ScienceAsia 35 (2009): 203–210
R ESEARCH
ARTICLE
doi: 10.2306/scienceasia1513-1874.2009.35.203
On the normal approximation of the number of vertices
in a random graph
Kritsana Neammanee∗ , Angkana Suntadkarn
Department of Mathematics, Faculty of Science, Chulalongkorn University, Phyathai Road, Patumwan,
Bangkok 10330, Thailand
∗
Corresponding author, e-mail: [email protected]
Received 22 Sep 2008
Accepted 9 May 2009
ABSTRACT: In this paper, we use Stein-Chen’s method to give a uniform bound on the normal approximation of the
number of vertices of a fixed degree in a random graph. This work corrects our results published previously.
KEYWORDS: Stein-Chen method, uniform bound, vertex
INTRODUCTION AND MAIN RESULTS
A random graph is a collection of points or vertices
with lines or degrees connecting pairs of them at
random. The study of random graphs has a long
history. A systematic study of random graphs began
with the influential work of Erdös and Rényi 1–3 and
it has since developed into a significant area of study
in modern discrete mathematics. There are many
applications of random graphs (see Refs. 4–7).
Let G(n, p) be a random graph on n labelled
vertices {1, 2, . . . , n} with
the edges added randomly
such that each of the n2 possible edges exists with
probability p, 0 < p < 1. In this study, the assumption
has been made that the presence of an edge between
two vertices is independent of the others.
Let G = (V (G), E(G)) be a graph where V (G)
and E(G) are the sets of vertices and edges of G,
respectively. The degree of a vertex v in graph G,
denoted by deg(v), is the number of edges incident
to v, i.e., deg(v) = |{w ∈ V (G) | vw ∈ E(G)}|.
Any vertex of degree zero is called an isolated vertex.
Let Sn be the number of vertices of a fixed degree
d > 0 in G(n, p). Then Sn = Y1 + Y2 + · · · + Yn
where
where q = 1 − p and E[Sn ] = nan . Also 8 ,
2
n
n−1
2
σn := Var(Sn ) =
(d − (n − 1)p)2
n−1
d
(E[Sn ])2
.
× p2d−1 (1 − p)2(n−d)−3 + E[Sn ] −
n
Stein introduced a new powerful technique for
obtaining the rate of convergence to the standard
normal distribution 9 . His approach was subsequently
extended to cover the convergence to the Poisson
distribution 10 . Stein’s method was applied to random
graphs by Barbour 11 .
The Poisson convergence has since been widely
taken up (see, e.g., Refs. 12–15). Barbour et al 16
proved that the distribution of Sn converges to the
Poisson distribution with parameter λ = nan if either
np → 0 and d > 2, or np is bounded away from 0
1
and (np)− 2 |d − np| → ∞. Later, Suntadkarn and
Neammanee 17 gave the rates of convergence in case
of d > 1 and p = 1/nδ for some δ > 0. They showed
that for A ⊂ {1, 2, . . . , n}, |P (Sn ∈ A) − Poiλ (A)|,
where
X e−λ λk
Poiλ (A) =
,
k!
k∈A
−(δ−1)(d−1)
(
1, if vertex i has degree d in G(n, p),
Yi =
0, otherwise,
for i = 1, 2, . . . , n. Note that an := E[Yi ] is given by
an = P (Yi = 1) =
n − 1 d n−1−d
p q
d
is O(n
) if δ > 1 and is O(n−d(1−δ) ) for
0 < δ < 1.
For the normal approximation, Barbour et al 8
proved that the distribution function of
Wn :=
Sn − nan
σn
converges to the standard normal distribution function
Z z
2
1
Φ(z) = √
e−t /2 dt.
2π −∞
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204
ScienceAsia 35 (2009)
For E[Sn ] → ∞ they gave the rate of convergence expressed in terms of E[Sn ] with respect to the Wasserstein metric d1 which is defined by d1 (X, Y ) = sup{|E[h(X)] − E[h(Y )]| :
0
supx∈R |h(x)| + supx∈R |h (x)| 6 1 for all bounded
test functions h with bounded derivative} for any
random variables X and Y . They showed that
C(d)
d1 (Wn , N (0, 1)) 6 p
E[Sn ]
where N (0, 1) is the standard normal random variable
and C(d) is a positive constant depending on d. In this
article we consider the uniform distance
δn := sup |P (Wn 6 z) − Φ(z)|.
z∈R
1/2
Note that δn = O({d1 (Wn , N (0, 1))} ), in general 18 . In Barbour et al 8 the authors explicitly restricted their considerations to smooth test functions
but Raič 19 used the Lipschitz test function to modify
the proof of Barbour et al 8 for non-smooth test functions at the cost of a boundedness condition.
