GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 6, 1999, 501-516
LIMIT DISTRIBUTION OF THE MEAN SQUARE
DEVIATION OF THE GASSER–MÜLLER
NONPARAMETRIC ESTIMATE OF THE REGRESSION
FUNCTION
R. ABSAVA AND E. NADARAYA
Abstract. Asymptotic distribution of the mean square deviation of
the Gasser–Müller estimate of the regression curve is investigated.
The testing of hypotheses on regression function is considered. The
asymptotic behaviour of the power of the proposed criteria under
contiguous alternatives is studed.
Let {Ynk , k = 1, . . . , n}n≥1 be a sequence of arrays of random variables
defined as follows:
k
Ynk = g(xnk ) + Znk , xnk = , k = 1, . . . , n, n ≥ 1,
n
where g(x), x ∈ [0, 1], is the unknown real-valued function to be estimated
by given observations Ynk ; Znk , k = 1, . . . , n, is a sequence of arrays of
independent random variables identically distributed in each array such
2
that EZnk = 0, EZnk
= σ 2 , k = 1, . . . , n, n ≥ 1.
Let us consider the estimate of the function g(x) from [1]:
gn (x) =
n
X
Wni (x)Yni ,
i=1
where
Wni = bn−1
Z
∆i
K
x − t‘
bn
dt,
x ∈ [0, 1],
∆i =
hi − 1 i i
, .
n n
Here K(x) is some kernel, bn is a sequence of positive numbers tending to 0.
Assume that the kernel K(x) has a compact support and satisfies the
conditions:
1991 Mathematics Subject Classification. 62G07.
Key words and phrases. Limit distribution, consistency, power of criteria, contiguous
alternatives.
501
c 1999 Plenum Publishing Corporation
1072-947X/99/1100-0501$16.00/0 
502
R. ABSAVA AND E. NADARAYA
1◦ . supp(K)
⊂ [−τ, τ ], 0 < τ < ∞, sup |K| < ∞,
R
K(u)du = 1, K(−x) = K(x),R
R
2◦ . K(u)uj du = 0, 0 < j < s,
K(u)us du 6= 0,
3◦ . K(x) has a bounded derivative in R = (−∞, ∞).
Denote by Fs the family of regression functions g(x), x ∈ [0, 1], hav(s)
ing
Pn derivatives of order up to s (s ≥ 2), g (x) is continuous. Note that
2
i=1 Wni (x) = 1 for x ∈ Ωn (τ ) = [τ bn , 1 − τ bn ] and Egn (x) = g(x) + O(bn )
◦
if
kernel K(x) satisfies condition 1 and g(x) ∈ F2 . On the other hand,
Pthe
n
6 1 for x ∈ [0, τ bn ) ∪ (1 − τ bn , 1] and it may happen that
i=1 Wni (x) =
g(1)
g(0)
Egn (x) 6→ g(x), for example,
PnEgn (0) → 2 (or Egn (1) → 2 ). If the estimate gn (x) is divided by i=1 Wni (x), then the proposed estimate gen (x)
becomes asymptotically unbiased and, moreover, Ee
gn (x) = g(x) + O(bn )
for x ∈ [0, τ bn ) ∪ (1 − τ bn , 1]. Hence the asymptotic behaviour of the estimate gn (x) near the boundary of the interval [0, 1] is worse than within the
interval Ωn (τ ). A phenomenon of such kind is known in the literature as
the boundary effect of the estimator gn (x) (see, for examle, [2]). It would
be interesting to investigate the limit behaviour of the distribution of the
mean square deviation of gn (x) from g(x) on the interval Ωn (τ ) and this is
the aim of the present article.
The method of proving the statements given below is based on the functional limit theorems for semimartingales from [3].
We will use the notation:
Un = nbn
Z
Ωn (τ )
(gn (x) − Egn (x))2 dx,
Qij ≡ Qij (n) =
ξnk =
k−1
X
Z
σn2 = 4σ 4 (nbn )2
n k−1
X
X
Q2ik (n),
k=2 i=1
Wni (x)Wnj (x)dx,
Ωn (τ )
ηik ≡ ηik (n) = 2nbn Qik Zni Znk σn−1 ,
k = 2, . . . , n,
ηik ,
ξn1 = 0,
ξnk = 0, k > n,
i=1
(n)
Fk
= σ(ω : Zn1 , Zn2 , . . . , Znk ),
(n)
where Fk is the σ-algebra generated by the random variables Zn1 , Zn2 , . . . ,
(n)
Znk , F0 = (∅, Ω).
