# Strong Consistency of Estimators in a Partially Linear Model with Asymptotically Almost Negatively Associated Errors

Authors: Yu Zhang; Xinsheng Liu; Mohamed Sief
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

This paper studies a heteroscedastic partially linear regression model in which the errors are asymptotically almost negatively associated (AANA, in short) random variables with not necessarily identical distribution and zero mean. Under some mild conditions, we establish the strong consistency of least squares estimators, weighted least squares estimators, and the ultimate weighted least squares estimators for the unknown parameter, respectively. In addition, the strong consistency of the estimator for nonparametric component is also investigated. The results derived in the paper include the corresponding ones of independent random errors and some dependent random errors as special cases. At last, two simulations are carried out to study the numerical performance of the strong consistency for least squares estimators and weighted least squares estimators of the unknown parametric and nonparametric components in the model.


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## Body

## 1. Introduction

Consider the following heteroscedastic partially linear regression model:(1)Yi=xiβ+gti+σiei,1≤i≤n,where σi2=fui, xi,ti,ui are known and nonrandom design points, β is an unknown parameter, f⋅ and g⋅ are unknown functions defined on a compact set A, and ei,1≤i≤n are random errors.Model (1) belongs to a kind of model called partially linear model which was introduced by Engle et al. [1] to analyse the relationship between temperature and electricity usage. Since then, many statisticians pay attention to studying partially linear regression models. Under the case of independent random errors, Hu et al. [2] studied the asymptotic normality of DHD estimators in a partially linear model; Hu [3] established the strong consistency and mean consistency of the estimators for β and g⋅ in model (1) with σi2=σ2; Gao et al. [4] established the asymptotic normality for the least squares estimators and weighted least squares estimators of β based on the family of nonparametric estimators for g⋅ and f⋅in model (1); Chen et al. [5] investigated the strong consistency of the estimators in model (1); and so on. Under the case of dependent random errors, Zeng and Liu [6] studied the asymptotic properties of the estimators for parametric and nonparametric parts in a partially linear model with NSD errors; Wang et al. [7] established the mean consistency, complete consistency, and uniform complete consistency of the estimators in a partially linear model based on φ-mixing errors. Wang et al. [8] and Wu and Wang [9] discussed the moment consistency and strong consistency for least squares estimators and weighted least squares estimators of β and f⋅ in a partially linear model with ρ˜-mixing errors. Pan et al. [10] obtained the mean consistency and complete consistency of the estimators for β and g⋅ in model (1) with σi2=σ2 under Lq mixingale errors; Hu [11] obtained the mean consistency and complete consistency of the estimators for β and g⋅ in model (1) with σi2=σ2 under linear time series errors; Liang and Jing [12] studied the asymptotic normality of the least squares estimators and the weighted least squares estimators in model (1) with martingale difference errors and linear process errors; Baek and Liang [13] investigated the strong consistency and asymptotic normality of the estimators in model (1) under negatively associated samples; Zhou et al. [14] derived the moment consistency of the estimators in model (1) with negatively associated errors; and so forth. For more studies on the asymptotic properties of the estimators in regression models, one can refer to [15, 16]. Asymptotically almost negatively associated sequences are widely used dependent sequences which include independent and negatively associated sequences as special cases. So, to study the limit properties of asymptotically almost negatively associated sequences has more theoretical significance and application value. The concept of asymptotically almost negatively associated sequences of random variables was introduced by Chandra and Ghosal [17] as follows.Definition 1.
A sequenceXn,n≥1 of random variables is called asymptotically almost negatively associated (AANA, in short) if there exists a non-negative sequence qn⟶0 as n⟶∞ such that(2)CovfXn,gXn+1,Xn+2,…,Xn+k≤qnVarfXnVargXn+1,Xn+2,…,Xn+k1/2,for all n,k≥1and for all coordinate-wise nondecreasing continuous functions f and g whenever the variances exist.
Chandra and Ghasal [17] pointed out that the family of AANA sequences contains negatively associated (NA, in short, see [18]) sequences (with qn=0,n≥1) and some more sequences of random variables which are not much deviated from being negatively associated. Two examples of AANA sequences which are not NA were constructed by Chandra and Ghosal: ξn=1+an2−1/2ηn+anηn+1 (see Chandra and Ghosal [17]) and ζn=ηn+anηn+1 (see [19]), where η1, η2, … are independent and identically distributed N0,1 random variables, an>0, and an⟶0 as n⟶∞.
Many applications of AANA sequences have been found. Chandra and Ghosal [17] derived the Kolmogorov-type inequality and Marcinkiewcz-Zygmund type strong laws of large numbers. An [20] studied the complete moment convergence of weighted sums for processes under AANA assumptions. Wang et al. [21] investigated the large deviation and Marcinkiewicz type strong law of large numbers for AANA sequences. Ko et al. [22] established the Hájeck-Rényi inequalities for AANA sequences. Shen and Wu [23] investigated the strong law of large numbers for AANA sequences. Yuan and An [24] provided some Rosenthal type inequalities for maximum partial sums of AANA sequences. Wang et al. [25] studied the complete convergence for weighted sums of arrays of rowwise AANA random variables. Xi et al. [26] investigated the LP convergence and complete convergence for weighted sums of AANA random variables. Chen et al. [27] obtained the strong laws of large numbers for the weighted sums of AANA sequences; Hu and Zhang [28] obtained the strong consistency of M-estimator in the linear regression model with AANA errors; Zhang et al. [29] established the weak consistency of M-estimator in the linear regression model with AANA errors; and so on.
However, we have not found the studies on the strong consistency of the estimators for parametric and nonparametric components in model (1) with AANA random errors in the literature. In this paper, we will consider the estimation problem for model (1) under the assumption that the errors are AANA sequences of random variables with not necessarily identical distribution and zero mean. The strong consistency of least squares estimators, weighted least squares estimators, and the ultimate weighted least squares estimators for β is derived, respectively, based on some mild conditions. In addition, the strong consistency of the estimators for g⋅ and f⋅ is also studied, respectively. These results extend and improve the corresponding ones for independent and identically distributed random errors and some dependent random errors.
The following concept of stochastic domination will be used in this work.Definition 2.
A sequenceXn,n≥1of random variables is said to be stochastically dominated by a random variable X if there exists a positive constant C such that(3)PXn>x≤CPX>x,for all x≥0 and n≥1.
The remainder of this paper is organized as follows. The least squares estimators, weight least squares estimators, and ultimate weighted least squares estimators ofβ based on the family of nonparametric estimators for g⋅ and f⋅ and some assumptions are introduced in Section 2. The main results are given in Section 3. We give some preliminary lemmas in Section 4. We provide the proofs of the main results in Section 5. Two simulations are presented in Section 6.
Throughout this paper, letC, C1, and C2 be positive constants whose values may vary at different places. ⟶a.s. stands for almost sure convergence, f+=max0,f, and f−=min0,f.


