A note on the use of V and U statistics in nonparametric models of ...

AISM (2006) 58: 389–406
DOI 10.1007/s10463-006-0034-z
Carlos Martins-Filho · Feng Yao
A <strong>note</strong> on the use of V and U statistics
in nonparametric models of regression
Received: 8 September 2004 / Revised: 12 April 2005 / Published online: 1 June 2006
© The Institute of Statistical Mathematics, Tokyo 2006
Abstract We establish the √ n asymptotic equivalence of V and U statistics when
the statistic’s kernel depends on n. Combined with a lemma of B. Lee this result
provides conditions under which U statistics projections and V statistics are √ n
asymptotically equivalent. The use of this equivalence in nonparametric regression
models is illustrated with several examples; the estimation of conditional variances,
skewness, kurtosis and the construction of a nonparametric R-squared measure.
Keywords U statistics · V statistics · local linear estimation
1 Introduction
The use of U statistics (see Hoeffding 1948), as a means of obtaining asymptotic
properties of kernel based estimators in semiparametric and nonparametric
regression models is now common practice in econometrics. For example, Powell,
Stock and Stoker (1989) use U statistics to obtain the asymptotic properties of their
semiparametric index model estimator. Ahn and Powell (1993) use U statistics to
investigate the properties of a two stage semiparametric estimation of censored
selection models where the first stage estimator involves a nonparametric regression
estimator for the selection variable. Fan and Li (1996) use U statistics to study
C. Martins-Filho (B)
Department of Economics, Oregon State University,
Ballard Hall 303, Corvallis, OR 97331-3612 USA
E-mail: carlos.martins@orst.edu
Tel.:+1-541-737-1476
F. Yao
Department of Economics, University of North Dakota,
P.O. Box 8369, Grand Forks, ND 58203 USA
E-mail: feng.yao@mail.business.und.edu
Tel.: +1-701-7773360

390 C. Martins-Filho and F. Yao
a set of specification tests for nonparametric regression models and Zheng (1998)
uses U-statistics to provide a nonparametric kernel based test for parametric quantile
regression models. See also Kemp (2000) and D’Amico (2003) for more recent
uses.
The appeal for the use of U statistics derives from the fact that often estimators
or statistics of interest – T n – are expressed as linear combinations of nonparametric
regression estimators. Consider for example a nonparametric regression model
E(Y|X = x) = m(x) with observations {(y i ,x i )} n i=1 , T n = ∑ n
i=1 c i ˆm(x i ) where
c i ∈Rare nonstochastic and ˆm(x) is an arbitrary nonparametric estimator for
m(x). Since ˆm(x) can usually be written as ˆm(x) = ∑ n
j=1 w jn(x)y j , we can write
T n =
n∑
i=1
( )
n
c i w in (x i )y i + u
2 n = T 1n + T 2n ,
( ) −1
n ∑
where u n =
2 1≤i

V and U statistics in nonparametric models 391
where ∑ (n,k) de<strong>note</strong>s a sum over all subsets 1 ≤ i 1 0 and consequently to reach the desired conclusion it suffices to show
that nE((u n − v n ) 2 ) = o(1).Bythec r -inequality we can write
(
nE((u n − v n ) 2 ) ≤ kn 1 − P k
n ) 2 k−1
E(u 2
n k n ) + kn ∑
κ=1
(
n
κ
) 2
n −2k E((u (κ)
n )2 ).
(4)