In this paper, we use Stein’s method and a Lipschitz test function to correct the work on normal
approximation of Neammanee and Suntadkarn 20 . We
give a uniform bound of the normal approximation of
the number of vertices of a fixed degree in a random
graph. The following theorem is our main result.
Theorem 1 For d > 0, 0 < β < 1, and a positive
integer r0 > β/(1−β) there exists a positive constant
C(d, r0 ) such that, for large n,
sup |P (Wn 6 z) − Φ(z)| 6
z∈R
+
C(d, r0 )(1 + (np)r0 +1 )
σnβ
√
C(d)(1 + (np)r0 +1 ) n
r (1−β)+1
σn0
AUXILIARY RESULTS
For each i ∈ {1, 2, . . . , n}, let
Xi =
and for any Λ ⊂ {1, 2, . . . , n} we define
(
1, if vertex i has degree d in G(n, p)−{Λ},
(Λ)
Yi =
0, otherwise,
where G(n, p) − {Λ} is the random graph obtained
from G(n, p) by removing the vertices in Λ.
For i, j = 1, 2, . . . , n, let
(
Yi /σn ,
i = j,
Zij =
(i)
(Yj − Yj )/σn , i 6= j,
Zi =
W (i) =
n
X
l=1
l6=i
(1)
i = j,
; i 6= j,
l6=i,j
n
X
1
(i,j)
(i,j)
(Y
− E[Yl
]) − E[Vij ] − E[Zi ]
Wij =
σn l
l=1
l6=i,j
= W (i) − Vij
(i)
(2)
({i})
(i,j)
({i,j})
where Yj := Yj
and Yl
:= Yl
. Note
(i)
that W is independent of Xi , and Wij is independent of the pair (Xi , Zij ) (see Ref. 8) and
.
Wn =
n
X
Xi .
i=1
C(d, r0 )(1 + (np)r0 +1 )
σnβ
.
In the case of np = O(1) we have σn2 = O(n).
Hence, for β > 12 and r0 > β/(1 − β), there exists a
constant C(d, r0 ) such that δn = C(d, r0 )/σnβ . Here,
and throughout the paper, C(c1 , c2 , . . . , ck ) stands for
an absolute constant depending on c1 , c2 , . . . , ck .
In the next section we prove auxiliary results and
in the final section we introduce the Stein’s method
for normal approximation which we use in the proof
of main result in the last section.
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1
(i)
(i)
(Y − E[Yl ]) − E[Zi ]
σn l
= Wn − Zi ,

0,


 n
n
o
X
1
(i)
(i)
(i,j)
Vij =
Y
+
(Y
−
Y
)
j
l
l


 σn
l=1
Furthermore, if σ = O(n) then
z∈R
Zil ,
l=1
n
X
2
sup |P (Wn 6 z)−Φ(z)| 6
Yi − E[Yi ]
σn
The following propositions improve the results
of Propositions 2.1–2.3 in Ref. 20 for the case of
arbitrary p.
Proposition 1 For large n, we have σn2 > 12 nan .
Proof : From Proposition 2.1 of Ref. 20 we know that
σn2 > nan (1 − an ). If we can show that an 6 21 , then
the proposition is proved. From the fact that 1 − p 6
e−p , we have
1
n − 1 d n−1−d
e(1+d)p (np)d
an =
p q
6
· np 6
d
d!
e
2
for large n.
