The Expected Discounted Tax Payments on Dual Risk
Model under a Dividend Threshold
*
Zhang Liu1, Aili Zhang2#, Canhua Li1
1College of Sciences, Jiangxi Agricultural University, Nanchang, China 2School of Mathematics and Statistics, Nanjing Audit University, Nanjing, China
Email: liuzhang1006@163.com, #zhangailiwh@126.com
Received November 19, 2012; revised December 20, 2012; accepted January 3, 2013
Copyright © 2013 Zhang Liu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
ABSTRACT
In this paper, we consider the dual risk model in which periodic taxation are paid according to a loss-carry-forward sys- tem and dividends are paid under a threshold strategy. We give an analytical approach to derive the expression of gδ(u) (i.e. the Laplace transform of the first upper exit time). We discuss the expected discounted tax payments for this model and obtain its corresponding integro-differential equations. Finally, for Erlang (2) inter-innovation distribution, closed- form expressions for the expected discounted tax payments are given.
Keywords: Dual Risk Model; Expected Discounted Tax Payments; Dividend; Threshold Strategy
1. Introduction
Consider the surplus process of an insurance portfolio
R t u ct S t
t
0
u c0
1: i
i S t
(1.1)
which is dual to the classical Cramér-Lundberg model in risk theory that describes the surplus at time , where is the initial capital, the constant is the
rate of expenses, and
Y N t
is aggregate profits process with the innovation number process N t
1, 2,
T ibeing a renewal process whose inter-innovation times
i Λ have common distribution F. We also
assume that the innovation sizes , independent of i , forms a sequence of i.i.d. exponentially
distributed random variables with exponential parameter . There are many possible interpretations for this model. For example, we can treat the surplus as the amount of capital of a business engaged in research and development. The company pays expenses for research, and occasional profit of random amounts arises accord- ing to a Poisson process.
Y ii, 1
T i, 1
0
Due to its practical importance, the issue of dividend strategies has received remarkable attention in the lite- rature. De Finetti [1] considered the surplus of the com-
pany that is a discrete process and showed that the opti- mal strategy to maximize the expectation of the dis- counted dividends must be a barrier strategy. Since then, researches on dividend strategies has been carried out extensively. For some related results, the reader may consult the following publications therein: Bühlmann [2], Gerber [3], Gerber and Shiu [4,5], Lin et al. [6], Lin and pavlova [7], Dickson and Waters [8], Albrecher et al. [9], Dong et al. [10] and Ng [11]. Recently, quite a few inter- esting papers have been discussing risk models with tax payments of loss carry forward type. Albrecher et al. [12] investigated how the loss-carry forward tax payments affect the behavior of the dual process (1.1) with general inter-innovation times and exponential innovation sizes. More results can be seen in Albrecher and Hipp [13], Albrecher et al. [14], Ming et al. [15], Wang and Hu [16] and Liu et al. [17,18].
Now, we consider the model (1.1) under the additional assumption that tax payments are deducted according to a loss-carry forward system and dividends are paid under a threshold strategy. We rewrite the objective process as
R,b
t t, 0
. that is, the insurance company pays tax at rate
0,1
on the excess of each new record high of the surplus over the previous one; at the same time, dividends are paid at a constant rate whenever the surplus of an insurance portfolio is more than b and otherwise no dividends are paid. Then the surplus pro- cess of our model*Supposed by the Jiangxi Agricultural University Youth Science
Foun-dation (No. 09003326).
#Corresponding author.
,b , 0
, ,
,
, ,
0
,
max
, ( )
b b
b
b b
s t
b
R s
R t b
R s
R t
b
0
t R,b
0 u A
, ,
0
, ,
0
, ,
0
, ,
0
0
d max
d max
,
d max
d max
d d 1 d max
, d
d d 1 d
b b
s t
b b
s t
b b
s t
b b
s t
s t
R t S t R s
R t S t R s
b
R t S t R s
R t S t R s
c t S t R t S t
R t
c t S t R t S t
1
1
1
1
(1.2)
gfor , with . where 1 is the indi-
cator function of event A and R,b
t is the surplus
immediately before time . t
For practical consideration, we assume that the posi- tive safety loading condition
1 1 ,
c E Y E T
,b inf 0 : ,b 0
T t R t T,b
t
,0
e d
b T
t
D D t
0 (1.3)
holds all through this paper. The time of ruin is defined
as with if
,b 0
R t for all 0.
