Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 1, Article 2, 2006
AN INEQUALITY FOR THE ASYMMETRY OF DISTRIBUTIONS AND A
BERRY-ESSEEN THEOREM FOR RANDOM SUMMATION
HENDRIK KLÄVER AND NORBERT SCHMITZ
I NSTITUTE OF M ATHEMATICAL S TATISTICS
U NIVERSITY OF M ÜNSTER
E INSTEINSTR . 62
D-48149 M ÜNSTER , G ERMANY
schmnor@math.uni-muenster.de
Received 16 March, 2004; accepted 15 December, 2005
Communicated by S.S. Dragomir
A BSTRACT. We consider random numbers Nn of independent, identically distributed (i.i.d.)
PNn
random variables Xi and their sums i=1
Xi . Whereas Blum, Hanson and Rosenblatt [3]
proved a central limit theorem for such sums and Landers and Rogge [8] derived the corresponding approximation order, a Berry-Esseen type result seems to be missing. Using an inequality
for the asymmetry of distributions, which seems to be of its own interest, we prove, under the
assumption E|Xi |2+δ < ∞ for some δ ∈ (0, 1] and Nn /n → τ (in an appropriate sense), a
Berry-Esseen theorem for random summation.
Key words and phrases: Random number of i.i.d. random variables, Central limit theorem for random sums, Asymmetry of
distributions, Berry-Esseen theorem for random sums.
2000 Mathematics Subject Classification. 60E15, 60F05, 60G40.
1. I NTRODUCTION
One of the milestones of probability theory is the famous theorem of Berry-Esseen which
gives uniform upper bounds for the deviation from the normal distribution in the Central Limit
Theorem:
Let {Xn , n ≥ 1} be independent random variables such that
EXn = 0, EXn2 =: σn2 ,
n
X
2
sn :=
σi2 > 0,
Γ2+δ
:=
n
i=1
n
X
i=1
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
056-04
E|Xi |2+δ < ∞
2
H ENDRIK K LÄVER AND N ORBERT S CHMITZ
P
for some δ ∈ (0, 1] and Sn = ni=1 Xi , n ≥ 1. Then there exists a universal constant Cδ such
that
2+δ
Γn
S
n
≤ x − Φ(x) ≤ Cδ
,
sup P
sn
sn
x∈R
where Φ denotes the cumulative distribution function of a N (0, 1)-(normal) distribution (see
e.g. Chow and Teicher [5, p. 299]).
For the special case of identical distributions this leads to:
Let {Xn , n ≥ 1} be i.i.d. random variables with EXn = 0, EXn2 =: σ 2 > 0, E|Xn |2+δ =:
γ 2+δ < ∞ for some δ ∈ (0, 1]. Then there exists a universal constant cδ such that
S
cδ γ 2+δ
n
√ ≤ x − Φ(x) ≤ δ/2
sup P
.
n
σ
σ n
x∈R
Van Beek [2] showed that C1 ≤ 0.7975; bounds for other values Cδ are given by Tysiak [12],
e.g. C0.8 ≤ 0.812; C0.6 ≤ 0.863, C0.4 ≤ 0.950, C0.2 ≤ 1.076.
