相协样本下概率密度函数的调整经验似然推断.pdf

应用数学MATHEMATICA APPLICATA2024,37(1):171-179相协样本下概率密度函数的调整经验似然推断王宁宁,秦永松(广西师范大学数学与统计学院,广西 桂林 541006)摘要：本文研究相协样本下概率密度函数的调整经验似然推断,证明对数调整经验似然比统计量服从2分布,由此构造了相协样本下概率密度函数的调整经验似然置信区间.在有限样本情况下通过数值模拟,对比分析得到AEL的表现略优于EL和NA的表现.关键词：调整经验似然;概率密度函数;相协样本;置信区间中图分类号：O212.7AMS(2010)主题分类：62G05;62E20文献标识码：A文章编号：1001-9847(2024)01-0171-091.引言经验似然是Owen12在独立样本下提出的一种非参数统计推断方法,与经典方法(如正态逼近理论)以及当前比较流行的统计方法(如Bootstrap)相比,经验似然方法具有诸如:置信域的形状由数据决定、置信域具有Bartlett可修正性、无需构造枢轴量、具有保值域性和函数变换不变性等优点.正因为如此,许多统计学家将这一方法应用到各种统计模型及各种领域.如线性回归模型的统计推断34,分位数统计推断5、广义线性统计推断6等,此外CHEN7将经验似然方法与核方法结合构造精度更高的概率密度函数的置信区间等.然而,上述文献所提到的经验似然方法都针对于独立样本情形,而现实中相协样本情形经常出现,针对于相协样本,现有的研究较多采用分块经验似然方法进行推导.Kitamural8在-混合样本下首次运用分块经验似然方法构建参数的置信区间.此后,ZHANG9将分块经验似然用于构造负相协样本下非参数整体均值的置信区间、LEI和QIN10将分块经验似然用于构造负相协样本下总体分位数的置信区间、QIN等11将分块经验似然用于构造负相协样本下概率密度函数的置信区间等.为了保证经验似然统计量的存在,需要假定0在数据的凸包内,针对0向量可能不在数据的凸包内的问题,CHEN等12提出了调整经验似然.此后CHEN和HUANG13使用调整经验似然构造总体均值的置信区间并研究了其有限样本下的性质,ZHOU和JING14使用调整经验似然构造分位数的置信区间等.本文受到文12,15的启发,研究相协样本16下概率密度函数的调整经验似然推断问题,采用不分块的调整经验似然方法进行推导避免了分块技术的复杂性,并构造了相协样本下概率密度函数的调整经验似然置信区间.通过模拟得到,调整经验似然(AEL)的表现略优于经验似然(EL)和正态逼近(NA).本文结构如下:第2节将给出本文的主要结论,第3节将给出有限样本下的模拟结果,第4节将给出相关的引理及其证明,第5节将给出主要结论的证明.收稿日期：2022-12-15基金项目：国家自然科学基金(12061017);广西研究生教育创新计划项目(YJSCXP202104)作者简介：王宁宁,女,汉族,河南人,研究方向:非参数统计通讯作者：秦永松.172应用数学20242.主要结论设X1,X2,Xn是来自总体X的一组严平稳相协样本,f(x)为X的概率密度函数,给定x R,f(x)的核密度估计为f(x)=1nhni=1K(x Xih),其中,K为核函数,h:=hn为窗宽,且 limnhn=0.我们使用经验似然方法来构建f(x)的置信区间,令gn,i()=h1K(xXih),1 i n,其中=f(x).则f(x)的经验似然比统计量为R()=supp1,p2,pnni=1npi?ni=1pign,i()=0,pi 0,ni=1pi=1,通过拉格朗日乘数法可以得到pi=1/n(1+gn,i(),其中=()满足ni=1gn,i()1+gn,i()=0,则对数经验似然比统计量为W()=logR()=ni=1log(1+gn,i().在经验似然方法的基础上,根据文12,我们提出如下的调整经验似然,给定正常数an,定义gn,n+1=an gn,i,1 i n,其中 gn,i=n1ni=1gn,i,gn,i=gn,i()=h1K(x Xi)/h)，1 i n.则调整经验似然比统计量为R()=supp1,p2,pn,pn+1n+1i=1(n+1)pi?n+1i=1pign,i()=0,pi 0,n+1i=1pi=1,通过拉格朗日乘数法可以得到pi=1/(n+1)(1+gn,i(),其中=()满足n+1i=1gn,i()1+gn,i()=0,(2.1)则对数调整经验似然比统计量为W()=logR()=n+1i=1log(1+gn,i().下面我们列出一些正则条件,这些正则条件将会用在后面的定理中.通常设C为正常数,在不同情况下可能取到不同的数值.正则条件:(A1)(i)X1,Xn为严平稳相协序列(正相协或者负相协);(ii)E(X21)且j=1|Cov(X1,Xj+1)|0.