ARTICLE IN PRESS Statistics & Probability Letters 78 (2008) 548–556 www.elsevier.com/locate/stapro Empirical likelihood for the two-sample mean problem Yukun Liu, Changliang Zou, Runchu Zhang LPMC and Department of Statistics, School of Mathematical Sciences, Nankai University, Tianjin 300071, China Received 9 October 2006; received in revised form 14 May 2007; accepted 7 September 2007 Available online 6 October 2007 Abstract We apply empirical likelihood method to constructing confidence regions for the difference of the means of two ddimensional samples. It is shown that the empirical likelihood ratio test has an asymptotic chi-squared distribution. The Bartlett correction for the univariate case ðd ¼ 1Þ has been investigated by Jing [1995. Two-sample empirical likelihood method. Statist. Probab. Lett. 24, 315–319]. Unfortunately, the Bartlett correction obtained in that article was incorrect. In this article the correct Bartlett correction is found and its effectiveness is shown by a simulation study. r 2007 Elsevier B.V. All rights reserved. MSC: primary 62G05; secondary 62E20 Keywords: Empirical likelihood; Bartlett correction; Coverage accuracy; Two-sample problem 1. Introduction Owen (1988, 1990, 1991) proposed the method of empirical likelihood as an alternative to the bootstrap for constructing confidence regions in nonparametric problems. DiCiccio et al. (1991) showed that in a very general setting, the empirical likelihood method for constructing confidence regions is Bartlett correctable, which is a key advantage over the bootstrap method. That is, through Bartlett correction, the coverage error of the empirical likelihood confidence region is sharply reduced from Oðn1 Þ to Oðn2 Þ, where n is the sample size. Chen and Hall (1993) studied Bartlett correction of quantiles, and Chen (1993) established a Bartlett correction for regression with non-random predictors. For more about Bartlett correction, see Hall and La Scala (1990), Chen (1994), Jing and Wood (1996), Zhang (1996), Baggerly (1998), Corcoran (1998) and Lazar and Mykland (1999), etc. In this paper, we apply the empirical likelihood method to the two-sample problem. To be specific, let i:i:d i:i:d x1 ; x2 ; . . . ; xn1 F and xn1 þ1 ; xn2 þ2 ; . . . ; xn1 þn2 G, where x1 ; x2 ; . . . ; xn1 þn2 2 Rd for some positive integer d. We want to test the following hypothesis: Z H0 : d9 Z x dG x dF ¼ d0 2H 1 : dad0 Corresponding author. E-mail address: [email protected] (R. Zhang). 0167-7152/$ - see front matter r 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2007.09.006 (1) ARTICLE IN PRESS Y. Liu et al. / Statistics & Probability Letters 78 (2008) 548–556 549 for some known d0 2 Rd , or equivalently, to construct confidence regions for d. Jing (1995) studied this problem in the univariate case ðd ¼ 1Þ and gave a Bartlett correction. However, the Bartlett correction term obtained in that article was incorrect. Here we extend Jing’s work to general d-dimensional multivariate random samples and present the correct Bartlett correction. The remainder of this paper is organized as follows. We formulate the empirical likelihood and confidence regions for d in Section 2. The coverage accuracy and Bartlett correction are also studied in this section. The simulation results are presented in Section 3. All the technical proofs are put in the Appendix. 