For initial surplus u0, let be the present value of all dividends until ruin, and
b D
is the discount factor. Denote by the expec-
tation of , that is,
,
V u
u,
, b 0 .
V u b E D R (1.4)
It needs to be mentioned that we shall drop the sub- script whenever is zero.
The rest of this paper is organized as follows. In Sec-
tion 2, We derive the expression of u
,0,00
, : e u u
u
g u u E
(i.e. the Laplace transform of the first upper exit time). We also discuss the expected discounted tax payments for this model and obtain its satisfied integro-differential equa- tions. Finally, for Erlang (2) inter-innovation distribu- tion, closed-form expressions for the the expected dis- counted tax payments are given.
2. Main Results and Proofs
Let
u,0,u
denote the Laplace trans-
form of the upper exit time 0 , which is the time
until the risk process
R t tb
, > 0
starting with initial capital u
u0
up-crosses the level u0
b for thefirst time without leading to ruin before that event. In
0 , 0 : lim ,0 0
g u u u u
is the probability that particular,
R t tb
, > 0
up-crosses the level
the process u0 b
, 0
before ruin.
For general innovation waiting times distribution, one can derive the integral equations for g u u
u b
. When
,
0
1
0
0 0
0 0
, e d , e d e d .
u c u u c t
t y y
T
u u c t
g u u f t t g u c t y u y y
0 b u u
(2.1)
When ,
0
0
1 1
0
0 0
0
0 0
0 0
,
e d , e d e e d
, e d e .
u b c u u ct u b c b c
u u ct
t y t
T T
u b c
u b c t u b c
u b c t u b c y
g u u
f t t g u ct y u y f t t
g u ct y u y
(2.2)
It follows from Equation (2.1) and from Equation (2.2) that g u u, 0
is continuous on
as a function ofand that 0
0,u u
, 0
.
0 , 0
0,
, 0
g u g b u g b u
1 T
(2.3)
For certain distributions F , one can derive integro- differential equations for g u u
, 0
and V u
,b . Let usassume that the i.i.d innovation waiting times have a common generalized Erlang n
T
1 2
:
S
distribution, i.e. the
i’s are distributed as the sum of n independent and exponentially distributed r.v.’s n Λ n
with
i having exponential parameters
i 0.
The following theorem 2.1 gives the integro-differ- ential equations for g u u
, 0
when Ti’s have a gene- ralized Erlang
nI
distribution.
Theorem 2.1 Let and D denote the identity ope-
rator and differentiation operator respectively. Then
u u, 0
satisfies the following equation for 0 u b
0 0
1 ,
d e u u,
g u u
y
0 1
0 0
, e
n
k k k
u u
y c
g u y u
I D
(2.4)
and
0 0
1 ,
d e u u,
g u u
y
u b
, 0
0 1
0 0
, e
n
k k k
u u
y c
g u y u
I D
(2.5)
for .
u u as k
, 0
Proof First, we rewrite g g u u
1 d
i n k
T S S
when
with S00 in the surplus process (1.2)
1 , 0 , 0
g
with 0. Thus, we have u u g u u
0 u b
. When
,
0
1 0
0
,
e k , d ,
k u c
t
k k
g u u
g u c t u t
1, 2, , 1
k n
(2.6)
for Λ
, and
0
0
0 0
0 0
d , e d e d .
n
u c u u c t
t y y
u u c t
, e
n n
g u u t g u c t y u y y
b
(2.7)
By changing variables in from Equation (2.6) and from Equation (2.7), we have for 0 u ,
0
k 1
, 0
d ,0
, u k e k cu x
k g u u
c
g x u x1, 2, , 1
k Λn
(2.8)
for , and
0
0 0
0
, e d e d .