(A2)(i)核函数K满足:RK(u)du=1;K(u)C,u R;limu|u|K(u)=0;(ii)对于所有的u R,核函数的导函数dK(u)du=K(u)存在并且有界,即|K(u)|C,u R;(iii)RuK(u)du=0,0 Ru2K(u)du ;(iv)0 R|K(u)|3du 0为窗宽,且pn,qn和hn分别满足第 1 期王宁宁等：相协样本下概率密度函数的调整经验似然推断173(i)pnhn/n 1;(ii)pnhn 0和p2n/(nhn)0;(iii)1/h3nj=qn|Cov(X1,Xj+1)|0;(iv)nh5n 0.注2.1本文中条件(A1)-(A3)与文16中条件(A1)-(A3)完全相同.条件(A1)(v),(A2)(iii),(A3)(iv)与文16推论2.1中条件(b)相同.若f在x的邻域内有连续的一阶导函数,可用文16中的条件(a)代替(A1)(v),(A2)(iii),(A3)(iv).注2.2有许多核函数满足条件(A2),如Epanechnikov核函数:345(1 t2/5)I(|t|5),Biweight核函数:15167(1 t2/7)2I(|t|7)和Gaussian核函数:12et2/2等等.注2.3正如文16所讨论的那样,条件A3(i)表明qnkn/n 0和qn/pn 0.条件A3(ii)表明qnhn 0和nhn.令pn h1n,qn h2n(0 2 1 1),则kn nh1n,满足条件(A3)(i)(ii)时,nh21+1n.本文主要结论为以下内容.定理2.1设条件(A1)-(A3)成立且an=o(n),则当n 时有2W(f(x)d21,其中21表示自由度为1的卡方分布.注2.4定理2.1用于构造f(x)的调整经验似然置信区间.令Z满足P(21 Z)=,其中0 i)和V ar(Xi)=0.52(1 i n),取an=1,由此生成的Xi;1 i n是负相协序列(见文17),该情形下的模拟结果见表1.174应用数学2024表1多元正态分布随机样本下置信度为0.90和0.95时f(1)的置信区间的CP和AL1-nCPALAELCIELCINACIAELCIELCINACI0.901000.8840.8820.8710.33160.32830.32961500.8910.8890.8780.30760.30550.30622000.8990.8960.8840.28800.28680.28702500.9010.8960.8930.27430.27320.27350.951000.9330.9320.9240.39470.39060.39281500.9360.9340.9250.36560.36310.36432000.9400.9370.9310.34350.34180.34252500.9460.9450.9360.32530.32400.3245由表1得,在多元正态随机样本下,随着样本量的不断增大,置信区间的覆盖率逐渐增大且越来越接近于置信度,置信区间的平均长度逐渐减小.当n相同时,AEL方法的置信区间的覆盖率高于EL和NA两种方法的置信区间的覆盖率,虽然AEL方法置信区间的平均长度略大于EL和NA两种方法的置信区间的平均长度,但是相差非常小.因此综合分析可得,多元正态随机样本下,AEL的表现略优于EL和NA的表现.ii)多元t分布随机样本在本节模拟中,(X1,Xn)是多元t分布随机向量,且满足E(X1,Xn)=(1,1),Cov(Xi,Xj)=3(ji)1(1 i n,j i)和Var(Xi)=5/3(1 i n),取an=0.5,由此生成的Xi;1 i n是负相协序列(见文17),该情形下的模拟结果见表2.表2多元t分布随机样本下置信度为0.90和0.95时f(1)的置信区间的CP和AL1-nCPALAELCIELCINACIAELCIELCINACI0.901000.8320.8260.8000.20110.20010.19891500.8410.8400.8020.18120.18050.17962000.8470.8440.8210.16780.16740.16662500.8510.8490.8300.15790.15760.15690.951000.8910.8860.8570.23990.23850.23661500.9020.9010.8700.21640.21560.21412000.9050.9030.8760.20080.20030.19902500.9110.9080.8980.18650.18610.1850由表2得,在多元t分布随机样本下,随着样本量的不断增大,三种方法的置信区间的覆盖率均逐渐增大且越来越接近置信度,置信区间的平均长度均不断减小.当n相同时,AEL方法的置信区间的覆盖率高于EL和NA两种方法的置信区间的覆盖率,尽管其置信区间的平均长度也稍有增加,但与EL和NA的置信区间的平均长度相比相差甚微.综合分析可得在多元t分布随机样本下,AEL表现略好于EL和NA的表现.因此一般来说,在正相协随机样本下,AEL表现略好于EL和NA的表现.此外,通过对比两种不同分布随机样本下的模拟结果可得,当n取同一值时,多元正态分布样本下三种方法的置信区间的覆盖率都更高且更接近于置信度,因此可得三种方法在多元正态分布随机样本下的表现均优于在多元t分布随机样本下的表现.