ARTICLE IN PRESS
Statistics & Probability Letters 76 (2006) 1812–1820
www.elsevier.com/locate/stapro
Finite time ruin probability with heavy-tailed
insurance and financial risks
Yu Chen, Chun Su
Department of Statistics and Finance, University of Science and Technology of China, Hefei 230026, PR China
Received 29 April 2005; received in revised form 30 March 2006; accepted 24 April 2006
Available online 2 June 2006
Abstract
This paper studies the probability of ruin within a finite time for a discrete-time model, in which the insurance risk is
assumed to be heavy tailed. A precise asymptotic estimate for the finite-time ruin probability is established as the initial
capital increases, extending the corresponding result of Tang and Tsitsashvili [2003. Precise estimates for the ruin
probability in finite horizon in a discrete-time model with heavy-tailed insurance and financial risks. Stochastic Process.
Appl. 108, 299–325] to the subexponential case.
r 2006 Elsevier B.V. All rights reserved.
Keywords: Subexponentiality; Independent product; Ruin probability; Financial risk; Insurance risk
1. Introduction
Properties of ruin probability on a stochastic economic environment have been investigated by many
researchers, such as Norberg (1999), Nyrhinen (1999, 2001), Kalashnikov and Norberg (2002), Tang and
Tsitsashvili (2003, 2004) and so on.
Following the work of Nyrhinen (1999, 2001) and Tang and Tsitsashvili (2003, 2004), we consider the ruin
probability on a stochastic environment. Let a random variable (r.v.) X n be the net loss—the total claim
amount premium minus the total incoming—of the insurer at time n, and a positive r.v. Y n be the discount
factor from time n to time n 1, n ¼ 1; 2; . . . : Hence the discounted surplus of the insurance company
accumulated till the end of time n can be modelled by a discrete time stochastic process
S0 ¼ x;
Sn ¼ x n
X
i¼1
Xi
i
Y
Y j;
n ¼ 1; 2; . . . ,
j¼1
where x is the initial capital of the insurance company. The probability of ruin within finite time is defined by
cðx; nÞ ¼ P min Sk o0 ¼ PðU n 4xÞ; n ¼ 1; 2; . . . ,
(1.1)
0pkpn
Corresponding author. Tel.: +86 551 3600565; fax: +86 551 3600025.
E-mail address: [email protected] (Y. Chen).
0167-7152/$ - see front matter r 2006 Elsevier B.V. All rights reserved.
doi:10.1016/j.spl.2006.04.029
ARTICLE IN PRESS
Y. Chen, C. Su / Statistics & Probability Letters 76 (2006) 1812–1820
1813
P
Q
where U n ¼ maxf0; max1pkpn ki¼1 X i ij¼1 Y j g with U 0 ¼ 0. Recently, the ruin probabilities in (1.1) have been
investigated by Nyrhinen (1990, 2002) and Tang and Tsitsashvili (2003, 2004), among others.
The standing assumption of the paper is that the distribution of the net loss X n is heavy-tailed. The most
important class of heavy-tailed distribution is the subexponential class. By definition, a distribution function
(d.f.) F supported on ½0; 1Þ is said to be subexponential, denoted by F 2 S, if
F 2 ðxÞ
¼ 2,
x!1 F ðxÞ
lim
where F 2 denotes the 2-fold convolution of F. More generally, a d.f. F supported on ð1; 1Þ is still said to
be subexponential if F ðxÞ1ð0pxo1Þ is subexponential, where 1A denotes the indicator function of A. We also
introduce some closely related heavy-tailed classes, a d.f. F supported on ð1; 1Þ is said to be long-tailed,
denoted by F 2 L, if
F ðx þ 1Þ
¼ 1;
x!1 F ðxÞ
lim
a d.f. F supported on ð1; 1Þ is said to be dominatedly varying tailed, denoted by F 2 D, if
lim
x!1
F ðyxÞ
o1
F ðxÞ
for some (or, equivalently for all) y 2 ð0; 1Þ: For more details of heavy-tailed distributions, see Embrechts et al.
(1997).
