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An estimate of the remainder of a limit theorem
Statistics and Probability Letters 82 (2012) 478–487
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Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
An estimate of the remainder of a limit theorem Jianjun He Department of Mathematics, China Jiliang University, Xueyuan Street, Hangzhou 310018, China
article
abstract
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Let {X , Xn , n ≥ 1 } be a sequence of i.i.d. random variables with zero mean and finite ∞ n 2 2 1/2+α variance. Set Sn = ϵ), 0 < α < 1. k=1 Xk , EX = σ > 0, λα (ϵ) = n=1 P (|Sn | ≥ n 1/α In this paper, we discuss the rate of the approximation of σ cα by ϵ 1/α λα (ϵ) under suitable conditions, and extend the results of Klesov (1994), and He and Xie (in press), where cα = π −1/2 21/2α Γ ( 21 + 21α ). © 2011 Elsevier B.V. All rights reserved.
Article history: Received 29 November 2010 Received in revised form 8 November 2011 Accepted 8 November 2011 Available online 18 November 2011 MSC: 60F15 60G50 Keywords: The rate of approximation I.i.d. random variable Complete convergence
1. Introduction and main results Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables and set Sn = k=1 Xk , λα (ϵ) = and Robbins (1947) introduced the concept of complete convergence. They proved that if
n
EX = 0,
EX 2 = σ 2 < ∞,
∞
n =1
1
P (|Sn | ≥ n 2 +α ϵ). Hsu (1.1)
then for all ϵ > 0, ∞
P (|Sn | ≥ nϵ) < ∞.
n =1
Erdös (1978); Erdös (1950) proved the converse. Heyde (1975) proved that lim ϵ 2
ϵ→0
∞
P (|Sn | ≥ nϵ) = σ 2 ,
(1.2)
n =1
under the condition (1.1), which means that σ 2 can be approximated by ϵ 2 studied the rate of the approximation of σ 2 by ϵ 2 λ1/2 (ϵ), and proved that if EX = 0, EX 2 = σ 2 > 0,
E |X |3 < ∞,
∞
n =1
P (|Sn | ≥ nϵ) as ϵ → 0. Klesov (1994)
(1.3)
then lim ϵ
ϵ→0
3/2
σ2 λ1/2 (ϵ) − 2 ϵ
= 0.
E-mail addresses: [email protected], [email protected]. 0167-7152/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2011.11.009
(1.4)
J. He / Statistics and Probability Letters 82 (2012) 478–487
479
It is obvious that (1.4) can be expressed by
ϵ 2 λ1/2 (ϵ) − σ 2 = o(ϵ 1/2 ),
as ϵ → 0. 1
Recently, He and Xie (in press) showed that o(ϵ 2 ) can be replaced by O(ϵ), and discussed the case that the condition E |X |3 < ∞ was replaced by E |X |2+δ < ∞, where 0 < δ < 1. Gut and Steinebach (2011) extended the results of Klesov (1994). In fact, many authors considered various extensions of (1.2), such as Gut and Spˇataru (2000a,b, 2003) and Liu and Lin (2006). Chen (1978) proved the following result. Theorem A. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with zero mean, EX 2 = 1, and E |X |t < ∞. Suppose that r and t are two constants satisfying 2 ≤ t < 2r ≤ 2t. Then lim ϵ s(r −1)
ϵ→0
∞
nr −2 P (|Sn | > ϵ nr /t ) = K (r , t ),
n =1 2s(r −1)/2 Γ ((1+s(r −1))/2)
and K (r , t ) = . where s = (r −1)Γ (1/2) By letting r = 2 in Theorem A, it is easy to get the following result. 2t , 2r −t
Theorem A*. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with zero mean, EX 2 = 1, and E |X |4/(1+2α) < ∞, 0 < α ≤ 21 . Then lim ϵ 1/α
ϵ→0
where cα = π
∞
P (|Sn | > ϵ n1/2+α ) = cα ,
n =1
−1/2 1/2α
2
Γ ( 12 +
1 2α
).
Chen (1976) discussed the case of independent, nonidentically distributed random variables. The moment condition was replaced by EX 2 g (X ) < ∞, where g (x) is a real function satisfying the following conditions: (a) g (x) is nondecreasing on the interval (0, ∞), even on (−∞, ∞) and g (x) → ∞ as x → ∞; (b) g (xx) does not decrease on (0, ∞). Petrov (1982) generalized and refined the results of Chen (1976). By applying the results of Petrov (1982) to a sequence of i.i.d. random variables, we have Theorem B. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with zero mean, EX 2 = 1, EX 2 g (X ) < ∞, and ∞
1
√
n2α g ( n) n =1 where 0 < α ≤ lim ϵ 1/α
ϵ→0
1 . 2
< ∞,
Then
∞
P (|Sn | > ϵ n1/2+α ) = cα .
n =1
Inspired by the above results, the aim of this paper is to discuss the deviation of an approximation of σ 1/α cα by ϵ 1/α λα (ϵ), which give the analogous generalization of He and Xie (in press). Our main results are as follows: Theorem 1. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with EX = 0, EX 2 = σ 2 > 0 and E |X |3 < ∞. Then
ϵ 1/α λα (ϵ) − σ 1/α cα = O(ϵ 1/2α ), where
1 6
as ϵ → 0,
(1.5)
< α < 1.
Remark 1. Setting α = 12 , we have σ 1/α cα = 2π −1/2 σ 2 Γ ( 32 ) = σ 2 , that is ϵ 2 λ1/2 (ϵ) − σ 2 = O(ϵ), as ϵ → 0. The result of Klesov (1994) is improved. Theorem 2. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with EX = 0, EX 2 = σ 2 > 0, and E |X |2+δ < ∞. (1) If 0 < δ < 1 and 2(22−δ < α < 1, then +δ)
ϵ 1/α λα (ϵ) − σ 1/α cα = o(ϵ δ/2α ), (2) If
1 6
<α≤
1 2
and
2−4α 2α+1
as ϵ → 0.
(1.6)
< δ < 1, then
ϵ 1/α λα (ϵ) − σ 1/α cα = o(ϵ δ/2α ),
as ϵ → 0.
Remark 2. Setting α = 21 , we obtain the results of He and Xie (in press). If α = δ = (1994), and weaken the condition of Klesov (1994).
(1.7) 1 , 2
we obtain the results of Klesov
480
J. He / Statistics and Probability Letters 82 (2012) 478–487 1
Remark 3. Let An (α, ϵ) = I (|Sn | ≥ n 2 +α ϵ) for all n = 1, 2, . . . and N∞ (α, ϵ) =
n=1 An (α, ϵ), ϵ). Theorems 1 and 2 implies 1 1 limϵ→0 ϵ α EN∞ (α, ϵ) = σ α cα . In particular, when α = 21 , by Theorem 1, we have ϵ 2 EN∞ ( 12 , ϵ) − σ 2 = O(ϵ), then for sufficient small ϵ > 0 and sufficient large m, we can estimate σ 2 by ϵ 2 Nm ( 21 , ϵ).
where I (A) is the indicator of the event A. Then EN∞ (α, ϵ) =
∞
n =1
An (α, ϵ), Nm (α, ϵ) =
n=1 P (|Sn | ≥ n
∞
m
1 +α 2
2. Some lemmas To prove the main results, we give the following lemmas. Lemma 2.1. Let 0 < α < 1. Then for sufficiently small δ > 0, we have 2α ∞ √ √ √ √ e−n δ ≤ 2 δ. δ ≤ π − 2α δ 1−α
n =1
n
Proof. From the identity ∞
e −x
√ π,
√ dx = x
0
it is easy to obtain that, for all δ > 0, ∞
n2α δ
e −x
x
(n−1)2α δ
n =1
√ π.
