A Berry-Esseen bound for U-Statistics

Journal of Mathematical Sciences, Vol. 128, No. 1, 2005

A BERRY–ESSEEN BOUND FOR U -STATISTICS L. V. Gadasina

UDC 519.21

The rate of convergence in the central limit theorem for nondegenerate U -statistics of n independent random variables is investigated under minimal sufficient moment conditions on canonical functions of the Hoeffding representation. Bibliography: 12 titles.

1. Introduction Let X1 , . . . , Xn be independent random variables with values from a measurable space (X, B). Assume that n ≥ m ≥ 2 and consider an U -statistic
Un = Un (X1 , . . . , Xn ) = Φi1 ...im (Xi1 , . . . , Xim ), 1≤i1 0, then ∆n ≤ Am ( σn−2

n


Egj2 I(|gj | > σn ) + σn−3

j=1

+σn−1

m





E|gj |3 I(|gj | ≤ σn )

j=1

E|gIc |I(|gIc | > σn ) +

c=2 1≤i1 σn )

c=2 Iˇc

j=1

and

m



(2)

E|zj |3 )−1 . Without loss of generality, we may assume that h > 2m+3 . By the Esseen

j=1

inequality,

 2 2 h |Eeit(sn+vn +wn ) − e−t /2 |t−1 dt sup |P (sn + vn + wn < x) − Φ(x)| ≤ π x 0  h  2m+3 2 24 2 2 + √ h−1 ≤ |Eeitsn − e−t /2 |t−1 dt |Eeit(sn+vn +wn ) − Eeitsn |t−1 dt π π π 2π 0 0  2 h 24 24 + |Eeit(sn+vn +wn ) − Eeitsn |t−1 dt + √ h−1 = ε1 + ε2 + ε3 + √ h−1 . π 2m+3 π 2π π 2π To estimate ε1 , we use [6, Chapter XVI.5, Theorem 2] with T = h−1 instead of T = 8h−1 /9 as follows: ε1 ≤ π −1 σ ¯n−3

n
j=1

E|gjn +

m


hcjn |3 ≤≤ 16π −1 σ ¯n−3

c=2

+32π −1 (m − 1)2 m¯ σn−3 σn2

n


E|gjn |3

j=1 m



E|gIc |I(|gIc | > σn ).

c=2 Iˇc

The term containing h−1 can be estimated similarly. 2525

(k)

(k)

Proposition 1. Let γ1 , . . . , γn , k = 1, . . . , c, be a series of sequences consisting of independent identically (1) (c) distributed random variables such that γi1 , . . . , γic are independent for i1 = · · · = ic and, in addition, do not depend on X1 , . . . , Xn . Then E|




(1)

(c)

γi1 . . . γic gIc n |p ≤ 2(2−p)c

Iˇc

c 

(s)

E|γ1 |p

s=1




E|gIcn |p ,

1 ≤ p ≤ 2.

Iˇc

To prove this proposition, we need the following statement. Proposition 2. For any x ∈ R, the following inequalities hold: (1) |eix − 1| ≤ |x|; (2) |eix − 1| ≤ |x|p−1, 1 ≤ p ≤ 2; (3) |eix − 1 − ix| ≤ 2|x|p, 1 ≤ p ≤ 2; (4) a · b ≤ p−1 ap + q −1 bq , a > 0, b > 0, p−1 + q −1 = 1. Estimation of ε2 . We need the following lemma. Lemma 1. If 0 ≤ t ≤ h and 1 ≤ i1 < · · · < ic ≤ n for c = 1, . . . , m, then n 

|E exp(itzk )| ≤ exp(−t2 (1/3 − c2−(2m+9)/3))).

k=1,k=Ic

Proof. Expanding exp(itzk ) into the Taylor series, we see that 1 1 1 1 |E exp(itzk )| ≤ 1 − Ezk2 t2 + |t|3E|zk |3 ≤ 1 − t2 (Ezk2 − hE|zk |3 ) 2 6 2 3   1 1 ≤ exp − t2 (Ezk2 − hE|zk |3 ) , 2 3 where the last inequality follows from the inequality 1 − x ≤ e−x , which holds for all x ∈ R. Thus,   n 

