Edgeworth Expasio for Studetized Statistics Big-Yi JING Hog Kog Uiversity of Sciece ad Techology Qiyig WANG Australia Natioal Uiversity ABSTRACT Edgeworth expasios are developed for a geeral class of studetized statistics uder rather weak coditios. Applicatios of these results are give to obtai Edgeworth expasios for the distributios of the studetized U-statistics ad studetized L-statistics uder weaker momet coditios tha those available i the literature. AMS 1991 Subject Classificatios: 62E20, 60F15. Keywords ad Phrases: Studetized statistic, U-statistic, L-statistic, Edgeworth expasio. 0
1 Itroductio Suppose that we are iterested i the distributio of some statistic, T = T (X 1,..., X ), where X 1,, X is a sequece of idepedet ad idetically distributed (i.i.d.) real radom variables. Typically, oe ca use the delta method to show that T coverges i distributio to a ormal distributio. The rates of covergece to ormality is usually of the order 1/2 ad ca be described by Berry-Essee bouds. To get a better approximatio tha asymptotic ormality, oe ca develop higher-order Edgeworth expasios uder appropriate coditios. The theory of Edgeworth expasios dates back a log way. Of course, the simplest case is the Edgeworth expasio for the sample mea [c.f. Chapter 16 of Feller (1971)]. I recet years, a great deal of effort has bee devoted to derivig Edgeworth expasios for other classes of statistics such as fuctio of multivariate sample meas (Bhattacharya ad Rao (1976), Bai ad Rao (1991), or Hall (1992)), U-statistics, L-statistics ad others. (See Sectio 3 for more reviews.) O the other had, Edgeworth expasios for their studetized couterparts, such as Studet s t-statistic, studetized U-, ad L-statistics ad so o, have also gaied much mometum, partly due to their usefuless i statistical iferece (e.g., i costructig cofidece regios for populatio parameters, or testig hypothesis). It is worth poitig out that each of the methods for derivig Edgeworth expasios for the abovemetioed statistics was tailored to the idividual structures of these statistics. A geeral uifyig approach is to cosider symmetric statistics, which iclude all the abovemetioed statistics as special cases. See Lai ad Wag (1993), Betkus, Götze, va Zwet (1997) ad Putter ad va Zwet (1998) for istace. A quick glace at the literature reveals that the momet coditios i Edgeworth expasios for the studetized statistics are typically stroger tha the correspodig stadardized statistics; see Sectio 3 regardig U-statistics, L-statistics for example. This may ot be surprisig to us sice it is usually more difficult to hadle studetized statistics tha stadardized oes. Oe otable exceptio is the case of the sample mea, where the same third momet assumptio is eough for both the stadardized mea ad Studet s t-statistic, see Hall (1987) ad Betkus ad Götze (1996), for istace. This begs the questio whether the same pheomeo is also true for U-, L-statistics, ad other classes of statistics. I this paper we shall attempt to address this issue. However, istead of dealig with each idividual class of statistics separately, we shall cosider Edgeworth expasios for a very geeral class of so-called studetized statistics, ad the apply them to some special cases of iterest, e.g., U-, ad L-statistics. To be more precise, we shall cosider 1
studetized statistics of the form T /S (ad its slight variatios), where T = 1/2 α(x j ) + 3/2 β(x i, X j ) + V 1 S 2 1 = 1 + γ(x i, X j ) + V 2. ( 1) More detailed defiitios will be give i the ext sectio. Note that this class of studetized statistics T /S icludes a wide rage of statistics, much more so tha it first appears. For istace, symmetrical statistics belogs to this class by applyig the H- decompositio ad by takig S = 1. Similarly, the studetized symmetrical statistics cosidered i Putter ad vo Zwet (1998) are also i this class. Oe atural questio that oe might ask is why we cosider such a class of studetized statistics i the first place. Ideed, the research work o symmetric statistics have bee successful i dealig with stadardized statistics. However, we argue that the symmetrical statistics approach is yet to achieve the same kid of success whe it comes to studetized statistics. To illustrate why, let us cosider the Studet s t-statistic, which is clearly a symmetrical statistic, ad also a special case of studetized U-statistics; see Remark 3.1 as well. First we discuss Berry-Essee bouds for Studet s t-statistic. The geeral result of va Zwet (1984) o the symmetric statistics yields a Berry-Essee boud of order O( 1/2 ) provided E X 1 4 <. Friedrich (1989) improved this momet coditio to E X 1 10/3 <, which was later show by Betkus, Götze ad Zitikis (1994) to be optimal by usig this geeral approach. O the other had, it is well-kow that the optimal momet coditio for Studet s t-statistic is E X 1 3 < ; see Betkus ad Götze (1996) for istace. Clearly, the symmetrical statistics approach of va Zwet (1984) ad Friedrich (1989) fails to produce the best result i this case. Secodly, we look at Edgeworth expasios for Studet s t-statistic. Betkus, Götze ad vo Zwet (1997) give secod-order Edgeworth expasios of order O( 1 ). Applied to the Studet s t-statistic, Betkus et al (1997) obtaied a Edgeworth expasio with error term of size O( 1 ) (they used N istead of ). However, they eeded the existece of the (4 + ɛ)th momet o the populatio, which is clearly ot the optimal momet coditio i this case. I fact, the optimal momet coditio for the Studet s t-statistic is the fiite 4th momet, as show by Hall ad Jig (1995). The above Studet s t-statistic example shows that the geeral symmetric statistic approach may ot yield the optimal results for some studetized statistics. To get the optimal results, it seems very atural to look more closely at both the umerator ad the deomiator of the studetized statistics idividually. This is oe of the mai reasos why we cosider such a class of studetized statistics i this paper. As we have show 2
