kmkm+ϕ−1|ξ˜ic|21+τ+O1+∑m=1k∑i=kmkm+ϕ−1ξ˜ic2+O11+τ≤Cn−1+τnϕε⋅max1≤i≤n|ξ˜ic|2τ+nϕτ⋅max1≤i≤n|ξ˜ic|2τ+nϕτ=o1, p2=1/4 in Lemma 2, one can also deduce that(30)B4n=2n3/4⋅∑i=1nn−3/4δiξ˜icμi=On3/4logn=ona.s.,Noting from Lemma5 (a), (28), and (29), we have(31)B3n≤∑i=1nδi⋅max1≤i≤n∑j=1nδjWnjctiμj2=on1/2a.s.,(32)B5n≤2∑i=1nδiξ˜ic⋅max1⩽i⩽n∑j=1nδjWnjctiμj=on3/4a.s.,(33)B6n≤2∑i=1nδiμi−Eδiμi+∑i=1nEδiμi⋅max1≤i≤n∑j=1nδjWnjctiμj=on3/4a.s..Therefore, for (29)–(33), one can deduce that S1n2=B1n+onB1n+oB1na.s., which yields that(34)limn⟶∞B1nS1n2=limn⟶∞B1nB1n+oB1n=1a.s..Therefore, by Lemma5 (b), we can get that S1n−2≤Cn−1a.s..Step 2. From (A0), we find out ηi=εi−μiβ,1≤i≤n is a NA sequence with Eηi=0, supiEηip≤C⋅supiEeip+C⋅supiEμip<∞, for some p>4. we deduce that(35)n−1A1n=n−1∑i=1nδiξ˜icηi≔∑i=1nδiQinηi,max1≤i≤nQin≤n−1max1≤i≤nξ˜ic=on−1/2,∑i=1nQin2=n−2∑i=1nξ˜ic2=On−1,We haven−1A1n=on−1/4a.s. by Lemma 1, Remark 3, and Lemma 5.Step 3. From (A0), we know that μi,i=1,2,…,n is a sequence of independent random variables, ϵi,i=1,2,…,n is a sequence of NA random errors, independent of μi,i≥1, and Eμi=Eϵi=0. So Covμiεi,μjεj=0, then Corrμiϵi,μjϵj=0. From Definition 2, μiεi,i≥1 is a sequence of ρ-mixing random variable, and the mixing coefficients ρn=0. In the situation, we can know ρ-mixing is also a sequence of strong mixing, and we have 0≤αn≤ρn/4=0. Therefore, μiϵi,i≤1 is a sequence of strong mixing random variables with the mixing coefficients αn=0 from Definition 2 and 3. So, in Lemma 4, let δ=p−4>0 and an=n−3/4. Then we have ∑n=1∞∑i=1∞n−3/42logn4+δ/2<∞, and ∑n=1∞αnp−2/p<∞, So(36)1n∑i=1nμiϵi=n−1/4⋅n−3/4∑i=1nμiϵi=on−1/4a.s..Then,n−1A2n=on−1/4a.s.Step 4. we verify that n−1Akn=on−1/4a.s. for k=3,4,5,9,10,12. Meanwhile, from (A0)–(A3), Lemma 5, (28), and (29), one can achieve that(37)n−1A3n≤Cn∑i=1nδiμi2−Ξμ2β=On−1/2logna.s.,n−1A4n≤Cn∑i=1nδiξ˜icg˜ic≤Cn∑i=1nδiξ˜ic⋅max1≤i≤ng˜ic=on−1/4,n−1A5n≤Cn∑i=1nδiμ˜icg˜ic≤Cn∑i=1nμic⋅δjg˜ic+∑i=1nδjg˜ic⋅max1≤i≤n∑j=1nδjWnjctiμj≤Cn⋅∑i=1nμi−Eμi+∑i=1nEμi⋅max1≤i≤nδjg˜ic+on−−1/2=on−1/4a.s.We can prove thatn−1Akn=on−1/4a.s. for n=9,10,12 in a similar way.Step 5. We verify that n−1Akn=on−1/4a.s. for k=6,7,8,11. Firstly, ϵi,i=1,2,…,n is a stationary NA sequence with zero mean, let ϵi+=ϵi+ϵi/2 and ϵi−=ϵi+ϵi/2. It is easy to know that the sequence ϵi+,i=1,2,…,n and ϵi−,i=1,2,…,n are all NA sequence satisfying ϵi=ϵi++ϵi−. From Lemma 3 and (A0), one can get(38)1n∑i=1nϵi=1n∑i=1nϵi++1n∑i=1nϵi−=1n∑i=1nϵi+−Eϵi++1n∑i=1nϵi−−Eϵi−+1n∑i=1nEϵi=O1a.s..Meanwhile, by applying (A0) and (A3), Lemma1 taking α=4, one can obtain that(39)max1≤i≤n∑i=1nδjWnjtiεj=on−1/4a.s.,max1≤i≤n∑i=1nWnjtiεj=on−1/4a.s..From Lemma5, (29), and (39), we have(40)n−1A6n≤Cn∑i=1nδiξ˜ic⋅max1≤i≤n∑j=1nδjWnjctiεj=on−1/4a.s.,n−1A7n≤Cn∑i=1nδiεi⋅max1≤i≤n∑j=1nWnjctiδjμj=on−1/4a.s..The proof ofn−1Akn=on−1/4a.s. for n=8,11 is analogous. As a result, one completes the proof of Theorem 1 (a).Proof ofTheorem 1(b). From (9), one can write