Bazarova, Alina

Berkes, István

Horváth, Lajos

Alina Bazarova

István Berkes

Lajos Horváth

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Statistics & Probability Letters

On the central limit theorem for modulus trimmed sums

TUGraz OPEN Library

ON THE CENTRAL LIMIT THEOREM FOR MODULUS TRIMMEDSUMSALINA BAZAROVA1, ISTVAN BERKES2, AND LAJOS HORVATH3Abstract. We prove a functional central limit theorem for modulus trimmed i.i.d. variablesin the domain of attraction of a nonnormal stable law. In contrast to the corresponding resultunder ordinary trimming, our CLT contains a random centering factor which is inevitablein the nonsymmetric case. The proof is based on the weak convergence of a two-parameterprocess where one of the parameters is time and the second one is the fraction of truncation.1. IntroductionLet X1; X2; : : : be independent, identically distributed random variables in the domain ofattraction of a stable law G with parameter 0 <  < 2. That is, assume that the partialsums Sn =Pnk=1Xk satisfy(1.1) (Sn   bn)=an d ! Gwith suitable norming and centering sequences fang, fbng. The necessary and sucientcondition for (1.1) is that F , the distribution function of X1, satises(1.2) 1  F (x) + F ( x) = x L(x); x > 0and(1.3)1  F (x)1  F (x) + F ( x) ! p;F ( x)1  F (x) + F ( x) ! q (x!1)where L is a function slowly varying at 1 and p; q  0, p + q = 1. (See e.g. Feller [8].)In contrast to the case of nite variances, the contribution of extremal terms in the partialsums Sn is not negligible and dropping a single term can change the asymptotic behavior of1Graz University of Technology, Institute of Statistics, Kopernikusgasse 24, 8010 Graz, Austria. e-mail:bazarova@tugraz.at. Research supported by Austrian Science Fund (FWF) Grant W1230.2Graz University of Technology, Institute of Statistics, Kopernikusgasse 24, 8010 Graz, Austria. e-mail:berkes@tugraz.at. Research supported by Austrian Science Fund (FWF) Grant P24302-N18.3University of Utah, Department of Mathematics, Salt Lake City, UT 84112-0090, USA. e-mail:horvath@math.utah.edu. Research supported by NSF grant DMS 1305858.1the sum. Let Xn;1  Xn;2  : : :  Xn;n be the order statistics of (X1; X2; : : : ; Xn) and putfor d  1(1.4) S(d)n =n dXj=d+1Xn;j:For xed d, Le Page, Woodrofe and Zinn [11] determined the asymptotic distribution of thetrimmed sum S(d)n and Csorg}o, Horvath and Mason [6] proved that under(1.5) dn !1; dn=n! 0the trimmed sum S(dn)n , suitably centered and normalized, is asymptotically normal. Theseresults give a remarkable picture on the partial sum behavior of i.i.d. sequences in the domainof attraction of a non-normal stable law. They show that the contribution of dn extremalterms under (1.5) already gives the stable limit distribution of the total partial sum Sn andthe contribution of the remaining elements will be an asymptotically normal variable withmagnitude negligible compared with Sn.The previous results describe the eects of the extremal elements of an i.i.d. sample ontheir partial sum. Note, however, that other kinds of trimming lead to dierent phenomena.For 1  d  n let d;n denote the d-th largest of jX1j; : : : ; jXnj and let(1.6) (d)Sn =nXk=1XkIfjXkj  d;ng:If the distribution of X1 is continuous, then jX1j; jX2j; : : : are dierent with probability 1,and thus (d)Sn coincides with the usual modulus trimmed sum obtained by discarding fromSn the d  1 elements with the largest moduli. Grin and Pruitt [9] showed that if X1 hasa symmetric distribution, then (dn)Sn is asymptotically normal for any dn !1, dn=n! 0,but this is generally false in the nonsymmetric case. The purpose of this paper is to describethe asymptotic distribution of (dn)Sn in the general case. PutH(t) = P (jXj  t) and m(t) = EXIfjXj  tg;and let H 1(t) = inffx : H(x)  tg (0 < t < 1) denote the generalized inverse of H. Ourmain result is the following.Theorem 1.1. Let X1; X2; : : : be i.i.d. random variables with distribution function F satis-fying (1.2), (1.3) and assume that (1.5) holds. Then we have(1.7)1An[nt]Xi=1(XiIfjXij  d;ng  m(d;n)) D[0;1] ! W (t)where(1.8) A2n =2  d(H 1(d=n))2and W is the Wiener process.2Theorem 1.1 shows that allowing a random centering factor, the modulus trimmed CLTholds for continuous i.i.d. variables under exactly the same conditions as under ordinarytrimming. If F is not continuous, the sample (X1; : : : ; Xn) may contain equal elementswith positive probability; according to the denition in Grin and Pruitt [9], 'ties' betweenelements with equal moduli are broken according to the order in which the variables occur in(X1; : : : ; Xn). But no matter how we break the ties, it may happen that from a set of sampleelements with equal moduli some are discarded and others are not, which is rather unnaturalfrom the statistical point of view, since trimming is mainly used to improve the performanceof statistical procedures by removing large elements from the sample. The denition of (d)Snin (1.6) resolves this diculty and leads to satisfactory asymptotic results in the generalcase.Theorem 1.1 enables one to give, among others, change point tests for heavy tailed pro-cesses, while the standard CUSUM test fails under innite variances. A fairly precise char-acterization of the modulus trimmed CLT with nonrandom centering and norming factorswas given in Berkes and Horvath [1].Under additional technical assumptions on the distribution function of X1 and on thegrowth speed of dn, Theorem 1.1 was proved by Berkes et al. in [2] with a fairly complicatedargument. The proof of Theorem 1.1 is much simpler and extends to dependent samples aswell, as we will show in a subsequent paper. LetA^2n =nXi=1X2i IfjXij  d;ng  1n nXi=1XiIfjXij  d;ng!2:Berkes et al. [2] showed that under the conditions of Theorem 1.1 we have thatA^n=AnP ! 1and therefore Theorem 1.1 yields1A^n0@ [nt]Xi=1XiIfjXij  d;ng   [nt]nnXi=1XiIfjXij  d;ng1A D[0;1] ! B(t);where B(t) = W (t) tW (1) denotes a Brownian bridge. Hence standard CUSUM techniquescan be used to detect changes in the mean and/or location when in the case of observationswithout second moments, observations with modulus larger than d;n are excluded from thesample.LetUn(t; s) =[nt]Xi=1 XiIfjXij  sH 1(d=n)g   EXiIfjXij  sH 1(d=n)g(s  0; t  0):We will deduce (1.7) from the following two-dimensional limit theorem.3Theorem 1.2. Let X1; X2; : : : be i.i.d. random variables with distribution function F satis-fying (1.2), (1.3) and assume that (1.5) holds. Then1AnUn(t; s)  !W (t; s2 ) weakly in D[0; 1]D[1=2; 3=2];where An is dened by (1.8) and fW (x; y); x  0; y  0g is a two-parameter Wiener process.Note that by Kiefer [10] we haved;nH 1(d=n)P ! 