On the other hand, there exist also central limit theorems for random summation, e.g. the
theorem of Blum, Hanson and Rosenblatt [3] which generalizes previous results by Anscombe
[1] and Renyi [11]:
P
Let {Xn , n ≥ 1} be i.i.d. random variables with EXn = 0, Var Xn = 1, Sn := ni=1 Xi and
P
let {Nn , n ≥ 1} be N-valued random variables such that Nn /n −→ U where U is a positive
random variable. Then
√
d
P SNn / Nn −→ N (0, 1).
Therefore, the obvious question arises whether one can prove also Berry-Esseen type inequalities for random sums. A first result concerning the approximation order is due to Landers and
Rogge [8] for random variables Xn with E|Xn |3 < ∞; this was generalized by Callaert and
Janssen [4] to the case that E|Xn |2+δ < ∞ for some δ ∈ (0, 1]:
Let {Xn , n ≥ 1} be i.i.d. random variables with E Xn = µ, Var Xn = σ 2 > 0, and
E|Xn |2+δ < ∞ for some δ > 0. Let {Nn , n ≥ 1} be N-valued random variables, {εn , n ≥ 1}
positive real numbers with εn −→ 0 where, for n large, n−δ ≤ εn if δ ∈ (0, 1] and n−1 ≤ εn if
n→∞
δ ≥ 1. If there exists a τ > 0 such that
Nn
√
P − 1 > εn = O( εn ),
nτ
then
sup P
x∈R PNn
!
√
− Φ(x) = O( εn )
sup P
x∈R PNn
!
√
− Φ(x) = O( εn ).
i=1 (Xi − µ)
√
≤x
σ nτ
and
i=1 (Xi − µ)
√
≤x
σ Nn
Moreover, there exist several further results on convergence rates for random sums (see e.g. [7]
and the papers cited there); as applications, sequential analysis, random walk problems, Monte
Carlo methods and Markov chains are mentioned.
However, rates of convergence without any knowledge about the factors are of very limited
importance for applications. Hence the aim of this paper is to prove a Berry-Esseen type result
for random sums i.e. a uniform approximation with explicit constants. Obviously, due to the
dependencies on the moments of the Xn as well as on the asymptotic behaviour of the sequence
Nn such a result will necessarily be more complex than the original Berry-Esseen theorem.
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
http://jipam.vu.edu.au/
A N I NEQUALITY
FOR THE
A SYMMETRY
OF
D ISTRIBUTIONS AND A B ERRY-E SSEEN T HEOREM
FOR
R ANDOM S UMMATION
3
For the underlying random variables Xn we make the same assumption as in the special
version of the Berry-Esseen theorem:
(
Xn , n ≥ 1, are i.i.d. random variables with EXn = 0, V ar Xn = 1 and
(M)
:
γ 2+δ := E|Xn |2+δ < ∞ for some δ ∈ (0, 1].
Similarly as Landers and Rogge [8] or Callaert and Janssen [4] resp. we assume on the random
indices


