224 F. De Vylder, M. Goovaerts Insurance: Mathematics and Economics 26 2000 223–238
We denote by Ut, u the probability of nonruin before t corresponding to the initial risk reserve u≥0, and by U
n
t, u the corresponding conditional probability of nonruin for fixed N
t
= n. Then
U t, u = X
n≥0
P N
t
= nU
n
t, u. 1
The homogeneous model with fixed N
t
= n is defined by the probabilities PN
t
= n=1 and PN
t
= m=0 m6=n.
Then U
n
t, u is the nonconditional probability of ruin in the homogeneous model with fixed N
t
= n. In the
homogeneous model with fixed N
t
= n, the claim instants are T
1
T
2
· · · T
n
and the random vector T
1
, . . . , T
n
has a constant density on the subset W
tn
= { t
1
, . . . , t
n
|0 t
1
t
2
· · · t
n
t } of R
n
. The Lebesgue volume of W
tn
is t
n
n. Hence the density of T
1
, . . . , T
n
equals nt
n
on W
tn
. Then, by the foregoing assumptions, the distribution of the random vector T
1
, . . . , T
n
, X
1
, . . . , X
n
is completely specified in the homogeneous model with fixed N
t
= n.
2. Nonruin probability before time ttt in the homogeneous model
Let τ , y
1
, y
2
, . . . be any numbers. The numbers z =
1, z
1
, z
2
, . . . are defined recursively as follows: −
z
k+1
= z
y
k+1 k+1
k + 1 +
z
1
y
k+1 k
k + · · · +
z
k
y
k+1 1
1 k = 0, 1, 2, . . . .
2 The first numbers z
k
are z
= 1,
z
1
= − y
1
, z
2
= − 1
2 y
2 2
+ y
1
y
2
, z
3
= − 1
6 y
3 3
+ 1
2 y
1
y
3 2
− −
1 2
y
2 2
+ y
1
y
2
y
3
. For all k=1, 2, . . . , we consider the k-tuple integral
I
k
y
1
, . . . , y
k
, τ = Z
[y
k
,τ ]
dτ
k
Z
[y
k−1
,τ
k
]
dτ
k−1···
Z
[y
2
,τ
3
]
dτ
2
Z
[y
1
,τ
2
]
dτ
1
. 3
The integration domains are oriented, i.e. R
[a,b]
= − R
[b,a]
, if ba this agreement is not used in the applications of following Lemma because then 0 ≤ y
1
≤ · · · ≤ y
k
≤ t .
Lemma 1.
I
k
y
1
, . . . ,y
k
,τ = z
τ
k
k +
z
1
τ
k−1
k−1 + · · · +
z
k−1
τ
1
1 +
z
k
τ k= 1, 2, . . . .
4 If y
k
= ky k=1, 2, . . . , then 0=z
2
= z
3
=· · · and
I
k
y, 2y, . . . ,ky,τ = τ
k
k −
yτ
k−1
k − 1 k= 1, 2, . . . .
5
Proof.
I
1
y
1
,τ = Z
[y
1
,τ ]
dτ
1
= τ − y
1
= z
τ
1
1 +
z
1
τ .
Hence 4 is correct for k=1. Then the proof is completed by induction. Indeed, let 4 be correct for k. Then
F. De Vylder, M. Goovaerts Insurance: Mathematics and Economics 26 2000 223–238 225
I
k+1
y
1
, . . . ,y
k+1
,τ = Z
[y
k+1
,τ ]
I
k
y
1
, . . . ,y
k
,τ
k+1
dτ
k+1
= X
0≤i≤k
z
i
Z
[y
k+1
,τ ]
τ
k+1 k−i
k − i dτ
k+1
= X
0≤i≤k
z
i
Z
[y
k+1
,τ ]
dτ
k+1 k+1−i
k+1−i =
X
0≤i≤k
z
i
τ
k+1−i
k+1−i −
X
0≤i≤k
z
i
y
k+1 k+1−i
k+1−i =
X
0≤i≤k
z
i
τ
k+1−i
k+1−i +
z
k+1
= X
0≤i≤k+1
z
i
τ
k+1−i
k+1−i ,
i.e. 4 is correct for k+1. Let us now assume that y
k
= ky. Then z
2
=− 2y
2
+ 2y
2
= 0. By induction we assume that z
2
= z
3
= · · · = z
k
= 0.