## 2. Estimation and Assumptions

Assume thatYi,xi,ti∈A,ui∈A,1≤i≤n satisfy model (1) and Wnit=Wnit;t1,t2,…,tn is a measurable weight function on the compact set A. Denote x˜i=xi−∑j=1nWnjtixj, Y˜i=Yi−∑j=1nWnjtiYj, γi=1/σi2⋅1≤i≤n, S˜n2=∑i=1nx˜i2 and T˜n2=∑i=1nγix˜i2.For model (1), by Eei=0, one can get that gti=EYi−xiβ for 1≤i≤n. Hence, for any given β, we define an estimator of g⋅ given by(4)gnt,β=∑i=1nWnitYi−xiβ.To estimateβ, we seek to minimize(5)Qβ=∑i=1nYi−xiβ−gnt,β2.The minimum point ofQβ is found as(6)β^LS=∑i=1nx˜iY˜iS˜n2,where β^LS is called a least squares (LS, in short) estimator of β. When the random errors are heteroscedastic, we modify β^LS to a weighted least squares (WLS, in short) estimator:(7)β^WLS=∑i=1nγix˜iY˜iT˜n2.WhenEei2=1, we have fui=EYi−xiβ−gti2. Hence, the estimator of f⋅ can be defined by(8)f^nu=∑i=1nW^niuY˜i−x˜iβ^LS2,where W^niu=W^niu;u1,u2,…,un is a measurable weight function on A. In general, one assumes that min1≤i≤nf^nui>0. Therefore, the ultimate weighted least squares estimators (UWLS, in short) of β is(9)β^UWLS=∑i=1nγnix˜iY˜iU˜n2,where U˜n2=∑i=1nγnix˜i2, γni=1/f^nui, 1≤i≤n.From (4)–(9), we further define(10)g^nt=∑i=1nWnitYi−xiβ^LS,(11)g˜nt=∑i=1nWnitYi−xiβ^WLS,(12)g¯nt=∑i=1nWnitYi−xiβ^UWLS.To obtain our results, the following assumptions are sufficient.A1(i)
limn⟶∞S˜n2/n=C.(ii)
0<m0≤min1≤i≤nfui≤max1≤i≤nfui≤M0<∞.(iii)
g⋅ satisfies the first-order Lipschitz condition on compact subset A.(iv)
f⋅ and g⋅ are continuous functions on compact subset A.(v)
max1≤i≤nx˜i=On/logn.A2max1≤i≤n∑j=1nWnjti=O1, max1≤i,j≤nWnjti=On−1/2logn−1.A3(i)
max1≤i≤n∑j=1nWnjti−1=o1.(ii)
max1≤i≤n∑j=1nWnjtiIti−tj>δ=o1 for any δ>0.A4(i)
max1≤i≤n∑j=1nWnjti−1=On−1/4.(ii)
max1≤i≤n∑j=1nWnjtiIti−tj>αn−1/4=On−1/4 for someα>0.A5W^ni⋅ satisfies the assumptions A2 and A3 replacing ti and Wni by ui and W^ni, respectively.A6∑i=1∞qi<∞.Remark 1.
The assumptionsA1 are used in Chen et al. [5], Baek and Liang [13], Zhou et al. [30], and so forth. From (i) and (ii) of A1, it follows that(13)T˜n2n≤C,S˜n−2∑i=1nx˜i≤CandT˜n−2∑i=1nγix˜i≤C.Remark 2.
The following two weight functions satisfy the assumptionsA2–A4:(14)Wni′t=1Hn∫Si−1SiKt−sHnds,Wni″t=Kt−Ti/Hn∑j=1nKt−Ti/Hn,where Si=Ti−1+Ti/2, 1≤i≤n−1, S0=0, Sn=1, K⋅ is the Parzen–Rosenblatt kernel function (see [31]), and Hn is a bandwidth parameter.


## 3. Statement of Main Results

In this section, letei,i≥1 be an AANA sequence of random variables with zero mean and mixing coefficients qi,i≥1, which is stochastically dominated by a random variable e.Theorem 1.
Suppose that (i), (ii), (iv), and (v) ofA1, A2, A3, and A6 hold. If Ee2<∞, then(15)β^LS⟶a.s.β,(16)β^WLS⟶a.s.β.Remark 3.
Since independent sequences are special AANA sequences (see Chandra and Ghosal [17]), Theorem 1 extends and improves the corresponding results of Chen et al. [5] for identically distributed independent random errors to the case of not necessarily identically distributed AANA setting.Theorem 2.
Suppose that the conditions of Theorem1 are satisfied. Assume further that max1≤i≤n∑j=1nWnjtixj=O1; then,(17)max1≤i≤ng^nti−gti⟶a.s.0,(18)max1≤i≤ng˜nti−gti⟶a.s.0.Remark 4.
Whenqi≡0, AANA sequences are reduced to NA sequences (see Chandra and Ghosal [17]). Therefore, Theorems 1 and 2 also hold for not necessarily identically distributed NA random errors.Theorem 3.
LetEei2=1. Suppose that A1, A2, andA4–A6hold. If Ee4<∞, then(19)β^LS−β=On−1/4a.s.,(20)max1≤i≤nf^nui−fui⟶a.s.0,(21)β^UWLS⟶a.s.β.Remark 5.
As independent sequences are special AANA sequences, Theorem3 extends and improves the corresponding results of Chen et al. [5] for identically distributed independent random errors to the case of not necessarily identically distributed AANA random errors.Theorem 4.
Suppose that the conditions of Theorem3 are satisfied. Assume further that max1≤i≤n∑j=1nWnjtixj=O1; then,(22)max1≤i≤ng¯nti−gti⟶a.s.0.Remark 6.
As NA sequences are special AANA sequences withqi≡0, Theorems 3 and 4 also hold for not necessarily identically distributed NA random errors.


