- Email: [email protected]
On the distribution of surplus immediately before ruin under interest force
Statistics & Probability Letters 55 (2001) 329 – 338
On the distribution of surplus immediately before ruin under interest force Hailiang Yanga; ∗ , Lihong Zhangb a Department
of Statistics and Actuarial Science, The University of Hong Kong, Hong Kong, China of Mathematical Finance, Beijing University, Beijing, People’s Republic of China
b Department
Received February 2001; accepted August 2001
Abstract In this paper, we consider a compound Poisson model with a constant interest force for an insurance portfolio. We investigate the distribution of surplus process immediately before ruin in particular. Equations satis2ed by the distributions of surplus immediately before ruin and their Laplace transform have been obtained. Some special cases are also discussed c 2001 Elsevier Science B.V. All rights reserved and Lundberg-type bounds are presented. Keywords: Laplace transform; Integral equations; Renewal theory; Lundberg bound; Surplus immediately before ruin
1. Introduction Recently, people in actuarial science have started paying more attention to the severity of ruin. In Gerber et al. (1987), the probability that ruin occurs with an initial surplus u and that the de2cit at the time of ruin is ¡ y was considered. Later, in Dufresne and Gerber (1988), the distribution of the surplus immediately prior to ruin in the classical compound Poisson risk model was introduced. Similar results to those in Gerber et al. (1987) were obtained in the paper. Risk models with interest rate have recently received a remarkable amount of attention. In Sundt and Teugels (1995), a compound Poisson model with a constant interest force was considered. In Yang (1999), a discrete time risk model with a constant interest force was considered. By using martingale inequalities, both Lundberg-type inequality and non-exponential upper bounds for ruin probabilities were obtained. In Yang and Zhang (2001), the distribution of the severity of ruin for a compound Poisson model with a constant interest force was considered. In this paper, we consider the distribution of the surplus process immediately before ruin. Some similar problems as to those in Yang and Zhang (2001) are considered here and similar results are ∗ Corresponding author. Tel.: +86-852-2857-8322; fax: +86-852-2858-9041. E-mail address: [email protected] (H. Yang).
c 2001 Elsevier Science B.V. All rights reserved 0167-7152/01/$ - see front matter PII: S 0 1 6 7 - 7 1 5 2 ( 0 1 ) 0 0 1 6 3 - 8
330
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
obtained. However, the method used in Yang and Zhang (2001) does not seem to work here so we employ a slightly diDerent mathematical method. 2. The model Here, we consider the same model as that in Sundt and Teugels (1995). We assume that the premium which the insurance company receives is paid continuously with a constant rate of p. In addition to the premium income, the company also receives interest on its reserves with a constant force of . Let U (t) denote the value of the reserve at time t. From the above assumptions, it follows that dU (t) = p dt + U (t) dt − dX (t); N (t) where X (t) = j=1 Yj . N (t) denotes the number of claims occurring in an insurance portfolio in the time interval (0; t] whilst Yi denotes the amount of the ith claim. We assume that {N (t) : t ¿ 0} is a homogeneous Poisson process with an intensity of . We also assume that the claim amounts are independent of the claim number process, positive and mutually ∞ independent and identically distributed with the common distribution F. F satis2es F(0) = 0 and k = 0 xk dF(x). We denote = 1 . For the sake of convenience, we will drop the index when the interest force is zero. Let (u) denote the probability of ruin with an initial reserve u. That is, (U (t) ¡ 0) : (u) = P t¿0
In Yang and Zhang (2001), the distribution of surplus immediately after ruin, denoted by G (u; x), was studied. In this paper, we consider another interesting function, B (u; y). B (u; y) denotes the probability that ruin occurs beginning with the initial reserve u and that the surplus immediately prior to ruin is ¡ y, i.e. B (u; y) = P(T ¡ + ∞; 0 ¡ U (T −) 6 y | U (0) = u) =
(u)
− P(T ¡ + ∞; U (T −) ¿ y | U (0) = u);
(1)
where T is the ruin time. Note that lim B (u; y) =
y→+∞
(u):
3. Equations for B (u; y) 3.1. Integral equations for B (u; y) Theorem 1. B (u; y) =
where
u p 1 B (u − z; y)[ + (1 − F(z))] d z B (0; y) + p + u p + u 0 u y I(u6y) − (1 − F(v)) dv + (1 − I(u6y) ) (1 − F(v)) dv ; p + u 0 0
I{u¿y} =
1 0
if u ¿ y; otherwise
is an indicator function.