u x
y y
u x 0
0
, e n d
u x u
n c
n
g u u x
c
g x y u y y
u
(2.9)
Then, differentiating both sides of from Equation (2.8) and from Equation (2.9) with respect to , one gets
0
1
0
1 k , k , ,
k k
c
g u u g u u
I D
1, 2, , 1
k n
(2.10)
Λ , and
for
0
0 0
0 0
1 ,
, e d e .
n
n n
u u
u u y
c
g u u
g u y u y
I D
(2.11)
Using from Equation (2.10) and from Equation (2.11), we can derive from Equation (2.4) for g u u
, 0
on
0,b .Similar to from Equation (2.6) and Equation (2.7), we have for u b
0 1
0
k , ke k , 0 d e k 1 , 0 d ,
u b c u b c b c
t t
k k k
u b c
u u g u ct u t g b c t u b c u t
g (2.12)
for k1, 2, ,Λn1, and
0
0
0 0
0
0 0
0
,
, e d e e d
, e d e .
n n
u b c b c u u ct
t y u u ct t
n u b c
c
u b c t u b c y
g u u
t g u ct y u y t
g b c t u b c y u y
0
0
e n d
u b c
n
u b c t u b
(2.13)
Again, by changing variables in Equation (2.12) and Equation (2.13) and then differentiating them with re- spect to , we obtain for
0
0 0
0 0
1 ,
, e d e .
n
n n
u u
u u y
c
g u u
g u y u y
I D
u u b
, 0
k 1
, 0
,1 k
k k
c g u u
I D g u u
1, 2, , 1
k Λn
(2.14)
for , and
(2.15)
It needs to be mentioned that from the proof of Lemma 2.1, we know that
0 , 0
0,
, 0
,k k k 0
, 2,3, , .g u g b u g b u k Λn
Therefore, Equations (2.10), (2.11), (2.14) and (2.15) yield
0
0
1 ,
1 , ,
g b u
1
1 1, 2, , .
k
i i i
k
i i i
c
c g b u
k n
I D
I D
Λ
(2.16)
Remark 2.1 Using a similar argument to the one used
in Lemma 2.1, one can get that when the innovation waiting times follow a common generalized Erlang
n
,V u b
distribution, the expected discounted dividend payments satisfies the following integro-differential equa- tion (see Liu et al. [17]).
,d ,0
y
V u b
y u b
I D
1
0 1
, e
n
k k k
c
V u y b
(2.17)
and
1 ,
d n,
V u b
y B u b
1
0
, e
n
k k k
y c
V u y b
I D
(2.18)
with
1 1
1 ,
k k
k
i i j i j
1, 2, , .B k Λn (2.19)
In addition, the boundary conditions for V u b
, are as follows:
1 ,
, k,
V b b
V b b B
1
1 1
1, 2, , ,
k
i i i
k
i i i
c
c
k n
I D
I D
Λ
(2.20)
with Equation (2.19).
1 V 0 , b 0,
n
1
1, 2, , 1, k
i i i
c
k n
I DΛ
(2.21)
With the preparations made above, we can now derive analytic expressions of the expected -th moment of the accumulated discounted tax payments for the surplus
R,b
t t, 0
. We claim that the processprocess
R,b
t t, 0
shall up-cross the initial surplus level uat least once to avoid ruin. Now, let
: e uu g u E
u
(2.22)
denote the Laplace transform of the first upper exit time
, which is the time until the risk process
R t t
, 0
u
b starting with initial capital reaches a new record high for the first time without leading to ruin
0 : lim0
g
before that event. In particular, u g u
R t t, 0
is the probability that the process b reaches a new record high before ruin. Then the closed-form expression of the quantity g u can be calculated as follows.