2. Main results For problem (1), R let n ¼ n1 þ n2 and y ¼ n1 =n. In this paper, it is assumed that y ¼ n1 =n ! y0 2 ð0; 1Þ as n ! 1 and l0 ¼ x dF . Since then, unless otherwise stated, subscript i runs from 1 to ny, j from ny þ 1 to n and k from 1 to n, while all superscripts run from 1 toPd. Let yj ¼ xj d0 , ðp1 ; p2 ; . . . ; pny Þ and P ðqnyþ1 ; qnyþ2 ; . . . ; qn Þ be probability vectors (that is, pi ¼ 1, qj ¼ 1 and pi X0, qj X0). The empirical likelihood for d, evaluated at d0 , is defined as ( ) Y Y X X pi qj p ðx lÞ ¼ qj ðyj lÞ ¼ 0 Lðd0 Þ ¼ sup i i i i j j and the corresponding empirical log-likelihood ratio is defined as ^ lðd0 Þ ¼ 2 ln Lðd0 Þ=LðdÞ, ^ ¼ ðnyÞny ½nð1 yÞnð1yÞ . where LðdÞ The Lagrange multiplier method leads to pi ¼ 1 1 ; ny 1 þ y1 kt1 ðxi lÞ qj ¼ 1 1 nð1 yÞ 1 þ ð1 yÞ1 kt2 ðyj lÞ and the maximum log-likelihood ratio " # X X 1 t 1 t lnf1 þ y k1 ðxi lÞg þ lnf1 þ ð1 yÞ k2 ðyj lÞg . lðk1 ; k2 ; lÞ ¼ 2 i Let j ðk1 ; k2 ; l Þ be the solution to the following equations: 8 1 P xi l > > ¼ 0; > > 1 > ny 1 þ y kt1 ðxi lÞ > < yj l P 1 ¼ 0; > 1 t > > > nð1 yÞ 1 þ ð1 yÞ k2 ðyj lÞ > > : k þ k ¼ 0: 1 (2) 2 Then lðd0 Þ ¼ lðk1 ; k2 ; l Þ: First we present the nonparametric version of Wilks’ theorem in the two-sample problem. Theorem 1. Assume both F and G have finite second moments. Then, under null hypothesis, then lðd0 Þ has an asymptotic w2d distribution, that is Pðlðd0 Þow2d ðaÞÞ ¼ a þ Oðn1 Þ, where w2d ðaÞ is the a percentile of the w2d distribution. Obviously, we can construct confidence regions for d as follows: I a ¼ fd : lðdÞow2d ðaÞg with coverage error of order n1 , that is PðI a Þ ¼ a þ Oðn1 Þ. ARTICLE IN PRESS Y. Liu et al. / Statistics & Probability Letters 78 (2008) 548–556 550 The order Oðn1 Þ was first pointed out by DiCiccio et al. (1991). Besides, DiCiccio et al. (1991) also showed that for the smooth function model using Bartlett correction can reduce the coverage error of empirical likelihood confidence regions from order Oðn1 Þ to order Oðn2 Þ. Here we show the two-sample mean problem is also Bartlett correctable. We first present some notation. Let lk and fk , respectively, be the approximations of m and l to order Oðnðkþ1Þ=2 Þ, the expressions of which are given in the Appendix. Let V1 ¼ covðxÞ=y, V2 ¼ covðyÞ=ð1 yÞ, V ¼ V1 þ V2 and W ¼ V1 V1 V2 . In addition, define zi0 ¼ V1=2 ðxi l0 Þ; zj0 ¼ V1=2 ðyj l0 Þ; zj1 ¼ V1=2 ðyj l1 Þ; gt1 t2 tl ¼ G t1 t2 tl ¼ G t11 t2 tl zi1 ¼ V1=2 ðxi l1 Þ, 1 y ð1Þl Eðzti01 zti02 zti0l Þ þ l1 1 X t 1 t2 ð1Þl 1 tl