Directory UMM :Data Elmu:jurnal:I:Insurance Mathematics And Economics:Vol28.Issue1.2001:

A decomposition of the ruin probability for the

risk process perturbed by diffusion

q

Guojing Wang

Department of Mathematics, Suzhou University, Suzhou 215006, People’s Republic of China Received 1 November 1999; received in revised form 1 July 2000; accepted 19 September 2000

Abstract

In this paper, we consider the ruin probabilities (caused by oscillation or by a claim) of the classical risk process perturbed by diffusion and the risk process with return on investments. We will prove their twice continuous differentiability and derive the integro-differential equations satisfied by them. We will present the explicit expressions for them when the claims are exponentially distributed. © 2001 Elsevier Science B.V. All rights reserved.

Keywords: Risk process; Ruin probability; Integro-differential equation

1. Introduction

Let(Ω, F, P )be a complete probability space containing all the random variables we meet in the following. The classical risk process perturbed by diffusion is

R_t0₌u₊ct₊σ W_t0₋

Nt X

k=1

Zk. (1.1)

In (1.1),u, candσ are all positive constants,udenotes the initial capital of an insurance company andcthe rate of premium income;_{Nt}is a Poisson process with parameterλ > 0, it counts the total numbers of the claims in the interval(0, t];_{Zk},k≥ 1 is a non-negative sequence of i.i.d. random variables,Zk the amount of thekth claim,_{W_t0_}the standard Brownian motion, it stands for the uncertainty associated with the income of the insurance company at timet,R0_t the surplus of the insurance company at timet,_{Nt},{Zk}and{Wt0}are mutually independent. The process (1.1) is a homogenous strong Markov process.

We now follow the steps in Paulsen and Gjessing (1997) to introduce the risk process with deterministic return on investments. Our risk process is, in fact, a special case of (2.4) in Paulsen and Gjessing (1997).

The basic process is the surplus generating process (1.1). The deterministic return on investments generating process is

It =rt, (1.2)

q

Supported by NNSF (Grant No. 19971047) in China and Doctoral Foundation of Suzhou University. E-mail address: rgwang@suda.edu.cn (G. Wang).

0167-6687/01/$ – see front matter © 2001 Elsevier Science B.V. All rights reserved. PII: S 0 1 6 7 - 6 6 8 7 ( 0 0 ) 0 0 0 6 5 - 2

wherer is a constant, it stands for a constant rate of return on investments. The risk process with deterministic return on investments is then the solution of the linear stochastic differential equation

Rt =R0t +

Z t

0

Rs−dIs. (1.3)

By Paulsen and Gjessing (1997), the solution of (1.3) is Rt =exp{rt}

u₊

Z t

0

exp_{−rs_}dR_s0

, (1.4)

and (1.4) is also a homogenous strong Markov process. Relation (1.4) can be rewritten as

Rt =exp{rt}



u+

c

r(1−exp{−rt})+σ

Z t

0

exp_{−rs_}dW_s0₋

Nt X

k+1

exp_{−rTk}Zk



, (1.5)

where_{Tk},k≥1, is the sequence of jump times of{Nt}.

Dufresne and Gerber (1991) decompose the ruin probability of the risk process (1.1) into two parts: the ruin probability caused by oscillation and the ruin probability caused by a claim. They obtain explicit expressions for these two different kinds of ruin probability in the form of series, making the assumption that the ruin probabilities are twice differentiable. Motivated by this idea in Dufresne and Gerber (1991), we will decompose the ruin probability of the process in (1.4) correspondingly into two parts. We will prove the twice continuous differentiability of these two different kinds of ruin probability and present their explicit expressions when the claims are exponentially distributed. Note that the twice continuous differentiability ensures that the solutions of the integro-differential equations satisfied by the ruin probabilities respectively exist.

2. The classical risk process perturbed by diffusion

Leta >0, defineτa =inf{s:|Ws0| =a}. Forx ∈[−a, a], put

H (a, t, x)₌(2π t )−1/2

+∞

X

k=−∞

exp

−1

2t(x+4ka)

2

−exp

−1

2t(x−2a+4ka)

2_{, (2.1)}

h(a, t )₌ 1 2√2πat

−3/2

+∞

X

k=−∞

(4k₊1)exp

−a

2

2t(4k+1)

2

+(4k₋3)exp

−a

2

2t(4k−3)

2

−(4k₋1)exp

−a

2

2t(4k−1)

2_{. (2.2)}

It follows from Revuz and Yor (1991, pp. 105–106) thatP (W_s0_∈dx, τa> s)=H (a, s, x)dxandP (τa ∈ds)=

h(a, s)ds.

LetT_u0₌inf_{t _≥0_;R_t0<0_}and definingT_u0 _{= +∞}ifR_t0_≥0 for allt _≥ 0.T_u0is the time of ruin. The ruin probabilityΨ0(u)of the risk process (1.1) is defined byΨ0(u)₌P (inft≥0R0t <0). We haveΨ0(u)=P (Tu0<

+∞). Letµ ₌E[Z1] andF (z) =P (Z1≤z). Foru >0 we assume thatE[R0t]=u+(c−λµ)t >0, i.e., we assumec₋λµ >0, so that we haveΨ0(u) <1.