 Nn , n ≥ 1, are integer-valued random variables and ζn , n ≥ 1,
(R)
:


real numbers with limn→∞ ζn = 0 such that there exist
√
d, τ > 0 with P (| Nnτn − 1| > ζn ) ≤ d ζn .
As they are essential for applications, e.g. in sequential analysis, arbitrary dependencies between the indices and the summands are allowed.
2. A N I NEQUALITY FOR THE A SYMMETRY
OF
D ISTRIBUTIONS
A main tool for deriving explicit constants for the rate of convergence is an inequality which
seems to be of its own interest. For a smooth formulation we use (for the different values of
the moment parameter δ ∈ (0, 1]) some (technical) notation: For ϑ := 2/(1 + δ) and y ≥ 1 let
gδ (y) be defined by
p

2
2y
−
1
+
2y
y2 − 1
if δ = 1




n

p

 min 2ϑ y 4ϑ − 1 + 2 ϑ+1
2ϑ
2 y

2ϑ−1 y 4ϑ − 1 ,


o

p
8ϑ
4ϑ
8ϑ − 1
2y
−
1
+
2y
y
if δ ∈ 13 , 1
gδ (y) :=


n

p
ϑ+1

ϑ (2+2k )ϑ
(1+2k−1 )ϑ

2 y

min
2
y
−
1
+
2
2ϑ−1 y (2+2k )ϑ − 1,


o
i

p

k+1 )ϑ

1
1
(4+2k+1 )ϑ
(2+2k )ϑ
(4+2
2y
− 1 + 2y
y
−1
if δ ∈ 2k +1 , 2k−1 +1 , k ≥ 2
Theorem 2.1. Let X be a random variable with EX = 0, Var X = 1 and γ 2+δ := E|X|2+δ <
∞ for some δ ∈ (0, 1]. Then
P (X < 0) ≤ gδ (γ 2+δ )P (X > 0) and P (X > 0) ≤ gδ (γ 2+δ )P (X < 0).
Proof. Let X be a random variable as described above. Let X+ = max{X, 0}, X− = min{X, 0},
E(X+ ) = E(X− ) = α and P (X > 0) = p, P (X = 0) = r, P (X < 0) = 1 − r − p. Since
E(X) = 0, Var(X) = 1 it is obvious that p, 1 − r − p > 0. As α = pE(|X| |X > 0), we have
α
E(|X| |X > 0) = αp ; analogously E(|X| |X < 0) = 1−r−p
.
Applying Jensen’s inequality to the convex function f : [0, ∞) → [0, ∞), f (x) = x
yields
3+δ
3+δ
2
3+δ
3+δ
α 2
α
E(|X| 2 |X > 0) ≥
and E(|X| 2 |X < 0) ≥
.
p
1−r−p
3+δ
2
Defining βz := E|X|z for z > 0 we get
3+δ
3+δ
β 3+δ = pE |X| 2 |X > 0 + (1 − r − p)E |X| 2 |X < 0
2
≥α
3+δ
2
(1 − r − p)
1+δ
2
+p
(p(1 − r − p))
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
1+δ
2
1+δ
2
http://jipam.vu.edu.au/
4
H ENDRIK K LÄVER AND N ORBERT S CHMITZ
and, therefore,
α
(i)
3+δ
2
≤ β 3+δ
2
(p(1 − r − p))
(1 − r − p)
1+δ
2
1+δ
2
+p
1+δ
2
Since γ 2+δ < ∞, we can apply the Cauchy-Schwarz-inequality to |X|1/2 and |X|1+δ/2 and
obtain
2
3+δ
E|X| 2
≤ E|X| E|X|2+δ = 2αγ 2+δ ;
hence
2
2
α≥
(ii)
β 3+δ
2γ 2+δ
.
Combining (i) and (ii) we obtain

2  3+δ
2
1+δ
β
3+δ
(p(1 − r − p)) 2


2
≤ β 3+δ 1+δ


1+δ ;
2
2γ 2+δ
p 2 + (1 − r − p) 2
hence
1
1
x2 ≤ −
4 1−r
ϑ+1
1
a1
2
a2
with
p
1
2
− , ϑ=
,
1 − r 2
1+δ
Obviously x ∈ − 21 , 12 , ϑ ∈ [1, 2).
Since
1
+x
2
ϑ1
+
ϑ+2
a1 = β 3+δ
,
x=
(iii)
2
1
−x
2
ϑ1 !ϑ
a2 = (γ 2+δ )ϑ+1 .
xµ + y µ ≥ (x + y)µ ∀µ ∈ (0, 1], x, y ≥ 0
and 0 < 1 − r ≤ 1 it follows altogether that
s
ϑ−1
1
1
a1
(iv)
|x| ≤
1−
.
2
2
a2