(n)
Lemma 1 ([4], p. 179). The stochastic sequence (ξnk , Fk )k≥1 is a
martingale-difference.
Lemma 2. Suppose the kernel K(x) satisfies conditions 1◦ and 3◦ . If
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
nb2n → ∞, then
bn−1 σn2 → 2σ 4
and
Z
EUn = σ 2
Z
503
2τ
−2τ
K02 (u) du,
K 2 (u)du + O(bn ) + O
(1)
 1 ‘
,
nbn
(2)
where K0 = K ∗ K (the symbol ∗ denotes the convolution).
Proof. It is not difficult to see that
”X
•
n
n X
n
X
4
2
2
2
2
σn = 2σ (nbn )
Qki −
Qii = d1 (n) + d2 (n).
k=1 i=1
Here by the definition of Qki we have
X’Z
d1 (n) = 2σ 4 n2 b−2
n
∆i
i,j
where
Ψn (t1 , t2 ) =
Z
(3)
i=1
K
Ωn (τ )
Z
Ψn (t1 , t2 )dt1 dt2
∆j
x − t ‘
1
bn
K
x − t ‘
2
bn
“2
,
dx.
Using the mean value theorem, we can rewrite d1 (n) as
X
(1) (2)
Ψn2 (θi , θj ),
d1 (n) = 2σ 4 (nbn )−2
i,j
(1)
(2)
with θi ∈ ∆i , θi ∈ ∆j .
(i)
Let P (x) be the uniform distribution function on [0, 1] and Pn (x) the
(i)
(i)
empirical distribution function of the “sample” θ1 , . . . , θn , i = 1, 2, i.e.,
P
(i)
(i)
n
Pn (x) = n−1 k=1 I(−∞,x) (θk ), where IA (·) is the indicator of the set A.
Then d1 (n) can be written as the integral
Z 1Z 1
Ψ2n (x, y)dPn(1) (x)dPn(2) (y).
(4)
d1 (n) = 2σ 4 bn−2
0
0
It is easy to check that
ŒZ 1 Z 1
Z
Œ
2
(1)
(2)
Œ
(x)dP
(y)
−
Ψ
(x,
y)dP
n
n
n
Œ
0
0
where
ŒZ
Œ
I1 = ŒŒ
0
1
Z
0
1
1
0
Z
1
0
Œ
Œ
Ψ2n (x, y)dP (x)dP (y)ŒŒ ≤ I1 +I2 ,
Œ
Œ
Ψ2n (x, y)dPn(2) (y)[dPn(1) (x) − dP (x)]ŒŒ,
504
R. ABSAVA AND E. NADARAYA
ŒZ
Œ
Œ
I2 = Œ
1
0
Z
1
0
Ψ2n (x, y)dP (x)[dPn(2) (y)
Œ
Œ
− dP (y)]ŒŒ.
Furthermore, the integration by parts of the internal integral in I1 yields
Z 1’
Z 1
dPn(2) (y)
I1 ≤ 2
|Pn(1) (x) − P (x)| |Ψn (x, y)|bn−1 ×
0
0
Z
Œ
‘Œ Œ  t − y ‘Œ “

Œ
Œ (1) t − x Œ Œ
×
(5)
ŒK
Œ ŒK
Œdt dx.
b
b
n
n
Ωn (τ )
(i)
2
n
Since sup |Pn − P (x)| ≤
0≤x≤1
and |Ψn (x, y)| ≤ cbn , we have
I1 = O(bn /n).
Here and in what follows c is the positive constant varying from one formula
to another.
In the same manner we may show that
I2 = O(bn /n).
Therefore
Z
d1 (n) = 2σ 4 b−2
n
1
0
Z
1
Ψ2n (t1 , t2 )dt1 dt2 + O
0
It is easy to show also that
’ Z xb−1
Z
Z
n
4
d1 (n) = 2σ
K(u)K
(x−1)b−1
n
Ωn (τ ) Ωn (τ )
x − y
bn
‘
 1 ‘
.
nbn
−u du
“2
(6)
dxdy+O
x
Since [−τ, τ ] ⊂ [ x−1
bn , bn ] for all x ∈ Ωn (τ ), we have
Z
Z
 1 ‘
x − y ‘
d1 (n) = 2σ 4
K02
dx dy + O
.
bn
nbn
Ωn (τ ) Ωn (τ )
But it is easy to see that
Z 1’Z
−1
4
bn d1 (n) = 2σ
0
(1−y)/bn −τ
τ −y/bn
K02 (u) du
Therefore
b−1
n d1 (n)
We can directly verify that
Z
n Z
X
2 −3
b−1
|d
(n)|
≤
b
n
2
n
n
i=1
∆i
Z
→ 2σ
Z
∆i ∆i ∆i
4
Z
’Z
“
dy + O(bn ) + O
K02 (u) du.