Following Nyrhinen (1999, 2001), Tang and Tsitsashvili (2003), etc., we also need some general
assumptions:
P1 . The successive net losses X n , n ¼ 1; 2; . . . ; constitute a sequence of independent and identically distributed
(i.i.d.) r.v.’s with generic r.v. X and common d.f. F concentrated on ð1; 1Þ;
P2 . The reserve is currently invested into a risky asset which may earn negative interest rn at year n, and the
discount factor Y n ¼ ð1 þ rn Þ1 , n ¼ 1; 2; . . . ; also constitute a sequence of i.i.d. r.v.’s, with generic r.v. Y
and common d.f. G concentrated on ð0; 1Þ;
P3 . The two sequences fX n : n ¼ 1; 2; . . .g and fY n : n ¼ 1; 2; . . .g are mutually independent.
In the terminology of Norberg (1999), we call the generic r.v.’s X and Y as the insurance risk and financial
risk, respectively.
Here and throughout, all limiting relationships are for x ! 1 unless otherwise stated. For two positive
functions aðÞ and bðÞ, we symbolize their relation by aðxÞbðxÞ if they satisfy limx!1 aðxÞ=bðxÞ ¼ 1; and
aðxÞ ¼ OðbðxÞÞ if lim supx!1 aðxÞ=bðxÞo1:
One of most important results of Tang and Tsitsashvili (2003) is as follows:
Theorem A (Tang and Tsitsashvili (2003)). Suppose that the assumptions P1 , P2 and P3 hold simultaneously. If
(1) F 2 L \ D,
(2) EY p o1 for some large p40,
then it holds for each n ¼ 1; 2; . . . that
!
n
k
X
Y
cðx; nÞ
P X
Y i 4x .
k¼1
(1.2)
i¼1
It is well-known that
L\DS
(see Embrechts et al., 1997) is a proper inclusion, many important heavy-tailed distributions do not belong to
L \ D, such as Lognorm distribution and Weibull distribution and so on. It is natural to investigate whether
ARTICLE IN PRESS
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1814
(1.2) still holds if the condition F 2 L \ D in Theorem A is extended to F 2 S. And Theorem A only consider
the case that the tail of the insurance risk X is heavier than that of the financial risk Y, then it is of interest to
investigate whether (1.2) still holds if Y is heavier then X. Unfortunately, Theorem A does not answer this
question.
In this paper, we restrict ourselves to the case that the insurance risk X has an absolutely continuous
subexponential distribution. We derive the precise asymptotic estimates for the finite ruin probability cðx; nÞ
under a very relax condition on G. In particular, the result holds for the case even if the tail of the financial risk
Y is heavier than that of the insurance risk X. Our main results extend Theorem A to the whole subexponential
class.
We will give our main results in Section 2. We noted that subexponentiality of independent product tails is
the key to the proof of Theorem 2.1, thus we will first cite our previous result of subexponentiality of
independent product (see Su and Chen, 2006) and also give some lemmas to be used in the proof of Theorem
2.1 in Section 3. Finally we give the proof of Theorem 2.1.
2. Main results
Throughout the paper we restrict ourselves to absolutely continuous F 2 S. For convenience, we write
X ðF Þ 2 C, if the distribution F of r.v. X belongs to the class C. First we give our main results Theorem 2.1 as
follows, and will give the proof in Section 4.
Theorem 2.1. Suppose that the assumptions P1 , P2 and P3 hold simultaneously. If
(1) F is absolutely continuous with a density function f ðÞ such that xf ðxÞ is eventually non-increasing, F 2 S;
(2) for each fixed a40
lim
x!1
PðY 4x=aÞ
¼ 0,
PðXY 4xÞ
then it holds for each n ¼ 1; 2; . . . that
!
n
k
X
Y
cðx; nÞ
P X
Y j 4x .
k¼1
(2.1)
(2.2)
j¼1
Remark 2.1. It is pointed that the condition that F is absolutely continuous with a density function f ðÞ such
that xf ðxÞ is eventually non-increasing is not too restrictive. If F is absolutely continuous with a finite
expectation m, then
Z 1
m¼
xf ðxÞ dxo1,
0
thus we usually have
lim xf ðxÞ ¼ 0.
x!1
In fact, almost all useful subexponential distributions including the distributions with regularly varying tails,
Lognormal, Pareto, Weibull, Loggamma, etc., satisfying the conditions above.
We will give some sufficient conditions for (2.1) below. We will see that the condition (2.1) is very rather
relaxed and that essentially Theorem 2.1 generalize the results of Theorem A.
Proposition 2.1. Following the same notations of Theorem 2.1, if
lim
x!1
GðcxÞ
¼0
F ðxÞ
for some 0oco1, then condition (2.1) holds for any a40.