√ dx =
(2.1)
Moreover, for all integer n ≥ 2, we have
√
n2α δ
2α 2 δ (nα − (n − 1)α ) e−n δ ≤
√ 2α √ dx ≤ 2 δ (nα − (n − 1)α ) e−(n−1) δ .
e−x
(n−1)2α δ
x
(2.2)
When 0 < α < 1, x > 0, we have 1 + αx +
α(α − 1) 2
x2 ≤ (1 + x)α ≤ 1 + α x +
α(α − 1) 2
x2 +
α(α − 1)(α − 2) 6
x3 .
So α
α
α
n − (n − 1) = (n − 1)
1+
≤ α(n − 1)α−1 +
α
1 n−1
−1
α(α − 1) 2(n − 1)2−α
+
α(α − 1)(α − 2) . 6(n − 1)3−α
(2.3)
and α
α
n − (n − 1) = n
α
1−
≥ α nα−1 −
1−
1
α
n
α(α − 1) 2n2−α
+
α(α − 1)(α − 2) 6n3−α
.
By noting that when l − α > 1, lim
δ→0
2α ∞ e −n δ
n=1
nl−α
=
∞ 1 n =1
nl−α
,
from (2.2) and (2.3), for sufficiently small δ > 0, we can obtain ∞ n =2
n2α δ
(n−1)2α δ
2α 2α ∞ ∞ √ √ e−(n−1) δ e−(n−1) δ + α(α − 1 ) δ √ dx ≤ 2α δ (n − 1)1−α (n − 1)2−α x n =2 n =2 √ ∞ −(n−1)2α δ α(α − 1)(α − 2) δ e + 3 ( n − 1)3−α n =2
e −x
(2.4)
J. He / Statistics and Probability Letters 82 (2012) 478–487 2α ∞ ∞ √ √ e −n δ 1 ≤ 2α δ + α(α − 1 ) + o(1) δ 1−α 2−α
n=1
+ Since for all p > 1, 2(p−1) ≤ p+1
n=1
√ ∞ α(α − 1)(α − 2) δ 1 3
∞
1 n =1 n p
2α
∞ √ √ e −n δ π − 2α δ ≤ 1−α n =1
n
n
n3−α
p , p−1
δ
and
0
+ o(1) .
(2.5)
−x e√ dx x
x
0
n
√ √ = 2 δ + o( δ), by using (2.1) and (2.5), we have ∞ ∞ √ √ e−x 1 α−2 1 + + o( δ) √ + α(α − 1) δ 2−α 3−α
≤
δ
n =1
481
n =1
n
3
n =1
n
√ √ 3−α + α(α − 1)(α − 2) δ + o( δ) 2(1 − α) 3(2 − α) α(α − 3)(1 + 2α) √ ≤ 2+ δ
≤
2 + α(α − 1)
3−α
6
√ ≤ 2 δ.
(2.6)
Similarly, according to (2.1), (2.2) and (2.4), we can obtain the following inequality. 2α
∞ √ √ e −n δ ≥ π − 2α δ 1−α n =1
n
δ
2α 2α ∞ ∞ √ −δ √ α−2 e −n δ e−n δ − √ dx − 2α δ e − α(α − 1) δ 2−α 3−α
e−x
x
0
n=2
n
3
n=2
n
∞ √ √ √ √ 1 ≥ 2 δ − 2α δ + e−δ α(α − 1) δ − α(α − 1) δ 2−α n =1
n
−n 2 α δ
√ α(α − 1)(α − 2) √ e δ + o( δ) 3 −α 3 n n =2 ∞ ∞ √ 1 α−2 1 ≥ δ 2 − 2α + α(α − 1) − α(α − 1) − 2−α 3−α ∞
+
√ α(α − 1)(α − 2) δ
n =1
n
3
n =1
n
√ + o( δ)
− 3 √ (α − 1)(α − 2)(3 − α) 3−α α(α − 1)(α − 2)(4 − α) − α(α − 1) + ≥ δ 6(2 − α) 3 2(1 − α) 2 √ 1 3 15 ≥ 1 + (1 − α) α− δ + 6
√ ≥
2
δ.
4
(2.7)
Together with (2.6) and (2.7), this implies 2α ∞ √ √ √ √ e −n δ δ ≤ π − 2α δ ≤ 2 δ. 1−α
n =1
n
And now the proof of Lemma 2.1 is complete.
Lemma 2.2. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. Gaussian random variables with EX = 0 and EX 2 = σ 2 > 0. Then there exist constants 0 < C1 < C2 < ∞, such that for sufficiently small ϵ > 0, we have C1 ϵ 1+1/α ≤ ϵ 1/α λα (ϵ) − σ 1/α cα +
ϵ 1/α 2
≤ C2 ϵ 1+1/α ,
where 0 < α < 1. Proof. We prove Lemma 2.2 in two case: σ 2 = 1 and σ 2 ̸= 1. (a) If σ 2 = 1, then
λα (ϵ) = P (|Sn | ≥ n1/2+α ϵ) ∞ ∞ 2 2 = e−t /2 dt π n =1 n α ϵ
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J. He / Statistics and Probability Letters 82 (2012) 478–487
=
∞ ∞
2
π
=
(k+1)α ϵ kα ϵ
n=1 k=n
2 e−t /2 dt
α ∞ 2 (n+1) ϵ −t 2 /2 ne dt . π n=1 nα ϵ
Note that cα =
=
∞
1
√ 22α π
x−1/2+1/2α e−x dx
0
∞ 2 21/2−1/2α t 1/α e−t /2 dt √ 22α π 0 ∞ 1
2
=
2 t 1/α e−t /2 dt
π
=
0
α ∞ 2 (n+1) ϵ
π
nα ϵ
n =1
t
1/α −t 2 /2
e
dt +
2
π
ϵ
2 t 1/α e−t /2 dt .
0
So
ϵ
1/α
λα (ϵ) − cα =
First, we estimate
In =
π
(n+1)α ϵ nα ϵ
(n+1)α ϵ nα ϵ
α ∞ 2 (n+1) ϵ
n =1
nα ϵ
(nϵ
(nϵ 1/α − t 1/α )e−t
1/α
2 /2
−t
1/α
)e
−t 2 /2
dt −
2
π
ϵ
2 t 1/α e−t /2 dt .
0
dt. Letting
(nϵ 1/α − t 1/α )dt ,
we can get
α (n + 1)α+1 − nα+1 ϵ 1+1/α α+1 α α+1 α 1 1 1+1/α α+1 1+1/α α+1 −1 −ϵ n 1+ −1 . =ϵ n 1+ n α+1 n
In = nϵ 1/α ((n + 1)α ϵ − nα ϵ) −
Note that
α α(α − 1)(α − 2) α(α − 1)(α − 2)(α − 3) 1 1+ + + + ≤ 1+ n 2n2 3!n3 4!n4 n α α(α − 1) α(α − 1)(α − 2) ≤1+ + + , n 2n2 3!n3 α
α(α − 1)
and 1+
α+1 n
+
α(α + 1) 2n2
+
α(α − 1)(α + 1) ≤ 3!n3
1+
≤ 1+ +
1
α+1
n
α+1
+
α(α + 1)
4!n4
we can obtain
α(α − 1)(α − 2) In ≤ n ϵ + + n 2n2 3!n3 α α + 1 α(α + 1) α(α + 1)(α − 1) − nα+1 ϵ 1+1/α + + α+1 n 2n2 3!n3 α α(α − 1) = ϵ 1+1/α − 1−α − , 2−α α+1 1+1/α
2n
α
α(α − 1)
3n
+
α(α − 1)(α + 1) 3!n3
2n2 α(α − 1)(α − 2)(α + 1) n
,
(2.8)
J. He / Statistics and Probability Letters 82 (2012) 478–487
483
and
α(α − 1)(α − 2) α(α − 1)(α − 2)(α − 3) + n 2n2 3!n3 4!n4 α α + 1 α(α + 1) α(α + 1)(α − 1) α(α + 1)(α − 1)(α − 2) − nα+1 ϵ 1+1/α + + + α+1 n 2n2 3!n3 4!n4 α(α − 1) α(α − 1)(α − 2) α − . = ϵ 1+1/α − 1−α − 2−α 3−α
In ≥ nα+1 ϵ 1+1/α
α
+
α(α − 1)
2n
+
3n
8n
So
ϵ 1+1/α −
α 2n1−α
−
α(α − 1) 3n2−α
−
α(α − 1)(α − 2) 8n3−α
≤ In ≤ ϵ 1+1/α −
α 2n1−α
−
α(α − 1)
3n2−α
.
