The asymptotic behaviour of the power of the proposed criteria under contiguous alternatives is studed

LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION OF THE GASSER–M ¨ULLER

NONPARAMETRIC ESTIMATE OF THE REGRESSION FUNCTION

R. ABSAVA AND E. NADARAYA

Abstract. Asymptotic distribution of the mean square deviation of the Gasser–M¨uller 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:

Ynk=g(xnk) +Znk, xnk= k

n, k= 1, . . . , n, n≥1,

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 thatEZnk= 0, EZ_nk² =σ²,k= 1, . . . , n,n≥1.

Let us consider the estimate of the functiong(x) from [1]:

gn(x) = Xn

i=1

Wni(x)Yni, x∈[0,1], where

Wni=b⁻_n¹ Z

∆i

Kx−t bn

‘dt, ∆i=hi−1 n , i

n i.

HereK(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:

1991Mathematics Subject Classification. 62G07.

Key words and phrases. Limit distribution, consistency, power of criteria, contiguous alternatives.

501

1072-947X/99/1100-0501$16.00/0 c1999 Plenum Publishing Corporation

1^◦. supp(K)R ⊂[−τ, τ], 0< τ <∞, sup|K|<∞, K(u)du= 1, K(−x) =K(x),

2^◦. R

K(u)u^jdu= 0, 0< j < s, R

K(u)u^sdu6= 0, 3^◦. K(x) has a bounded derivative inR= (−∞,∞).

Denote by Fs the family of regression functions g(x), x ∈ [0,1], hav- ing derivatives of order up to s (s ≥2), g^(s)(x) is continuous. Note that Pn

i=1Wni(x) = 1 forx∈Ωn(τ) = [τ bn,1−τ bn] andEgn(x) =g(x) +O(b²_n) if the kernelK(x) satisfies condition 1^◦andg(x)∈F2. On the other hand, Pn

i=1Wni(x) 6= 1 for x ∈ [0, τ bn)∪(1−τ bn,1] and it may happen that Egn(x)6→g(x), for example,Egn(0)→ ^g(0)₂ (orEgn(1)→ ^g(1)₂ ). If the es- timategn(x) is divided byPn

i=1Wni(x), then the proposed estimate egn(x) becomes asymptotically unbiased and, moreover, Eegn(x) = g(x) +O(bn) forx∈[0, τ bn)∪(1−τ bn,1]. Hence the asymptotic behaviour of the esti- mategn(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 estimatorgn(x) (see, for examle, [2]). It would be interesting to investigate the limit behaviour of the distribution of the mean square deviation ofgn(x) fromg(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 func- tional limit theorems for semimartingales from [3].

We will use the notation:

Un=nbn

Z

Ωn(τ)

(gn(x)−Egn(x))²dx, σ²_n= 4σ⁴(nbn)² Xn

k=2 kX−1

i=1

Q²_ik(n), Qij ≡Qij(n) =

Z

Ωn(τ)

Wni(x)Wnj(x)dx, ηik≡ηik(n) = 2nbnQikZniZnkσ_n⁻¹, ξnk=

k−1

X

i=1

ηik, k= 2, . . . , n, ξn1= 0, ξnk= 0, k > n, Fk⁽ⁿ⁾=σ(ω:Zn1, Zn2, . . . , Znk),

whereFk⁽ⁿ⁾is theσ-algebra generated by the random variablesZn1,Zn2, . . ., Znk,F0⁽ⁿ⁾= (∅,Ω).

Lemma 1 ([4], p. 179). The stochastic sequence (ξnk,Fk⁽ⁿ⁾)k≥1 is a martingale-difference.

Lemma 2. Suppose the kernel K(x) satisfies conditions 1^◦ and 3^◦. If

nb²_n → ∞, then

b⁻_n¹σ²_n→2σ⁴ Z 2τ

−2τ

K₀²(u)du, (1)

and

EUn =σ² Z

K²(u)du+O(bn) +O 1 nbn

‘

, (2)

whereK0=K∗K (the symbol ∗denotes the convolution).