 1 ‘
.
nbn
 1 ‘
.
nbn
(7)
Œ  x − t ‘  x − t ‘Œ
Œ
1
2 Œ
K
ŒK
Œdx×
b
b
n
n
Ωn (τ )
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
×
Z
“
Œ  y − t ‘  y − t ‘Œ
1
Œ
3
4 Œ
K
.
ŒK
Œdy dt1 dt2 dt3 dt4 ≤ c
b
b
nb
n
n
n
Ωn (τ )
505
(8)
Statement (1) immediately follows directly from (3), (7), and (8).
Let us now show that (2) holds. We have
Z 1
x − t‘
σ2
K2
Dgn (x) = 2
dPn (x),
nbn 0
bn
where
n
X
Pn (x) = n−1
I(−∞,x) (θk ),
k=1
θk ∈ ∆k , k = 1, . . . , n.
Applying the same argument as in deriving (6), we find
Z 1
x − t‘
 1 ‘
σ2
Dgn (x) = 2
K2
dt + O
.
nbn 0
bn
(nbn )2
(9)
x
Furthermore, taking into account that [−τ, τ ] ⊂ [ x−1
bn , bn ] for all x ∈ Ωn (τ )
by (9) we can write
Z
σ2
K 2 (u) du + O((nbn )−2 ).
Dgn (x) =
nbn
Thus
EUn = σ
2
Z
K 2 (u) du + O(bn ) + O
 1 ‘
.
nbn
Theorem 1. Suppose the kernel K(x) satisfies conditions 1◦ and 3◦ . If
4
sup EZn1
< ∞ and nbn2 → ∞, then
n≥1
d
bn−1/2 (Un − σ 2 θ1 )σ −2 θ2−1 −→ξ,
where
θ1 =
Z
d
K 2 (u) du,
θ22 = 2
Z
K02 (u) du.
By the symbol −→ we denote the convergence in distribution, ξ is a
random variable distributed according to the standard normal distribution.
Proof. Note that
Un − EUn
= Hn(1) + Hn(2) ,
σn
where
n
n
X
nbn X 2
2
(Z − EZni
Hn(1) =
ξnk , Hn(2) =
)Qii .
σn i=1 ni
k=1
506
R. ABSAVA AND E. NADARAYA
(2)
Hn converges to zero in probability. Indeed,
DHn(2) =
n
n
2 X
(nbn )2 X
4
2
4 (nbn )
2
4 |d2 (n)|
Q
EZ
= sup EZn1
EZ
≤
sup
Qii
.
ni ii
n1
2
2
σn i=1
σn i=1
σn2
n≥1
n≥1
(2) P
(2)
This and (8) yield DHn = O(1/(nσn2 )) = O(1/(nbn )). Hence Hn −→0.
P
(By the symbol −→ we denote the convergence in probability).
(1) d
Now let us prove the convergence Hn −→ξ. For this it is sufficient verify
the validity of Corollaries 2 and 6 of Theorem 2 from [3]. We will show
that the asymptotic normality conditions from the corresponding statement
(n)
are fulfilled by the sequence {ξnk , Fk }k≥1 which is a square-integrable
martingale-difference according to Lemma
Pn 1. 2
A direct calculation shows that
k=1 Eξnk = 1. Now the asymptotic
normality will take place if for n → ∞ we have
n
X
k=1
(n)
P
2
I(|ξnk | ≥ ) Fk−1 ]−→0
E[ξnk
(10)
and
n
X
k=1
P
2
−→1.
ξnk
(11)
But, as shown in [3], the validity of (11) together with the condition
P
sup |ξnk |−→0 implies the statement (10) as well.
1≤k≤n
Since for  > 0
P
ˆ
n
X
‰
4
sup |ξnk | ≥  ≤ −4
,
Eξnk
1≤k≤n
k=1
(1) d
to prove Hn −→ξ we have to verify only (11), since relation (12) given
below is fulfiled.