Denote byΨ_d0(u)the ruin probability caused by oscillation andΨ_s0(u)the ruin probability caused by a claim. We have (see (1.5) in Dufresne and Gerber (1991))

IfF (z)has continuous density function on [0,_+∞), thenP (_∪+∞_k=1{R0_T

k =0})=0, and therefore

Ψ_d0(u)₌P (T_u0<_+∞, R_T00

u =0), (2.4)

Ψ_s0(u)₌P (T_u0<_+∞, R_T00

u <0). (2.5)

From (2.4) and (2.5) we see that Ψ_d0(u)₌

0, u <0,

1, u₌0, (2.6)

Ψ_s0(u)₌

1, u <0,

0, u₌0. (2.7)

Theorem 2.1. Letu >0, supposeF (z)has continuous density function on [0,_+∞), then the probabilityΨ_d0(u)

satisfies the following integral equation:

Ψ_d0(u)₌1 2

Z _+∞

0

(Ψ_d0(ct)₊Ψ_d0(2u₊ct))exp_{−λt_}hu σ, t

dt

+

Z _+∞

0

λexp_{−λs_}ds

Z u/σ

−u/σ

Hu σ, s, x

dx

Z u+cs+σ x 0

Ψ_d0(u₊cs₊σ x₋z)dF (z). (2.8) Proof. Denote byAdthe event of ruin due to oscillation. LetFt =σ{Rs0, s≤t}. DefineMtbyMt =E[I (Ad)|Ft]. Note that_{Mt, t ≥ 0}is aFt-martingale. PutT =τu/σ∧T1, we haveP (T < +∞)≤ P (T1 <+∞)=1. By optional stopping theorem together with the homogeneous strong Markov process ofR_t0, we get

Ψ_d0(u)₌EM0=E[MT]=E[E[I (Ad)|FT]]=E[Ψd0(R0T)]. (2.9) Therefore

Ψ_d0(u)₌E[Ψ_d0(R_T0)]₌E[Ψ_d0(u₊cτu/σ+σ Wτ0u/σ)I (τu/σ < T1)]

+E[Ψ_d0(u₊cT1+σ WT01 −Z1)I (τu/σ ≥T1)]=I1+I2. (2.10)

Exactly similar to the calculation ofI₁0andI₂0in Theorem 3.1 in Wang and Wu (2000) we get I1=

1 2

Z +∞

0

(Ψ_d0(ct)₊Ψ_d0(2u₊ct))exp_{−λt_}hu σ, t

dt, (2.11)

I2=

Z _+∞

0

λexp_{−λs_}ds

Z u/σ

−u/σ

Hu σ, s, x

dx

Z u+cs+σ x 0

Ψ_d0(u₊cs₊σ x₋z)dF (z), (2.12) respectively. Formula (2.8) now follows from (2.9)–(2.12). The proof is completed. h

Remark 2.1. The result of Theorem 2.1 can be interpreted and proved by a heuristic argument. To see this, we denote byAu_dthe event_{ruin occurs due to oscillation_|R0₀₌u_}. Since fort _∈(0, T ),R_t0>0, i.e. no ruin occurs up to timeT, conditioned on_{τu/σ =s < T1}, we conclude thatAu_doccurs if and only ifAu_d+cs−uorAu_d+cs+uoccurs. The probabilities ofAu_d+cs−uandA_du+cs+uareΨ_d0(cs)andΨ_d0(2u₊cs), respectively. Note thatP (W_s0_{= −}u/σ )₌ P (W_s0₌u/σ )₌ 1₂, summing all overswe get the equality (2.11).

Suppose that the first jump of the sample function of the process R_t0 occurs at timet and has magnitude z. Since no ruin occurs up to timeT, conditioned on_{T1=t < τu/σ, Wt0=x}, we see thatAudoccurs if and only if

z_≤u₊ct₊σ xandAu_d+ct+σ x−zoccurs. The probability of the eventA_du+ct+σ x−zisΨ_d0(u₊cs₊σ x₋z). Summing all over(t, x, z)_∈(0,_+∞)[₋u/σ, u/σ](0, u₊ct₊σ x₋z] we get the equality (2.12). Therefore, Theorem 2.1 holds. Theorem 2.2. SupposeF (z)has continuous density function on [0,_+∞), thenΨ_d0(u)is twice continuously differ-entiable in the interval(0,_+∞).

Proof. For any ε0 > 0 it is sufficient to show that Ψd0(u) is twice continuously differentiable in the interval

(ε0,+∞). Replacingτu/σ byτε0/σ in the proof of Theorem 2.1, then for anyu∈(ε0,+∞)we have

Ψ_d0(u)₌1 2

Z +∞

0

(Ψ_d0(u₋ε0+ct)+Ψ_d0(u+ε0+ct))exp{−λt}h

ε₀

σ, t

dt

+

Z +∞

0

λexp_{−λs_}ds

Z ε0/σ −ε0/σ

Hε0 σ, s, x

dx

Z u+cs+σ x

0

Ψ_d0(u₊cs₊σ x₋z)dF (z). (2.13) By changing the variable of integration, we can moveuin theΨ_d0(u)in the integrands on the right-hand side of (2.13) intoHorh. Note that the density functionh(a, t )is twice continuously differentiable intand, forx ₆₌0,H (a, t, x) is twice continuously differentiable inxand satisfies limt↓0Hx′(a, t, x)=limt↓0Hx′′(a, t, x)=0. Therefore, by the dominated convergence theorem and equality (2.13) we can verify thatΨ_d0(u)is twice continuously differentiable

in the interval(ε0,+∞). h

Theorem 2.3. Letu > 0, Suppose F (z)has continuous density function on [0,_+∞), thenΨ_d0(u) satisfies the following integro-differential equation:

1 2σ

2_Ψ0′′

d (u)+cΨ 0′

d (u)=λΨ 0 d(u)−λ

Z u

0

Ψ_d0(u₋z)dF (z). (2.14)

Proof. The proof is exactly similar to that of Theorem 3.3 in Wang and Wu (2000). h

Similar to Theorems 2.2 and 2.3, we have the following theorem.