392 C. Martins-Filho and F. Yao
We will deal with the two terms on the righthand side of inequality (4) separately.
We first <strong>note</strong> that by the representation theorem for U statistics (Serfling, 1980; p.
180) u (κ)
n = 1 n!
∑p W n
(κ)(Z
i 1
,... ,Z in ) where
W (κ)
n (Z i 1
,... ,Z in ) = 1 γ κ
(φ κ,n (Z i1 ,... ,Z iκ ) + φ κ,n (Z iκ+1 ,... ,Z i2κ ) +···
+φ κ,n (Z iγκ κ−κ+1
,...,Z iγκ κ
)),
γ κ = [n/κ] is the greatest integer ≤ n/κ and ∑ p
de<strong>note</strong>s the sum over all permutations
(i 1 ,...,i n ) of {1, 2,...,n} and similarly u n = 1 ∑
n! p W n(Z i1 ,... ,Z in )
where W n (Z i1 ,... ,Z in ) = 1 γ (ψ n(Z i1 ,... ,Z ik ) + ψ n (Z ik+1 ,... ,Z i2k ) + ··· +
ψ n (Z iγk−k+1 ,...,Z iγk )) and γ = [n/k] is the greatest integer ≤ n/k. Now, for the
first term on the righthand side of inequality (4), since n k −Pk
n = O(nk−1 ) we have
( )
n 1 − P k
n 2 ( )
n = O(n −1 ) and n 1 − P n 2 ((
k
k
n E(u
2 k n )=O(n −1 1
∑
)E
n! p W n(Z i1 ,...,
Z in ) ) ) 2
. By Minkowski’s inequality
⎛(
E ⎝
1
n!
) ⎞ 2
∑
W n (Z i1 ,... ,Z in )
p
⎠ ≤ 1
(n!) 2 ( ∑
p
= E(|W n (Z i1 ,... ,Z in )| 2 ),
(
E(|Wn (Z i1 ,... ,Z in )| 2 ) ) ) 2
1/2
where the last equality follows since {Z i } i=1,2,... is an i.i.d. sequence. By definition
of W n and another use of Minkowski’s inequality we have
⎛
⎞
γ∑
E(|W n (Z i1 ,... ,Z in )| 2 ) ≤ γ −2 (
⎝ E(ψ
2
n (Z ijk−k+1 ,... ,Z ijk )) ) 1/2
⎠
j=1
= E(ψ n (Z i1 ,... ,Z ik )) = o(n),
( )
where the last equality follows by assumption.We therefore conclude that n 1− P k
n 2
n k
E(u 2 n ) = o(1).
(
n
For the second term on inequality (4), <strong>note</strong> first that n
κ)
−k = O(n κ−k ) and
(( ) 2
n
since κ ∈{1,... ,k− 1}, max κ n κ−k = n −1 and therefore n
κ)
−k =O(n −2 ).
Once again, by Minkowski’s inequality we have E((u (κ)
n )2 ) ≤ E(|W n (κ)(Z
i 1
,
... ,Z in )| 2 ). Now, given the definition of W n
(κ) and the fact that {φ κ,n (Z ijκ−κ+1
,... ,Z ijκ )} j=1,2,... is an i.i.d. sequence of random variables we have by another
application of Minkowski’s inequality that E(|W n
(κ)(Z
i 1
,... ,Z in )| 2 ) ≤ E((φ κ,n (Z i1
,... ,Z iκ )) 2 ). Let l κ de<strong>note</strong> the number of terms in ∑ {ν 1 ,... ,ν κ ,k} , then by the c r
inequality
2

V and U statistics in nonparametric models 393
E((φ κ,n (Z i1 ,... ,Z iκ )) 2 )
⎛⎛
⎞2⎞
= E ⎝⎝
∑ k!
∏ κ
{ν 1 ,... ,ν κ ,k} j=1 ν j ! ψ ⎟ ⎟
n ×(Z i1 ,... ,Z i1 ,... ,Z iκ ,... ,Z iκ ) ⎠ ⎠
} {{ } } {{ }
ν 1 − times
ν κ − times
⎛
⎞
∑ (k!) 2
≤ l κ
( ∏ κ
{ν 1 ,... ,ν κ ,k} j=1 ν j !) E ⎜
⎝ψ 2 2 n (Z ⎟
i 1
,... ,Z i1 ,... ,Z iκ ,... ,Z iκ ) ⎠ .
} {{ } } {{ }
ν 1 − times
ν κ − times
Now, since by assumption E ( ψn 2(Z i 1
,... ,Z ik ) ) = o(n) for all 1 ≤ i 1 ,...,i k
≤ n, k ≤ n,, then it is true that for all κ ∈{1,... ,k − 1} and all ν j ≥ 1 such
that k = ∑ κ
j=1 ν j we have E ( ψn 2(Z i 1
,... ,Z i1 ,... ,Z iκ ,... ,Z iκ ) ) = o(n). Then
E(|W (κ)
n
∑k−1
(
n
n
κ
κ=1
(Z i 1
,... ,Z in )| 2 ) = o(n) and consequently E((u (κ))2
) = o(n).As a result,
) 2
n −2k E((u (κ)
n
∑k−1
)2 )≤nO(n −2 )o(n)
κ=1
l κ
∑
n
{ν 1 ,... ,ν κ ,k}
(k!) 2
( ∏κ
j=1 ν j !) 2
=o(1),
which completes the proof.
⊓⊔
Theorem 1 can be proved using the following two conditions which are implied
by E ( ψn 2(Z i 1
,... ,Z ik ) ) = o(n) for all 1 ≤ i 1 ,...,i k ≤ n, k ≤ n: (a)
E ( ψn 2(Z i 1
,... ,Z ik ) ) = o(n) for all 1 ≤ i 1 < i 2 < ··· < i k ≤ n, k ≤ n;
(b) for all κ ∈ {1,... ,k − 1} and all ν j ≥ 1 such that k = ∑ κ
j=1 ν j that
define Z i1 ,... ,Z i1 ,... ,Z iκ ,... ,Z iκ E ( ψ
} {{ } } {{ }
n 2(Z i 1
,... ,Z i1 ,... ,Z iκ ,... ,Z iκ ) )
ν 1 − times
ν κ − times
= o(n 2(k−κ)−1 ). The next corollary establishes the √ n asymptotic equivalence
between V statistics and U statistics projections.
Corollary 1 Let {Z i } n i=1 be a sequence of i.i.d. random variables and u n and
v n be U and V statistics with kernel function ψ n (Z 1 ,... ,Z k ). In addition, let
û n = k n
∑ n
i=1 (ψ 1n(Z i ) − θ n ) + θ n , where ψ 1n (Z i ) = E (ψ n (Z 1 ,... ,Z k )|Z i ) and
θ n = E (ψ n (Z 1 ,... ,Z k )). IfE ( ψ 2 n (Z 1,... ,Z k ) ) = o(n) then √ n(v n −û n ) =
o p (1).
Proof From Theorem 1 we have √ n(v n − u n ) = o p (1), and from Lemma 2.1 in
Lee (1988) we have that √ n(u n −û n ) = o p (1). Hence, √ n(v n −û n ) = o p (1). ⊓⊔
3 Some applications in regression models
We provide applications of Theorem 1 and its corollary around the following regression
model,
y t = m(x t ) + ε t , (5)