1:Since the limit process in Theorem 1.2 has continuous trajectories a.s., Billingsley [4], p.144-145 implies that1AnUn(t; d;n=H 1(d=n))D[0;1] ! W (t; 1)which is exactly the functional CLT in (1.7), sinceW (t; 1) is a Wiener process. Thus Theorem1.1 is a consequence of Theorem 1.2.2. Proof of Theorem 1.2Given a process Y (s; t) dened on a rectangle H = [a; b]  [c; d], let Y (H) denote theincrement of Y over H.Lemma 2.1. Let fYn(t; s); n  1g be processes dened on a rectangle [a; b] [c; d]  [0;1)2and assume that for some  > 0(2.1) EjYn(B)jjYn(C)j  (B)(C);where  denotes area and B and C are rectangles of the form [t1; t2]  [s1; s2] having onecommon edge, but otherwise disjoint. Then the sequence fYn(t; s); n  1g is tight. If everyYn(t; s) is piecewise constant in t, i.e. there exists a nite set Hn  [a; b] such that Yn(t; s)is constant in the left closed intervals determined by the elements of Hn [ fag [ fbg, then itsuces to verify (2.1) for rectangles [t1; t2] [s1; s2] where t1; t2 2 Hn.This is a special case of a general tightness condition due to Bickel and Wichura, see [3],Theorem 3.As is shown in Csorg}o et al. [7], Proposition A.3, the conditions of Theorem 1.2 implythat H 1(t) = t 1=`(t) (0 < t < 1), where ` is slowly varying at 0. Then by (1.8) we have(2.2) A2n 2  d (n=d)2=`2(d=n) as n!1where an  bn means an=bn ! 1 as n!1.4Lemma 2.2. If the conditions of Theorem 1.2 are satised, then for any p  2 and anyxed 0  a < b <1 we have(2.3) EjX1jpIfaH 1(d=n) < jX1j  bH 1(d=n)g  p  (bp    ap )`p(d=n)(n=d)(p )=as n!1. Also, if b > 0, then(2.4) EjX1jIfjX1j  bH 1(d=n)g =8><>:O((n=d)(1 )=`(d=n)) if  < 1;O((n=d)") if  = 1;O(1) if  > 1for any " > 0.Proof. Assume rst p  2, 0 < a < b <1. Clearly the left hand side of (2.3) equals Z bH 1(d=n)aH 1(d=n)tpdH(t) =Z H(aH 1(d=n))H(bH 1(d=n))H 1(u)pdu:(2.5)(Note that H is non-increasing and thus the left hand side of (2.5) is nonnegative.) Since His regularly varying with exponent  , we haveH(aH 1(d=n))  a (d=n); H(bH 1(d=n))  b (d=n) as n!1:Thus using the uniform convergence theorem for regularly varying functions (see e.g. [5],Theorem 1.5.2; note that we actually need the analogous result for regular variation at 0),we see that for n ! 1 we have, uniformly for all u in the interval of integration of thesecond integral in (2.5),H 1(u) = u 1=`(u)  u 1=`(d=n):Thus the integral equals(2.6) (1 + o(1))Z (1+o(1))a (d=n)(1+o(1))b (d=n)u p=`p(d=n) du;which yields the right hand side of (2.3) after a simple calculation, since p 6= . If a = 0,then the upper limit in the integral on the right hand side of (2.5) and thus also in (2.6)becomes H(0) = 1 and by using Theorem 1.5.11 of [5] we get the right hand side of (2.3)with a = 0.In the case of (2.4), instead of the integral in (2.6) we get(2.7)Z 1(1+o(1))b (d=n)u 1=`(u) du:By Proposition 1.3.6(i) in [5], p. 16 we have `(u) = O(u ") as u ! 0 for any " > 0 whichshows that for  > 1 the integralR 10u 1=`(u) du converges and thus the expression (2.7)is O(1). Using the same estimate for `(u) for  = 1 we get the second bound in (2.4).5Finally, for  < 1 Theorem 1.5.11 of [5] yields the rst bound in (2.4), completing the proofof Lemma 2.2.Proof of Theorem 1.2. Let  (t; s) denote the limit process in Theorem 1.2 and putQn =1AnMXm=1JXj=1m;jUn([tm 1; tm] [sj 1; sj])andZ =MXm=1JXj=1m;j ([tm 1; tm] [sj 1; sj])for all M  1; J  1, real coecients m;j, 1=2  s1 < s2 < : : : < sJ  3=2, 0 < t1 < : : : <tM = 1, t0 = s0 = 0. Clearly, Z is a centered normal r.v. and(2.8) EZ2 =MXm=1JXj=12m;j(s2 j   s2 j 1 )(tm   tm 1):We claim that(2.9) Qnd ! Z for all considered values of M;J; m;j; tm; sj:Since the processes Un and   are equal to 0 on the boundary of the rst quadrant, we