# A Berry-Esseen Type Bound in Kernel Density Estimation for Negatively Associated Censored Data

Authors: Qunying Wu; Pingyan Chen
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We discuss the kernel estimation of a density function based on censored data when the survival and the censoring times form the stationary negatively associated (NA) sequences. Under certain regularity conditions, the Berry-Esseen type bounds are derived for the kernel density estimator and the Kaplan-Meier kernel density estimator at a fixed pointx.


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## Body

## 1. Introduction

Let{Ti;i≥1} be a sequence of the true survival times. The random variables (r.v.s.) are not assumed to be mutually independent; it is assumed, however, that they have a common unknown continuous marginal distribution function (d.f.) F(x)=P(Ti≤x) and density function f(x). Let the r.v.s. Ti be censored on the right by the censoring r.v.s. Yi, so that one observes only (Zi,δi), where
(1)Zi=min(Ti,Yi)≔Ti∧Yi,δi=I(Ti≤Yi),i=1,…,n,
here and in the sequel, and I(A) is the indicator random variable of the event A. In this random censorship model, the censoring times Yi, i=1,…,n, are assumed to have the common d.f. G(y); they are also assumed to be independent of the r.v.s. Ti′s. Following the convention in the survival literature, we assume that both Xi and Yi are nonnegative random variables. In contrast to statistics for complete data, we observe only the pairs (Zi,δi), i=1,…,n, and the estimators are based on these pairs.The following nonparametric estimation of the distribution functionsF and G due to Kaplan and Meier [1] is widely used to estimate F and G on the basis of the data (Zi,δi):
(2)F^n(x)=1-∏k=1n‍(1-δ(k)n-k+1)I(Z(k)≤x)=1-∏k=1n‍(n-kn-k+1)I(δ(k)=1,Z(k)≤x),G^n(x)=1-∏k=1n‍(n-kn-k+1)I(δ(k)=0,Z(k)≤x),
where Z(1)≤Z(2)≤⋯≤Z(n) denote the order statistics of Z1,Z2,…,Zn and δ(i) is the concomitant of Z(i).We introduce the kernel density estimator(3)fn(x)=1nhn∑i=1n‍K(Zi-xhn)δi1-G(Zi),
where 0<hn→0 are bandwidths and K is some kernel function. When G is known, (3) can be used to estimate the common density of the lifetimes. However, in most practical cases G is unknown and must be replaced by the Kaplan-Meier estimator G^n, so the Kaplan-Meier kernel density estimator of the f is defined by
(4)f^n(x)=1nhn∑i=1n‍K(Zi-xhn)δi1-G^n(Zi).There is an extensive literature on the Kaplan-Meier estimator for censored independent observations. We refer to papers by Földes and Rejtő [2], Gu and Lai [3], Gill [4], and Sun and Zhu [5]. Sun and Zhu obtained the following Berry-Esseen bound for i.i.d. censored sequences.Theorem A.
LetK be a bounded probability kernel function with compact support [-1,1] satisfying for integer r≥2,
(5)1j!∫‍ujK(u)du={1,j=0,0,j=1,2,…,r-1,cr≠0,j=r.Letf be r-order continuously differentiable and let G be continuously differentiable in a neighborhood of x with f(x)>0 for x<τL. Then
(6)supy∈ℝ|P((nhn)1/2[f^n(x)-f(x)]≤yσ(x))-Φ(y)|=O(bn),
where Φ(·) denotes the standard normal distribution function, bn=(nhn)-1/2+n1/2hnr+1/2+hn1/4 and σ2(x)=(f(x)/(1-G(x)))∫‍K2(t)dt.However, the censored dependent data appear in a number of applications. For example, repeated measurements in survival analysis follow this pattern; see Kang and Koehler [6]. In the context of censored time series analysis, Shumway et al. [7] considered (hourly or daily) measurements of the concentration of a given substance subject to some detection limits, thus being potentially censored from the right. Lecoutre and Ould-Said [8], Cai [9], and Liang and Uña-Álvarez [10] studied the convergence for the stationary α-mixing data. However, the convergence for the NA data has not been reported.The main purpose of this paper is to study the kernel density estimator and the Kaplan-Meier kernel estimator of a density function based on censored data when the survival and the censoring times form the stationary NA (see the following definition) sequences. Under certain regularity conditions, the Berry-Esseen type bounds are derived for the kernel density estimator and the Kaplan-Meier kernel estimator at a fixed pointx.Definition 1.