i our paper, this approach does lead to Edgeworth expasios uder rather weak ad atural coditios for studetized statistics tha usig the symmetric statistics approach. I particular, applicatio of our geeral results to studetized U- ad L-statistics leads to weaker ad more atural momet coditios tha previous work. Furthermore, applicatio to Studet s t-statistics lead to the optimal momet coditio as well (this was also show by Hall (1987)). The layout of the preset paper is as follows. I Sectio 2, we shall first itroduce a geeral class of studetized statistics ad the derive their Edgeworth expasios uder rather weak coditios. I Sectio 3, the results obtaied i Sectio 2 are applied to obtai Edgeworth expasios for the distributios of the studetized U-statistics ad studetized L-statistics. Proofs of the mai theorems are give i Sectio 4. Fially, some techical details are deferred to Sectio 5. Throughout this paper, we use C, C 1, C 2, ad also A 1, A 2, to deote some positive costats, which are idepedet of, ad may or may ot deped o the uderlyig distributio i questio. The same costat may resume differet values at each occurrece. For a set B, write I (B) as its idicator fuctio. Furthermore, we deote the stadard ormal distributio fuctio by Φ(x) ad its kth derivative by Φ (k) (x), ad φ(x) = Φ (1) (x). Fially, we itroduce the followig otatio for simplicity of presetatio,,, 1 <k 1<k i j i, i j i j k i,j,k=1 i j,j k,k i, (i) j<k j<k,j,k i. 2 Mai results Let X, X 1,, X be a sequece of i.i.d. real radom variables. Let α(x), β(x, y), γ(x, y), ζ(x, y) ad η(x, y, z) be some real-valued Borel measurable fuctios of x, y ad z. Furthermore, let V i V i (X 1,, X ), i = 1, 2, be real-valued fuctios of {X 1,, X }. Defie the statistic ad a ormalizig statistic T = 1/2 α(x j ) + 3/2 β(x i, X j ) + V 1 (2.1) S 2 = 1 + 1 ( 1) γ(x i, X j ) + V 2. (2.2) Uder appropriate coditios, the domiat term i the studetized statistic T /S is 1/2 α(x j ), which coverges i distributio to a ormal distributio as teds 3
to ifiity by the Cetral Limit Theorem. I the followig theorem, we shall derive secod-order Edgeworth expasios with error size o( 1/2 ) uder rather weak ad atural coditios. THEOREM 2.1. Assume that a) β(x, y) ad γ(x, y) are symmetric i their argumets. b) Eα(X 1 ) = 0, Eα 2 (X 1 ) = 1, E α(x 1 ) 3 <, lim t Ee itα(x 1 ) < 1; E[β(X 1, X 2 ) X 1 ] = 0, Eβ 2 (X 1, X 2 ) < ; Eγ(X 1, X 2 ) = 0, E γ(x 1, X 2 ) 3/2 <. c) P ( V j o( 1/2 ) ) = o ( 1/2), j = 1, 2. The we have where x P (T /S x) F (1) (x) = o ( 1/2), (2.3) F (1) (x) = Φ(x) Φ(3) (x) 6 ( Eα 3 (X 1 ) + 3Eα(X 1 )α(x 2 )β(x 1, X 2 ) ) xφ(2) (x) 2 Eα(X 1)γ(X 1, X 2 ). I Theorem 2.1, the fuctios β(x, y) ad γ(x, y) are assumed to be symmetric i their argumets. However, i some applicatios, it is ofte easier to use the ext theorem, which removes the symmetry restrictio o β(x, y) ad γ(x, y). To describe the theorem, we first defie T = 1/2 α(x j ) + 3/2 ζ(x i, X j ) + V 1, (2.4) i j S 2 = 1 + 3 η(x i, X j, X k ) + V 2. (2.5) i j k The followig theorem establishes a secod-order Edgeworth expasio for the distributio of T / S. THEOREM 2.2. Assume that a) Eα(X 1 ) = 0, Eα 2 (X 1 ) = 1, E α(x 1 ) 3 Ee <, lim itα(x 1 ) t < 1; E[ζ(X 1, X 2 ) X t ] = 0, t=1,2, Eζ 2 (X 1, X 2 ) < ; Eη(X 1, X 2, X 3 ) = 0, E η(x 1, X 2, X 3 ) 3/2 <. b) P ( V j o( 1/2 ) ) = o ( 1/2), j = 1, 2. 4
The we have where with x P ( T / S ) F (2) (x) = o ( 1/2), (2.6) F (2) (x) = Φ(x) Φ(3) (x) 6 ( Eα 3 (X 1 ) + 3Eα(X 1 )α(x 2 ) ζ(x 1, X 2 ) ) xφ(2) (x) 4 Eα(X 1) η(x 1, X 2, X 3 ), ζ(x 1, X 2 ) = ζ(x 1, X 2 ) + ζ(x 2, X 1 ), η(x 1, X 2, X 3 ) = η(x 1, X 2, X 3 ) + η(x 1, X 3, X 2 ) + η(x 2, X 1, X 3 ) +η(x 2, X 3, X 1 ) + η(x 3, X 1, X 2 ) + η(x 3, X 2, X 1 ). Remark 2.1. If we take γ(x, y) = 0 i Theorems 2.1, the we obtai the correspodig secod-order Edgeworth expasios for the stadardized statistics. I this sese, we have uified the treatmet for both types of statistics ad study them uder the same framework. Clearly, uless γ(x 1, X 2 ) i the studetized statistics T /S requires stroger coditios tha those imposed o α(x 1 ) ad β(x 1, X 2 ), the the secod-order Edgeworth expasios for both stadardized ad studetized statistics will hold uder the same set of coditios. This is oe of the appealig features of adoptig studetized statistics approach i this paper. Similar commets are also true for Theorem 2.2. 3 Applicatios I this sectio, we apply the mai results preseted i Sectio 2 to two well-kow examples, amely, the studetized U- ad L-statistics ad studetized fuctios of the sample mea. As ca be see later, applicatios of Theorem 2.1 ad 2.2 to these statistics lead to secod-order Edgeworth expasios uder weaker coditios tha ay other previous results. 3.1 Studetized U-Statistics. Let h(x, y) be a real-valued Borel measurable fuctio, symmetric i its argumets with Eh(X 1, X 2 ) = θ. Defie a U-statistic of degree 2 with kerel h(x, y) by U = 2 ( 1) 5 h(x i, X j ),
ad pose that g(x j ) = E (h(x i, X j ) θ X j ), σg 2 = V ar (g(x 1 )), R 2 = 4( 1)( 2) 2 1 h(x i, X j ) U 1 i=1 Note that 1 R 2 is the jackkife estimator of 4σ 2 g. Defie the distributios of the stadardized ad studetized U-statistic respectively by j i G 1 (x) = P ( (U θ)/(2σ g ) x ), G 2 (x) = P ( (U θ)/r x ). It is well-kow that G 1 (x) ad G 2 (x) coverge to the stadard ormal distributio fuctio Φ(x) provided Eh 2 (X 1, X 2 ) < ad σ 2 g > 0 [see Hoeffdig (1948) ad Arvese (1969), respectively]. Berry-Essee bouds have also bee studied by various authors. For stadardized U-statistics, Berry-Essee bouds have bee ivestigated by Grams ad Serflig (1973), Bickel (1974) ad Cha ad Wierma (1977), Callaert ad Jasse (1978), va Zwet (1984), Friedrich (1989), Betkus, Götze ad Zitikis (1994). For studetized U-statistics, Berry-Essée bouds were give by Callaert ad Veraverbeke (1981), Zhao (1983), Helmers (1985), ad Putter ad va Zwet (1998). Edgeworth expasios for U-statistics have also bee itesively studied i recet years. Here, we shall oly focus o the secod-order Edgeworth expasios with error size o( 1/2 ). For stadardized U-statistics, Bickel, Götze ad vo Zwet (1986) showed that where G 1 (x) F 1 (x) = o( 1/2 ), x 2. F 1 (x) = Φ(x) φ(x) 6 σg 3 ( x 2 1 ) { Eg 3 (X 1 ) + 3Eg(X 1 )g(x 2 )h(x 1, X 2 ) }, uder the coditios that σ 2 g > 0, the d.f. of g(x 1 ) is olattice, E g(x 1 ) 3 < ad E h(x 1, X 2 ) 2+ɛ < for some ɛ > 0. For studetized U-statistics, Helmers (1991) showed that G 2 (x) F 2 (x) = o( 1/2 ), x (the defiitio of F 2 (x) will be give i Theorem 3.1 below) uder the coditios that σ 2 g > 0, the d.f. of g(x 1 ) is olattice ad E h(x 1, X 2 ) 4+ɛ < for some ɛ > 0. Maesoo (1995, 1996, 1997) exteded Helmers result to arbitrary degree r ad fuctio 6