## 4. Preliminary Lemmas

To prove the results of this paper, the following lemmas are needed.Lemma 1 (see [24]).
IfXi,i≥1 is an AANA sequence with mixing coefficients qi,i≥1, then fiXi,i≥1 is still an AANA sequence with mixing coefficients qi,i≥1, where f1,f2,…are nondecreasing or nonincreasing functions.Lemma 2 (see [28]).
LetXi,i≥1 be an AANA sequence of random variables with zero mean and mixing coefficients qi,i≥1. If Xi≤bi, t>0 and tmax1≤i≤nbi≤1a.s. Then,(23)P∑i=1nXi≥ε≤Cexp−tε∑i=1n−1qi+2exp−tε+t2∑i=1nEXi2,for any ε>0.Lemma 3.
LetXi,i≥1 be an AANA sequence of random variables with zero mean and mixing coefficients qi,i≥1. Assume that Xi≤ba.s. for i≥1 and ∑i=1∞qi<∞; then,(24)P∑i=1nXi≥ε≤Cexp−ε222∑i=1nEXi2+bε,for any ε>0.Proof.
From Lemma2 and ∑i=1∞qi<∞, it follows that(25)P∑i=1nXi≥ε≤Cexp−tε+t2∑i=1nEXi2.
Taket=ε/2∑i=1nEXi2+bε; then, bt≤1. By (25), we derive that(26)P∑i=1nXi≥ε≤Cexp−tε+t2∑i=1nEXi2=Cexp−ε22∑i=1nEXi2+bε+ε2∑i=1nEXi22∑i=1nEXi2+bε2=Cexp−ε222∑i=1nEXi2+bε2−2∑i=1nEXi22∑i=1nEXi2+bε.
Since2−2∑i=1nEXi2/2∑i=1nEXi2+bε≥1, we have(27)P∑i=1nXi≥ε≤Cexp−ε222∑i=1nEXi2+bε.
This completes the proof of Lemma3.Lemma 4 (see [21]).
LetXi,i≥1 be an AANA sequence of random variables with zero mean, mixing coefficients qi,i≥1, and ∑i=1∞q2i<∞. If ∑i=1∞VarXi<∞, then(28)∑i=1∞Xi−EXiconverges a.s.Lemma 5.
LetXi,i≥1 be an AANA sequence of random variables with mixing coefficients qi,i≥1 and ∑i=1∞q2i<∞. Denote Xic=XiIXi≤c for some c>0. Suppose that(29)∑i=1∞PXi>c<∞,(30)∑i=1∞EXicconverges,(31)∑i=1∞VarXic<∞.
Then,(32)∑i=1∞Xiconverges a.s.Proof.
DenoteX¯ic=XiIXi≤c+cIXi>c−cIXi<−c. By Lemma 1, we know that X¯ic,i≥1is still an AANA sequence of random variables with mixing coefficients qi,i≥1. By (29) and (30), we have(33)∑i=1∞EX¯ic=∑i=1∞EXic−c∑i=1∞PXi<−c+c∑i=1∞PXi>cconverges.
Note that(34)EX¯ic2=EXic2+c2PXi>c,EX¯ic2=EXic2+c2PXi>c+2cPXi>cEXic−2cPXi<−cEXic−2c2PXi>cPXi<−c.
Hence,(35)VarX¯ic=EX¯ic2−EX¯ic2=VarXic+Qic,where Qic=2cPXi<−cEXic+2c2PXi>cPXi<−c−2cPXi>cEXic.
By (30), we obtain that(36)EXic≤C<∞a.s.
Hence,(37)Qic≤2cC+cPXi>c.
Thus, it follows from (29) and (31) that(38)∑i=1∞VarX¯ic<∞.
By Lemma4, we derive that(39)∑i=1∞X¯ic−EX¯icconverges a.s.
Hence, it follows from (33) that(40)−∞<∑i=1∞X¯ic<∞a.s.
By (29), we obtain that(41)∑i=1∞PXi≠X¯ic=∑i=1∞PXi>c<∞.
Hence, it follows from Borel–Cantelli lemma that(42)Xi=X¯ica.s.
Therefore, (32) follows from (40) and (42).
This completes the proof of Lemma5.Lemma 6.
LetXi,i≥1 be an AANA sequence of random variables with mixing coefficients qi,i≥1 and ∑i=1∞q2i<∞, which is stochastically dominated by a random variable X. For any ε>0, denote(43)Xi′=−εi1/pIXi<−εi1/p+XiIXi≤εi1/p+εi1/pIXi>εi1/p,Xi″=Xi−Xi′=Xi+εi1/pIXi<−εi1/p+Xi−εi1/pIXi>εi1/p.
IfEXp<∞ for some p>0, then(44)∑i=1∞Xi″<∞a.s.Proof.
By Lemma1, we know that Xi′,i≥1 and Xi″,i≥1 are still AANA random variables with mixing coefficientsqi,i≥1. Denote(45)Xi″=Xi−Xi′=Xi+εi1/pIXi<−εi1/p+Xi−εi1/pIXi>εi1/p≜X″i−+X″i+,and then Xi″=X″i+−X″i−. By Lemma 4, we know that X″i+,i≥1 and X″i−,i≥1 are still AANA sequences of random variables with mixing coefficientsqi,i≥1. Since Xi is stochastically dominated by a random variable X, for some fixed d>0, we have(46)PX″i+>d=PXi−εi1/p>d≤PXi>εi1/p≤PXip>εpi≤CPXp>εpi.
Hence,(47)∑i=1∞PX″i+>d≤C∑i=1∞PXp>εpi.
DenoteAi=Xp>εpi for i≥1 and A0=Xp≥0. Then,(48)0≤εp∑i=1nPAi−nPAnεp=∑i=1nεpi−1PAi−1−PAi≤∑i=1∞EXpIAi−1−Ai=EXp<∞.
Sincelimn⟶∞nPAnεp≤limn⟶∞EXpIXp>εpn=0, we have(49)∑i=1∞PAi<∞.
Thus,(50)∑i=1∞PX″i+>d≤C∑i=1∞PXp>εpi=C∑i=1∞PAi<∞.
Denote(51)X″i+d≜X″i+IX″i+≤d=Xi−εi1/pIXi>εi1/pIXi−εi1/p≤d=Xi−εi1/pI0<Xi−εi1/p≤d.
Then,(52)∑i=1∞EX″i+d≤∑i=1∞d⋅EI0<Xi−εi1/p≤d≤∑i=1∞d⋅EIXi>εi1/p≤d∑i=1∞PXip>εpi≤C∑i=1∞PXp>εpi<∞,(53)Var∑i=1∞X″i+d≤∑i=1∞EX″i+d2=∑i=1∞EXi−εi1/p2I0<Xi−εi1/p≤d≤d2∑i=1∞EIXi>εi1/p≤d2∑i=1∞PXip>εpi,≤C∑i=1∞PXp>εpi<∞.
Hence, it follows from (50), (52), (53), and Lemma 5 that(54)∑i=1∞X″i+converges a.s.
Similar to the proof of (54), we can get that(55)∑i=1∞X″i−converges a.s.
Therefore,(56)∑i=1∞Xi″=∑i=1∞X″i+−X″i−=∑i=1∞X″i+−∑i=1∞X″i−<∞a.s.
This completes the proof of Lemma6.Lemma 7 (see [32]).
LetXn,n≥1be a sequence of random variables which is stochastically dominated by a random variable X. For any a>0 and β>0, the following two statements hold:(57)EXnβIXn≤a≤C1EXβIX≤a+aβPX>a,EXnβIXn>a≤C2EXβIX>a,where C1 and C2 are positive constants. Thus,(58)EXnβ≤CEXβ,where C is a positive constant.Lemma 8.
LetXi,i≥1 be an AANA sequence of random variables with mixing coefficientsqi,i≥1 and ∑i=1∞q2i<∞, which is stochastically dominated by a random variable X with EXp<∞ for 0<p<2. Then,(59)n−1/p∑i=1nXi−EXi⟶a.s.0.Proof.
Denote(60)Xn′=−n1/pIXn≤−n1/p+XnIXn<n1/p+n1/pIXn≥n1/p,and then we obtain by EXp<∞ that(61)∑n=1∞PXn≠Xn′=∑n=1∞PXn≥n1/p≤C∑n=1∞PX≥n1/p≤CEXp<∞.
Hence, it follows by the Borel–Cantelli lemma that(62)Xn=Xn′a.s.
Thus, to prove (59), it suffices to show that(63)n−1/p∑i=1nXi′−EXi′⟶a.s.0.
By Lemma1, we know that Xn′−EXn′,n≥1 is still an AANA sequence of random variables. Hence, by Lemma 4 and Kronecker lemma, to prove (63), we only need to show that(64)∑n=1∞VarXn′n1/p<∞.
By the Markov inequality, Lemma4, and EXp<∞, we have(65)∑n=1∞VarXn′n1/p≤∑n=1∞EXn′2n2/p,≤C∑n=1∞n2/pPX≥n1/p+EX2IX<n1/pn2/p,≤C+C∑n=1∞n−2/p∑k=1nEX2Ik−1≤Xp<k,=C+C∑k=1∞∑n=k∞n−2/pEX2Ik−1≤Xp<k,≤C+C∑k=1∞k−2/p+1EXpk2−p/pIk−1≤Xp<k,=C+C∑k=1∞EXpIk−1≤Xp<k<∞.
Therefore, (59) follows.
This completes the proof of Lemma8.