(2)
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
331
Proof. We condition on the 2rst claim time T1 and the amount of the 2rst claim Y1 . Given that T1 = t and Y1 = z, the reserve just before the occurrence of the claim is uet + p(et − 1)=. Then,
eT1 − 1 T1 B (u; y) = E B ue + p − Y1 ; y I T eT1 −1 (ue 1 +p 6y)
eT1 − 1 eT1 − 1 − Y1 ; y Y1 6 ueT1 + p = E B ueT1 + p eT1 − 1 P Y1 6 ueT1 + p I T eT1 −1 (ue 1 +p 6y)
eT1 − 1 eT1 − 1 T1 T1 − Y1 ; y Y1 ¿ ue + p + E B ue + p eT1 − 1 P Y1 ¿ ueT1 + p I T eT1 −1 (ue 1 +p 6y) =
0
+∞
e
−t
uet +p
et −1
0 y+p ln( u+p )
+ I(u6y)
0
e
−t
et − 1 B uet + p − z; y dF(z) dt
+∞
uet +p
et −1
dF(z) dt:
By using the substitution s = uet + p(et − 1)=, we obtain B (u; y) = (p + u)
=
u
+∞
(p + s)
+ (p + u)= I(u6y)
u
y
−=−1
0
s
B (s − z; y) dF(z) ds
(s + p)−=−1 (1 − F(s)) ds:
Assume that F is continuous, diDerentiation of the above expression gives u @B (u; y) B (u − z; y) dF(z) − = B (u; y) − I(u6y) (1 − F(u)): @u p + u p + u 0 p + u For u 6 y, integrating both sides of (3) from 0 to u, we have u p 1 B (u; y) = B (u − z; y)[ + (1 − F(z))] d z B (0; y) + p + u p + u 0 u − (1 − F(v)) dv: p + u 0 From (4), we obtain (2) in the case where u 6 y.
(3)
(4)
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H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
For u ¿ y, integrating both sides of (3) from y to u, we have u B (v; y) dv (p + u)B (u; y) − (p + y)B (y; y) − =
u
y
B (v; y) dv −
u
y
y
v
0
B (v − z; y) dF(z) dv;
the second term on the right-hand side of the above equation equals u v B (v − z; y) dF(z) dv y
=
0
y
=
u
u
0
B (v − z; y)F(z)|vz=0 dv + B (u − z; y)F(z) d z +
0
y
y
0
(p + u)B (u; y) − (p + y)B (y; y) − y
= so
u
u
0
B (v; y) dv −
0
u
v
@B (v − z; y) F(z) d z dv @v
[ − B (y − z; y)]F(z) d z;
so we have
=
u
u
y
B (v; y) dv
B (u − z; y)F(z) d z +
(B (u − z; y)(1 − F(z))) d z −
0
y
0
y
B (y − z; y)F(z) d z
B (y − z; y)(1 − F(z)) d z;
u p + y 1 B (u; y) = B (u − z; y)[ + (1 − F(z))] d z B (y; y) + p + u p + u 0 y 1 − B (y − z; y)[ + (1 − F(z))] d z: p + u 0
(5)
From the de2nition of B (u; y), we know that B (u; y) is a continuous function of u ¿ 0 for a given value of y. Let u = y in (4), we then have y p 1 B (y; y) = B (y − z; y)( + (1 − F(z))) d z B (0; y) + p + y p + y 0 y − (1 − F(v)) dv: p + y 0 Plugging B (y; y) into (5), B (u; y) =
u p 1 B (u − z; y)( + (1 − F(z))) d z B (0; y) + p + u p + u 0 y − (1 − F(v)) dv: p + u 0
This is (2) for u ¿ y.
(6)
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
333
3.2. The case = 0 Let F1 be the equilibrium distribution of F, given by 1 x (1 − F(v)) dv: F1 (x) = 0 Denote the moments of the equilibrium distribution by 'k = From Sundt and Teugels (1995), we know that 1 k+1 'k = : k +1
+∞ 0
xk dF1 (x).
Denote the Laplace transformation of F1 by +∞ ((s) = e−su dF1 (u) = (0 (s); 0
(y (s) =
+∞
y
e−su dF1 (u):
We have the following theorem. Theorem 2. When the adjustment coe
u→∞
where F ∗ (x) =
0
x
F ∗ (y) − (=p)F1 (y) ; (=p)(−( (−R) − p=)
(8)
Rz e (1 − F(z)) d z: p
Proof. In the case of u 6 y, limu→+∞ eRu F(u; y) = limu→∞ eRu (u). From Grandell (1991), we have 1 − =p (u) ∼ e−Ru : (=p)(−( (−R) − p=) This proves that Eq. (7) is true. For u ¿ y, from Dickson (1992), we know that 1 − G(0; y) F(u; y) = G(u − y; y) − ( (u − y) − (u)) 1 − (0)
for u ¿ y;
where G(u; y) denotes the distribution of the surplus at ruin. So 1 − G(0; y) Ru ( (u − y) − (u)) eRu : lim e F(u; y) = lim G(u − y; y) − u→∞ u→∞ 1 − (0) Using the asymptotic result for G(u; y), obtained in Yang and Zhang (2001), and the result of Grandell (1991), we obtain (8). Remark. We can also prove this theorem by directly using Eq. (4).