Theorem 2.4. SupposeF (z)has continuous density function on [0,_+∞), thenΨ_s0(u)is twice continuously differ-entiable in the interval(0,_+∞).

Theorem 2.5. Letu > 0, supposeF (z)has continuous density function on [0,_+∞), thenΨ_s0(u) satisfies the integro-differential equation

1 2σ

2_Ψ0′′

s (u)+cΨ0 ′

s (u)=λΨs0(u)−λ

Z u

0

Ψ_s0(u₋z)dF (z)₋λ(1₋F (u)). (2.15) Proposition 2.1. SupposeF (z)has continuous density function on [0,_+∞), thenΨ_d0(u)is continuous on [0,_+∞). Proof. By Theorem 2.2, it is sufficient to show thatΨ_d0(0+)₌Ψ_d0(0). SinceP (T0

0+ =T00=0)=1 (see Lemma 4.1 in Wang and Wu (2000)) andR_t0is right continuous, by the dominated convergence theorem we get

lim u_↓0Ψ

0

d(u)=lim u_↓0E[I (T

0

u <+∞, R0T0

u =0)]=E[limu_↓0I (T

0

u <+∞, RT00

u =0)]

=E[I (T₀0<_+∞, R0₀₌0)]₌1₌Ψ_d0(0).

Similar to Proposition 2.1, we have the following proposition.

Proposition 2.2. SupposeF (z)has continuous density function on [0,_+∞), thenΨ_s0(u)is continuous on [0,_+∞).

3. The case with deterministic return on investments

LetBt =σR₀texp{−rs}dWs0,Btis Itˇo’s stochastic integral, its variance process ishBit =(σ2/2r)(1−exp{−2rt}). Letv(s)₌inf_{t:_hB_it > s}, then

v(s)₌ 1 2rln

_σ2

σ2_−2rs

. (3.1)

Fors < σ2/2r, setWs =Bv(s), it is well known thatWs is a local Brownian motion starting from the origin and running for an amount of timeσ2/2r.

LetTu = inf{t ≥ 0 : Rt < 0}and definingTu = +∞ifRt ≥ 0 for allt ≥ 0. Denote byΨ (u) the ruin probability of the risk process (1.4), thenΨ (u) ₌P (Tu < +∞). Foru > 0 we assumeE[Rt] = exp{rt}[u+

(1/r)(1₋exp_{−rt_})(c₋λµ)]>0 (see Lemma 2.1 in Wang and Wu (1998)), i.e., we assumec₋λµ >0 so that we haveΨ (u) <1.

For the risk process (1.4) we denote byΨd(u)andΨs(u)the ruin probability caused by oscillation and the ruin probability caused by a claim, respectively. We still have

Ψ (u)₌Ψd(u)+Ψs(u). (3.2)

SupposeF (z)has continuous density function on [0,_+∞), similar to (2.4) and (2.5) we have

Ψd(u)=P (Tu<+∞, RTu =0), (3.3)

Ψs(u)=P (Tu<+∞, RTu<0). (3.4)

From (3.3) and (3.4) we get that Ψd(u)=

0, u <0,

1, u₌0, (3.5)

Ψs(u)=

1, u <0,

0, u₌0. (3.6)

Theorem 3.1. Letu > 0, supposeF (z)has continuous density function on [0,_+∞), then the probabilityΨd(u)

satisfies the following integral equation:

Ψd(u)=

Z _+∞

0

λexp_{−λs_}ds

Z u

−u

H (u, v−1(s), x)dx

×

Z exp{rs}[u+x+(c/r)(1−exp{−rs})]

0

Ψd

exp_{rs_}hu₊x₊c

r(1−exp{−rs})−zexp{−rs}

i

dF (z)

+1

2

Z σ2/2r 0

h

Ψd

exp_{rv(s)_}h2u₊c

r(1−exp{−rv(s)})

i

+Ψd

c

r(exp{rv(s)} −1)

i

Proof. Letτ_u0₌inf_{v(s):_|Bv(s)| =u},τu =inf{s:|Bv(s)| =u}, thenτu0=v(τu). PutT =T1∧τu0. Similar to formula (2.9) we have

Ψd(u)=E[Ψd(RT)]. (3.8)

Thus,

Ψd(u)=E[Ψd(RT)]=E[Ψd(RT1)I (T1≤τ

0

u)]+E[Ψd(Rτ0

u)I (T1> τ

0 u)]=I

1 1+I

1

2. (3.9)

We now present the expressions forI1andI2.

I₁1₌E

Ψd

exp_{rT1}

u₊

Z T1

0

exp_{−rs_}dR_s0

I (T1≤τu0)

=EhΨd

exp_{rT1}

h

u₊c

r(1−exp{−rT1})+BT1

i

−Z1

I (T1≤τu0)

i

,

and

I₂1₌E

"

Ψd exp{rτu0}

"

u₊

Z τu0

0

exp_{−rs_}dR_s0

#!

I (T1> τu0)

#

=EhΨd

exp_{rτ_u0_}hu₊c

r(1−exp{−rτ

0 u})+Bτ0

u i

I (T1> τu0)

i

.