For large values of |x| this estimation is rather poor, so we notice, furthermore, that
ϑ1 !ϑ
ϑ1 1
1
≥ 2ϑ−1 (1 − 4x2 )1/2 ;
+
−x
+x
2
2
hence
1
|x| ≤
2
(v)
s
a1
a2
2
s
a2
a1
1−
.
From (iv) and (v) it follows that
(vi)
ϑ+1
p
a2
≤ 2ϑ − 1 + 2 2
1−r−p
a1
a2
2ϑ−1
a1
−1
and
(vii)
p
≤2
1−r−p
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
a2
a1
2
a2
−1+2
a1
s
a2
a1
2
− 1.
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A N I NEQUALITY
FOR THE
A SYMMETRY
OF
D ISTRIBUTIONS AND A B ERRY-E SSEEN T HEOREM
FOR
R ANDOM S UMMATION
5
4
3+δ
Now we estimate β 3+δ . Due to Jensen’s inequality we have β 3+δ
≤ σ 2 = 1. Let δ = 1.
2
2
≤ 2 + δ, so β 5−δ exists. Due
Then β 3+δ = σ 2 = 1 and so aa21 = (γ 3 )2 . Let δ ∈ 13 , 1 . Then 5−δ
2
2
2
to the Cauchy-Schwarz-inequality we have
1 = (σ 2 )2 ≤ β 3+δ β 5−δ ;
2
hence
β 3+δ ≥
2
1
β 5−δ
2
1
≥
.
5−δ
(γ 2+δ ) 2(2+δ)
2
altogether
a2
≤ (γ 2+δ )4ϑ .
a1
1
≤ 2 + δ; hence β2+ 1−δ exists. Due to
Let k ∈ N, k ≥ 2. For δ ≥ 2k +1 it follows that 2 + 1−δ
2k
2k
the Cauchy-Schwarz-inequality we have
1 = (σ 2 )2 ≤ β2+ 1−δ β2− 1−δ
2k
and
β2− 1−δ
j+1
2 j
=
2
β2− 1−δ
j+1
2 2j−1
2
for j ∈ {1, . . . , k − 1}. This yields
2k−1
β 3+δ ≥ β2− 1−δ
≥
2
2k
2k
2j−1
≤ β2− 1−δ
β
2
j
2
1
2k−1 ≥
1
(γ 2+δ )
β2+ 1−δ
2k
h
a2
1
1
2+δ (2k +2)ϑ
Altogether we get a1 ≤ (γ )
for δ ∈ 2k +1 , 2k−1 +1 .
Combining this with (vi) and (vii) we obtain the assertion.
2k+1 +1−δ
2(2+δ)
.
Remark 2.2. For each δ ∈ (0, 1] equality holds in Theorem 2.1 iff P X = 12 (ε1 + ε−1 ) (where
εx denotes the Dirac measure in x).
Proof. (i) Let X be a real random variable with P X = 12 (ε1 +ε−1 ). Then E(X) = 0, Var(X) =
1, γ 2+δ = 1 and so gδ (γ 2+δ ) = 1 for all δ ∈ (0, 1]. Since P (X < 0) = P (X > 0) = 12 we get
equality.
(ii) In Theorem 2.1 we have
β 23+δ
! 3+δ
2
2
2γ 2+δ
≤α
3+δ
2
≤
(p(1 − r − p))
(1 − r − p)
1+δ
2
1+δ
2
+p
1+δ
2
β 3+δ .
2
In the first “≤” there is equality iff |X|1/2 and |X|1+δ/2 are linearly dependent P -almost surely,
i.e. P (|X| ∈ {0, c}) = 1 for some c > 0. As E(X) = 0, we obtain P X = p(εc + ε−c ) + (1 −
2p)ε0 . So the inequality is sharp iff gδ (γ 2+δ ) = 1. With 1 = E(X 2 ) = 2c2 p we obtain
γ 2+δ = 2pc2+δ =
1
.
(2p)δ/2
The functions hδ : (0, 12 ] → R, δ ∈ (0, 1], defined by hδ (p) = gδ (2p)1δ/2 , are strictly de
creasing. Due to p ∈ 0, 21 and hδ 12 = 1 we get p = 12 , r = 0 and c = 1, therefore
P X = 12 (ε1 + ε−1 ).
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
http://jipam.vu.edu.au/
6
H ENDRIK K LÄVER AND N ORBERT S CHMITZ
Since the Central Limit Theorem is concerned with sums of random variables (instead of
single variables) we need a corresponding generalization of Theorem 2.1.
Corollary 2.3. Under assumption (M),