of studetized U-statistics. More recetly, Putter ad vo Zwet (1998) weakeed the (4 + ɛ)-th momet coditio of Helmers (1991) to E h(x 1, X 2 ) 3+ɛ <. However, this momet coditio still appears stroger tha eeded. I the ext theorem, we show that this momet coditio ca further be reduced to E h(x 1, X 2 ) 3 <. THEOREM 3.1. Suppose that σ 2 g > 0, E h(x 1, X 2 ) 3 < ad lim t Ee itg(x 1 ) < 1. The we have where G 2 (x) F 2 (x) = o( 1/2 ), x F 2 (x) = Φ(x) + φ(x) 6 σg 3 { (2x 2 + 1)Eg 3 (X 1 ) + 3(x 2 + 1)Eg(X 1 )g(x 2 )h(x 1, X 2 ) }. Before we prove the theorem, we shall make several remarks. Remark 3.1. If h(x, y) = 1(x + y), (U 2 θ)/r reduces to the well-kow Studet s t-statistic. Hall (1987) cosidered the Edgeworth expaios for Studet s t-statistic uder miimal coditios while Hall ad Jig (1995) derived Berry-Essee bouds for Edgeworth expasios of error size O( 1 ) uder the optimal fourth momet coditio. Remark 3.2. From Bickel, Götze ad vo Zwet (1986), it is kow that the momet coditio E h(x 1, X 2 ) 3 < is certaily ot optimal i obtaiig secod-order Edgeworth expasios with error size o( 1/2 ) for stadardized U-statistics. Judgig from the proof of this paper, it seems that it is ot optimal for studetized U-statistics either. I fact, it remais a ope questio as to what the optimal momet coditios are for both cases. As metioed earlier, the optimal momet coditio i Berry-Essee bouds for stadardized U-statitics are: E g(x 1 ) 3 < ad E h(x 1, X 2 ) 5/3 < ad σg 2 > 0. It would be iterestig to see whether these are sufficiet for establishig secod-order Edgeworth expasios with error size o( 1/2 ) for stadardized ad studetized U-statistics. Proof of Theorem 3.1. Similar to (A 3 ) i Callaert ad Veraverbeke (1981) (also see Serflig (1980)), we may rewrite (U θ) R = (U θ)/(2σ g ) 1/2 R /(2σ g ) T S, 7
where with T = 1/2 α(x i ) + 3/2 ζ(x i, X j ) + V 1, i=1 i j = 1 + ( 1) 1 ( 2) 2 η(x i, X j, X k ) + V 2 S 2 = 1 + 3 i j k i j k η(x i, X j, X k ) + V 3, α(x j ) = σ 1 g g(x j ), ζ(x i, X j ) = (2σ g ) 1 [h(x i, X j ) θ g(x i ) g(x j )], η(x i, X j, X k ) = σ 2 g [h(x i, X j ) θ] [h(x i, X k ) θ] 1, V 1 = 2 3/2 ( 1) 1 ζ(x i, X j ), V 2 = 2 2 + Q 1 + Q 2, V 3 = V 2 + Q 3, Q 1 = 2σ 2 g (h(x ( 1)( 2) 2 i, X j ) θ) 2, ( 1)σ 2 g Q 2 = (U ( 2) 2 θ) 2, Q 3 = ( 2)2 + 4( 1) 3 ( 1)( 2) 2 i j k η(x i, X j, X k ). By the properties of coditioal expectatio, it ca be easily show that for 1 i j k 3, Eα(X 1 )α(x 2 )ζ(x i, X j ) = 1 2 σ 3 g Eg(X 1 )g(x 2 )h(x 1, X 2 ), if 1 i j 2, Eα(X 1 )η(x i, X j, X k ) = σ 3 g Eg 3 (X 1 ), if i = 1, j k i, These estimates, together with the followig relatios = σ 3 g Eg(X 1 )g(x 2 )h(x 1, X 2 ), if i = 2 or 3, j k i. Φ (2) (x) = xφ(x), Φ (3) (x) = (x 2 1)φ(x), imply that (see Theorem 2.2 for the defiitio of ζ(x 1, X 2 ) ad η(x 1, X 2, X 3 )) F (2) (x) = Φ(x) Φ(3) (x) 6 ( Eα 3 (X 1 ) + 3Eα(X 1 )α(x 2 ) ζ(x 1, X 2 ) ) xφ(2) (x) 4 Eα(X 1) η(x 1, X 2, X 3 ) = Φ(x) + φ(x) 6 σ 3 g { (2x 2 + 1)Eg 3 (X 1 ) + 3(x 2 + 1)Eg(X 1 )g(x 2 )h(x 1, X 2 ) }. 8
O the other had, the coditio (a) of Theorem 2.2 ca be easily checked. Therefore, by Theorem 2.2, Theorem 3.1 follows if we ca prove P ( V 1 1/2 (log ) 1) = o( 1/2 ), (3.1) P ( Q j 1/2 (log ) 1) = o( 1/2 ), for j = 1, 2, 3. (3.2) We oly prove (3.2) for j = 1. Proofs of others are similar ad thus omitted. Let h (X i, X j ) = [h(x i, X j ) θ] 2 E [h(x 1, X 2 ) θ] 2. We have h (X i, X j ) = h (X j, X i ) ad E h (X i, X j ) 3/2 <. Therefore, it follows from part (b) of Lemma 5.2 that for 4, P ( Q 1 1/2 (log ) 1) 2 P (h(x i, X j ) θ) 2 ( 1) 1 4 σ2 g 1/2 (log ) 1 P E(h(X 1, X 2 ) θ) 2 2 + h (X i, X j ) ( 1) 1 4 σ2 g 1/2 (log ) 1 = o( 1/2 ). The proof of Theorem 3.1 is thus complete. 3.2 Studetized L-statistics Let X 1,, X be i.i.d. real radom variables with distributio fuctio F. Defie F to be the empirical distributio, i.e., F (x) = 1 I (Xi x). Let J(t) be a real-valued fuctio o [0, 1] ad T (G) = xj(g(x)) dg(x). The statistic T (F ) is called a L- statistic (see Chapter 8 of Serflig (1980)). Write σ 2 σ 2 (J, F ) = J (F (s)) J (F (t)) F (s t) [1 F (s t)] dsdt, where s t = mi{s, t} ad s t = max{s, t}. Clearly, a atural estimate of σ 2 is give by σ 2 σ 2 (J, F ). Now let us defie the distributios of the stadardized ad studetized L-statistic T (F ) respectively by H 1 (x) = P ( σ 1 (T (F ) T (F )) x ), H 2 (x) = P ( σ 1 (T (F ) T (F )) x ). It is well-kow that H 1 (x) ad H 2 (x) coverge to the stadard ormal distributio fuctio Φ(x) provided E X 1 2 <, σ 2 > 0 ad some smoothess coditios o J(t) 9
(see Serflig (1980) ad Helmers, Jasse ad Serflig (1990) for refereces). Berry- Essee bouds for stadardized L-statistics were give by Helmers (1977), vo Zwet (1984), Helmers, Jasse ad Serflig (1990) ad so o while Berry-Essee bouds for studetized L-statistics were also give by Helmers (1982). Earlier work o Edgeworth expasios for stadardized L-statistics icludes Helmers (1982), Lai ad Wag (1993) amog others. I fact, some of these papers are cocered with third-order Edgeworth expasios with error size O( 1 ). O the other had, secodorder Edgeworth expasios of error size o( 1/2 ) for studetized L-statistics is give i Putter ad vo Zwet (1998) uder some smoothess coditios o J(t) ad the momet coditio E X 1 3+ɛ for some ɛ > 0. However, this momet coditio still appears stroger tha eeded. I the ext theorem, we show that this momet coditio ca be weakeed eve further. THEOREM 3.2. Assume that (a). J (t) is bouded o t [0, 1], (b). σ 2 > 0 ad [F (t)(1 F (t))] 1/3 dt <. (c). lim t Ee ity < 1, where Y = J(F (s)) ( I(X1 s) F (s) ) ds. The we have where H 2 (x) F2(x) = o( 1/2 ), x F2(x) = Φ(x) + 1 ( 6 3a σ Φ(1) (x) + 3(b + c)x σ 3 with J 0 (t) = J(F (t)), ad a = J 0(x)F (x)[1 F (x)]dx; b = J 0 (x)j 0 (y)j 0 (z)k(x, y, z)dxdydz, K(x, y, z) = [F (x y z) F (x y)f (z)] [(1 F (x y)] c = +F (x y)f (x y) [F (z) 1] ; J 0(x)J 0 (y)j 0 (z)k (x, y, z)dxdydz, K (x, y, z) = F (x y) [1 F (x y)] [F (y z) F (y)f (z)] ; κ = J 0 (x)j 0 (y)j 0 (z) {F (x y z) + U} dxdydz, Φ (2) (x) + κ + 3d ) Φ (3) (x) σ 3 U = F (x)f (y z) F (y)f (x z) F (z)f (x y) + 2F (x)f (y)f (z); d = J 0 (x)j 0 (y)j 0(z)V (x, z)v (y, z)dxdydz, V (s, t) = F (s t) F (s)f (t). 10