(2.3)
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Proof. Only consider the case that G is unbounded, i.e., GðxÞ40 for any x40. It is easy to see that for any
a40,
lim sup
x!1
Gðx=aÞ
Gðx=aÞ
1
Gðx=aÞ
p lim sup R 1
lim sup
p
¼ 0,
PðXY 4xÞ
GðacÞ x!1 F ðx=acÞ
x!1
ac F ðx=yÞ dGðyÞ
which implies that limx!1 Gðx=aÞ=HðxÞ ¼ 0 for any a40:
&
By Proposition 2.1, we know the precise estimate of ruin probability (2.2) holds in the case that the tail of
the insurance risk X is heavier than that of the financial risk Y, particularly, for the case that Y has a light tail.
In order to demonstrate that many popular distributions satisfying the condition (2.1), we introduce the
definition of the class R1 , which was first applied to the ruin theory by Tang and Tsitsashvili (2004). We say
a distribution F is said to have a rapid variation, denoted by F 2 R1 ; if
F ðtxÞ
¼ 0 for any t41.
x!1 F ðxÞ
lim
For the detailed properties of the class R1 , we refer the reader to Bingham et al. (1987, Chapter 2.4),
Embrechts et al. (1997, Chapter 3.3) or Tang and Tsitsashvili (2004).
By the definition of the class R1 , it follows immediately:
Proposition 2.2. Suppose G 2 R1 , F is unbounded, then the condition (2.1) holds for any a40.
Proof. For any fixed a40, by the definition of R1 we have
lim sup
x!1
Gðx=aÞ
Gðx=aÞ
1
Gðx=aÞ
p lim sup R 1
p
¼ 0,
lim sup
PðXY 4xÞ
F
ð2aÞ
Gðx=yÞ
dF
ðyÞ
Gðx=2aÞ
x!1
x!1
2a
we know the condition (2.1) holds for any a40.
&
It is well-known that all power moments of G are finite if G 2 R1 . And class R1 includes not only some
heavy-tailed distributions such as the Lognormal distribution, Weibull distribution, etc. but also almost all
light-tailed distributions. From Proposition 2.2, result (2.2) holds if only the financial risk Y has a rapid
varying tail, whether the tail of the insurance risk X is heavier than that of the financial risk Y or not.
3. Some lemmas
Subexponentiality of independent product plays an important role in time series, risk theory and stochastic
process, which was pointed by Cline and Samorodnitsky (1994). The results about independent product are
fewer, see Embrechts and Goldie (1980, 1982), Cline and Samorodnitsky (1994), Tang and Tsitsashvili (2003,
2004), Su and Chen (2006), etc. Su and Chen (2006) have proved the following proposition, which is crucial for
proving Theorem 2.1.
Proposition 3.3. Let X and Y be independent non-negative r.v.’s with distribution F and G, respectively. Suppose
that F is absolutely continuous with a density function f ðÞ and xf ðxÞ is eventually non-increasing. If F 2 S, and
for each fixed a40,
lim
Gðx=aÞ
x!1 PðXY 4xÞ
¼ 0,
(3.1)
then XY 2 S.
In what follows we give the equivalent condition of the condition (3.1), which will be used many times
throughout this paper.
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Proposition 3.4. The relation (3.1) holds for each a40 if and only if the relation
lim
Gðx=aðxÞÞ
x!1 PðXY 4xÞ
¼0
(3.2)
holds for some positive function aðÞ satisfying aðxÞ " 1 and x=aðxÞ " 1.
Proof. The sufficiency holds trivially, so we only need to prove the necessity. By (3.1), for any nX1, choosing
an satisfying 2n pan o2nþ1 , then there exists xn such that when x4xn ,
Gðx=an Þ
p2n .
PðXY 4xÞ
en ¼ maxf2nþ1 x
en1 ; xn g; nX2, define
Choosing x~ 1 ¼ maxf1; x1 g; x
(
1;
0oxoe
x1 ;
aðxÞ ¼
en pxoe
xnþ1 ; nX1:
an ; x
By the above definition we know aðxÞ ! 1. If 0oxoe
x1 , x=aðxÞ ¼ x.
en pxoe
If x
xnþ1 ,
en Xe
x=aðxÞ ¼ x=an X2ðnþ1Þ x
xn1 ! 1
as n ! 1
and
Gðx=aðxÞÞ
Gðx=an Þ
¼
p2n ;
PðXY 4xÞ PðXY 4xÞ
nX1.