Random variablesX1,X2,…,Xn, n≥2 are said to be negatively associated (NA) if for every pair of disjoint subsets A1 and A2 of {1,2,…,n},
(7)cov(f1(Xi;i∈A1),f2(Xj;j∈A2))≤0,
where f1 and f2 are increasing for every variable (or decreasing for every variable) such that this covariance exists. A sequence of random variables {Xi;i≥1} is said to be NA if every finite subfamily is NA.Obviously, if{Xi;i≥1} is a sequence of NA random variables and {fi;i≥1} is a sequence of nondecreasing (or nonincreasing) functions, then {fi(Xi);i≥1} is also a sequence of NA random variables.This definition was introduced by Joag-Dev and Proschan [11]. Statistical test depends greatly on sampling. The random sampling without replacement from a finite population is NA but is not independent. NA sampling has wide applications such as those in multivariate statistical analysis and reliability theory. Because of the wide applications of NA sampling, the limit behavior of NA random variables has received more and more attention recently. One can refer to Joag-Dev and Proschan [11] for fundamental properties, Matuła [12] for the three-series theorem, and Wu and Jiang [13, 14] for the strong convergence.


## 2. Main Results

In what follows, letL be the d.f. of the Zi’s, L¯∶=1-L. Since the sequences {Tn;n≥1} and {Yn;n≥1} are independent, it follows that L=1-F¯G¯∶=1-(1-F)(1-G).Define (possibly infinite) timesτF, τG, and τL by
(8)τF=inf{y;F(y)=1},τG=inf{y;G(y)=1},τL=inf{y;L(y)=1}.
Then, τL=τF∧τG.We give the following four lemmas, which are helpful in proving our theorems.Lemma 2 (Chang and Rao, [15]).
LetX and Y be random variables, then for any a>0(9)supy∈ℝ|P(X+Y≤y)-Φ(y)|≤supy∈ℝ|P(X≤y)-Φ(y)|+a2π+P(|Y|>a),
here and in the sequel, where Φ(·) denotes the standard normal distribution function.Lemma 3 (Su et al. [16, Theorem 1]).
Let{Xi;i≥1} be a sequence of NA r.v.s. with zero means and 𝔼|Xi|p<∞, i=1,2,… and p≥2. Then for Sn=∑i=1n‍Xi,
(10)𝔼|Sn|p≤cp(∑i=1n‍𝔼|Xi|p+(∑i=1n‍𝔼Xi2)p/2),
where cp>0 depends only on p.Lemma 4.
Let{Xi;i≥1} be a sequence of NA r.v.s. with continuous d.f. F, and let Fn(x)∶=(1/n)∑i=1n‍I(Xi<x) be the empirical d.f. based on the segments X1,…,Xn. Then
(11)supx∈ℝ|Fn(x)-F(x)|=O(n-1/2ln1/2n)a.s.Proof.
Similar to the proof of Lemma 4 in Yang [17], we can prove Lemma 4.Lemma 5 (Wu and Chen [18, Theorem 1.3]).