haveUn(tm; sj) =MXm=1JXj=1Un([tm 1; tm] [sj 1; sj])and the same relation holds for  . Thus (2.9) implies1AnMXm=1JXj=1m;jUn(tm; sj)d !MXm=1JXj=1m;j (tm; sj)for arbitrary real coecients m;j and this, by the Cramer-Wold device, implies the conver-gence of the nite-dimensional distributions in Theorem 1.2.Relation (2.9) can be written equivalently as(2.10)1AnnXk=1(zk;n   Ezk;n) d ! N(0; EZ2);wherezk;n =JXj=1m;jXkIfsj 1H 1(d=n) < jXkj  sjH 1(d=n)g; [ntm 1] + 1  k  [ntm]:6Since the terms in the last sum are random variables with disjoint support, we get fromLemma 2.2Ez2k;n = (1+on(1))2  (n=d)(2 )=`2(d=n)JXj=12m;j(s2 j  s2 j 1 ); [ntm 1]+1  k  [ntm]and similarlyEz4k;n = (1+on(1))4  (n=d)(4 )=`4(d=n)JXj=14m;j(s4 j  s4 j 1 ); [ntm 1]+1  k  [ntm]:Thus using d = dn !1 we get by a simple calculation(2.11) limn!1Pnk=1 Ez4k;n Pnk=1 Ez2k;n2 = 0:On the other hand, the previous asymptotics for Ez2k;n and the statement of Lemma 2.2 forp = 1 implyE2jzk;nj = on(1)Ez2k;n; 1  k  nand thus by Minkowski's inequality(2.12) Ejzk;n   Ezk;nj2 = (1 + on(1))Ez2k;n; Ejzk;n   Ezk;nj4 = (1 + on(1))Ez4k;n:Thus (2.11) remains valid if we replace zk;n with zk;n   Ezk;n. Further by (2.2) and (2.8)nXk=1Ez2k;n = (1 + on(1))2  n(n=d)(2 )=`2(d=n)MXm=1JXj=12m;j(s2 j   s2 j 1 )(tm   tm 1)= (1 + on(1))A2nEZ2:The last relation, together with (2.11), (2.12) and Ljapunov's CLT for triangular arrays,implies (2.10).Next we prove tightness in Theorem 1.2. Consider two pairs of sets B11 = [t1; t] [s1; s],B12 = [t1; t]  [s; s2] and B11 = [t1; t]  [s1; s], B21 = [t; t2]  [s1; s], where t1 < t < t2,s1 < s < s2. In view of Lemma 2.1, it suces to show that(2.13) E 1AnUn(B11)2  1AnUn(Bij)2  C(B11)(Bij);holds for each ij 2 f12; 21g with some constant C > 0. Moreover, since Un(t; s) is constanton intervals k=n  t < (k + 1)=n, by the last statement of Lemma 2.1 we may assume thatnt; nt1 and nt2 are all integers. Using the independence of the Xj's, relation (2.2), Lemma2.2 and the fact that the function x2  has a bounded derivative on [1=2; 3=2], we getE 1AnUn(B11)2  1AnUn(B21)27= E 1AnntXi=nt1+1 XiIfs1H 1(d=n) < jXij  sH 1(d=n)g  mi!2 1Annt2Xi=nt+1 XiIfs1H 1(d=n) < jXij  sH 1(d=n)g  mi!2= E 1AnntXi=nt1+1 XiIfs1H 1(d=n) < jXij  sH 1(d=n)g  mi!2 E 1Annt2Xi=nt+1 XiIfs1H 1(d=n) < jXij  sH 1(d=n)g  mi!2 1A4n ntXi=nt1+1EX2i Ifs1H 1(d=n) < jXij  sH 1(d=n)g!(2.14) nt2Xi=nt+1EX2i Ifs1H 1(d=n) < jXij  sH 1(d=n)g! C1(t  t1)(t2   t)(s2    s2 1 )2  C2(t  t1)(t2   t)(s  s1)2= C2(B11)(B21);wheremi = mi(s1; s) = EXiIfs1H 1(d=n) < jXij  sH 1(d=n)gand C1; C2 are positive constants. On the other hand,E 1AnUn(B11)2  1AnUn(B12)2=1A4nE ntXi=nt1+1XiIfs1H 1(d=n) < jXij  sH 1(d=n)g  m(s1;s)i!2 ntXi=nt1+1XiIfsH 1(d=n) < jXij  s2H 1(d=n)g  m(s;s2)i!2=1A4nE ntXi=nt1+1(X(s1;s)i  m(s1;s)i )!2 ntXi=nt1+1(X(s;s2)i  m(s;s2)i )!2(2.15)where we putX(u;v)i = XiIfuH 1(d=n) < jXij  vH 1(d=n)g; m(u;v)i = EX(u;v)i :Expanding the product expectation in (2.15), we get the sum of all expressions(2.16) E(X(s1;s)i  m(s1;s)i )(X(s1;s)j  m(s1;s)j )(X(s;s2)k  m(s;s2)k )(X(s;s2)`  m(s;s2)` );8where nt1 + 1  i; j; k; `  nt. By the independence of the X 's, the product expectation in(2.16) equals 0 if one of the i; j; k; ` diers from the other three. Thus it suces to estimatethe contribution of the terms where i; j; k; ` are pairwise equal, or all are equal. Assume rstthat i = j, k = ` and i 6= k; the other cases i = k; j = `, i 6= j and i = `; j = k, i 6= j can behandled similarly as the case