Pn
2 P
−→1. It is sufficient to verify that
Now we will establish k=1 ξnk
E
n
X
k=1
as n → ∞, i.e., due to
E
n
X
k=1
2
ξnk
‘2
Pn
k=1
=
2
ξnk
−1
‘2
→0
2
Eξnk
=1
n
X
k=1
4
Eξnk
+2
X
1≤k1 <k2 ≤n
2
ξ 2 → 1.
Eξnk
1 nk2
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
507
Pn
4
First show that k=1 Eξnk
→ 0 as n → ∞. By virtue of the definitions of
we
can
write
ξnk and ηij
n
X
4
Eξnk
= In(1) + In(2) ,
k=1
where
In(1) =
n
k−1
X
16(nbn )4 X
4
4
EZnj
Qjk ,
EZ
nk
σn4
j=1
k=2
In(2) =
n k−1
48(nbn ) X X
2
2
EZni
EZnj
Q2ik Q2jk .
σn4
4
k=2 i6=j
4
On the other hand, since sup EZn1
< ∞, |Qij | ≤ cbn−1 n−2 and bn−1 σn2 →
n≥1
(1)
(2)
σ 4 θ22 , we have In = O((nbn )−2 ), In = O(1/(nσn4 )) = O(1/(nb2n )). Hence
n
X
k=1
4
Eξnk
→ 0,
n → ∞.
(12)
Now let us establish that
X
2
1≤k1 <k2 ≤n
2
ξ2 → 1
Eξnk
1 nk2
as n → ∞. The definition of ξni yields
(1)
(2)
(3)
(4)
2
2
ξnk
= Bk1 k2 + Bk1 k2 + Bk1 k2 + Bk1 k2 ,
· ξnk
2
1
where
(1)
(2)
(3)
(4)
Bk1 k2 = σ2 (k1 )σ2 (k2 ), Bk1 k2 = σ2 (k1 )σ1 (k2 ),
Bk1 k2 = σ1 (k1 )σ2 (k2 ), Bk1 k2 = σ1 (k1 )σ1 (k2 ),
σ1 (k) =
X
ηik ηjk ,
σ2 (k) =
k−1
X
2
.
ηik
i=1
i6=j
Therefore
2
X
2
ξ2 =
Eξnk
1 nk2
A(i)
n =2
X
1≤k1 <k2 ≤n
A(i)
n ,
i=1
1≤k1 <k2 ≤n
where
4
X
(i)
EBk1 k2 ,
i = 1, 2, 3, 4.
508
R. ABSAVA AND E. NADARAYA
(3)
Let us consider the term An . According to the definition of ηij we have
2
Eηik
η η = 0,
2 sk1 tk1
6 t, k1 < k2 .
s=
Thus
An(4) = 0.
(13)
(2)
4
−2
< ∞ and |Qij | ≤ cb−1
To estimate the term An , note that sup EZn1
.
n n
n≥1
So we obtain
|A(2)
n |
n k1 −1
 1 ‘
1
nb2n X X
.
Q2ik1 ≤ c 2 = O
≤c 4
σn
nσn
nbn
i=1
(14)
k1 =2
(1)
(1)
Now we will verify that An → 1 as n → ∞. For this let us represent An
in the form
An(1) = Dn(1) + Dn(2) ,
where
Dn(1)
Dn(2)
=2
=2
’ X
X
1≤k1 <k2 ≤n
(1)
Bk 1 k 2
k1 <k2
−
’ kX
1 −1
2
Eηik
1
“’ kX
2 −1
j=1
i=1
1 −1
X ’ kX
2
Eηik
1
i=1
k1 <k2
2
Eηjk
2
“’ kX
2 −1
j=1
“
,
2
Eηjk
2
““
.
From the definition of σn2 it follows that
Dn(1) = 1 −
n ’ k−1
X
X
k=2
2
Eηik
i=1
“2
,
where
n ’ k−1
X
X
2
Eηik
i=1
k=2
“2
“2
n ’ k−1
 1 ‘
 nb ‘4 X
X
1
n
.
Q2ik
≤c 4 =O
≤c
σn
nσn
nbn2
i=1
k=2
Therefore
Dn(1) → 1
as n → ∞.
(15)
(2)
Next we will show that Dn → 0 as n → ∞. It is easy to verify that
Dn(2) = 2
1 −1
X ” kX
k1 <k2
i=1
2
2
cov(ηik
)+
, ηik
1
2
kX
1 −1
i=1
•
2
2
)
cov(ηik
,
η
k1 k 2 .