Proof. It is not difficult to see that σ_n² = 2σ⁴(nbn)²

”Xn k=1

Xn i=1

Q²_ki− Xn i=1

Q²_ii

•

=d1(n) +d2(n). (3) Here by the definition ofQki we have

d1(n) = 2σ⁴n²b⁻_n²X

i,j

’ Z

∆i

Z

∆j

Ψn(t1, t2)dt1dt2

“2

, where

Ψn(t1, t2) = Z

Ωn(τ)

Kx−t1

bn

‘

Kx−t2

bn

‘ dx.

Using the mean value theorem, we can rewrited1(n) as d1(n) = 2σ⁴(nbn)⁻²X

i,j

Ψ²_n(θ⁽¹⁾_i , θ⁽²⁾_j ), withθ⁽¹⁾_i ∈∆i,θ_i⁽²⁾∈∆j.

Let P(x) be the uniform distribution function on [0,1] and Pn⁽ⁱ⁾(x) the empirical distribution function of the “sample” θ⁽ⁱ⁾₁ , . . . , θn⁽ⁱ⁾, i = 1,2, i.e., Pn⁽ⁱ⁾(x) =n⁻¹Pn

k=1I(−∞,x)(θ_k⁽ⁱ⁾),whereIA(·) is the indicator of the setA.

Thend1(n) can be written as the integral d1(n) = 2σ⁴b⁻_n²

Z 1 0

Z 1 0

Ψ²_n(x, y)dP_n⁽¹⁾(x)dP_n⁽²⁾(y). (4) It is easy to check that

ŒŒ

ŒŒ Z 1

0

Z 1 0

Ψ²_n(x, y)dP_n⁽¹⁾(x)dP_n⁽²⁾(y)− Z 1

0

Z 1 0

Ψ²_n(x, y)dP(x)dP(y)

ŒŒ

ŒŒ≤I1+I2, where

I1=

ŒŒ

ŒŒ Z 1

0

Z 1 0

Ψ²_n(x, y)dP_n⁽²⁾(y)[dP_n⁽¹⁾(x)−dP(x)]

ŒŒ

ŒŒ,

I2=

ŒŒ

ŒŒ Z 1

0

Z 1 0

Ψ²_n(x, y)dP(x)[dP_n⁽²⁾(y)−dP(y)]

ŒŒ

ŒŒ.

Furthermore, the integration by parts of the internal integral inI1 yields I1≤2

Z 1 0

dP_n⁽²⁾(y) Z 1

0

’

|P_n⁽¹⁾(x)−P(x)| |Ψn(x, y)|b⁻_n¹×

× Z

Ωn(τ)

ŒŒ

ŒK⁽¹⁾t−x bn

‘ŒŒŒ

ŒŒ

ŒKt−y bn

‘ŒŒŒdt

“

dx. (5)

Since sup

0≤x≤1|Pn⁽ⁱ⁾−P(x)| ≤ n² and|Ψn(x, y)| ≤cbn, we have I1=O(bn/n).

Here and in what followscis the positive constant varying from one formula to another.

In the same manner we may show that I2=O(bn/n).

Therefore

d1(n) = 2σ⁴b⁻_n² Z 1

0

Z 1 0

Ψ²_n(t1, t2)dt1dt2+O 1 nbn

‘

. (6)

It is easy to show also that d1(n) = 2σ⁴

Z

Ωn(τ)

Z

Ωn(τ)

’ Z xb⁻_n¹ (x−1)b⁻n¹

K(u)Kx−y bn −u‘

du

“2

dxdy+O 1 nbn

‘ . Since [−τ, τ]⊂[^x_b⁻¹

n ,_b^x

n] for allx∈Ωn(τ), we have d1(n) = 2σ⁴

Z

Ωn(τ)

Z

Ωn(τ)

K₀²x−y bn

‘

dx dy+O 1 nbn

‘ . But it is easy to see that

b⁻_n¹d1(n) = 2σ⁴ Z 1

0

’ Z (1−y)/bn−τ τ−y/bn

K₀²(u)du

“

dy+O(bn) +O 1 nbn

‘ . Therefore

b⁻_n¹d1(n)→2σ⁴ Z

K₀²(u)du. (7)

We can directly verify that b⁻_n¹|d2(n)|≤n²b⁻_n³

Xn i=1

Z

∆i

Z

∆i

Z

∆i

Z

∆i

’ Z

Ωn(τ)

ŒŒ

ŒKx−t1

bn

‘

Kx−t2

bn

‘ŒŒŒdx×

× Z

Ωn(τ)

ŒŒ

ŒKy−t3

bn

‘

Ky−t4

bn

‘ŒŒŒdy

“

dt1dt2dt3dt4≤c 1

nbn. (8) Statement (1) immediately follows directly from (3), (7), and (8).