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ScienceAsia 35 (2009)
Proposition 2 For i, j
∈
{1, 2, . . . , n} and
r1 , r2 , r3 ∈ N, there exists a positive constant
C(d, r2 , r3 ) such that for large n,
E[|Xir1 Zir2 Vijr3 |] 6
r2 +r2 −1
C(d, r2 , r3 )an (1 + (np)
σ r1 +r2 +r3
)
Since
(Yi − µ) 6 1 ,
|Xi | = σ n σn
. we get
Proof : We follow the proof of Proposition 2.2 of
Ref. 20 to show that for r, r1 , r2 , . . . , rm ∈ N and
for distinct j1 , j2 , . . . , jm which are not equal to i we
have
(i)
(i)
(i)
E[|Yj1 − Yj1 |r1 |Yj2 − Yj2 |r2 · · · |Yjm − Yjm |rm ]
pm−1 an h
np i
pm−1 an
6
d+
6 C(d)(1 + np)
(n − 1)
q
n
(3)
and
r #
" n
X
(i) E (Yj − Yj )
1
E[|Zir2 Vijr3 |]
σnr1
2
3
o r r+r
n
o r r+r
1 n
2
3
2
3
6 r1 E[|Zi |r2 +r3 ]
E[|Vij |r2 +r3 ]
σn
an
6 C(d, r2 , r3 )(1 + (np)r2 +r3 −1 ) r1 +r2 +r3
σn
E[|Xir1 Zir2 Vijr3 |] 6
= C(d)(1 + np)an 1 + np + · · · + (np)r−1
Proposition 3 For r1 ∈ N ∪ {0}, r2 , r3 ∈ N and
i, j ∈ {1, 2, . . . , n}, there exists a positive constant
C(d, r3 ) such that
2r−1
6 r
σn
an
.
nσnr1 +r2 +r3
Proof : From the proof of Proposition 2.3 of Ref. 20
we can see that
(4)
(i)
(i)
E[|(Yj )r1 (Yj − Yj )r2 |] 6 C(d)an p
Hence
1
E[|Zir |] = r E
σn
from (5) and (6).
r2 r3
E[|Xir1 Zij
Vij |] 6 C(d, r3 )(1+(np)r3 )
j=1
j6=i
6 C(d, r)(1 + (np)r−1 )an .
(7)
and
#
"
n
X
(i) r (Yj − Yj )) (Yi +
r #)
" n
X
(i)
E[|Yir |] + E (Yj − Yj )
j=1
j6=i
2r−1 r−1
E[Y
]
+
C(d,
r)(1
+
(np)
)a
i
n
σnr
an
= C(d, r)(1 + (np)r−1 ) r .
(5)
σn
6
Again for distinct l1 , l2 , . . . , lm which are not equal to
i, j, we have
#
" m
Y
pm−1 an
(i)
(i,j)
|Ylk − Ylk |rk 6 C(d)(1+np)
E n
k=1
and
#
" n
X
(i)
(i,j) r E (Yl −Yl
) 6 C(d, r)(1+(np)r−1 )an .
l=1
l6=i,j
(i)
(i)
E[|Vijr |] 6 C(d, r)(1 + (np)r−1 )
an
.
σnr
(i,j) r2
(i)
= 1) 6
(6)
(i)
(i,j) r3
) (Yl2 − Yl2
)
m
(i,j)
· · · (Ylm − Ylm )rm+1 |] 6 C(d)(1 + np)
(
From this and the fact that E[Yj ] = P (Yj
C(d)an (see Ref. 20) we have,
(i)
(i)
E[|(Yj − Yj )r1 (Yl1 − Yl1
j=1
j6=i
p an
.
n
Hence
r2 r3
E[|Xir1 Zij
Vij |]
"
1
(i)
6 r1 +r2 +r3 E (Yj − Yj )r2
σn
#
n
n
or3 X
(i)
(i)
(i,j)
Yj +
(Yl − Yl
) l=1
l6=i,j
(
6
C(d, r3 )
(i)
(i)
E[|(Yj − Yj )r2 (Yj )r3 |]
σnr1 +r2 +r3
#)
"
n
X
(i) r2
(i)
(i,j) r3 + E (Yj − Yj ) [
(Yl − Yl
)] l=1
l6=i,j
C(d, r3 ) an 6 r1 +r2 +r3 an p + (1 + (np)r3 )
n
σn
an
r3
6 C(d, r3 )(1 + (np) ) r1 +r2 +r3 .
nσn
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206
ScienceAsia 35 (2009)
where
STEIN’S METHOD FOR NORMAL
APPROXIMATION
Stein’s method 9 relies on the elementary differential
equation
0
f (w) − wf (w) = h(w) − N h
In order to use (8) to obtain the bound of normal
approximation, many authors 20–22 , choose the test
function h = Iz where
(
1, w 6 z,
Iz (w) =
0, w > z.