1
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
But
2
Eηik
η2 ≤ c
1 ik2
Similarly,
 nb ‘4
n
σn
2
Q2ik1 Qik
≤c
2
509
1
.
n4 σn4
2
= O(n−2 σn−2 ).
Eηij
Therefore
2
2
) = O((nσn )−4 ).
cov(ηik
, ηik
2
1
Furthermore, since
P
1≤k1 <k2 ≤n (k1
Dn(2) = O
(16)
− 1) = O(n3 ), equation (16) implies
 1 ‘
 1 ‘
=
O
.
nσn4
nbn2
(17)
Thus, according to (15) and (17),
A(1)
n =1+O
(4)
 1 ‘
.
nb2n
(18)
Finally, we will show that An → 0 as n → ∞. Again, by the definition of
ηij we can write
|A(4)
n |
Œ
Œ
1 −1
ΠX kX
Œ
Œ
Œ
= 4Œ
Eηsk1 ηtk1 ηsk2 ηtk2 Œ ≤
k1 <k2 t<s
Œ
 nb ‘4 ”Œ X
Œ
Œ
n
≤c
Qsk1 Qsk2 Qtk1 Qtk2 Π+
Œ
σn
s,t,k1 ,k2
Υ
Œ ŒX
ŒX
Œ Œ
Œ
Œ
2
+Œ
Qks
Qkt Qst Qks Qss Π=
Q2kt Œ + Œ
k,s,t
k,s,t
 nb ‘4 ‚
ƒ
n
|En(1) | + |En(2) | + |En(3) | .
=c
σn
(19)
By the same argument as in (4), it can be shown that
X
(2)
(1) (2)
En(1) = n−7 b−8
Ψn (θs(1) , θk )Ψn (θt , θk ) ×
n
×
Z
s,t,k
1
0
(1)
Ψn (θs(1) , u)Ψn (θt , u)dPn(1) (u).
Hence, integrating by parts and taking into account that sup |Ψn (t1 , t2 )| ≤
cbn and sup |K (1) (u)| < ∞, we obtain
Z 1X
(2)
(1) (2)
Ψn (θs(1) , θk )Ψn (θt , θk ) ×
En(1) = n−7 b−8
n
0 s,t,k
510
R. ABSAVA AND E. NADARAYA
(1)
×Ψn (θs(1) , u)Ψn (θt , u))du = O
 1 ‘
.
n5 b4n
(20)
In the same manner, we can rewrite (20) as
Z 1Z 1Z 1Z 1
−4 −8
(1)
Ψn (z, u)Ψn (z, t)Ψn (y, u) ×
E n = n bn
0
0
0
0
×Ψn (y, t) du dt dy dz + O(n−5 b−4
n ).
It is easy to verify that
Z 1Z 1Z 1Z 1
n−4 bn−8
|Ψn (z, u)Ψn (z, t)Ψn (y, u)Ψn (y, t)| du dt dy dz ≤
0
0
0
Z
Z0
Œ  x − v ‘Œ
Œ
Œ
−4 −2
≤ cn bn
Œ dx dv ≤ cn−4 b−1
ŒK0
n .
b
n
Ωn (τ ) Ωn (τ )
Thus
 nb ‘4
n
σn
En(1) = O(bn ) + O
Furthermore, it is not difficillt to show
(nbn )4 σn−4 En(2) = O
and
(nbn )4 σn−4 En(3) = O
Therefore (19), (21), and (22) imply
A(4)
n →0
 1 ‘
.
nb2n
(21)
 1 ‘
nb2n
 1 ‘
.
nbn2
as n → ∞.
(22)
(23)
Combining relations (12), (13), (14), (18), and (23), we obtain
’X
“2
n
2
E
ξnk
− 1 → 0 as n → ∞.
k=1
Therefore
Un − EUn d
−→ ξ.
σn
But, due to Lemma 2, we have EUn = σ 2 θ1 + O(bn ) + O((nbn )−1 ) and
4 2
2
b−1
n σn → σ θ2 , and hence we obtain
−1/2
bb
 U − σ2 θ ‘
d
1
n
−→ ξ.
σ 2 θ2
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
511
Let us denote by
Tn = nbn
Z
Ωn (τ )
[gn (x) − g(x)]2 dx.