## 5. Proofs of the Main Results

By (1), (6), (7), and (9), we derive that(66)β^LS−β=S˜n−2∑i=1nx˜iε˜i+∑i=1nx˜ig˜i,(67)β^WLS−β=T˜n−2∑i=1nγix˜iε˜i+∑i=1nγix˜ig˜i,(68)β^UWLS−β=T˜n−2∑i=1nγnix˜iε˜i+∑i=1nγnix˜ig˜i,where ε˜i=εi−∑j=1nWnjtiεj, g˜i=gti−∑j=1nWnjtigtj, and εi=σiei.Proof of Theorem 1.
We prove (16) first. By (67), we can get that(69)β^WLS−β=T˜n−2∑i=1nγix˜iεi−∑i=1nγix˜i∑j=1nWnjtiεj+∑i=1nγix˜ig˜i,≜Δ1n−Δ2n+Δ3n.
Hence, to prove (16), it suffices to show that(70)Δin⟶a.s.0i=1,2,3.
Firstly, we prove(71)Δ1n⟶a.s.0.
In view of (69), we have(72)Δ1n=T˜n−2∑i=1nγix˜iσiei=∑i=1nT˜n−2γix˜iσiei≜∑i=1nbniei.
From (ii) and (v) ofA1 and (13), it follows that(73)∑i=1nbni2=∑i=1nγix˜iγiσi2T˜n4≤CT˜n2≤Cn,max1≤i≤nbni≤max1≤i≤nx˜in1/2max1≤i≤nγiσin1/2T˜n2≤Cn−1/2logn−1.
Denote(74)ei′=−εi1/2Iei<−εi1/2+eiIei≤εi1/2+εi1/2Iei>εi1/2,ei″=ei−ei′=ei+εi1/2Iei<−εi1/2+ei−εi1/2Iei>εi1/2,for any ε>0. Write eni=bniei′−Eei′=bni+ei′−Eei′+bni−ei′−Eei′≜eni1+eni2, 1≤i≤n; then, we know by Lemma 1 that eni1,1≤i≤n and eni2,1≤i≤nare still AANA sequences with mixing coefficients qi,i≥1 and zero mean. By (ii) and (iv) of A1 and Lemma 7, we have(75)max1≤i≤nenik≤2εn1/2max1≤i≤nbni±≤2εn1/2max1≤i≤nbni≤Cεlogn−1,∑i=1nEenik2≤∑i=1nbni±2Eei′2≤∑i=1nbni2Eei′2≤C∑i=1nbni2Ee2≤C∑i=1nbni2≤Cn−1,for k=1,2.
Hence, for anyμ>0 and sufficiently large n, by Lemma 4, we have(76)P∑i=1neni>μ=P∑i=1neni1+eni2>μ≤P∑i=1neni1>μ2+P∑i=1neni2>μ2,≤Cexp−μ2/422C/n+C1εμlogn−1,≤Cexp−μ2C2εμlogn−1,≤Cn20<ε<μ2C2.
Hence, it follows from the Borel–Cantelli lemma that(77)∑i=1neni⟶a.s.0.
By Lemma6, we obtain that(78)∑i=1∞ei″<∞a.s.
By (73) and (78), we derive that(79)∑i=1nbniei″≤∑i=1nbniei″≤max1≤i≤nbni∑i=1nei″=On−12logn−1a.s,(80)∑i=1nbniEei″≤∑i=1nbniEei″≤max1≤i≤nbni∑i=1nEeiIei>εi1/2,≤Cn−1/2logn−1∑i=1nε−1i−1/2Eei2Iei>εi1/2≤Cn−1/2logn−1∑i=1ni−1/2≤Clogn−1.
Hence, combining (77), (79), and (80), we have(81)I1n=∑i=1nbniei′+ei″=∑i=1neni+∑i=1nbniei″−∑i=1nbniEei″,≤∑i=1neni+∑i=1nbniei″+∑i=1nbniEei″⟶a.s.0.
Thus, (71) follows.
Secondly, we prove(82)Δ2n⟶a.s.0.
In view of (69), we have(83)Δ2n=T˜n−2∑i=1nγix˜i∑j=1nWnjtiεj=∑j=1n∑i=1nT˜n−2γix˜iWnjtiσjej≜∑j=1nbnj′ej.
From (ii) ofA1, A2, and (13), it follows that(84)∑j=1nb′nj2≤∑j=1nmax1≤j≤nσj2max1≤i,j≤nWnjti2T˜n−2∑i=1nγix˜i2≤Clogn−2,max1≤j≤nbnj′≤max1≤j≤nσjmax1≤i,j≤nWnjtiT˜n−2∑i=1nγix˜i≤Cn−12logn−1.
Hence, similar to the proof (71), one can derive (82).
Finally, we prove(85)Δ3n⟶a.s.0.
In view of (69), we have(86)Δ3n=T˜n−2∑i=1nγix˜ig˜i.
By (iv) ofA1, A3, and max1≤i≤n∑j=1nWnjti=O1, we have(87)max1≤i≤ng˜i=max1≤i≤ngti−∑j=1nWnjtigtj≤max1≤i≤ngti∑j=1nWnjti−1,+max1≤i≤n∑j=1nWnjtigti−gtjIti−tj>δ+max1≤i≤n∑j=1nWnjtigti−gtjIti−tj≤δ⟶0,as n⟶∞.
Hence, it follows from (13) that(88)Δ3n≤T˜n−2∑i=1nγix˜ig˜i≤max1≤i≤ng˜iT˜n−2∑i=1nγix˜i⟶0asn⟶∞.
Thus, (85) follows. Therefore, (16) follows form (71), (82), and (85).
The proof of (15) is similar to that of (16); thus, we omit the details here.