Let us now show that (2) holds. We have Dgn(x) = σ²

nb²_n Z 1

0

K²x−t bn

‘

dPn(x), where

Pn(x) =n⁻¹ Xn

k=1

I(−∞,x)(θk), θk ∈∆k, k= 1, . . . , n.

Applying the same argument as in deriving (6), we find Dgn(x) = σ²

nb²_n Z 1

0

K²x−t bn

‘

dt+O 1 (nbn)²

‘

. (9)

Furthermore, taking into account that [−τ, τ]⊂[^x_b⁻¹

n ,_b^x

n] for allx∈Ωn(τ) by (9) we can write

Dgn(x) = σ² nbn

Z

K²(u)du+O((nbn)⁻²).

Thus

EUn=σ² Z

K²(u)du+O(bn) +O 1 nbn

‘ .

Theorem 1. Suppose the kernelK(x) satisfies conditions1^◦ and3^◦. If sup

n≥1

EZ_n1⁴ <∞ andnb²_n→ ∞, then

b⁻_n^1/2(Un−σ²θ1)σ⁻²θ⁻₂¹−→^d ξ, where

θ1= Z

K²(u)du, θ²₂= 2 Z

K₀²(u)du.

By the symbol −→^d we denote the convergence in distribution, ξ is a random variable distributed according to the standard normal distribution.

Proof. Note that

Un−EUn

σn

=H_n⁽¹⁾+H_n⁽²⁾, where

H_n⁽¹⁾ = Xn

k=1

ξnk, H_n⁽²⁾= nbn

σn

Xn

i=1

(Z_ni² −EZ_ni²)Qii.

Hn⁽²⁾ converges to zero in probability. Indeed, DH_n⁽²⁾=(nbn)²

σ²_n Xn i=1

EZ_ni⁴Q²_ii≤sup

n≥1

EZ_n1⁴ (nbn)² σ_n²

Xn i=1

Q²_ii = sup

n≥1

EZ_n1⁴ |d2(n)| σ_n² . This and (8) yield DHn⁽²⁾ = O(1/(nσ_n²)) = O(1/(nbn)). Hence Hn⁽²⁾

−→P 0.

(By the symbol−→^P we denote the convergence in probability).

Now let us prove the convergenceHn⁽¹⁾ d

−→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 are fulfilled by the sequence {ξnk,Fk⁽ⁿ⁾}k≥1 which is a square-integrable martingale-difference according to Lemma 1.

A direct calculation shows that Pn

k=1Eξ_nk² = 1. Now the asymptotic normality will take place if for n→ ∞we have

Xn

k=1

E[ξ_nk² I(|ξnk| ≥)Fk⁽ⁿ⁾−1]−→^P 0 (10) and

Xn k=1

ξ²_nk−→^P 1. (11)

But, as shown in [3], the validity of (11) together with the condition sup

1≤k≤n|ξnk|−→^P 0 implies the statement (10) as well.

Since for >0 Pˆ

sup

1≤k≤n|ξnk| ≥‰

≤⁻⁴ Xn k=1

Eξ⁴_nk,

to prove Hn⁽¹⁾

−→d ξ we have to verify only (11), since relation (12) given below is fulfiled.

Now we will establishPn

k=1ξ_nk² −→^P 1. It is sufficient to verify that EXⁿ

k=1

ξ_nk² −1‘2

→0 asn→ ∞,i.e., due toPn

k=1Eξ²_nk= 1 EXⁿ

k=1

ξ_nk² ‘2

= Xn

k=1

Eξ_nk⁴ + 2 X

1≤k1<k2≤n

Eξ_nk² ₁ξ²_nk2 →1.