Remark 3.3. Uder the same coditios as i Theorem 3.2, similarly to the proof of Theorem 3.2, we ca show that where H 1 (x) F1(x) = o( 1/2 ), x F1(x) = Φ(x) + 1 ( 6 3a σ Φ(1) (x) + κ + 3d ) Φ (3) (x). σ 3 Remark 3.4. I Coditio (b), we assumed that [F (t)(1 F (t)] 1/3 dt <. This is weaker tha the coditio E X 1 3+δ <. To show that, we first ote that, by applyig Markov s iequality, we have F (t)(1 F (t)) E X 1 3+ɛ / t 3+ɛ. Thus [F (t)(1 F (t)] 1/3 dt 1dt + [F (t)(1 F (t)] 1/3 dt t 1 <. t >1 2 + ( E X 1 3+ɛ) 1/3 t 1 ɛ/3 dt t >1 I fact, the coditio [F (t)(1 F (t)] 1/3 dt < is almost the same as the momet coditio E X 1 3 <. Proof of Theorem 3.2. For abbreviatio, we further itroduce the followig otatio, J (t) = J(F (t)), Z(s, t, F ) = F (s t)(1 F (s t)), ξ(x i, X j ) = σ 2 J 0 (s)j 0 (t) ( I (Xi s t)i (Xj >s t) Z(s, t, F ) ) dsdt ϕ(x i, X j, X k ) = σ 2 J 0(s)J 0 (t) [ I (Xi t) F (t) ] I (Xj s t)i (Xk >s t) dsdt. Recallig (see Lemma B of Serflig, 1980, p.265), we may write t T (F ) T (F ) = [K 1 (F (x)) K 1 (F (x))]dx, K 1 (t) = J(u)du, 0 (T (F ) T (F )) σ = (T (F ) T (F )) /σ σ/σ T / S, where T = 1/2 α(x j ) + 3/2 ζ(x i, X j ) + 1/2 Eζ(X 1, X 1 ) + V 1, i j 2 S = 1 + 3 η(x i, X j, X k ) + V 2, i j k 11
with α(x j ) = σ 1 J 0 (t) [ I (Xj t) F (t) ] dt, ζ(x i, X j ) = 1 2 σ 1 J 0(t) [ I (Xi t) F (t) ] [ I (Xj t) F (t) ] dt, η(x i, X j, X k ) = ξ(x i, X j ) + ϕ(x i, X j, X k ) V 1 = 1/2 (Q 1 + Q 2 ), V 2 = Q 3 + Q 4 + Q 5 ad Q i, i = 1,, 5, are defied by Q 1 = 2 Q 2 A(J)σ 1 Q 3 = 2σ 2 Q 4 = σ 2 [ζ(x j, X j ) Eζ(X 1, X 1 )], (3.3) F (t) F (t) 3 dt, (3.4) [J (s) J 0 (s) J 0(s)(F (s) F (s))] J 0 (t)z(s, t, F )dsdt, (3.5) [J (s) J 0 (s)] [J (t) J 0 (t)] Z(s, t, F )dsdt, (3.6) Q 5 = 3 [ξ(x j, X k ) + ϕ(x j, X j, X k ) + ϕ(x k, X j, X k )] j k 1 σ 2 F (s t) [1 F (s t)] dsdt. (3.7) We would like to apply Theorem 2.2. The coditio (a) of Theorem 2.2 ca be easily checked. Next let us check the coditio (b). By Lemma 5.6 i the Appedix, we have that P ( V 1 3 1/2 (log ) 1) P ( Q 1 + Q 2 3 1 (log ) 1) = o( 1/2 ), P ( V 2 3 1/2 (log ) 1) P ( Q 3 + Q 4 + Q 5 3 1/2 (log ) 1) = o( 1/2 ). These estimates imply the coditio (b) of Theorem 2.2. Now, by usig Theorem 2.2, we obtai that where x ( P 1 S ( T 1/2 Eζ(X 1, X 1 ) ) ) x F (x) = o( 1/2 ), (3.8) F (x) = Φ(x) Φ(3) (x) 6 ( Eα 3 (X 1 ) + 3Eα(X 1 )α(x 2 ) ζ(x 1, X 2 ) ) xφ(2) (x) 4 Eα(X 1) η(x 1, X 2, X 3 ), ζ(x 1, X 2 ) = ζ(x 1, X 2 ) + ζ(x 2, X 1 ), η(x 1, X 2, X 3 ) = η(x 1, X 2, X 3 ) + η(x 1, X 3, X 2 ) + η(x 2, X 1, X 3 ) +η(x 2, X 3, X 1 ) + η(x 3, X 1, X 2 ) + η(x 3, X 2, X 1 ). 12
It follows from (3.8) ad similar method used i proof of Theorem 2.1 that P ( T / S x ) 1/2 Φ (1) (x)eζ(x 1, X 1 ) F (x) = o( 1/2 ) (3.9) x I terms of (3.9), Theorem 3.2 follows from the followig tedious but simple calculatios: Eζ(X 1, X 1 ) = 1 J 2σ 0(x)F (x)(1 F (x))dx; Eα 3 (X 1 ) = 1 J σ 3 0 (x)j 0 (y)j 0 (z) {F (x y z) + U} dxdydz, Eα(X 1 )α(x 2 )ζ(x 1, X 2 ) = Eα(X 1 )α(x 2 )ζ(x 2, X 1 ) = 1 J 2σ 3 0 (x)j 0 (y)j 0(z)V (x, z)v (y, z)dxdydz, Eα(X 1 )ξ(x 1, X 2 ) = Eα(X 1 )ξ(x 1, X 3 ) = 1 J σ 3 0 (x)j 0 (y)j 0 (z)w (x, y, z)(1 F (x y))dxdydz, W (x, y, z) = F (x y z) F (x y)f (z), Eα(X 1 )ξ(x 2, X 1 ) = Eα(X 1 )ξ(x 3, X 1 ) = 1 σ 3 J 0 (x)j 0 (y)j 0 (z)f (x y)f (x y)(f (z) 1)dxdydz, Eα(X 1 )ξ(x 2, X 3 ) = Eα(X 1 )ξ(x 3, X 2 ) = 0; Eα(X 1 )ϕ(x 1, X 2, X 3 ) = Eα(X 1 )ϕ(x 1, X 3, X 2 ) = 1 σ 3 J 0(x)J 0 (y)j 0 (z)f (x y)(1 F (x y))v (y, z)dxdydz, V (y, z) = F (y z) F (y)f (z), Eα(X 1 )ϕ(x 2, X 1, X 3 ) = Eα(X 1 )ϕ(x 2, X 3, X 1 ) = 0, Eα(X 1 )ϕ(x 3, X 1, X 2 ) = Eα(X 1 )ϕ(x 3, X 2, X 1 ) = 0. The proof of Theorem 3.2 is complete. 3.3 Studetized fuctios of the sample mea Let X 1,, X be i.i.d. real radom variables with EX 1 = µ ad V ar(x 1 ) = σ 2 <. Let f be a real-valued fuctio differetiable i a eighborhood of µ with f (µ) 0. Thus the asymptotic variace of f(x) is give by σ 2 f = (f (µ)) 2 σ 2. Deote the sample mea ad sample variace by X = 1 i=1 X i ad σ 2 = 1 i=1 (X i X) 2. The a obvious estimate of σ f is simply f (X) σ. I this paper, however, we shall use a alterative estimate, i.e., the jackkife variace estimate give by σ f 2 = 1 ( ) ( 2 f(x (j) (j) 1 ) ) f(x), where X = X i X j. 1 i=1 Defie the distributios of the stadardized ad studetized f(x) respectively by L 1 (x) = P ( σ 1 f (f(x) f(µ)) x), L 2 (x) = P ( σ 1 f (f(x) f(µ)) x). 13
Asymptotic properties of L 1 (x) (e.g., the asymptotic ormality, Berry-Essée boud ad Edgeworth expasio) have bee well studied (see Bhattacharya ad Ghosh (1976) for istace). O the other had, Miller (1964) showed that σ 2 f is a cosistet estimator of σ 2 f ad hece proved that L 2 (x) follows the asymptotic stadard ormal distributio. Applyig the results of Bai ad Rao (1991), a secod-order Edgeworth expasios for L 1 (x) with error size o( 1/2 ) holds uder the miimal momet coditio E X 1 3 <. O the other had, Putter ad vo Zwet (1998) gave Edgeworth expasios for L 2 (x) uder the coditio that E X 1 3+ɛ < for some ɛ > 0. I this sectio, we hope to weake this coditio eve further. THEOREM 3.3. Assume that f (3) (x) is bouded i a eighborhood of µ ad f (µ) 0, E X 1 3 < ad lim t Ee itx 1 < 1. The we have where F 2 (x) = Φ(x) + x L 2 (x) F 2 (x) = o ( 1/2), σf (u) 2 f (u) Φ(1) (x) + φ(x) ( 6 (2x 2 + 1) E(X 1 µ) 3 + 3σf ) (u). σ 3 f (u) PROOF. Usig Taylor s expasio ad otig that X (j) X = (X X j )/( 1), we ca get σ 2 f = ( 1) { ( ) f (µ) X (j) X + ( f (X) f (µ) ) ( ) X (j) X + 1 ) ( } 2 2 (η 2 f j X (j) + (1 η j )X X (j) X) (for 0 η j 1) = f 2 (µ) (X j X) 2 + 2f (µ)f (µ) (X j X) 2 (X µ) + W 2 1 1 = f 2 (µ) (X j µ) 2 + 2f (µ)f (µ) (X j µ) 2 (X µ) + W 3 1 1 = σ 2 f 2 (µ) + 1 { f 2 (µ)((x 2 j µ) 2 σ 2 )+ j k +2f (µ)f (µ)(x j µ) 2 (X k µ) } + W 4, where W 4 A(f) (K 1 + K 2 + K 3 + K 4 ) with K i beig defied by K 1 = (X µ) 2 + X µ 3 + (X µ) 4, K 2 = 1 ( (Xj µ) 2 + X 2 j µ 3), 14
K 3 = 1 3 K 4 = 1 (X µ)2 Similarly, by Taylor s expasio, we have (X j µ) 4, (X j µ) 2. f(x) f(µ) = f (µ)(x µ) + 1 2 f (µ)(x µ) 2 + W 5, where W 6 A(f) ( K 5 + K 6 ) with where K 5 = 1 2 = 1 f (µ)(x j µ) + 1 f (µ)(x 2 2 j µ)(x k µ) j k + σ2 f (u) + W 6, (3.10) 2 ( (Xj µ) 2 σ 2) ad K 6 = X µ 3. I order to apply Theorem 2.1, we rewrite σ f 1 (f(x) f(µ)) = T / S, where T = σf (u) 2 f (u) + 1/2 S 2 = 1 + 1 ( 1) α(x j ) = (X j µ)/σ, β(x i, X j ) = f (µ) σf (µ) (X i µ)(x j µ), η(x i, X j ) = (X j µ) 2 σ 2 V 1 = V 2 = σf (u) W 6, α(x i ) + 3/2 β(x i, X j ) + V 1, i=1 (η(x i, X j ) + η(x j, X i )) + V 2, σ 2 + f (µ) σ 2 f (µ) (X i µ) 2 (X j µ), 1 σ 2 f 2 (u) W 4 + 1 2 ( 1) (η(x i, X j ) + η(x j, X i )). The coditios (a) ad (b) i Theorem 2.1 ca be easily checked. Next let us first assume that the coditio (c) i Theorem 2.1 holds, i.e., these exists a δ 0 such that P ( V j δ 1/2) = o ( 1/2), for j = 1, 2. (3.11) Uder this assumptio, by usig Theorem 2.1, we have that ( ( x P 1 S T σf ) ) (u) 2 x F f 2 (x) = o( 1/2 ), (3.12) (u) 15