This completes the proof of Theorem1.Proof of Theorem 2.
We prove (18) first. In view of (11), we have(89)g˜nti−gti=∑j=1nWnjtixjβ+gtj+σjej−xjβ^WLS−gti=∑j=1nWnjtixjβ−β^WLS−gti−∑j=1nWnjtigtj+∑j=1nWnjtiσjej.
Hence,(90)max1≤i≤ng˜nti−gti≤max1≤i≤nβ−β^WLS∑j=1nWnjtixj+g˜i+∑j=1nWnjtiσjej≜I1n+I2n+I3n.
From (16) and max1≤i≤n∑j=1nWnjtixj=O1, it follows that(91)I1n⟶a.s.0.
By (87), we have(92)I2n⟶a.s.0.
Next, we will prove(93)I3n⟶a.s.0.
Denote(94)∑j=1nWnjtiσjej≜∑j=1ncnjej.
By (ii) ofA1 and A2, we derive that(95)∑j=1ncnj2=∑j=1nWnj2tiσj2≤Cmax1≤j≤nWnjti∑j=1nWnjti,≤Cn−12logn−1,max1≤j≤ncnj≤Cn−12logn−1.
Similar to the proof of (71), we can get (93). Therefore, (18) follows from (90)–(93).
The proof of (17) is similar to that of (18); thus, we omit the details here.
This completes the proof of Theorem2.Proof of Theorem 3.
Firstly, we prove (19). In view of (66), we have(96)β^LS−β=S˜n−2∑i=1nx˜iε˜i−∑i=1nx˜i∑j=1nWnjtjεj+∑i=1nx˜ig˜i≜J1n+J2n+J3n.
By (ii) ofA1, A4, and max1≤i≤n∑j=1nWnjti=O1, we have(97)max1≤i≤ng˜i=max1≤i≤ngti−∑j=1nWnjtigtj≤max1≤i≤ngti∑j=1nWnjti−1+max1≤i≤n∑j=1nWnjtigti−gtjIti−tj>αn−14+max1≤i≤n∑j=1nWnjtigti−gtjIti−tj≤αn−1/4,=On−1/4.
By (13), we obtain that(98)J3n≤max1≤i≤ng˜i∑i=1nx˜iS˜n2=On−1/4.
Note that(99)J1n=∑i=1nS˜n−2x˜iσiei≜∑i=1ndniei,J2n=∑j=1nS˜n−2∑i=1nx˜iWnjtiσjej≜∑j=1nd¯njej.
Denote(100)e¯′i=−εi1/4Iei<−εi1/4+eiIei≤εi1/4+εi1/4Iei>εi1/4,e″¯i=ei−e′¯i=ei+εi1/4Iei<−εi1/4+ei−εi1/4Iei>εi1/4,for any ε>0. Write(101)e′ni=dnie′¯i−Ee′¯i=dni+e′¯i−Ee′¯i+dni−e¯′i−Ee¯′i≜e′ni1+e′ni2,for 1≤i≤n, and(102)e¯nj=d¯nje′¯j−Ee′¯j=d¯nj+e′¯j−Ee′¯j+d¯nj−e′¯j−Ee′¯j≜e¯nj1+e¯nj2,for 1≤j≤n.
Then, we know by Lemma1 that e′ni1,1≤i≤n, e′ni2,1≤i≤n, e¯nj1,1≤j≤n, and e¯nj2,1≤j≤n are still AANA sequences with mixing coefficients qi,i≥1 and zero mean. Note that(103)max1≤i≤ndni=max1≤i≤nS˜n−2x˜iσi≤Cmax1≤i≤nx˜innS˜n2≤Cn−12logn−1,∑i=1ndni2≤Cn,max1≤j≤nd¯nj=max1≤i,j≤nWnjtimax1≤j≤nσjS˜n−2∑i=1nx˜i≤Cn−12logn−1,∑j=1nd¯nj2≤∑j=1nS˜n−4σj2∑i=1nx˜i2∑i=1nWnj2ti≤S˜n−2∑i=1n∑j=1nσj2Wnj2ti,≤S˜n−2max1≤i≤nσj2max1≤i,j≤nWnjti∑i=1n∑j=1nWnjti,≤Cn−12logn−1.
By Lemma7, we obtain that(104)Eei4≤CEe4<∞.
Hence, similar to the proof of (78), we can easily derive that ∑i=1∞e″¯i<∞a.s.
Since(105)∑i=1ndniEe″¯i≤max1≤i≤ndni∑i=1nEeiIei>εi1/4≤Cn−12logn−1∑i=1ni−3/4ε−3Eei4Iei>εi1/4,≤Cn−1/2logn−1n1/4=on−1/4.
Hence, similar to the proof of (71), we can easily obtain that(106)J1n≤∑i=1neni′+∑i=1ndnie″¯i+∑i=1ndniEe″¯i=on−14a.s.,(107)J2n≤∑j=1ne¯nj+∑j=1nd¯nje″¯j+∑j=1nd¯njEe″¯j=On−1/4a.s.
Therefore, (19) follows from (96), (98), (106), and (107).
Secondly, we prove (20). In view of (8), we have(108)max1≤j≤nf^nuj−fuj≤max1≤j≤n∑i=1nW^niujεi2−fuj+2max1≤j≤n∑i=1nW^niujεi∑j=1nWnjtiεj+max1≤j≤n∑i=1nW^niuj∑j=1nWnjtiεj2+2max1≤j≤n∑i=1nW^niujg˜iε˜i+2max1≤j≤n∑i=1nW^niujx˜iε˜iβ−β^LS+β−β^LS2max1≤j≤n∑i=1nW^niujx˜i2+2β−β^LSmax1≤j≤n∑i=1nW^niujx˜ig˜i+max1≤j≤n∑i=1nW^niujg˜i2≜B1n+B2n+B3n+B4n+B5n+B6n+B7n+B8n.
