50 G. Wang Insurance: Mathematics and Economics 28 2001 49–59
where r 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
R
t
= R
t
+ Z
t
R
s
−
dI
s
. 1.3
By Paulsen and Gjessing 1997, the solution of 1.3 is R
t
= exp{rt} u +
Z
t
exp{−rs} dR
s
, 1.4
and 1.4 is also a homogenous strong Markov process. Relation 1.4 can be rewritten as R
t
= exp{rt}
u +
c r
1 − exp{−rt} + σ Z
t
exp{−rs} dW
s
−
N
t
X
k+1
exp{−rT
k
}Z
k
, 1.5
where {T
k
}, k ≥ 1, is the sequence of jump times of {N
t
}. 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
Let a 0, define τ
a
= inf{s : |W
s
| = a}. For x ∈ [−a, a], put H a, t, x = 2πt
−12 +∞
X
k=−∞
exp −
1 2t
x + 4ka
2
− exp −
1 2t
x − 2a + 4ka
2
, 2.1
ha, t = 1
2 √
2π at
−32 +∞
X
k=−∞
4k + 1exp −
a
2
2t 4k + 1
2
+4k − 3exp −
a
2
2t 4k − 3
2
− 4k − 1exp −
a
2
2t 4k − 1
2
. 2.2
It follows from Revuz and Yor 1991, pp. 105–106 that P W
s
∈ dx, τ
a
s = H a, s, x dx and P τ
a
∈ ds = ha, s ds.
Let T
u
= inf{t ≥ 0; R
t
0} and defining T
u
= +∞ if R
t
≥ 0 for all t ≥ 0. T
u
is the time of ruin. The ruin probability Ψ
u of the risk process 1.1 is defined by Ψ u = P inf
t ≥0
R
t
0. We have Ψ u = P T
u
+∞. Let µ = E[Z
1
] and F z = P Z
1
≤ z. For u 0 we assume that E[R
t
] = u + c − λµt 0, i.e., we assume c − λµ 0, so that we have Ψ
u 1. Denote by Ψ
d
u the ruin probability caused by oscillation and Ψ
s
u the ruin probability caused by a claim. We have see 1.5 in Dufresne and Gerber 1991
Ψ u = Ψ
d
u + Ψ
s
u. 2.3
G. Wang Insurance: Mathematics and Economics 28 2001 49–59 51
If F z has continuous density function on [0, +∞, then P ∪
+∞ k=1
{R
T
k
= 0} = 0, and therefore Ψ
d
u = P T
u
+∞, R
T
u
= 0, 2.4
Ψ
s
u = P T
u
+∞, R
T
u
0. 2.5
From 2.4 and 2.5 we see that Ψ
d
u = 0, u 0,
1, u = 0,
2.6 Ψ
s
u = 1, u 0,
0, u = 0.
2.7
Theorem 2.1. Let u 0, suppose F z has continuous density function on [0, +∞, then the probability Ψ
d
u satisfies the following integral equation:
Ψ
d
u = 1
2 Z
+∞
Ψ
d
ct + Ψ
d
2u + ctexp{−λt}h u
σ , t
dt +
Z
+∞
λ exp{−λs} ds Z
uσ −uσ
H u
σ , s, x
dx Z
u+cs+σ x
Ψ
d
u + cs + σ x − z dF z. 2.8
Proof. Denote by A
d
the event of ruin due to oscillation. Let F
t
= σ {R
s
, s ≤ t}. Define M
t
by M
t
= E[I A
d
|F
t
]. Note that {M
t
, t ≥ 0} is a F
t
-martingale. Put T = τ
uσ
∧ T
1
, we have P T +∞ ≤ P T
1
+∞ = 1. By optional stopping theorem together with the homogeneous strong Markov process of R
t
, we get Ψ
d
u = EM = E[M
T
] = E[E[I A
d
|F
T
]] = E[Ψ
d
R
T
]. 2.9
Therefore Ψ
d
u = E[Ψ
d
R
T
] = E[Ψ
d
u + cτ
uσ
+ σ W
τ
uσ
I τ
uσ
T
1
] +E[Ψ
d
u + cT
1
+ σ W
T
1
− Z
1
I τ
uσ
≥ T
1
] = I
1
+ I
2
. 2.10
Exactly similar to the calculation of I
1
and I
2
in Theorem 3.1 in Wang and Wu 2000 we get I
1
= 1
2 Z
+∞
Ψ
d
ct + Ψ
d
2u + ctexp{−λt}h u
σ , t
dt, 2.11
I
2
= Z
+∞
λ exp{−λs} ds Z
uσ −uσ
H u
σ , s, x
dx Z
u+cs+σ x
Ψ
d
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 by A
u d
the event {ruin occurs due to oscillation |R = u}. Since for t ∈ 0, T , R
t
0, i.e. no ruin occurs up to time T , conditioned on {τ
uσ
= s T
1
}, we conclude that A
u d
occurs if and only if A
u+cs−u d
or A
u+cs+u d
occurs. The probabilities of A
u+cs−u d
and A
u+cs+u d
are Ψ
d
cs and Ψ
d
2u + cs, respectively. Note that P W
s
= −uσ = P W
s
= uσ =
1 2
, summing all over s we get the equality 2.11.