(9) (u), in
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H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
3.3. Solution of the integral equation Following Sundt and Teugels (1995), we introduce the following auxiliary function: B (u; y) − B (0; y) A (u; y) = : 0 − B (0; y) Eq. (2), in terms of A (u; y) and F1 (x), can be written as u u pA (u; y) + uA (u; y) = A (v; y) dv + A ∗ F1 (u; y) − (1 − F(z)) d z 0
+ Let
+1 (s; y) =
+∞
0
B (0; y)
I{u¿y}
0
0
y
(1 − F(v)) dv + (1 − I{u¿y} )
0
u
(1 − F(v)) dv : (10)
e−sv dA (v; y);
where y is regarded as a parameter. The 2rst order diDerential equation for the function, +1 (s; y), can be obtained by taking the Laplace transform of (10) respect to u: −+1 (s; y) + (p − ((s))+1 (s; y) = (((s) − (y (s)) − ((s); (11) B (0; y) the initial condition of the above equation is given by lims→+∞ +1 (s; y) = 0. Similar to Sundt and Teugels (1995), the solution of equation (11) is given by +∞ −1 −{,(v)−,(s)}= (((v) − (y (v)) +1 (s; y) = − ((v) dv e B (0; y) s = where
,(s) :=
s
0
+∞
0
e−
w 0
[p−((s+z)] d z
((s + w) − (y (s + w) − ((s + w) dw; B (0; y)
(12)
(p − ((v)) dv:
By letting s = 0 we have
−1 +∞ +∞ −,(v)= −,(v)= B (0; y) = e ((v) dv e (((v) − (y (v)) dv + 0 0
=
1 + +∞
0
+∞
0
e−
w 0
e−
w 0
[p−((z)] d z
[p−((z)] d z
−1 ((w) dw
(((w) − (y (w)) dw
From Sundt and Teugels (1995), we know
+∞ −1 z − 0 (p−((w)) dw e dz : (0) = 1 − p 0
:
(13)
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
From Yang and Zhang (2001), we have
+∞ −1 z − 0 (p−((w)) dw G (0; y) = p e dz 0
+∞ z
e
0
(p−((w)) dw 0
(((z) − e
335
zy
(y (z)) d z ;
so, when = 0, F(0; y) = G(0; y), but for ¿ 0, G (0; y) = B (0; y). Hence, the reLection principle is not valid when = 0. In addition, the relationships among G(u; y), F(u; y) and (u) in Dickson (1992), Willmot and Lin (1998) are not valid when = 0. 4. B (u; y) with zero initial reserve The main result of this section is the following theorem. Theorem 3.
−1 -2 '11 (y) 1 1 1 1 + F1 (y) + -'11 (y) R(-) − e 2 p − '1 '1 '1
6 B (0; y) 6
1 + '1
-2
e2 1 R(-) − +
−1
-2
where R(-) is the well-known Mills’ ratio given by R(-) = e 2 y and '11 (y) = 0 z dF1 (z). Proof. Let A1 :=
+∞
0
e−
w 0
(p−((z)) d z
((w) dw =
+∞
0
e
w 0
F1 (y) ; p − +∞ -
(((z)−p) d z
(14)
x2 e− 2 d x; and - = ( − p)=( '1 )
((w) dw:
(15)
Using the property that ((s) is a strictly decreasing convex function, we obtain +∞ +∞ w w 1 e 0 ((1−'1 z)−p) d z (1 − '1 w) dw 6 A1 6 e 0 (−p) d z dw = : p − 0 0 Using a similar method as to that in Sundt and Teugels (1995) (or Yang and Zhang, 2001), we have the following inequality for A1 : -2 e2 1 1 R(-) − + 6 A1 6 : (16) '1 p − Write
A2 :=
+∞
0
e−
w 0
(p−((z)) d z
(((w) − (y (w)) dw:
Again, using ((s), (y (s), and ((s) − (y (s) are strictly decreasing convex functions, we have
+∞ y w z−p) d z (−' 1 A2 ¿ F1 (y) − w e0 u dF1 (u) dw 0
=
0
0
+∞
e
−
2
(w−(p−)='1 ) (p−) + 2' 2(1='1 ) 1
2
(F1 (y) − w'11 ) dw
(17)
336
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
and
A2 6
+∞
e
0
2 0
(−p) d z
F1 (y) dw =
F1 (y) : p −
So we have the following inequality for A2 : -2 '11 (y) 1 1 F1 (y) F1 (y) + -'11 (y) : R(-) − e 2 6 A2 6 '1 '1 '1 p −
(18)
Notice that (13) can be rearranged as −1
1 B (0; y) = A1 + A2 : From this, the theorem is proved. Similar to Sundt and Teugels (1995), we can expand B (0; y) as a McLaurin series with respect to the variable, . We have B (0; y) =