Next, we will prove(109)B1n⟶a.s.0.
Sinceei=ei++ei− and ei2=ei+2+ei−2, then Eei2=Eei+2+Eei−2=1. We know by Lemma 1 that ei+2 and ei−2 are still AANA sequences with mixing coefficients qi,i≥1. Observe that(110)B1n≤max1≤j≤n∑i=1nW^niujfuiei+2−Eei+2+max1≤j≤n∑i=1nW^niujfuiei−2−Eei−2+max1≤j≤n∑i=1nW^niujfui−fuj≜B11n+B12n+B13n.
Denoteηi=ei±2−Eei±2, 1≤i≤n; then Eηi=0, Eηi2≤Eei4≤Ee4<∞, and ηi is still an AANA sequence with mixing coefficients qi,i≥1. Similar to the proof of (71), one can derive that(111)B11n⟶a.s.0,B12n⟶a.s.0.
By (iv) ofA1 and A5, we have(112)B13n⟶a.s.0.
Therefore, (109) follows.
Note that(113)B2n≤2max1≤j≤n∑i=1nW^niujσieimax1≤i≤n∑j=1nWnjtiσjej.
Similar to the proof of (71), one can easily obtain that(114)max1≤j≤n∑i=1nW^niujσiei=O1,(115)max1≤i≤n∑j=1nWnjtiσjej⟶a.s.0.
Hence,(116)B2n⟶a.s.0.
Under the conditions of Theorem3, by (13), (19), (97), (114), and (115), we derive that(117)B3n≤max1≤i≤n∑j=1nWnjtiεj2max1≤j≤n∑i=1nW^niuj⟶a.s.0,(118)B4n≤2max1≤j≤n∑i=1nW^niujg˜iεi+∑i=1nW^niujg˜i∑k=1nWnktiεk≤2max1≤i≤ng˜imax1≤j≤n∑i=1nW^niujεi+max1≤i≤ng˜imax1≤i≤n∑k=1nWnktiεkmax1≤j≤n∑i=1nW^niuj⟶a.s.0,(119)B5n≤2β^LS−βmax1≤j≤n∑i=1nW^ni2ujε˜i2∑i=1nx˜i21/2⟶a.s.0,(120)B6n≤β^LS−β2max1≤i,j≤nW^niuj∑i=1nx˜i2≤Clogn−1⟶a.s.0,(121)B7n≤2β^LS−βmax1≤i≤ng˜imax1≤i,j≤nW^niuj∑i=1nx˜i≤Clogn−1⟶a.s.0,(122)B8n≤max1≤i≤ng˜i2max1≤j≤n∑i=1nW^niuj⟶a.s.0.
Thus, (20) follows from (108), (109), and (116)–(122).
Lastly, we prove (21). By (20) and (ii) of A1, we have, for sufficiently large n,(123)0<m1≤min1≤i≤nf^nui≤max1≤i≤nf^nui≤M1<∞,(124)U˜n2n≤C,∑i=1nγnix˜iU˜n2≤C.
It follows from (68) that(125)β^UWLS−β=nU˜n21n∑i=1nγnix˜iε˜i+1U˜n2∑i=1nγnix˜ig˜i≜nU˜n2L1n+L2n.
By (87) and (125), one can obtain(126)L2n≤max1≤i≤ng˜i∑i=1nγnix˜iU˜n2⟶a.s.0,(127)L1n≤n−1∑i=1nγni−γix˜iε˜i+n−1γix˜iε˜i≤n−1∑i=1nγni−γix˜iεi+n−1∑i=1nγni−γix˜i∑j=1nWnjtiεj+n−1γix˜iε˜i≜L11n+L12n+L13n.
From (13), (20), (109), (115), and Lemma 8, it follows that(128)L11n≤max1≤i≤nγni−γiσin−1∑i=1nx˜iei≤max1≤i≤nfui−f^nuifuif^nui1n∑i=1nx˜i21n∑i=1nei21/2,=max1≤i≤nfui−f^nuifuif^nui1n∑i=1nx˜i21n∑i=1nei+2+1n∑i=1nei−21/2⟶a.s.max1≤i≤nfui−f^nuifuif^nui1n∑i=1nx˜i21n∑i=1nEei+2+Eei−21/2,=max1≤i≤nfui−f^nuifuif^nui1n∑i=1nx˜i2⋅11/2⟶a.s.0,(129)L12n≤max1≤i≤n∑j=1nWnjtiεjmax1≤i≤nγni−γin−1∑i=1nx˜i⟶a.s.0.
By (13), (69), (71), and (82), we have(130)L13n=n−1T˜n2Δ1n−Δ2n⟶a.s.0.
Hence, by (127)–(130), we have(131)L1n⟶a.s.0.
Therefore, (21) follows from (125), (126), and (131).
This completes the proof of Theorem 3.Proof of Theorem 4.
In view of (12), we have(132)g¯nti−gti=∑j=1nWnjtixjβ+gtj+σjej−xjβ^UWLS−gti=∑j=1nWnjtixjβ−β^UWLS−gti−∑j=1nWnjtigtj+∑j=1nWnjtiσjej.
Hence,(133)max1≤i≤ng¯nti−gti≤max1≤i≤nβ−β^UWLS∑j=1nWnjtixj+g˜i+∑j=1nWnjtiσjej.
From (20) and max1≤i≤n∑j=1nWnjtixj=O1, it follows that(134)max1≤i≤nβ−β^UWLS∑j=1nWnjtixj⟶a.s.0.
Therefore, (22) follows from (92), (93), and (134).
This completes the proof of Theorem 4.