正相协样本在本节模拟中,使用模型Xi=Xi+Xi+1产生随机数样本,其中Xi,i 1为独立同分布随机数列且Xi N(1,1),1 i n或Xi 21,1 i n,取an=1,由此产生第 1 期王宁宁等：相协样本下概率密度函数的调整经验似然推断175的Xi,i 1为(正)相协序列(见文16).这两种情形下的模拟结果分别见表3和表4.表3Xi N(1,1)情形下置信度为0.90和0.95时f(1)的置信区间的CP和AL1-nCPALAELCIELCINACIAELCIELCINACI0.901000.8830.8800.8720.20800.20590.20491500.8920.8900.8760.18650.18520.18442000.8930.8910.8780.17160.17080.17002500.8960.8930.8800.16180.16110.16050.951000.9360.9310.9080.24910.24660.24501500.9390.9370.9310.22450.22300.22172000.9430.9420.9250.20630.20520.20402500.9480.9470.9360.19340.19260.1915表4Xi 21情形下置信度为0.90和0.95时f(1)的置信区间的CP和AL1-nCPALAELCIELCINACIAELCIELCINACI0.901000.8630.8580.8500.23810.23570.23551500.8660.8640.8540.20220.20090.20062000.8710.8680.8560.18280.18180.18162500.8810.8760.8650.16590.16880.16860.951000.9070.9040.8840.28000.27720.27671500.9270.9260.9130.24340.24170.24142000.9350.9340.9140.21720.21610.21572500.9390.9360.9270.20080.20000.1997由表3、表4得,从置信区间覆盖率来看,随着样本量的不断增大,两种随机分布样本下置信区间的覆盖率均不断增加且越来越接近置信度,当n相同时,AEL的置信区间的覆盖率高于EL和NA的置信区间的覆盖率.从置信区间的平均长度来看,两种随机分布样本下置信区间的平均长度均随着样本量的增加不断减小.尽管AEL方法在置信区间的覆盖率增大的同时其平均长度也稍有增加,但其与EL和NA的置信区间的平均长度相比相差甚微.综合分析可得,正相协样本下,AEL的表现略好于EL和NA.此外,通过两表对比可得,当n取相同值时,正态分布样本下三种方法对的置信区间的覆盖率更高且更接近于置信度,且置信区间的平均长度也更短,由此可得三种方法在正态分布随机样本下的表现优于在卡方分布随机样本下的表现.综上可得,相协样本下(正相协或负相协),对于概率密度函数置信区间的估计,AEL方法的表现略好于EL和NA两种方法的表现,且三种方法均在正态分布随机样本下表现得更好.此外,通过模拟结果发现,在相协样本下文中提到的三种方法得到的置信区间的覆盖率一致低于名义覆盖水平1-,尝试增大样本量进行模拟发现(模拟结果不再列出),置信区间的覆盖率有所改善且偶尔出现高于置信度1-的情形,但是与样本量较小时模拟结果相比,样本量增大时AEL方法的表现优于EL和NA的表现的程度有所下降.因此,AEL方法更适合在样本量较小时使用,在样本容量较大时,EL和AEL的表现相差不大.在独立样本情形下,Tsao18讨论了EL置信区间的覆盖率常常低于名义覆盖水平的情况,并且从理论上分析了出现这一现象的原因,相协样本情况下EL置信区间出现低覆盖的原因尚需进一步研究.176应用数学20244.引理以下引理将用于后面定理的证明,其中引理4.1-4.3的证明见文15的引理4.1-4.3的证明.引理4.11)设条件(A1)(i)-(iv),(A2)(i)(ii),(A3)(i)-(iii)成立,则对于f的连续点x有nh(fn(x)Efn(x)dN(0,2(x),(4.1)其中,2(x)=f(x)RK2(u)du.2)设条件(A1)-(A3)成立,则nh(fn(x)f(x)dN(0,2(x).引理4.2设条件(A1)(i)(v)和(A2)(i)成立.则对于任意的l n有h1E(Kl(x X1h)f(x)RKl(u)du.引理4.3设条件(A1)-(A3)成立,则n1hni=1gn,i(f(x)dN(0,2(x),n1hni=1g2n,i(f(x)=2(x)+op(1),(4.2)max1in|gn,i(f(x)|=op(1n1h).引理4.4设条件(A1)-(A3)成立,则(n1h)3/2ni=1|gn,i(f(x)|3=op(1).证由Cr-不等式得(n1h)3/2ni=1|gn,i(f(x)|3 C(nh)1/2(nh)1ni=1|Kh(x Xi)|3+n1/2h3/2f3(x).令M(x)=|K(x)|3/K3,其中K3=R|K(x)|3dx,则函数M(x)满足正则条件(A2)(i)(ii).