1 1 |E exp(itzk )| ≤ exp − t2 ( Ezk2 − h E|zk |3 ) 2 3 k=1, k=Ic

k=Ic

k=Ic

  1 2 1 1 = exp − t2 ( − Ezi21 − · · · − Ezi2c + hE|zi1 |3 + · · · + hE|zic |3 ) . 2 3 3 3 In addition, E|zi |3 ≥ 0 and Ezi2 ≤ (E|zi |3 )2/3 ≤ h−2/3 ≤ 2−2(m+3)/3 . This proves the lemma. Consider the integrand in ε2 : Eeit(sn +vn +wn ) − Eeitsn = Eeitsn (eit(vn+wn ) − eitvn + eitvn − 1) = Eeit(sn +vn ) (eitwn − 1) + Eeitsn (eitvn − 1) = Eeit(sn+vn ) (eitwn − 1 − itwn ) + (itEeitsn wn + itEeitsn (eitvn − 1)wn + Eeitsn (eitvn − 1) = L1 + L2 + L3 + L4 . To estimate L1 , we apply the third inequality of Proposition 2 and then Proposition 1 as follows: |L1 | ≤ 2|t|pE|wn |p ≤ 2(2−p)m+1 (m − 1)p−1 σ¯n−p |t|p

m



E|gIcn |p ,

c=2 Iˇc

Now we estimate L2 : |L2 | ≤ σ ¯n−1 |t|

m

c=2 Iˇc

2526

|E exp(it


k=Ic

zk )eitzi1 . . . eitzic gIc n |

1 ≤ p ≤ 2.

=σ ¯n−1 |t|

m



|E exp(it

c=2 Iˇc




zk )| · |E(eitˆzi1 − 1) . . . (eitˆzic − 1)gIc n |.

k=Ic

The last inequality holds since the gIc n are degenerate. By Lemma 1, ¯n−1 |t|m+1 exp(−t2 (1/3 − m2−(2m+9)/3 )) |L2 | ≤ σ

m



E|ˆ zi1 . . . zˆic gIc n |.

c=2 Iˇc

Now we pass to L3 . The H¨older inequality with 1 ≤ p ≤ 2 and q = p/(p − 1) implies that |L3 | ≤ |t|(E|eitvn − 1|q )1/q (E|wn |p)1/p . By Propositions 1 and 2, 2 1 |t|E|vn| + |t|p E|wn|p q p m
m


2 m 1 (2−p)m −1 p−1 p −p ≤ 3 |t|¯ σn E|gIc |I(|gIc | > σn ) + 2 (m − 1) |t| σ ¯n E|gIc n |p . q p c=3 ˇ c=2 ˇ |L3 | ≤ 21/q |t|1+1/q (E|vn |)1/q (E|wn |p )1/p ≤

Ic

Ic

Finally, |L4 | ≤ |t|E|vn | ≤ σ¯n−1 |t|

m−1


m



E|gIc Jd n | ≤ 3m σ ¯n−1 |t|

d=2 c=d+1 Jˇd

m



E|gIc |I(|gIc | > σn ).

c=3 Iˇc

Estimation of ε3 . Decompose the integrand as follows: Eeit(sn +vn +wn ) − Eeitsn = (Eeit(sn +wn ) − Eeitsn ) + (Eeit(sn +vn +wn ) − Eeit(sn+wn ) ) = I1 + I2 . We use a randomization method (see, e.g., [1,2]). Fix t ∈ [2m+3 , h] and consider a sequence of mutually independent and independent of X1 , . . . , Xn random variables α1 , . . . , αn with the Bernoulli distribution and such that P (αj = 1) = 1 − P (αj = 0) = f(t) = 9 · 2m−1 t−2 log(t), j = 1, . . . , n. Note that f(t) ∈ [0, 1] for all t. The random variables Xj , j = 1, . . . , n, can be represented in the following form: d

¯ j + (1 − αj )X ˜j , Xj = αj X

j = 1, . . . .n,

¯j and X ˜j are independent copies of Xj for all j = 1, . . . , n. In this case, where X d

sn =

n


αj z¯j +

j=1

n
(2) ˜ ¯ (1 − αj )˜ zj = s(1) n (X) + sn (X). j=1

Since the functions gIc n are additive and homogeneous, wn = σ ¯n−1 d

c m



Wck =

c=2 k=0

where ¯ ck = σ W ¯n−1







k 

αls

ˇ k Iˇc ,I c =Lk s=1 L k k

and d

vn =

m−1


m
d


V¯cdk = σ ¯n−1







k 

ˇ k Jˇd ,J d =Lk s=1 L k k

Note that Wck consists of

¯ ck , W

c=2 k=0 c 

¯ Lk , X ˜ I \L ), (1 − αis )gIc n (X c k

(6)

s=k+1

V¯cdk = σ¯n−1

d=2 c=d+1 k=0

where

c m



m−1


m
d


Vcdk ,

d=2 c=d+1 k=0

αls

d 

¯ Lk , X ˜ J \L ). (1 − αjs )gIc Jd n (X d k

s=k+1

    c d summands and Vcdk consists of summands. k k 2527

Lemma 2. Denote the sequence α1 , . . . , αn by α. Then E|E[eitsn |α]|2 ≤ t−3·2 (1)