Then exactly similar to the calculations ofI1andI2in Theorem 2.1 in Wang and Wu (1998), we can verify that

I₁1andI₂1are equal to the first term and the second term on the right-hand side of (3.7), respectively. The proof is

completed. h

Similar to Theorem 2.2, we have the following theorem.

Theorem 3.2. SupposeF (z)has continuous density function on [0,_+∞), thenΨd(u)is twice continuously

differ-entiable in(0,_+∞).

Theorem 3.3. Suppose F (z) has continuous density function on [0,_+∞), then Ψd(u) satisfies the following

integro-differential equation:

1 2σ

2_Ψ′′

d(u)+(ru+c)Ψd′(u)=λΨd(u)−λ

Z u

0

Ψd(u−z)dF (z), u >0. (3.10)

Proof. Letε, t, M >0 such thatε < u < MandT_tε,M ₌inf_{s >0 : exp_{rs_}[u₊(c/r)(1₋exp_{−rs_})₊Bs]∈/

(ε, M)_{} ∧}t. Note thatΨ_d′(u)andΨ_d′′(u)are bounded on [ε, M] and thereforeRS∧Ttε,M

0 Ψd′(exp{rv}[u+(c/r)(1− exp_{−rv_})₊Bv])exp{rv}dBvis a martingale. PutT =Ttε,M∧T1, similar to the equality (3.8) we have

Ψd(u)=E[Ψd(R_Tε,M

Therefore, by a standard use of Itˇo’s formula, we get Ψd(u)=exp{−λt}E

h

Ψd

exp_{rTε,M_{t }}hu₊c

r(1−exp{−rT

ε,M

t })+BT_tε,M

ii

+

Z t

0

λexp_{−λs_}nEhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M i

I (T_sε,M < s)i

+

Z +∞

0

EhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M i

−zI (T_sε,M ₌s)idF (z)

ds

=exp_{−λt_}

(

Ψd(u)+E

Z Tsε,M

0

Ψ_d′exp_{rs_}hu₊c

r(1−exp{−rs})+Bs

i

×exp_{rs_}hu₊c

r(1−exp{−rs})+c+rexp{rs}Bs

i +σ 2 2 Ψ ′′ d

exp_{rs_}hu₊c

r(1−exp{−rs})+Bs

i ds + Z t 0

λexp_{−λs_}nEhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M i

I (T_sε,M < s)i

+

Z +∞

0

EhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M i

−zI (T_sε,M ₌s)idF (z)

ds. Dividing bytgives

1₋exp_{−λt_}

t Ψd(u)=exp{−λt}

(

E

"

1 t

Z Tsε,M

0

Ψ_d′exp_{rs_}hu₊c

r(1−exp{−rs})+Bs

i

×exp_{rs_}hu₊c

r(1−exp{−rs})+c+rexp{rs}Bs

i +σ 2 2 Ψ ′′ d

exp_{rs_}hu₊c

r(1−exp{−rs})+Bs

i

ds

+1_t

Z t

0

λexp_{−λs_}

×nEhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M i

I (T_sε,M < s)i

+

Z +∞

0

EhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M s })

+B

Tsε,M i

−zI (T_sε,M ₌s)idF (z)ods.

Lettingt _→0 we get that (3.10) holds in(ε, M)and therefore in(0,_+∞). The proof is completed. h

Similar to Theorems 3.2 and 3.3 and Propositions 2.1 and 2.2, we have the following theorem.

Theorem 3.4. SupposeF (z)has continuous density function on [0,_+∞), thenΨs(u)is twice continuously

differ-entiable in(0,_+∞).

Theorem 3.5. Suppose F (z) has continuous density function on [0,_+∞), then Ψs(u) satisfies the following

integro-differential equation:

1 2σ

2_Ψ′′

s(u)+(ru+c)Ψs′(u)=λΨs(u)−λ

Z u

0

Ψs(u−z)dF (z)−λ(1−F (u)), u >0. (3.12)

Proposition 3.2. Suppose F (z) has continuous density function on [0,_+∞), then Ψs(u) is continuous on [0,_+∞).

4. Examples

This section will present the explicit expression for the ruin probability caused by oscillation and the ruin probability caused by a claim when the claims are exponentially distributed.

SupposeF (z)has continuous density function on [0,_+∞), from Sections 2 and 3, we can derive thatΨ_d0(u), Ψ_s0(u), Ψd(u) and Ψs(u) are, respectively, the bounded continuous solutions of the following boundary value problems:

1 2σ

2_Ψ0′′

d (u)+cΨ0 ′

d (u)=λΨd0(u)−λ

Z u

0

Ψ_d0(u₋z)dF (z), u >0, (4.1)

Ψ_d0(0)₌1, Ψ_d0(_+∞)₌0, (4.2)

1 2σ

2_Ψ0′′

s (u)+cΨ 0′

s (u)=λΨ 0 s(u)−λ

Z u

0

Ψ_s0(u₋z)dF (z)₋λ(1₋F (u)), u >0, (4.3)

Ψ_s0(0)₌0, Ψ_s0(_+∞)₌0, (4.4)

1 2σ

2_Ψ′′

d(u)+(ru+c)Ψd′(u)=λΨd(u)−λ

Z u

0

Ψd(u−z)dF (z), u >0, (4.5)

Ψd(0)=1, Ψd(+∞)=0, (4.6)

1 2σ

2_Ψ′′

s(u)+(ru+c)Ψs′(u)=λΨs(u)−λ

Z u

0

Ψs(u−z)dF (z)−λ(1−F (u)), u >0, (4.7)