where F 2 (x) = Φ(x) Φ (3) (x) 6 ( Eα 3 (X 1 ) + 3Eα(X 1 )α(x 2 )β(x 1, X 2 ) ) xφ(2) (x) 2 Eα(X 1)(η(X 1, X 2 ) + η(x 2, X 1 )) It follows from (3.12) ad similar method used i proof of Theorem 2.1 that x P ( T / S x ) σf (u) 2 f (u) Φ(1) (x) F 2 (x) = o( 1/2 ) (3.13) I terms of (3.13), Theorem 3.3 follows easily from the followig simple calculatios: Eα 3 (X 1 ) = E(X 1 u) 3, Eα(X σ 3 1 )α(x 2 )β(x 1, X 2 ) = σf (u) f (u) Eα(X 1 )η(x 1, X 2 ) = 0, Eα(X 1 )η(x 2, X 1 ) = E(X 1 u) 3 Φ (2) (x) = xφ(x), Φ (3) (x) = (x 2 1)φ(x). σ 3 + σf (u) f (u). I the remaider of this sectio, let us show (3.11). It suffices to show that that there exists a δ 0 such that P ( K j δ 1/2) = o ( 1/2), for j = 1, 2, 3, 4, (3.14) P ( K j δ 1) = o ( 1/2), for j = 5, 6, (3.15) 1 P 2 (η(x i, X j ) + η(x j, X i )) ( 1) δ 1/2 = o( 1/2 ). (3.16) We oly prove (3.14) for j = 4. The others are similar but more simpler by Lemmas 5.2-5.3 i the Appedix ad details are omitted. Usig Lemma 5.3 i the Appedix, we have P ( K 4 1/2 (log ) 1) = P 1 (X µ)2 (X j µ) 2 1/2 (log ) 1 P ( (X µ) 2 4σ 2 1 (log ) ) + P 1 (X j µ) 2 1/2 (σ 2 log ) 2 /4 = P ( 1/2 X µ 2σ(log ) 1/2) +P 1 [ (Xj µ) 2 σ 2] 1 4 1/2 (σ 2 log ) 2 σ 2 o ( 1/2) + P 1 [ (Xj µ) 2 σ 2] C (for large eough) = o ( 1/2). This proves (3.14) for j = 4. The proof of Theorem 3.3 is complete. 16
4 Proof of Theorems I this sectio, we give the proofs of the mai theorems. Some lemmas eeded i the proofs of mai theorems will be relegated to the Appedix. Proof of Theorem 2.1. Notice that for ay radom variables X, Y, Z 1, Z 2 ad ay costats C 1, C 2 > 0, P P ( ) X + Z1 x 1 + Y + Z2 ( ) X + Z1 x 1 + Y + Z2 P P ( ) X C1 x + 1 + Y + C2 ( ) X C1 x 1 + Y + C2 2 P ( Z j C j ), (4.1) 2 P ( Z j C j ), (4.2) we ca replace V j by ±δ = o ( 1/2). To simplify the otatio, we further assume V j = 0. It will be clear that this assumptio will ot affect the proof of mai results. We ext tur back to the proof of Theorem 2.1. Let γ 1 (x) = Eγ (x, X 1 ), γ 2 (x, y) = γ(x, y) γ 1 (x) γ 1 (y). It is easy to show that where S 2 = 1 + 1 1 γ 1 (X j ) + γ 2 (X i, X j ) ( 1) = 1 + Z + R, (4.3) Z = 1 γ 1 (X j ), R = By usig elemetary iequality: for u 1/9, we fid that if Z + R 1/9, the 1 + 1 2 (Z + R ) 1 3 1 ( 1) 1 + u 2 u2 6 (1 + u)1/2 1 + u 2 + u2 6, ( Z 2 + R 2 γ 2 (X i, X j ). ) S = (1 + Z + R ) 1/2 1 + 1 2 (Z + R ) + 1 3 ( Z 2 + R 2 ). (4.4) Put (s) = 1 2 Z + 1 2 R + s Z 2. 17
The from (4.3) ad (4.4), we have P (T /S x) P (T /S x, Z + R 1/9) + P ( Z + R 1/9) ( { P T x 1 + (1/3) + 1 2 R + 1 }) 3 R2 + P ( Z + R 1/9) Similarly, we get P ( T x { 1 + (1/3) + 3/5}) ( 1 +P ( Z + R 1/9) + P 2 R + 1 ) 3 R2 3/5. (4.5) P (T /S x) P ( T x { 1 + ( 1/3) 3/5}) ( 1 P ( Z + R 1/9) P 2 R 1 ) 3 R2. (4.6) Usig Jese s iequality, we ca easily see that E γ 1 (X 1 ) 3/2 < ad E γ 2 (X 1, X 2 ) 3/2 <. (4.7) Furthermore, we ote that 2R is a degeerate U-statistic of order 1 (whose defiitio is give i the Appedix). The it follows from Lemmas 5.2-5.3 that ( 1 P 2 R ± 1 ) 3 R2 3/5 P ( R 1) + P ( R 3/10 ) = o( 1/2 ), P ( Z + R 1/9) P ( Z 1/18) + P ( R 1/18) = o ( 1/2). These iequalities, together with (4.5) ad (4.6), imply that (2.3) holds if P (T x (1 + (s) + A )) F (1) (x) = o( 1/2 ) (4.8) x where A 1 3 3/5 ad s 1/3. To prove (4.8), we let ς j (x) = ( 1 2 γ 1(X j ) + s ) γ2 1(X j ), ψ ij (x) = β(x i, X j ) x (2sγ 1 (X i )γ 1 (X j ) + 1 2 γ 2(X i, X j ) Elemetary calculatio shows that P (T x (1 + (s) + A )) = P 1 α(x j ) + x ς j (x) + 1 ψ 3/2 ij (x) x (1 + A ). Recallig (4.7), it is easy to check that all coditios of Lemma 5.4 are satisfied with s 1/3, V = 3/5, ad ξ 1 (X j ) = α(x j ), ξ 2 (X j ) = 1 2 γ 1(X j ), ξ 3 (X j ) = sγ1(x 2 j ), ( ϕ 1 (X i, X j ) = β(x i, X j ), ϕ 2 (X i, X j ) = 2sγ 1 (X i )γ 1 (X j ) + 1 ) 2 γ 2(X i, X j ). ). 18
Therefore, (4.8) follows immediately from Lemma 5.4 ad the well-kow result: Eα(X 1 )γ 1 (X 1 ) = EE [α(x 1 )γ(x 1, X 2 ) X 1 ] = Eα(X 1 )γ(x 1, X 2 ). Proof of Theorem 2.1 is thus complete. Proof of Theorem 2.2. We first reduce the statistic S 2 to the form of (2.2). To do this, let us itroduce symmetric fuctio η(x, y, z) = η(x, y, z) + η(x, z, y) + η(y, x, z) +η(y, z, x) + η(z, x, y) + η(z, y, x), ad kerels η (1), η (2), ad η (3) defied recursively by the equatios η (1) (x 1 ) = E η(x 1, X 2, X 3 ), η (2) (x 1, x 2 ) = 2 E η(x 1, x 2, X 3 ) η (1) (x j ), η (3) (x 1, x 2, x 3 ) = 3 η(x 1, x 2, x 3 ) η (1) (x j ) η (2) (x i, x j ). 13 By applyig Hoeffdig s decompositio for U-statistics (see, e.g., Lee (1990, page 25)), it is easy to show that 6 η(x i, X j, X k ) ( 1)( 2) i j k ( ) 1 = η(x i, X j, X k ) 3 <k = 3 ( ) 1 η (1) (X i ) + 3 η (2) (X i, X j ) + H (3) 2 = i=1 6 ( 1) where γ(x i, X j ) = η (2) (X i, X j ) + 1 2 ( η(1) (X i ) + η (1) (X j )) ad H (3) = I terms of (4.9), we ca rewrite S 2 as where S 2 = 1 + γ(x i, X j ) + H (3), (4.9) ( ) 1 η (3) (X i, X j, X k ). 3 <k 1 ( 1) V2 3 2 = V 2 3 ( 1)( 2) 19 <k γ(x i, X j ) + V 2, η(x i, X j, X k ) + 1 6 H(3).
Similarly, we get T = 1/2 α(x j ) + 3/2 β(x i, X j ) + V 1, where β(x, y) = ζ(x, y) = ζ(x, y)+ζ(y, x), which is a symmetric fuctio of its argumets. By Jese s iequality, it is easy to show that E γ(x 1, X 2 ) 3/2 AE η(x 1, X 2, X 3 ) 3/2 A 1 E η(x 1, X 2, X 3 ) 3/2 <, E η (3) (X 1, X 2, X 3 ) 3/2 AE η(x 1, X 2, X 3 ) 3/2 A 1 E η(x 1, X 2, X 3 ) 3/2 <. Hece, by otig ( ) 1 3 <k η(x i, X j, X k ) is a degeerate U-statistic of order 0 ad is a degeerate U-statistic of order 2, it follows from Lemma 5.2 that H (3) Therefore, from Theorem 2.1 ad P ( V 2 o( 1/2 )) = o( 1/2 ). Eα(X 1 )γ(x 1, X 2 ) = 1 2 Eα(X 1) η(x 1, X 2, X 3 ), we have the desired result. The proof of Theorem 2.2 is complete. 5 Appedix I this sectio, we give several Lemmas that are commetary of the proofs of mai results. Let X 1, X 2,, be i.i.d. radom variables ad g k (x 1,, x k ) be symmetric Borelmeasurable real fuctio of its argumets. Defie U (g k ) = ( ) 1 g k (X i1,, X ik ). k 1i 1 < <i k U (g k ) is called a degeerate U-statistic of order m if Eg k (x 1,, x m, X m+1,, X k ) = 0. Lemma 5.1. Assume that U (g k ) is a degeerate U-statistic of order m ad E g k (X 1,, X k ) p <. The, E U (g k ) p C (m+1)(1 p), for 1 p 2, E U (g k ) p C 1 (m+1)p/2, for p 2. 20