Theorem 2. Suppose g(x) ∈ Fs , s ≥ 2 and K(x) satisfies conditions
4
< ∞, nbn2 → ∞ and nb2s
1◦ − 3◦ . If, in addition, sup EZn1
n → 0, then
n≥1
d
(Tn − σ 2 θ1 )σ −2 θ2−1 −→ ξ.
b−1/2
n
Proof. Note that
(1)
Tn = Un + d(1)
n + dn ,
where
d(1)
n = 2nbn
Z
[gn (x) − Egn (x)] [Egn (x) − g(x)]dx,
Z
= nbn
[Egn (x) − g(x)]2 dx.
Ωn (τ )
d(2)
n
Ωn (τ )
It is easy to verify that
Egn (x) − g(x) =
= b−1
n
+bn−1
Z
1
with
But
Z
K
0
gen (x) =
b−1
n
n Z
X
i=1
1
K
0
∆i
x − t‘
x − t‘
bn
K
bn
x − t‘
i‘
dx g
− g(x) =
bn
n
[e
gn (t) − g(t)]dt +
[g(t) − g(x)]dt,
n
i‘
X
I∆i (x),
g
n
i=1
∆i =
x ∈ Ωn (τ ),
(24)
hi − 1 i i
, .
n n
Œ  ‘
Œ
Œ
Πi
sup |e
− g(x)Œ =
gn (x) − g(x)| ≤ max sup Œg
1≤i≤n x∈∆i
n
0≤x≤1
Œ
Œ
1‘
Œi
Œ
= max sup Œ − xŒ |g 0 (τi )| = O
, τi ∈ ∆i .
1≤i≤n x∈∆i n
n
Therefore the second term in (24) is O( n1 ). Since g(x) ∈ Fs and K(x)
satisfies conditions 1◦ –3◦ , we have
Œ
ŒZ 1 
Œ
Œ
x − t‘
Œ
[g(t) − g(x)]dtŒŒ =
sup Œ
K
b
n
x∈Ωn (τ )
0
512
R. ABSAVA AND E. NADARAYA
ŒZ
Œ
Œ
sup Œ
bsn
=
(s − 1)! x∈Ωn (τ )
τ
−τ
Z
0
1
(1 − t)
s−1 (s)
g
Thus from (24) and (25) it follows that
Œ
Œ
(x + tubn )u K(u)duŒŒ ≤ cbsn . (25)
s

1‘
sup |Egn (x) − g(x)| ≤ c bsn +
.
n
x∈Ωn (τ )
(26)
p 
€ 2s p
b−1/2
d(2)
bn + bs+1/2
+ n−1 bn .
n
n ≤ c nbn
n
(27)
Therefore
On the other hand, (9) and (26) yield
šZ
1/2
(1)
|
≤
E(gn (x) − Egn (x))2 dx ×
E|d
b−1/2
nb
n
n
n
×
Z
Ωn (τ )
Ωn (τ )
(Egn (x) − g(x))2 dx
›1/2
√
≤ c nbs .
(28)
Finally, the statement of Theorem 2 follows directly from Theorem 1, (27)
and (28).
Using Theorem 2, it is easy to solve the problem of testing the hypothesis
about g(x). Given σ 2 , it is required to verify the hypothesis
H0 : g(x) = g0 (x),
x ∈ Ωn (τ ).
The critical region is defined approximately by the inequality
Z
Tn0 = nbn (gn (x) − g0 (x))2 dx ≥ qn (α),
(29)
where
p
€

qn (α) = σ 2 θ1 + λα bn θ2 ,
θ1 =
Z
K 2 (u)du,
θ22 = 2
Z
K02 (u)du,
and λα is the quantile of level 1 − α of the standard normal distribution,
i.e.,
Z u
 x2 ‘
dx.
Φ(λα ) = 1 − α, Φ(u) = (2π)−1/2
exp −
2
−∞
−1/2
Suppose now that σ 2 is unknown. We will use the bn
of variance σ 2 (see, for example, [5]):
Sn2 =
n−1
X
1
(Yi+1 − Yi )2 ,
2(n − 1) i=1
-consistent estimate
Yi ≡ Yni .
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
513
Indeed,
ESn2 =
n−1
n−1
 i ‘i2
X
X h i − 1‘
1
1
Ei2 +
g
−g
,
2(n − 1) i=1
2(n − 1) i=1
n
n
where i = Zi+1 − Zi , Zi ≡ Zni . Moreover, since E2i = 2σ 2 , we can write
ESn2 = σ 2 +
n−1
1
1 X (1)
(g (τi ))2 = σ 2 + O(n−2 ).