E|E[exp(it

m−1

,

(2)

E|E[eitsn |α]|2 ≤ t3·2

αk zk )|α]|2 ≤ t−3·2

m/3−1

m−1

e−t

(22m/3 −3c/4)

2

/3

,

,

k=Ic

and E|E[exp(it




(1 − αk )zk )|α]|2 ≤ t3·2

m/3−1

(22m/3 −3c/4)

exp(−t2 (1/3 − c2−2(m+3)/3 )).

k=Ic

˘ n such that the X ˘ j are independent copies of Xj for ˘1 , . . . , X Proof. Consider a sequence of random variables X ˘ all j = 1, . . . , n. Then z¯j , z˜j , and zj (Xj ) = z˘j are independent and identically distributed for all j = 1, . . . , n. It follows that n n

E|E[exp(it αj z¯j )|α]|2 = EE[exp(it αj (¯ zj − z˘j ))|α] j=1

= E(exp(it

n


j=1

αj (¯ zj − z˘j ))) =

j=1

n 

E exp(itαj (¯ zj − z˘j )).

j=1

Expand exp(itαj (¯ zj − z˘j )) into the Taylor series: 1 1 zj − z˘j ))| ≤ 1 − t2 E(αj (¯ zj − z˘j ))2 + |t|3 E(αj |¯ zj − z˘j |)3 . |E exp(itαj (¯ 2 6 Using the inequality E|¯ zj − z˘j |3 ≤ 4E|zj |3 , we see that 2 2 zj − z˘j ))| ≤ 1 − t2 f(t)(Ezj2 − hE|zj |3 ) ≤ exp(−t2 f(t)(Ezj2 − hE|zj |3 )). |E exp(itαj (¯ 3 3 Hence, n n
(1) m−1 2
E|E[eitsn |α]|2 ≤ exp(−t2 f(t)( Ezj2 − h E|zj |3 )) = exp(−3 · 2m−1 t2 t−2 log(t)) = t−3·2 . 3 j=1

j=1

This proves the first inequality of the lemma. We proceed similarly to prove the second inequality: (2) m−1 2 1 E|E[eitsn |α]|2 ≤ exp(−t2 (1 − f(t)) ) = t3·2 e−t /3 . 3

The last two inequalities are obtained similarly to Lemma 1: E|E[exp(it


k=Ic

c c 1
2 2
αk zk )|α]|2 ≤ exp(−t2 f(t)( − Ezis + h E|zis |3 )) 3 s=1 3 s=1

m/3−1 1 (22m/3 −3c/4) ≤ exp(−t2 f(t)( − c2−2(m+3)/3 )) = t−3·2 . 3 The proof of Lemma 2 is completed.

Decompose I1 as follows: (2) I1 = E exp(it(s(1) n + sn +

c m



(2) ¯ ck )) − E exp(it(s(1) W n + sn ))

c=2 k=0

(2) = E exp(it(s(1) n + sn +

m
c=2

2528

¯ cc )) − E exp(it(s(1) + s(2) )) W n n

(2) +E exp(it(s(1) n + sn +

m m

(2) ¯ cc )) ¯ cc + W ¯ c0 ))) − E exp(it(s(1) W (W + s + n n c=2

(2) +E exp(it(s(1) n + sn +

c m



c=2

¯ ck )) − E exp(it(s(1) + s(2) + W n n

c=2 k=0

m


¯ cc + W ¯ c0 ))) = I11 + I12 + I13 . (W

c=2

Now we transform the first summand: (2) I11 = E exp(it(s(1) n + sn ))(exp(it

m


¯ cc ) − 1 − it W

c=2

(2) +itE exp(it(s(1) n + sn ))

m


m


¯ cc ) W

c=2

¯ cc = I111 + I112 . W

c=2

For a fixed α, the random variable inequality

(2) sn

(1)

does not depend on sn m


|I111| ≤ 2|t|p(m − 1)p−1

and Wcc . Hence, Proposition 2 implies the

(2)