that,(41)g^nct−gt=∑j=1nWnjctδjξjβ−β^c+∑j=1nWnjctδjgtj−gt+∑j=1nWnjctδjϵj+∑j=1nWnjctδjμjβ−β^c−∑j=1nWnjctδjμjβ:=F1nt+F2nt+F3nt+F4nt+F5nt.Therefore, we only need to prove thatFknt=on−1/4 a.s. for k=1,2,…,5. From (A0)–(A3), Theorem 1 (a), Lemma 5, (29), and (39), for every t∈0,1 and any a>0, one can get,(42)F1nt≤|β−β^c|⋅max1≤j≤n|Wnjctδj|⋅max1≤m≤n∑k=1mvjk+∑j=1nWnjctδjhtj=on−1/4a.s.,F2nt≤∑j=1nδjWnjct⋅gtj−gt⋅Itj−t>a⋅n−1/4+∑j=1nδjWnjct⋅gtj−gt⋅Itj−t<a⋅n−1/4≤C⋅a⋅n−1/4=on−1/4.The proof ofn−1Fknt=on−1/4a.s. for n=3,4,5 is analogous. Thus, the proof of Theorem 1 (b) is completed.Proof ofTheorem 2(a). From (10) and (11), we write that,(43)β^I−β=S2n−2∑i=1nδiξ˜iεi−μiβ−∑i=1n1−δiξ˜igti−g^ncti+∑i=1nξ˜iξi+μi1−δiβ^c−β−∑i=1n∑j=1nWnjtiδjξ˜iεj+∑i=1n∑j=1nWnjtiδjξ˜iμjβ+∑i=1n∑j=1nWnjti1−δjξ˜igtj−g^nctj−∑i=1n∑j=1nWnjti1−δjξ˜iξj+μjβ^c−β+∑i=1nξ˜ig˜ic+∑i=1nδiμiεi−∑i=1n∑j=1nWnjtiδiμjεi−∑i=1nδiμi2−Ξμ2β+∑i=1n∑j=1nWnjtiδiμiμjβ−∑i=1n1−δiμ˜igti−g^ncti+∑i=1n1−δiμ˜iξi+μiβ^c−β−∑i=1n∑j=1nWnjtiδjμiεj+∑i=1n∑j=1nWnjtiWnktiδjμkεj+∑i=1n∑j=1nWnjtiδjμiμjβ−∑i=1n∑j=1n∑k=1nWnjtiWnktiδjμjμkβ+∑i=1n∑j=1nWnjti1−δjμ˜igtj−g^nctj−∑i=1n∑j=1nWnjti1−δjμ˜iξj+μjβ^c−β+∑i=1nμ˜ig˜ic≔S2n−2∑k=121Dkn.Similar to Step 1 in the proof of Theorem1 (a), one can get S2n−2≤Cn−1a.s. Therefore, we just need to verify that n−1Dkn=on−1/4 a.s. for k=1,2,…,21. From (A0)–(A4), Lemma 2–5, Theorem 1(a), (29)–(39), we have,(44)n−1D1n=n−1∑i=1nδiξ˜iεi−μiβ=on−1/4a.s.,n−1D2n≤n−1∑i=1n1−δiξ˜i⋅max1≤i≤ngti−g^ncti=on−1/4a.s.,n−1D3n=n−1∑i=1nξ˜i21−δiβ^c−β+n−1∑i=1nξ˜i1−δi⋅max1≤i≤n∑j=1nWnjtiξjβ^c−β+n−1∑i=1nξ˜iμi1−δiβ^c−β=on−1/4a.s.,n−1D4n≤n−1∑i=1nξ˜i⋅max1≤i≤n∑j=1nδjWnjtiεj=on−1/4a.s.n−1D7n=n−1∑i=1nξ˜i⋅max1≤i≤n∑j=1nWnjti⋅max1≤j≤n∑j=1nWnjtihtj+vj⋅β^c−β+n−1∑i=1nξ˜i⋅max1≤i≤n∑j=1nWnjti1−δjμj⋅β^c−β=on−1/4a.s.,n−1D14n=n−1∑i=1nμ˜i1−δiξi+μiβ^c−β≤n−1∑i=1n1−δiξiμi⋅β^c−β+n−1∑i=1n1−δiμi2⋅β^c−β+n−1∑i=1nξi⋅max1≤i≤n∑j=1nWnjtiμj⋅β^c−β+n−1∑i=1nμi⋅max1≤i≤n∑j=1nWnjtiμj⋅β^c−β=on−1/4a.s.,n−1D19n≤n−1∑i=1nμ˜i⋅max1≤i≤n∑j=1n1−δjWnjti⋅max1≤i≤ngti−g^ncti≤n−1∑i=1nμi−Eμi+∑i=1nEμi−∑i=1n∑j=1nWnjtiμj⋅on−1/4=on−1/4a.s..The proof ofn−1Dkn=on−1/4a.s. for n=5,6,8,…,13,15,…,18,20,21 is analogous. Thus, Theorem 2 (a) is completed.Proof ofTheorem 2(b). From (11), one can write that,(45)g^nIt−gt=∑j=1nWnjtδjξjβ−β^c+∑j=1nWnjtδjgtj−g^nctj+∑j=1nWnjtδjϵj−∑j=1nWnjtξjβ−β^c−∑j=1nWnjtμjβ−β^c−∑j=1nWnjtδjμjβ^c+∑j=1nWnjtg^nctj−gtj+∑j=1nWnjtgtj−gt+∑j=1nWnjtξjβ−β^I+∑j=1nWnjtμjβ−β^I:=∑k=110Gkn.Therefore, we only need to prove thatGknt=on−1/4 a.s. for k=1,2,…,10. From (A0)–(A4), Lemma 5, (29), (39), Theorem 1, and Theorem 2 (a), we can get,(46)G1nt≤β−β^c⋅∑j=1nWnjtδjξj≤β−β^c⋅∑j=1nWnjtδjhtj+vj≤on−1/4⋅∑j=1nWnjt+max1≤j≤nWnjtδj⋅max1≤m≤n∑k=1mvjk=on−1/4a.s..G2nt≤max1≤j≤ngtj−g^nctj⋅∑j=1nWnjtδj=on−1/4a.s.,G8nt≤∑j=1nWnjtgtj−gt≤∑j=1nWnjtgtj−gt⋅Itj−t>a⋅n−1/4+∑j=1nWnjtgtj−gt⋅Itj−t<a⋅n−1/4=on−1/4a.s..The proof ofn−1Gknt=on−1/4a.s. for n=3,…,7,9,10 is analogous. Thus, the proof of Theorem 2 (b) is completed.Proof ofTheorem 3. The proof of Theorem 3 is similar to the proof of Theorem 2.