Ψs(0)=0, Ψs(+∞)=0. (4.8)

LetF (z) ₌ 1₋exp_{−αz_},(α > 0), then the above boundary value problems can be, in turn, reduced to the following problems:

1 2σ

2_Ψ0′′′

d (u)+(12ασ 2

+c)Ψ_d0′′(u)₊(αc₋λ)Ψ_d0′(u)₌0, u >0, (4.9) Ψ_d0(0)₌1, Ψ_d0(_+∞)₌0, 1₂σ2Ψ_d0′′(0+)₊cΨ_d0′(0+)₌λ, (4.10)

1 2σ

2_Ψ0′′′

s (u)+(12ασ 2

+c)Ψ_s0′′(u)₊(αc₋λ)Ψ_s0′(u)₌0, u >0, (4.11) Ψ_s0(0)₌0, Ψ_s0(_+∞)₌0, 1₂σ2Ψ_s0′′(0+)₊cΨ_s0′(0+)_{= −}λ, (4.12)

1 2σ

2_Ψ′′′

d (u)+(12ασ 2

+ru₊c)Ψ_d′′(u)₊(αru₊αc₊r₋λ)Ψ_d′(u)₌0, u >0, (4.13) Ψd(0)=1, Ψd(+∞)=0, 1₂σ2Ψd′′(0+)+cΨd′(0+)=λ, (4.14)

1 2σ

2_Ψ′′′

s (u)+(12ασ 2

+ru₊c)Ψ_s′′(u)₊(αru₊αc₊r₋λ)Ψ_s′(u)₌0, u >0, (4.15) Ψs(0)=0, Ψs(+∞)=0, 1₂σ2Ψs′′(0+)+cΨs′(0+)= −λ. (4.16) Since the solutions of the above boundary value problems are all unique, we can get the explicit expressions for Ψ_d0(u), Ψ_s0(u), Ψd(u)andΨs(u).

It is easy to solve the problem (4.9) and (4.10) and the problem (4.11) and (4.12). Note that

(1₂ασ2₊c)2> (₂1ασ2₊c)2₋2σ2(αc₋λ)₌(1₂ασ2₋c)2₊2λσ2>0. (4.17) The solution of (4.9) and (4.10) is that

Ψ_d0(u)₌1₊c1exp{−λ1u} +c2exp{−λ2u}, (4.18)

where

λ1=

(1₂ασ2₊c)₋

q

(1₂ασ2_+c)2_−2σ2_(αc−λ)

σ2 , (4.19)

λ2=

(1₂ασ2₊c)₊

q

(1₂ασ2_+c)2_−2σ2_(αc−λ)

σ2 , (4.20)

c1=

σ2λ2₂₋2cλ2−2λ

σ2_(λ2

2−λ21)−2c(λ2−λ1)

, (4.21)

c2= −

σ2λ2₁₊2cλ1+2λ

σ2_(λ2

2−λ21)−2c(λ2−λ1)

. (4.22)

The solution of (4.11) and (4.12) is

Ψ_s0(u)₌c1₁exp_{−λ1u} +c21exp{−λ2u}, (4.23)

where

c1₁₌ 2λ

σ2_(λ2 2−λ

2

1)−2c(λ2−λ1)

, (4.24)

c1₂₌ −2λ σ2_(λ2

2−λ21)−2c(λ2−λ1)

. (4.25)

We now consider the boundary value problem (4.13) and (4.14). Let β ₌c

r − ασ2

2r , (4.26)

U (a, b, x)₌ 1 Γ (a)

Z +∞

0

exp_{−xt_}ta−1(1₊t )b−a−1dt, a >0, x >0, (4.27) F (a, b, x)₌ Γ (b)

Γ (b₋a)Γ (a)

Z 1

0

exp_{xt_}ta−1(1₋t )b−a−1dt, b > a >0. (4.28) Put

¯

Q′₁(x)₌expn₋αx₊ r

σ2(x+β)

2o_Uλ 2r,

1 2,

r

σ2(x+β)

2_{, (4.29)}

¯

Q′₂(x)₌(x₊β)expn₋αx₊ r

σ2(x+β)

2o_F1 2 +

λ 2r,

3 2,

r

σ2(x+β)

2_{. (4.30)}

Remark 4.1. F (a, b, x)denotes the standard confluent hypergeometric function andU (a, b, x)indicates its second form (see Slater (1960, p. 5) for details).F (a, b, x)andQ(a, b, x)are two useful functions in our discussion. Paulsen

and Gjessing (1997) use them to express the non-ruin probability of (1.4). Wang and Wu (1999) use them to express the distribution of the surplus of (1.4) at the time of ruin. Here we will use them to express ourΨd(u)andΨs(u).