394 C. Martins-Filho and F. Yao
where, {(y t ,x t )} n t=1 is a sequence of i.i.d. random variables, y t,x t ∈Rwith joint
density q(y,x), E(ε t |x t ) = 0 and V(ε t |x t ) = σ 2 (x t ). This is similar to the regression
model considered by Fan and Yao (1998), with the exception that here the
observations are i.i.d. rather than strictly stationary. Our applications all involve
establishing the asymptotic distribution of statistics that are constructed as linear
combinations of a first stage nonparametric estimator of m(x). As the first stage
estimator for m(x) we consider the local linear estimator of Fan (1992), i.e., for
x ∈R, we obtain ˆm(x) =ˆα where
n∑
( ˆα, ˆβ) = argmin α,β (y t − α − β(x t − x)) 2 K
t=1
( )
xt − x
,
h n
where K(·) is a density function and 0

V and U statistics in nonparametric models 395
(a) Given E(y 2 t f 2 (x t ,y t )) < ∞ then
I n = 1 n
n∑
∫
ε t f(x t , y)g(y|x t )dy + 1 2 h2 n σ K 2 E(m(2) (x t )f (x t ,y t ))
t=1
where g(y|x) de<strong>note</strong>s the conditional density of y given x.
(b) Given E(y 4 t f 2 (x t ,y t )) < ∞ then
+o p (n − 1 2 ) + op (h 2 n ), (6)
J n = σ 2 K h2 n
n
n∑
ε t m (2) (x t )E(f (x t ,y t )|x 1 ,... ,x n )
t=1
+ 1 4 h4 n σ 4 K E((m(2) (x t )) 2 f(x t ,y t ))
+o p (n − 1 2 ) + op (h 3 n ), (7)
(c) Given E(y 6 t f 2 (x t ,y t )) < ∞ then
L n = o p (h 3 n ) (8)
(d) Given E(y 8 t f 2 (x t ,y t )) < ∞ then
M n = o p (h 4 n ). (9)
Proof Here we prove parts (a)–(c). The proof of (d) is similar to that of (c) and is
omitted. Let ∑ (p) h i 1 ...i p
de<strong>note</strong> the sum of h i1 ...i p
over the permutations of i 1 ...i p ,
i.e., ∑ (p) h i 1 ...i p
is the sum of p! terms given by h i1 i 2 ...i p
+ h i2 i 1 ...i p
+···+h ip i p−1 ...i 1
.
(a) Note that,
ˆm(x t ) − m(x t ) =
1
nh n g(x t )
n∑
K
k=1
1
+
2nh n g(x t )
( )
xk − x t
ε k
h n
n∑
K
k=1
( )
xk − x t
m (2) (x kt )(x k − x t ) 2 + w(x t ),
h n
where w(x t ) = ˆm(x t )−m(x t )− 1 ∑ n
nh n g(x t ) k=1 K(x k−x t
h n
)yk ∗ and y∗ k = ε k+ 1 2 m(2) (x kt )
(x k −x t ) 2 with x kt = λx t +(1−λ)x k for some λ ∈ (0, 1). Hence, I n = I 1n +I 2n +I 3n ,
where
I 1n = 1
n 2 h n
n∑
n∑
K
t=1 k=1
( )
xk − x t
h n
1
ε k f(x t ,y t )
g(x t )