Hence (8) becomes
f 0 (w) − wf (w) = Iz (w) − Φ(z)
and the unique solution fz of (9) is
(√
2
2πew /2 Φ(w)[1 − Φ(z)],
fz (w) = √
2
2πew /2 Φ(z)[1 − Φ(w)],
(9)
w 6 z,
w > z.
(10)
Substituting any random variable W for w in (9), we
get
Z
1
2ε
Z
0
fz,ε
(w) =
z+ε
ft (w) dt
z−ε
and
(8)
where f : R → R is a continuous and piecewise
continuously differentiable function, h is a bounded
test function with bounded derivative, and
Z ∞
2
1
h(t)e−t /2 dt.
N h := √
2π −∞
1
fz,ε (w) =
2ε
z+ε
ft0 (w) dt.
z−ε
Note that 19
√
sup |fz,ε (x)| 6
x∈R
ε>0
2π
,
4
f (x) − f (y) z,ε
z,ε
sup 6 1.
x−y
x,y∈R
x6=y
ε>0
Raič 19 used a test function Iz,ε to give a bound in
the following theorem. However, his work cannot
be applied to a random graph since our Vij is not
bounded.
Theorem 2 (Raič 19 ) Let Wn be a decomposed random variable defined by
X
Wn =
Xi ,
i∈I
E[Xi ] = 0, i ∈ I, E[Wn2 ] = 1, Wn = W (i) + Zi ,
where W (i) is independent of Xi , and
X
Zi =
Zij ,
i ∈ I, Ki ⊂ I
j∈Ki
E[fz0 (W ) − W fz (W )] = P (W 6 z) − Φ(z).
Hence, to bound |P (W 6 z) − Φ(z)|, it suffices to
bound E[fz0 (W ) − W fz (W )]. But in our work, we
choose the Lipschitz test function


1,
w < z − ε,


w−z−ε
Iz,ε (w) = −
, z − ε 6 w < z + ε,

2ε

0,
w > z + ε,
where ε > 0 is fixed. Observe that
Z z+ε
1
Iz,ε (w) =
It (w) dt
2ε z−ε
N Iz,ε
Z
z+ε
z−ε
1
N It dt =
2ε
Z
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i ∈ I,
j ∈ Ki ,
where Wij is independent of the pair (Xi , Zij ). Suppose that |Xi | 6 Ai , |Zik | 6 Bik , |Vik | 6 Cik ,
0
|Zi + Vik | 6 Cik
for some constants Ai , Bik , Cik ,
0
and Cik . Then
sup |P (Wn 6 z) − Φ(z)|
z∈R
X
X X
0
6 13.7
Ai Bi2 +
Ai Bik (6.8Cik +9.3Cik
)
where Bi :=
i∈I k∈Ki
X
Bik .
k∈Ki
z+ε
Φ(t) dt.
To prove this theorem, Raič 19 showed that for all
ε > 0,
z−ε
|P (Wn 6 z) − Φ(z)|
From (8) we have
E[Iz,ε (W ) − N Iz,ε ] =
= Wij + Vij ,
i∈I
and
1
=
2ε
W
(i)
0
E[fz,ε
(W )
− W fz,ε (W )]
ε
6 A1 + A2 + A3 + B1 + B2 + B3 + √ , (11)
2π
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ScienceAsia 35 (2009)
where
A1 :=
A3 6 C
X
E[|gz,ε (W (i) + θ1 Zi ) − gz,ε (W (i) )|
n X
n X
i∈I
+ E[|Zi + Vij ||Wn |]E[|Xi Zij |] .