### 6.2. Proof of Asymptotic Normality

Proof ofTheorem 4(a). From (8), one can write that β^c−β:=S1n−2∑K=112Akn. Let An=A1n+A2n+A3n=∑i=1nδiξ˜ic+μiϵi−μiβ+Ξμ2β. In order to prove S1n−2β^c−β/Σ1n⟶DN0,1, we just prove An/Σ1n⟶DN0,1 and Akn/Σ1n⟶P0,k=4,5,…,10.Step 1. We prove An/Σ1n⟶DN0,1. Assuming that ωi=δiξ˜ic+μiϵi−μiβ+Ξμ2β, Zni=ωi/Σ1n. By (A0), Lemma 5, Σ1n2≥Cn, We can deduce that EZni=0, Var∑i=1NZni=1, EZni2+δ<∞. Because ϵi is a NA sequence with Eϵi=0, ϵi−μiβ is a NA sequence with Eϵi−μiβ=0. And ξ˜ic+μi is a sequence of independent random variables. By Definition 2, we know ξ˜ic+μiϵi−μiβ is a sequence of ρ-mixing random variable, and the mixing coefficients ρn=0. In this situation, we can know ρ-mixing is also a sequence of strong mixing, and we have 0≤αn≤ρn/4=0. Therefore, ξ˜ic+μiϵi−μiβ is a sequence of strong mixing random variables with the mixing coefficients αn=0. Now we employ Bernstein’s big block and small block procedure. Assuming k=n/ϕ+φ and let(47)Πnm=∑i=kmkm+ϕ−1Zni,Πnm′=∑i=lmlm+φ−1Zni,Πnk+1′=∑i=kϕ+φ+1nZni,where km=m+1ϕ+φ+1, lm=m+1ϕ+φ+ϕ+1, and m=1,2,…,k. Then(48)AnΣ1n:=Θn=∑m=1kΠnm+∑m=1kΠnm′+Πnk+1′:=Θn′+Θn″+Θn′″.Thus, we only need to proveEΘn″2⟶0, EΘn′″2⟶0, and Θn′⟶DN0,1. Applying Lemma 7, from (A0)–(A3) and (A7), it follows that(49)EΘn′′2≤Cn⋅E∑m=1k∑i=lmlm+φ−1δiξ˜icεi−δiξ˜icμiβ+δiμiεi−δiμi2−Ξμ2β2≤Ckφn⋅max1≤i≤nξ˜ic2+C1n⋅E∑m=1k∑i=lmlm+φ−1μiεi2+C2n⋅∑m=1k∑i=lmlm+φ−1Eμi2−Ξμ22≤C⋅φϕ−11/2max1≤i≤nhti+vi+max1≤i≤n∑j=1nWnjtiξj+C1=o1,(50)EΘn′″2≤Cn⋅E∑i=kϕ+φ+1nδiξ˜icεi−δiξ˜icμiβ+δiμiεi−δiμi2−Ξμ2β2≤Cϕn⋅max1≤i≤nξ˜ic2+C1ϕn=o1.Now we check thatΘn′⟶DN0,1. It suffices to show that(51)VarΘn′⟶1,Eexpit∑m=1kΠnm−∏m=1kEexpitΠnm⟶0,Unω=∑m=1kEΠnm2⋅IΠnm>ω⟶0,∀ω>0.By (50) and (51), EΘn2=1 and(52)EΘn′2=EΘn−Θn″+Θn′″2=1+EΘn″+Θn′″2−2EΘnΘn″+Θn′″.It is easy to verify that(53)VarΘn′−1⩽EΘn′′+Θn′′′2+2EΘnΘn′′+2EΘnΘn′′′⩽2EΘn″2+EΘn′′′2+2EΘn″21/2+2EΘn′″21/2=o1.Using Lemma8 and (A7), we have(54)Eexpit∑m=1kΠnm−∏m=1kEexpitΠnm⟶0.Next, becauseαn=0, we can assume λ>p/p−4 in Lemma 7. Setting r=21+τ, η=p−4/2−2τ where 0<τ≤λp−4−p/4λ+p<p−4/4. Obviously, r+η=p/2, and(55)pp−4<p1+τp−4−4τ=rr+η2η<λ.Applying Lemma7 and (A8), for small 0<ε<τ, we find that(56)Unω=∑m=1kEΠnm2⋅IΠnm>ω≤C∑m=1kEΠnmr=C∑m=1kE∑i=kmkm+ϕ−1Znir≤Cn−r/2∑m=1kϕε∑i=kmkm+ϕ−1Eωir+∑i=kmkm+ϕ−1‖ωi‖p22r/2≤Cn−1+τϕε∑m=1k∑i=kmkm+ϕ−1ξ˜ic21+τ+O1+∑m=1k∑i=kmkm+ϕ−1ξ˜ic2+O11+τ≤Cn−1+τnϕε⋅max1≤i≤nξ˜ic2τ+nϕτ⋅max1≤i≤nξ˜ic2τ+nϕτ=o1.Thus, we knowA1n/Σ1n⟶DN0,1.Step 2. We prove Akn/Σ1n⟶P0,k=4,5,…,12.