Gðx=aðxÞÞ
!0
PðXY 4xÞ
&
So
as x ! 1:
Tang and Tsitsashvili (2003, Lemma 3.2) proved the following result:
Lemma 3.1. Let F 1 and F 2 be two d.f.’s concentrated on ð1; 1Þ. If F 1 2 S, F 2 2 S, and F 2 ðxÞ ¼ OðF 1 ðxÞÞ,
then F 1 F 2 2 S and
F 1 F 2 ðxÞF 1 ðxÞ þ F 2 ðxÞ.
(3.3)
Lemma 3.2. Let F 1 be a d.f. supported on ð1; 1Þ and F 2 be a finite non-negative measure. If F 1 2 L and
F 1 F 2 , then F 1 F 1 F 2 F 2 .
Proof. See Corollary 2.5 in Cline (1987).
&
Lemma 3.3. Let F 1 and F 2 be two d.f.’s concentrated on ð1; 1Þ. Suppose F 1 2 S; F 2 2 S, and
F 1 F 2 F 1 þ F 2 , then F 1 F 2 2 S.
Proof. First it is easy to see F 1 F 2 2 L; then by Lemma 3.2, we have
ðF 1 F 2 Þ2 F 1 F 1 þ 2F 1 F 2 þ F 2 F 2
2F 1 þ 2F 1 F 2 þ 2F 2
2F 1 F 2 .
Then we have F 1 F 2 2 S by the definition of the class S.
&
Lemma 3.4. Suppose that F is absolutely continuous with a density function f ðÞ such that xf ðxÞ is eventually nonincreasing, fY ; Y n ; nX1g is i.i.d. r.v.’s series with the generic distribution
G and independent of X. If F 2 S, and
Q
(3.1) holds for each fixed a40, then for any nX1 we have X nj¼1 Y j 2 S.
ARTICLE IN PRESS
Y. Chen, C. Su / Statistics & Probability Letters 76 (2006) 1812–1820
Proof. It suffices by Proposition 3.3 to show
Q
Pð nj¼1 Y j 4x=cÞ
Qn
¼0
lim
x!1 PðX
j¼1 Y j 4xÞ
1817
(3.4)
holds for any nX1 and each c40.
By the given conditions we know the above equation holds for n ¼ 1, and it follows immediately that
XY 1 2 S by Proposition 3.3. Now assuming (3.4) holds for n ¼ k; kX1 by induction, we turn to prove (3.4)
also holds for n ¼ k þ 1. By Proposition 3.4, there exists some aðxÞ satisfying aðxÞ " 1 and x=aðxÞ " 1 such
that (3.2) holds. Thus for any x40,
Pð
Qkþ1
j¼1 Y j 4x=cÞ
Qkþ1
PðX j¼1 Y j 4xÞ
R1
Qk
0 Pð j¼1 Y j 4x=ctÞGðdtÞ
¼ R1
Qk
j¼1 Y j 4x=tÞGðdtÞ
0 PðX
R aðxÞ Qk
Pð j¼1 Y j 4x=ctÞGðdtÞ
0
p R aðxÞ
Q
PðX kj¼1 Y j 4x=tÞGðdtÞ
0
R1
þ
Qk
aðxÞ Pð j¼1 Y j 4x=ctÞGðdtÞ
R1
Qk
j¼1 Y j 4x=tÞGðdtÞ
0 PðX
:¼I 1 þ I 2 .
For I 1 , by induction it holds that
Pð
Qk
j¼1 Y j 4x=ctÞ
I 1 p sup
Qk
0otpaðxÞ PðX
j¼1 Y j 4x=tÞ
Q
Pð kj¼1 Y j 4y=cÞ
¼ sup
! 0.
Qk
yXx=aðxÞ PðX
j¼1 Y j 4yÞ
For I 2 , from (3.2) it holds that
R1
I2 ¼ R 1
0
¼
p
Pð
Qk
j¼1 Y j 4x=ctÞGðdtÞ
pR1
Q
PðXY 1 4x=tÞ dPð kj¼1 Y j ptÞ
1
aðxÞ
GðaðxÞÞ
PðXY 4x=tÞ dPð
Qk
j¼1 Y j ptÞ
GðaðxÞÞ
Rx Q
PðXY 4xÞPð j¼1 Y j 41Þ þ 0 Pð kj¼1 Y j 4x=tÞ dPðXY ptÞ
Qk
GðaðxÞÞ
Qk
PðXY 4xÞPð
This completes the proof.
j¼1 Y j 41Þ
! 0.