z z z þ i0 i0 i0 yl1 ny i ð1 yÞl1 nð1 yÞ 1 Eðztj01 ztj02 ztj0l Þ, ð1 yÞl1 X ztj01 ztj02 ztj0l gt1 t2 tl , j X 1 1 X t 1 t2 ð1Þl 1 ¼ l1 zi1 zi1 zti1l þ ztj11 ztj12 ztj1l gt1 t2 tl , l1 nð1 yÞ ny y ð1 yÞ i j then by the three steps given in the Appendix we can finally show that lðd0 Þ ¼ D1 þ D2 þ Op ðn5=2 Þ, n (3) where D1 ¼ G r G r Grs G r G s þ 23 gruv G r Gu Gv þ G rs G st G r Gt þ 23 Gruv G r G u G v 2grst G ru G s G u G t þ grst gruv G s G t G u G v 12 grstu Gr Gs Gt G u , r s ruv ruv r u v 2 D2 ¼ ðGrs Grs 1 ÞG G þ 3 ðG 1 G ÞG G G . Here we use the summation convention according to which, if an index occurs more than once in an expression, summation over the index is understood. Note that D1 is similar to the expansion of one-sample empirical log-likelihood ratio, compared with the terms given in DiCiccio et al. (1991). However, D2 is a new and unique term for the two-sample case. The existence of this term is because under H0 the true means of x’s and y’s are not known except for their difference. When we approximate the empirical log-likelihood ratio statistic up to order Oðn3=2 Þ, we have to use their estimates to some order instead. Therefore it is somewhat different from one-sample problem, where under the corresponding null hypothesis the population mean is known. Taking expectations of D1 and D2 and ignoring terms of order Oð1=nÞ, nED1 ¼ d 1 ruv ruv 1 g g þ grruu ; 3n 2n nED2 ¼ 1 trðV1=2 WV1=2 Þ, nyð1 yÞ we can obtain the Bartlett correction, hn i x ¼ n ðED1 þ ED2 Þ 1 d 1 1 1 trðV1=2 WV1=2 Þ. ¼ gruv gruv þ grruu þ 3d 2d dyð1 yÞ ð4Þ The first two terms of x will reduce to the Bartlett correction given in Jing (1995) when d ¼ 1. However, Jing (1995) ignored D2 in Eq. (3) therefore the last term in (4) was missing in the Bartlett correction he gave. Thus, theoretically speaking, by using the Bartlett correction given by Theorem 2 of Jing (1995), the coverage accuracy of order Oðn2 Þ cannot be attained. Formally, we have the following theorem. ARTICLE IN PRESS Y. Liu et al. / Statistics & Probability Letters 78 (2008) 548–556 551 Theorem 2. Assume F and G have finite fourth moments, if H0 is true, then Pðlðd0 Þow2d ðaÞð1 þ x=nÞÞ ¼ a þ Oðn2 Þ, (5) where x is given by (4). Based on Eq. (3), the proofs of Theorems 1 and 2 are similar to DiCiccio et al. (1991), hence are omitted here. Now we have the adjusted confidence regions J a ¼ fd : lðdÞow2d ðaÞð1 þ x=nÞg, which have coverage error of order Oðn2 Þ. pffiffiffi In practice, we can substitute the moment estimates for the unknown gt1 t2 tl , V and W to obtain a nconsistent estimate x^ (under the condition that F and G have finite eighth moments). By utilizing the parity property of the polynomials in Edgeworth expansion (see Barndorff-Nielsen and Hall, 1988 and Hall and La Scala, 1990, Section 3.3), we can show that replacing x with x^ in (5) does not affect Theorem 2. 