396 C. Martins-Filho and F. Yao
I 2n = h n
2n 2
I 3n = 1 n
n∑
t=1 k=1
n∑
( )( )
xk − x t xk − x 2
t
K
m (2) 1
(x kt )f (x t ,y t )
h n g(x t )
h n
n∑
w(x t )f (x t ,y t ).
t=1
We treat each term separately. Observe that I 1n can be written as
I 1n = 1 n∑ n∑
( ( )
1 xk − x t
1
K ε
2n 2
k f(x t ,y t )
h
t=1 k=1 n h n g(x t )
+ 1 ( )
)
xt − x k
1
K ε t f(x k ,y k )
h n h n g(x k )
⎛ ⎞
= 1 n∑ n∑
⎝ ∑ h
2n 2
tk
⎠ = 1 n∑ n∑
ψ
2n 2
n (Z t ,Z k ) = 1 2 v n,
t=1 k=1 (2)
t=1 k=1
where v n is a V statistic with Z t = (x t ,y t ) and ψ n (Z t ,Z k ) a symmetric kernel.
Hence, by Corollary 1, if E ( ψn 2(Z t,Z k ) ) = o(n) for all t,k then √ n(v n −û n ) =
o p (1). First, <strong>note</strong> that for t ≠ k 1 n E ( ψn 2(Z t,Z k ) ) = 2 n E(h2 tk ) + 2 n E(h tkh kt ) and
since E(εk 2|x 1,... ,x n ) = σ 2 (x k ), we have from the proposition of Prakasa-Rao
that
1
n E(h2 tk ) = 1 E
nh 2 n
( ( )
K 2 xk −x t
h n
)
σ 2 (x k )E(f 2 1
(x t ,y t )|x 1 ,... ,x n ) → 0,
g 2 (x t )
provided that nh n → ∞. Similarly it can be shown that E( 1 n h tkh kt ) = o(1)
and therefore 1 n E(ψ2 n (Z t,Z k )) = o(1). Fort = k it suffices to verify that 4K2 (0)
(
)
nh 2 n
E σ 2 (x t ) f 2 (x t ,y t )
g 2 (x t
= o(1), but this follows directly given our assumptions. Now,
)
since E(ε t |x 1 ,... ,x n ) = 0, θ n = E (ψ n (Z t ,Z k )) = 0 and û n = 2 ∑ n
n t=1 ψ 1n(Z t ).
Therefore,
ψ 1n (Z t ) = 1 ∫ ∫ ( )
xk − x t
ε t K f(x k ,y k )g(y k |x k )dx k dy k
h n h
∫
n
= ε t f(x t , y)g(y|x t )dy + o(1),
where g(y|x) = q(y,x)
g(x) . Consequently,
I 1n = 1 n∑
∫
ε t f(x t , y)g(y|x t )dy + o p (n − 1 2 ) (10)
n
t=1
given that 1 n
∑ n
t=1 ε t = O p (n − 1 2 ). Now, <strong>note</strong> that
I 2n = 1
4n 2
n∑
n∑ ∑
t=1 k=1 (2)
h tk = 1
4n 2
n∑
t=1 k=1
n∑
ψ n (Z t ,Z k ) = 1 4 v n,

V and U statistics in nonparametric models 397
where h tk = h n K( x k−x t
h n
)( x k−x t
h n
) 2 m (2) (x kt )f (x t ,y t ) 1
g(x t ) . By Corollary 1 if E(ψ2 n (Z t,
Z k )) = o(n) for all t,k, then √ n(v n −û n ) = o p (1). Hence we focus on û n . Note
that, û n = 2 ∑ n
n t=1 ψ 1n(Z t ) − θ n , where ψ 1n (Z t ) = E(h tk |Z t ) + E(h kt |Z t ) and
E(û n ) = θ n = 2E(h tk ). Hence, by the proposition of Prakasa-Rao,
) (ûn
E = 2 ∫ ∫ ∫ ( )( )
xk − x t xk − x 2
t
K
m (2) (x
h 2 kt )
n h n h n h n
×f(x t ,y t ) g(x k)
g(x t ) q(x t,y t )dx k dx t dy t
−→ 2σ 2 K E(m(2) (x t )f (x t ,y t )).
In addition, <strong>note</strong> that
( ) 1
V û
h 2 n = 4 nV (E(h
n n 2 h 4 tk |Z t ) + E(h kt |Z t ))
n
= 4 (
E(E(htk |Z
nh 4 t )) 2 +E(E(h kt |Z t )) 2
n
+2E(E(h tk |Z t )E(h kt |Z k ))−θn) 2 .
Under our assumptions, it is straightforward to show that
1
E(E(h tk |Z t )) 2
nh 4 n
= 1 E
nh 2 n
(
as n →∞. Similarly,
(
as n →∞. Hence, V
f(x t ,y t ) 2 1
g 2 (x t ) E2 (
K
1
nh 4 n
1
h 2 n
(
xk − x t
h n
)(
xk − x t
h n
) 2
m (2) (x kt )|Z t
))
→0
E(E(h kt |Z t )) 2 → 0 and 1
nh 4 n
E(E(h tk |Z t )E(h kt |Z t )) → 0
û n
)
→ 0. By Markov’s inequality we conclude that
û n = 2h 2 n σ K 2 E(m(2) (x t )f (x t ,y t )) + o p (h 2 n ) and by Corollary 1,
I 2n = 1 2 h2 n σ K 2 E(m(2) (x t )f (x t ,y t )) + o p (h 2 n ). (11)
Finally, verification that E(ψn 2(Z t,Z k )) = 2E(h 2 tk ) + 2E(h tkh kt ) = o(n) for all
t ≠ k follows directly from our assumptions and use of the proposition of Prakasa-Rao,
and for the case where t = k we have that E(ψn 2(Z t,Z k )) = 0. From
Martins-Filho and Yao’s (2003) Lemma 2 and Theorem 1
|w(x t )|=
∣ ˆm(x t) − m(x t ) −
1
nh n g(x t )
n∑
K
k=1
( )
xk − x t
h n
y ∗ k
∣ = O p(R n,2 (x t ))
and sup xt ∈G |R n,2(x t )|=o p (h 2 n ), hence sup x t ∈G |w(x t)| =o p (h 2 n ). As a result we
have that |I 3n |≤o p (h 2 n ) 1 ∑ n
n t=1 |f(x t,y t )|=o p (h 2 n ), since E(f 2 (x t ,y t )) < ∞.
Therefore, I 3n = 1 ∑ n
n t=1 w(x t)f (x t ,y t ) = o p (h 2 n ). Combining this result with