(57)A4nΣ1n≤Cn∑i=1nδiξ˜icg˜ic≤Cn∑i=1nδih˜icg˜ic+∑i=1nδivig˜ic+∑i=1nδi∑j=1nWnjctivjg˜ic,≤Cnmax1≤i≤nδin⋅max1≤i≤nh˜ic⋅max1≤i≤ng˜ic+max1≤m≤n∑i=1mvki⋅max1≤i≤ng˜ic+max1≤j≤n∑j=1nWnjcti⋅max1≤i≤ng˜ic⋅max1≤m≤n∑j=1mvkj=o1,EA5nΣ1n2≤CnE∑i=1nδiμig˜ic2+E∑i=1nδig˜ic∑j=1nWnjctiμj2=on−1/2,A6nΣ1n≤Cn∑i=1n∑j=1nδiδjWnjctiξ˜icεj≤Cn∑i=1nδih˜ic+vi−∑k=1nWnkctivk⋅∑j=1nδjWnjctiεj=o1a.s.,EA7nΣ1n2≤CnE∑i1=1n∑i2=1n∑j1=1n∑j2=1nδi1δi2δj1δj2Wnj1cti1Wnj2cti2εi1εi2μj1μj2≤Cn∑i1=1n∑i2=1n∑j=1nδi1δi1δj2Wnjcti1Wnjcti2Covεi1,εi2Ξμ2=On−1.The proof ofAkn/Σ1n⟶P0,k=8,cdots,12 is similar. Thus, the proof of Theorem 4 (a) is completed.Proof ofTheorem 4(b). From (9), one can write that,(58)g^nct−gt=∑j=1nWnjctδjϵj−μjβ+∑j=1nWnjctδjgtj−gt+∑j=1nWnjctδjξj+∑j=1nWnjctδjμjβ−β^c≔F1nt+F2nt+F3nt+F4ntβ−β^c.From the proof of Theorem4 (a), we know that β−β^c=Opn−1/2. Because ϵi−μiβ is a NA sequence with Eϵi−μiβ=0. From the condition of Theorem 4 (b), we can know nΓ1n2t⟶∞, max1≤i≤nWnjctδj=on−1/2log−1n=oΓ1nt, and ∑j=1nWnjct2δj2=on−1/2log−1n=oΓ1n2t. From Lemma 6, we can verify F1nt/Γ1nt⟶DN0,1. Thus, we need only to prove F2nt=On−1/2, F3nt=O1, and F4nt=Op1. For ∀b>0,(59)F3nt≤∑j=1nWnjctδjξj≤∑j=1nWnjctδjhtj+vj≤max1≤j≤nhtj⋅∑j=1nWnjctδj+max1≤j≤nWnjctδj⋅max1≤m≤n∑j=1mvkj=O1,EF4nt2≤∑j=1nWnjctδj2Ξμ2=On−1/2log−1n.The proof ofF2nt=On−1/2 is similar. Thus, the proof of Theorem 4 (b) is completed.Proof ofTheorem 5(a). From (10) and (11), one can write that,(60)β^I−β=S2n−2∑i=1nδiξ˜i+U+Vξ˜ic+1+U+Vμiϵi−μiβ+1+U+VΞμ2β+U+V∑k=210Akn−∑i=1n1−δiξ˜igti−g^ncti+∑i=1n∑j=1nξ˜i1−δiWnjtiξj+∑i=1nξ˜i1−δiμiβ^c−β−∑i=1n∑j=1nWnjtiδjξ˜iϵj+∑i=1n∑j=1nWnjtiδjξ˜iμjβ+∑i=1n∑j=1nWnjti1−δjξ˜igtj−g^nctj−∑i=1n∑j=1nWnjti1−δjξ˜iξj+μjβ^c−β+∑i=1nξ˜ig˜ic−∑i=1n∑j=1nWnjtiδiμjϵi+∑i=1n∑j=1nWnjtiδiμiμjβ−∑i=1n1−δiμ˜igti−g^ncti+∑i=1n1−δiμi2−Eμi2+∑i=1n1−δiξiμi−∑i=1n1−δiξi∑j=1nWnjtiμj−∑i=1n1−δiμi∑j=1nWnjtiμjβ^c−β−∑i=1n∑j=1nWnjtiδjμiϵj+∑i=1n∑j=1n∑k=1nWnjtiWnktiδjμkϵj+∑i=1n∑j=1nWnjtiδjμiμjβ−∑i=1n∑j=1n∑k=1nWnjtiWnktiδjμjμkβ+∑i=1n∑j=1nWnjti1−δjμ˜igtj−g^nctj−∑i=1n∑j=1nWnjti1−δjμ˜iξj+μjβ^c−β+∑i=1nμ˜ig˜ic:=S2n−2∑k=120Dkn.Similar to the proof of Theorem4 (a), We can prove D1n/Σ2n⟶DN0,1, and D2n/Σ2n=op1. We only need to prove D2n/Σ2n=op1,k=3,…,20.(61)D3nΣ2n≤Cnmax1≤i≤n1−δin⋅max1≤i≤nh˜i⋅max1≤i≤ngti−g^ncti+max1≤m≤n∑i=1mvki⋅max1≤i≤ngti−g^ncti+max1≤i≤n∑j=1nWnjti⋅max1≤i≤ngti−g^ncti⋅max1≤m≤n∑j=1mvkj=o1a.s.,D4n=P1n+P2nβ^c−βP1nΣ2n≤Cn⋅∑i=1nh˜i+vi−∑k=1nWnktivk⋅max1≤i≤n1−δimax1≤i≤n∑j=1nWnjtihtj+vj=on1/2P2nΣ2n≤Cn∑i=1nξ˜i1−δiμi=On1/4logn,D8n=∑i=1nξ˜i∑j=1nWnjti1−δjξj+∑i=1nξ˜i∑j=1nWnjti1−δjμjβ^c−β≔Q1n+Q2nβ^c−βQ1nΣ2n≤Cn⋅∑i=1nhi˜+vi−∑k=1nWnktivk,max1≤i≤n∑j=1nWnjti1−δjξj=On1/2EQ2nΣ2n2≤Cn∑i=1n∑j=1nξ˜iWnjti1−δj2Ξμ2=O1.The proof ofDkn=op1 for k=5,6,7,9,…,20 is analogous. Thus, the proof of Theorem 5 (a) is completed.Proof ofTheorem 5(b). The proof of Theorem 5 (b) is similar.Proof ofTheorem 6. The proof of Theorem 6 is similar.