Following the steps to solve the differential equation (2.16) in Paulsen and Gjessing (1997) we can get the general solution of (4.13) as follows:

Ψd(u)=c02+c21Q1(u)+c22Q2(u), (4.31)

where Q1(u)=

Z u

0

¯

Q′₁(x)dx, (4.32)

Q2(u)=

Z u

0

¯

Q′₂(x)dx. (4.33)

To find the solution of (4.13) and (4.14) we should determine the arbitrary constantsc2₀, c2₁andc₂2according to boundary value conditions (4.14). Lettingδ ₌(r/σ2)β2, some calculations give that

c2₀₌1, (4.34)

c2_{1= −}k2+λQ2(+∞)

C , (4.35)

c2₂₌ k1+λQ1(+∞)

C , (4.36)

where

C₌k2Q1(u)−k1Q2(u), (4.37)

k1=

c₋1 2ασ

2

−rβ

exp_{−δ_}U

_λ

2r, 1 2, δ

−1

2λβexp{−δ}U

1₊ λ 2r,

3 2, δ

, (4.38)

k2=

₁

2σ 2

+cβ₋1 2αβσ

2

−1

2rβ 2

exp_{−δ_}F

₁

2+ λ 2r,

3 2, δ

+1

3(λ+r)β 2_exp

{−δ_}F

₃

2+ λ 2r,

5 2, δ

. (4.39)

Therefore, we get the explicit expression Ψd(u)=1−

k2+λQ2(+∞)

C Q1(u)+

k1+λQ1(+∞)

C Q2(u). (4.40)

Similar to solving the boundary value problem (4.13) and (4.14) we can get the solutionΨs(u)of boundary value problem (4.15) and (4.16) as follows:

Ψs(u)=

λQ2(+∞)

C Q1(u)−

λQ1(+∞)

C Q2(u). (4.41)

Acknowledgements

The author would like to thank the referee for truly useful suggestions to improve the earlier version of the paper.

References

Dufresne, F., Gerber, H.U., 1991. Risk theory for the compound Poisson process that is perturbed by diffusion. Insurance: Mathematics and Economics 10, 51–59.

Paulsen, J., Gjessing, H.K., 1997. Ruin theory with stochastic return on investments. Advances in Applied Probability 29, 965–985. Revuz, D., Yor, M., 1991. Continuous Martingales and Brownian Motion. Springer, Berlin.

Slater, L.J., 1960. Confluent Hypergeometric Functions. Cambridge University Press, London.

Wang, G., Wu, R., 1998. Ruin theory for risk process with return on investments. Insurance: Mathematics and Economics, submitted for publication.

Wang, G., Wu, R., 1999. Notes on risk process with return on investments. Insurance: Mathematics and Economics, submitted for publication. Wang, G., Wu, R., 2000. Some distributions for classical risk process that is perturbed by diffusion. Insurance: Mathematics and Economics

Proof. Letτ_u0₌inf_{v(s):_|Bv(s)| =u},τu =inf{s:|Bv(s)| =u}, thenτu0=v(τu). PutT =T1∧τu0. Similar to

formula (2.9) we have

Ψd(u)=E[Ψd(RT)]. (3.8)

Thus,

Ψd(u)=E[Ψd(RT)]=E[Ψd(RT1)I (T1≤τ 0

u)]+E[Ψd(Rτ0

u)I (T1> τ 0

u)]=I

1 1+I

1

2. (3.9)

We now present the expressions forI1andI2.

I₁1₌E

Ψd

exp_{rT1}

u₊

Z T1

0

exp_{−rs_}dR_s0

I (T1_≤τ_u0)

=EhΨd

exp_{rT1}

h

u₊c

r(1−exp{−rT1})+BT1

i −Z1

I (T1≤τu0) i

,

and

I₂1₌E

"

Ψd exp{rτu0} "

u₊

Z τu0

0

exp_{−rs_}dR_s0

#!

I (T1> τu0) #

=EhΨd

exp_{rτ_u0_}hu₊c

r(1−exp{−rτ 0

u})+Bτ0 u

i

I (T1> τu0) i

.

Then exactly similar to the calculations ofI1andI2in Theorem 2.1 in Wang and Wu (1998), we can verify that I₁1andI₂1are equal to the first term and the second term on the right-hand side of (3.7), respectively. The proof is

completed. h

Similar to Theorem 2.2, we have the following theorem.

Theorem 3.2. SupposeF (z)has continuous density function on [0,_+∞), thenΨd(u)is twice continuously

differ-entiable in(0,_+∞).

Theorem 3.3. Suppose F (z) has continuous density function on [0,_+∞), then Ψd(u) satisfies the following

integro-differential equation:

1 2σ

2_Ψ′′

d(u)+(ru+c)Ψd′(u)=λΨd(u)−λ Z u

0

Ψd(u−z)dF (z), u >0. (3.10)

Proof. Letε, t, M >0 such thatε < u < MandT_tε,M ₌inf_{s >0 : exp_{rs_}[u₊(c/r)(1₋exp_{−rs_})₊Bs]∈/

(ε, M)_{} ∧}t. Note thatΨ_d′(u)andΨ_d′′(u)are bounded on [ε, M] and thereforeRS∧Ttε,M

0 Ψd′(exp{rv}[u+(c/r)(1−

exp_{−rv_})₊Bv])exp{rv}dBvis a martingale. PutT =Ttε,M∧T1, similar to the equality (3.8) we have Ψd(u)=E[Ψd(R_Tε,M

Therefore, by a standard use of Itˇo’s formula, we get Ψd(u)=exp{−λt}E

h

Ψd

exp_{rTε,M_{t }}hu₊c

r(1−exp{−rT

ε,M

t })+BT_tε,M ii

+ Z t

0

λexp_{−λs_}nEhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M

i

I (T_sε,M < s)i

+ Z +∞

0

EhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M

i

−zI (T_sε,M ₌s)idF (z)

ds

=exp_{−λt_}

(

Ψd(u)+E

Z Tsε,M

0

Ψ_d′exp_{rs_}hu₊c

r(1−exp{−rs})+Bs

i

×exp_{rs_}hu₊c

r(1−exp{−rs})+c+rexp{rs}Bs

i +σ 2 2 Ψ ′′ d

exp_{rs_}hu₊c

r(1−exp{−rs})+Bs

i ds + Z t 0

λexp_{−λs_}nEhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M

i

I (T_sε,M < s)i

+ Z +∞

0

EhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M

i

−zI (T_sε,M ₌s)idF (z)

ds. Dividing bytgives

1₋exp_{−λt_}

t Ψd(u)=exp{−λt}

(

E

"

