Some distributions for classical risk process that is
perturbed by diffusion
qGuojing Wang
a,b,∗, Rong Wu
baDepartment of Physics, Hebei University, Baoding 071002, China bDepartment of Mathematics, Nankai University, Tianjin 300071, China
Received December 1998; received in revised form July 1999
Abstract
In this paper we discuss the classical risk process that is perturbed by diffusion. We prove some properties of the supremum distribution of the risk process before ruin when ruin occurs and the surplus distribution at the time of ruin. We present the simple and explicit expression for these distributions when the claims are exponentially distributed. ©2000 Published by Elsevier Science B.V. All rights reserved.
Keywords: Risk process; Ruin probability; Supremum distribution; Surplus distribution at the time of ruin; Integro-differential equation
1. Introduction
Let(,J, P )be a complete probability space containing all objects defined in the following. We consider the classical risk process that is perturbed by diffusion
Rt =u+ct+σ Wt −
Nt
X
k=1
Zk, (1.1)
whereudenotes the initial capital,cthe premium income,{Wt :t≥0}is a standard Brownian motion,{Nt :t ≥0}
is a Poisson process with parameterλ >0, it counts the number of the claims in the interval(0, t],{Zk, k ≥1}is
a non-negative sequence of i.i.d. random variables,Zk denotes the amount of thekth claim.Rt is the surplus of an
insurance company at time t. We assume throughout the paper that{Nt},{Wt}and{Zk}are independent. Denote
the distribution function of the claims byF and their mean value byµ. LetT1, T2, . . . be the occurrence times of
the claims and setT0=0.
The model (1.1) is first introduced by Gerber (1970) and further studied by many authors during the last few years, such as Dufresne and Gerber (1991), Veraverbeke (1993), Furrer and Schmidli (1994), Schmidli (1995), Zhang
q
Supported by NNSF (grant no. 16971047) and DSF in China.
(1997) and so on. Many results on ruin probability and other ruin problems have been obtained by above-mentioned works. The purpose of this paper will be to discuss the supremum distribution of the risk process before ruin when ruin occurs and the surplus distribution at the same time of ruin. We first prove some properties of these two distributions and then give some special examples for those distributions we consider. The ideas and methods we used in this paper are motivated by Grandell (1991), Paulsen (1993) and Paulsen and Gjessing (1997), can be used to the risk model with return on investment, see Wang and Wu (1998).
In Section 2, we introduce the supremum distribution of the risk process before ruin when ruin occurs. We first derive the integral equation satisfied by it. Then by using the integral equation we prove its differentiability. Finally we give the integro-differential equation satisfied by it. In the last part of this section we define the supremum distribution of the risk process before ruin.
In Section 3, we discuss the surplus distribution at the time of ruin. We also derive the integral equation and the integro-differential equation satisfied by it. The contents in this section are parallel those in Section 2.
In Section 4, we get the simple and explicit expression for the ruin probability, the supremum distribution of the risk process before ruin when ruin occurs and the surplus distribution at the time of ruin by solving boundary value problems with same differential equation of third order but with different boundary values with each other when the claims are exponentially distributed. For general expressions of these distributions, we can refer to Dufresne and Gerber (1991 ) for the series expression of ruin probability and to Zhang (1997) for the series expression of the surplus distribution at the time of ruin.
2. Supremum distribution before ruin
LetTu = inf{t ≥ 0 : Rt < 0}andTu = +∞ if Rt ≥ 0 for all t ≥ 0. ThenTu is the time of ruin. The
Fort ≥0 letθt be the shift operators fromto itself defined byRs(θtω)=Rs+t(ω). For finite stopping time T,
we define the mapθT fromto itself byθT(ω)=θt(ω)ifT (ω)=t(see Revuz and Yor (1991), p. 34 and p. 97).
Clearly, we haveRt◦θT =Rt+T.
Theorem 2.1. Letx > u >0,thenG(u, x)satisfies the following integral equation:
G(u, x)=exp{−λt0}
We now compute the expectation on the right-hand side of equality (2.6).
E[G(RT, x)]=EG Rt0, x
By assumption of independence we have
I1=EG u+ct0+σ Wt0, x
By Proposition 2.8.3 in Port and Stone (1978) we get
P Wτa =a, τa∈dt
=P Wτa = −a, τa∈dt
By equality (2.9) we get
Formula (2.5) now follows from equalities (2.6),(2.7) and (2.8) and (2.10) and (2.11).