## 6.1. Proof of Strong Consistency

Proof ofTheorem 1(a). According to (8), one can write that,(26)β^c−β=S1n−2∑i=1nδiξ˜ic+μ˜icϵ˜ic−μ˜icβ+δiΞμ2β+∑i=1nδiξ˜icg˜ic+∑i=1nδiμ˜icg˜ic=S1n−2∑i=1nδiξ˜icϵi−μiβ+∑i=1nδiμiϵi−∑i=1nδiμi2−Ξμ2β+∑i=1nδiξ˜icg˜ic+∑i=1nδiμ˜icg˜ic−∑i=1n∑j=1nδiδjWnjctiξ˜icϵj−∑i=1n∑j=1nδiδjWnjctiϵiμj−∑i=1n∑j=1nδiδjWnjctiμiϵj+∑i=1n∑j=1nδiδjWnjctiξ˜icμjβ+2∑i=1n∑j=1nδiδjWnjctiμiμjβ+∑i=1n∑j=1n∑k=1nδiδjδkWnjctiWnkctiμjϵk−∑i=1n∑j=1n∑k=1nδiδjδkWnjctiWnkctiμjμkβ:=S1n−2∑k=112Akn.Hence, in order to prove Theorem1 (a), according to (26), we just deduce that S1n−2≤Cn−1a.s. and n−1Akn=on−1/4a.s. for k=1,2,…,12.Step 1. We first prove S1n−2≤Cn−1a.s.. One can write S1n2 as follows:(27)S1n2=∑i=1nδiξ˜ic2+∑i=1nδiμi2−Ξμ2+∑i=1nδi∑j=1nδjWnjctiμj2+2∑i=1nδiξ˜icμi−2∑i=1nδiξ˜ic∑j=1nδjWnjctiμj−2∑i=1nδiμi∑j=1nδjWnjctiμj:=B1n+B2n+B3n+B4n+B5n+B6n.By Lemma5 (b), we have n−1B1n⟶Σ1. Hence, it suffices to verify that Bkn=on a.s. for k=2,3,…,6. Let r=p/2>2, p1=1/2, p2=1/2, and δ=0 in Lemma 2 and by (A0), we can verify that(28)∑i=1nϖi−Eϖi=n1/2⋅∑i=1nn−1/2ϖi−Eϖi=On1/2logna.s.,where ϖi are independent random variables satisfying Eϖi=0 and sup1≤i≤nEϖip/2<∞. Therefore, we obtain B2n=On1/2logn=on a.s. from (A0) and (28). On the other hand, taking α=4, p>4 in Lemma 1, we have(29)max1≤i≤n∑j=1nδjWnjctiϖj=on−1/4a.s.,max1≤i≤n∑j=1nWnjtiϖj=on−1/4a.s.,where ϖi are independent random variables satisfying Eϖi=0 and sup1≤i,k≤nEϖip<∞. By (A0) and Lemma 5, taking r=4, Unω=∑m=1kEΠnm2⋅IΠnm>ω≤C∑m=1kEΠnmr=C∑m=1kE∑i=kmkm+ϕ−1Znir≤Cn−r/2∑m=1kϕε∑i=kmkm+ϕ−1Eωir+∑i=kmkm+ϕ−1ωip/22r/2≤Cn−1+τϕε∑m=1k∑i=kmkm+ϕ−1|ξ˜ic|21+τ+O1+∑m=1k∑i=kmkm+ϕ−1ξ˜ic2+O11+τ≤Cn−1+τnϕε⋅max1≤i≤n|ξ˜ic|2τ+nϕτ⋅max1≤i≤n|ξ˜ic|2τ+nϕτ=o1, p2=1/4 in Lemma 2, one can also deduce that(30)B4n=2n3/4⋅∑i=1nn−3/4δiξ˜icμi=On3/4logn=ona.s.,Noting from Lemma5 (a), (28), and (29), we have(31)B3n≤∑i=1nδi⋅max1≤i≤n∑j=1nδjWnjctiμj2=on1/2a.s.,(32)B5n≤2∑i=1nδiξ˜ic⋅max1⩽i⩽n∑j=1nδjWnjctiμj=on3/4a.s.,(33)B6n≤2∑i=1nδiμi−Eδiμi+∑i=1nEδiμi⋅max1≤i≤n∑j=1nδjWnjctiμj=on3/4a.s..Therefore, for (29)–(33), one can deduce that S1n2=B1n+onB1n+oB1na.s., which yields that(34)limn⟶∞B1nS1n2=limn⟶∞B1nB1n+oB1n=1a.s..Therefore, by Lemma5 (b), we can get that S1n−2≤Cn−1a.s..Step 2. From (A0), we find out ηi=εi−μiβ,1≤i≤n is a NA sequence with Eηi=0, supiEηip≤C⋅supiEeip+C⋅supiEμip<∞, for some p>4. we deduce that(35)n−1A1n=n−1∑i=1nδiξ˜icηi≔∑i=1nδiQinηi,max1≤i≤nQin≤n−1max1≤i≤nξ˜ic=on−1/2,∑i=1nQin2=n−2∑i=1nξ˜ic2=On−1,We haven−1A1n=on−1/4a.s. by Lemma 1, Remark 3, and Lemma 5.Step 3. From (A0), we know that μi,i=1,2,…,n is a sequence of independent random variables, ϵi,i=1,2,…,n is a sequence of NA random errors, independent of μi,i≥1, and Eμi=Eϵi=0. So Covμiεi,μjεj=0, then Corrμiϵi,μjϵj=0. From Definition 2, μiεi,i≥1 is a sequence of ρ-mixing random variable, and the mixing coefficients ρn=0. In the situation, we can know ρ-mixing is also a sequence of strong mixing, and we have 0≤αn≤ρn/4=0. Therefore, μiϵi,i≤1 is a sequence of strong mixing random variables with the mixing coefficients αn=0 from Definition 2 and 3. So, in Lemma 4, let δ=p−4>0 and an=n−3/4. Then we have ∑n=1∞∑i=1∞n−3/42logn4+δ/2<∞, and ∑n=1∞αnp−2/p<∞, So(36)1n∑i=1nμiϵi=n−1/4⋅n−3/4∑i=1nμiϵi=on−1/4a.s..Then,n−1A2n=on−1/4a.s.Step 4. we verify that n−1Akn=on−1/4a.s. for k=3,4,5,9,10,12. Meanwhile, from (A0)–(A3), Lemma 5, (28), and (29), one can achieve that(37)n−1A3n≤Cn∑i=1nδiμi2−Ξμ2β=On−1/2logna.s.,n−1A4n≤Cn∑i=1nδiξ˜icg˜ic≤Cn∑i=1nδiξ˜ic⋅max1≤i≤ng˜ic=on−1/4,n−1A5n≤Cn∑i=1nδiμ˜icg˜ic≤Cn∑i=1nμic⋅δjg˜ic+∑i=1nδjg˜ic⋅max1≤i≤n∑j=1nδjWnjctiμj≤Cn⋅∑i=1nμi−Eμi+∑i=1nEμi⋅max1≤i≤nδjg˜ic+on−−1/2=on−1/4a.s.We can prove thatn−1Akn=on−1/4a.s. for n=9,10,12 in a similar way.Step 5. We verify that n−1Akn=on−1/4a.s. for k=6,7,8,11. Firstly, ϵi,i=1,2,…,n