(2.9)
Obviously 2α 2 e−n ϵ /2 In ≤
by (2.9) and (2.10),
(n+1)α ϵ nα ϵ
(n+1)α ϵ nα ϵ
Since for all t > 0, 1 −
t2 2
(nϵ 1/α − t 1/α )e−t
(nϵ 1/α − t 1/α )e−t
≤ e−t
2 /2
2 /2
2α 2 dt ≤ e−(n+1) ϵ /2 In .
2
π
(2.10)
dt is estimated. Next we consider the second term on the right hand of (2.8).
≤ 1, we can get ϵ
α αϵ 3+1/α ϵ 1+1/α − ≤ π α+1 2(1 + 3α) 2
2 /2
2 t 1/α e−t /2 dt ≤
0
2 α ϵ 1+1/α . π α+1
(2.11)
2 By (2.8)–(2.11), letting δ = ϵ2 in Lemma 2.1, we have
ϵ
1/α
∞ 2 −n2α ϵ 2 /2 e In −
λα (ϵ) − cα ≥
π
n=1
2 α ϵ 1+1/α π α+1
∞ α(1 − α) α(α − 1)(α − 2) 1+1/α 2 −n2α ϵ 2 /2 α ≥ − ϵ − e − 1−α + π n=1 2n 3n2−α 8n3−α 1 2α 2 2α 2 ∞ ∞ e −n ϵ / 2 ϵα ϵ2 2α(1 − α)ϵ 1+1/α e−n ϵ /2 = −√ α + √ 2 n=1 n1−α n2−α π 3 2π n =1
2 α ϵ 1+1/α π α+1
2α 2
∞ α(α − 1)(α − 2) 2α e−n ϵ /2 − √ √ ϵ 1+1/α 3−α n 4 2π (α + 1 ) 2π n =1 ∞ ∞ √ ϵ 2α(1 − α) 1+1/α 1 3(α − 2) 1 ϵ 1/α ϵ + + o(1) ≥ √ √ √ − π + n2−α 8 n3−α 2 π 3 2π 2 n =1 n=1 2α − √ ϵ 1+1/α (α + 1) 2π ϵ 1/α ϵ 1+1/α 1 α(3 − α) α(α − 1)(3 − α) 2α + √ + + − ≥− 2 2 3 4 α+1 2π
− ϵ 1+1/α
≥−
ϵ 1/α 2
+ C1 ϵ 1+1/α ,
(2.12)
where C1 =
=
√
1
12 2π (α + 1)
√
1
12 2π (α + 1)
(−3α 4 + 5α 3 + 11α 2 − 15α + 6) 2 15 59 3 3 2α + 3α (1 − α) + 11 α − + > 0. 22
44
Noting
(1 + x)α = 1 + α x +
α(α − 1) 2 α(α − 1) . . . (α − m + 1) m α(α − 1) . . . (α − m) xm+1 x + ··· + x + , 2! m! (m + 1)! (1 + θ x)m+1−α
484
J. He / Statistics and Probability Letters 82 (2012) 478–487
where x ∈ (−1, +∞), θ ∈ (0, 1), m ∈ Z + , for 0 < α < 1, we have
1+
1
α
+ 1−
n+1
α+1
1 n+1
≥2−
1 n+1
α2 α(α − 1) α(α − 1)2 (α − 2) − + , (n + 1)2 2(n + 1)3 12(n + 1)4
+
and
1+
1
α+1
+ 1−
n+1
1
α+1
n+1
α(α + 1) α(α + 1)(α − 1)(α − 2) + (n + 1)2 12(n + 1)4 4−α α(α + 1)(α − 1)(α − 2)(α − 3) 1 − 1+ . 5!(n + 1)5 n
≤ 2+
Since In+1 = (n + 1)ϵ 1+1/α ((n + 2)α − (n + 1)α ) −
α ϵ 1+1/α (n + 2)α+1 − (n + 1)α+1 , α+1
we have In+1 − In = ϵ 1+1/α (n + 1)(n + 2)α − (2n + 1)(n + 1)α + nα+1
α ϵ 1+1/α (n + 2)α+1 − 2(n + 1)α+1 + nα+1 α+1 α α+1 1 1 1 1+1/α α+1 =ϵ (n + 1) 1+ + 1− −2+ n+1 n+1 n+1 α+1 α+1 α 1 1 1+1/α α+1 − ϵ (n + 1) + 1− −2 1+ α+1 n+1 n+1 α2 α(α − 1) α(α − 1)2 (α − 2) ≥ ϵ 1+1/α (n + 1)α+1 − + (n + 1)2 2(n + 1)3 12(n + 1)4 α α(α + 1)(α − 1)(α − 2) 1+1/α α+1 α(α + 1) − ϵ (n + 1) + α+1 (n + 1)2 12(n + 1)4 4−α α 2 (α − 1)(α − 2)(α − 3) 1 − 1+ 4 −α 5!(n + 1) n α(α − 1) α(α − 1)(α − 2) α 2 (α − 1)(α − 2)(α − 3) = ϵ 1+1/α − − + . 2(n + 1)2−α 12(n + 1)3−α 5!n4−α −
(2.13)
From (2.8), (2.9), (2.10) and (2.13), we have
ϵ
1/α
λα (ϵ) − cα ≤
∞ 2 −(n+1)2α ϵ 2 /2 e In −
π
n=1
2
π
ϵ
2 t 1/α e−t /2
0
α(α − 1)(α − 2) α(α − 1) + π n=1 π n=1 2(n + 1)2−α 12(n + 1)3−α ∞ ϵ 2 α 2 (α − 1)(α − 2)(α − 3) 2 2 − ϵ 1+1/α − t 1/α e−t /2 dt + o(ϵ 1+1/α ) π n =1 5!n4−α π 0 ∞ 2 − n2α ϵ 2 2 ≤ e In + C ϵ 1+1/α + o(ϵ 1+1/α ) π n=1 ∞ 2 1+1/α −n2α ϵ 2 /2 α α(α − 1) ≤ ϵ e − 1−α − + Dϵ 1+1/α 2−α π 2n 3n n =1
≤
≤−
2
∞
ϵ 1/α 2
−(n+1)2α ϵ 2 /2
e
+ C2 ϵ 1+1/α .
In + 1 + ϵ
1+1/α
∞ 2
(2.14)
J. He / Statistics and Probability Letters 82 (2012) 478–487
485
(b) If σ 2 ̸= 1, then
∞ ϵ |Sn | P ≥ n1/2+α σ σ n =1 1/α (ϵ/σ ) 1/α 1+1/α ≤σ cα − + C2 (ϵ/σ )
ϵ 1/α λα (ϵ) = ϵ 1/α
2
= σ 1/α cα −
ϵ
1/α
+
2
C2
σ
ϵ 1+1/α .
Similarly, we can obtain
ϵ 1/α
ϵ 1/α λα (ϵ) ≥ σ 1/α cα −
+
2
C1
σ
ϵ 1+1/α .
Thus we have proved both in case (a) and case (b), there exist constants 0 < C1 < C2 < ∞, such that C1 ϵ 1+1/α ≤ ϵ 1/α λα (ϵ) − σ 1/α cα + And Lemma 2.2. is now proved.