Let{Tn;n≥1} and {Yn;n≥1} be two sequences of NA r.v.s. Suppose that the sequences {Tn;n≥1} and {Yn;n≥1} are independent. Then for any 0<τ<τL,
(12)sup0≤t≤τ|F^n(t)-F(t)|=O(n-1/2ln1/2n)a.s.In order to formulate our main results, we now list some assumptions.(A1)
{Yi;i≥1} and {Ti;i≥1} are two sequences of stationary NA random variables, and {Yi} and {Ti} are independent.(A2)
Suppose thatx<τL, f(x)>0, and f and G have bounded derivative in a neighborhood of x.(A3)
For all integersj≥1, the conditional distribution Tj+1, given T1=x1, has a density fj(·|x1), and for all x∈ℝ, fj(x2|x1)≤M0 for x1,x2∈U(x) and some M0>0, where U(x) represents a neighborhood of x.(A4)
The kernelK is a bounded derivative function with K(u)=0 for |u|>1 and ∫-11‍K(u)du=1.(A5)
Letpn, qn, and kn=[n/(pn+qn)] be positive integers with
(13)limn→∞kn=∞,limn→∞pnknn=1,limn→∞pnhn=0,limn→∞nhn=∞,limn→∞hn-3u(qn)=0,whereu(n)∶=∑j=n∞‍|Cov(T1,Tj+1)|.Remark 6.
(A5) Implies limn→∞(knqn/n)=0 and limn→∞(pn/n)=0.
Letσn2(x)=nhnVar(fn(x)), σ2(x)=(f(x)/(1-G(x)))∫-11‍K2(t)dt.Theorem 7.
Suppose that(A1)–(A5) are satisfied; then
(14)|σn2(x)-σ2(x)|=O(an),
where a1n=(qnkn/n)+(qnknu(pn)/nhn3), a2n=pn/n, a3n=u(qn)/hn3, an=pnhn+a1n1/2+a2n1/2+a3n→0.
Consider the following:(15)supy∈ℝ|P(fn(x)-𝔼fn(x)Varfn(x)≤y)-Φ(y)|=O(bn),
where bn=1/(nhn)1/2+a1n1/3+a2n1/3+a3n1/3→0.
Furthermore, if(16)limn→∞n1/2hn3/2=0,
then
(17)supy∈ℝ|P(fn(x)-f(x)Varfn(x)≤y)-Φ(y)|=O(bn+n1/2hn3/2).Theorem 8.
Assume that the conditions of Theorem7 hold. Then
(18)supy∈ℝ|P(f^n(x)-𝔼f^n(x)Varfn(x)≤y)-Φ(y)|=O(bn+(hnlnn)1/4).
Furthermore, if (16) holds, then
(19)supy∈ℝ|P(f^n(x)-f(x)Varfn(x)≤y)-Φ(y)|=O(bn+(hnlnn)1/4+n1/2hn3/2).


## 3. Proofs

Proof of Theorem7.
We observe that, by (3),
(20)(nhn)1/2fn(x)=∑i=1n‍1(nhn)1/2K(Zi-xhn)δi1-G(Zi)∶=∑i=1n‍Zni∶=Sn.
Letkm=(m-1)(pn+qn)+1, lm=(m-1)(pn+qn)+pn+1, m=1,2,…,kn, where
(21)Unm=∑i=kmkm+pn-1‍Zni,Unm′=∑i=lmlm+qn-1‍Zni,Un,kn+1′=∑i=kn(pn+qn)+1n‍Zni,Sn′=∑m=1kn‍Unm,Sn′′=∑m=1kn‍Unm′,Sn′′′=Un,kn+1′,
and then
(22)Sn=Sn′+Sn′′+Sn′′′.
By (20),
(23)σn2(x)=VarSn=Var(Sn′+Sn′′+Sn′′′)=VarSn′+VarSn′′+VarSn′′′+2Cov(Sn′,Sn′′)+2Cov(Sn′,Sn′′′)+2Cov(Sn′′,Sn′′′).