¯ cc|p |α]|. E|E[eitsn |α] · E[|W

c=2

Using the H¨ older inequality, Proposition 1 and Lemma 2, we see that |I111 | ≤ Am σ ¯n−p |t|p

m
c
 (2) (E|E[eitsn |α]|2)1/2 ( E|αis |2p)1/2 E|gIc n |p c=2 Iˇc

≤ Am σ¯n−p |t|p+3·2

m−2

s=1

e−t

2

/6

f(t)

m



E|gIc n |p ,

1 ≤ p ≤ 2.

c=2 Iˇc

Now we estimate I112 : σn−1 |I112| ≤ |t|¯

m






(2)

E|E[eitsn |α] · E[exp(it

c=2 Iˇc

αk z¯k )|α]

c 

αis E[

s=1

k=Ic

c 

eitαis z¯is gIc n |α]|.

s=1

Similarly to the case of L2 , we get the estimate |I112| ≤

|t|¯ σn−1

m



its(2) n

E|E[e

|α] · E[exp(it

c=2 Iˇc




αk z¯k )|α]

c 

α is

s=1

k=Ic

×E[(eitαi1 zˆi1 − 1) · . . . · (eitαic zˆic − 1)gIc n |α]|. By the H¨older inequality, Proposition 2, and Lemma 2, ¯n−1 |I112 | ≤ |t|m+1 σ

m

(2) (E|E[eitsn |α]|2)1/2 c=2 Iˇc

(E|E[exp(it




αk zk )|α]|2)1/2 (

Eα4is )1/2 × E|ˆ zi1 · . . . · zˆic gIc n |

s=1

k=Ic

≤ Am σ ¯n−1 |t|3m/4(1+3·2

c 

m/3−1

)+1 −t2 /6

e

logm/2 (t)

m



E|ˆ zi1 · . . . · zˆic gIc n |.

c=2 Iˇc

We decompose I12 as follows: (2) I12 = E exp(it(s(1) n + sn ))(exp(it

m
c=2

¯ cc ) − 1)(exp(it W

m


¯ c0 ) − 1) W

c=2

2529

m m

(2) ¯ ¯ c0 ) W W +E exp(it(s(1) + s ))(exp(it( )) − 1 − it c0 n n c=2 (2) +itE exp(it(s(1) n + sn ))

m


c=2

¯ c0 = I121 + I122 + I123 . W

c=2

The H¨older inequality with 1 ≤ p ≤ 2 and q = p/(p − 1) and Propositions 1 and 2 imply that |I121 | ≤ (E| exp(it

m


¯ cc ) − 1|p)1/p (E| exp(it W

c=2

m


¯ c0 ) − 1|q )1/q W

c=2

m m

¯ cc|p )1/p ( ¯ c0 |p )1/q ≤ 21/q |t|1+p/q (m − 1)p−1 ( E|W E|W c=2

≤ Am |t|pf(t)2/p σ ¯n−p

c=2

m



E|gIc n |p ,

1 ≤ p < 2.

c=2 Iˇc (1)

(2)

For a fixed α, sn does not depend on Wc0 and sn . The same reasoning as in the case of I111 shows that |I122 | ≤ Am σ¯n−p |t|p−3·2

m−2

m



E|gIcn |p ,

1 ≤ p ≤ 2.

c=2 Iˇc

We estimate I123 similarly to I112 : σn−1 I123 ≤ |t|¯

m



(1)

E|E[eitsn |α] · E[exp(it

c=2 Iˇc c 




(1 − αk )zk |α]

k=Ic

(1 − αis ) × E[(eit(1−αi1 )ˆzi1 − 1) · . . . · (eit(1−αic )ˆzic − 1)gIc n |α]|

s=1

≤ |t|1−9m2

m/3−4

e−t

2

(1/6−m2−2m/3−3 ) −1 σ ¯n

m



E|ˆ zi1 · . . . · zˆic gIc n |.

c=2 Iˇc

For I13 , we use the following representation: (2) I13 = E exp(it(s(1) n + sn +

m


¯ c0 ))(exp(it W

c=2

(exp(it

c−1 m



c−1 m



¯ ck ) − 1 − it W

m


c=2 m


¯ c0 )) W

c=2

c=2 k=1

+

¯ cc ) − 1) W

(2) ¯ ck ) − 1) + E exp(it(s(1) W n + sn +

c=2 k=1

(exp(it

m


c−1 m



¯ ck ) + itE exp(it(s(1) + s(2) W n n

c=2 k=1

¯ c0 )) W

c=2

c−1 m



¯ ck = I131 + I132 + I133 , W

c=2 k=1

where the values I131 and I132 can be estimated as above. (s) (s) To estimate I133 , we need a series of randomizations. Let η1 , . . . , ηn , s = 1, . . . , m − 1, be a series of independent identically distributed random variables that are independent of α and of X1 , . . . , Xn and have the Bernoulli distributions (s)

P (ηj 2530

(s)

= 1) = 1 − P (ηj

= 0) = 1/2,

j = 1, . . . , n, s = 1, . . . , m − 1.