## 6. Numerical Simulations

In this section, we will investigate the numerical performance of the strong consistency for the least squares estimatorsβ^LS and g^nt and weighted least squares estimators β^WLS and g˜nt with AANA random errors by two simulated examples.Two AANA sequences are given as follows:(135)Sequence1:ei=1+ai2−1/2ηi+aiηi+1,Sequence2:εi=ηi+aiηi+1,where η1, η2, ⋯ are independent and identically distributed N0,1 random variables and ai=1/i2. The two sequences have been proved to be AANA sequences but not NA sequences (see [17, 19]).Next, we will apply the two sequences to the following two simulation examples, respectively.


### 6.1. Simulated Example 1

We will simulate a heteroscedastic partially linear model(136)Yi=xiβ+gti+σiei,whereβ=3.5, gt=sin2πt, σi=fui1/2=1+0.5sinπui1/2, ui=xi=−1i⋅i/n, 1≤i≤n, and the random errors are given by Sequence 1.In particular, we take the weight functionWni⋅ as the following nearest neighbor weight function (see [11]). Without loss of generality, let A=0,1 and ti=i/n1≤i≤n. For each t∈A, we rewrite(137)t1−t,t2−t,…,tn−t,as follows:(138)tR1t−t≤tR2t−t≤⋯≤tRnt−t.Takekn=n0.4 and define the nearest neighbor weight function(139)Wnit=1kn,ifti−t≤tRkn−t,0,otherwise.The sample sizes are taken asn=200, 500, 1000, 1500, 2000, and 2500, respectively, and each case is repeated for 1000 times and the average values of β^LS, g^nt, β^WLS, and g˜ntare calculated as the estimators, respectively. Then, we calculate the corresponding absolute errors β^LS−β, g^nt−gt, β^WLS−β, and g˜nt−gt and the root mean square errors (RMSEs) of β^LS, g^nt, β^WLS, and g˜nt, respectively. The above results are presented in Tables 1 and 2, and the curves of gt, g^nt, and g˜nt are provided in Figures 1–2, respectively.Table 1
The LS estimators ofβ and gt and the absolute errors and RMSEs of β^LS and g^nt with β=3.5 and gt=sin2πt.
tnβ^LSβ^LS−βRMSE (β^LS)g^ntg^nt−gtRMSEg^nt20/n2003.38900.11100.13080.55850.02930.03455003.57900.07900.09310.22780.02090.024610003.55430.05430.06390.11080.01450.017115003.53540.03540.04170.07490.00880.010420003.47920.02080.02450.05770.00510.006025003.50910.00910.01070.04780.00240.002940/n2003.60780.10780.12700.91720.03380.03985003.57610.07610.08970.45640.02530.029910003.44860.05140.06060.22960.01910.022515003.53260.03260.03850.15440.01240.014620003.51790.01790.02110.11710.00830.009725003.49360.00640.00750.09640.00390.0046Table 2
The WLS estimators ofβ and gt and the absolute errors and RMSEs of β^WLS and g˜nt with β=3.5 and gt=sin2πt.
tnβ^WLSβ^WLS−βRMSEβ^WLSg˜ntg˜nt−gtRMSEg˜nt20/n2003.60750.10750.12660.55940.02840.03355003.42360.07640.09010.22850.02010.023710003.55260.05260.06200.11130.01400.016515003.46570.03430.04050.07520.00850.010020003.47990.02010.02370.05780.00500.005925003.50890.00890.01050.04790.00230.002740/n2003.39560.10440.12300.91830.03090.03865003.57380.07380.08690.45720.02320.029010003.45120.04980.05870.23020.01740.021715003.53160.03160.03720.15480.01130.014120003.51740.01740.02050.11740.00750.009425003.50610.00610.00720.09650.00360.0045Figure 1
Curves ofgt=sin2πt and g^nt with β=3.5 and n=1500.Figure 2
Curves ofgt=sin2πt and g˜nt with β=3.5 and n=1500.


### 6.2. Simulated Example 2

We will simulate a heteroscedastic partially linear model(140)Yi=xiβ+gti+σiεi,where β=2.5, gt=cosπt, σi=fui1/2=1−0.5sinπui1/2, ui=xi=−1i⋅i/n, 1≤i≤n, and the random errors are given by Sequence 2.By the same estimating methods as model (6), we obtain the estimators of β and gnt in model (140) under different sample size n, g^nt−gt, β^WLS−β, and g˜nt−gt and the RMSE of β^LS, g^nt, β^WLS, and g˜nt, respectively. The above results are provided in Tables 3 and 4, and the curves of gt, g^nt, and g˜nt are presented in Figures 3 and 4, respectively.Table 3
The LS estimators ofβ and gt and the absolute errors and RMSEs of β^LS and g^nt with β=2.5 and gt=cosπt.
tnβ^LSβ^LS−βRMSEβ^LSg^ntg^nt−gtRMSEg^nt20/n2002.39260.10740.12650.91940.03170.03735002.57360.07360.08670.96890.02320.027410002.54990.04990.05880.98120.01690.019915002.53190.03190.03760.98800.01110.013120002.48340.01660.01960.99310.00640.007525002.50520.00520.00610.99690.00280.003240/n2002.60870.10870.12800.77610.03300.03885002.57660.07660.09030.94410.02450.028910002.44820.05180.06110.97430.01780.021015002.53310.03310.03900.98500.01140.013520002.51830.01830.02160.99080.00720.008525002.49320.00680.00800.99550.00330.0039Table 4
The WLS estimators ofβ and gt and the absolute errors and RMSEs of β^WLS and g˜nt with β=2.5 and gt=cosπt.
tnβ^WLSβ^WLS−βRMSEβ^WLSg˜ntg˜nt−gtRMSEg˜nt20/n2002.39590.10410.12250.92030.03070.03625002.42870.07130.08390.96950.02260.026610002.54830.04830.05690.98160.01640.019315002.53100.03100.03640.98830.01080.012720002.51610.01610.01900.99340.00610.007225002.50510.00510.00600.99700.00270.003140/n2002.60530.10530.12380.77710.03190.03665002.42580.07420.08730.94480.02370.027910002.55020.05020.05910.97480.01730.020315002.53200.03200.03770.98540.01110.013120002.48230.01770.02080.99100.00700.008225002.49340.00660.00770.99560.00320.0037Figure 3
Curves ofgt=cosπt and g^nt with β=2.5 and n=1500.Figure 4
Curves ofgt=cosπt and g˜nt with β=2.5 and n=1500.It can be seen from Tables1–4 that regardless of the values of t, the absolute errors of both the least squares estimators and weighted least squares estimators decrease gradually as the sample size n has good effects. The simulation results show the strong consistency of least squares estimators β^LS and g^nt and weighted least squares estimators β^WLS and g˜nt in model (1) with AANA random errors. The simulation results also show that the absolute error of weighted least squares estimator is smaller than that of least squares estimator. Moreover, consistency is the basic standard that all estimators should meet, and it is the necessary condition to measure whether an estimator is feasible. AANA sequences are widely used dependent sequences which include independent and NA sequences as special cases. Therefore, to study the consistency of estimators in regression models with AANA errors is of considerable significance.