We first estimate VarSn′, VarSn′′, and VarSn′′′. Obviously, (A1) implies that {Unm} and {Zni} are stationary; thus,
(24)VarSn′=Var(∑m=1kn‍Unm)=∑m=1kn‍VarUnm+2∑1≤i<j≤kn‍Cov(Uni,Unj)=knVarUn1+2∑1≤i<j≤kn‍Cov(Uni,Unj)=knpnVarZn1+2kn∑1≤i<j≤pn‍Cov(Zni,Znj)+2∑1≤i<j≤kn‍Cov(Uni,Unj)∶=In1+In2+In3.
From(A1), (A2), and (A4), we obtain
(25)VarZn1=1nhnVarK(Z1-xhn)δ11-G(Z1)=1nhn{𝔼[K2(Z1-xhn)δ1(1-G(Z1))2]-[𝔼K(Z1-xhn)δ11-G(Z1)]2}=1nhn∬K2(min(u,v)-xhn)×I(u<v)(1-G(min(u,v)))2dF(u)dG(v)-1nhn[∬K(min(u,v)-xhn)×I(u<v)1-G(min(u,v))dF(u)dG(v)]2=1nhn∫∫u<vK2(u-xhn)1(1-G(u))2dF(u)dG(v)-1nhn[∫∫u<vK(u-xhn)11-G(u)dF(u)dG(v)]2=1nhn{[∫K(u-xhn)f(u)du]2∫K2(u-xhn)f(u)1-G(u)du-[∫K(u-xhn)f(u)du]2}=1n{[∫-11‍K(u)f(x+uhn)du]2∫-11K2(u)f(x+uhn)1-G(x+uhn)du-hn[∫-11‍K(u)f(x+uhn)du]2}(26)=O(1n).
Hence, by (A5), In1=O(1).
Fori<j and x<τL, by (A1)-(A4),
(27)|Cov(Zni,Znj)|=|Cov(Zn1,Zn,j-i+1)|=1nhn|Cov(K(Z1-xhn)δ11-G(Z1),K(Zj-i+1-xhn)δj-i+11-G(Zj-i+1))|≤cnhn{+(𝔼K(Z1-xhn))2}𝔼|K(T1-xhn)K(Tj-i+1-xhn)|+(𝔼K(Z1-xhn))2}≤cnhn{+(∫‍K(u-xhn)f(u)du)2}∬|K(u-xhn)K(v-xhn)|×f(u)fj-i(v∣u)dudv+(∫‍K(u-xhn)f(u)du)2}≤chnn{(∫-11‍K(u)f(x+hnu)du)2∬-11|K(u)K(v)|dudv+(∫-11‍K(u)f(x+hnu)du)2}=O(hnn).
Therefore, by(A5),
(28)|In2|=2kn|∑1≤i<j≤pn‍Cov(Zni,Znj)|=O(knpn2hnn)=O(pnhn)⟶0.
By (A1), (A2), (A4), and Lemma 2.3 of Zhang [19], for l≥1,
(29)|Cov(Un1,Un,l+1)|=|∑i=1pn‍∑j=l(pn+qn)+1l(pn+qn)+pn‍Cov(Zni,Znj)|=|∑i=1pn‍{∑j=1i-1‍Cov(Zn,i-j+1,Zn,l(pn+qn)+1)+∑j=ipn‍Cov(Zn1,Zn,l(pn+qn)+j-i+1)}|=|∑r=1pn‍(pn-r+1)Cov(Zn1,Zn,l(pn+qn)+r)i+∑r=1pn-1‍(pn-r)Cov(Zn,r+1,Zn,l(pn+qn)+1)|≤pn∑r=l(pn+qn)l(pn+qn)+pn-1‍|Cov(Zn1,Zn,r+1)|+pn∑r=l(pn+qn)-(pn-1)l(pn+qn)-1‍|Cov(Zn1,Zn,r+1)|=pn∑r=l(pn+qn)-(pn-1)l(pn+qn)+pn-1‍|Cov(Zn1,Zn,r+1)|=pnnhn∑r=l(pn+qn)-(pn-1)l(pn+qn)+pn-1|Cov(δr+11-G(Zr+1))K(Z1-xhn)‍×δ11-G(Z1),K(Zr+1-xhn)×δr+11-G(Zr+1))|≤cpnnhn3∑r=l(pn+qn)-(pn-1)l(pn+qn)+pn-1‍|Cov(T1,Tr+1)|.