−'1 F1 (y) + + o(): p p(p − )
(19)
5. Exponential claim size In this section, we will try to evaluate expression (13) in the case of an exponential claim size. First, we have −1 +∞ B (0; y) = e−(pv− log(1+v))= (1 + v)−1 dv + 0 ×
+∞
= + then −1
+1 (s; y) =
−1
=
e
0
+∞
e
0
+∞
s
−(pv− log(1+v))=
e
(1 + v)
(=)−1
(1 + v)
s
s
e
p − (v−s)
+∞
dv
−1
(1 + v)
p e− s (v−s) (1
(1 − e
−1
+∞
e
− p v
−1
(1 + v)
1 + v 1 + s
−1
+ v)
−(v+1=)y
0
−[p(v−s)− log(1+v)=(1+s)]=
+∞
1 − B (0; y)
− p v
−1
(=)
1 + v 1 + s
) dv
(=)−1
(1 + v)
(1 − e
−(v+1=)y
1 − e−(v+1=)y − 1 dv B (0; y)
1 − 1 dv B (0; y)
(=) e
−vy −y=
e
dv
) dv ; (20)
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
=
+∞
0
+
p
e
−sx −x=
e
+∞
0
e
1+ x p
(=)−1
p
y x+y y −s(x+ p x+y) −( + p x) e
337
1 −1 B (0; y)
1+
x p
dx
(=)−1 dx
1 : B (0; y)
(21)
Using the substitution x+
y x + y = t; p
we have +1 (s; y) =
+∞
0
−
e−st e
+∞
y
t −
e−st e
1+ t −
x p
1+
(=)−1
(t − y) p + y
p
1 −1 B (0; y)
(=)−1
dt
1 dt: p + y B (0; y)
(22)
Remember that +1 (s; y) was introduced as a Laplace transformation of A (u; y) respect to u. Hence, by inverting, we 2nd that (=)−1
u t 1 − − 1 dt 1+ x e A (u; y) = p p B (0; y) 0 −
y
u
e
t −
(t − y) 1+ p + y
(=)−1
1 I{u¿y} dt; p + y B (0; y)
(23)
where I{u¿y} is the indicator function.
6. Lundberg bound Following the arguments in Sundt and Teugels (1995), we are also able to obtain the Lundberg-type inequality. Same as in Sundt and Teugels (1995), we assume that ((s) has −. as an abscissa of convergence. So for any u ¿ 0 and for s ∈ I := (−.; 0), we have +∞ +1 (s; y) ¿ e−sx dA (x; y) ¿ e−su {1 − A (u; y)}; u
so we obtain the ChernoD-type estimate 1 − A (u; y) 6 exp inf {su + log +1 (s; y)} : s∈I
(24)
Let s (u; y) denote the solution of the following equation: u+1 (s; y) = − +1 (s; y); and we call |s (u; y)| the adjustment function.
(25)
338
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
Similar to Sundt and Teugels (1995), write ([=B (u; y)]F1 (y) − p) u = ; we have the following result:
(26)
Theorem 4. When the initial surplus u satis?es u ¿ u ; then 1−A (u; y) satis?es a Lundberg-type inequality: (=B (0; y))[((s (u; y)) − (y (s (u; y))] − ((s (u; y)) us (u; y) e : 1 − A (u; y) 6 u + p − ((s (u; y)) Acknowledgements The work described in this paper was partially supported by a grant from the Research Grants Council of the Hong Kong Special Administrative Region, China (Project No. HKU 7168=98H). References Dickson, D.C.M., 1992. On the distribution of the surplus prior to ruin. Insurance Math. Econom. 11, 191–207. Dufresne, F., Gerber, H.U., 1988. The surpluses immediately before and at ruin, and the amount of the claim causing ruin. Insurance Math. Econom. 7, 193–199. Gerber, H.U., Goovaerts, M.J., Kaas, R., 1987. On the probability and severity of ruin. ASTIN Bull. 17, 151–163. Grandell, J., 1991. Aspects of Risk Theory. Springer, New York. Sundt, B., Teugels, J.L., 1995. Ruin estimates under interest force. Insurance Math. Econom. 16, 7–22. Willmot, G.E., Lin, X., 1998. Exact and approximate properties of the distribution of surplus before and after ruin. Insurance Math. Econom. 23, 91–110. Yang, H., 1999. Non-exponential bounds for ruin probability with interest eDect included. Scand. Actuar. J. 99, 66–79. Yang, H., Zhang, L., 2001. On the distribution of surplus immediately after ruin under interest force. Insurance: Mathematics and Economics 29, 247–255.