|Xi Zi |] ,
X X
E[|gz,ε (W (i) ) − gz,ε (Wik )|
A2 :=
From (1), (5), (7), and the facts that Wij is independent of (Xi , Zij ) and that σn2 E[Xi2 ] = E[(Yi −
an )2 ] 6 an , we have
i∈I k∈Ki
|Xi Zik |] ,
A3 :=
X X
E[|Wn Xi |Zi2 ] = E[|(W (i) + Zi )Xi |Zi2 ]
E[|gz,ε (Wn ) − gz,ε (Wik )|
i∈I k∈Ki
6 E[|W (i) Xi |Zi2 ] + E[|Xi Zi3 |]
o 12
n
o 21 n
6 E[(W (i) )2 Xi2 ]
E[Zi4 ]
E[|Xi Zik |]] ,
B1 :=
X
E[|Iz,ε (W (i) ) − Iz,ε (W (i) + θ1 Zi )|
C(d)(1 + (np)2 )an
σn4
n
o 12 n
o 12
= E[(W (i) )2 ]E[Xi2 ]
E[Zi4 ]
i∈I
+
|Xi Zi |] ,
X X
B2 :=
E[|Iz,ε (Wij ) − Iz,ε (Wik + Vik )|
i∈I k∈Ki
C(d)(1 + (np)2 )an
σn4
n
o 12 n
o 12
6 C E[Wn2 + Zi2 ]E[Xi2 ]
E[Zi4 ]
|Xi Zik |] ,
B3 :=
X X
+
E[|Iz,ε (Wn ) − Iz,ε (Wik )|
i∈I k∈Ki
E[|Xi Zik |]] ,
in which gz,ε (x) = fz,ε (x)x, and θ1 is a random
variable whose value is in [0, 1]. Then he used the
boundness of Xi , Zik , Vik and Zi + Vik to complete
his result. In our work, we also prove our result by
using (11).
C(d)(1 + (np)2 )an
σn4
C(d)(1 + (np)2 )an
6
.
σn3
+
Hence, by (5) and (7),
PROOF OF MAIN RESULT
A1 6
In this section, we give the proof of Theorem 1. Using
the argument of Raič 19 , we can show that (11) holds.
Step 1 We will show that
A1 + A2 + A3 6
n X
√
C(d)(1 + (np)r0 +1 ) n
r (1−β)+1
σn0
.
E[|Xi |Zi2 ] + E[|Wn Xi |Zi2 ] ,
From (1) and (5), we have E[|Wij |] 6 E[|Wn |] +
E[|Vij |] + E[|Zi |] 6 C(d). Therefore,
+ E[|Wij Xi Zij Vij I(|Vij | > 1/σnβ )|]
1
6 β E[|Wij |]E[|Xi Zij |] + σnr0 β E[|Wij Xi Zij Vijr0 +1 |]
σn
C(d)an
6
nσn2+β
n
o 12 n
o 12
2 2(r0 +1)
+ σnr0 β E[Wij2 ]E[Xi2 ]
E[|Zij
Vij
|]
i=1
6
A2 6 C
n X
n X
E[|Xi Zij Vij |]
i=1 j=1
j6=i
√
+
(12)
6E[|Wij Xi Zij Vij I(|Vij | 6 1/σnβ )|]
σnβ
Following the argument of equations (5.17) and (5.18)
of Raič 19 , we can show that
A1 6 C
C(d)(1 + (np)2 )
.
σn
E[|Wij Xi Zij Vij |]
C(d)(1 + (np)2 )
+
E[|Zi + Vij |]E[|Xi Zij |]
i=1 j=1
j6=i
2π
E[|Wij + Vij ||Xi Zij Vij |]
2
C(d)an
nσn2+β
+
C(d)(1 + (np)r0 +1 )an
.
√ r0 (1−β)+3
nσn
Hence,
A2 6
C(d)(1 + (np)2 )
σnβ
+
√
C(d)(1 + (np)r0 +1 ) n
r (1−β)+1
σn0
.
(13)
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ScienceAsia 35 (2009)
From (3), (5), (6), (7), and E[Wn2 ] = 1,
A3 6
C(d)(1 + (np)2 )
.
σn
(14)
From (12–14),
A1 + A2 + A3 6
side by B11 and B12 , respectively. By Proposition 2,
"
n
σnr0 β X
B11 6
E |Xi Zi2 Vijr0 |
2ε i=1
Z
#
1
× I[Ξi (θ),|Vij |> 1β ] dθ
0
σn
C(d)(1 + (np)2 )
+
σnβ
√
C(d)(1 + (np)r0 +1 ) n
r (1−β)+1
σn0
6
6
.
6
In the remainder of the proof we let ε > 0.
C(d, r0 )(1 + (np)r0 +1 )nan
r (1−β)+3
εσn0
C(d, r0 )(1 + (np)r0 +1 )
εσnβ+1
.