&
Under the standard assumption above, Tang and Tsitsashvili (2003) showed that
d
U n ¼ V n;
n ¼ 1; 2; . . . ,
d
where ¼ denotes equality in distribution and V n ’s are determined by a Markov chain
V 0 ¼ 0;
V n ¼ Y n maxf0; X n þ V n1 g;
n ¼ 1; 2; . . . .
(3.5)
Under the model (1.1), we have for n ¼ 1; 2; . . . that
cðx; nÞ ¼ PðV n 4xÞ ¼ PððX n þ V n1 ÞY n 4xÞ;
x40.
(3.6)
Lemma 3.5. Under the same conditions of Theorem 2.1, then we have that V n Y 2 S for all nX1, where V n ; nX1
is defined by (3.5).
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Proof. By Lemma 3.4, we know XY 1 Y 2 2 S, that is V 1 Y 2 2 S. Now we assume by induction that
V k1 Y k 2 S, kX2, then we will prove V k Y kþ1 2 S: By (3.5) we know for any x40,
!
k
k
þ1
Y
X
PðV k Y kþ1 4xÞ ¼ PððX k þ V k1 ÞY k Y kþ1 4xÞ ¼ P
Xj
Y i 4x
k
X
¼P
Xj
kþ1
Y
Yi þ X
i¼j
j¼2
!
kþ1
Y
Y i 4x
j¼1
i¼j
¼ P V k1 Y k þ X
i¼1
kþ1
Y
!
Y i 4x ,
i¼1
P
Q
d
the last equation follows by kj¼2 X j kþ1
i¼j Y i ¼ V k1 Y k .
1 . Pð0oY p1Þ ¼ 1, for any lpk; kX1;
!
!!
k
þ1
l
Y
Y
P X
Y i 4x ¼ O P X
Y i 4x
.
i¼1
i¼1
Qkþ1
It follows that PðX i¼1 Y i 4xÞ ¼ OðPðV k1 Y k 4xÞÞ; kX2.
2 . PðY 41Þ40, thus for any lpk; kX1;
!
!
!
Z 1
k
þ1
l
k
Y
Y
Y
P X
Y i 4x ¼
P X
Y i 4x=t dP
Y i pt
0
i¼1
Z
i¼1
1
P X
X
1
l
Y
Y i 4x=t dP
i¼1
XP X
l
Y
i¼lþ1
!
k
Y
!
Y i pt
i¼lþ1
!
Y i 4x P
i¼1
k
Y
!
Y i 41 .
i¼lþ1
So it follows that
P X
l
Y
!
Y i 4x
¼O P X
i¼1
k
þ1
Y
!!
Y i 4x
,
i¼1
Q
Y i 4xÞÞ; kX2.
that is PðV k1 Y k 4xÞ ¼ OðPðX kþ1
Qkþ1i¼1
By Lemma 3.4 we know X i¼1 Y i 2 S, thus applying Lemma 3.1 we know V k Y kþ1 2 S. This completes
the proof. &
4. The proof of Theorem 2.1
The proof of Theorem 2.1 can be formulated into two parts according to PðY 41Þ ¼ 0 and PðY 41Þ40,
respectively.
1 Pð0oY p1Þ ¼ 1. By (3.6), it is clear that
cðx; 1Þ ¼ PðV 1 4xÞ ¼ PðX 1 Y 1 4xÞ;
x40.
(4.1)
From Proposition 3.3 we know V 1 2 S and (2.2) holds for n ¼ 1. It is trivial that PðV 1 4xÞ ¼ OðF ðxÞÞ for
x40.
Now we assume by induction that (2.2) holds for n ¼ m, V m 2 S; mX1 and
PðV m 4xÞ ¼ OðF ðxÞÞ;
x40.
Then we need to show the above three facts hold with n ¼ m þ 1. By the assumption V m 2 S and F 2 S,
applying Lemma 3.1 we know for x40,
PðV mþ1 4xÞ ¼ PððX mþ1 þ V m ÞY mþ1 4xÞpPðX mþ1 þ V m 4xÞ
PðX mþ1 4xÞ þ PðV m 4xÞ ¼ OðF ðxÞÞ.