On the distribution of surplus immediately before ruin under interest force Hailiang Yanga; ∗ , Lihong Zhangb a Department
of Statistics and Actuarial Science, The University of Hong Kong, Hong Kong, China of Mathematical Finance, Beijing University, Beijing, People’s Republic of China
b Department
Received February 2001; accepted August 2001
Abstract In this paper, we consider a compound Poisson model with a constant interest force for an insurance portfolio. We investigate the distribution of surplus process immediately before ruin in particular. Equations satis2ed by the distributions of surplus immediately before ruin and their Laplace transform have been obtained. Some special cases are also discussed c 2001 Elsevier Science B.V. All rights reserved and Lundberg-type bounds are presented. Keywords: Laplace transform; Integral equations; Renewal theory; Lundberg bound; Surplus immediately before ruin
1. Introduction Recently, people in actuarial science have started paying more attention to the severity of ruin. In Gerber et al. (1987), the probability that ruin occurs with an initial surplus u and that the de2cit at the time of ruin is ¡ y was considered. Later, in Dufresne and Gerber (1988), the distribution of the surplus immediately prior to ruin in the classical compound Poisson risk model was introduced. Similar results to those in Gerber et al. (1987) were obtained in the paper. Risk models with interest rate have recently received a remarkable amount of attention. In Sundt and Teugels (1995), a compound Poisson model with a constant interest force was considered. In Yang (1999), a discrete time risk model with a constant interest force was considered. By using martingale inequalities, both Lundberg-type inequality and non-exponential upper bounds for ruin probabilities were obtained. In Yang and Zhang (2001), the distribution of the severity of ruin for a compound Poisson model with a constant interest force was considered. In this paper, we consider the distribution of the surplus process immediately before ruin. Some similar problems as to those in Yang and Zhang (2001) are considered here and similar results are ∗ Corresponding author. Tel.: +86-852-2857-8322; fax: +86-852-2858-9041. E-mail address: [email protected] (H. Yang).
c 2001 Elsevier Science B.V. All rights reserved 0167-7152/01/$ - see front matter PII: S 0 1 6 7 - 7 1 5 2 ( 0 1 ) 0 0 1 6 3 - 8
330
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
obtained. However, the method used in Yang and Zhang (2001) does not seem to work here so we employ a slightly diDerent mathematical method. 2. The model Here, we consider the same model as that in Sundt and Teugels (1995). We assume that the premium which the insurance company receives is paid continuously with a constant rate of p. In addition to the premium income, the company also receives interest on its reserves with a constant force of . Let U (t) denote the value of the reserve at time t. From the above assumptions, it follows that dU (t) = p dt + U (t) dt − dX (t); N (t) where X (t) = j=1 Yj . N (t) denotes the number of claims occurring in an insurance portfolio in the time interval (0; t] whilst Yi denotes the amount of the ith claim. We assume that {N (t) : t ¿ 0} is a homogeneous Poisson process with an intensity of . We also assume that the claim amounts are independent of the claim number process, positive and mutually ∞ independent and identically distributed with the common distribution F. F satis2es F(0) = 0 and k = 0 xk dF(x). We denote = 1 . For the sake of convenience, we will drop the index when the interest force is zero. Let (u) denote the probability of ruin with an initial reserve u. That is, (U (t) ¡ 0) : (u) = P t¿0
In Yang and Zhang (2001), the distribution of surplus immediately after ruin, denoted by G (u; x), was studied. In this paper, we consider another interesting function, B (u; y). B (u; y) denotes the probability that ruin occurs beginning with the initial reserve u and that the surplus immediately prior to ruin is ¡ y, i.e. B (u; y) = P(T ¡ + ∞; 0 ¡ U (T −) 6 y | U (0) = u) =
(u)
− P(T ¡ + ∞; U (T −) ¿ y | U (0) = u);
(1)
where T is the ruin time. Note that lim B (u; y) =
y→+∞
(u):
3. Equations for B (u; y) 3.1. Integral equations for B (u; y) Theorem 1. B (u; y) =
where
u p 1 B (u − z; y)[ + (1 − F(z))] d z B (0; y) + p + u p + u 0 u y I(u6y) − (1 − F(v)) dv + (1 − I(u6y) ) (1 − F(v)) dv ; p + u 0 0
I{u¿y} =
1 0
if u ¿ y; otherwise
is an indicator function.