First show thatPn

k=1Eξ⁴_nk→0 asn→ ∞. By virtue of the definitions of ξnkandηij we can write

Xn k=1

Eξ⁴_nk=I_n⁽¹⁾+I_n⁽²⁾, where

I_n⁽¹⁾ =16(nbn)⁴ σ⁴_n

Xn

k=2

EZ_nk⁴

k−1

X

j=1

EZ_nj⁴ Qjk,

I_n⁽²⁾ =48(nbn)⁴ σ⁴_n

Xn

k=2 kX−1

i6=j

EZ_ni²EZ_nj² Q²_ikQ²_jk. On the other hand, since sup

n≥1

EZ_n1⁴ < ∞, |Qij| ≤ cb⁻_n¹n⁻² and b⁻_n¹σ²_n → σ⁴θ²₂,we have In⁽¹⁾ =O((nbn)⁻²),In⁽²⁾=O(1/(nσ⁴_n)) =O(1/(nb²_n)).Hence

Xn

k=1

Eξ⁴_nk→0, n→ ∞. (12)

Now let us establish that

2 X

1≤k1<k2≤n

Eξ_nk² ₁ξ²_nk2 →1 asn→ ∞.The definition ofξni yields

ξ_nk² ₁·ξ_nk² ₂ =B⁽¹⁾_k1k2+B_k⁽²⁾_1k2+B_k⁽³⁾_1k2+B_k⁽⁴⁾_1k2, where

B_k⁽¹⁾_1k2=σ2(k1)σ2(k2), B_k⁽²⁾_1k2 =σ2(k1)σ1(k2), B_k⁽³⁾_1k2=σ1(k1)σ2(k2), B_k⁽⁴⁾_1k2 =σ1(k1)σ1(k2),

σ1(k) =X

i6=j

ηikηjk, σ2(k) =

k−1

X

i=1

η_ik². Therefore

2 X

1≤k1<k2≤n

Eξ_nk² ₁ξ_nk² ₂ = X4 i=1

A⁽ⁱ⁾_n , where

A⁽ⁱ⁾_n = 2 X

1≤k1<k2≤n

EB_k⁽ⁱ⁾_1k2, i= 1,2,3,4.

Let us consider the termA⁽³⁾n .According to the definition of ηij we have Eη²_ik2ηsk1ηtk1 = 0, s6=t, k1< k2.

Thus

A⁽⁴⁾_n = 0. (13)

To estimate the termA⁽²⁾n , note that sup

n≥1

EZ_n1⁴ <∞and|Qij| ≤cb⁻_n¹n⁻². So we obtain

|A⁽²⁾_n | ≤cnb²_n σ⁴_n

Xn k1=2

kX1−1 i=1

Q²_ik1 ≤c 1

nσ_n² =O 1 nbn

‘

. (14)

Now we will verify thatA⁽¹⁾n →1 asn→ ∞. For this let us representA⁽¹⁾n

in the form

A⁽¹⁾_n =D_n⁽¹⁾+D⁽²⁾_n , where

D_n⁽¹⁾= 2 X

1≤k1<k2≤n

’kX1−1 i=1

Eη²_ik1

“’kX2−1 j=1

Eη²_jk2

“ ,

D_n⁽²⁾= 2’ X

k1<k2

B_k⁽¹⁾_1k2− X

k1<k2

’kX1−1 i=1

Eη²_ik1

“’kX2−1 j=1

Eη_jk²₂

““

. From the definition ofσ_n² it follows that

D_n⁽¹⁾= 1− Xn

k=2

’kX−1 i=1

Eη²_ik

“2

, where

Xn

k=2

’^kX−1 i=1

Eη_ik²

“2

≤cnbn

σn

‘4Xn k=2

’^kX−1 i=1

Q²_ik

“2

≤c 1

nσ_n⁴ =O 1 nb²_n

‘ . Therefore

D⁽¹⁾_n →1 as n→ ∞. (15)

Next we will show thatD⁽²⁾n →0 asn→ ∞.It is easy to verify that D⁽²⁾_n = 2 X

k1<k2

”kX1−1 i=1

cov(η²_ik1, η_ik²₂) +

kX1−1 i=1

cov(η_ik²₁, η²_k1k2)

• .