2+δ 
n
1 X
γ 2+δ − 1
Xi  ≤
+ n2−δ/2
E  √
δ/2
n
n
i=1
holds, and therefore,
P
n
X
!
≤ gδ
Xi < 0
i=1
n
X
γ 2+δ − 1
2−δ/2
+n
P
nδ/2
!
Xi > 0 .
i=1
Proof.
2+δ 
3  2+δ
3
n
n
X
X
Xi = 
Xi 
i=1
i=1


≤
n
X
|Xi |3 + 3
i=1
≤
n
X
|Xk2 Xi | +
k,i=1
k6=i
2+δ
|Xi |
i=1
n
X
+3
n
X
 2+δ
3
n
X

|Xi Xk Xl |
i,k,l=1
i6=k6=l6=i
2+δ
|Xk2 Xi | 3
k,i=1
k6=i
n
X
+
|Xi Xk Xl |
2+δ
3
.
i,k,l=1
i6=k6=l6=i
Since the Xi are independent and, due to E|Xi |2 = 1 and Jensen’s inequality, E|Xi |α ≤ 1 for
α ≤ 2, we obtain
2+δ
n
1 X
1
γ 2+δ − 1
2+δ
3
√
E
Xi ≤ 1+δ/2 (nγ
+ (n − n)) =
+ n2−δ/2 .
δ/2
n
n
n
i=1
P
P
2
From E √1n ni=1 Xi = 0, E √1n ni=1 Xi = 1 and Theorem 2.1 the assertion follows.
We need a bound which does not depend on the number of summands. For this aim we
use for n < κ := (4cδ γ 2+δ )2/δ the asymmetry inequality of Corollary 2.3 and for n ≥ κ the
Berry-Esseen bound (the choice of κ as boundary between “small” and “large” n is somewhat
arbitrary).
Corollary 2.4. Under Assumption (M)
!
n
X
P
Xi < 0 ≤ (fδ (γ 2+δ ) − 1)P
i=1
where
n
X
!
Xi > 0 ,
i=1
4
y−1
−1
fδ (y) := max 3, gδ (y), gδ
+ (4cδ y) δ
+ 1.
4cδ
Proof. (i) Let n < κ. Then Corollary 2.3 yields
!
2+δ
n
X
γ
−1
2−δ/2
P
Xi < 0 ≤ gδ
+n
P
nδ/2
i=1
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
n
X
!
Xi > 0 .
i=1
http://jipam.vu.edu.au/
A N I NEQUALITY
FOR THE
A SYMMETRY
OF
D ISTRIBUTIONS AND A B ERRY-E SSEEN T HEOREM
For the function h : [1, ∞) → [1, ∞) defined by h(y) :=
δ
h (y) =
2
00
δ
1+
2
(γ
2+δ
− 1)y
γ 2+δ −1
y δ/2
δ
+ 2−
2
−2−δ/2
FOR
R ANDOM S UMMATION
7
+ y 2−δ/2 we obtain
δ
1−
y −δ/2 > 0,
2
i.e., h is convex. Hence the maximum of
h : {1, . . . , bκc} → [1, ∞)
is attained either for n = 1 or for n = bκc. Since gδ is strictly increasing this yields
2+δ
2+δ
γ
−1
γ
−1
2−δ/2
2+δ
2+δ 4/δ−1
gδ
+n
≤ max gδ (γ ), gδ
+ (4cδ γ )
nδ/2
4cδ γ 2+δ
for all n < κ.
(ii) Let n ≥ κ, i.e. nδ/2 ≥ 4cδ γ 2+δ . Then the (special case of the) theorem of Berry-Esseen
yields
!
n
X
γ 2+δ
1
1
≤
c
P
X
<
0
−
i
δ δ/2 ≤
2
n
4
i=1
and, therefore,
n
X
P
!,
Xi < 0
n
X
P
i=1
!
Xi > 0
≤
i=1
3/4
= 3.
1/4
Combining (i) and (ii) we obtain the assertion.
3. S OME F UTHER I NEQUALITIES
The next inequality represents a quantification of Lemma 7 by Landers and Rogge [8].
Lemma 3.1. Under Assumption (M)
P
min
p≤n≤q
n
X
!
Xi ≤ r
−P
i=1
max
p≤n≤q
n
X
!
Xi ≤ r
i=1
p
≤ fδ (γ 2+δ ) P
X
Xi ≤ r,
i=1
q
X
!
Xi ≥ r
+P
i=1
p
X
i=1
Xi ≥ r,
q
X
!!
Xi ≤ r
i=1
for all p, q ∈N s.t. p < q and for all r ∈ R.
Proof. Using the same notation (A, α, β, Ak ) as Landers and Rogge [8, p. 280], we have to
prove that
P (A) ≤ fδ (γ 2+δ )(α + β).
P
(i) First we show that P (A ∩ { pi=1 Xi ≤ r}) ≤ fδ (γ 2+δ ) · α; for this it is sufficient to prove
that
) ( q
)!
( p
X
X
Xi ≤ r
≤ (fδ (γ 2+δ ) − 1) · α.
Xi ≤ r ∩
P A∩
i=1
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
i=1
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8