Thus, by(A1) and (A5),
(30)|In3|=|2∑1≤i<j≤kn‍Cov(Uni,Unj)|=|2∑l=1kn-1‍(kn-l)Cov(Un1,Un,l+1)|≤cpnknnhn3∑l=1kn-1‍∑r=l(pn+qn)-(pn-1)l(pn+qn)+pn-1‍|Cov(T1,Tr+1)|≤cpnknn1hn3∑r=qn∞‍|Cov(T1,Tr+1)|=O(u(qn)hn3)=O(a3n)⟶0.
Therefore, by the combination of (A5), (24), (26), (28), and (30),
(31)VarSn′=O(1)+O(pnhn)+O(a3n)=O(1).
Similarly,(32)VarSn′′=knVarUn1′+2∑1≤i<j≤kn‍Cov(Uni′,Unj′)=knqnVarZn1+2kn∑1≤i<j≤qn‍Cov(Zni,Znj)+2∑1≤i<j≤kn‍Cov(Uni′,Unj′)=O(qnknn+qn2knhnn+qnknu(pn)nhn3)=O(qnknn+qnknu(pn)nhn3)=O(a1n).
By (26), (27), (A1), (A5), and n-kn(pn+qn)≤pn+qn≤2pn,
(33)VarSn′′′=Var(∑i=kn(pn+qn)+1n‍Zni)=(n-kn(pn+qn))Var(Zn1)+2∑kn(pn+qn)+1≤i<j≤n‍Cov(Zni,Znj)=O(pnn+pn2hnn)=O(pnn)=O(a2n).
By (25), (A2), and (A4), (34)|nVarZn1-σ2(x)|=|∫-11‍K2(u)(f(x+uhn)1-G(x+uhn)-f(x)1-G(x))du-hn[∫-11‍K(u)f(x+uhn)du]2|≤|∫-11‍K2(u)(1-G(x))(f(x+uhn)-f(x))+f(x)(G(x+uhn)-G(x))(1-G(x+uhn))(1-G(x))du|+hn[∫-11‍K(u)f(x+uhn)du]2=O(hn).Note that |Cov(X,Y)|≤(VarXVarY)1/2 for any random variables X and Y; from (31)–(33),
(35)|Cov(Sn′,Sn′′)|=O(a1n1/2),|Cov(Sn′,Sn′′′)|=O(a2n1/2),|Cov(Sn′′,Sn′′′)|=O(a1n1/2a2n1/2).
Therefore, from the combination of (23) and (31)–(34), it follows that
(36)|σn2(x)-σ2(x)|=|nVarZn1-σ2(x)+O(pnhn+a3n+a1n1/2+a2n1/2)|=O(pnhn+a3n+a1n1/2+a2n1/2)=O(an).
Thus, (14) holds.
Now, we prove (15). Let S~n=(Sn-𝔼Sn)/σn(x), S~n′=(Sn′-𝔼Sn′)/σn(x), S~n′′=(Sn′′-𝔼Sn′′)/σn(x), S~n′′′=(Sn′′′-𝔼Sn′′′)/σn(x). Then, S~n=S~n′+S~n′′+S~n′′′. According to Lemma 2, (14), (20), (32), and (33), we have
(37)supy∈ℝ|P(fn(x)-𝔼fn(x)Varfn(x)≤y)-Φ(y)|=supy∈ℝ|P(S~n′+S~n′′+S~n′′′≤y)-Φ(y)|≤supy∈ℝ|P(S~n′≤y)-Φ(y)|+a1n1/32π+P(S~n′′>a1n1/3)+a2n1/32π+P(S~n′′′>a2n1/3)=supy∈ℝ|P(S~n′≤y)-Φ(y)|+O(a1n1/3+a2n1/3).