When u b g u , g u u,
0 u b
0
, e y , d ,
V u u g u V u y u y
. When , using
a simple sample path argument, we immediately have,
or, equivalently
0
,
.
e y , d
V u u g u
V u y u y
0 0
(2.23)
Let and define
1 ,
0 ,
inf : max ,
n t n Rb t s tRb s
n
(2.24)
to be the -th taxation time point. Thus,
,,
, , 1
1
, :
: e
1
n
n b
n
n u
n
u b n b n T
n
M u b E D
E R R
1 n
(2.25) denotes the -th moment of the accumulated discounted tax payments over the life time of the surplus process
R,b
t t, 0
We will consider a recursive formula of .
,n M u b
0 u b
in the following theorem 2.2.
Theorem 2.2 When
, we have
1 1 d 1 1 1 d 1 1 1 d1 1
, e e d e d ,
b s b
n n n
u b s
b
g t t g t t g t t
n n
n n
b n u n
M s M s
n
1
u b g u s s
g s g s
nM and
and when u b , we have
1 d
e d .
s n u
g t t 1 1 1 , 1 n n n u n
M u b
M s
n
g u s
g s
(2.27)Proof Given that the after-tax excess of the surplus
level over u at time u is exponentially distributed with mean
1
0
u
due to the memoryless property of the exponential distribution. By a probabilistic argument, one easily shows that for
e d d 1 n n x x 1 , 0 1 , , 1 1 e 1 n x n x u n u M u bg u E D u x
g u E D
x x u x u (2.28) Differentiating with respect to yields
1 , 1 1 , 1 , 1 1 e 1 1 1 , 1 , . 1 n n n n n n x u u n n n n n n nM u b
g u
g u M u
g u
n
g u
E D x
g u
g u M u
g u
g u
n
M u b
g u
1 d n bx u x
b
0 u b (2.29)
When , we have
1 d 1 de d .
n
n
g t t
g t t
C s
u b
1 1 1 1 , e 1 b u b s n n b n u n
M u b g u
M s n g s
(2.30)When , the general solution of Equation (3.20) can be expressed as
1 d 1 d e d s n u tg t t
1 1 1 1 , e 1 n u g t n n n n n u n M
M u b g u
g M s n s g s
(2.31)Due to the facts that
n g0 , we have for u b
1 d 1
1 1
,
e d .
1
s n u n
g t t n
n
u n
M u b
M s
n
g u s
g s
(2.32)Now, it remains to determine the unknown constant C
in Equation (3.20). The continuity of M u b
, on band Equation (3.22) lead to
1 d 1
1
1
e d .
1
s n b
g t t
n b n M s n C s g s
(2.33)Plugging Equation (2.33) into Equation (2.30), we ar
rive at Equation (2.26). □
The special case n1 leads to an expression for the
ex total
pected discounted sum of tax payments over the life time of the risk process
1 1 d1 , e d .
1
s
u g t t
u
M u b g u s
(2.34)for all u0.
3. Explicit Results for Erlang(2) Innovation
In ume that Wi’s are Erlang(2) dis-
Waiting Times
this section, we asstributed with parameters 1 and 2. We also assume
that 12 (without loss f genera ty). Exam 3.1 Note that
o li ple
0 0 0 0, e d e
, .
u u y
g u y u y
g u u
I D (3.1)Applying the operator
0
uu
ID
to Equations (2.4)an
d (2.5) gives
0
0
1
1 , , ,
0 ,
k k k
c
n
g u u g u u
u b
I D I D
(3.2) and
0
0
1
1 , , ,
,
n
k k k
c g u u g u u
u b
I D I D
(3.3) The characteristic equation for Equation (3.2) is
2 c 1 1
k k k
r r
(3.4)without loss of generality, we assume that 12. We
know that Equation (3.4) ha and r3 which satisfies
s three real roots, say r r1, 2
1 2
1 2
3 .
