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
Let B
t
= σ R
t
exp{−rs} dW
s
, B
t
is Itˇo’s stochastic integral, its variance process is hBi
t
= σ
2
2r1−exp{−2rt}. Let vs = inf{t : hBi
t
s}, then vs =
1 2r
ln σ
2
σ
2
− 2rs .
3.1 For s σ
2
2r, set W
s
= B
vs
, it is well known that W
s
is a local Brownian motion starting from the origin and running for an amount of time σ
2
2r. Let T
u
= inf{t ≥ 0 : R
t
0} and defining T
u
= +∞ if R
t
≥ 0 for all t ≥ 0. Denote by Ψ u the ruin probability of the risk process 1.4, then Ψ u = P T
u
+∞. For u 0 we assume E[R
t
] = exp{rt}[u + 1r1 − exp{−rt}c − λµ] 0 see Lemma 2.1 in Wang and Wu 1998, i.e., we assume c − λµ 0 so that
we have Ψ u 1. For the risk process 1.4 we denote by Ψ
d
u and Ψ
s
u the ruin probability caused by oscillation and the ruin probability caused by a claim, respectively. We still have
Ψ u = Ψ
d
u + Ψ
s
u. 3.2
Suppose F z has continuous density function on [0, +∞, similar to 2.4 and 2.5 we have Ψ
d
u = P T
u
+∞, R
T
u
= 0, 3.3
Ψ
s
u = P T
u
+∞, R
T
u
0. 3.4
From 3.3 and 3.4 we get that Ψ
d
u = 0, u 0,
1, u = 0,
3.5 Ψ
s
u = 1, u 0,
0, u = 0.
3.6
Theorem 3.1. 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 = Z
+∞
λ exp{−λs} ds Z
u −u
H u, v
−1
s, x dx ×
Z
exp{rs}[u+x+cr1−exp{−rs}]
Ψ
d
exp{rs} h
u + x + c
r 1 − exp{−rs} − z exp{−rs}
i dF z
+ 1
2 Z
σ
2
2r
h Ψ
d
exp{rvs} h
2u + c
r 1 − exp{−rvs}
i + Ψ
d
c r
exp{rvs} − 1 i
×exp{−λvs}hu, s ds. 3.7
54 G. Wang Insurance: Mathematics and Economics 28 2001 49–59
Proof. Let τ
u
= inf{vs : |B
vs
| = u}, τ
u
= inf{s : |B
vs
| = u}, then τ
u
= vτ
u
. Put T = T
1
∧ τ
u
. Similar to formula 2.9 we have
Ψ
d
u = E[Ψ
d
R
T
]. 3.8
Thus, Ψ
d
u = E[Ψ
d
R
T
] = E[Ψ
d
R
T
1
I T
1
≤ τ
u
] + E[Ψ
d
R
τ
u
I T
1
τ
u
] = I
1 1
+ I
1 2
. 3.9
We now present the expressions for I
1
and I
2
. I
1 1
= E Ψ
d
exp{rT
1
} u +
Z
T
1
exp{−rs} dR
s
I T
1
≤ τ
u
= E h
Ψ
d
exp{rT
1
} h
u + c
r 1 − exp{−rT
1
} + B
T
1
i − Z
1
I T
1
≤ τ
u
i ,
and I
1 2
= E Ψ
d
exp{rτ
u
} u +
Z
τ
u
exp{−rs} dR
s
I T
1
τ
u
= E h
Ψ
d
exp{rτ
u
} h
u + c
r 1 − exp{−rτ
u
} + B
τ
u
i I T
1
τ
u
i .
Then exactly similar to the calculations of I
1
and I
2
in Theorem 2.1 in Wang and Wu 1998, we can verify that I
1 1
and I
1 2
are 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. Suppose F z has continuous density function on [0, +∞, then Ψ
d
u is twice continuously differ- entiable in 0, +∞.
Theorem 3.3. 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 + ru + cΨ
′ d
u = λΨ
d
u − λ Z
u
Ψ
d
u − z dF z, u 0.