Letξnm, m=1,2,…,kn be independent random variables with the same distribution as U~nm∶=(Unm-𝔼Unm)/σn(x) for m=1,2,…,kn. Put Hn=∑m=1kn‍ξnm, Bn2=∑m=1kn‍VarU~nm=∑m=1kn‍Varξnm=VarHn. Obviously,
(38)supy∈ℝ|P(S~n′≤y)-Φ(y)|≤supy∈ℝ|P(S~n′≤y)-P(Hn≤y)|+supy∈ℝ|Φ(yBn)-Φ(y)|+supy∈ℝ|P(Hn≤y)-Φ(yBn)|∶=J1n+J2n+J3n.
Note that VarSn=σn2(x) and Bn2=(VarSn′-In3)/σn2(x) from (20) and (24). By (14), (30), (32), and (33),
(39)J2n=supy∈ℝ|Φ(yBn)-Φ(y)|≤|Bn2-1|=|VarSn′-In3-VarSnσn2(x)|≤c|VarSn′-VarSn|+c|In3|≤cVar(Sn′′+Sn′′′)+2|Cov(Sn′,Sn′′+Sn′′′)|+O(a3n)=O(a1n1/2+a2n1/2+a3n)⟶0.
Note that ξnm, m=1,2,…,kn, are independent random variables, and Bn2=VarHn. Therefore, by Bn→1 (from (39)), (14), and Berry-Esseen inequality (cf. Petrov [20, page 154, Theorem 5.7]), there exists some constant c>0 such that
(40)J3n=supy∈ℝ|P(HnBn≤y)-Φ(y)|≤c∑m=1kn𝔼|ξnm|3Bn3≤c∑m=1kn‍𝔼|Unm|3=∑m=1kn‍𝔼|∑i=kmkm+pn-1‍Zni|3.
Similar to (26), we can get 𝔼|Zn1|3=O(1/n3/2hn1/2) and 𝔼Zn12=O(1/n). It is easy to see from Property P7 of Joag-Dev and Proschan [11] that {Zn;n≥1} is also sequence of NA r.v.s., so by using Lemma 3, we have
(41)J3n≤c∑m=1kn‍[∑i=kmkm+pn-1‍𝔼|Zni|3+(∑i=kmkm+pn-1‍𝔼Zni2)3/2]≤ckn(pn1n3/2hn1/2+(pnn)3/2)=O(1(nhn)1/2+a2n1/2).
Assume thatφ(t) and ψ(t) are the characteristic functions of S~n′ and Hn, respectively. By Esseen inequality (cf. Petrov [20, page 146, Theorem 5.3]), for any T>0, there exists some constant c>0 such that
(42)J1n=supy∈ℝ|P(S~n′≤y)-P(Hn≤y)|≤∫-TT‍|φ(t)-ψ(t)t|dt+Tsupy∈ℝ∫|u|≤(c/T)‍|P(Hn≤u+y)-P(Hn≤y)|du∶=J1n′+J1n′′.
By Theorem 10 in Newman [21], (14), and (30),
(43)|φ(t)-ψ(t)=|𝔼exp(it∑m=1kn‍U~nm)-∏m=1kn‍𝔼exp(itU~nm)|≤2∑1≤i<j≤kn‍t2|Cov(U~ni,U~nj)|≤c∑1≤i<j≤kn‍t2|Cov(Uni,Unj)|=t2O(a3n).
Therefore,
(44)J1n′=O(a3nT2).
On applying (39)–(41), we have
(45)supy∈ℝ|P(Hn≤u+y)-P(Hn≤y)|≤supy∈ℝ[|P(HnBn≤u+yBn)-Φ(u+yBn)|+|P(HnBn≤yBn)-Φ(yBn)|+|Φ(u+yBn)-Φ(yBn)|]=O(1(nhn)1/2+a2n1/2+|u|).
Thus,
(46)J1n′′=O(1(nhn)1/2+a2n1/2+1T).