¯ j can be represented in the following way: We denote this series by η. In this case, X d (1) (1) (1) (1) ¯j = X ηj Yj + (1 − ηj )Y¯j , (1)

where Yj

(1) ¯j for all j = 1, . . . , n. Further, and Y¯j are independent copies of X (1) d

Yj

(2)

(2)

(2)

(2)

+ (1 − ηj )Y¯j ,

= ηj Yj

... (m−1) d

Yj (s+1)

where Yj

(m)

= ηj

(m)

(m)

+ (1 − ηj

Yj

(m)

)Y¯j

,

(s+1) (s) and Y¯j are independent copies of Yj for s = 2, . . . , m − 1. In this case,

d

(1)

(2)

(1) ¯ ) + sn1 (Y¯ (1) ) = s(1) n (X) = sn1 (Y

n


(1)

(1)

ηj αj zj (Yj ) +

n
(1) (1) (1 − ηj )αj zj (Y¯j ).

j=1

j=1

For s = 1, . . . , m − 1, (1)

(2)

(1)

(2)

(s+1) ) + sns+1 (Y¯ (s+1) ). s(1) ns = sns+1 + sns+1 = sns+1 (Y

Set (1)

(2)

sn0 := s(1) n

and sn0 := s(2) n .

The equalities (2) d

(1)

sn0 + sn0 =

k


(1) d

s(2) nr + snk =

r=0

s


(1)

s(2) nr + snk ,

s = 1, . . . , k,

r=0

hold, where (1) snk (Y¯ (k) ) =

k n 


(s)

(k)

ηl αl zl (Yl

)

l=1 s=1

and ¯ (r) ) = s(2) nr (Y

n r−1


(s)

(r)

(r)

ηl (1 − ηl )αl zl (Y¯l

).

l=1 s=1

Therefore, d ¯ ck = W

k






c 

ˇ k ,Lk ∈Ic s=k+1 b1 =0 Iˇc L

×

b1 

ηr(1) s

s=1

k 

1


1

b1 

αls

ˇ b ,Rb ∈Lk s=1 R 1 1

(1) (1) (1) ˜I \L ) (1 − ηls )gIc nRb1 (YRb , Y¯Lk \Rb , X c k




ˇb L ˇ k ,Lk =Rb Iˇc ,I c =Lk b1 =0 R 1 b b 1 k k b

×

k 

1

s=b1 +1

k



d

=




(1 − αis )

ηr(1) s

s=1

1

k 

1

c 

(1 − αis )

k 

αls

s=1

Rb1 s=k+1

(1) (1) (1) ˜ I \L ). (1 − ηls )gIc nRb1 (YRb , Y¯Lk\Rb , X c k 1

1

s=b1 +1

¯ ck into summands of three types: We decompose W d ¯ ck = W

k


¯ ckb1 = W ¯ ck0 (Y¯ (1), X ˜I \L ) + W ¯ ckk (Y (1), X ˜I \L ) W c k c k Lk Lk

b1 =0

2531

+

k−1


¯ ckb (Y (1) , Y¯ (1) ˜ W 1 Rb Lk \Rb , XIc \Lk ). 1

1

b1 =1

Further, k−1


k−1


d ¯ ckb = W 1

d

k−1


¯ ckB b (Y (2) , Y¯ (2) ¯ (1) ˜ W 1 2 Rb Rb \Rb , YLk \Rb , XIc \Lk ) 2

b1 =1

=

b1


k−1 1 −1
b


k−1 1 −1
b


¯ ckB b + W ¯ ckB 0 ) + (W 1 1 1

b1 =1

+




¯ ckBk−1 bk−1 + W ¯ ckBk−1 0 ) = (W

where




WckBj bj =


j

 b j−1

i=1

¯ ckB b + W ¯ ckB 0 ) + . . . (W 2 2 2

k−1





¯ ckBj bj + W ¯ ckBj 0 ), (W

j−1 

(




··· 2

j

2

k 

ηr(i) (1 − ηr(j) )) · · · s s

s=bj +1 i=1





ˇ k Iˇc ˇ b1 ,Rb1 =( Rbs−1 ) L R b1 k b b bs

ˇ b ˇ bj−1 bj−1 R ,Rb =Rbj j Rb

ηr(i) s

1

j=1 1≤bj