1 t

Z Tsε,M

0

Ψ_d′exp_{rs_}hu₊c

r(1−exp{−rs})+Bs

i

×exp_{rs_}hu₊c

r(1−exp{−rs})+c+rexp{rs}Bs

i +σ 2 2 Ψ ′′ d

exp_{rs_}hu₊c

r(1−exp{−rs})+Bs

i

ds

+1_t

Z t

0

λexp_{−λs_}

×nEhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M

s })+BTsε,M

i

I (T_sε,M < s)i

+ Z +∞

0

EhΨd

exp_{rTε,M_{s }}hu₊c

r(1−exp{−rT

ε,M s }) +B

Tsε,M

i

−zI (T_sε,M ₌s)idF (z)ods.

Lettingt _→0 we get that (3.10) holds in(ε, M)and therefore in(0,_+∞). The proof is completed. h

Similar to Theorems 3.2 and 3.3 and Propositions 2.1 and 2.2, we have the following theorem.

Theorem 3.4. SupposeF (z)has continuous density function on [0,_+∞), thenΨs(u)is twice continuously

differ-entiable in(0,_+∞).

Theorem 3.5. Suppose F (z) has continuous density function on [0,_+∞), then Ψs(u) satisfies the following

integro-differential equation:

1 2σ

2_Ψ′′

s(u)+(ru+c)Ψs′(u)=λΨs(u)−λ Z u

0

Ψs(u−z)dF (z)−λ(1−F (u)), u >0. (3.12)

Proposition 3.2. Suppose F (z) has continuous density function on [0,_+∞), then Ψs(u) is continuous on

[0,_+∞).

4. Examples

This section will present the explicit expression for the ruin probability caused by oscillation and the ruin probability caused by a claim when the claims are exponentially distributed.

SupposeF (z)has continuous density function on [0,_+∞), from Sections 2 and 3, we can derive thatΨ_d0(u), Ψ_s0(u), Ψd(u) and Ψs(u) are, respectively, the bounded continuous solutions of the following boundary value

problems: 1 2σ

2_Ψ0′′

d (u)+cΨ0 ′

d (u)=λΨd0(u)−λ Z u

0

Ψ_d0(u₋z)dF (z), u >0, (4.1)

Ψ_d0(0)₌1, Ψ_d0(_+∞)₌0, (4.2)

1 2σ

2_Ψ0′′

s (u)+cΨ

0′

s (u)=λΨ

0

s(u)−λ Z u

0

Ψ_s0(u₋z)dF (z)₋λ(1₋F (u)), u >0, (4.3)

Ψ_s0(0)₌0, Ψ_s0(_+∞)₌0, (4.4)

1 2σ

2_Ψ′′

d(u)+(ru+c)Ψd′(u)=λΨd(u)−λ Z u

0

Ψd(u−z)dF (z), u >0, (4.5)

Ψd(0)=1, Ψd(+∞)=0, (4.6)

1 2σ

2_Ψ′′

s(u)+(ru+c)Ψs′(u)=λΨs(u)−λ Z u

0

Ψs(u−z)dF (z)−λ(1−F (u)), u >0, (4.7)

Ψs(0)=0, Ψs(+∞)=0. (4.8)

LetF (z) ₌ 1₋exp_{−αz_},(α > 0), then the above boundary value problems can be, in turn, reduced to the following problems:

1 2σ

2_Ψ0′′′

d (u)+(12ασ 2

+c)Ψ_d0′′(u)₊(αc₋λ)Ψ_d0′(u)₌0, u >0, (4.9) Ψ_d0(0)₌1, Ψ_d0(_+∞)₌0, 1₂σ2Ψ_d0′′(0+)₊cΨ_d0′(0+)₌λ, (4.10)

1 2σ

2_Ψ0′′′

s (u)+(12ασ 2

+c)Ψ_s0′′(u)₊(αc₋λ)Ψ_s0′(u)₌0, u >0, (4.11) Ψ_s0(0)₌0, Ψ_s0(_+∞)₌0, 1₂σ2Ψ_s0′′(0+)₊cΨ_s0′(0+)_{= −}λ, (4.12)

1 2σ

2_Ψ′′′

d (u)+(12ασ 2

+ru₊c)Ψ_d′′(u)₊(αru₊αc₊r₋λ)Ψ_d′(u)₌0, u >0, (4.13) Ψd(0)=1, Ψd(+∞)=0, 1₂σ2Ψd′′(0+)+cΨd′(0+)=λ, (4.14)

1 2σ

2_Ψ′′′

s (u)+(12ασ 2

+ru₊c)Ψ_s′′(u)₊(αru₊αc₊r₋λ)Ψ_s′(u)₌0, u >0, (4.15) Ψs(0)=0, Ψs(+∞)=0, 1₂σ2Ψs′′(0+)+cΨs′(0+)= −λ. (4.16)

Since the solutions of the above boundary value problems are all unique, we can get the explicit expressions for Ψ_d0(u), Ψ_s0(u), Ψd(u)andΨs(u).

