S. Cheng et al. Insurance: Mathematics and Economics 26 2000 239–250 241
Note that f0, y, t; u can be different from 0 only if t=1 and u=0; obviously f
0, y, 1; 0 = qpy + 1. 2.6
Let v be a discount factor 0v1. Our primary goal is to explore the “discounted” probability f x, y; u =
∞
X
t = 1
v
t
f x, y, t ; u. 2.7
Using the function fx, y; u, we can calculate the expected discounted value of a “penalty” which is due at ruin and may depend on the surpluses just before and at ruin. We may also view v as a parameter; then 2.7 is the formula
of a probability generating function.
3. Lundberg’s fundamental equation
We seek a number r0 such that {v
t
r
− U t
} is a martingale. This is the condition that
E [v
t
r
− U t
| U
0 = u] = r
− u
. 3.1
Because E
[r
− U t
| U
0 = u] = r
− u−t
{ qE
[r
X
] + 1 − q}
t
, 3.2
the martingale condition is φ r = v
− 1
, 3.3
where φ r =
qE [r
X− 1
] + 1 − qr
− 1
, r
0. 3.4
We call 3.3 Lundberg’s fundamental equation to honor the Swedish actuary Lundberg, who had pointed out that the compound Poisson version of 3.3 is “fundamental to the whole of collective risk theory” Lundberg, 1932, p.
144. Because φr is a strictly convex function, Eq. 3.3 has at most two roots. Also, φ1=1v
− 1
, and φr tends to ∞ for r→0+; hence, Eq. 3.3 has a solution r=ρ∈0, 1. Furthermore, under some regularity conditions for the
tail of the probability function px, Eq. 3.3 has another solution r=R1. For a first application, let x be an integer, xU0=u, and consider
T
x
= inf{t : U t ≥ x},
3.5 the first time the surplus rises to the level x. Note that the inequality in definition 3.5 can be replaced by an equality,
because the process {Ut} is skip-free upwards. We claim that
E [v
T
x
I T
x
∞|U 0 = u] = ρ
x−u
, 3.6
where I denotes the indicator function, i.e. IA=1 if A is true and IA=0 if A is false. To show this, we observe that {v
t
ρ
− U t
} is a positive martingale that is bounded above by ρ
−x
for tT
x
. By applying the optional sampling theorem
, we have ρ
− u
= E
[v
T
x
ρ
− U T
x
I T
x
∞|U 0 = u] = E[v
T
x
I T
x
∞|U 0 = u]ρ
− x
, 3.7
from which 3.6 follows.
242 S. Cheng et al. Insurance: Mathematics and Economics 26 2000 239–250
For a second application, suppose that 3.3 has the other root R1. Then {v
t
R
− U t
} is a positive martingale that
is bounded above by 1 for tT. By the optional sampling theorem, we obtain R
− u
= E
[v
T
R
− U T
I T ∞|U 0 = u].
3.8 In general, it is difficult to simplify the expression on the right-hand side of 3.8, because T and UT are not
independent, and the distribution of UT is unknown. There are two noteworthy exceptions. One is the case where all claims are of size 2, X≡
≡2. Then UT=0, provided u0. Hence it follows from 3.8 that, for u=1, 2, 3,. . . , E
[v
T
I T ∞|U 0 = u] = R
− u
, 3.9
where R =
1 + p
1 − 4q1 − qv
2
2qv 3.10
by 3.3. The formula for u=0 will be given in formula 4.8, which is a general result. We note that 3.10 can be found in Feller 1968, p. 350 The other exception is the case of a geometric claim amount distribution,
px = 1 − cc
x− 1
, x =
1, 2, 3, . . . 3.11
Then, given T∞, −UT is independent of T, and 3.8 becomes R
− u
= E
[v
T
I T ∞|U 0 = u]E[R
− U T
| T ∞
]. 3.12
Because, given T∞, UT has the geometric distribution 1 − cc
x
, x =
0, 1, 2, . . . we have
E [R
− U T
| T ∞
] = 1 − c
1 − cR .
3.13 It follows from 3.12 and 3.13 that
E [v
T
I T ∞|U 0 = u] =
1 − cR 1 − c
R
− u
. 3.14
Here, 3.3 leads also to a quadratic equation, with which we can determine R.
Remarks.
1. Assume that the second root R of Lundberg’s equation 3.3 exists. Consider both roots as functions of v, ρ=ρv and R=R
v. Note that φ1=1 and φ
′
1 = qE[X] − 1, 3.15
which is the negative of the security loading. If the security loading is negative, then φ
′
10. Hence the convex function φ has its minimum in the interval
0, 1, from which it follows that: lim
v→ 1
ρv 1 = lim
v→ 1
Rv. 3.16
If the security loading is zero, then φ
′
1=0. Thus φ has its minimum at 1 and lim
v→ 1
ρv = 1 = lim
v→ 1
Rv. 3.17
S. Cheng et al. Insurance: Mathematics and Economics 26 2000 239–250 243
If the security loading is positive, then φ
′
10, and lim
v→ 1
ρv = 1 lim
v→ 1
Rv. 3.18
The corresponding relationships for the compound Poisson case can be found in Section 3 of Gerber and Shiu
1999. 2. In formulas such as 3.6 and 3.8, the indicator random variables IT
x
∞ and IT∞ can be dropped,
because v 1. However, leaving them in facilitates the discussion of limiting results for v→1. From 3.6 we
have Pr[T
x
∞|U 0 = u] = ρ1
x−u
. 3.19
It then follows from 3.17 and 3.18 that, if the security loading is nonnegative, the right-hand side of 3.19 is
1, i.e. the surplus will rise to the level x with certainty. If the security loading is negative, then ρ11 according to
3.16, and hence 3.19 is now not a trivial result. On the other hand, as v→1 3.8 becomes R
1
− u
= E
[R1
− U T
I T ∞|U 0 = u].
3.20 If the security loading is nonpositive, it follows from
3.16 and 3.17 that R1=1, and hence T∞ with probability
1, i.e. ruin is certain. If the security loading is positive, then R11 according to 3.18, and ψu =
Pr[T ∞|U 0 = u] = R
1
− u
E [R1
− U T
| T ∞, U
0 = u] .
3.21
4. The key formula