But

Eη²_ik1η_ik²₂ ≤cnbn

σn

‘4

Q²_ik1Q²_ik2 ≤c 1 n⁴σ_n⁴. Similarly,

Eη_ij² =O(n⁻²σ⁻_n²).

Therefore

cov(η²_ik1, η²_ik2) =O((nσn)⁻⁴). (16) Furthermore, sinceP

1≤k1<k2≤n(k1−1) =O(n³), equation (16) implies D⁽²⁾_n =O 1

nσ_n⁴

‘

=O 1 nb²_n

‘

. (17)

Thus, according to (15) and (17),

A⁽¹⁾_n = 1 +O 1 nb²_n

‘

. (18)

Finally, we will show thatA⁽⁴⁾n →0 as n→ ∞.Again, by the definition of ηij we can write

|A⁽⁴⁾_n |= 4

ŒŒ

ŒŒ X

k1<k2

kX1−1 t<s

Eηsk1ηtk1ηsk2ηtk2

ŒŒ

ŒŒ≤

≤cnbn

σn

‘4”ŒŒŒ X

s,t,k1,k2

Qsk1Qsk2Qtk1Qtk2

ŒŒ

Œ+ +ŒŒŒX

k,s,t

Q²_ksQ²_ktŒŒŒ+ŒŒŒX

k,s,t

QktQstQksQss

ŒŒ

Œ

•

=

=cnbn

σn

‘4‚

|E_n⁽¹⁾|+|E_n⁽²⁾|+|E_n⁽³⁾|ƒ

. (19)

By the same argument as in (4), it can be shown that E_n⁽¹⁾=n⁻⁷b⁻_n⁸X

s,t,k

Ψn(θ_s⁽¹⁾, θ_k⁽²⁾)Ψn(θ_t⁽¹⁾, θ_k⁽²⁾)×

× Z 1

0

Ψn(θ_s⁽¹⁾, u)Ψn(θ_t⁽¹⁾, u)dP_n⁽¹⁾(u).

Hence, integrating by parts and taking into account that sup|Ψn(t1, t2)| ≤ cbn and sup|K⁽¹⁾(u)|<∞,we obtain

E_n⁽¹⁾=n⁻⁷b⁻_n⁸ Z 1

0

X

s,t,k

Ψn(θ_s⁽¹⁾, θ_k⁽²⁾)Ψn(θ_t⁽¹⁾, θ_k⁽²⁾)×

×Ψn(θ⁽¹⁾_s , u)Ψn(θ⁽¹⁾_t , u))du=O 1 n⁵b⁴_n

‘

. (20)

In the same manner, we can rewrite (20) as E_n⁽¹⁾=n⁻⁴b⁻_n⁸

Z 1 0

Z 1 0

Z 1 0

Z 1 0

Ψn(z, u)Ψn(z, t)Ψn(y, u)×

×Ψn(y, t)du dt dy dz+O(n⁻⁵b⁻_n⁴).

It is easy to verify that n⁻⁴b⁻_n⁸

Z 1 0

Z 1 0

Z 1 0

Z 1

0 |Ψn(z, u)Ψn(z, t)Ψn(y, u)Ψn(y, t)|du dt dy dz≤

≤cn⁻⁴b⁻_n² Z

Ωn(τ)

Z

Ωn(τ)

ŒŒ

ŒK0

x−v bn

‘ŒŒŒdx dv≤cn⁻⁴b⁻_n¹. Thus

nbn

σn

‘4

E_n⁽¹⁾=O(bn) +O 1 nb²_n

‘

. (21)

Furthermore, it is not difficillt to show

(nbn)⁴σ_n⁻⁴E_n⁽²⁾=O 1 nb²_n

‘

and

(nbn)⁴σ_n⁻⁴E⁽³⁾_n =O 1 nb²_n

‘

. (22)

Therefore (19), (21), and (22) imply

A⁽⁴⁾_n →0 as n→ ∞. (23)

Combining relations (12), (13), (14), (18), and (23), we obtain E

’Xn k=1

ξ_nk² −1

“2

→0 as n→ ∞. Therefore

Un−EUn

σn

−→d ξ.