ϵ 1/α
≤ C2 ϵ 1+1/α .
2
3. Proofs of theorems Proof of Theorem 1. Under the condition (1.4), from the non-uniform estimate of central limit theorem by Nagaev (1965), we have for every x ∈ R,
3 P √1 Sn < x − Φ (x) ≤ √ AE |X | , nσ nσ 3 (1 + |x|)3
(3.1)
where A is an absolute positive constant, and Φ (x) is the standard normal distribution function. Since P (|Sn | ≥ n
1/2+α
ϵ) =
∞
2
π
nα ϵ/σ
2 e−t /2 dt + Rn
where Rn = Φ
nα ϵ
−P
σ
1
√
nσ
Sn ≤
nα ϵ
σ
+P
1
√
nσ
Sn ≤
−n α ϵ σ
−Φ
−n α ϵ σ
.
By (3.1), we have
|Rn | ≤ √
2AE |X |3
3 nσ 3 1 + σϵ nα
.
(3.2)
So
λα (ϵ) =
∞
2
π
n =1
∞
nα ϵ/σ
2 e−t /2 dt +
∞
Rn .
n =1
By Lemma 2.2,
ϵ 1/α
∞ n =1
2
π
∞ nα ϵ/σ
ϵ 2 e−t /2 dt − σ 1/α cα = −
1/α
2
+ O(ϵ 1+1/α ).
And in consequence,
ϵ 1α λα (ϵ) − σ 1/α cα = −
ϵ 1/α 2
1
+ϵα
∞
Rn + O(ϵ 1+1/α ).
(3.3)
n =1
In order to prove Theorem 1, we only need to show that ∞ n =1
Rn = O(ϵ −1/2α ).
(3.4)
486
J. He / Statistics and Probability Letters 82 (2012) 478–487
Setting ∞
σ2 ϵ 1/α
Rn =
n =1
∞
Rn +
n=1
n=
Rn =: I1 + I2 .
(3.5)
σ 2 +1 ϵ 1/α
It follows from (3.2) that
σ2 ϵ 1/α
2AE |X |3 √ = O(ϵ −1/2α ). σ3 n n=1 ∞ 1 Since for 3α + 21 > 1, n=1 n3α+ 1/2 < ∞, we have I1 ≤
∞
I2 ≤
C
σ 2 +1 ϵ 1/α
n3α+1/2 ϵ 3
(3.6)
≤ C ϵ −1/2α .
(3.7)
Combining (3.5) together with (3.6), (3.7), we can obtain (3.4). And now the proof of Theorem 1 is complete.
Proof of Theorem 2. By Osipov and Petrov (1967) inequality there exists a bounded and decreasing function ψ(u) on the interval (0, ∞), such that limu→∞ ψ(u) = 0 and
√ P √1 Sn < x − Φ (x) ≤ ψ( n(1 + |x|)) . nδ/2 (1 + |x|)2+δ nσ
(3.8)
By a process similar to the proof of (3.3), we can obtain
ϵ 1/α λα (ϵ) − σ 1/α cα = −
ϵ 1/α
+ ϵ 1/α
2
∞
R∗n + O(ϵ 1+1/α ),
(3.9)
n=1
where ∗
|Rn | ≤
√
n(1 + nα ϵ/σ )
2ψ
.
nδ/2 (1 + nα ϵ/σ )2+δ
Since for both the case of 0 < δ < 1, 2(22−δ < α < 1 and the case of +δ) ∞
1
n2α+αδ+δ/2 n =1
1 6
< α ≤ 12 ,
2−4α 2α+1
< δ < 1, we have
< ∞.
Then ∞
|R∗n | ≤ ψ
2 n= σ1/α +1 ϵ
σ ϵ 1/2α
∞ n=
ϵ
σ 2 +1 1/α
C n2α+αδ+δ/2 ϵ 2+δ
σ 1 ≤ C ψ 1/2α ϵ −2−δ (ϵ − α )−2α−αδ−δ/2+1 ϵ = o(ϵ δ/2α−1/α ).
By (3.9), we have
ϵ 1/α λα (ϵ) − σ 1/α cα = ϵ 1/α
σ2 ϵ 1/α
R∗n + o(ϵ δ/2α ).
n=1
In order to prove Theorem 2, it is only to prove
ϵ 1/α
σ2 ϵ 1/α
n =1
R∗n = o(ϵ δ/2α ) as ϵ → 0.
(3.10)
J. He / Statistics and Probability Letters 82 (2012) 478–487
487
√ Since limu→∞ ψ(u) = 0, for any η > 0, there exists a natural number N0 , such that ψ( n) < η whenever n > N0 . Hence
ϵ 1/α
σ2 ϵ 1/α
|R∗n | = ϵ 1/α
N0
|R∗n | + ϵ 1/α
n=1
n =1
σ2 ϵ 1/α
|R∗n |
n=N0 +1
≤ ϵ 1/α
N0
√
n−δ/2 ψ( n) + ηϵ 1/α
n=1
σ2 ϵ 1/α
n−δ/2
n=N0 +1
≤ ϵ 1/α M + ηϵ δ/2α , where M is a constant. So
ϵ 1/α
σ2 ϵ 1/α
R∗n = o(ϵ δ/2α ).
(3.11)
n =1
From (3.10) and (3.11), we have
ϵ 1/α λα (ϵ) − σ 1/α cα = o(ϵ δ/2α ). This completes the proof of Theorem 2.
(3.12)
Acknowledgments The author is very grateful to the referees for carefully reading the manuscript and for valuable comments, which improved the presentation of this paper. The author also thanks Professor Hira Koul very much for efficient work and useful suggestions. References Chen, R., 1976. A remark on the strong law of large numbers. Proc. Amer. Math. Soc. 61, 112–116. Chen, R., 1978. A remark on the tail probability of a distribution. J. Multivariate Anal. 8, 328–333. Erdös, P., 1949. On a theorem of Hsu and Robbins. Ann. Math. Statist. 20, 286–291. Erdös, P., 1950. Remark on my paper on a theorem of Hsu and Robbins. Ann. Math. Statist. 21, 138. Gut, A., Spˇataru, A., 2000a. Precise asymptotics in the Baum–Katz and Davis law of large numbers. J. Math. Anal. Appl. 248, 233–246. Gut, A., Spˇataru, A., 2000b. Precise asymptotics in the law of the iterated logarithm. Ann. Probab. 28, 1870–1883. Gut, A., Spˇataru, A., 2003. Precise asymptotics in some strong limit theorem for multidimensionally indexed random variables. J. Multivariate Anal. 86, 398–422. Gut, A., Steinebach, J., 2011. Convergence rates in precise asymptotics. UUDM. Report. He, J.J., Xie, T.F., 2012. Asymptotics property for some series of probability. Acta Math. Applacatae Sin. (in press). Heyde, C.C., 1975. A supplement to the strong law of large numbers. J. Appl. Probab. 12, 903–907. Hsu, P.L., Robbins, H., 1947. Complete convergence and the law of large numbers. Proc. Natl. Acad. Sci. USA 33, 25–31. Klesov, O.I., 1994. On the convergence rate in a theorem of Heyde. Theory Probab. Math. Statist. 49, 83–87. Liu, W.D., Lin, Z.Y., 2006. Precise asymptotics for a new kind of complete moment convergence. Statist. Probab. Lett. 76, 1787–1799. Nagaev, S.V., 1965. Some limit theorems for large deviation. Theory Probab. Appl. 10, 214–235. Osipov, L.V, Petrov, V.V., 1967. On an estimate of the remainder term in the central limit theorem. Theory Probab. Appl. 12, 281–286. Petrov, V.V., 1982. A limit theorem for sums of independent nonidentically distributed random variables. J. Math. Sci. 20, 2232–2235.