Theorem 2.2. SupposeF (z) has continuous density function on [0,+∞). ThenG(u, x) is twice continuously differentiable inuin interval(0, x).
Proof. For arbitraryε0 > 0, it is sufficient to show thatG(u, x)is twice continuously differentiable in u in
in-terval (ε0, x−ε0). If we set t0 < (1/2c)ε0 and a < (1/2σ )ε0 (note that t0 and a do not depend on u), we
still have formula (2.5). By changing the variable of integration, we can moveuinG’s in the integrand on the right-hand side of equality (2.5) into H or h. Then by the fine properties H and h we can verify that the theorem
holds.
Therefore, by Itô formula, we get
Lettingt →0 gives formula (2.12).
We now consider the distributionŴ(u, x)defined byŴ(u, x)=P (sup0≤t <T
Indeed, since Rt is a process with stationary and independent increments and E[Rt] = (c−λu)t > 0, thus
limt↑+∞Rt = +∞ P-a.e. Therefore,
Ŵ(u, x)is the probability that the supremum value of the risk process before ruin reaches or surpasses the level x. We call it the supremum distribution of the risk process before ruin.
3. Surplus distribution at the time of ruin
We consider the surplus distribution at the time of ruinD(u, y)defined by
D(u, y)=P Tu<+∞, RTu≥ −y
, y >0. (3.1)
D(u, y)=I[0,y](−u), u≤0,
D(+∞, y)=0. (3.2)
Theorem 3.1. Letu >0, thenD(u, y)satisfies the following integral equation
D(u, y)=1
By equality (3.4), we get
D(u, y)=EhDRτu/σ∧T1, y
By equality (2.9) we get
I10=E[D(u+cτu/σ+σ Wτu/σ, y)I (τu/σ < T1)]=
Similar to (2.11) we have
I20=E
Formula (3.3) now follows from equalities (3.4), (3.5), (3.6) and (3.7). Corresponding to Theorem 2.2, we have the following Theorem 3.2.
Theorem 3.3. Letu > 0, supposeF (z)has continuous density function on [0,+∞). ThenD(u, x)satisfies the
Similar to (2.15) we have
1−exp{−λt}
This section will give the explicit expressions for the ruin probability, the supremum distribution of the risk process before ruin and the surplus distribution at the time of ruin when the claims are exponentially distributed. Corresponding to Lemma 4.2 and Propositions 2.2 and 2.3 in Wang and Wu (1998) we have the following Lemma 4.1 and Proposition 4.1 and 4.2, respectively:
Lemma 4.1. P (T0+=T0=0)=1.
Proposition 4.1. SupposeF (z)has continuous density function on [0,+∞), then8(u)is continuous on [0,+∞).
Proposition 4.2. SupposeF (z)has continuous density function on [0,+∞), thenG(u, x)is continuous on [0, x].
Proposition 4.3. Lety >0. SupposeF (z)has continuous density function on [0,+∞), thenD(u, y)is continuous inuon [0,+∞).
Proof. By Theorem 3.2, it is sufficient to show thatD(0+, y)=D(0, y). By Lemma 4.1, we have limn
→+∞I (T1/n< +∞)=1 P — a.e. By Lemma 4.1 and the law of iterated logarithm for Brownian motion we get limn→+∞RT1/n≥
−y P-a.e., i.e., limn→+∞I (RT1/n ≥ −y)=1 P-a.e.
Thus by Fatou Lemma we get
1≥ lim
Following the proof of (2.12) we can prove that formula (2.1) in Dufresne and Gerber (1991 ) holds. Therefore, it follows from Proposition 4.1 that8(u)is the bounded continuous solution of the following boundary value problem:
(1/2)σ28′′u(u)+c8′u(u)=λ8(u)−λR0u8(u−z)dF (z), u >0,
8(0)=0 8(+∞)=limu↑+∞8(u)=1.