(2)
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
331
Proof. We condition on the 2rst claim time T1 and the amount of the 2rst claim Y1 . Given that T1 = t and Y1 = z, the reserve just before the occurrence of the claim is uet + p(et − 1)=. Then,
eT1 − 1 T1 B (u; y) = E B ue + p − Y1 ; y I T eT1 −1 (ue 1 +p 6y)
eT1 − 1 eT1 − 1 − Y1 ; y Y1 6 ueT1 + p = E B ueT1 + p eT1 − 1 P Y1 6 ueT1 + p I T eT1 −1 (ue 1 +p 6y)
eT1 − 1 eT1 − 1 T1 T1 − Y1 ; y Y1 ¿ ue + p + E B ue + p eT1 − 1 P Y1 ¿ ueT1 + p I T eT1 −1 (ue 1 +p 6y) =
0
+∞
e
−t
uet +p
et −1
0 y+p ln( u+p )
+ I(u6y)
0
e
−t
et − 1 B uet + p − z; y dF(z) dt
+∞
uet +p
et −1
dF(z) dt:
By using the substitution s = uet + p(et − 1)=, we obtain B (u; y) = (p + u)
=
u
+∞
(p + s)
+ (p + u)= I(u6y)
u
y
−=−1
0
s
B (s − z; y) dF(z) ds
(s + p)−=−1 (1 − F(s)) ds:
Assume that F is continuous, diDerentiation of the above expression gives u @B (u; y) B (u − z; y) dF(z) − = B (u; y) − I(u6y) (1 − F(u)): @u p + u p + u 0 p + u For u 6 y, integrating both sides of (3) from 0 to u, we have u p 1 B (u; y) = B (u − z; y)[ + (1 − F(z))] d z B (0; y) + p + u p + u 0 u − (1 − F(v)) dv: p + u 0 From (4), we obtain (2) in the case where u 6 y.
(3)
(4)
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H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
For u ¿ y, integrating both sides of (3) from y to u, we have u B (v; y) dv (p + u)B (u; y) − (p + y)B (y; y) − =
u
y
B (v; y) dv −
u
y
y
v
0
B (v − z; y) dF(z) dv;
the second term on the right-hand side of the above equation equals u v B (v − z; y) dF(z) dv y
=
0
y
=
u
u
0
B (v − z; y)F(z)|vz=0 dv + B (u − z; y)F(z) d z +
0
y
y
0
(p + u)B (u; y) − (p + y)B (y; y) − y
= so
u
u
0
B (v; y) dv −
0
u
v
@B (v − z; y) F(z) d z dv @v
[ − B (y − z; y)]F(z) d z;
so we have
=
u
u
y
B (v; y) dv
B (u − z; y)F(z) d z +
(B (u − z; y)(1 − F(z))) d z −
0
y
0
y
B (y − z; y)F(z) d z
B (y − z; y)(1 − F(z)) d z;
u p + y 1 B (u; y) = B (u − z; y)[ + (1 − F(z))] d z B (y; y) + p + u p + u 0 y 1 − B (y − z; y)[ + (1 − F(z))] d z: p + u 0
(5)
From the de2nition of B (u; y), we know that B (u; y) is a continuous function of u ¿ 0 for a given value of y. Let u = y in (4), we then have y p 1 B (y; y) = B (y − z; y)( + (1 − F(z))) d z B (0; y) + p + y p + y 0 y − (1 − F(v)) dv: p + y 0 Plugging B (y; y) into (5), B (u; y) =
u p 1 B (u − z; y)( + (1 − F(z))) d z B (0; y) + p + u p + u 0 y − (1 − F(v)) dv: p + u 0
This is (2) for u ¿ y.
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H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
333
3.2. The case = 0 Let F1 be the equilibrium distribution of F, given by 1 x (1 − F(v)) dv: F1 (x) = 0 Denote the moments of the equilibrium distribution by 'k = From Sundt and Teugels (1995), we know that 1 k+1 'k = : k +1
+∞ 0
xk dF1 (x).
Denote the Laplace transformation of F1 by +∞ ((s) = e−su dF1 (u) = (0 (s); 0
(y (s) =
+∞
y
e−su dF1 (u):
We have the following theorem. Theorem 2. When the adjustment coe
u→∞
where F ∗ (x) =
0
x
F ∗ (y) − (=p)F1 (y) ; (=p)(−( (−R) − p=)
(8)
Rz e (1 − F(z)) d z: p
Proof. In the case of u 6 y, limu→+∞ eRu F(u; y) = limu→∞ eRu (u). From Grandell (1991), we have 1 − =p (u) ∼ e−Ru : (=p)(−( (−R) − p=) This proves that Eq. (7) is true. For u ¿ y, from Dickson (1992), we know that 1 − G(0; y) F(u; y) = G(u − y; y) − ( (u − y) − (u)) 1 − (0)
for u ¿ y;
where G(u; y) denotes the distribution of the surplus at ruin. So 1 − G(0; y) Ru ( (u − y) − (u)) eRu : lim e F(u; y) = lim G(u − y; y) − u→∞ u→∞ 1 − (0) Using the asymptotic result for G(u; y), obtained in Yang and Zhang (2001), and the result of Grandell (1991), we obtain (8). Remark. We can also prove this theorem by directly using Eq. (4).