But, due to Lemma 2, we have EUn = σ²θ1+O(bn) +O((nbn)⁻¹) and b⁻_n¹σ_n²→σ⁴θ₂²,and hence we obtain

b⁻_b^1/2Un−σ²θ1

σ²θ2

‘ _d

−→ξ.

Let us denote by

Tn=nbn

Z

Ωn(τ)

[gn(x)−g(x)]²dx.

Theorem 2. Suppose g(x) ∈ Fs, s ≥ 2 and K(x) satisfies conditions 1^◦−3^◦. If, in addition,sup

n≥1EZ_n1⁴ <∞, nb²_n → ∞andnb^2s_n →0, then b⁻_n^1/2(Tn−σ²θ1)σ⁻²θ₂⁻¹−→^d ξ.

Proof. Note that

Tn =Un+d⁽¹⁾_n +d⁽¹⁾_n , where

d⁽¹⁾_n = 2nbn

Z

Ωn(τ)

[gn(x)−Egn(x)] [Egn(x)−g(x)]dx, d⁽²⁾_n =nbn

Z

Ωn(τ)

[Egn(x)−g(x)]²dx.

It is easy to verify that Egn(x)−g(x) =b⁻_n¹

Xn i=1

Z

∆i

Kx−t bn

‘dx gi n

‘−g(x) =

=b⁻_n¹ Z 1

0

Kx−t bn

‘

[egn(t)−g(t)]dt+ +b⁻_n¹

Z 1 0

Kx−t bn

‘

[g(t)−g(x)]dt, x∈Ωn(τ), (24) with

e gn(x) =

Xn i=1

gi n

‘

I∆i(x), ∆i=hi−1 n , i

n i

. But

sup

0≤x≤1|egn(x)−g(x)| ≤ max

1≤i≤n sup

x∈∆i

ŒŒ

Œgi n

‘−g(x)ŒŒŒ=

= max

1≤i≤n sup

x∈∆i

ŒŒ

Œi

n−xŒŒŒ|g⁰(τi)|=O1 n

‘

, τi∈∆i.

Therefore the second term in (24) is O(_n¹). Since g(x) ∈ Fs and K(x) satisfies conditions 1^◦–3^◦,we have

sup

x∈Ωn(τ)

ŒŒ

ŒŒ Z 1

0

Kx−t bn

‘

[g(t)−g(x)]dt

ŒŒ

ŒŒ=

= b^s_n

(s−1)! sup

x∈Ωn(τ)

ŒŒ

ŒŒ Z τ

−τ

Z 1 0

(1−t)^s−1g^(s)(x+tubn)u^sK(u)du

ŒŒ

ŒŒ≤cb^s_n.(25) Thus from (24) and (25) it follows that

sup

x∈Ωn(τ)|Egn(x)−g(x)| ≤c b^s_n+ 1

n

‘

. (26)

Therefore

b⁻_n^1/2d⁽²⁾_n ≤c€ nb^2s_np

bn+b^s+1/2_n +n⁻¹p bn

. (27) On the other hand, (9) and (26) yield

b⁻_n^1/2E|d⁽¹⁾_n | ≤nb^1/2_n

š Z

Ωn(τ)

E(gn(x)−Egn(x))²dx×

× Z

Ωn(τ)

(Egn(x)−g(x))²dx

›1/2

≤c√

nb^s. (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 aboutg(x). Givenσ², it is required to verify the hypothesis

H0:g(x) =g0(x), x∈Ωn(τ).

The critical region is defined approximately by the inequality T_n⁰=nbn

Z

(gn(x)−g0(x))²dx≥qn(α), (29) where

qn(α) =σ²€ θ1+λα

pbnθ2

, θ1= Z

K²(u)du, θ₂²= 2 Z

K₀²(u)du, and λα is the quantile of level 1−αof the standard normal distribution, i.e.,

Φ(λα) = 1−α, Φ(u) = (2π)^−1/2 Z u

−∞

exp

−x² 2

‘dx.

Suppose now thatσ²is unknown. We will use theb⁻n^1/2-consistent estimate of varianceσ²(see, for example, [5]):

S_n²= 1 2(n−1)

nX−1 i=1

(Yi+1−Yi)², Yi≡Yni.