3. A simulation study Here we report a simulation study designed to evaluate the performance of the proposed empirical likelihood confidence regions. Twenty thousand pairs of simulated data sets of various n1 and n2 are generated from two distributions, respectively, where x’s are drawn from the standard exponential distribution and y’s from a chi-squared distribution with 3 degrees of freedom. The nominal coverage level is a ¼ 0:95. Table 1 shows the coverage percentage comparisons for constructing the confidence regions. The four values in each cell are the coverages of the confidence regions obtained by studentized bootstrap approach (see Davison and Hinkley, 1997), unadjusted empirical likelihood, adjusted empirical likelihood with the Bartlett correction ^ respectively. As given by Jing (1995) and adjusted empirical likelihood with the Bartlett correction given by x, we can expected, the unadjusted empirical likelihood has similar performance to the bootstrap approach because they both have theoretical coverage errors of order Oðn1 Þ. We can also clearly see that the Bartlett correction we give works uniformly better than that given by Jing (1995), although the latter may improve the efficiency of the ordinary empirical likelihood. Remark. As a referee pointed out, though the confidence regions I a and their adjusted counterparts J a , respectively, have coverage error of order Oðn1 Þ and Oðn2 Þ, they both have one-sided coverage error of order Oðn1=2 Þ, which means the Bartlett correction does not improve the one-sided coverage accuracy. To tackle this problem, the standard multivariate normal approximation to the distribution of the signed root of the logempirical likelihood ratio statistic can be applied. For general d-dimensional case, this method can improve the one-sided coverage accuracy from order Oðn1=2 Þ to order Oðn1 Þ. However, it is rather lengthy and hence Table 1 The coverage percentage comparisons n2 ¼ 10 n2 ¼ 20 n2 ¼ 30 n1 ¼ 10 0.890 0.888 0.900 0.904 0.919 0.919 0.927 0.930 0.924 0.926 0.932 0.935 n1 ¼ 20 0.886 0.883 0.896 0.899 0.922 0.920 0.927 0.930 0.928 0.931 0.935 0.937 n1 ¼ 30 0.891 0.884 0.897 0.899 0.919 0.919 0.927 0.929 0.930 0.933 0.938 0.940 ARTICLE IN PRESS Y. Liu et al. / Statistics & Probability Letters 78 (2008) 548–556 552 is omitted. We only present the result for the case d ¼ 1 instead, for the corresponding result in one-sample case see DiCiccio and Romano (1989). When a scalar d is of interest (d ¼ 1), the one-sided coverage accuracy can even be improved to order Oðn3=2 Þ. Here we give a brief illustration. Since all the variables are scalar, the unbold letterform is taken here. Denote the following scalar moment estimates: X 1 X 1 x¯ 1 ¼ xi ; x¯ 2 ¼ xj , ny i nð1 yÞ j X 1 X 1 ðxi x¯ 1 Þ2 ; v^2 ¼ ðxj x¯ 2 Þ2 ; v^ ¼ v^1 þ v^2 , v^1 ¼ 2 2 ny i nð1 yÞ j " # X 1 X 1 3=2 3 3 ðxi x¯ 1 Þ ðxj x¯ 2 Þ , g^ 3 ¼ v^ ny3 i nð1 yÞ3 j " # X 1 X 1 2 4 4 ðxi x¯ 1 Þ þ ðxj x¯ 2 Þ , g^ 4 ¼ v^ ny4 i nð1 yÞ4 j ! 