i = j = k = ` below. Then Xi and Xk are independent, andthus using Lemma 2.2, the product expectation (2.16) becomesEh(X(s1;s)i  m(s1;s)i )2(X(s;s2)k  m(s;s2)k )2i= E(X(s1;s)i  m(s1;s)i )2E(X(s;s2)k  m(s;s2)k )2(2.17) E(X(s1;s)i )2E(X(s;s2)k )2 2(2  )2 (s2    s2 1 )(s2 2   s2 )`4(d=n)(n=d)(4 2)= C3(s  s1)(s2   s)`4(d=n)(n=d)(4 2)=:The number of such pairs (i; k) is at most (nt   nt1)2 and thus dividing by A4n and using(2.2) we get that the contribution of such terms (2.16) is not greater thanC4(t  t1)2(s  s1)(s2   s) = C4(B11)(B12):Consider now the case i = j = k = `. In this case (2.16) becomes, expanding and introducingnew letters to lighten the notations,Eh(X(s1;s)i  m(s1;s)i )2(X(s;s2)i  m(s;s2)i )2i= E(  m(1))2(  m(2))2(2.18)= E22   2m(2)E2 + (m(2))2E2   2m(1)E2 + 4m(1)m(2)E  2m(1)(m(2))2E + (m(1))2E2   2(m(1))2m(2)E + (m(1))2(m(2))2;where = X(s1;s)i ;  = X(s;s2)i ; m(1) = E; m(2) = E:Clearly  and  have disjoint support and thus  = 0, showing that the rst, second, fourthand fth term of the last sum in (2.18) are equal to 0. Thus the sum equals(m(2))2E2   2m(1)(m(2))2E + (m(1))2E2   2(m(1))2m(2)E + (m(1))2(m(2))2= (m(2))2E2   2(m(1))2(m(2))2 + (m(1))2E2   2(m(1))2(m(2))2 + (m(1))2(m(2))2:By the Cauchy-Schwarz inequality we have (m(1))2  E2, (m(2))2  E2 and thus theabsolute value of the last sum is at most 7E(2)E(2), which, apart from the coecient, isexactly the third expression in (2.17), leading to the same estimate as there. The numberof choices for i in (2.18) is nt   nt1  (nt   nt1)2, so for the contribution of all terms in(2.18) we get the same estimate as for (2.17), i.e. C5(B11)(B12). Thus we proved (2.13)for Bij = B12 and the proof of Theorem 1.2 is completed.9References[1] Berkes, I. and Horvath, L.: The central limit theorem for sums of trimmed variables with heavy tails.Stochastic Process. Appl. 122 (2012), 449{465.[2] Berkes, I., Horvath, L. and Schauer, J.: Asymptotics of trimmed CUSUM statistics. Bernoulli 17 (2011),1344{1367.[3] Bickel, P. J. and Wichura, M. J.: Convergence criteria for multiparameter stochastic processes and someapplications. Ann. Math. Statist. 42 (1971), 1656{1670.[4] Billingsley, P.:. Convergence of probability measures. Wiley, New York 1968.[5] Bingham, N.H., Goldie, C. M., Teugels, J.L.: Regular variation. Cambridge University Press, 1987.[6] Csorg}o, S., Horvath, L. and Mason, D.: What portion of the sample makes a partial sum asymptoticallystable or normal ? Z. Wahrschein. verw. Gebiete 72 (1986), 1{16.[7] Csorg}o, S., Csorg}o, M., Horvath, L. and Mason, D.: Normal and stable convergence of integral functionsof the empirical distribution function. Ann. Probability 14 (1986), 86-118.[8] Feller, W.: An Introduction to probability theory and its applications, Vol II, Second Edition. Wiley,1971.[9] Grin, P. S. and Pruitt, W. E.: The central limit problem for trimmed sums. Math. Proc. CambridgePhilos. Soc., 102 (1987), 329{349.[10] Kiefer, J.: Iterated logarithm analogues for sample quantiles when pn ! 0. In: Proceedings of theSixth Berkeley Symposium on Mathematical Statistics and Probability, Vol. I, pp. 227{244. Universityof California Press, 1972.[11] Le Page, R., Woodrofe, M. and Zinn, J.: Convergence to a stable distribution via order statistics. Ann.Prob. 9 (1981), 624{632.10

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On the central limit theorem for modulus trimmed sums

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