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Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
An estimate of the remainder of a limit theorem Jianjun He Department of Mathematics, China Jiliang University, Xueyuan Street, Hangzhou 310018, China
article
abstract
info
Let {X , Xn , n ≥ 1 } be a sequence of i.i.d. random variables with zero mean and finite ∞ n 2 2 1/2+α variance. Set Sn = ϵ), 0 < α < 1. k=1 Xk , EX = σ > 0, λα (ϵ) = n=1 P (|Sn | ≥ n 1/α In this paper, we discuss the rate of the approximation of σ cα by ϵ 1/α λα (ϵ) under suitable conditions, and extend the results of Klesov (1994), and He and Xie (in press), where cα = π −1/2 21/2α Γ ( 21 + 21α ). © 2011 Elsevier B.V. All rights reserved.
Article history: Received 29 November 2010 Received in revised form 8 November 2011 Accepted 8 November 2011 Available online 18 November 2011 MSC: 60F15 60G50 Keywords: The rate of approximation I.i.d. random variable Complete convergence
1. Introduction and main results Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables and set Sn = k=1 Xk , λα (ϵ) = and Robbins (1947) introduced the concept of complete convergence. They proved that if
n
EX = 0,
EX 2 = σ 2 < ∞,
∞
n =1
1
P (|Sn | ≥ n 2 +α ϵ). Hsu (1.1)
then for all ϵ > 0, ∞
P (|Sn | ≥ nϵ) < ∞.
n =1
Erdös (1978); Erdös (1950) proved the converse. Heyde (1975) proved that lim ϵ 2
ϵ→0
∞
P (|Sn | ≥ nϵ) = σ 2 ,
(1.2)
n =1
under the condition (1.1), which means that σ 2 can be approximated by ϵ 2 studied the rate of the approximation of σ 2 by ϵ 2 λ1/2 (ϵ), and proved that if EX = 0, EX 2 = σ 2 > 0,
E |X |3 < ∞,
∞
n =1
P (|Sn | ≥ nϵ) as ϵ → 0. Klesov (1994)
(1.3)
then lim ϵ
ϵ→0
3/2
σ2 λ1/2 (ϵ) − 2 ϵ
= 0.
E-mail addresses: [email protected], [email protected]. 0167-7152/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2011.11.009
(1.4)
J. He / Statistics and Probability Letters 82 (2012) 478–487
479
It is obvious that (1.4) can be expressed by
ϵ 2 λ1/2 (ϵ) − σ 2 = o(ϵ 1/2 ),
as ϵ → 0. 1
Recently, He and Xie (in press) showed that o(ϵ 2 ) can be replaced by O(ϵ), and discussed the case that the condition E |X |3 < ∞ was replaced by E |X |2+δ < ∞, where 0 < δ < 1. Gut and Steinebach (2011) extended the results of Klesov (1994). In fact, many authors considered various extensions of (1.2), such as Gut and Spˇataru (2000a,b, 2003) and Liu and Lin (2006). Chen (1978) proved the following result. Theorem A. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with zero mean, EX 2 = 1, and E |X |t < ∞. Suppose that r and t are two constants satisfying 2 ≤ t < 2r ≤ 2t. Then lim ϵ s(r −1)
ϵ→0
∞
nr −2 P (|Sn | > ϵ nr /t ) = K (r , t ),
n =1 2s(r −1)/2 Γ ((1+s(r −1))/2)
and K (r , t ) = . where s = (r −1)Γ (1/2) By letting r = 2 in Theorem A, it is easy to get the following result. 2t , 2r −t
Theorem A*. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with zero mean, EX 2 = 1, and E |X |4/(1+2α) < ∞, 0 < α ≤ 21 . Then lim ϵ 1/α
ϵ→0
where cα = π
∞
P (|Sn | > ϵ n1/2+α ) = cα ,
n =1
−1/2 1/2α
2
Γ ( 12 +
1 2α
).
Chen (1976) discussed the case of independent, nonidentically distributed random variables. The moment condition was replaced by EX 2 g (X ) < ∞, where g (x) is a real function satisfying the following conditions: (a) g (x) is nondecreasing on the interval (0, ∞), even on (−∞, ∞) and g (x) → ∞ as x → ∞; (b) g (xx) does not decrease on (0, ∞). Petrov (1982) generalized and refined the results of Chen (1976). By applying the results of Petrov (1982) to a sequence of i.i.d. random variables, we have Theorem B. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with zero mean, EX 2 = 1, EX 2 g (X ) < ∞, and ∞
1
√
n2α g ( n) n =1 where 0 < α ≤ lim ϵ 1/α
ϵ→0
1 . 2
< ∞,
Then
∞
P (|Sn | > ϵ n1/2+α ) = cα .
n =1
Inspired by the above results, the aim of this paper is to discuss the deviation of an approximation of σ 1/α cα by ϵ 1/α λα (ϵ), which give the analogous generalization of He and Xie (in press). Our main results are as follows: Theorem 1. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with EX = 0, EX 2 = σ 2 > 0 and E |X |3 < ∞. Then
ϵ 1/α λα (ϵ) − σ 1/α cα = O(ϵ 1/2α ), where
1 6
as ϵ → 0,
(1.5)
< α < 1.
Remark 1. Setting α = 12 , we have σ 1/α cα = 2π −1/2 σ 2 Γ ( 32 ) = σ 2 , that is ϵ 2 λ1/2 (ϵ) − σ 2 = O(ϵ), as ϵ → 0. The result of Klesov (1994) is improved. Theorem 2. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. random variables with EX = 0, EX 2 = σ 2 > 0, and E |X |2+δ < ∞. (1) If 0 < δ < 1 and 2(22−δ < α < 1, then +δ)
ϵ 1/α λα (ϵ) − σ 1/α cα = o(ϵ δ/2α ), (2) If
1 6
<α≤
1 2
and
2−4α 2α+1
as ϵ → 0.
(1.6)
< δ < 1, then
ϵ 1/α λα (ϵ) − σ 1/α cα = o(ϵ δ/2α ),
as ϵ → 0.
Remark 2. Setting α = 21 , we obtain the results of He and Xie (in press). If α = δ = (1994), and weaken the condition of Klesov (1994).
(1.7) 1 , 2
we obtain the results of Klesov
480
J. He / Statistics and Probability Letters 82 (2012) 478–487 1
Remark 3. Let An (α, ϵ) = I (|Sn | ≥ n 2 +α ϵ) for all n = 1, 2, . . . and N∞ (α, ϵ) =
n=1 An (α, ϵ), ϵ). Theorems 1 and 2 implies 1 1 limϵ→0 ϵ α EN∞ (α, ϵ) = σ α cα . In particular, when α = 21 , by Theorem 1, we have ϵ 2 EN∞ ( 12 , ϵ) − σ 2 = O(ϵ), then for sufficient small ϵ > 0 and sufficient large m, we can estimate σ 2 by ϵ 2 Nm ( 21 , ϵ).
where I (A) is the indicator of the event A. Then EN∞ (α, ϵ) =
∞
n =1
An (α, ϵ), Nm (α, ϵ) =
n=1 P (|Sn | ≥ n
∞
m
1 +α 2
2. Some lemmas To prove the main results, we give the following lemmas. Lemma 2.1. Let 0 < α < 1. Then for sufficiently small δ > 0, we have 2α ∞ √ √ √ √ e−n δ ≤ 2 δ. δ ≤ π − 2α δ 1−α
n =1
n
Proof. From the identity ∞
e −x
√ π,
√ dx = x
0
it is easy to obtain that, for all δ > 0, ∞
n2α δ
e −x
x
(n−1)2α δ
n =1
√ π.