is a stationary NA sequence with zero mean, let ϵi+=ϵi+ϵi/2 and ϵi−=ϵi+ϵi/2. It is easy to know that the sequence ϵi+,i=1,2,…,n and ϵi−,i=1,2,…,n are all NA sequence satisfying ϵi=ϵi++ϵi−. From Lemma 3 and (A0), one can get(38)1n∑i=1nϵi=1n∑i=1nϵi++1n∑i=1nϵi−=1n∑i=1nϵi+−Eϵi++1n∑i=1nϵi−−Eϵi−+1n∑i=1nEϵi=O1a.s..Meanwhile, by applying (A0) and (A3), Lemma1 taking α=4, one can obtain that(39)max1≤i≤n∑i=1nδjWnjtiεj=on−1/4a.s.,max1≤i≤n∑i=1nWnjtiεj=on−1/4a.s..From Lemma5, (29), and (39), we have(40)n−1A6n≤Cn∑i=1nδiξ˜ic⋅max1≤i≤n∑j=1nδjWnjctiεj=on−1/4a.s.,n−1A7n≤Cn∑i=1nδiεi⋅max1≤i≤n∑j=1nWnjctiδjμj=on−1/4a.s..The proof ofn−1Akn=on−1/4a.s. for n=8,11 is analogous. As a result, one completes the proof of Theorem 1 (a).Proof ofTheorem 1(b). From (9), one can write that,(41)g^nct−gt=∑j=1nWnjctδjξjβ−β^c+∑j=1nWnjctδjgtj−gt+∑j=1nWnjctδjϵj+∑j=1nWnjctδjμjβ−β^c−∑j=1nWnjctδjμjβ:=F1nt+F2nt+F3nt+F4nt+F5nt.Therefore, we only need to prove thatFknt=on−1/4 a.s. for k=1,2,…,5. From (A0)–(A3), Theorem 1 (a), Lemma 5, (29), and (39), for every t∈0,1 and any a>0, one can get,(42)F1nt≤|β−β^c|⋅max1≤j≤n|Wnjctδj|⋅max1≤m≤n∑k=1mvjk+∑j=1nWnjctδjhtj=on−1/4a.s.,F2nt≤∑j=1nδjWnjct⋅gtj−gt⋅Itj−t>a⋅n−1/4+∑j=1nδjWnjct⋅gtj−gt⋅Itj−t<a⋅n−1/4≤C⋅a⋅n−1/4=on−1/4.The proof ofn−1Fknt=on−1/4a.s. for n=3,4,5 is analogous. Thus, the proof of Theorem 1 (b) is completed.Proof ofTheorem 2(a). From (10) and (11), we write that,(43)β^I−β=S2n−2∑i=1nδiξ˜iεi−μiβ−∑i=1n1−δiξ˜igti−g^ncti+∑i=1nξ˜iξi+μi1−δiβ^c−β−∑i=1n∑j=1nWnjtiδjξ˜iεj+∑i=1n∑j=1nWnjtiδjξ˜iμjβ+∑i=1n∑j=1nWnjti1−δjξ˜igtj−g^nctj−∑i=1n∑j=1nWnjti1−δjξ˜iξj+μjβ^c−β+∑i=1nξ˜ig˜ic+∑i=1nδiμiεi−∑i=1n∑j=1nWnjtiδiμjεi−∑i=1nδiμi2−Ξμ2β+∑i=1n∑j=1nWnjtiδiμiμjβ−∑i=1n1−δiμ˜igti−g^ncti+∑i=1n1−δiμ˜iξi+μiβ^c−β−∑i=1n∑j=1nWnjtiδjμiεj+∑i=1n∑j=1nWnjtiWnktiδjμkεj+∑i=1n∑j=1nWnjtiδjμiμjβ−∑i=1n∑j=1n∑k=1nWnjtiWnktiδjμjμkβ+∑i=1n∑j=1nWnjti1−δjμ˜igtj−g^nctj−∑i=1n∑j=1nWnjti1−δjμ˜iξj+μjβ^c−β+∑i=1nμ˜ig˜ic≔S2n−2∑k=121Dkn.Similar to Step 1 in the proof of Theorem1 (a), one can get S2n−2≤Cn−1a.s. Therefore, we just need to verify that n−1Dkn=on−1/4 a.s. for k=1,2,…,21. From (A0)–(A4), Lemma 2–5, Theorem 1(a), (29)–(39), we have,(44)n−1D1n=n−1∑i=1nδiξ˜iεi−μiβ=on−1/4a.s.,n−1D2n≤n−1∑i=1n1−δiξ˜i⋅max1≤i≤ngti−g^ncti=on−1/4a.s.,n−1D3n=n−1∑i=1nξ˜i21−δiβ^c−β+n−1∑i=1nξ˜i1−δi⋅max1≤i≤n∑j=1nWnjtiξjβ^c−β+n−1∑i=1nξ˜iμi1−δiβ^c−β=on−1/4a.s.,n−1D4n≤n−1∑i=1nξ˜i⋅max1≤i≤n∑j=1nδjWnjtiεj=on−1/4a.s.n−1D7n=n−1∑i=1nξ˜i⋅max1≤i≤n∑j=1nWnjti⋅max1≤j≤n∑j=1nWnjtihtj+vj⋅β^c−β+n−1∑i=1nξ˜i⋅max1≤i≤n∑j=1nWnjti1−δjμj⋅β^c−β=on−1/4a.s.,n−1D14n=n−1∑i=1nμ˜i1−δiξi+μiβ^c−β≤n−1∑i=1n1−δiξiμi⋅β^c−β+n−1∑i=1n1−δiμi2⋅β^c−β+n−1∑i=1nξi⋅max1≤i≤n∑j=1nWnjtiμj⋅β^c−β+n−1∑i=1nμi⋅max1≤i≤n∑j=1nWnjtiμj⋅β^c−β=on−1/4a.s.,n−1D19n≤n−1∑i=1nμ˜i⋅max1≤i≤n∑j=1n1−δjWnjti⋅max1≤i≤ngti−g^ncti≤n−1∑i=1nμi−Eμi+∑i=1nEμi−∑i=1n∑j=1nWnjtiμj⋅on−1/4=on−1/4a.s..The proof ofn−1Dkn=on−1/4a.s. for n=5,6,8,…,13,15,…,18,20,21 is analogous. Thus, Theorem 2 (a) is completed.Proof ofTheorem 2(b). From (11), one can write that,(45)g^nIt−gt=∑j=1nWnjtδjξjβ−β^c+∑j=1nWnjtδjgtj−g^nctj+∑j=1nWnjtδjϵj−∑j=1nWnjtξjβ−β^c−∑j=1nWnjtμjβ−β^c−∑j=1nWnjtδjμjβ^c+∑j=1nWnjtg^nctj−gtj+∑j=1nWnjtgtj−gt+∑j=1nWnjtξjβ−β^I+∑j=1nWnjtμjβ−β^I:=∑k=110Gkn.Therefore, we only need to prove thatGknt=on−1/4 a.s. for k=1,2,…,10. From (A0)–(A4), Lemma 5, (29), (39), Theorem 1, and Theorem 2 (a), we can get,(46)G1nt≤β−β^c⋅∑j=1nWnjtδjξj≤β−β^c⋅∑j=1nWnjtδjhtj+vj≤on−1/4⋅∑j=1nWnjt+max1≤j≤nWnjtδj⋅max1≤m≤n∑k=1mvjk=on−1/4a.s..G2nt≤max1≤j≤ngtj−g^nctj⋅∑j=1nWnjtδj=on−1/4a.s.,G8nt≤∑j=1nWnjtgtj−gt≤∑j=1nWnjtgtj−gt⋅Itj−t>a⋅n−1/4+∑j=1nWnjtgtj−gt⋅Itj−t<a⋅n−1/4=on−1/4a.s..The proof ofn−1Gknt=on−1/4a.s. for n=3,…,7,9,10 is analogous. Thus, the proof of Theorem 2 (b) is completed.Proof ofTheorem 3. The proof of Theorem 3 is similar to the proof of Theorem 2.