For a proof of Lemma 5.1, see Theorems 2.1.3 ad 2.1.4 i Koroljuk ad Borovskich (1994). Lemma 5.2. Assume that U (g k ) is a degeerate U-statistic of order m. (a) If E g k (X 1,, X k ) 2 <, the P ( U (g k ) C 1 1/2) = o( 1/2 ), for m 1 (b) If E g k (X 1,, X k ) 3/2 <, the P ( U (g k ) C 2 1/2 (log ) 1) = o( 1/2 ), for m = 0, P ( U (g k ) C 3 3/10) = o( 1/2 ), for m = 1, P ( U (g k ) C 4 1/2 (log ) 1) = o( 1/2 ), for m = 2. (c) If E g k (X 1,, X k ) 3 <, the for all m 0 ad k > 0, P ( U (g k ) C 5 1/4 (log ) k) = o( 1/2 ). By applyig Markov s iequality ad Lemma 5.1, proof of Lemma 5.2 is simple ad will be omitted. Lemma 5.3. If E X 1 <, the there exists a δ 0 such that P 1 3/2 X j δ = o( 1/2 ). (5.1) If EX 1 = 0 ad E X 1 3/2 <, the there exists a δ 0 such that P 1 X j δ = o( 1/2 ). (5.2) If E X 1 3/4 <, the P 1 2 X j C 1 = o( 1/2 ). (5.3) If EX 1 = 0, EX 2 1 = 1 ad E X 1 3 <, the P x 2 10 log 1 X j x = o( 1/2 ). (5.4) 3 21
Proof. To prove (5.1), we let δ 3 = max { E X 1 I ( X1 1/4 ), 2 1/2 E X 1 }. Uder the coditio that E X 1 <, it follows that δ 0 ad P 1 X 2 j δ 1/2 P ( X 1 3/2 ) + P 1 ( ) 3/2 Xj I ( Xj 3/2 ) EX j I ( Xj 3/2 ) δ 1/2 E X 1 1/2 E X 1 I ( X1 3/2 ) + 8 ( EX 2 2 δ 2 1 I ( X1 1/4 ) + EX1I 2 ( 1/4 < X 1 )) 3/2 o( 1/2 8 ) + (E X 1 ) 2 EX2 1I ( X1 1/4 ) + 8δ 1/2 = o ( 1/2). This implies (5.1). Similarly, we ca prove (5.2) ad (5.3). To prove (5.4), by usig o-uiform Berry-Essée theorem for sums of idepedet radom variables (cf. Chapter V of Petrov, 1975) x (1 + x 3 ) P 1 X j x Φ(x) A 1/2 E X 1 3 (5.5) ad 1 Φ(x) 1 2π e x2 /2 2, for x 1, (5.6) 1 + x 3 we have that for 3 ad x 2 10 log, P 1 X j x /3 1 Φ( x /3) + C 1/2 (log ) 1 = o( 1/2 ). Similarly, we have that for 3 ad x 2 10 log, P 1 X j x /3 Φ( x /3) + C 1/2 (log ) 1 = o( 1/2 ). Therefore, (5.4) follows. The proof of Lemma 5.3 is complete. Next we preset Lemma 5.4, which is of its ow idepedet iterest. Lemma 5.4. Let ξ j (x) ad ϕ j (x, y) be real Borel measurable fuctios ad symmetric i their argumets. Let V V (X 1,..., X ) be a real Borel-measurable fuctio of X 1,..., X. Assume that (D1) Eξ 1 (X 1 ) = 0, Eξ 1 (X 1 ) 2 = 1, E ξ 1 (X 1 ) 3 <, Ee lim itξ 1 (X 1 ) t < 1. (D2) Eξ 2 (X 1 ) = 0, E ξ 2 (X 1 ) 3/2 <, E ξ 3 (X 1 ) 3/4 < ; (D3) E(ϕ j (X 1, X 2 ) X 1 ) = 0, j = 1, 2; 22
E ϕ 1 (X 1, X 2 ) 2 <, E ϕ 2 (X 1, X 2 ) 3/2 <. (D4) P ( V o( 1/2 )) = o( 1/2 ). The, as, where x P (K (x) x(1 + V )) F (3) (x) = o( 1/2 ), (5.7) ς j = ξ 2 (X j ) + 1 ξ 3(X j ), ψ ij (x) = ϕ 1 (X i, X j ) + x ϕ 2 (X i, X j ); K (x) = 1 ξ 1 (X j ) + x ς j + 1 ψ 3/2 ij (x) F (3) (x) = Φ(x) Φ(3) (x) 6 ( Eξ 3 1 (X 1 ) + 3Eξ 1 (X 1 )ξ 1 (X 2 )ϕ 1 (X 1, X 2 ) ) + xφ(2) (x) Eξ 1 (X 1 )ξ 2 (X 1 ). Proof. Without loss of geerality, we assume that For, if ot, we ca defie The we have ϕ 2 (X i, X j ) 4 2 for all i, j. (5.8) ϕ 3 (X i, X j ) = ϕ 2 (X i, X j )I ( ϕ2 2 ) Eϕ 2 (X i, X j )I ( ϕ2 2 ), ϕ 4 (X i, X j ) = ϕ 3 (X i, X j ) E(ϕ 3 (X i, X j ) X i ) E(ϕ 3 (X i, X j ) X j ), ψ ij (x) = ϕ 1 (X i, X j ) + x ϕ 4 (X i, X j ). 1 ψ 3/2 ij (x) = 1 3/2 ψ ij (x) + xr, say. Write δ 2 = E ϕ 2 (X 1, X 2 ) 3/2 I ( ϕ2 2 ). By Markov s iequality ad usig the assumptio E [ϕ 2 (X 1, X 2 ) X 1 ] = 0, we get P ( R δ / ) = P 1 [ϕ 2 2 (X i, X j ) ϕ 4 (X i, X j )] δ / 4 δ 1 E ϕ 2 (X 1, X 2 ) I ( ϕ2 2 ) (4 δ 1 ) 1 E ϕ 2 (X 1, X 2 ) 3/2 I ( ϕ2 2 ) 4δ 1/2. 23
It is easy to show that δ satisfyig 0 ad that ϕ 4 (x, y) is a real-valued symmetric fuctio E(ϕ 4 (X 1, X 2 ) X 1 ) = 0, E ϕ 4 (X 1, X 2 ) 3/2 <, ϕ 4 (X i, X j ) 4 2. Therefore, if (5.8) does ot hold, we ca replace ϕ 2 (X i, X j ) by ϕ 4 (X i, X j ), ad V by V R i the proof. I view of (4.1) ad (4.2), we may further assume V = 0 for coveiece. It is obvious that this assumptio does ot affect the proof of the mai results. We tur back to the proof of the mai result. Write The we have ξ 2(X j ) = ξ 2 (X j )I ( ξ2 (X j ) /(1+x 2 )), ξ 3(X j ) = ξ 3 (X j )I ( ξ3 (X j ) 2 /(1+x 2 )), ς j = ξ 2(X j ) + 1 ξ 3(X j ), K(x) = 1 ξ 1 (X j ) + x ς j + 1 ψ 3/2 ij (x). P (K (x) x) F (3) (x) x P (K (x) x) F (3) (x) + P (K(x) x) F (3) (x) x 2 10 log x 2 10 log + x 2 10 log P (K(x) x) P (K (x) x) P 1 + P 2 + P 3, say, where P 1 = P (K (x) x) F (3) (x), (5.9) P 2 = x 2 10 log P (K(x) x) F (3) (x), (5.10) x 2 10 log P 3 = x 2 10 log P (K(x) x) P (K (x) x). (5.11) Therefore, it suffices to show P i = o( 1/2 ), i = 1, 2, 3. (5.12) First, we shall prove P 1 = o( 1/2 ). Note that P (K (x) x) F (3) (x) x (10 log ) 1/2 24
P (K (x) x) + 1 F (3) (x) x (10 log ) 1/2 x (10 log ) 1/2 P 1 ξ 1 (X j ) x + P 1 ς j 1 x (10 log ) 1/2 3 3 + P 1 ψ x 1 3/2 ij (x) x + 1 F (3) (x) 3 x (10 log ) 1/2 P 11 + P 12 + P 13 + P 14. Usig Lemmas 2-3 ad the iequality (5.6), we have P 11 = o( 1/2 ); P 12 P 1 ξ 2 (X j ) 1 + P 1 6 ξ 2 3 (X j ) 1 = o( 1/2 ); 6 P 13 P 1 3/2 ϕ 1 (X i, X j ) 1 + P 1 ϕ 6 2 2 (X i, X j ) 1 = o( 1/2 ); 6 P 14 = o( 1/2 ). This implies that x (10 log ) 1/2 P (K (x) x) F (3) (x) = o( 1/2 ). Similarly, we ca show that Thus, P 1 = o( 1/2 ). x (10 log ) 1/2 P (K (x) x) F (3) (x) = o( 1/2 ). where Secodly, let us show that P 3 = o( 1/2 ). We write P 3 = x 2 10 log P (K(x) x) P (K (x) x) Ω 1 + Ω 2 + Ω 3, Ω 0 = P (K(x) x) P (K (x) x), x 2 1 Ω 1 = P ( K (x) x, ς j ςj, for some j ), 1x 2 10 log Ω 2 = P ( K(x) x, ς j ςj, for some j ). 1x 2 10 log We shall show ow that Ω i = o( 1/2 ), i = 0, 1, 2. 25
First cosider the term Ω 0. It is easy to see that P ( ς j ς j It follows from (5.13) that ) P ( ξ2 (X 1 ) /(1 + x 2 ) ) + P ( ξ 3 (X 1 ) 2 /(1 + x 2 ) ) 1 + x 3/2 + x 3 3/2 ( E ξ2 (X 1 ) 3/2 I ( ξ2 (X 1 ) /(1+x 2 )) + E ξ 3 (X 1 ) 3/4 I ( ξ3 (X 1 ) 2 /(1+x 2 ))). (5.13) Ω 0 P (ς j ςj) = o( 1/2 ). (5.14) x 2 1 Next we ivestigate the term Ω 1. Without loss of geerality, we assume that x 1. The i view of (5.13) ad idepedece of X k, we obtai ( 1 P 1x 2 10 log 1x 2 10 log P = o( 1/2 ) + = o( 1/2 ), P k=1 ( 1 ξ 1 (X k ) x/3, ς j ς j, k=1 ) ( 1 ξ 1 (X j ) 1/6 1x 2 10 log ξ 1 (X k ) x/3, ς j ς j P + 1x 2 10 log 1 k=1 k j P for some j ) 1 k=1 k j ) ξ 1 (X k ) x/6 P ( ) ς j ςj where we have used the followig estimate: for all 1 j, P 1 ξ 1 (X k ) x/6 Cx 3. k=1 k j From this ad P 12 = o( 1/2 ) ad P 13 = o( 1/2 ), we get Ω 1 P 1 ψ x 2 1 3/2 ij (x) x/3 + P 1 ς j 1/3 ( 1 + P 1x 2 10 log k=1 = P 12 + P 13 + P 1x 2 10 log = o( 1/2 ). Similarly, we have that ξ 1 (X k ) x/3, ς j ς j, ( 1 k=1 ξ 1 (X k ) x/6, ς j ςj for some j ξ 1 (X k ) x/3, ς j ς j, ) for some j ) (5.15) Ω 2 = o( 1/2 ). (5.16) 26