(16)
To bound B12 we use the concentration inequality
Step 2 We will show that
C(d, r0 )(1 + (np)r0 +1 )
B1 6
εσn
C(d)(1 + np)
+
σn
n
σnr0 β X
E[|Xi Zi2 Vijr0 |]
2ε i=1
δn +
1
σnβ
b−a
P (a 6 Wn 6 b) 6 2δn + √
2π
(17)
(see Raič 19 ). Note that from (1), (2), (17) and
Chebyshev’s inequality we have
P (a 6 Wij 6 b)
where δn = supz∈R |P (Wn 6 z) − Φ(z)|. From the
fact that 19 ,
Z
y−x 1
Iz,ε (x) − Iz,ε (y) =
I[z−ε6(1−θ)x+θy6z+ε] dθ
2ε 0
(15)
+ P (a 6 Wn − (Zi + Vij ) 6 b, |Zi + Vij | >
6 P (a −
1
σnβ
6 Wn 6 b +
+ P (|Zi + Vij | >
where
(
1, w ∈ A
IA (w) =
0, otherwise
we have
B1 =
= P (a 6 Wn − (Zi + Vij ) 6 b, |Zi + Vij | 6
n
X
E[|Iz,ε (W (i) ) − Iz,ε (W (i) + θ1 Zi )||Xi Zi |]
i=1
" Z
#
n
1
1 X
2
6
E I[Ξi (θ)] dθ|Xi Zi |
0
2ε i=1
Z
#
"
n
1
1 X
6
E |Xi Zi2 | I[Ξi (θ),|Vij |> 1β ] dθ
0
2ε i=1
σn
Z
#
"
n
1
1 X
2 +
E |Xi Zi | I[Ξi (θ),|Vij |6 1β ] dθ
0
2ε
σn
i=1
where Ξi (θ) denotes z − ε 6 W (i) + θθ1 Zi 6 z + ε,
and we denote the two terms on the last right-hand
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1
σnβ
2
1
σnβ
1
σnβ
1
σnβ
)
)
)
)
b−a
6 2δn + √
+√
+ E|Zi + Vij |r0 σnr0 β
2π
2πσnβ
b−a
2
6 2δn + √
+√
2π
2πσnβ
an
+ C(d, r0 )(1 + (np)r0 −1 ) β
σn
b − a C(d, r0 )(1 + (np)r0 −1 )
6 2δn + √
+
.
(18)
2π
σnβ
From (5), (7), (18), and the fact that Wij is independent of (Xi , Zij ), we have
"
n
1 X
B12 =
E |Xi Zi2 |
2ε i=1
Z
1
× I[z−ε6Wij +Vij +θθ1 Zi 6z+ε,|Zi |6 1β ,|Vij |6 1β ] dθ
0
σn
σn
#
Z 1
+ I[z−ε6Wij +Vij +θθ1 Zi 6z+ε,|Zi |> 1β ,|Vij |6 1β ] dθ
σ
σ
n
n
0
209
ScienceAsia 35 (2009)
"
#
Z 1
n
1 X
2
6
E |Xi Zi | I[z−ε− 2β 6Wij 6z+ε+ 2β ] dθ
2ε i=1
σn
σn
0
"
n
n
1 XX
E Xi Zij Vij
+
2ε i=1 j=1
j6=i
n
σnr0 β X
E[|Xi Zir0 +2 |]
2ε i=1
"
n
1 X
6
E |Xi Zi2 |
2ε i=1
+
×P z−ε−
2
×
0
6
6 Wij 6 z + ε +
σnβ
2 #
I[z−ε6Wij +θVij 6z+ε,|Vij |6
+
B1 6
C(d, r0 )(1 + (np)
εσn
C(d)(1 + np)
.
+
σn
δn +
1 σnβ
Step 3 We will show that
B2 6
C(d, r0 )(1 + (np)r0 +1 ) +
εσnβ
C(d)(1 + np)
σnβ
δn +
1 σnβ
.