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Now we prove (2.2) holds for n ¼ m þ 1, for x40,
Z 1
cðx; m þ 1Þ ¼
PðX mþ1 þ V m 4x=tÞGðdtÞ
Z
0
1
ðPðX mþ1 4x=tÞ þ PðV m 4x=tÞÞGðdtÞ
0
Z
m
1X
PðX mþ1 Y 4tÞ þ
0
¼
m
þ1
X
P X
k¼1
k
Y
P X
k¼1
!
k
Y
ð4:2Þ
!
Y j 4x=t GðdtÞ
j¼1
Y j 4x .
j¼1
From (4.2), we have for x40 that
PðV mþ1 4xÞ ¼ PððX mþ1 þ V m ÞY mþ1 4xÞPðX mþ1 Y mþ1 4xÞ þ PðV m Y mþ1 4xÞ.
This with Lemmas 3.4 and 3.5 implies X mþ1 Y mþ1 2 S and V m Y mþ1 2 S, thus applying Lemma 3.3 we obtain
V mþ1 2 S.
2 PðY 41Þ40. In this case it is obvious that F ðxÞ ¼ OðPðV 1 4xÞÞ for x40. Similar to 1 , we have V 1 2 S
and (2.2) holds for n ¼ 1.
Now assume by induction that (2.2) holds for n ¼ m, V m 2 S; mX2 and F ðxÞ ¼ OðPðV m 4xÞÞ for x40. By
the assumption of the induction, we have for x40,
PðX mþ1 þ V m 4xÞPðX mþ1 4xÞ þ PðV m 4xÞ.
(4.3)
By (4.3), for x40 we have
Z 1 x
P X mþ1 þ V m 4 GðdyÞ
PðV mþ1 4xÞ ¼ PððX mþ1 þ V m ÞY mþ1 4xÞ ¼
y
0
Z 1 x
P X mþ1 þ V m 4 GðdyÞ
X
y
1
XPðX mþ1 þ V m 4xÞGð1Þ
ðPðX mþ1 4xÞ þ PðV m 4xÞÞGð1Þ
XPðV m 4xÞGð1Þ
X XPðV 1 4xÞGð1Þm XF ðxÞGð1Þm ,
that is F ðxÞ ¼ OðPðV mþ1 4xÞÞ.
Finally, we will prove (2.2) holds for n ¼ m þ 1 and V mþ1 2 S. By Lemma 3.1,
cðx; m þ 1Þ ¼ PðV mþ1 4xÞ ¼ PððX mþ1 þ V m ÞY mþ1 4xÞ
Z x=aðxÞ Z 1 ! x
þ
¼
P X mþ1 þ V m 4 GðdyÞ
y
0
x=aðxÞ
:¼J 1 þ J 2 .
From (4.3) and (3.2),
Z
x=aðxÞ
J1 ¼
PðX mþ1 þ V m 4x=yÞGðdyÞ
0
Z
x=aðxÞ
ðPðX mþ1 4x=yÞ þ PðV m 4x=yÞÞGðdyÞ
0
ð4:4Þ
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1820
Z
Z
1
!
1
PðX mþ1 4x=yÞ þ
x=aðxÞ
0
m
X
k¼1
k¼1
P X
k
Y
P X
k¼1
¼ PðX mþ1 Y mþ1 4xÞ þ
m
þ1
X
m
X
!
P X
kþ1
Y
Y i 4x
i¼1
!
k
Y
!!
Y i 4x=y
i¼1
x
þ Oð1ÞG
aðxÞ
GðdyÞ
Y i 4x .
i¼1
On the other hand,
x
J 2 ¼ Oð1ÞG
¼ oð1ÞPðXY 1 4xÞ.
aðxÞ
Following (4.4), for x40,
PðV mþ1 4xÞ ¼ PððX mþ1 þ V m ÞY mþ1 4xÞ
¼ PðX mþ1 Y mþ1 4xÞ þ PðV m Y mþ1 4xÞ þ Oð1ÞG
x
aðxÞ
PðX mþ1 Y mþ1 4xÞ þ PðV m Y mþ1 4xÞ.
By Lemmas 3.4 and 3.5 we know X mþ1 Y mþ1 2 S and V m Y mþ1 2 S, thus by applying Lemma 3.3 it follows
that V mþ1 2 S.
Combining 1 and 2 , the proof is completed. &
Acknowledgements
We are very grateful to Dr. Tang Qihe for his comments on the present paper. Research was supported by
National Science Foundation of China (no. 10371117); research was also supported by Special Foundation of
USTC.
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