r
c c
1 2
0
r r
c c
With c replace c in Equation (3.4), e get the charact equation of Equation (3.3), whose roots are
4, 5
r r and r6 with
w eristic
ted by deno
1 2
4 0 5 6
r r r
c c
1 2
c c
Thus, we have
3
3e ,0 ,
r u
1 20 1 2
, er u er u
g u u c c c u b (3.5)
an
5 6
6 r u, ,
d
40 4 5
, er u er u e
g u u c c c u b
2 3 4 5 6
, , , , ,c c c c are arbitrary constants. To de- e arbitrary constants, we insert Equatio
and (3.6) into Equation (2.3) and obtain
c
3
1 2 4
5 6
1 2 3 4
5 6
e e e
e e 0.
r b
r b r b r b
r b r b
e c c c c
c c
1
k
(3.6)
where c c1
termine th ns (3.5)
1 c2c30, (3.7)
and
(3.8)
Apply Equation (2.10) together with Equations (2.3) and (3.5) when , we get
1 1 2 2 3 3 0.
r c r c r c (3.9) Insert Equation (3.5) into Equation (2.4), we have
1 0 2 0 3 0
1 2 3
1 2 3
er u er u er u 1.
c c c
r r r
(3.10)
In addition, plugging Equations (3.5) and (3.6) into Equation (2.16) yields
1
2
35 6
4
1 1 2 2 3 3
4 4 5 5 6 6
e e e
e e e 0,
r b
r b r b
r b r b
r b
c r c c r c c r c
cr c cr c cr c
(3.11)
and
3
1 2 4
5 6
1 2 3 4
1 2 3 4
5 6
5 6
e e e e
e e 0.
r b
r b r b r b
r b r b
c c c c
r r r r
c c
r r
(3.12)
Some calculations give
1 0 2 0 3 0
1 0 2 0 3 0
1 0 2 0 3 0
3 2
1
3 2 1 3 2 1
1 2 3
1 3
2
3 2 1 3 2 1
1 2 3
2 1
3
3 2 1 3 2 1
1 2 3
1 2 1 3 5 6 3 1 1 2
4
,
e e e
,
e e e
,
e e e
e , , e ,
r u r u r u
r u r u r u
r u r u r u
r b r b
c
r r
r r r
r r
c r r r r
r r
r r r
r r
c
r r r r r r
r r r
r r r r r r r r r
c
3 2
r r
r r r r
1
1 0 2 0 3 0
3 2 1
5 1 0
5 6 2 3 1 1 5 6
3 2 1 3 2 1 6 5 4 6 5 4
1 2 3 4 5 6
1 3 4 6 1 3 1 2 4 6 3 2 1 1 4 6
5
3 2
1
, e , ,
,
e e e e
e , , e , , e , ,
e e
r b
r u r u r u
r b
r b r b r b
r b r u
r r r r r r
r r r r r r r r r r r r
r r r r r r
r r r r r r r r r r r r r
r r
c
r
4
2 1
c
r r
c
2 0 3 0
3 2 1
6 1 0 2 0 3 0
1 3 2 1 6 5 4 6 5 4
2 3 4 5 6
1 3 4 5 3 1 1 2 4 5 2 3 1 1 4 5
3 2 1 3 2 1 6 5
1 2 3 4
,
e e
e , , e , , e , ,
e e e e
r u r u
r b r b r b
r u r u r u
r r r r r r r r r r
r r r r r
r r r r r r r r r r r r r
r r r r r r r r
r r r r
1 2
6
r b
r r
c c
4 6 5 4
5 6
,
r r r r
r r
with g u
. By Equations (3.6) and (3.13), we have foru b
1 , ,
k j
i j k
k
c r r
r r r
cr c r
r
i k
i j
j i
c r cr
r r
(3.14)
Remark 3.1 Now, we give the explicit results for
4
5
63
1 2
4 5 6
3 2 1 3 2 1
1 2 3
,
e e e
,
e e e
r u r u
r u
r u
r u r u
g u g u u
l b l b l b
r r r r r r
r r r
(3.15)
with
3 2 1
4
3 2 1
5
1 2 1 3 5 6 3 1 1 2 5 6 2 3 1 1 5 6
6 5 4 6 5 4
4 5 6
1 1 3 4 6 1 3 1 2 4 6 3 2 1 1 4 6
6 5 4 6 5
4 5
e , , e , , e , ,
, e
e , , e , , e , ,
e
r b r b r b
r b
r b r b r b
r b
r r r r r r r r r r r r r r r
l b
r r r r r r
c
r r r
r r r r r r r r r r r r r r r
r r r r r r
c
r r
4
2 5
l b
3
2
1
6
1 2 1 3 4 5 3 1 1 2 4 5 2 3
6
6 5 4 6 5 4
4 5 6
e , , e , ,
e
r b r b r
r b
r r r r r r r r r r r r
l b
r r r r r r
c
r r r
u b
, using the explicit expressions of
4 6
1 1 4 5 ,
e , , .