H ENDRIK K LÄVER AND N ORBERT S CHMITZ
P
But due to the independence of the Ak and qi=k+1 Xi
) ( q
)!
( p
X
X
Xi ≤ r ∩
Xi ≤ r
P A∩
i=1
i=1
q−1
X
≤
P (Ak )P
k=p+1
q
X
!
Xi ≤ 0
i=k+1
q−1
≤ (fδ (γ 2+δ ) − 1)
X
q
X
P (Ak )P
k=p+1
!
Xi ≥ 0
i=k+1
according to Corollary 2.4
≤ (fδ (γ 2+δ ) − 1) · α.
(ii) Similarly, it follows that
A∩
P
( k
X
)!
Xi > r
≤ fδ (γ 2+δ ) · β.
i=1
(i) and (ii) yield the assertion.
A thorough examination of the proof of Lemma 8 of Landers and Rogge [8] allows a generalization and quantification of their result:
Lemma 3.2. Under Assumption (M),
P
Pn+k
n
X
≥
t
≤
X
≤
t,
(i) P
i
i
i=1
i=1
(ii) P
P
n
i=1
Xi ≥ t,
Pn+k
i=1
Xi ≤ t ≤
2cδ γ 2+δ
nδ/2
+
√1
2π
2cδ γ 2+δ
nδ/2
+
√1
2π
q
q
k
.
n
k
.
n
Proof. (ii) follows from (i) by replacing Xi by −Xi . Analogously to the proof of [8], Lemma 8,
we obtain
!
n
n+k
X
X
P
Xi ≤ t,
Xi ≥ t
i=1
i=1
!
Z k
t
1 X
t
−Φ √ − √
xi Q(dx1 , . . . , dxk )
Φ √
n
n
n i=1
r Z k
k 1 X 1
xi Q(dx1 , . . . , dxk )
+√
√
2π n k i=1 r
k
1
,
+√
2π n
2cδ γ 2+δ
≤
+
nδ/2
≤
2cδ γ 2+δ
nδ/2
≤
2cδ γ 2+δ
nδ/2
since
v
2
u k
k
1 X
u 1 X
t E √
Xi ≤ E √
Xi = 1.
k
k
i=1
i=1
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
http://jipam.vu.edu.au/
A N I NEQUALITY
A SYMMETRY
FOR THE
OF
D ISTRIBUTIONS AND A B ERRY-E SSEEN T HEOREM
4. A B ERRY-E SSEEN T HEOREM
FOR
FOR
R ANDOM S UMMATION
9
R ANDOM S UMS
As an important application of these inequalities we state our main result:
Theorem 4.1 (Berry-Esseen type result for random sums). Let {Xn , n ≥ 1} be i.i.d. random
variables with EXn = 0, Var Xn = 1 and γ 2+δ := E|Xn |2+δ < ∞ for some δ ∈ (0, 1];
let {Nn , n ∈ N} be integer-valued random variables and {ζn , n ∈ N} real numbers with
limn→∞ ζn = 0 such that there exist d, τ > 0 with
p
Nn
P − 1 > ζn ≤ d ζn .
nτ
Then1
!
Nn
1 X
√
Xi ≤ t − Φ(t)
nτ i=1
sup P
t∈R (i)
1
1
+p
δ/2
bnτ c
2πebnτ c
s
p
2 + 1/τ
1
2+δ
+ 2fδ (γ )
max
, ζn + 2d ζn .
π
n
≤ cδ γ 2+δ (1 + 22+δ/2 fδ (γ 2+δ ))
(ii) sup P
t∈R !
Nn
1 X
√
Xi ≤ t − Φ(t)
Nn i=1
1
1
p
+
bnτ cδ/2
2πebnτ c
s
p
2 + 1/τ
1
2+δ
+ 2fδ (γ )
max
, ζn + (3d + 1) ζn
π
n
≤ cδ γ 2+δ (1 + 22+δ/2 fδ (γ 2+δ ))
for all n ∈ N s.t.
Since
nτ
2
nτ
2
− nτ ζn ≥ 1.
− nτ ζn −→ ∞ there exists an n0 s.t.
n→∞
nτ
2
− nτ ζn ≥ 1 for all n ≥ n0 .
Proof. (i) Let n ∈ N fulfill nτ
− nτ ζn ≥ 1 and define as Landers and Rogge [8, p. 271]
2
√
bn (t) := t nτ and In := {k ∈ N : bnτ − nτ ζn c ≤ k ≤ bnτ + nτ ζn c}.
Due to the assumption on Nn we have
p
P (Nn ∈
/ In ) ≤ d ζn .
For
(
An (t) :=
max
k∈In
k
X
)
Xi ≤ bn (t) ,
(
Bn (t) :=
i=1
min
k∈In
k
X
)
Xi ≤ bn (t)
i=1
follows (see Landers and Rogge [8, p. 272]) for each t ∈ R
P (An (t) ∩ {Nn ∈ In }) ≤ P
Nn
X
!
Xi ≤ bn (t), Nn ∈ In
≤ P (Bn (t))
i=1
1
bxc := max{n ∈ N : n ≤ x}.
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
http://jipam.vu.edu.au/
10
H ENDRIK K LÄVER
AND
N ORBERT S CHMITZ
and