1 v^1 1 v^21 v^2 v^22 þ þ ; t^1 ¼ 2 . t^1 ¼ y 1y v^ y 1 y v^ Let R0 ¼ n1=2 sgnf½d0 ðx¯ 2 x¯ 1 Þgl 1=2 ðd0 Þ be the signed root of the log-empirical likelihood ratio statistic. ^ will have a standard normal distribution to order Oðn3=2 Þ, and the normal ^ Then n1=2 ðR0 n1 aÞ=ð1 þ 12 n1 bÞ approximation can be used to set one-sided confidence limits with coverage error of the same order, where a^ ¼ 16 g^ 3 ; ^ ^2 ^ b^ ¼ 56 g^ 4 47 36 g3 þ t1 2t2 . The approximate upper ð1 aÞ confidence limit of d obtained in this way is n x¯ 2 x¯ 1 þ n1=2 v^1=2 z1a 16 n1=2 g^ 3 ð2z21a þ 1Þ 5 o 2 1^ ^ ^ þn1 z31a 14 g^ 4 29 g^ 23 12 t^2 þ z1a 12 þ t t g^ 4 13 g , 1 2 3 24 2 which has coverage level 1 a þ Oðn3=2 Þ. In fact, this method also has two-sided coverage accuracy of order Oðn2 Þ. A detailed technical report is available from the authors. Acknowledgements The authors are grateful to the editor and an anonymous referee for their valuable comments that make an important improvement to this article. This work was supported by the NNSF of China Grant No. 10571093, the SRFDP of China Grant No. 20050055038 and the NSF of Tianjin Grant No. 07JCYBJC04300. Zhang’s research was also supported by the Visiting Scholar Program at Chern Institute of Mathematics. Appendix First, we need the following lemma which establishes the orders of l ; k1 , and k2 . Lemma 1. Under the condition of Theorem 1, we have l ¼ l0 þ Op ðn1=2 Þ; k1 ¼ k2 ¼ Op ðn1=2 Þ. Proof. The proof is similar to that of Theorem 1 of Owen (1990), hence is omitted. The following three steps establish lk and fk for k ¼ 1; 2; 3, respectively. & ARTICLE IN PRESS Y. Liu et al. / Statistics & Probability Letters 78 (2008) 548–556 553 P P 1 1 Step 1: Denote C 11 ¼ ny i ðxi l0 Þ and C 12 ¼ nð1yÞ j ðyj l0 Þ. Based on the first equation of (2), using the standard expansion method, we have ! 1 X 1 X t ðxi l Þ ðxi l Þðxi l Þ V1 k1 V1 k1 þ Op ðn1 Þ ¼ 0. ny i ny2 i 1=2 It follows that k1 ¼ V1 Þ. Similarly, from the second equation of (2), we have 1 ðC 11 l Þ þ Op ðn 1 1=2 k2 ¼ V2 ðC 12 l Þ þ Op ðn Þ. Then substituting k1 and k2 into the last equation of (2), it leads to 1 l1 ¼ WðV1 1 C 11 þ V2 C 12 Þ, f1 ¼ V1 ðC 11 C 12 Þ V1 D1 . Step 2: Denote " # 1 X t C 21 ¼ ðxi l0 Þðxi l0 Þ V1 f1 , ny2 i C 31 ¼ 1 X t ½f1 ðxi l0 Þ2 ðxi l0 Þ, 3 ny i " C 22 # X 1 t ¼ ðyj l0 Þðyj l0 Þ V2 f1 , nð1 yÞ2 j C 32 ¼ X 1 ½ft1 ðyj l0 Þ2 ðyj l0 Þ. 3 nð1 yÞ j Ignoring terms of higher order, a three-term Taylor expansion of the first equation of (2) yields " # 1 X 1 X t ðxi l Þ ðxi l0 Þðxi l0 Þ V1 f1 V1 k1 ny i ny2 i 1 X t þ 3 ½f1 ðxi l0 Þ2 ðxi l0 Þ þ Op ðn3=2 Þ ¼ 0. ny i That is C 11 l þ C 21 V1 k1 þ C 31 þ Op ðn3=2 Þ ¼ 0. By the second equation of (2), similar arguments yield 3=2 Þ, k1 ¼ V1 1 ðC 11 þ C 21 þ C 31 l Þ þ Op ðn 3=2 Þ. k2 ¼ V1 2 ðC 12 þ C 22 þ C 32 l Þ þ Op ðn Then by the last equation of (2), it leads to 1 l2 ¼ W½V1 1 ðC 11 þ C 21 þ C 31 Þ þ V2 ðC 12 þ C 22 þ C 32 Þ, f2 ¼ V1 ½ðC 11 C 12 Þ þ ðC 21 C 