√ dx =
(2.1)
Moreover, for all integer n ≥ 2, we have
√
n2α δ
2α 2 δ (nα − (n − 1)α ) e−n δ ≤
√ 2α √ dx ≤ 2 δ (nα − (n − 1)α ) e−(n−1) δ .
e−x
(n−1)2α δ
x
(2.2)
When 0 < α < 1, x > 0, we have 1 + αx +
α(α − 1) 2
x2 ≤ (1 + x)α ≤ 1 + α x +
α(α − 1) 2
x2 +
α(α − 1)(α − 2) 6
x3 .
So α
α
α
n − (n − 1) = (n − 1)
1+
≤ α(n − 1)α−1 +
α
1 n−1
−1
α(α − 1) 2(n − 1)2−α
+
α(α − 1)(α − 2) . 6(n − 1)3−α
(2.3)
and α
α
n − (n − 1) = n
α
1−
≥ α nα−1 −
1−
1
α
n
α(α − 1) 2n2−α
+
α(α − 1)(α − 2) 6n3−α
.
By noting that when l − α > 1, lim
δ→0
2α ∞ e −n δ
n=1
nl−α
=
∞ 1 n =1
nl−α
,
from (2.2) and (2.3), for sufficiently small δ > 0, we can obtain ∞ n =2
n2α δ
(n−1)2α δ
2α 2α ∞ ∞ √ √ e−(n−1) δ e−(n−1) δ + α(α − 1 ) δ √ dx ≤ 2α δ (n − 1)1−α (n − 1)2−α x n =2 n =2 √ ∞ −(n−1)2α δ α(α − 1)(α − 2) δ e + 3 ( n − 1)3−α n =2
e −x
(2.4)
J. He / Statistics and Probability Letters 82 (2012) 478–487 2α ∞ ∞ √ √ e −n δ 1 ≤ 2α δ + α(α − 1 ) + o(1) δ 1−α 2−α
n=1
+ Since for all p > 1, 2(p−1) ≤ p+1
n=1
√ ∞ α(α − 1)(α − 2) δ 1 3
∞
1 n =1 n p
2α
∞ √ √ e −n δ π − 2α δ ≤ 1−α n =1
n
n
n3−α
p , p−1
δ
and
0
+ o(1) .
(2.5)
−x e√ dx x
x
0
n
√ √ = 2 δ + o( δ), by using (2.1) and (2.5), we have ∞ ∞ √ √ e−x 1 α−2 1 + + o( δ) √ + α(α − 1) δ 2−α 3−α
≤
δ
n =1
481
n =1
n
3
n =1
n
√ √ 3−α + α(α − 1)(α − 2) δ + o( δ) 2(1 − α) 3(2 − α) α(α − 3)(1 + 2α) √ ≤ 2+ δ
≤
2 + α(α − 1)
3−α
6
√ ≤ 2 δ.
(2.6)
Similarly, according to (2.1), (2.2) and (2.4), we can obtain the following inequality. 2α
∞ √ √ e −n δ ≥ π − 2α δ 1−α n =1
n
δ
2α 2α ∞ ∞ √ −δ √ α−2 e −n δ e−n δ − √ dx − 2α δ e − α(α − 1) δ 2−α 3−α
e−x
x
0
n=2
n
3
n=2
n
∞ √ √ √ √ 1 ≥ 2 δ − 2α δ + e−δ α(α − 1) δ − α(α − 1) δ 2−α n =1
n
−n 2 α δ
√ α(α − 1)(α − 2) √ e δ + o( δ) 3 −α 3 n n =2 ∞ ∞ √ 1 α−2 1 ≥ δ 2 − 2α + α(α − 1) − α(α − 1) − 2−α 3−α ∞
+
√ α(α − 1)(α − 2) δ
n =1
n
3
n =1
n
√ + o( δ)
− 3 √ (α − 1)(α − 2)(3 − α) 3−α α(α − 1)(α − 2)(4 − α) − α(α − 1) + ≥ δ 6(2 − α) 3 2(1 − α) 2 √ 1 3 15 ≥ 1 + (1 − α) α− δ + 6
√ ≥
2
δ.
4
(2.7)
Together with (2.6) and (2.7), this implies 2α ∞ √ √ √ √ e −n δ δ ≤ π − 2α δ ≤ 2 δ. 1−α
n =1
n
And now the proof of Lemma 2.1 is complete.
Lemma 2.2. Let {X , Xn , n ≥ 1} be a sequence of i.i.d. Gaussian random variables with EX = 0 and EX 2 = σ 2 > 0. Then there exist constants 0 < C1 < C2 < ∞, such that for sufficiently small ϵ > 0, we have C1 ϵ 1+1/α ≤ ϵ 1/α λα (ϵ) − σ 1/α cα +
ϵ 1/α 2
≤ C2 ϵ 1+1/α ,
where 0 < α < 1. Proof. We prove Lemma 2.2 in two case: σ 2 = 1 and σ 2 ̸= 1. (a) If σ 2 = 1, then
λα (ϵ) = P (|Sn | ≥ n1/2+α ϵ) ∞ ∞ 2 2 = e−t /2 dt π n =1 n α ϵ
482
J. He / Statistics and Probability Letters 82 (2012) 478–487
=
∞ ∞
2
π
=
(k+1)α ϵ kα ϵ
n=1 k=n
2 e−t /2 dt
α ∞ 2 (n+1) ϵ −t 2 /2 ne dt . π n=1 nα ϵ
Note that cα =
=
∞
1
√ 22α π
x−1/2+1/2α e−x dx
0
∞ 2 21/2−1/2α t 1/α e−t /2 dt √ 22α π 0 ∞ 1
2
=
2 t 1/α e−t /2 dt
π
=
0
α ∞ 2 (n+1) ϵ
π
nα ϵ
n =1
t
1/α −t 2 /2
e
dt +
2
π
ϵ
2 t 1/α e−t /2 dt .
0
So
ϵ
1/α
λα (ϵ) − cα =
First, we estimate
In =
π
(n+1)α ϵ nα ϵ
(n+1)α ϵ nα ϵ
α ∞ 2 (n+1) ϵ
n =1
nα ϵ
(nϵ
(nϵ 1/α − t 1/α )e−t
1/α
2 /2
−t
1/α
)e
−t 2 /2
dt −
2
π
ϵ
2 t 1/α e−t /2 dt .
0
dt. Letting
(nϵ 1/α − t 1/α )dt ,
we can get
α (n + 1)α+1 − nα+1 ϵ 1+1/α α+1 α α+1 α 1 1 1+1/α α+1 1+1/α α+1 −1 −ϵ n 1+ −1 . =ϵ n 1+ n α+1 n
In = nϵ 1/α ((n + 1)α ϵ − nα ϵ) −
Note that
α α(α − 1)(α − 2) α(α − 1)(α − 2)(α − 3) 1 1+ + + + ≤ 1+ n 2n2 3!n3 4!n4 n α α(α − 1) α(α − 1)(α − 2) ≤1+ + + , n 2n2 3!n3 α
α(α − 1)
and 1+
α+1 n
+
α(α + 1) 2n2
+
α(α − 1)(α + 1) ≤ 3!n3
1+
≤ 1+ +
1
α+1
n
α+1
+
α(α + 1)
4!n4
we can obtain
α(α − 1)(α − 2) In ≤ n ϵ + + n 2n2 3!n3 α α + 1 α(α + 1) α(α + 1)(α − 1) − nα+1 ϵ 1+1/α + + α+1 n 2n2 3!n3 α α(α − 1) = ϵ 1+1/α − 1−α − , 2−α α+1 1+1/α
2n
α
α(α − 1)
3n
+
α(α − 1)(α + 1) 3!n3
2n2 α(α − 1)(α − 2)(α + 1) n
,
(2.8)
J. He / Statistics and Probability Letters 82 (2012) 478–487
483
and
α(α − 1)(α − 2) α(α − 1)(α − 2)(α − 3) + n 2n2 3!n3 4!n4 α α + 1 α(α + 1) α(α + 1)(α − 1) α(α + 1)(α − 1)(α − 2) − nα+1 ϵ 1+1/α + + + α+1 n 2n2 3!n3 4!n4 α(α − 1) α(α − 1)(α − 2) α − . = ϵ 1+1/α − 1−α − 2−α 3−α
In ≥ nα+1 ϵ 1+1/α
α
+
α(α − 1)
2n
+
3n
8n
So
ϵ 1+1/α −
α 2n1−α
−
α(α − 1) 3n2−α
−
α(α − 1)(α − 2) 8n3−α
≤ In ≤ ϵ 1+1/α −
α 2n1−α
−
α(α − 1)
3n2−α
.