(9) (u), in
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H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
3.3. Solution of the integral equation Following Sundt and Teugels (1995), we introduce the following auxiliary function: B (u; y) − B (0; y) A (u; y) = : 0 − B (0; y) Eq. (2), in terms of A (u; y) and F1 (x), can be written as u u pA (u; y) + uA (u; y) = A (v; y) dv + A ∗ F1 (u; y) − (1 − F(z)) d z 0
+ Let
+1 (s; y) =
+∞
0
B (0; y)
I{u¿y}
0
0
y
(1 − F(v)) dv + (1 − I{u¿y} )
0
u
(1 − F(v)) dv : (10)
e−sv dA (v; y);
where y is regarded as a parameter. The 2rst order diDerential equation for the function, +1 (s; y), can be obtained by taking the Laplace transform of (10) respect to u: −+1 (s; y) + (p − ((s))+1 (s; y) = (((s) − (y (s)) − ((s); (11) B (0; y) the initial condition of the above equation is given by lims→+∞ +1 (s; y) = 0. Similar to Sundt and Teugels (1995), the solution of equation (11) is given by +∞ −1 −{,(v)−,(s)}= (((v) − (y (v)) +1 (s; y) = − ((v) dv e B (0; y) s = where
,(s) :=
s
0
+∞
0
e−
w 0
[p−((s+z)] d z
((s + w) − (y (s + w) − ((s + w) dw; B (0; y)
(12)
(p − ((v)) dv:
By letting s = 0 we have
−1 +∞ +∞ −,(v)= −,(v)= B (0; y) = e ((v) dv e (((v) − (y (v)) dv + 0 0
=
1 + +∞
0
+∞
0
e−
w 0
e−
w 0
[p−((z)] d z
[p−((z)] d z
−1 ((w) dw
(((w) − (y (w)) dw
From Sundt and Teugels (1995), we know
+∞ −1 z − 0 (p−((w)) dw e dz : (0) = 1 − p 0
:
(13)
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
From Yang and Zhang (2001), we have
+∞ −1 z − 0 (p−((w)) dw G (0; y) = p e dz 0
+∞ z
e
0
(p−((w)) dw 0
(((z) − e
335
zy
(y (z)) d z ;
so, when = 0, F(0; y) = G(0; y), but for ¿ 0, G (0; y) = B (0; y). Hence, the reLection principle is not valid when = 0. In addition, the relationships among G(u; y), F(u; y) and (u) in Dickson (1992), Willmot and Lin (1998) are not valid when = 0. 4. B (u; y) with zero initial reserve The main result of this section is the following theorem. Theorem 3.
−1 -2 '11 (y) 1 1 1 1 + F1 (y) + -'11 (y) R(-) − e 2 p − '1 '1 '1
6 B (0; y) 6
1 + '1
-2
e2 1 R(-) − +
−1
-2
where R(-) is the well-known Mills’ ratio given by R(-) = e 2 y and '11 (y) = 0 z dF1 (z). Proof. Let A1 :=
+∞
0
e−
w 0
(p−((z)) d z
((w) dw =
+∞
0
e
w 0
F1 (y) ; p − +∞ -
(((z)−p) d z
(14)
x2 e− 2 d x; and - = ( − p)=( '1 )
((w) dw:
(15)
Using the property that ((s) is a strictly decreasing convex function, we obtain +∞ +∞ w w 1 e 0 ((1−'1 z)−p) d z (1 − '1 w) dw 6 A1 6 e 0 (−p) d z dw = : p − 0 0 Using a similar method as to that in Sundt and Teugels (1995) (or Yang and Zhang, 2001), we have the following inequality for A1 : -2 e2 1 1 R(-) − + 6 A1 6 : (16) '1 p − Write
A2 :=
+∞
0
e−
w 0
(p−((z)) d z
(((w) − (y (w)) dw:
Again, using ((s), (y (s), and ((s) − (y (s) are strictly decreasing convex functions, we have
+∞ y w z−p) d z (−' 1 A2 ¿ F1 (y) − w e0 u dF1 (u) dw 0
=
0
0
+∞
e
−
2
(w−(p−)='1 ) (p−) + 2' 2(1='1 ) 1
2
(F1 (y) − w'11 ) dw
(17)
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H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
and
A2 6
+∞
e
0
2 0
(−p) d z
F1 (y) dw =
F1 (y) : p −
So we have the following inequality for A2 : -2 '11 (y) 1 1 F1 (y) F1 (y) + -'11 (y) : R(-) − e 2 6 A2 6 '1 '1 '1 p −
(18)
Notice that (13) can be rearranged as −1
1 B (0; y) = A1 + A2 : From this, the theorem is proved. Similar to Sundt and Teugels (1995), we can expand B (0; y) as a McLaurin series with respect to the variable, . We have B (0; y) =
−'1 F1 (y) + + o(): p p(p − )