Choosing T=a3n-1/3, then by (42)–(46),
(47)J1n=O(1(nhn)1/2+a2n1/2+a3n1/3).
Therefore, the combination of (37)–(39), (41), (47), and (15) holds.
Finally, we prove (17). By Lemma 2 and (15), for any a>0,
(48)supy∈ℝ|P(fn(x)-f(x)Varfn(x)≤y)-Φ(y)|=supy∈ℝ|P(fn(x)-𝔼fn(x)Varfn(x)+𝔼fn(x)-f(x)Varfn(x)≤y)-Φ(y)|≤supy∈ℝ|P(fn(x)-𝔼fn(x)Varfn(x)≤y)-Φ(y)|+a2π+P(|𝔼fn(x)-f(x)|Varfn(x)>a)=O(bn)+a2π+P(|𝔼fn(x)-f(x)|Varfn(x)>a).
Applying (14), (A3), (A4), and differential mean value theorem, there exists a constant 0<θ<1, such that
(49)|𝔼fn(x)-f(x)|Varfn(x)≤cnhn|1hn𝔼K(Z1-xhn)δ11-G(Z1)-f(x)|=cnhn|∫-11‍K(u)(f(x+uhn)-f(x))du|=cnhn∫-11‍|K(u)uhnf′(x+θuhn)|du=O(n1/2hn3/2).
Hence, there exists a constant M sufficiently large such that |𝔼fn(x)-f(x)|/Varfn(x)<Mn1/2hn3/2. Let a=Mn1/2hn3/2 in (48); then P(|𝔼fn(x)-fn(x)|/Varfn(x)>a)=P(ϕ)=0. Therefore, by (48), (16) holds.Proof of Theorem8.
Using (15) and Lemma 2,
(50)supy∈ℝ|P(f^n(x)-𝔼f^n(x)Varfn(x)≤y)-Φ(y)|=supy∈ℝ|P(fn(x)-𝔼fn(x)Varfn(x)i+f^n(x)-fn(x)-𝔼(f^n(x)-fn(x))Varfn(x)≤y)-Φ(y)f^n(x)-fn(x)-𝔼(f^n(x)-fn(x))Varfn(x)|≤supy∈ℝ|P(fn(x)-𝔼fn(x)Varfn(x)≤y)-Φ(y)|+(hnlnn)1/42π+P(|f^n(x)-fn(x)-𝔼(f^n(x)-fn(x))|Varfn(x)>(hnlnn)1/4|f^n(x)-fn(x)-𝔼(f^n(x)-fn(x))|Varfn(x))≤O(bn+(hnlnn)1/4)+c𝔼((nhn)1/2|f^n(x)-fn(x)|)(hnlnn)1/4.
LetLn(x)=n-1∑i=1n‍I(Zi≤x) be the empirical d.f. of L. Then, by (2),
(51)Ln(x)=1-(1-F^n(x))(1-G^n(x)).
Thus, by Lemmas4 and 5, for τ<τL,
(52)sup0≤x≤τ|G^n(x)-G(x)|≤sup0≤x≤τ11-F^n(x)×{|Ln(x)-L(x)|+(1-G(x))|F^n(x)-F(x)|}=O(lnnn).
Using (14), we get
(53)𝔼((nhn)1/2|f^n(x)-fn(x)|)≤1nhn∑i=1n‍𝔼{×|G^n(Zi)-G(Zi)|(1-G(Zi))(1-G^n(Zi))δi}|K(Zi-xhn)|×|G^n(Zi)-G(Zi)|(1-G(Zi))(1-G^n(Zi))δi}≤clnnhn|𝔼K(T1-xhn)|=clnnhnhn∫-11‍|K(t)|f(x+hnt)dt=O(hnlnn).
Therefore, (18) holds from (50) and (53).
Using (18), similar to the proof of (17), we can prove (19). This completes the proof of Theorem 8.



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*Source: 541250-2013-08-28.xml*