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
The following result is for U0=u=0. It is the discrete counterpart of formula 3.12 in Gerber and Shiu 1997 and formula 3.3 in Gerber and Shiu 1998a.
Theorem 1.
For x= 0, 1, 2,. . . , y=0, 1, 2,. . . ,
f x, y; 0 = qvρ
x
px + y + 1.
4.1
Proof.
By 2.7, the left-hand side of 4.1 is
∞
X
t = 1
v
t
f x, y, t ; 0.
4.2 According to definition 2.5, fx, y, t; 0 is the following product of three probabilities:
Pr[U 1 0, U 2 0, . . . , U t − 2 0, U t − 1 = x|U 0 = 0]qpx + y + 1. 4.3
It follows from a simple duality argument which can be visualized by rotating the graph of the surplus process from time 0 to time t−1 by 180
◦
Feller, 1971, Section XII.2 that the conditional probability in 4.3 is the same as Pr[U 1 x, U 2 x, . . . , U t − 2 x, U t − 1 = x|U 0 = 0].
4.4 Multiplying 4.3 with v
t
and summing over t, recalling that T
x
is the first passage time of the surplus process at level x, and applying 3.6 with u=0, we have
244 S. Cheng et al. Insurance: Mathematics and Economics 26 2000 239–250
∞
X
t = 1
v
t
Pr[U 1 0, U 2 0, . . . , U t − 2 0, U t − 1 = x|U 0 = 0]qpx + y + 1 =
v qpx + y +
1
∞
X
t = 1
v
t − 1
Pr[U 1 x, U 2 x, . . . , U t − 2 x, U t − 1 = x|U 0 = 0] =
v qpx + y +
1E[v
T
x
| U
0 = 0] = vqpx + y + 1ρ
x
, 4.5
which is indeed the right-hand side of 4.1. As a check for 4.1, we consider the special case of x=0,
f 0, y; 0 = qvpy + 1,
which can be verified using 2.6. We note that DeVylder and Goovaerts 1998 have given an interesting proof of the compound Poisson version of Theorem 1 using the Lagrange expansion formula. Also, Theorem 1 can be
generalized to the case where there are m types of individual claims, which are independent and at most one of which can occur in a time period; for such results in the compound Poisson model, see Gerber and Shiu 1999.
Option-pricing applications of the compound Poisson version of Theorem 1 can be found in Gerber and Shiu 1998b, 1999.
As an illustration of Theorem 1, we calculate the expected present value of 1, payable at the time of ruin, if u=0: E
[v
T
I T ∞|U 0 = 0] =
∞
X
x= ∞
X
y=
f x, y; 0 = qv
∞
X
x=
ρ
x ∞
X
k=x+ 1
pk 4.6
by 4.1. By interchanging the order of summation and then noting that ρ is a solution of 3.3, we obtain
∞
X
x=
ρ
x ∞
X
k=x+ 1
pk = 1
1 − ρ
∞
X
k= 1
1 − ρ
k
pk = 1
1 − ρ 1 − E[ρ
X
] = 1
1 − ρ 1
q −
ρ qv
. 4.7
Substituting 4.7 in 4.6 yields E
[v
T
I T ∞|U 0 = 0] =
v − ρ 1 − ρ
. 4.8
Note that 4.8 can be interpreted as the probability generating function of the time-of-ruin random variable T. An explicit expansion of the right-hand side of 4.8 in powers of v is possible in special cases such as the two examples
in the last section; see also Section XI.3 of Feller 1968, where our v is s. Note that formula 4.8 can be rewritten as
E [v
T
I T ∞|U 0 = 0] = 1 −
1 − v 1 − ρ
, 4.9
the compound Poisson version of which is formula 3.9 in Gerber and Shiu 1998a and formula 4.11 in Gerber and Shiu 1999.
Let us first assume a negative loading. Then ruin is certain, i.e. IT∞≡ ≡1. Consider ρ as a function of v,
ρ=ρ v. Then ρ11 by 3.16. Differentiating 4.8 and setting v=1, we obtain
E [T |U 0 = 0] =
1 1 − ρ1
. 4.10
Note that, as the security loading tends to zero, the quantity ρ1 tends to 1 according to 3.17, and hence the expectation of T goes to infinity; in the special case of a symmetric random walk X≡
≡2 in our model, this is the well-known result that the time of return to the origin has an infinite expectation.
S. Cheng et al. Insurance: Mathematics and Economics 26 2000 239–250 245
For the remainder of this section, we consider the more important case where the security loading is positive. Setting v=1 in 4.6 yields an expression for the probability of ruin for u=0:
ψ 0 = q
∞
X
x= ∞
X
k=x+ 1
pk = qE
[X]. 4.11
This is formula 7 in Gerber 1988. In terms of the relative security loading θ , formula 4.11 is ψ
0 = 1
1 + θ ,
4.12 which is a classical result in the continuous-time compound Poisson model and can be viewed as an extension of
the ballot theorem Feller, 1968, p. 69. As a check, let us verify 4.11 starting with 4.8. Using the rule of Bernoulli–Hôpital, we obtain
ψ 0 = 1 −
1 ρ
′
1 .
4.13 To determine ρ
′
1, we replace r by ρv in 3.3: φ ρv =
1 v
. 4.14
Using the chain rule, we differentiate with respect to v and set v=1 to see that φ
′
ρ 1ρ
′
1 = −1. It follows from 3.18 and 3.15 that
ρ
′
1 = −
1 φ
′
1 =
1 1 − qE[X]
. 4.15
Substituting 4.15 in 4.13 we obtain 4.11 again. Let us now calculate the expected time of ruin, given that ruin occurs. First we differentiate 4.6 with respect to
v and set v=1. This yields
E [TIT ∞|U 0 = 0] = q
∞
X
x= ∞
X
k=x+ 1
pk + ρ
′
1
∞
X
x=
x
∞
X
k=x+ 1
pk
= ψ
0 + qρ
′
1E[XX − 1] 2
. After dividing by ψ0 and applying 4.11 and 4.15, we obtain the desired result:
E [T |T ∞, U 0 = 0] = 1 +
E [XX − 1]
2E[X]1 − qE[X] .
4.16 Hence, somewhat surprisingly, the conditional expected time of ruin, given that ruin occurs, is the shorter, the larger
the security loading. This result, which is in contradistinction to 4.10, can be interpreted as follows. The larger the security loading, the faster the surplus is expected to grow as time passes. A larger surplus, in turn, means a
smaller chance for ruin. Thus, given that ruin is to occur, it must occur earlier in some sense for the case of a larger security loading.
We can use 4.9 to obtain higher moments of T. For n=1, 2, 3,. . . , the factorial moment E
[T T − 1 . . . T − n + 1I T ∞|U 0 = 0] 4.17
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