3.10
Proof. Let ε, t, M 0 such that ε u M and T
ε,M t
= inf{s 0 : exp{rs}[u + cr1 − exp{−rs} + B
s
] ∈
ε, M} ∧ t. Note that Ψ
′ d
u and Ψ
′′ d
u are bounded on [ε, M] and therefore R
S∧T
ε,M t
Ψ
′ d
exp{rv}[u + cr1 − exp{−rv} + B
v
]exp{rv} dB
v
is a martingale. Put T = T
ε,M t
∧ T
1
, similar to the equality 3.8 we have Ψ
d
u = E[Ψ
d
R
T
ε,M t
∧T
1
]. 3.11
G. Wang Insurance: Mathematics and Economics 28 2001 49–59 55
Therefore, by a standard use of Itˇo’s formula, we get Ψ
d
u = exp{−λt}E h
Ψ
d
exp{rT
ε,M t
} h
u + c
r 1 − exp{−rT
ε,M t
} + B
T
ε,M t
ii +
Z
t
λ exp{−λs} n
E h
Ψ
d
exp{rT
ε,M s
} h
u + c
r 1 − exp{−rT
ε,M s
} + B
T
ε,M s
i I T
ε,M s
s i
+ Z
+∞
E h
Ψ
d
exp{rT
ε,M s
} h
u + c
r 1 − exp{−rT
ε,M s
} + B
T
ε,M s
i − z
I T
ε,M s
= s i
dF z ds
= exp{−λt} Ψ
d
u + E Z
T
ε,M s
Ψ
′ d
exp{rs} h
u + c
r 1 − exp{−rs} + B
s
i ×
exp{rs} h
u + c
r 1 − exp{−rs} + c + r exp{rs}B
s
i +
σ
2
2 Ψ
′′ d
exp{rs} h
u + c
r 1 − exp{−rs} + B
s
i ds
+ Z
t
λexp{−λs} n
E h
Ψ
d
exp{rT
ε,M s
} h
u + c
r 1 − exp{−rT
ε,M s
} + B
T
ε,M s
i I T
ε,M s
s i
+ Z
+∞
E h
Ψ
d
exp{rT
ε,M s
} h
u + c
r 1 − exp{−rT
ε,M s
} + B
T
ε,M s
i − z
I T
ε,M s
= s i
dF z ds.
Dividing by t gives 1 − exp{−λt}
t Ψ
d
u = exp{−λt} E
1 t
Z
T
ε,M s
Ψ
′ d
exp{rs} h
u + c
r 1 − exp{−rs} + B
s
i ×
exp{rs} h
u + c
r 1 − exp{−rs} + c + r exp{rs}B
s
i +
σ
2
2 Ψ
′′ d
exp{rs} h
u + c
r 1 − exp{−rs} + B
s
i ds
+ 1
t Z
t
λ exp{−λs} ×
n E
h Ψ
d
exp{rT
ε,M s
} h
u + c
r 1 − exp{−rT
ε,M s
} + B
T
ε,M s
i I T
ε,M s
s i
+ Z
+∞
E h
Ψ
d
exp{rT
ε,M s
} h
u + c
r 1 − exp{−rT
ε,M s
} +B
T
ε,M s
i − z
I T
ε,M s
= s i
dF z o
ds. Letting t → 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. Suppose F z has continuous density function on [0, +∞, then Ψ
s
u is twice continuously differ- entiable in 0, +∞.
Theorem 3.5. Suppose F z has continuous density function on [0, +∞, then Ψ
s
u satisfies the following integro-differential equation:
1 2
σ
2
Ψ
′′ s
u + ru + cΨ
′ s
u = λΨ
s
u − λ Z
u
Ψ
s
u − z dF z − λ1 − F u, u 0.
3.12
Proposition 3.1. Suppose F z has continuous density function on [0, +∞, then Ψ
d
u is continuous on [0, +∞.
56 G. Wang Insurance: Mathematics and Economics 28 2001 49–59
Proposition 3.2. Suppose F z has continuous density function on [0, +∞, then Ψ
s
u is continuous on [0, +∞.
4. Examples