398 C. Martins-Filho and F. Yao
(10) and (11) proves part (a).
For part (b) <strong>note</strong> that
J n = 1 n∑ n∑ n∑
( )
1
f(x
n 3 h 2 t ,y t )
n
g 2 (x
t=1 k=1 l=1
t ) K xk − x t
K
h n
n∑ n∑ n∑
(
+ h2 n
1
4n 3
g 2 (x t ) f(x xk − x t
t,y t )K
t=1 k=1 l=1
(
xl − x t
×K
h n
h n
)(
xl − x t
h n
) 2
m (2) (x kt )m (2) (x lt )
+ 1 n∑
w 2 (x t )f (x t ,y t )
n
t=1
+ 1 n∑ n∑ n∑
( )
1
n 3 g 2 (x
t=1 k=1 l=1 t ) f(x xk − x t
t,y t )K
h n
( )( )
xl − x t xl − x 2
t
×K
m (2) (x lt )ε k
+ 2
n 2 h n
n∑
h n
t=1 k=1
h n
n∑ 1
w(x t )
g(x t ) f(x t,y t )K
+ h n∑ n∑
n
1
w(x
n 2
t )
g(x
t=1 k=1
t ) f(x t,y t )K
= J 1n + J 2n + J 3n + J 4n + J 5n + J 6n .
We examine each term separately.
J 1n = 1 1
n∑ n∑ n∑ ∑
h
6 n 3
tkl = 1 1
6 n 3
t=1 k=1 l=1 (3)
n∑
( )
xk − x t
ε k
h n
(
xk − x t
n∑
h n
t=1 k=1 l=1
1
( )
xl − x t
ε k ε l
h n
)(
xk − x t
h n
) 2
)(
xk − x t
h n
) 2
m (2) (x kt )
n∑
ψ n (Z t ,Z k ,Z l ) = 1 6 v n,
where h tkl = 1 K( x k−x t
h 2 n h n
)K( x l−x t
h n
)ε k ε l f(x t ,y t )
g 2 (x t ) . By Corollary 1, if E(ψ2 n (Z t,
Z k ,Z l )) = o(n) for all t,k,l then √ n(v n −û n ) = o p (1). Since ψ 1n (Z t ) = 0
and θ n = 0, we have that û n = 0 and therefore √ nI 1n = o p (1). To verify that
1
n E(ψ2 n (Z t,Z k ,Z l )) = o(1) for t ≠ k ≠ l we <strong>note</strong> that 1 n E(ψ2 n (Z t,Z k ,Z l )) =
4
n (3E(h2 tkl ) + 2E(h tklh ktl ) + 2E(h tkl h ltk ) + 2E(h ktl h ltk )). Under A1–A4, repeated
application of the proposition of Prakasa-Rao to each of these expectations shows
that each approaches zero as n →∞. Also, if t = k but t ≠ l we have that
E(ψn 2(Z t,Z t ,Z l )) = o(n) and when t = k = l, E(ψn 2(Z t,Z t ,Z t )) = o(n 2 )
directly from our assumptions provided nh 3 n → 0.
J 2n = 1
24n 3
n∑
n∑
n∑ ∑
t=1 k=1 l=1 (3)
)( x k−x t
h n
h tkl = 1
24n 3
n∑
) 2 K( x l−x t
h n
)( x l−x t
n∑
n∑
t=1 k=1 l=1
ψ n (Z t ,Z k ,Z l ) = 1
24 v n,
where h tkl = h 2 n K(x k−x t
h n
By Corollary 1 √ n(v n −û n ) = o p (1) provided that E(ψ n (Z t ,Z k ,Z l )) = o(n)
h n
) 2 m (2) (x kt )m (2) (x lt )
g 2 (x t ) f(x t,y t ).
1