(2.9)
Obviously 2α 2 e−n ϵ /2 In ≤
by (2.9) and (2.10),
(n+1)α ϵ nα ϵ
(n+1)α ϵ nα ϵ
Since for all t > 0, 1 −
t2 2
(nϵ 1/α − t 1/α )e−t
(nϵ 1/α − t 1/α )e−t
≤ e−t
2 /2
2 /2
2α 2 dt ≤ e−(n+1) ϵ /2 In .
2
π
(2.10)
dt is estimated. Next we consider the second term on the right hand of (2.8).
≤ 1, we can get ϵ
α αϵ 3+1/α ϵ 1+1/α − ≤ π α+1 2(1 + 3α) 2
2 /2
2 t 1/α e−t /2 dt ≤
0
2 α ϵ 1+1/α . π α+1
(2.11)
2 By (2.8)–(2.11), letting δ = ϵ2 in Lemma 2.1, we have
ϵ
1/α
∞ 2 −n2α ϵ 2 /2 e In −
λα (ϵ) − cα ≥
π
n=1
2 α ϵ 1+1/α π α+1
∞ α(1 − α) α(α − 1)(α − 2) 1+1/α 2 −n2α ϵ 2 /2 α ≥ − ϵ − e − 1−α + π n=1 2n 3n2−α 8n3−α 1 2α 2 2α 2 ∞ ∞ e −n ϵ / 2 ϵα ϵ2 2α(1 − α)ϵ 1+1/α e−n ϵ /2 = −√ α + √ 2 n=1 n1−α n2−α π 3 2π n =1
2 α ϵ 1+1/α π α+1
2α 2
∞ α(α − 1)(α − 2) 2α e−n ϵ /2 − √ √ ϵ 1+1/α 3−α n 4 2π (α + 1 ) 2π n =1 ∞ ∞ √ ϵ 2α(1 − α) 1+1/α 1 3(α − 2) 1 ϵ 1/α ϵ + + o(1) ≥ √ √ √ − π + n2−α 8 n3−α 2 π 3 2π 2 n =1 n=1 2α − √ ϵ 1+1/α (α + 1) 2π ϵ 1/α ϵ 1+1/α 1 α(3 − α) α(α − 1)(3 − α) 2α + √ + + − ≥− 2 2 3 4 α+1 2π
− ϵ 1+1/α
≥−
ϵ 1/α 2
+ C1 ϵ 1+1/α ,
(2.12)
where C1 =
=
√
1
12 2π (α + 1)
√
1
12 2π (α + 1)
(−3α 4 + 5α 3 + 11α 2 − 15α + 6) 2 15 59 3 3 2α + 3α (1 − α) + 11 α − + > 0. 22
44
Noting
(1 + x)α = 1 + α x +
α(α − 1) 2 α(α − 1) . . . (α − m + 1) m α(α − 1) . . . (α − m) xm+1 x + ··· + x + , 2! m! (m + 1)! (1 + θ x)m+1−α
484
J. He / Statistics and Probability Letters 82 (2012) 478–487
where x ∈ (−1, +∞), θ ∈ (0, 1), m ∈ Z + , for 0 < α < 1, we have
1+
1
α
+ 1−
n+1
α+1
1 n+1
≥2−
1 n+1
α2 α(α − 1) α(α − 1)2 (α − 2) − + , (n + 1)2 2(n + 1)3 12(n + 1)4
+
and
1+
1
α+1
+ 1−
n+1
1
α+1
n+1
α(α + 1) α(α + 1)(α − 1)(α − 2) + (n + 1)2 12(n + 1)4 4−α α(α + 1)(α − 1)(α − 2)(α − 3) 1 − 1+ . 5!(n + 1)5 n
≤ 2+
Since In+1 = (n + 1)ϵ 1+1/α ((n + 2)α − (n + 1)α ) −
α ϵ 1+1/α (n + 2)α+1 − (n + 1)α+1 , α+1
we have In+1 − In = ϵ 1+1/α (n + 1)(n + 2)α − (2n + 1)(n + 1)α + nα+1
α ϵ 1+1/α (n + 2)α+1 − 2(n + 1)α+1 + nα+1 α+1 α α+1 1 1 1 1+1/α α+1 =ϵ (n + 1) 1+ + 1− −2+ n+1 n+1 n+1 α+1 α+1 α 1 1 1+1/α α+1 − ϵ (n + 1) + 1− −2 1+ α+1 n+1 n+1 α2 α(α − 1) α(α − 1)2 (α − 2) ≥ ϵ 1+1/α (n + 1)α+1 − + (n + 1)2 2(n + 1)3 12(n + 1)4 α α(α + 1)(α − 1)(α − 2) 1+1/α α+1 α(α + 1) − ϵ (n + 1) + α+1 (n + 1)2 12(n + 1)4 4−α α 2 (α − 1)(α − 2)(α − 3) 1 − 1+ 4 −α 5!(n + 1) n α(α − 1) α(α − 1)(α − 2) α 2 (α − 1)(α − 2)(α − 3) = ϵ 1+1/α − − + . 2(n + 1)2−α 12(n + 1)3−α 5!n4−α −
(2.13)
From (2.8), (2.9), (2.10) and (2.13), we have
ϵ
1/α
λα (ϵ) − cα ≤
∞ 2 −(n+1)2α ϵ 2 /2 e In −
π
n=1
2
π
ϵ
2 t 1/α e−t /2
0
α(α − 1)(α − 2) α(α − 1) + π n=1 π n=1 2(n + 1)2−α 12(n + 1)3−α ∞ ϵ 2 α 2 (α − 1)(α − 2)(α − 3) 2 2 − ϵ 1+1/α − t 1/α e−t /2 dt + o(ϵ 1+1/α ) π n =1 5!n4−α π 0 ∞ 2 − n2α ϵ 2 2 ≤ e In + C ϵ 1+1/α + o(ϵ 1+1/α ) π n=1 ∞ 2 1+1/α −n2α ϵ 2 /2 α α(α − 1) ≤ ϵ e − 1−α − + Dϵ 1+1/α 2−α π 2n 3n n =1
≤
≤−
2
∞
ϵ 1/α 2
−(n+1)2α ϵ 2 /2
e
+ C2 ϵ 1+1/α .
In + 1 + ϵ
1+1/α
∞ 2
(2.14)
J. He / Statistics and Probability Letters 82 (2012) 478–487
485
(b) If σ 2 ̸= 1, then
∞ ϵ |Sn | P ≥ n1/2+α σ σ n =1 1/α (ϵ/σ ) 1/α 1+1/α ≤σ cα − + C2 (ϵ/σ )
ϵ 1/α λα (ϵ) = ϵ 1/α
2
= σ 1/α cα −
ϵ
1/α
+
2
C2
σ
ϵ 1+1/α .
Similarly, we can obtain
ϵ 1/α
ϵ 1/α λα (ϵ) ≥ σ 1/α cα −
+
2
C1
σ
ϵ 1+1/α .
Thus we have proved both in case (a) and case (b), there exist constants 0 < C1 < C2 < ∞, such that C1 ϵ 1+1/α ≤ ϵ 1/α λα (ϵ) − σ 1/α cα + And Lemma 2.2. is now proved.