It is easy to solve the problem (4.9) and (4.10) and the problem (4.11) and (4.12). Note that

(1₂ασ2₊c)2> (1₂ασ2₊c)2₋2σ2(αc₋λ)₌(1₂ασ2₋c)2₊2λσ2>0. (4.17) The solution of (4.9) and (4.10) is that

Ψ_d0(u)₌1₊c1exp_{−λ1u_{} +}c2exp_{−λ2u_}, (4.18) where

λ1=

(1₂ασ2₊c)₋

q

(1₂ασ2_+c)2_−2σ2_(αc−λ)

σ2 , (4.19)

λ2=

(1₂ασ2₊c)₊

q

(1₂ασ2_+c)2_−2σ2_(αc−λ)

σ2 , (4.20)

c1=

σ2λ2₂₋2cλ2−2λ σ2_(λ2

2−λ21)−2c(λ2−λ1)

, (4.21)

c2= −

σ2λ2₁₊2cλ1+2λ σ2_(λ2

2−λ21)−2c(λ2−λ1)

. (4.22)

The solution of (4.11) and (4.12) is

Ψ_s0(u)₌c1₁exp_{−λ1u} +c21exp{−λ2u}, (4.23) where

c1₁₌ 2λ

σ2_(λ2 2−λ

2

1)−2c(λ2−λ1)

, (4.24)

c1₂₌ −2λ

σ2_(λ2

2−λ21)−2c(λ2−λ1)

. (4.25)

We now consider the boundary value problem (4.13) and (4.14). Let β ₌c

r − ασ2

2r , (4.26)

U (a, b, x)₌ 1 Γ (a)

Z +∞

0

exp_{−xt_}ta−1(1₊t )b−a−1dt, a >0, x >0, (4.27) F (a, b, x)₌ Γ (b)

Γ (b₋a)Γ (a)

Z 1 0

exp_{xt_}ta−1(1₋t )b−a−1dt, b > a >0. (4.28) Put

¯

Q′₁(x)₌expn₋αx₊ r

σ2(x+β)

2o_Uλ 2r,

1 2,

r

σ2(x+β)

2_{, (4.29)}

¯

Q′₂(x)₌(x₊β)expn₋αx₊ r

σ2(x+β)

2o_F1 2 +

λ 2r,

3 2,

r

σ2(x+β)

2_{. (4.30)}

Remark 4.1. F (a, b, x)denotes the standard confluent hypergeometric function andU (a, b, x)indicates its second form (see Slater (1960, p. 5) for details).F (a, b, x)andQ(a, b, x)are two useful functions in our discussion. Paulsen

and Gjessing (1997) use them to express the non-ruin probability of (1.4). Wang and Wu (1999) use them to express the distribution of the surplus of (1.4) at the time of ruin. Here we will use them to express ourΨd(u)andΨs(u).

Following the steps to solve the differential equation (2.16) in Paulsen and Gjessing (1997) we can get the general solution of (4.13) as follows:

Ψd(u)=c02+c21Q1(u)+c22Q2(u), (4.31) where

Q1(u)=

Z u

0

¯

Q′₁(x)dx, (4.32)

Q2(u)₌

Z u

0

¯

Q′₂(x)dx. (4.33)

To find the solution of (4.13) and (4.14) we should determine the arbitrary constantsc2₀, c2₁andc₂2according to boundary value conditions (4.14). Lettingδ ₌(r/σ2)β2, some calculations give that

c2₀₌1, (4.34)

c2_{1= −}k2+λQ2(+∞)

C , (4.35)

c2₂₌ k1+λQ1(+∞)

C , (4.36)

where

C₌k2Q1(u)−k1Q2(u), (4.37)

k1=

c₋1 2ασ

2

−rβ

exp_{−δ_}U

_λ

2r, 1 2, δ

−1

2λβexp{−δ}U

1₊ λ 2r,

3 2, δ

, (4.38)

k2=

₁

2σ 2

+cβ₋1 2αβσ

2

−1

2rβ 2

exp_{−δ_}F

₁

2+ λ 2r,

3 2, δ

+1

3(λ+r)β 2_exp

{−δ_}F

₃

2+ λ 2r,

5 2, δ

. (4.39)

Therefore, we get the explicit expression Ψd(u)=1−

k2+λQ2(+∞)

C Q1(u)+

k1+λQ1(+∞)

C Q2(u). (4.40)

Similar to solving the boundary value problem (4.13) and (4.14) we can get the solutionΨs(u)of boundary value

problem (4.15) and (4.16) as follows: Ψs(u)=

λQ2(_+∞)

C Q1(u)−

λQ1(_+∞)

C Q2(u). (4.41)

Acknowledgements

The author would like to thank the referee for truly useful suggestions to improve the earlier version of the paper.

References

Dufresne, F., Gerber, H.U., 1991. Risk theory for the compound Poisson process that is perturbed by diffusion. Insurance: Mathematics and Economics 10, 51–59.

Paulsen, J., Gjessing, H.K., 1997. Ruin theory with stochastic return on investments. Advances in Applied Probability 29, 965–985. Revuz, D., Yor, M., 1991. Continuous Martingales and Brownian Motion. Springer, Berlin.

Slater, L.J., 1960. Confluent Hypergeometric Functions. Cambridge University Press, London.

Wang, G., Wu, R., 1998. Ruin theory for risk process with return on investments. Insurance: Mathematics and Economics, submitted for publication.

Wang, G., Wu, R., 1999. Notes on risk process with return on investments. Insurance: Mathematics and Economics, submitted for publication. Wang, G., Wu, R., 2000. Some distributions for classical risk process that is perturbed by diffusion. Insurance: Mathematics and Economics