Combiig (5.14)-(5.16), we have show that P 3 = o( 1/2 ). Fially, we shall prove P 2 = o( 1/2 ). Write ad defie Y j (x) = ξ 1 (X j ) + x ( ς j Eς j), σ(x) 2 = EY1(x), 2 θ (x) = x σ (x) (1 Eς 1). { ( L (y) = EΦ y Y ) } 1(x) Φ(y) 1 σ (x) 2 Φ(2) (y), K (x) = 1 1 σ (x) Y j(x) + 1 3/2 1 σ (x) ψ ij(x), E (y) = Φ(y) + L (y) Φ(3) (y) 2 σ 3 (x) EY 1(x)Y 2 (x)ψ 12 (x), E (y) = Φ(y) Φ(3) (y) 6 σ 3 (x) The we have P (K(x) x) F (3) (x) x 2 10 log = P ( K (x) θ (x) ) F (3) (x) x 2 10 log P ( K (x) y ) E (y) + x 2 10 log y + x 2 10 log E(θ (x)) F (3) = I 1 + I 2 + I 3, say. Therefore, (5.10) follows if ( EY 3 1 (x) + 3EY 1 (x)y 2 (x)ψ 12 (x) ). (x) x 2 10 log E (y) E(y) y I j = o( 1/2 ), j = 1, 2, 3. (5.17) Next we tur to the proof of (5.17). Uder the coditio (D2), we obtai that for all x 2 10 log, Eξ 2(X 1 ) E ξ 2 (X 1 ) I ( ξ2 (X 1 ) /(1+x 2 )) = o ( ) 1 + x, (5.18) ( ) α 3/2 E ξ2(x 1 ) α E ξ 2 (X 1 ) α I ( ξ2 (X 1 ) ) + E ξ2 (X 1 + x 2 1 ) 3/2 I ( ξ2(x1) ) ( ) α 3/2 = o, for α > 3/2. (5.19) 1 + x 2 Similarly, we have E ξ 3(X 1 ) α = o ( 2 1 + x 2 ) α 3/4, for α > 3/4. (5.20) 27
Recallig ς j = ξ 2(X j ) + 1 ξ 3(X j ), it follows from (5.18)-(5.20) that for all x 2 10 log, Eς 1 = Eξ 2(X 1 ) + 1 ( ) Eξ 3(X 1 ) 1 + x = o, (5.21) E ς1 E ξ2(x 1 ) + 1 E ξ 3(X 1 ) = O(1), (5.22) ( ) 2 ( ) 2 ( x x E(ς 1) 2 2 E(ξ 2(X 1 )) 2 + 1 ) ( ) 1 + x 2 E(ξ 3(X 1 )) 2 = o,(5.23) ( ) 3 ( ) 3 ( x x E ς 1 3 6 E ξ 2(X 1 ) 3 + 1 ) 3 E ξ 3(X 1 ) 3 = o(1), (5.24) By usig (5.21) - (5.24), together with Hölder s iequality, we get if x 2 10 log, the σ(x) 2 = 1 + 2x ( ) 1 + x Eξ 1 (X 1 )ξ 2 (X 1 ) + o, (5.25) 1 σ (x) = 1 x ( ) 1 + x Eξ 1 (X 1 )ξ 2 (X 1 ) + o, (5.26) EY 3 1(x) = Eξ 1 (X 1 ) 3 + o(1), (5.27) E Y 1 (x) 3 = O(1), (5.28) EY 1 (x)y 2 (x)ψ 12 (x) = Eξ 1 (X 1 )ξ 1 (X 2 )ϕ 1 (X 1, X 2 ) + o(1). (5.29) We oly check (5.29) i details. I fact, let µ j (x) = x ( ς j Eς j), the EY 1 (x)y 2 (x)ψ 12 (x) = EY 1 (x)y 2 (x)ϕ 1 (X 1, X 2 ) + x EY 1 (x)y 2 (x)ϕ 2 (X 1, X 2 ) = Eξ 1 (X 1 )ξ 1 (X 2 )ϕ 1 (X 1, X 2 ) + E {ξ 1 (X 1 )µ 2 (x) + µ 1 (x)y 2 (x)} ϕ 1 (X 1, X 2 ) + x EY 1 (x)y 2 (x)ϕ 2 (X 1, X 2 ) It follows from (5.24) that E µ 1 (x) 3 = o(1). Therefore, by otig idepedece of X k, (5.29) follows from (5.28) ad the followig estimates: E {ξ 1 (X 1 )µ 2 (x) + µ 1 (x)y 2 (x)} ϕ 1 (X 1, X 2 ) 3 ( E µ 1 (x) 3) 1/3 ( E ξ 1 (X 1 ) 3 + E Y 1 (x) 3) 1/3 ( E ϕ 1 (X 1, X 2 ) 3/2) 2/3 = o(1), E Y 1 (x)y 2 (x)ϕ 2 (X 1, X 2 ) ( E Y 1 (x) 3) 2/3 ( E ϕ 2 (X 1, X 2 ) 3/2) 2/3 = O(1). We tur back to the proof of (5.17). We first prove (5.17) for j = 3. Recallig defiitio of E(y) ad F (3) (x), it suffices to show Φ(θ (x)) Φ(x) xφ(2) (x) Eξ 1 (X 1 )ξ 1 (X 2 ) = o( 1/2 ) (5.30) x 2 10 log 28
Φ (3) (θ (x)) Φ (3) (x) = O( 1/2 ) (5.31) x 2 10 log EY 3 1(x) Eξ x 2 10 log σ(x) 3 1 (X 1 ) 3 = o(1) (5.32) EY 1 (x)y 2 (x)ψ 12 (x) Eξ x 2 10 log σ(x) 3 1 (X 1 )ξ 1 (X 2 )ϕ 1 (X 1, X 2 ) = o(1) (5.33) We oly check (5.30). Others are simple by usig (5.25)-(5.29). Recallig θ (x) = x 2 10 log, θ (x) = x ad hece for sufficietly large, x (1 σ Eς (x) 1), by usig (5.21) ad (5.26), we get that for all ( 1 x ) Eξ 1 (X 1 )ξ 1 (X 2 ) x/2 θ (x) 3x/2. + o ( 1 + x 2 ) From these estimates ad Taylor s expasio, it follows that there exists 1/2 δ 3/2 such that for all x 2 10 log, Φ(θ (x)) = Φ(x) + (θ (x) x)φ(x) + (θ (x) x) 2 Φ (2) (δx) 2 = Φ(x) x2 φ(x) Eξ 1 (X 1 )ξ 2 (X 1 ) + o( 1/2 )f(x)φ(x/2) where f(x) is a polyomial of x. By otig Φ (2) (x) = xφ(x), (5.30) follows easily. We fiish the proof of (5.17) for j = 3. We use Lemma 5.5 to prove (5.17) for j = 1. I terms of (5.22) ad (5.26), we have that for ay fixed β > 0, Y1(x) Eeit x 2 10 log t β Y 1 (x) E x 2 10 log ( 1 + β β x 2 10 log σ(x) Ee itξ 1(X 1 ) σ (x) ξ 1(X 1 ) σ (x) 1 x E ς 1 Eς 1 ) = o(1). (5.34) Notig that lim t Ee itξ 1 (X 1 ) < 1, it follows from (5.34) that for all sufficietly large ad all x 2 10 log, there exists a positive absolute costat δ such that lim t Ee it Y 1 (x) σ(x) < 1 δ. Therefore, by usig Lemma 5.5 with V (X j ) = Y j(x) σ (x), W (X i, X j ) = ψ ij(x) σ (x), 29
we obtai that for all sufficietly large ad all x 2 10 log (recallig Remark 5.1), P ( K (x) y ) E (y) y A δ 1 ( E ψ12 (x) 2 σ 2 (x) + E Y j(x) 3 ) 2/3 (E Y j (x) 3 ) 2 log + A σ(x) 3 1 1. (5.35) σ(x) 6 Recallig (5.8), we have ϕ 2 (X 1, X 2 ) 4 2 ad hece ( ) 2 x E ψ 12 (x) 2 2Eϕ 2 1(X 1, X 2 ) + 2 Eϕ 2 2(X 1, X 2 ) 2Eϕ 2 1(X 1, X 2 ) + 8x 2 E ϕ 2 (X 1, X 2 ) 3/2. (5.36) I terms of (5.25), (5.28), (5.35) ad (5.36), simple calculatio shows that I 1 = P ( K (x) y ) E (y) = o( 1/2 ). x 2 10 log y This fiishes the proof of (5.17) for j = 1. To prove (5.17) for j = 2, we ote that y = y E (y) E(y) { ( EΦ y Y ) } 1(x) Φ(y) σ (x) 6 σ(x) 3 Φ(3) (y) 1 2 Φ(2) (y) + EY 3 1(x) A 1 σ 3 (x) E Y 1(x) 3 I ( Y1 (x) σ (x)) + A 2 σ 4 (x) E Y 1(x) 4 I ( Y1 (x) σ (x)), where the last iequality follows from Theorem 3.2 of Hall (1982). It follows from (5.25) that for sufficietly large ad all x 2 10 log, 1/2 < σ (x) < 3/2. (5.37) It follows from (5.21) that for sufficietly large ad all x 2 10 log, Y 1 (x) = ξ 1(X 1 ) + x (ς 1 Eς1) 1 + ξ 1 (X 1 ) + ξ 2 (X 1 ) 1/2 + ξ 3 (X 1 ) 1/4 = κ(x 1 ), say. Notig (5.37) ad Eκ 3 (X 1 ) <, we get for sufficietly large, I 2 = E (y) E(y) x 2 10 log y ( C 1/2 x 2 10 log ( C 1/2 Eκ 3 (X 1 )I (κ(x1 ) 1/4 ) + 1 = o( 1/2 ). E Y 1 (x) 3 I ( Y1 (x) 1/4 ) + 1 E Y 1 (x) 4 I ( Y1 (x) 1/4 ) ) 1/4 Eκ3 (X 1 ) 30 )