2(n − 1) n2 i=1
To calculate the variance, it is sufficient to note that
E(Sn2 − σ 2 )2 =
=
n−1
X
X
1
1
E2i E2j =
E4i +O(n−2 )−σ 2 +
2
2
4(n − 1) i=1
4(n − 1)
i6=j
1
4(n − 1)2
n−1
X
E4i +
i=1
=
(n − 2)(n − 3)
(2σ 2 )2 − σ 4 + O(n−1 ) =
4(n − 1)2
n−1
X
1
E4 + O(n−1 ).
4(n − 1)2 i=1 i
4
< ∞, this yields E(Sn2 − σ 2 )2 = O(n−1 ). Therefore
Since sup EZn1
n≥1
P
(Sn2 − σ 2 ) −→ 0.
b−1/2
n
Corollary. Under the conditions of Theorem 2
 T 0 − S2 θ ‘
d
n 1
n
b−1/2
−→ ξ.
n
Sn2 θ2
This corollary enables us to construct a test for verifying
H0 : g(x) = g0 (x).
The critical region is defined approximately by the inequality Tn0 ≥ qen (α),
where qen (α) is obtained from qn (α) by using Sn2 instead of σ 2 .
Consider now the asymptotic properties of test (29) (i.e., the asymptotic
behaviour of the power function as n → ∞). First, let us study the question
whether the corresponding test is consistent.
Theorem 3. Under the conditions of Theorem 2
Πn (g1 ) = PH1 (Tn0 ≥ qn (α)) → 1,
n → ∞,
i.e., the test defined by (29) is consistent under any alternatives
Z 1
H1 : g(x) = g1 (x) 6= g0 (x), ∆ =
(g1 (x) − g0 (x))2 dx > 0.
0
514
R. ABSAVA AND E. NADARAYA
Proof. Denote
’
Z
Zn (g1 ) = b−1/2
nb
n
n
Ωn (τ )
“
(gn (x) − g1 (x))2 dx − σ 2 θ1 σ −2 θ2−1 .
It is easy to show that
ˆ
‰
Πn (g1 ) = PH1 Zn (g1 ) ≥ −nbn1/2 (θ2−1 σ −2 ∆ + op (1)) .
Since Zn (g1 ) is asymptoticaly normally distributed with parameters (0, 1)
1/2
under hypothesis H1 , nbn → ∞ and ∆ > 0, we have Πn (g1 ) → 1 as
n → ∞.
Thus under any fixed alternative the power of test (29) tends to 1 as n →
∞. Nevertheless, if the alternative hypothesis varies with n and becomes
“closer” to the null hypothesis H0 , the power of the test may not converge
to 1 depending on the rate at which the alternative approaches the null
hypothesis. In our case the sequence of “close” alternatives has the form
Hn : gen (x) = g0 (x) + γn φ(x) + o(γn ).
(30)
Theorem 4. Suppose g0 (x) and ϕ(x) are from Fs , but K(x) satisfies
4
< ∞. If bn = n−δ , γn = n−1/2+δ/4 , 1/2s <
coditions 1◦ –3◦ and sup EZn1
n≥1
δ < 1/2, then under alternatives Hn the statistic
b−1/2
(Tn0 − θ1 σ 2 )σ −2 θ2−1
n
has the limiting normal distribution with parameters ( σ21θ2
R1
0
ϕ2 (x)dx, 1).
Proof. Let us represent Tn0 as the sum
Z
Z
Tn0 = nbn
(e
(gn (x) − gen (x))2 dx + nbn
gn (x) − g0 (x))2 dx +
Ωn (τ )
Ωn (τ )
Z
(gn (x) − gen (x))(e
+2nbn
gn (x) − g0 (x))dx = Tn1 + A1 (n) + A2 (n).
Ωn (τ )
It is easy to check that
b−1/2
A1 (n) =
n
Z
1
ϕ2 (u)du + O(n−δ ).
0
Let us introduce the random variable
Z
(gn (x) − E1 gn (x))ϕ(x) dx.
dn =
Ωn (τ )
(31)
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
515
Here E1 denotes the mathematical expectation under the hypothesis Hn .
We can derive the inequality
•
”
Z
1/2
−1/2
|E1 gn (x) − gen (x)|ϕ(x) dx =
bn E|A2 (n)| ≤ nbn γn E|dn | +
Ωn (τ )
=
nbn1/2 γn E|dn |
+ O(nγn bns+1/2 ).