## 6.1. Simulated Example 1

We will simulate a heteroscedastic partially linear model(136)Yi=xiβ+gti+σiei,whereβ=3.5, gt=sin2πt, σi=fui1/2=1+0.5sinπui1/2, ui=xi=−1i⋅i/n, 1≤i≤n, and the random errors are given by Sequence 1.In particular, we take the weight functionWni⋅ as the following nearest neighbor weight function (see [11]). Without loss of generality, let A=0,1 and ti=i/n1≤i≤n. For each t∈A, we rewrite(137)t1−t,t2−t,…,tn−t,as follows:(138)tR1t−t≤tR2t−t≤⋯≤tRnt−t.Takekn=n0.4 and define the nearest neighbor weight function(139)Wnit=1kn,ifti−t≤tRkn−t,0,otherwise.The sample sizes are taken asn=200, 500, 1000, 1500, 2000, and 2500, respectively, and each case is repeated for 1000 times and the average values of β^LS, g^nt, β^WLS, and g˜ntare calculated as the estimators, respectively. Then, we calculate the corresponding absolute errors β^LS−β, g^nt−gt, β^WLS−β, and g˜nt−gt and the root mean square errors (RMSEs) of β^LS, g^nt, β^WLS, and g˜nt, respectively. The above results are presented in Tables 1 and 2, and the curves of gt, g^nt, and g˜nt are provided in Figures 1–2, respectively.Table 1
The LS estimators ofβ and gt and the absolute errors and RMSEs of β^LS and g^nt with β=3.5 and gt=sin2πt.
tnβ^LSβ^LS−βRMSE (β^LS)g^ntg^nt−gtRMSEg^nt20/n2003.38900.11100.13080.55850.02930.03455003.57900.07900.09310.22780.02090.024610003.55430.05430.06390.11080.01450.017115003.53540.03540.04170.07490.00880.010420003.47920.02080.02450.05770.00510.006025003.50910.00910.01070.04780.00240.002940/n2003.60780.10780.12700.91720.03380.03985003.57610.07610.08970.45640.02530.029910003.44860.05140.06060.22960.01910.022515003.53260.03260.03850.15440.01240.014620003.51790.01790.02110.11710.00830.009725003.49360.00640.00750.09640.00390.0046Table 2
The WLS estimators ofβ and gt and the absolute errors and RMSEs of β^WLS and g˜nt with β=3.5 and gt=sin2πt.
tnβ^WLSβ^WLS−βRMSEβ^WLSg˜ntg˜nt−gtRMSEg˜nt20/n2003.60750.10750.12660.55940.02840.03355003.42360.07640.09010.22850.02010.023710003.55260.05260.06200.11130.01400.016515003.46570.03430.04050.07520.00850.010020003.47990.02010.02370.05780.00500.005925003.50890.00890.01050.04790.00230.002740/n2003.39560.10440.12300.91830.03090.03865003.57380.07380.08690.45720.02320.029010003.45120.04980.05870.23020.01740.021715003.53160.03160.03720.15480.01130.014120003.51740.01740.02050.11740.00750.009425003.50610.00610.00720.09650.00360.0045Figure 1
Curves ofgt=sin2πt and g^nt with β=3.5 and n=1500.Figure 2
Curves ofgt=sin2πt and g˜nt with β=3.5 and n=1500.


## 6.2. Simulated Example 2

We will simulate a heteroscedastic partially linear model(140)Yi=xiβ+gti+σiεi,where β=2.5, gt=cosπt, σi=fui1/2=1−0.5sinπui1/2, ui=xi=−1i⋅i/n, 1≤i≤n, and the random errors are given by Sequence 2.By the same estimating methods as model (6), we obtain the estimators of β and gnt in model (140) under different sample size n, g^nt−gt, β^WLS−β, and g˜nt−gt and the RMSE of β^LS, g^nt, β^WLS, and g˜nt, respectively. The above results are provided in Tables 3 and 4, and the curves of gt, g^nt, and g˜nt are presented in Figures 3 and 4, respectively.Table 3
The LS estimators ofβ and gt and the absolute errors and RMSEs of β^LS and g^nt with β=2.5 and gt=cosπt.
tnβ^LSβ^LS−βRMSEβ^LSg^ntg^nt−gtRMSEg^nt20/n2002.39260.10740.12650.91940.03170.03735002.57360.07360.08670.96890.02320.027410002.54990.04990.05880.98120.01690.019915002.53190.03190.03760.98800.01110.013120002.48340.01660.01960.99310.00640.007525002.50520.00520.00610.99690.00280.003240/n2002.60870.10870.12800.77610.03300.03885002.57660.07660.09030.94410.02450.028910002.44820.05180.06110.97430.01780.021015002.53310.03310.03900.98500.01140.013520002.51830.01830.02160.99080.00720.008525002.49320.00680.00800.99550.00330.0039Table 4
The WLS estimators ofβ and gt and the absolute errors and RMSEs of β^WLS and g˜nt with β=2.5 and gt=cosπt.
tnβ^WLSβ^WLS−βRMSEβ^WLSg˜ntg˜nt−gtRMSEg˜nt20/n2002.39590.10410.12250.92030.03070.03625002.42870.07130.08390.96950.02260.026610002.54830.04830.05690.98160.01640.019315002.53100.03100.03640.98830.01080.012720002.51610.01610.01900.99340.00610.007225002.50510.00510.00600.99700.00270.003140/n2002.60530.10530.12380.77710.03190.03665002.42580.07420.08730.94480.02370.027910002.55020.05020.05910.97480.01730.020315002.53200.03200.03770.98540.01110.013120002.48230.01770.02080.99100.00700.008225002.49340.00660.00770.99560.00320.0037Figure 3
Curves ofgt=cosπt and g^nt with β=2.5 and n=1500.Figure 4
Curves ofgt=cosπt and g˜nt with β=2.5 and n=1500.It can be seen from Tables1–4 that regardless of the values of t, the absolute errors of both the least squares estimators and weighted least squares estimators decrease gradually as the sample size n has good effects. The simulation results show the strong consistency of least squares estimators β^LS and g^nt and weighted least squares estimators β^WLS and g˜nt in model (1) with AANA random errors. The simulation results also show that the absolute error of weighted least squares estimator is smaller than that of least squares estimator. Moreover, consistency is the basic standard that all estimators should meet, and it is the necessary condition to measure whether an estimator is feasible. AANA sequences are widely used dependent sequences which include independent and NA sequences as special cases. Therefore, to study the consistency of estimators in regression models with AANA errors is of considerable significance.



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