## 6.2. Proof of Asymptotic Normality

Proof ofTheorem 4(a). From (8), one can write that β^c−β:=S1n−2∑K=112Akn. Let An=A1n+A2n+A3n=∑i=1nδiξ˜ic+μiϵi−μiβ+Ξμ2β. In order to prove S1n−2β^c−β/Σ1n⟶DN0,1, we just prove An/Σ1n⟶DN0,1 and Akn/Σ1n⟶P0,k=4,5,…,10.Step 1. We prove An/Σ1n⟶DN0,1. Assuming that ωi=δiξ˜ic+μiϵi−μiβ+Ξμ2β, Zni=ωi/Σ1n. By (A0), Lemma 5, Σ1n2≥Cn, We can deduce that EZni=0, Var∑i=1NZni=1, EZni2+δ<∞. Because ϵi is a NA sequence with Eϵi=0, ϵi−μiβ is a NA sequence with Eϵi−μiβ=0. And ξ˜ic+μi is a sequence of independent random variables. By Definition 2, we know ξ˜ic+μiϵi−μiβ is a sequence of ρ-mixing random variable, and the mixing coefficients ρn=0. In this situation, we can know ρ-mixing is also a sequence of strong mixing, and we have 0≤αn≤ρn/4=0. Therefore, ξ˜ic+μiϵi−μiβ is a sequence of strong mixing random variables with the mixing coefficients αn=0. Now we employ Bernstein’s big block and small block procedure. Assuming k=n/ϕ+φ and let(47)Πnm=∑i=kmkm+ϕ−1Zni,Πnm′=∑i=lmlm+φ−1Zni,Πnk+1′=∑i=kϕ+φ+1nZni,where km=m+1ϕ+φ+1, lm=m+1ϕ+φ+ϕ+1, and m=1,2,…,k. Then(48)AnΣ1n:=Θn=∑m=1kΠnm+∑m=1kΠnm′+Πnk+1′:=Θn′+Θn″+Θn′″.Thus, we only need to proveEΘn″2⟶0, EΘn′″2⟶0, and Θn′⟶DN0,1. Applying Lemma 7, from (A0)–(A3) and (A7), it follows that(49)EΘn′′2≤Cn⋅E∑m=1k∑i=lmlm+φ−1δiξ˜icεi−δiξ˜icμiβ+δiμiεi−δiμi2−Ξμ2β2≤Ckφn⋅max1≤i≤nξ˜ic2+C1n⋅E∑m=1k∑i=lmlm+φ−1μiεi2+C2n⋅∑m=1k∑i=lmlm+φ−1Eμi2−Ξμ22≤C⋅φϕ−11/2max1≤i≤nhti+vi+max1≤i≤n∑j=1nWnjtiξj+C1=o1,(50)EΘn′″2≤Cn⋅E∑i=kϕ+φ+1nδiξ˜icεi−δiξ˜icμiβ+δiμiεi−δiμi2−Ξμ2β2≤Cϕn⋅max1≤i≤nξ˜ic2+C1ϕn=o1.Now we check thatΘn′⟶DN0,1. It suffices to show that(51)VarΘn′⟶1,Eexpit∑m=1kΠnm−∏m=1kEexpitΠnm⟶0,Unω=∑m=1kEΠnm2⋅IΠnm>ω⟶0,∀ω>0.By (50) and (51), EΘn2=1 and(52)EΘn′2=EΘn−Θn″+Θn′″2=1+EΘn″+Θn′″2−2EΘnΘn″+Θn′″.It is easy to verify that(53)VarΘn′−1⩽EΘn′′+Θn′′′2+2EΘnΘn′′+2EΘnΘn′′′⩽2EΘn″2+EΘn′′′2+2EΘn″21/2+2EΘn′″21/2=o1.Using Lemma8 and (A7), we have(54)Eexpit∑m=1kΠnm−∏m=1kEexpitΠnm⟶0.Next, becauseαn=0, we can assume λ>p/p−4 in Lemma 7. Setting r=21+τ, η=p−4/2−2τ where 0<τ≤λp−4−p/4λ+p<p−4/4. Obviously, r+η=p/2, and(55)pp−4<p1+τp−4−4τ=rr+η2η<λ.Applying Lemma7 and (A8), for small 0<ε<τ, we find that(56)Unω=∑m=1kEΠnm2⋅IΠnm>ω≤C∑m=1kEΠnmr=C∑m=1kE∑i=kmkm+ϕ−1Znir≤Cn−r/2∑m=1kϕε∑i=kmkm+ϕ−1Eωir+∑i=kmkm+ϕ−1‖ωi‖p22r/2≤Cn−1+τϕε∑m=1k∑i=kmkm+ϕ−1ξ˜ic21+τ+O1+∑m=1k∑i=kmkm+ϕ−1ξ˜ic2+O11+τ≤Cn−1+τnϕε⋅max1≤i≤nξ˜ic2τ+nϕτ⋅max1≤i≤nξ˜ic2τ+nϕτ=o1.Thus, we knowA1n/Σ1n⟶DN0,1.Step 2. We prove Akn/Σ1n⟶P0,k=4,5,…,12.