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GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 6, 1999, 501-516
LIMIT DISTRIBUTION OF THE MEAN SQUARE
¨
DEVIATION OF THE GASSER–MULLER
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:
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
K(u)us du 6= 0,
2◦ . K(u)uj du = 0, 0 < j < s,
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 ,
ηik , k = 2, . . . , n,
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
n X
X
4
2
2
2
2
σn = 2σ (nbn )
Qii = d1 (n) + d2 (n).
Qki −
k=1 i=1
Here by the definition of Qki we have
XZ
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
Z 1 Z 1
Z
(2)
(1)
2
(y)
−
(x)dP
Ψ
(x,
y)dP
n
n
n
0
0
where
Z
I1 =
0
1
Z
0
1
1
0
Z
1
0
Ψ2n (x, y)dP (x)dP (y) ≤ I1 +I2 ,
Ψ2n (x, y)dPn(2) (y)[dPn(1) (x) − dP (x)],
504
R. ABSAVA AND E. NADARAYA
Z
I2 =
1
0
Z
1
0
Ψ2n (x, y)dP (x)[dPn(2) (y)
− dP (y)].
Furthermore, the integration by parts of the internal integral in I1 yields
Z 1
Z 1
dPn(2) (y)
|Pn(1) (x) − P (x)| |Ψn (x, y)|bn−1 ×
I1 ≤ 2
0
0
Z
t − y
(1) t − x
×
(5)
K
K
dt 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
Ψ2n (t1 , t2 )dt1 dt2 + O
0
It is easy to show also that
Z xb−1
Z
Z
n
4
d1 (n) = 2σ
Ωn (τ ) Ωn (τ )
1
K(u)K
(x−1)b−1
n
x − y
bn
1
.
nbn
−u du
2
(6)
dxdy+O
1
.
nbn
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 1Z
4
−1
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
(n)|
≤
|d
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
(7)
x − t x − t
1
2
K
K
dx ×
b
b
n
n
Ωn (τ )
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
×
Z
y − t y − t
1
3
4
.
K
dy dt1 dt2 dt3 dt4 ≤ c
K
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 (nbn )
2
4 |d2 (n)|
4
2
EZ
Q
= sup EZn1
≤
sup
EZ
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 | ≥ ǫ) Fk−1 ]−→0
E[ξnk
(10)
and
n
X
k=1
P
2
ξnk
−→1.
(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 ǫ > 0
P
n
X
4
sup |ξnk | ≥ ǫ ≤ ǫ−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
2
ξnk
−1
Pn
2
=1
Eξnk
=
4
Eξnk
+2
k=1
n
X
k=1
2
→0
X
1≤k1
LIMIT DISTRIBUTION OF THE MEAN SQUARE
¨
DEVIATION OF THE GASSER–MULLER
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:
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
K(u)us du 6= 0,
2◦ . K(u)uj du = 0, 0 < j < s,
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 ,
ηik , k = 2, . . . , n,
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
n X
X
4
2
2
2
2
σn = 2σ (nbn )
Qii = d1 (n) + d2 (n).
Qki −
k=1 i=1
Here by the definition of Qki we have
XZ
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
Z 1 Z 1
Z
(2)
(1)
2
(y)
−
(x)dP
Ψ
(x,
y)dP
n
n
n
0
0
where
Z
I1 =
0
1
Z
0
1
1
0
Z
1
0
Ψ2n (x, y)dP (x)dP (y) ≤ I1 +I2 ,
Ψ2n (x, y)dPn(2) (y)[dPn(1) (x) − dP (x)],
504
R. ABSAVA AND E. NADARAYA
Z
I2 =
1
0
Z
1
0
Ψ2n (x, y)dP (x)[dPn(2) (y)
− dP (y)].
Furthermore, the integration by parts of the internal integral in I1 yields
Z 1
Z 1
dPn(2) (y)
|Pn(1) (x) − P (x)| |Ψn (x, y)|bn−1 ×
I1 ≤ 2
0
0
Z
t − y
(1) t − x
×
(5)
K
K
dt 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
Ψ2n (t1 , t2 )dt1 dt2 + O
0
It is easy to show also that
Z xb−1
Z
Z
n
4
d1 (n) = 2σ
Ωn (τ ) Ωn (τ )
1
K(u)K
(x−1)b−1
n
x − y
bn
1
.
nbn
−u du
2
(6)
dxdy+O
1
.
nbn
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 1Z
4
−1
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
(n)|
≤
|d
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
(7)
x − t x − t
1
2
K
K
dx ×
b
b
n
n
Ωn (τ )
LIMIT DISTRIBUTION OF THE MEAN SQUARE DEVIATION
×
Z
y − t y − t
1
3
4
.
K
dy dt1 dt2 dt3 dt4 ≤ c
K
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 (nbn )
2
4 |d2 (n)|
4
2
EZ
Q
= sup EZn1
≤
sup
EZ
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 | ≥ ǫ) Fk−1 ]−→0
E[ξnk
(10)
and
n
X
k=1
P
2
ξnk
−→1.
(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 ǫ > 0
P
n
X
4
sup |ξnk | ≥ ǫ ≤ ǫ−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
2
ξnk
−1
Pn
2
=1
Eξnk
=
4
Eξnk
+2
k=1
n
X
k=1
2
→0
X
1≤k1