令 mn(x)=1nhni=1M(xXih),则 mn(x)为f(x)的另一个核估计且满足引理4.1成立的条件,故由(4.1)式可得nh(mn(x)E mn(x)dN(0,2(x),其中2(x)=f(x)RM2(x)dx.由此可得 mn(x)E mn(x)p0.另由引理4.2得E mn(x)=1K3hE?K(x X1h)?31K3f(x)R|K(u)|3du,则 mn(x)p(1/K3)f(x)R|K(u)|3du,即1nhni=1?K(x Xih)?3pf(x)R|K(u)|3du,由条件(A2)(iv)可知,0 R|K(u)|3du 0知,n1ni=1g2n,i=2(x)+op(1)h h12(x)(1 )依概率成立,其中 0且充分小.令an=o(n),由(5.1)式可得1+Zn(gn,i+Op(n3/2h1/2an)h2(x)(1 )1=Op(n1/2h1/2).由上式得=Op(n1/2h1/2),则=Op(n1/2h1/2).2)令Vn,i=Vn,i(f(x)=n1ni=1g2n,i(f(x),则0=1nn+1i=1gn,i1+gn,i=1nn+1i=1gn,i1nn+1i=1(gn,i)21+gn,i gn,i(1 n1an)1+ZnVn,i,由于(1+Zn)0,在上式两边同时乘以(1+Zn)得0 (1+Zn)gn,i n1(1+Zn)an gn,i Vn,i=gn,i Vn,i+op(n1/2h1/2),则当n 时,=gn,i(Vn,i)1+op(n1h).3)最后,我们将2W(f(x)通过Taylor展开得2W(f(x)=2n+1i=1log1+gn,i(f(x)=2n+1i=1(gn,i12(gn,i)2)+2n+1i=1O(gn,i)3,其中2n+1i=1O(gn,i)3=23ni=1g3n,iO(1)+23g3n,n+1O(1)=op(1),故2W(f(x)=2n+1i=1log1+gn,i(f(x)=2n gn,i n2Vn,i+op(1),将=gn,i(Vn,i)1+op(n1h)代入上式得2W(f(x)=n(gn,i)2(Vn,i)1+op(1)=(n1hni=1gn,i)2(n1hni=1g2n,i)1+op(1),由此得出,当n 时,2W(f(x)收敛到21分布.故定理2.1成立.定理2.2的证明注意到gn,i=gn,i(),i=1,n,且同样定义 gn,i和gn,n+1.根据大数定律,当n ,(gn,i)2(f(x)2=2 0依概率成立.由引理4.3可得max1in|gn,i()|max1in|gn,i(f(x)|+|f(x)|=Op(n1/2h1/2),178应用数学2024 gn,i()=gn,i(f(x)+f(x)=Op(1).对于正常数M,令=n2/3h1/3 gn,iM,由条件(A3)(ii)知,当n 时nh .取an=o(n2/3),令 i=gn,i(),则 max1in|i|max1in|gn,i()|=Op(nh)1/6)=op(1),|n+1|=?gn,n+1()?=op(1),由此可得max1in+1|i|=op(1),(5.2)由(5.2)式和Taylor展开式log(1+i)=i 2i/2+i得n+1i=1log(1+i)=n+1i=1 i12n+1i=1 2i+n+1i=1 i=ni=1 i12ni=1 2i+n+1i=1 i+n+112 2n+1=ni=1 i12ni=1 2i+n+1i=1 i+op(1),(5.3)当n 时,P(|i|B|i|3,1 i n+1)1,其中B 0.且由引理4.3和引理4.4分别得ni=1g2n,i()=ni=1(K(x Xi)/h)h f(x)+f(x)2=Op(nh1),(5.4)ni=1|gn,i()|3=ni=1?K(x Xi)/h)h f(x)+f(x)?3=op(n3/2h3/2),(5.5)(5.4)式和(5.5)式分别表明?ni=1 2i?=op(1)和?n+1i=1 i?Bni=1|i|3+B|n+1|3=op(1).则(5.3)式可表示为n+1i=1log(1+i)=ni=1 i+op(1)=ni=1gn,i()+op(1)=(nh)1/3M(gn,i)2+op(1)=(nh)1/3(2+op(1)M+op(1)=(nh)1/32M+op(nh)1/3),利用最大化问题的对偶性可得W()=sup(n+1i=1log(1+gn,i)n+1i=1log(1+gn,i),因为M可任意大,则对于=f(x),当n 时,依概率趋于1有2(nh)1/3W()成立.对于2(nh)1/3W()的证明与2(nh)1/3W()的证明相似.定理2.2证毕.