b r
r r r
(3.16)
For 0 V b b
, in Liu et al. [17], we obtain
3
3
2 1
3 2 1 3 3 2 1
,
e , d
e e e
e e e
y
r u
r u r u
r u
r u r u
V u u g u
V u y u y
cr r
r r r r r r
r r
r r r l r r
(3.17)
h
1 2
1 2
0
5 6 6 5
3 2 1 3
6 5
1
l r l2
r cr
wit
3 3
2
6 5
,5 ,6
,5,6 , 1, 2,3,
i
i i
l i i
r r
(3.18)
where
2 , ,
1 6,
k j i i k j j i
i j k
i j
c r r r c r cr r cr c r r
r r r
r r r
i j k
,
k
k
and
3
j i
i j
i j
cr r c r
r r
r r
, i rj,1 i j 6.
oint out expone llow the same steps to get the
explicit ex
We p that when the innovation times are ntially distributed, one can fo pressions of g u
, which coincide with the results in Albrecher et ale 3.2 (The expected discounted tax payments.) Following from Equat
ve for 0 u b,
. (2008).
Exampl ion (2.34) of Theorem 2.2 and Remark
3
1 2
3
1 2
1 2
1 2
5 6 6 5
3 2 1 3 2 1
6 5
1
1 3 2 2 1 3 3 2 1
5 6 6 5
3 2 1 3 2
6 5
1 3 2 2 1 3 3 2
e e e
,
1 e e e
e e
exp 1
1 e e
r u
r u r u
r u
r u r u
r t r t
r t r t
cr r cr r
r r r r r r
r r
M u b
l r r l r r l r r
cr r cr r
r r r r r
r r
l r r l r r l r
3
1 e
r t r
3
3
1 2
3
1 2
1
1
5 6 6 5
3 2 1 3 2 1
6 5
1 3 2 2 1 3 3 2 1
5 6 6 5
3 2 1 3
6 5
d d e
e e e
1 e e e
e e
exp 1
1
b s
r t
u u
r u
r u r u
r u
r u r u
r t r
t s
r
cr r cr r
r r r r r r
r r
l r r l r r l r r
cr r cr r
r r r r
r r
(3.19)
3 2
3
1 2
5 6
4
3
1 2
2 1
1 3 2 2 1 3 3 2 1
4 5 6
3 2 1 3 2 1
1 2 3
e
d d
e e e
e e e
exp 1 d .
1 e e e
r t t
b
r t
r t r t
b u
r t r t
r t
r t
r t r t
b
r r
t s
l r r l r r l r r
l b l b l b
t
r r r r r r
r r r
And, for u b , we have
4
4 5
5 6
6 4
4 5
5 6
61
e e e e e e
, exp 1 d d .
1
r u r u r t r t
r u s r t
r u
l b l b l b l b l b l b
3 3
1 2 1 2
3 2 1 3 2 1 3 2 1 3 2 1
1 2 3 1 2 3
1
e er u er u u u er t er t er t
M u b t s
r r r r r r r r r r r r
r r r r r r
Then we can get that when Wi’s are Erlang (2) dis-
tributed with pa 1
(3.20)rameters and 2, the expresses of
g u can Equations (3.15) and (3.17) and the expecte tax payments can be given by Equation (3.20).
4. Acknowledgements
The author would like to thank Professor Ruixing Ming and Professor Guiying Fang for their useful discussions and valuable suggestions.
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