V and U statistics in nonparametric models 399
for all t,k,l. As above, verification of this condition is straightforward given our
assumptions. Note that û n = 3 ∑ n
n t=1 ψ 1n(Z t ) − 2θ n with E(û n ) = θ n . Hence,
(
1
E(û
h 4 n ) = 6 ( )( )
xk −x t xk − x 2 ( )( )
t xl −x t xl −x 2
t
E K
K
m (2) (x
n h 2 kt )
n h n h n h n h n
)
× m (2) 1
(x lt )
g 2 (x t ) f(x t,y t )
and by the proposition of Prakasa-Rao we have, 1 E(û
h 4 n )→6σK 4 E((m(2) (x t )) 2 f(x t ,
n
y t )) and V( 1 û
h 4 n ) → 0. By Markov’s inequality we conclude that J 2n = 1
n 4 h4 n σ K
4
E((m (2) (x t )) 2 f(x t ,y t )) + o p (h 4 n ) + o p(n −1/2 ). J 3n = 1 ∑ n
n t=1 w2 (x t )f (x t ,y t ) =
o p (h 4 n ) follows directly from the analysis of the I 3n in part (a). Now,
J 4n = 1
6n 3
n∑
n∑
where h tkl = K( x k−x t
h n
n∑ ∑
t=1 k=1 l=1 (3)
h tkl = 1
6n 3
n∑
n∑
t=1 k=1 l=1
n∑
ψ n (Z t ,Z k ,Z l ) = 1 6 v n,
)K( x l−x t
h n
)( x l−x t
h n
) 2 m (2) 1
(x lt )ε k g 2 (x t ) f(x t,y t ). Once again given
our assumptions it can be verified that E(ψn 2(Z t,Z k ,Z l )) = o(n) for all t,k,l,
therefore we have √ n(v n −û n ) = o p (1), where û n = 3 ∑ n
n t=1 ψ 1n(Z t ) − 2θ n .
Given that in this case θ n = 0, we have
û n = 6 n
(
n∑
ε t E K
t=1
( )
xt −x k
K
h n
( )( ) )
xl −x k xl −x 2
k
m (2) 1
(x lk )
h n g 2 (x k ) f(x k,y k )|Z t ,
h n
Under A1–A4, the proposition of Prakasa Rao gives
( ( ) ( )( ) )
1 xt − x k xl − x k xl − x 2
k
E K K
m (2) 1
(x
h 2 lk )
n h n h n h n g 2 (x k ) f(x k,y k )|Z t →
σK 2 m(2) (x t )E(f (x t ,y)|x 1 ,... ,x n ).
∑
Hence, û n = 6h2 n n
n t=1 ε tm (2) (x t )σK 2 E(f(x t,y)|x 1 ,... ,x n )+o p (n − 1 2 ), given that
6
∑ n
n t=1 ε t = O p (n − 1 2 ). Consequently, J 4n = σ K 2 ∑ h2 n n
n t=1 ε tm (2) (x t )E(f (x t ,y)
|x 1 ,... ,x n ) + o p (n − 1 2 ).
|J 5n | = ∣ 2 ∑ (
n
n t=1 w(x 1
∑ ) n
t)
nh n g(x t ) k=1 K(x k−x t
h n
)ε k f(x t ,y t ) ∣ and from
Martins-Filho and Yao (2003) Theorem 1 we have sup xt ∈G |w(x t)|
= o p (h 2 n ) and sup 1
∑
∣
n
x t ∈G ∣
nh n g(x t ) k=1 K(x k−x t ∣∣
h n
)ε k = op (h n ). Hence, |J 5n |
≤ 2 ∑ n
n t=1 |f(x t,y t )|o p (h 3 n ) = O p(1)o p (h 3 n ) = o p(h 3 n ) since 2 ∑ n
n t=1 |f(x t,y t )|
= O p (1), which gives J 5n = o p (h 3 n ). Finally, from Martins-Filho and Yao (2003)
Theorem 1 we have
h n
n∑
( )( ) xk − x t xk − x 2 t
sup
K
m (2) (x kt )
∣ng(x t )
h n ∣ = O p(h 2 n ) (12)
x t ∈G
k=1
h n

