246 S. Cheng et al. Insurance: Mathematics and Economics 26 2000 239–250
is the nth derivative of −1−v[1−ρv] with respect to v at v=1. Because ρv −
1 v −
1 =
∞
X
k= 1
1 k
ρ
k
1v − 1
k− 1
, 4.18
the factorial moment is expressed in terms of ρ
k
1 up to k=n+1. We remark that in the continuous time model, the moments of the time to ruin have been examined by Picard and Lefèvre 1998.
5. Recursive formulas
For an arbitrary initial surplus u0, it is possible to determine fx, y; u by a recursive procedure. Given U0=u, we consider the first time t, t≥1, when Ut≤u. The “discounted” probability of this event with the surplus being of
the amount u−y at that time is gy =
∞
X
x=
f x, y; 0 = qv
∞
X
x=
ρ
x
px + y + 1
5.1 by 4.1. For u0, we evaluate fx, y; u by conditioning on the first time when the surplus drops to or below its
initial value u. We need to distinguish two cases. If ux, the event that ruin occurs at this time makes no contribution to the quantity fx, y; u. Hence we have
f x, y; u =
u− 1
X
z=
f x, y; u − zgz, u = x +
1, x + 2, . . . 5.2
If u≤x, the event that ruin occurs at this time contributes the quantity fx−u, y+u; 0 to fx, y; u. Thus f x, y; u =
u− 1
X
z=
f x, y; u − zgz + f x − u, y + u; 0,
u = 1, 2, . . . , x.
5.3 Formulas 5.2 and 5.3 are discrete analogs of 2.11 and 2.10 in Gerber and Shiu 1997. By 4.1
f x − u, y + u; 0 = qvρ
x−u
px + y + 1 = ρ
− u
f x, y; 0,
u ≤ x. Hence we can combine 5.2 and 5.3 as
f x, y; u =
u− 1
X
z=
f x, y; u − zgz + f x, y; 0ρ
− u
I u ≤ x, 5.4
which corresponds to 4.1 of Gerber and Shiu 1997. Here, u=1, 2, 3,. . . , x=1, 2, 3,. . . , and y=0, 1, 2,. . . For given values of x and y, we can calculate fx, y; u recursively by 5.4. The following result gives a more
elegant way to obtain fx, y; u:
Theorem 2. Let h
u be defined as the solution of hu =
u− 1
X
z=
hu − zgz + ρ
− u
, u =
1, 2, 3, . . . 5.5
Then, for x= 1, 2, 3,. . . , and y=0, 1, 2,. . . ,
f x, y; u = f x, y; 0[hu − ρ
− x
hu − xI u x ],
u = 1, 2, 3, . . .
5.6
S. Cheng et al. Insurance: Mathematics and Economics 26 2000 239–250 247
Proof.
For each positive integer x, we introduce a function of u, γ x; u, which is defined as the solution of the renewal equation
γ x; u =
u− 1
X
z=
γ x; u − zgz + ρ
− u
I u ≤ x, u =
1, 2, 3, . . . 5.7
Comparing 5.7 with 5.4, we gather that f x, y; u = f x, y;
0γ x; u. 5.8
Thus to prove 5.6, we have to show that γ x; u = hu − ρ
− x
hu − xI u x. 5.9
Because the solution of 5.7 is unique, it suffices to verify that the function on the right-hand side of 5.9 satisfies 5.7. Hence we consider
u− 1
X
z=
[hu − z − ρ
− x
hu − z − xI u − z x ]gz + ρ
− u
I u ≤ x, u =
1, 2, 3, . . . If u≤x, this is
u− 1
X
z=
hu − zgz + ρ
− u
= hu = hu − ρ
− x
hu − xI u x. If ux, this is
u− 1
X
z=
hu − zgz − ρ
− x
u−x− 1
X
z=
hu − z − xgz = [hu − ρ
− u
] − ρ
− x
[hu − x − ρ
− u−x
] =
hu − ρ
− x
hu − xI u x.
Remarks.
1. With 5.5 and 5.1, the function hu can be determined recursively for u=1, 2, 3,. . . Alternatively, Eq. 5.5 can be solved in terms of generating functions.
2. For u=1, 2, 3,. . . , the quantity 1hu has a probabilistic interpretation. For given positive integers u and x, u≤x, it follows from
5.6 that f x, y;
0 = 1
hu f x, y; u.
5.10 Suppose that U
0=0. Before ruin can occur with UT−1=x, x≥u0, the surplus process {Ut} must neces-
sarily have attained the level u because the premium is 1 per unit time. From this observation and formula
5.10 we see that 1hu can be interpreted as the expected present value of a contingent payment of 1 that is made when the surplus process, with initial value
0, attains the level u for the first time, provided that ruin has not occurred by then. The compound Poisson version of
5.10 is 6.26 of Gerber and Shiu 1998a. 3. It follows from the interpretation above that
1 h
1 =
1 − qv. 5.11
248 S. Cheng et al. Insurance: Mathematics and Economics 26 2000 239–250
On the other hand, we obtain from 5.5 and 5.1 that
h 1 = h1g0 + ρ
− 1
= h
1qv
∞
X
x=
ρ
x
px + 1 + ρ
− 1
. 5.12
These two formulas can be reconciled because ρ satisfies Lundberg’s equation 3.3.
4. Assuming a positive security loading, we consider the limiting case v=1 and hence ρ=1. Then it follows from 5.5 that hucan be interpreted as the expected number of “weak” record lows of the surplus process
{Ut} before ruin, with U
0=u. There is a close connection between hu and the probability of ruin ψu. Observe that the number of future record lows of the process
{Ut} has at any time a geometric distribution with mean 1[1−ψ0]. The expected number of record lows before ruin is the expected number of record lows starting
from time zero minus the expected number of future record lows from the time of ruin. Thus we have hu =
1 − ψu 1 − ψ0
. 5.13
Substituting 5.13 in 5.6, we obtain a discrete version of a formula by Dickson 1992, which is for the
compound Poisson model. In that context, Dickson’s formula has been generalized by Gerber and Shiu 1997,
1998a and Theorem 2 suggests that a generalization is also possible in the compound binomial model.
6. Asymptotic formulas