52 G. Wang Insurance: Mathematics and Economics 28 2001 49–59
Suppose that the first jump of the sample function of the process R
t
occurs at time t and has magnitude z. Since no ruin occurs up to time T , conditioned on {T
1
= t τ
uσ
, W
t
= x}, we see that A
u d
occurs if and only if z ≤ u+ct+σ x and A
u+ct+σ x−z d
occurs. The probability of the event A
u+ct+σ x−z d
is Ψ
d
u+cs+σ x−z. Summing all over t, x, z ∈ 0, +∞[−uσ, uσ ]0, u + ct + σ x − z] we get the equality 2.12. Therefore, Theorem 2.1 holds.
Theorem 2.2. Suppose F z has continuous density function on [0, +∞, then Ψ
d
u is twice continuously differ- entiable in the interval 0, +∞.
Proof. For any ε 0 it is sufficient to show that Ψ
d
u is twice continuously differentiable in the interval ε
, +∞. Replacing τ
uσ
by τ
ε σ
in the proof of Theorem 2.1, then for any u ∈ ε , +∞ we have
Ψ
d
u = 1
2 Z
+∞
Ψ
d
u − ε + ct + Ψ
d
u + ε + ctexp{−λt}h
ε σ
, t dt
+ Z
+∞
λ exp{−λs} ds Z
ε σ
−ε σ
H ε
σ , s, x
dx Z
u+cs+σ x
Ψ
d
u + cs + σ x − z dF z. 2.13
By changing the variable of integration, we can move u in the Ψ
d
u in the integrands on the right-hand side of 2.13 into H or h. Note that the density function ha, t is twice continuously differentiable in t and, for x 6= 0, H a, t, x
is twice continuously differentiable in x and satisfies lim
t ↓0
H
′ x
a, t, x = lim
t ↓0
H
′′ x
a, t, x = 0. Therefore, by the dominated convergence theorem and equality 2.13 we can verify that Ψ
d
u is twice continuously differentiable in the interval ε
, +∞.
h
Theorem 2.3. Let u 0, Suppose F z has continuous density function on [0, +∞, then Ψ
d
u satisfies the following integro-differential equation:
1 2
σ
2
Ψ
′′
d
u + cΨ
′
d
u = λΨ
d
u − λ Z
u
Ψ
d
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. Suppose F z has continuous density function on [0, +∞, then Ψ
s
u is twice continuously differ- entiable in the interval 0, +∞.
Theorem 2.5. Let u 0, suppose F z has continuous density function on [0, +∞, then Ψ
s
u satisfies the integro-differential equation
1 2
σ
2
Ψ
′′
s
u + cΨ
′
s
u = λΨ
s
u − λ Z
u
Ψ
s
u − z dF z − λ1 − F u. 2.15
Proposition 2.1. Suppose F z has continuous density function on [0, +∞, then Ψ
d
u is continuous on [0, +∞.
Proof. By Theorem 2.2, it is sufficient to show that Ψ
d +
= Ψ
d
0. Since P T
+
= T = 0 = 1 see Lemma
4.1 in Wang and Wu 2000 and R
t
is right continuous, by the dominated convergence theorem we get lim
u↓0
Ψ
d
u = lim
u↓0
E[I T
u
+∞, R
T
u
= 0] = E[lim
u↓0
I T
u
+∞, R
T
u
= 0] = E[I T
+∞, R = 0] = 1 = Ψ
d
0. This ends the proof.
h
G. Wang Insurance: Mathematics and Economics 28 2001 49–59 53
Similar to Proposition 2.1, we have the following proposition.
Proposition 2.2. Suppose F z has continuous density function on [0, +∞, then Ψ
s
u is continuous on [0, +∞.
3. The case with deterministic return on investments