22 Þ þ ðC 31 C 32 Þ V1 ðD1 þ D2 þ D3 Þ. Step 3. Similar to Steps 1 and 2, we can obtain 2 k1 ¼ V1 1 ðC 11 þ C 21 þ C 31 þ C 41 l Þ þ Op ðn Þ, 2 k2 ¼ V1 2 ðC 12 þ C 22 þ C 32 þ C 42 l Þ þ Op ðn Þ, ARTICLE IN PRESS Y. Liu et al. / Statistics & Probability Letters 78 (2008) 548–556 554 where " C 21 # 1 X t ¼ ðxi l1 Þðxi l1 Þ V1 f2 , ny2 i " C 22 # X 1 t ¼ ðyj l1 Þðyj l1 Þ V2 f2 , nð1 yÞ2 j C 31 ¼ 1 X t ðf2 ðxi l1 ÞÞ2 ðxi l1 Þ, ny3 i C 32 ¼ X 1 ðft2 ðyj l1 ÞÞ2 ðyj l1 Þ, 3 nð1 yÞ j C 41 ¼ C 42 ¼ 1 X t ðf1 ðxi l0 ÞÞ3 ðxi l0 Þ, 4 ny i X 1 ðft1 ðyj l0 ÞÞ3 ðyj l0 Þ. 4 nð1 yÞ j It follows that 1 l3 ¼ WV1 1 ½ðC 11 þ C 21 Þ þ ðC 31 þ C 41 Þ þ WV2 ½ðC 12 þ C 22 Þ þ ðC 32 þ C 42 Þ, f3 ¼ V1 ½ðC 11 C 12 Þ þ ðC 21 C 22 Þ þ ðC 31 C 32 Þ þ ðC 41 C 42 Þ V1 ðD1 þ D2 þ D3 þ D4 Þ. From above formulae, we may obtain the following expansion for lðd0 Þ=n: X lðd0 Þ 2 X t 2 ¼ f3 ðxi l3 Þ ft ðy l3 Þ n ny i nð1 yÞ j 3 j X 1 X t 1 f3 ðxi l1 Þðxi l1 Þt f3 ft3 ðyj l1 Þðyj l1 Þt f3 2 ny i nð1 yÞ2 j X 2 1 X t 2 1 þ ðf2 ðxi l1 ÞÞ3 ðft2 ðyj l1 ÞÞ3 3 3 3 ny i 3 nð1 yÞ j X 1 1 X t 1 1 ðf1 ðxi l0 ÞÞ4 ðft1 ðyj l0 ÞÞ4 þ Op ðn5=2 Þ 4 4 2 ny i 2 nð1 yÞ j 2 1 2I 1 I 2 þ I 3 I 4 þ Op ðn5=2 Þ. 3 2 Note that from the first two equations of (2), we have ! 1 X t C 11 l3 ðxi l1 Þðxi l1 Þ V1 f3 V1 f3 þ C 31 þ C 41 þ Op ðn2 Þ ¼ 0 ny2 i and C 12 l3 ! X 1 t ðyj l1 Þðyj l1 Þ V1 f3 V1 f3 þ C 32 þ C 42 þ Op ðn2 Þ ¼ 0. nð1 yÞ2 j ð6Þ ARTICLE IN PRESS Y. Liu et al. / Statistics & Probability Letters 78 (2008) 548–556 555 Consequently, 1 X ðxi l1 Þðxi l1 Þt f3 ¼ C 11 l3 þ C 31 þ C 41 þ Op ðn2 Þ, ny2 i X 1 ðyj l1 Þðyj l1 Þt f3 ¼ C 12 l3 þ C 32 þ C 42 þ Op ðn2 Þ. 2 nð1 yÞ j According to above formulae, we can obtain I 1 ¼ ft3 ðC 11 C 12 Þ ¼ Dt1 V1 ðD1 þ D2 þ D3 þ D4 Þ, I 2 ¼ ft3 ðC 11 C 12 þ C 31 C 32 þ C 41 C 42 Þ ¼ ðD1 þ D3 þ D4 Þt V1 ðD1 þ D2 þ D3 þ D4 Þ, I 3 ¼ ft2 ðC 31 C 32 Þ ¼ ðD3 Þt V1 ðD1 þ D2 þ D3 Þ, I 4 ¼ ft1 ðC 41 C 42 Þ ¼ Dt1 V1 D4 . Also, by noting that D1 ¼ Op ðn1=2 Þ; D3 ¼ Op ðn1 Þ; D2 ¼ Op ðn1 Þ; D3 ¼ Op ðn1 Þ; D2 ¼ Op ðn1 Þ; D2 D2 ¼ Op ðn3=2 Þ, D3 D3 ¼ Op ðn3=2 Þ; D4 ¼ Op ðn3=2 Þ, after some algebra, (6) yields lðd0 Þ ¼ Dt1 V1 D1 þ Dt1 V1 D2 þ 23 Dt1 V1 D3 n 13 Dt2 V1 D3 13 Dt3 V1 D3 þ 12 Dt1 V1 D4 þ Op ðn5=2 Þ. ð7Þ Using the definitions given in Section 2 and the facts G t1 t2 tl ¼ Op ðn1=2 Þ and Gt11 t2 tl ¼ Op ðn1=2 Þ, we have s st t suv u v 2 ðV1=2 D2 Þr ¼ G rs 1 ðG G G þ g G G Þ þ Op ðn Þ, s t ruv u v rst s u v tuv ðV1=2 D3 Þr ¼ G rst 2grst Gs Gtu G u þ Op ðn2 Þ, 1 G G þ g G G þ 2g G G G g where for a vector P, Pr means its rth component. Therefore, Dt1 V1 D1 ¼ G r G r ; Dt2 V1 D3 ¼ grst Gru G u G s Gt , s r rs st t suv r rs u v 5=2 Dt1 V1 D2 ¼ Gr Grs Þ, 1 G þ G G G G g G G G G þ Op ðn r s t ruv r u v rst tuv r s u v Dt1 V1 D3 ¼ G rst 1 G G G þ g G G G þ 2g g G G G G 2grst G tu G s G u G r þ Op ðn5=2 Þ, Dt3 V1 D3 ¼ grst gruv G s G t G u G v ; Dt1 V1 D4 ¼ grstu G r G s G t G u . 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