(57)A4nΣ1n≤Cn∑i=1nδiξ˜icg˜ic≤Cn∑i=1nδih˜icg˜ic+∑i=1nδivig˜ic+∑i=1nδi∑j=1nWnjctivjg˜ic,≤Cnmax1≤i≤nδin⋅max1≤i≤nh˜ic⋅max1≤i≤ng˜ic+max1≤m≤n∑i=1mvki⋅max1≤i≤ng˜ic+max1≤j≤n∑j=1nWnjcti⋅max1≤i≤ng˜ic⋅max1≤m≤n∑j=1mvkj=o1,EA5nΣ1n2≤CnE∑i=1nδiμig˜ic2+E∑i=1nδig˜ic∑j=1nWnjctiμj2=on−1/2,A6nΣ1n≤Cn∑i=1n∑j=1nδiδjWnjctiξ˜icεj≤Cn∑i=1nδih˜ic+vi−∑k=1nWnkctivk⋅∑j=1nδjWnjctiεj=o1a.s.,EA7nΣ1n2≤CnE∑i1=1n∑i2=1n∑j1=1n∑j2=1nδi1δi2δj1δj2Wnj1cti1Wnj2cti2εi1εi2μj1μj2≤Cn∑i1=1n∑i2=1n∑j=1nδi1δi1δj2Wnjcti1Wnjcti2Covεi1,εi2Ξμ2=On−1.The proof ofAkn/Σ1n⟶P0,k=8,cdots,12 is similar. Thus, the proof of Theorem 4 (a) is completed.Proof ofTheorem 4(b). From (9), one can write that,(58)g^nct−gt=∑j=1nWnjctδjϵj−μjβ+∑j=1nWnjctδjgtj−gt+∑j=1nWnjctδjξj+∑j=1nWnjctδjμjβ−β^c≔F1nt+F2nt+F3nt+F4ntβ−β^c.From the proof of Theorem4 (a), we know that β−β^c=Opn−1/2. Because ϵi−μiβ is a NA sequence with Eϵi−μiβ=0. From the condition of Theorem 4 (b), we can know nΓ1n2t⟶∞, max1≤i≤nWnjctδj=on−1/2log−1n=oΓ1nt, and ∑j=1nWnjct2δj2=on−1/2log−1n=oΓ1n2t. From Lemma 6, we can verify F1nt/Γ1nt⟶DN0,1. Thus, we need only to prove F2nt=On−1/2, F3nt=O1, and F4nt=Op1. For ∀b>0,(59)F3nt≤∑j=1nWnjctδjξj≤∑j=1nWnjctδjhtj+vj≤max1≤j≤nhtj⋅∑j=1nWnjctδj+max1≤j≤nWnjctδj⋅max1≤m≤n∑j=1mvkj=O1,EF4nt2≤∑j=1nWnjctδj2Ξμ2=On−1/2log−1n.The proof ofF2nt=On−1/2 is similar. Thus, the proof of Theorem 4 (b) is completed.Proof ofTheorem 5(a). From (10) and (11), one can write that,(60)β^I−β=S2n−2∑i=1nδiξ˜i+U+Vξ˜ic+1+U+Vμiϵi−μiβ+1+U+VΞμ2β+U+V∑k=210Akn−∑i=1n1−δiξ˜igti−g^ncti+∑i=1n∑j=1nξ˜i1−δiWnjtiξj+∑i=1nξ˜i1−δiμiβ^c−β−∑i=1n∑j=1nWnjtiδjξ˜iϵj+∑i=1n∑j=1nWnjtiδjξ˜iμjβ+∑i=1n∑j=1nWnjti1−δjξ˜igtj−g^nctj−∑i=1n∑j=1nWnjti1−δjξ˜iξj+μjβ^c−β+∑i=1nξ˜ig˜ic−∑i=1n∑j=1nWnjtiδiμjϵi+∑i=1n∑j=1nWnjtiδiμiμjβ−∑i=1n1−δiμ˜igti−g^ncti+∑i=1n1−δiμi2−Eμi2+∑i=1n1−δiξiμi−∑i=1n1−δiξi∑j=1nWnjtiμj−∑i=1n1−δiμi∑j=1nWnjtiμjβ^c−β−∑i=1n∑j=1nWnjtiδjμiϵj+∑i=1n∑j=1n∑k=1nWnjtiWnktiδjμkϵj+∑i=1n∑j=1nWnjtiδjμiμjβ−∑i=1n∑j=1n∑k=1nWnjtiWnktiδjμjμkβ+∑i=1n∑j=1nWnjti1−δjμ˜igtj−g^nctj−∑i=1n∑j=1nWnjti1−δjμ˜iξj+μjβ^c−β+∑i=1nμ˜ig˜ic:=S2n−2∑k=120Dkn.Similar to the proof of Theorem4 (a), We can prove D1n/Σ2n⟶DN0,1, and D2n/Σ2n=op1. We only need to prove D2n/Σ2n=op1,k=3,…,20.(61)D3nΣ2n≤Cnmax1≤i≤n1−δin⋅max1≤i≤nh˜i⋅max1≤i≤ngti−g^ncti+max1≤m≤n∑i=1mvki⋅max1≤i≤ngti−g^ncti+max1≤i≤n∑j=1nWnjti⋅max1≤i≤ngti−g^ncti⋅max1≤m≤n∑j=1mvkj=o1a.s.,D4n=P1n+P2nβ^c−βP1nΣ2n≤Cn⋅∑i=1nh˜i+vi−∑k=1nWnktivk⋅max1≤i≤n1−δimax1≤i≤n∑j=1nWnjtihtj+vj=on1/2P2nΣ2n≤Cn∑i=1nξ˜i1−δiμi=On1/4logn,D8n=∑i=1nξ˜i∑j=1nWnjti1−δjξj+∑i=1nξ˜i∑j=1nWnjti1−δjμjβ^c−β≔Q1n+Q2nβ^c−βQ1nΣ2n≤Cn⋅∑i=1nhi˜+vi−∑k=1nWnktivk,max1≤i≤n∑j=1nWnjti1−δjξj=On1/2EQ2nΣ2n2≤Cn∑i=1n∑j=1nξ˜iWnjti1−δj2Ξμ2=O1.The proof ofDkn=op1 for k=5,6,7,9,…,20 is analogous. Thus, the proof of Theorem 5 (a) is completed.Proof ofTheorem 5(b). The proof of Theorem 5 (b) is similar.Proof ofTheorem 6. The proof of Theorem 6 is similar.



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