(19)
5. Exponential claim size In this section, we will try to evaluate expression (13) in the case of an exponential claim size. First, we have −1 +∞ B (0; y) = e−(pv− log(1+v))= (1 + v)−1 dv + 0 ×
+∞
= + then −1
+1 (s; y) =
−1
=
e
0
+∞
e
0
+∞
s
−(pv− log(1+v))=
e
(1 + v)
(=)−1
(1 + v)
s
s
e
p − (v−s)
+∞
dv
−1
(1 + v)
p e− s (v−s) (1
(1 − e
−1
+∞
e
− p v
−1
(1 + v)
1 + v 1 + s
−1
+ v)
−(v+1=)y
0
−[p(v−s)− log(1+v)=(1+s)]=
+∞
1 − B (0; y)
− p v
−1
(=)
1 + v 1 + s
) dv
(=)−1
(1 + v)
(1 − e
−(v+1=)y
1 − e−(v+1=)y − 1 dv B (0; y)
1 − 1 dv B (0; y)
(=) e
−vy −y=
e
dv
) dv ; (20)
H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
=
+∞
0
+
p
e
−sx −x=
e
+∞
0
e
1+ x p
(=)−1
p
y x+y y −s(x+ p x+y) −( + p x) e
337
1 −1 B (0; y)
1+
x p
dx
(=)−1 dx
1 : B (0; y)
(21)
Using the substitution x+
y x + y = t; p
we have +1 (s; y) =
+∞
0
−
e−st e
+∞
y
t −
e−st e
1+ t −
x p
1+
(=)−1
(t − y) p + y
p
1 −1 B (0; y)
(=)−1
dt
1 dt: p + y B (0; y)
(22)
Remember that +1 (s; y) was introduced as a Laplace transformation of A (u; y) respect to u. Hence, by inverting, we 2nd that (=)−1
u t 1 − − 1 dt 1+ x e A (u; y) = p p B (0; y) 0 −
y
u
e
t −
(t − y) 1+ p + y
(=)−1
1 I{u¿y} dt; p + y B (0; y)
(23)
where I{u¿y} is the indicator function.
6. Lundberg bound Following the arguments in Sundt and Teugels (1995), we are also able to obtain the Lundberg-type inequality. Same as in Sundt and Teugels (1995), we assume that ((s) has −. as an abscissa of convergence. So for any u ¿ 0 and for s ∈ I := (−.; 0), we have +∞ +1 (s; y) ¿ e−sx dA (x; y) ¿ e−su {1 − A (u; y)}; u
so we obtain the ChernoD-type estimate 1 − A (u; y) 6 exp inf {su + log +1 (s; y)} : s∈I
(24)
Let s (u; y) denote the solution of the following equation: u+1 (s; y) = − +1 (s; y); and we call |s (u; y)| the adjustment function.
(25)
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H. Yang, L. Zhang / Statistics & Probability Letters 55 (2001) 329 – 338
Similar to Sundt and Teugels (1995), write ([=B (u; y)]F1 (y) − p) u = ; we have the following result:
(26)
Theorem 4. When the initial surplus u satis?es u ¿ u ; then 1−A (u; y) satis?es a Lundberg-type inequality: (=B (0; y))[((s (u; y)) − (y (s (u; y))] − ((s (u; y)) us (u; y) e : 1 − A (u; y) 6 u + p − ((s (u; y)) Acknowledgements The work described in this paper was partially supported by a grant from the Research Grants Council of the Hong Kong Special Administrative Region, China (Project No. HKU 7168=98H). References Dickson, D.C.M., 1992. On the distribution of the surplus prior to ruin. Insurance Math. Econom. 11, 191–207. Dufresne, F., Gerber, H.U., 1988. The surpluses immediately before and at ruin, and the amount of the claim causing ruin. Insurance Math. Econom. 7, 193–199. Gerber, H.U., Goovaerts, M.J., Kaas, R., 1987. On the probability and severity of ruin. ASTIN Bull. 17, 151–163. Grandell, J., 1991. Aspects of Risk Theory. Springer, New York. Sundt, B., Teugels, J.L., 1995. Ruin estimates under interest force. Insurance Math. Econom. 16, 7–22. Willmot, G.E., Lin, X., 1998. Exact and approximate properties of the distribution of surplus before and after ruin. Insurance Math. Econom. 23, 91–110. Yang, H., 1999. Non-exponential bounds for ruin probability with interest eDect included. Scand. Actuar. J. 99, 66–79. Yang, H., Zhang, L., 2001. On the distribution of surplus immediately after ruin under interest force. Insurance: Mathematics and Economics 29, 247–255.