400 C. Martins-Filho and F. Yao
and |J 6n |≤ 1 ∑ n
n t=1 |f(x t,y t )|o p (h 2 n )O p(h 2 n ) = O p(1)o p (h 4 n ) = o p(h 4 n ). Combining
the orders of all six terms gives the desired result.
∑ n
k=1 K (
xk −x t
1
2nh n g(x t )
∑ n
k=1 K (
xk −x t
h n
)
m (2) (x kt )
)
(c) Let a tn = 1
nh n g(x t )
h n
ε k , b tn =
(x k − x t ) 2 , c tn = w(x t ). Then ( ˆm(x t ) − m(x t )) 3 = atn 3 + b3 tn + c3 tn + 3a tnbtn 2 +
3a tn ctn 2 +3a2 tn b tn+3atn 2 c tn+3btn 2 c tn+3b tn ctn 2 +6a tnb tn c tn and n −1 ∑ n
t=1 ( ˆm(x t)−
m(x t )) 3 f(x t ,y t ) = ∑ 10
i=1 L in, where L 1n = n −1 ∑ n
t=1 a3 tn f(x t,y t ),... ,L 10n =
n −1 ∑ n
t=1 6a tnb tn c tn f(x t ,y t ). Since
∣ sup 1
n∑
( ) ∣
xk − x ∣∣∣∣
t
x t
K ε k = o p (h n ),
nh n g(x t ) h n
k=1
then |L 1n |≤o p (h 3 n ) × 1 ∑ n
n t=1 |f(x t,y t )|=o p (h 3 n ). By (12), |L 2n| ≤O p (h
∑ 6 n )n−1
n
t=1 |f(x t,y t )|=O p (h 6 n ) since E(f 2 (x t ,y t )) < ∞. |L 3n |≤o p (h 6 n ) follows
directly from sup xt ∈G|w(x t )|=o p (h 2 n ).
|L 4n |≤ 3 n∑
1
n∑
( ) ∣
xk − x ∣∣∣∣
t
K ε k
n ∣nh n g(x t ) h n
t=1
(
×
∣
1
2nh n g(x t )
k=1
n∑
K
k=1
×|f(x t ,y t )|≤o p (h n )O p (h 4 n ) 3 n
( )
xk − x t
(x k − x t ) 2 m (2) (x kt )
h n
) 2
∣ ∣∣∣∣∣
n∑
|f(x t ,y t )|=o p (h 5 n )
from (12) and the analysis of J 5n in part (b). Similarly, we obtain |L 5n |, |L 10n |≤
o p (h 5 n ), |L 6n|, |L 7n |≤o p (h 4 n ) and |L 8n|, |L 9n |≤o p (h 6 n ). Hence, we have L n =
o p (h 3 n ).
⊓⊔
We now use the results in the Lemma to establish the asymptotic distribution of
several statistics of interest.
t=1
3.1 Estimating conditional variance
Consider first a special case of the model (5), where ε t |x t ∼ N(0,σ 2 ). Then a
natural estimator of σ 2 is
ˆσ 2 = 1 n∑
(y t −ˆm(x t )) 2 = 1 n∑ (
ε
2
n
n t +2(m(x t )−ˆm(x t ))ε t +(m(x t )−ˆm(x t )) 2) .
t=1
t=1
The difficulty in dealing with such expressions lies in the average terms involving
(m(x t ) −ˆm(x t )) and (m(x t ) −ˆm(x t )) 2 , since the first term is an average of an
i.i.d. sequence. By using Corollary 1 and Lemma 1 we have a convenient way to
establish their asymptotic properties. This is shown in the next theorem.
Theorem 2 Assume that in model (5), ε t |x t ∼ N(0,σ 2 ) and E(yt 4 )

V and U statistics in nonparametric models 401
Proof Note that ˆσ 2 = 1 ∑ n
(
n t=1 ε
2
t + 2(m(x t ) −ˆm(x t ))ε t + (m(x t ) −ˆm(x t )) 2) .
Now, letting f(x t ,y t )=−2ε t in part (a) of Lemma 1 we have 1 ∑ n
n t=1 ε ∫
t f(xt ,y t )
g(y t |x t )dy t = 0, since E(ε t |x 1 ,... ,x n ) = 0 and 1 2 h2 n σ K 2 E ( m (2) (x t )f (x t ,y t ) ) = 0.
Hence, 1 ∑ n
n t=1 2(m(x t) −ˆm(x t ))ε t = o p (n − 1 2 ) + o p (h 2 n ). By part (b) of Lemma 1
with f(x t ,y t ) = 1wehave
1
4 h4 n σ K 4 E((m(2) (x t )) 2 f(x t ,y t )) = O p (h 4 n )
and
σK 2 n∑ h2 n
ε t m (2) (x t )E(f (x t ,y)|x 1 ,... ,x n )
n
t=1
= σ K 2 n∑ h2 n
ε t m (2) (x t ) = O p (n − 1 2 h
2
n
n ) = o p (n − 1 2 )
t=1
by the central limit theorem for i.i.d. sequences. Hence, 1 ∑ n
n t=1 (m(x t)− ˆm(x t )) 2 =
o p (n − 1 2 ) + o p (h 3 n ) provided E(y4 t )

402 C. Martins-Filho and F. Yao
(b) If E(y 8 t )

V and U statistics in nonparametric models 403
Since E(yt 6)