This implies (5.17) for j = 2. Now, we fiish the proof of (5.10). Proof of Lemma 5.4 ow is complete. Lemma 5.5. Let V (x) ad W (x, y) be real Borel-measurable fuctios ad W (x, y) be symmetric i its argumets. Assume that there exists a 0 such that for all 0, The, for all 0, x (D1) EV (X 1 ) = 0, EV 2 (X 1 ) = 1, τ > 0, (D2) E[W (X 1, X 2 ) X 1 ] = 0. P 1 V (X j ) + 1 W 3/2 (X i, X j ) x E (x) A τ 1 (λ + ρ ) 2/3 log + A 1 (ρ ) 2 1. (5.38) where A ad A 1 are absolute costats, λ = EW 2 (X 1, X 2 ), ρ = E V (X 1 ) 3, τ = 1 { Ee itv (X 1 ) : 1 /(4ρ ) t 1/6}, E (x) = Φ(x) + L (x) Φ(3) (x) 2 EV (X 1 )V (X 2 )W (X 1, X 2 ), { ( L (x) = EΦ x 1 ) } V (X 1 ) Φ(x) 1 2 Φ(2) (x). Remark 5.1. If the distributio of V (X 1 ) is o-lattice ad the Cramér coditio that τ lim Ee itv (X 1 ) t < 1 is satisfied, the τ > 0. It is clear that we ca replace τ by 1 τ i (5.38) if τ < 1. O the other had, we ote that the result give i Lemma 5.5 is ot optimal, but it is eough for our propose. This result ca ot be obtaied from Betkus, Götze ad va Zwet (1997). Proof. Without loss of geerality, we assume λ < ad ρ <. Write γ(t) = Ee itv(x 1)/, f (t) = 1 + (γ(t) 1) + t2 e t2 /2, 2 ϕ (t) = t 2 B e t2 /2, B = 1 2 EV (X 1 )V (X 2 )W (X 1, X 2 ) L (x) = B Φ (3) (x), S m = 1 m V (X j ),,m = 1 m 1 3/2 [ i=1 j=i+1 ] W (X i, X j ). 31
Simple calculatio shows that e itx d (Φ(x) + L (x)) = f (t), e itx dl (x) = itϕ (t). (5.39) From (5.39) ad Essee s smoothig lemma (Petrov, 1975), it follows that (otig ρ 1) P 1 V (X j ) + 1 W x 3/2 (X i, X j ) x E (x) 1 Ee it(s+,) f (t) itϕ (t) dt + A 2/3 de (x) t 2/3 t x dx 4 I j + A 1 2/3 (λ + ρ ), (5.40) where I 1 = I 2 = I 3 = I 4 = 1 t 1/10 1 t 1/10 t 1/10 Ee it(s+,) Ee its ite, e its dt, t Ee its f (t) dt, t E, e its ϕ (t) dt, 1/10 t 2/3 1 t Ee it(s+,) dt. I the followig, we estimate the each term of (5.40). First we estimate I 1. I terms of the coditio (D2), it ca be easily show that E 2,m A 2 mλ. (5.41) Thus, we have I 1 1 t E ( ) 2 2 t 1/10, dt A 2/3 λ. Secodly we estimate I 2. Usig similar argumets to the proof of Lemma 5.2.1 i Hall (1982), we have I 2 A ( (EV 2 (X 1 ) ) 2 + ( E V (X 1 ) 3) 2 ) A 1 (ρ ) 2. Thirdly we estimate I 3. Similar to the proof of Bickel et al (1986), we fid that EW (X i, X j )e it(v(x i)+v (X j ))/ = t2 EV (X 1 )V (X 2 )W (X 1, X 2 ) + θ (1) ij (t) (5.42) 32
where ( otig e ix 1 ix 2 x 3/2 ad e ix 1 x ) θ (1) ij (t) 2 4 ( EW (X i, X j ) e itv(x i)/ 1 itv ) (X i ) (e itv (X j )/ 1 ) + t ( EW (X i, X j )V (X i ) e itv(x j)/ 1 itv ) (X j ) ( t ) 5/2 { E W (X 1, X 2 ) V (X 1 ) 3/2 V (X 2 ) +E W (X 1, X 2 ) V (X 1 ) V (X 2 ) 3/2} ( ) 5/2 ( ) 5/2 t t (λ ρ ) 1/2 8 (λ + ρ ). Together (5.42) with the followig relatios (see Petrov, 1975) γ(t) e t2 /3, γ (t) e t2 /2 16ρ t 3 e t2 /3 (5.43) for t /(4ρ ), we obtai that if t 1/10 ad is sufficietly large, where EW (X i, X j )e its = γ 2 (t)ew (X i, X j )e it(v(x i)+v (X j ))/ = t2 γ 2 (t) EV (X 1 )V (X 2 )W (X 1, X 2 ) + θ (1) ij (t)γ 2 (t) = t2 e t2 /2 EV (X 1 )V (X 2 )W (X 1, X 2 ) + θ (2) ij (t) (5.44) θ (2) ij (t) θ (1) ij (t) e t2 /4 + λ1/2 t 2 γ 2 (t) e t2 /2 A ( 5/4 (λ + ρ ) + 3/2 λ 1/2 ρ ) ( 1 + t 6 ) e t2 /4. From (5.44), it follows that I 3 A ( 3/4 (λ + ρ ) + 1 λ 1/2 ρ ) 2A ( 3/4 (λ + ρ ) + 1 (ρ ) 2). Fially we estimate I 4. Put I view of (5.41), we have that,m =,,m = 1 1 3/2 i=m j=i+1 W (X i, X j ). Ee it(s+,) Ee it(s+,m ) ite,m e it(s+,m) A t 2 2 mλ. 33
This iequality, together with idepedece of X k, implies that for ay 1 m, Ee it(s+,) γ m 1 (t) + A 1 1/2 λ 1/2 t γ m 2 (t) + A 2 t 2 2 mλ, (5.45) where we use the estimate: E,m (E,m 2 ) 1/2. Choosig m = [ ] 6 log t + 2, it follows from (5.45) ad (5.43) that 2 1/10 t /(4ρ ) 1 Ee it(s+,) dt A 2/3 (1 + λ ). (5.46) t O the other had, otig that τ > 0, it follows that whe /(4ρ ) t 2/3, γ(t) 1 τ e τ. (5.47) I this case, choosig m = [ ] 4 log τ + 2, it follows from (5.45) ad (5.47) that 1 Ee it(s+,) dt A τ 1 /(4ρ) t 2/3 2/3 (log ) λ. (5.48) t Substitutig the above estimate for I j s ito (5.40), we get the required result. We fiish the proof of Lemma 5.5. Lemma 5.6. Assume that [F (t)(1 F (t)] 1/3 dt <. The we have P ( Q j 1 (log ) 1) = o( 1/2 ), for j = 1, 2, (5.49) P ( Q j 1/2 (log ) 1) = o( 1/2 ), for j = 3, 4, 5, (5.50) where Q j for j = 1,, 5 are defied as i (3.3) (3.7). Proof. (i). First cosider Q 1. Sice ζ(x i, X i ) = (2σ) 1 J 0(t) [ I (Xi t) F (t) ] 2 dt, the E ζ(x i, X i ) 3 1 ( 8 σ 3 J 0(t) 3 E (I(Xit) F (t) ) ) 2 3 dt t [ C(J) E (I(Xis) F (s) ) 2 ( I (Xit) F (t) ) 2 ( I (Xi v) F (v) ) 2 ] dsdtdv C(J) ( [ E ( I (Xit) F (t) ) 6 ] 1/3 dt ) 3 C(J) ( = C(J) C. ( [ E ( I (Xit) F (t) ) 2 ] 1/3 dt ) 3 ) 3 F 1/3 (t) (1 F (t)) 1/3 dt 34
Applyig Lemma 5.1 i the Appedix with p = 3, k = 1 ad m = 0, we get E Q 1 3 C 3/2. Therefore, P ( Q 1 1 (log ) 1) (log ) 3 E Q 1 3 C 3/2 (log ) 3 = o( 1/2 ). (ii). Let us deal with the term Q 2. Note that for k 2, we have It follows that E F (t) F (t) k A k/2 E I{X 1 t} F (t) k P ( Q 2 1 (log ) 1) ( log ) 2 E Q 2 2 A k/2 E I{X 1 t} F (t) 2 = A k/2 F (t)(1 F (t)). (5.51) ( ) 2 = C( log ) 2 σ 2 E F (t) F (t) 3 dt = C( log ) 2 σ 2 E { F (s) F (s) 3 F (t) F (t) 3} dsdt C( log ) 2 σ 2 ( (E F (s) F (s) 6) 1/2 ds ) 2 C( log ) 2 3 σ 2 ( = o( 1/2 ). ) 2 F 1/2 (s)(1 F (s)) 1/2 ds (iii). Next we cosider the term Q 3. I view of the iequality Z(s, t, F ) [F (s)(1 F (s))] 1/2 [F (t)(1 F (t))] 1/2, (5.52) we have Q 3 σ 2 J 0 (x)j 0 (y) (F (s) F (s)) 2 Z(s, t, F )dsdt x,y A(J)σ 2 (F (s) F (s)) 2 ds F 1/2 (t)(1 F (t)) 1/2 dt = A(J)σ 2 Q 6 Q 7. (5.53) where Q 6 = (F (s) F (s)) 2 ds, Q 7 = F 1/2 (t)(1 F (t)) 1/2 dt. It follows from this ad (5.51) that ( ) 3 EQ 3 6 = E (F (s) F (s)) 2 ds = E { (F (s) F (s)) 2 (F (t) F (t)) 2 (F (v) F (v)) 2} dsdtdv ( (E F (s) F (s) 6) ) 1/3 3 ds A 3 ( F 1/3 (s)(1 F (s)) 1/3 ds) 3. (5.54) 35
Similarly, we have ( EQ 3 7 = E = ) 3 F 1/2 (t)(1 F (t)) 1/2 dt E { F 1/2 (s)(1 F (s)) 1/2 F 1/2 (t)(1 F (t)) 1/2 F 1/2 (v)(1 F (v)) 1/2} dsdtdv (E { F 3/2 (s)(1 F (s)) 3/2}) 1/3 ( E { F 3/2 (t)(1 F (t)) 3/2}) 1/3 = ( E { F 3/2 (v)(1 F (v)) 3/2}) 1/3 dsdtdv ( (E { F 3/2 (t)(1 F (t)) 3/2}) ) 1/3 3 dt ( ) 3 (E {F (t)(1 F (t))}) 1/3 dt ( F 1/3 (s)(1 F (s)) 1/3 ds) 3, (5.55) where i the secod last step, we have used the iequality E[F (t)(1 F (t))] F (t)(1 F (t)). Combiig (5.53) (5.55) ad applyig Markov s iequality, we have P ( Q 3 1/2 (log ) 1) ( 1/2 log ) 3/2 E Q3 3/2 A 3/2 (J)σ ( 3 1/2 log ) 3/2 ( ) EQ 3 1/2 ( ) 6 EQ 3 1/2 7 ( ) 3 C 3/4 (log ) 3/2 [F (t)(1 F (t)] 1/3 dt = o( 1/2 ). (5.56) get (iv). Next we cosider the term Q 4. Usig (5.52) ad Cauchy-Swartze iequality, we Q 4 = σ 2 [J (s) J 0 (s)] [J (t) J 0 (t)] Z(s, t, F )dsdt, σ 2 J 0(x) 2 F (s) F (s) F (t) F (t) Z(s, t, F )dsdt x B(J)σ 2 F (s) F (s) F (t) F (t) [F (s)(1 F (s))] 1/2 [F (t)(1 F (t))] 1/2 dsdt ( ) 2 = B(J)σ 2 F (t) F (t) [F (t)(1 F (t))] 1/2 dsdt ( ) ( ) B(J)σ 2 (F (s) F (s)) 2 ds F (t)(1 F (t))dt ( ) ( ) B(J)σ 2 (F (s) F (s)) 2 ds F 1/2 (t)(1 F (t)) 1/2 dt = B(J)σ 2 Q 6 Q 7, 36