bnτ c
P (An (t)) ≤ P 
X

Xi ≤ bn (t) ≤ P (Bn (t)).
i=1
Using the Berry-Esseen theorem and the result (3.3) of Petrov [10, p. 114], we obtain


bnτ c
X
sup P 
Xi ≤ bn (t) − Φ(t)
t∈R i=1


r
r
r
bnτ c
X
1
nτ
nτ
nτ
−Φ t
≤ sup P  p
Xi ≤ t
+ sup Φ t
− Φ(t)
bnτ c
bnτ c t∈R
bnτ c
t∈R bnτ c i=1
≤
1
cδ γ 2+δ
p
+
.
bnτ cδ/2
2πebnτ c
For p(n) := bnτ − nτ ζn c, q(n) := bnτ + nτ ζn c we obtain from Lemma 3.1
P (Bn (t)) − P (An (t))
 
≤ fδ (γ 2+δ ) · P 
p(n)
X
Xi ≤ bn (t) ≤
i=1
q(n)
X


Xi  + P 
i=1
p(n)
X
Xi ≥ bn (t) ≥
i=1
q(n)
X

Xi  .
i=1
According to Lemma 3.2 it follows that


s
p(n)
q(n)
2+δ
X
X
1
2cδ γ
q(n) − p(n)
sup P 
Xi ≤ bn (t) ≤
Xi  ≤
+√
δ/2
(p(n))
p(n)
2π
t∈R
i=1
i=1
s
21+δ/2 cδ γ 2+δ
2 + 1/τ
1
≤
+
max
, ζn ,
(nτ )δ/2
π
n
since p(n) ≥ nτ − nτ ζn − 1 ≥ nτ /2 and, therefore,
s
q(n) − p(n)
≤
p(n)
s
2nτ ζn + 1
=
nτ /2
r
4ζn +
2
;
nτ
analogously


s
p(n)
q(n)
1+δ/2
2+δ
X
X
2
c
γ
2 + 1/τ
1
δ


sup P
Xi ≥ bn (t) ≥
Xi ≤
+
max
, ζn .
δ/2
(nτ
)
π
n
t∈T
i=1
i=1
J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
http://jipam.vu.edu.au/
A N I NEQUALITY
FOR THE
A SYMMETRY OF D ISTRIBUTIONS AND A B ERRY-E SSEEN T HEOREM
FOR
R ANDOM S UMMATION
11
Altogether we obtain
!
Nn
X
Xi ≤ bn (t) − Φ(t)
sup P
t∈R
i=1
!
Nn
X
≤ sup P
Xi ≤ bn (t), Nn ∈ In − Φ(t) + P (Nn ∈
/ In )
t∈R
i=1
≤ P (Bn (t)) − P (An (t)) +
≤ 2fδ γ 2+δ
21+δ cδ γ 2+δ
(nτ )δ/2
p
cδ γ 2+δ
1
p
+
+
2d
ζn
bnτ cδ/2
2πebnτ c
s
!
2 + 1/τ
1
+
max
, ζn
π
n
p
1
cδ γ 2+δ
p
+
+
2d
ζn
bnτ cδ/2
2πebnτ c
1
1
≤ cδ γ 2+δ (1 + 22+δ/2 fδ (γ 2+δ ))
+p
δ/2
bnτ c
2πebnτ c
s
p
2 + 1/τ
1
2+δ
+ 2fδ (γ )
max
, ζn + 2d ζn .
π
n
+
(ii) Applying Lemma 1 of Michel and Pfanzagl [9] for
r = ζn ,
Nn
1 X
f=√
Xi ,
nτ i=1
r
g=
Nn
nτ
q
√
and using the fact that Nnτn − 1 > ζn implies Nnτn − 1 > ζn , hence
r
!
N
p
p
Nn
n
− 1 > ζn ≤ d ζn ,
P − 1 > ζn ≤ P nτ
nτ
we obtain from part (i)
!
Nn
X
1
sup P √
Xi ≤ t − Φ(t)
Nn i=1
t∈T
1
1
+p
δ/2
bnτ c
2πebnτ c
s
p
2 + 1/τ
1
2+δ
+ 2fδ (γ )
max
, ζn + (3d + 1) ζn .
π
n
≤ cδ γ 2+δ (1 + 22+δ/2 fδ (γ 2+δ ))
R EFERENCES
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12
H ENDRIK K LÄVER
AND
N ORBERT S CHMITZ
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J. Inequal. Pure and Appl. Math., 7(1) Art. 2, 2006
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