404 C. Martins-Filho and F. Yao
Hence,
b 3n = n −1
n∑
t=1
ε 4 t − µ 4 − 4µ 3
n
n∑
ε t − 2σK 2 h2 n µ 3E ( m (2) (x t ) )
+o p (n −1/2 ) + o p (h 2 n ). (23)
µ
Combining (20) and (23), we write b 3n − b 4
4n (µ 2
= n −1 ∑ n
) 2 t=1 W t + A 2n where
W t = εt 4 + µ 4 − 4ε t µ 3 − 2 µ 4
µ 2
εt 2 and A 2n =−2h 2 n σ K 2 µ 3E(m (2) (x t )) + o p (n −1/2 ) +
o p (h 2 n ). Note that E(W t) = 0 and by the c r -inequality, V(W t ) = E(Wt 2)

V and U statistics in nonparametric models 405
√ n( ˆη 2 −η 2 −b 2n ) → d N(0,V(ξ t )) where ξ t = 1 (
ε
2
V(y t ) t −(1−η 2 )(y t −E(y t )) 2)
and b 2n = o p (h 2 n ).
(
)
Proof ˆη 2 − η 2 1
E(y
=− 1
∑ n β
n t=1 (y t −ȳ) 2 1 − β t −m(x t )) 2
2 V(y t
where β
)
1 = 1 ∑ n
n t=1 (y t −
ˆm(x t )) 2 − E(y t − m(x t )) 2 , β 2 = 1 ∑ n
n t=1 (y t − ȳ) 2 − V(y t ). First, <strong>note</strong> that by the
law of large numbers 1 ∑ n
n t=1 y2 p
t → E(yt 2 ) and ȳ2 → p E(y t ) 2 , hence
(
) −1
1
n∑
(y t − ȳ) 2 p
→ V(yt ) −1 . (26)
n
t=1
1
For β 2 , ȳ 2 − (E(y t )) 2 = 1 ∑ n ∑ n
2 n 2 t=1 k=1 (y ty k + y k y t ) − E(m(x t )) 2 = 1 2 v n −
(E(m(x t ))) 2 . By Corollary 1, since 1 n E(ψ2 n (y t,y k )) = 4 n E((y2 t ))2 = o(1) for all
t,k, √ n(v n −û n ) = o p (1) where û n = 2 ∑ n
n t=1 (ψ 1n(Z t ))−θ n = 4 ∑ n
n t=1 y tE(m(x t ))−
2(E(m(x t ))) 2 . Hence, ȳ 2 − (E(y t )) 2 = 2 ∑ n
n t=1 y tE(m(x t )) − 2(E(m(x t ))) 2 +
o p (n − 1 2 ), and
E(y t − m(x t )) 2
β 2 = 1 n∑
{ E(σ 2 (x t ))
(yt 2 − E(yt 2 V(y t ) n V(y
t=1
t )
))
}
−2 E(σ2 (x t ))E(m(x t ))
(y t − E(m(x t ))) + o p (n − 1 2 ).
V(y t )
For β 1 , 1 ∑ n
n t=1 (y t−ˆm(x t )) 2 = 1 ∑ n
n t=1(
ε
2
t + 2(m(x t )−ˆm(x t ))ε t +(m(x t )−ˆm(x t )) 2) .
Similarly to the proof of Theorem 2, we use part (a) of Lemma 1 with f(x t ,y t )
=−2ε t to obtain 1 ∑ n
n t=1 2(m(x t)− ˆm(x t ))ε t = o p (n − 1 2 )+o p (h 2 n ). By part (b) with
f(x t ,y t ) = 1 we obtain 1 ∑ n
n t=1 (m(x t)− ˆm(x t )) 2 = o p (n − 1 2 )+o p (h 3 n ). Therefore,
1
∑ n
n t=1 (y t −ˆm(x t )) 2 = 1 ∑ n
n t=1 ε2 t + o p (n − 1 2 ) + o p (h 2 n ). Hence,
β 1 − β 2
E(y t − m(x t )) 2
V(y t )
= 1 n
n∑
t=1
(ε 2 t − E(σ 2 (x t )) − E(σ2 (x t ))
V(y t )
+ 2E(σ2 (x t ))E(m(x t ))
(y t − E(m(x t )))
V(y t )
+o p (n − 1 2 ) + op (h 2 n )
= 1 n∑
ζ t + o p (n − 1 2 ) + op (h 2 n
n
).
t=1
(y 2 t − E(y 2 t ))
Since E(ζ t ) = 0,V(ζ t ) = V(εt 2 − (1 − η 2 )(y t − E(y t )) 2 )

406 C. Martins-Filho and F. Yao
4 Conclusion
We have established the √ n asymptotic equivalence between V and U statistics
when their kernels depend on n, the sample size. We combine our result with a
result of Lee (1988) to obtain the √ n asymptotic equivalence between V statistics
and U statistics projections. We provide a number of examples and illustrations
on how our results can be used in nonparametric kernel estimation. The list of
examples is obviously not exhaustive as our results can be used in much broader
contexts.
Acknowledgements
We would like to thank Professor Katuomi Hirano and two anonymous referees
for comments and suggestions.
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