B2 =
2εσnβ
n
E[|Iz,ε (Wij ) − Iz,ε (Wij + Vij )||Xi Zij |]
n
×
I[z−ε−
0
0
#
1
I[z−ε6Wij +θVij 6z+ε] dθ
"
n
n
1 XX
6
E |Xi Zij Vij |
2ε i=1 j=1
j6=i
Z
×
0
1
β
σn
6Wij 6z+ε+
1
β
σn
]
dθ
C(d, r0 )(1 + (np)r0 +1 )
+
1
2εσnβ
εσnβ+1
"
n X
n
X
E |Xi Zij |
i=1 j=1
j6=i
×P z−ε−
1
1 #
6 Wij 6 z + ε + β
σnβ
σn
h
C(d)(1 + np)
C(d, r0 )(1 + (np)r0 ) i
6
δn + ε +
β
εσn
σnβ
C(d, r0 )(1 + (np)r0 +1 ) 1 δn + β
6
β
εσn
σn
C(d)(1 + np)
+
.
σnβ
B3 6
C(d)(1 + np) εσnβ
δn + ε +
C(d, r0 ) σnβ
.
n X
n
X
E[|Xi Zij |] 6 C(d)(1 + np).
(20)
By (5), (6), (15) and (17) we have
j6=i
×
#
i=1 j=1
j6=i
"
1 XX
E |Xi Zij Vij |
2ε i=1 j=1
Z
E Xi Zij
i=1 j=1
j6=i
1
Z
6
"
From Proposition 1 we have
i=1 j=1
j6=i
6
dθ
Step 4 We will show that
By using (15), Proposition 1, Proposition 3 and
(18) we have
n X
n
X
n X
n
X
1
From (16) and (19) we have
)
]
n
n
σnr0 β X X
E[|Xi Zij Vijr0 +1 |]
2ε i=1 j=1
+
r0 +1
1
β
σn
j6=i
σnβ
C(d, r0 )(1 + (np)r0 +1 )σnr0 β nan
εσnr0 +3
h
C(d)(1 + np)
C(d, r0 )(1 + (np)r0 ) i
6
δn + ε +
εσn
σnβ
C(d, r0 )(1 + (np)r0 +1 ) 1 6
δn + β
εσn
σn
C(d)(1 + np)
.
(19)
+
σn
#
1
Z
#
1
I[z−ε6Wij +θVij 6z+ε,|Vij |>
1
β
σn
]
dθ
E[|Iz,ε (Wn ) − Iz,ε (Wij )|]
"
#
Z 1
1
6 E |Zi + Vij | I[z−ε6Wn −θ(Zi +Vij )6z+ε] dθ
2ε
0
"
1
6 E |Zi + Vij |
2ε
#
Z 1
× I[z−ε6Wn −θ(Zi +Vij )6z+ε,|Zi +Vij |6 1β ] dθ
0
σn
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210
ScienceAsia 35 (2009)
"
1
+ E |Zi + Vij |
2ε
Z 1
× I[z−ε6Wn −θ(Zi +Vij )6z+ε,|Zi +Vij |>
0
6
"Z
1
2εσnβ
1
β
σn
dθ
]
#
1
E
#
I[z−ε−
0
1
β
σn
6Wn 6z+ε+
1
β
σn
]
dθ
σnr0 β
E[|Zi + Vij |r0 +1 ]
2ε
1
1
1 =
P
z
−
ε
−
6
W
6
z
+
ε
+
n
2εσnβ
σnβ
σnβ
C(d, r0 )
+
εσnβ+1
C 1 C(d, r0 )
6 β δn + ε + β +
εσn
σn
εσnβ+1
C
C(d, r0 )
6 β δn + ε +
.
εσn
σnβ
+
From this fact, and (20) we have
B3 6
C(d)(1 + np) εσnβ
δn + ε +
C(d, r0 ) σnβ
.
We now deduce our main result. From (11) and
Steps 1–4 we have
δn = sup |P (Wn 6 z) − Φ(z)|
z∈R
6
C(d, r0 )(1 + (np)r0 +1 ) +
εσnβ
C(d)(1 + (np)2 )
σnβ
δn +
1 σnβ
C
ε
+ β +√ .
2π
σn
for all ε > 0. Hence we can choose ε = C(d, r0 )(1 +
(np)r0 +1 )/2σnβ such that for large n,
B1 + B2 + B3 6
C(d, r0 )(1 + (np)r0 +1 )
σnβ
.
This fact and Step 1 complete the proof.
Acknowledgements: The authors would like to thank
the referee for many helpful remarks which have led us to
improve the presentation and the Thailand Research Fund
for financial support (RGJ2550).
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