参考文献：1 OWEN A B.Empirical likelihood ratio confidence intervals for a single functionalJ.Biometrika,1988,75(2):237-249.2 OWEN A B.Empirical likelihood ratio confidence regionsJ.The Annals of Statistics,1990,18(1):90-120.3 OWEN A B.Empirical likelihood for linear modelsJ.The Annals of Statistics,1991,19(4):1725-1747.4 CHEN S X.On the accuracy of empirical likelihood confidence regions for linear regression modelJ.Annals of the Institute of Statistical Mathematics,1993,45(4):621-637.第 1 期王宁宁等：相协样本下概率密度函数的调整经验似然推断1795 CHEN S X,HALL P.Smoothed empirical likelihood confidence intervals for quantilesJ.The Annalsof 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properties of the adjusted empirical likelihoodJ.Journal ofNonparametric Statistics,2013,25(1):147-159.14 ZHOU W,JING B Y.Adjusted empirical likelihood method for quantilesJ.Annals of the Instituteof Statistical Mathematics,2003,55(4):689-703.15 XIONG X,LIN Z.Empirical likelihood inference for probability density functions under associationJ.Journal of Statistical Planning and Inference,2012,142(4):986-992.16 ROUSSAS G G.Asymptotic normality of the kernel estimate of a probability density function underassociationJ.Statistics and Probability Letters,2000,50(1):1-12.17 JOAG-DEV K,PROSCHAN F.Negative association of random variables with applicationsJ.TheAnnals of Statistics,1983,11(1):286-295.18 TSAO M.Bounds on coverage probabilities of the empirical likelihood ratio confidence regionsJ.The Annals of Statistics,2004,32(3):1215-1221.Adjusted Empirical Likelihood Inference of ProbabilityDensity Function Under Associated SamplesWANG Ninging,QIN Yongsong(School of Mathematics and Statistics,Guangxi Normal University,Guilin 541006,China)Abstract:In this paper,we study the adjusted empirical likelihood inference of the probabilitydensity function under the associated samples.We prove that the log-adjusted empirical likelihood ratiostatistic obeys the 2distribution,and use this conclusion to construct the adjusted empirical likelihoodconfidence interval of the probability density function under the associated samples.Through simulationand comparative analysis,we find that the performance of AEL is slightly better than that of EL and NA.Key words:Adjusting empirical likelihood;Probability density function;Associated sample;Con-fidence region