But
E|dn | ≤ σ
šX
Z
n ’
bn−1
i=1
Ωn (τ )
’Z
K
∆i
“2 ›1/2
x − t‘ “
≤ cn−1/2 .
dt )ϕ(x) dx
bn
Therefore
€
2−(4s+1)δ 
4
bn−1/2 E|A2 (n)| ≤ c n−δ/4 + n
→ 0.
(32)
−1/2
Referring to the proof of Theorem 2 it is easy to verify that bn (Tn0 −
σ 2 θ1 )σ −2 θ2−1 is asymptotically normally distributed with parameters (0,1).
Hence, from (31) and (32) we conclude that the theorem is valid.
Remark 1. It follows from Theorem 4 that more closer alternatives of
form (30) (i.e., under γn n1/2−δ/4 → 0) are not distinguished from H0 by
this test (i.e., PHn (Tn0 ≥ qn (α)) → α), and for more remote alternatives (i.e.,
under γn n1/2−δ/4 → ∞) the corresponding test preserves the consistency
property (i.e., PHn (Tn0 ≥ qn (α)) → 1).
Thus the local behaviour of the power PHn (Tn0 ≥ qn (α)) is
’
“
Z 1
1
2
PHn (Tn ≥ qn (α)) → 1 − Φ λα − 2
ϕ (u) du .
(33)
σ θ2 0
R1
Since 0 ϕ2 (u)du > 0 and is equal to zero if and only if ϕ(u) = 0, it follows
from (33) that the test for testing the hypothesis H0 : g(x) = g0 (x) against
the alternative of form (30) is asymptotically strictly unbiased.
Remark 2. Theorem 4 is analogous to Theorem 6.1 from the book by
J. D. Hart [6] with the only difference that [6] deals with the statistic based
on the Priestley–Chao estimate [7].
Remark 3. If in the interval [0, 1] we choose points xnk in a more general
manner, i.e.,
Z
xnk
p(u) du = k/n,
k = 1, . . . , n,
0
where p(x) > 0, x ∈ [0, 1], is the probability density satisfying certain
conditions of smoothness, then Theorems 1–4 remain valid provided that
the parameters θ1 and θ2 are changed appropriately.
516
R. ABSAVA AND E. NADARAYA
Remark 4. If instead of K(x) we consider its modification Kq,r (x) from
[2], then we can give the hypothesis H0 on the entire interval [0, 1]. For the
corresponding estimate we have gn∗ (x) ≡ gn (x), x ∈ Ωn , while the relation
“
’Z 1
Z
P
∗
2
2
1/2
(gn − g) dx −
(gn − g) dx −→0
nbn
0
Ωn
will be proved in our forthcoming paper.
s
2
Remark 5. If in Theorem 4 we set δ = 2s+1
, then γn = n− 2s+1 . By
Yu. Ingster’s results [8] the test Tn is minimax consistent with respect to
alternatives of form (30).
Acknowledgement
The authors express their thanks to the referee for his careful reading of
the paper and comments that led to adding Remarks 2–5.
References
1. T. Gasser and H. Müller, Kernel estimation of regression functions.
Smoothing technique for curve estimation. Eds. T. Gasser and M. Rosenblatt. Lecture Notes in Math. 757, 23–68, Springer, Berlin etc., 1979.
2. J. Hart and T. Wehrly, Kernel regression when the boundary is large.
J. Amer. Statist. Assoc. 87(1992), No. 420, 1018–1022.
3. R. S. Lipster and A. N. Shiryaev, A functional limit theorem for
semimartingales. (Russian) Teor. Veroyatnost. i Primenen. 25(1980), No.
4, 683–703.
4. E. A. Nadaraya, Nonparametric estimation of probarility densities and
regression curves. Kluver Academic Publishers, Dordrecht, 1989.
5. K. A. Brownlee, Statistical theory and methodology in science and
engineering. Wiley, New York, 1960.
6. J. D. Hart, Nonparametric smoothing and lack-of-fit tests. Springer,
New York, 1997.
7. M. B. Priestley and M. T. Chao, Nonparametric function fitting. J.
Roy. Stat. Soc. Ser. B 34(1972), 385–392.
8. Yu. I. Ingster, Asymptotically minimax hypothesis testing for nonparametric alternatives, I. Math. Methods Statist. 2(1993), No. 2, 84–114.
(Received 20.05.1998)
Authors’ address:
I. Javakhishvili Tbilisi State University
2, University St., Tbilisi 380043
Georgia