Indeed,

ES_n²= 1 2(n−1)

n−1

X

i=1

E²_i + 1 2(n−1)

n−1

X

i=1

hgi−1 n

‘−gi n

‘i2

, wherei=Zi+1−Zi,Zi≡Zni.Moreover, sinceE²_i = 2σ²,we can write

ES_n²=σ²+ 1 2(n−1)

1 n²

n−1

X

i=1

(g⁽¹⁾(τi))²=σ²+O(n⁻²).

To calculate the variance, it is sufficient to note that E(S_n²−σ²)²= 1

4(n−1)²

nX−1 i=1

E⁴_i+O(n⁻²)−σ²+ 1 4(n−1)²

X

i6=j

E²_iE²_j=

= 1

4(n−1)²

nX−1

i=1

E⁴_i +(n−2)(n−3)

4(n−1)² (2σ²)²−σ⁴+O(n⁻¹) =

= 1

4(n−1)²

nX−1

i=1

E⁴_i +O(n⁻¹).

Since sup

n≥1

EZ_n1⁴ <∞,this yields E(S_n²−σ²)²=O(n⁻¹).Therefore b⁻_n^1/2(S_n²−σ²)−→^P 0.

Corollary. Under the conditions of Theorem 2 b⁻_n^1/2T_n⁰−S_n²θ1

S_n²θ2

‘ _d

−→ξ.

This corollary enables us to construct a test for verifying H0:g(x) =g0(x).

The critical region is defined approximately by the inequality T_n⁰≥eqn(α), whereqen(α) is obtained fromqn(α) by usingS_n² instead ofσ².

Consider now the asymptotic properties of test (29) (i.e., the asymptotic behaviour of the power function asn→ ∞). First, let us study the question whether the corresponding test is consistent.

Theorem 3. Under the conditions of Theorem 2 Πn(g1) =PH1(T_n⁰≥qn(α))→1, n→ ∞, i.e., the test defined by (29)is consistent under any alternatives

H1:g(x) =g1(x)6=g0(x), ∆ = Z 1

0

(g1(x)−g0(x))²dx >0.

Proof. Denote Zn(g1) =b⁻_n^1/2

’ nbn

Z

Ωn(τ)

(gn(x)−g1(x))²dx−σ²θ1

“

σ⁻²θ⁻₂¹. It is easy to show that

Πn(g1) =PH1

ˆZn(g1)≥ −nb^1/2_n (θ₂⁻¹σ⁻²∆ +op(1))‰ .

Since Zn(g1) is asymptoticaly normally distributed with parameters (0,1) under hypothesis H1, nb^1/2n → ∞ and ∆ > 0, we have Πn(g1) → 1 as n→ ∞.

Thus under any fixed alternative the power of test (29) tends to 1 asn→

∞. Nevertheless, if the alternative hypothesis varies with n and becomes

“closer” to the null hypothesisH0,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:egn(x) =g0(x) +γnφ(x) +o(γn). (30) Theorem 4. Suppose g0(x) and ϕ(x) are from Fs, but K(x) satisfies coditions 1^◦–3^◦ andsup

n≥1

EZ_n1⁴ <∞. Ifbn =n^−δ,γn =n^−1/2+δ/4,1/2s <

δ <1/2, then under alternativesHn the statistic b⁻_n^1/2(T_n⁰−θ1σ²)σ⁻²θ⁻₂¹

has the limiting normal distribution with parameters(_σ2¹θ2

R1

0 ϕ²(x)dx,1).

Proof. Let us representT_n⁰ as the sum T_n⁰=nbn

Z

Ωn(τ)

(gn(x)−gen(x))²dx+nbn

Z

Ωn(τ)

(egn(x)−g0(x))²dx+ +2nbn

Z

Ωn(τ)

(gn(x)−egn(x))(gen(x)−g0(x))dx=T_n¹+A1(n) +A2(n).

It is easy to check that

b⁻_n^1/2A1(n) = Z 1

0

ϕ²(u)du+O(n^−δ). (31) Let us introduce the random variable

dn = Z

Ωn(τ)

(gn(x)−E1gn(x))ϕ(x)dx.