ϵ 1/α
≤ C2 ϵ 1+1/α .
2
3. Proofs of theorems Proof of Theorem 1. Under the condition (1.4), from the non-uniform estimate of central limit theorem by Nagaev (1965), we have for every x ∈ R,
3 P √1 Sn < x − Φ (x) ≤ √ AE |X | , nσ nσ 3 (1 + |x|)3
(3.1)
where A is an absolute positive constant, and Φ (x) is the standard normal distribution function. Since P (|Sn | ≥ n
1/2+α
ϵ) =
∞
2
π
nα ϵ/σ
2 e−t /2 dt + Rn
where Rn = Φ
nα ϵ
−P
σ
1
√
nσ
Sn ≤
nα ϵ
σ
+P
1
√
nσ
Sn ≤
−n α ϵ σ
−Φ
−n α ϵ σ
.
By (3.1), we have
|Rn | ≤ √
2AE |X |3
3 nσ 3 1 + σϵ nα
.
(3.2)
So
λα (ϵ) =
∞
2
π
n =1
∞
nα ϵ/σ
2 e−t /2 dt +
∞
Rn .
n =1
By Lemma 2.2,
ϵ 1/α
∞ n =1
2
π
∞ nα ϵ/σ
ϵ 2 e−t /2 dt − σ 1/α cα = −
1/α
2
+ O(ϵ 1+1/α ).
And in consequence,
ϵ 1α λα (ϵ) − σ 1/α cα = −
ϵ 1/α 2
1
+ϵα
∞
Rn + O(ϵ 1+1/α ).
(3.3)
n =1
In order to prove Theorem 1, we only need to show that ∞ n =1
Rn = O(ϵ −1/2α ).
(3.4)
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J. He / Statistics and Probability Letters 82 (2012) 478–487
Setting ∞
σ2 ϵ 1/α
Rn =
n =1
∞
Rn +
n=1
n=
Rn =: I1 + I2 .
(3.5)
σ 2 +1 ϵ 1/α
It follows from (3.2) that
σ2 ϵ 1/α
2AE |X |3 √ = O(ϵ −1/2α ). σ3 n n=1 ∞ 1 Since for 3α + 21 > 1, n=1 n3α+ 1/2 < ∞, we have I1 ≤
∞
I2 ≤
C
σ 2 +1 ϵ 1/α
n3α+1/2 ϵ 3
(3.6)
≤ C ϵ −1/2α .
(3.7)
Combining (3.5) together with (3.6), (3.7), we can obtain (3.4). And now the proof of Theorem 1 is complete.
Proof of Theorem 2. By Osipov and Petrov (1967) inequality there exists a bounded and decreasing function ψ(u) on the interval (0, ∞), such that limu→∞ ψ(u) = 0 and
√ P √1 Sn < x − Φ (x) ≤ ψ( n(1 + |x|)) . nδ/2 (1 + |x|)2+δ nσ
(3.8)
By a process similar to the proof of (3.3), we can obtain
ϵ 1/α λα (ϵ) − σ 1/α cα = −
ϵ 1/α
+ ϵ 1/α
2
∞
R∗n + O(ϵ 1+1/α ),
(3.9)
n=1
where ∗
|Rn | ≤
√
n(1 + nα ϵ/σ )
2ψ
.
nδ/2 (1 + nα ϵ/σ )2+δ
Since for both the case of 0 < δ < 1, 2(22−δ < α < 1 and the case of +δ) ∞
1
n2α+αδ+δ/2 n =1
1 6
< α ≤ 12 ,
2−4α 2α+1
< δ < 1, we have
< ∞.
Then ∞
|R∗n | ≤ ψ
2 n= σ1/α +1 ϵ
σ ϵ 1/2α
∞ n=
ϵ
σ 2 +1 1/α
C n2α+αδ+δ/2 ϵ 2+δ
σ 1 ≤ C ψ 1/2α ϵ −2−δ (ϵ − α )−2α−αδ−δ/2+1 ϵ = o(ϵ δ/2α−1/α ).
By (3.9), we have
ϵ 1/α λα (ϵ) − σ 1/α cα = ϵ 1/α
σ2 ϵ 1/α
R∗n + o(ϵ δ/2α ).
n=1
In order to prove Theorem 2, it is only to prove
ϵ 1/α
σ2 ϵ 1/α
n =1
R∗n = o(ϵ δ/2α ) as ϵ → 0.
(3.10)
J. He / Statistics and Probability Letters 82 (2012) 478–487
487
√ Since limu→∞ ψ(u) = 0, for any η > 0, there exists a natural number N0 , such that ψ( n) < η whenever n > N0 . Hence
ϵ 1/α
σ2 ϵ 1/α
|R∗n | = ϵ 1/α
N0
|R∗n | + ϵ 1/α
n=1
n =1
σ2 ϵ 1/α
|R∗n |
n=N0 +1
≤ ϵ 1/α
N0
√
n−δ/2 ψ( n) + ηϵ 1/α
n=1
σ2 ϵ 1/α
n−δ/2
n=N0 +1
≤ ϵ 1/α M + ηϵ δ/2α , where M is a constant. So
ϵ 1/α
σ2 ϵ 1/α
R∗n = o(ϵ δ/2α ).
(3.11)
n =1
From (3.10) and (3.11), we have
ϵ 1/α λα (ϵ) − σ 1/α cα = o(ϵ δ/2α ). This completes the proof of Theorem 2.
(3.12)
Acknowledgments The author is very grateful to the referees for carefully reading the manuscript and for valuable comments, which improved the presentation of this paper. The author also thanks Professor Hira Koul very much for efficient work and useful suggestions. References Chen, R., 1976. A remark on the strong law of large numbers. Proc. Amer. Math. Soc. 61, 112–116. Chen, R., 1978. A remark on the tail probability of a distribution. J. Multivariate Anal. 8, 328–333. Erdös, P., 1949. On a theorem of Hsu and Robbins. Ann. Math. Statist. 20, 286–291. Erdös, P., 1950. Remark on my paper on a theorem of Hsu and Robbins. Ann. Math. Statist. 21, 138. Gut, A., Spˇataru, A., 2000a. Precise asymptotics in the Baum–Katz and Davis law of large numbers. J. Math. Anal. Appl. 248, 233–246. Gut, A., Spˇataru, A., 2000b. Precise asymptotics in the law of the iterated logarithm. Ann. Probab. 28, 1870–1883. Gut, A., Spˇataru, A., 2003. Precise asymptotics in some strong limit theorem for multidimensionally indexed random variables. J. Multivariate Anal. 86, 398–422. Gut, A., Steinebach, J., 2011. Convergence rates in precise asymptotics. UUDM. Report. He, J.J., Xie, T.F., 2012. Asymptotics property for some series of probability. Acta Math. Applacatae Sin. (in press). Heyde, C.C., 1975. A supplement to the strong law of large numbers. J. Appl. Probab. 12, 903–907. Hsu, P.L., Robbins, H., 1947. Complete convergence and the law of large numbers. Proc. Natl. Acad. Sci. USA 33, 25–31. Klesov, O.I., 1994. On the convergence rate in a theorem of Heyde. Theory Probab. Math. Statist. 49, 83–87. Liu, W.D., Lin, Z.Y., 2006. Precise asymptotics for a new kind of complete moment convergence. Statist. Probab. Lett. 76, 1787–1799. Nagaev, S.V., 1965. Some limit theorems for large deviation. Theory Probab. Appl. 10, 214–235. Osipov, L.V, Petrov, V.V., 1967. On an estimate of the remainder term in the central limit theorem. Theory Probab. Appl. 12, 281–286. Petrov, V.V., 1982. A limit theorem for sums of independent nonidentically distributed random variables. J. Math. Sci. 20, 2232–2235.