(I)
From Theorem 2.3, boundary conditions (2.2) and Proposition 4.2, it follows thatG(u, x)is the bounded continuous solution of the following boundary value problem
(1/2)σ2G′′u(u, x)+cG′u(u, x)=λG(u, x)−λRu
0 G(u−z, x)dF (z), 0< u < x,
G(0, x)=0, x >0 G(x, x)=9(x), (II)
and from Theorem 3.3, boundary conditions (3.2) and Proposition 4.3 it follows that D(u, y) is the bounded continuous solution of the following boundary value problem
(1/2)σ2Du′′(u, y)+cDu′(u, y)=λD(u, y)−λR0uD(u−z, y)dF (z)−λRuu+y dF (z), u >0,
D(0, y)=1, D(+∞, y)=limu↑+∞D(u, y)=0.
(III)
Remark 4.1. If the bounded continuous solution of (II) is unique, then by comparing the boundary value problems
(I) and (II) we get
G(u, x)= 8(u)
8(x)9(x). (4.1)
In fact, we have the following proposition.
Proposition 4.4. IfF (z)has continuous density function on [0,+∞), then (4.1) holds.
Proof. Note that8(u)andG(u, x)solves (I) and (II), respectively. LetTux,ε=inf{t :Rt ∈/(ε, x−ε)}. Assumef (u)
solves (I), thenf (RTx,ε
u ,∧t)is a martingale. This impliesf (u)=E[f (RTux,ε∧t)]. Lettingt → +∞showsf (u)= E[f (RTx,ε
u )]. Lettingε → 0 givesf (u) = E[f (RTux,0∧t)] = f (x)P (RTux,0 = x). This givesP (RTux,0 = x) =
Ŵ(u, x) = f (u)/f (x). Lettingx → +∞showsf (u) = 8(u). Letf (u, x)solve (II), thenf (RTx,ε
u ∧t, x) is a
martingale. We havef (u, x) = E[f (RTx,ε
u ∧t, x)]. Lettingt → +∞showsf (u, x) = E[f (RTux,ε, x)]. Letting ε→0 givesf (u, x)=E[f (R
Tux,0, x)]=9(x)P (RTux,0 =x)=9(x)8(u)/8(x). This ends the proof. Remark 4.2. In the proof of Proposition 4.4 we get the formulaŴ(u, x)=8(u)/8(x). It can also been proved by
a different “martingale approach” (see (Zhang, 1997)).
WhenF (z)=1−exp{−αt}α >0, the boundary value problems (I),(II) and (III) can be reduced to the boundary value problems (Ia),(IIa) and (IIIa) as follows respectively:
(1/2)σ28′′′(u)+ (1/2)ασ2 +c
8′′(u)+(αc−λ)8′(u)=0, u >0,
8(0)=0, 8(+∞)=limu↑+∞8(u)=1, (1/2)σ28′′(0+)+c8′(0)=0,
. (Ia)
(1/2)σ2G′′u′(u, x)+ (1/2)ασ2+cG′′u(u, x)+(αc−λ)G′u(u, x)=0,0< u < x,
G(0)=0, G(x, x)=9(x), (1/2)σ2G′′u(0+, x)+cG′u(0, x)=0, (IIa)
and
(1/2)σ2Du′′′(u, y)+ (1/2)ασ2+cD′′u(u, y)+(αc−λ)Du′(u, y)=0, u >0, D(0, y)=1, D(+∞, y)=limu↑+∞D(u, y)=0, (1/2)σ2Du′′(0+, y)+cDu′(0+, y) =λexp{−αy}
(IIIa)
and the solutions of (Ia),(IIa) and (IIIa) are all unique. Note that
1. The solution of boundary value problem (Ia) is
2. The solution of the boundary value problem (IIIa) is
D(u, y)=c3exp{−λ1u} +c4exp{−λ2u} u≥0 (4.8)
3. From equality (4.1) we get that the solution of (IIa) is
G(u, x)=8(u)
8(x)9(x)=
1−c1exp{−λ1u} −c2exp{−λ2u}
1−c1exp{−λ1x} −c2exp{−λ2x}
(c1exp{−λ1x} +c2exp{−λ2x}). (4.11)
4. From Remark 4.2 and formula (4.3) we get
Ŵ(u, x)= 1−c1exp{−λ1u} −c2exp{−λ2u}
1−c1exp{−λ1x} −c2exp{−λ2x}
. (4.12)
Acknowledgements
References
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