Then by 2, −
z
k+1
= z
y
k+1 k+1
k+1 +
z
1
y
k+1 k
k =
y
k+1
k
k+1
k+1 −
y
k+1
k
k
k =
We recall that X
1
, X
2
,. . . are the i.i.d. claim amounts with distribution function F and that u≥0 is the initial risk reserve. We define
Y
k
= X
1
+ · · · + X
k
− u
+
k= 1, 2, . . . , 6
where x
+
= x, if x≥0 and x
+
= 0, if x0. The random variables X
k
are nonnegative. Hence Y
k
= X
1
+ · · · + X
k
, if u=0. The random variables Z
= 1, Z
1
, Z
2
, . . . are defined by induction as follows: −
Z
k+1
= Z
Y
k+1 k+1
k+1 +
Z
1
Y
k+1 k
k + · · · +
Z
k
Y
k+1 1
1 k= 0, 1, 2, . . . .
7 We notice that this is relation 2 with capital letters everywhere. The first random variables Z
k
are Z
= 1,
Z
1
= − Y
1
= − X
1
− u
+
, Z
2
= −
1 2
Y
2 2
+ Y
1
Y
2
= −
1 2
X
1
+ X
2
− u
+ 2
+ X
1
− u
+
X
1
+ X
2
− u
+
, Z
3
= −
1 6
Y
3 3
+
1 2
Y
1
Y
3 2
−
1 2
Y
2 2
+ Y
1
Y
2
Y
3
= −
1 6
X
1
+ X
2
+ X
3
− u
+ 3
+
1 2
X
1
− u
+
X
1
+ X
2
+ X
3
− u
+ 2
− h
−
1 2
− X
1
+ X
2
− u
+ 2
+ X
1
− u
+
X
1
+ X
2
− u
+
i X
1
+ X
2
+ X
3
− u
+
. In Lemma 2, ϕX
1
, . . . , X
n
is any symmetrical function of X
1
, X
2
, . . . , X
n
such that all expectations involved in the proof, are finite. Later in this paper, the lemma is only used with the particular function
ϕX
1
, . . . ,X
n
= 1X
1
+ · · · + X
n
≤ u+ct,
where 1. is the indicator function of a proposition. Then all expectations are integrals with respect to dFx
1
· · · dFx
n
of a bounded function on a bounded integration domain of R
n
, and they are finite indeed.
Lemma 2. Let u=0 and let ϕX
1
, . . . , X
n
be a symmetrical function of X
1
, . . . , X
n
. Then E[Z
1
ϕX
1
, . . . ,X
n
] = − E Y
n
n ϕX
1
, . . . ,X
n
8
226 F. De Vylder, M. Goovaerts Insurance: Mathematics and Economics 26 2000 223–238
and E[Z
k
ϕX
1
, . . . ,X
n
] = 0 k= 2, 3, . . . ,n
9 if all involved expectations are finite.
Proof. The proof results from the adequate use of the following remark. Let us consider an expectation such as E[fX
1
, . . . , X
k
ϕX
1
, . . . , X
n
], where k≤n. By the i.i.d. assumption on X
1
, X
2
, . . . , we may replace X
1
, . . . , X
n
with any permutation Xi
1
, . . . , Xi
n
. But ϕX
1
, . . . , X
n
equals ϕXi
1
, . . . , Xi
n
by the symmetry assumption on ϕ. Hence
E[f X
1
, . . . ,X
k
ϕX
1
, . . . ,X
n
] =E[f Xi
1
, . . . ,Xi
k
ϕX
1
, . . . ,X
n
] 10
for any k different subscripts i
1
, . . . , i
k
in the set {1, 2, . . . , n}. We can also consider various subsets {i
1
, . . . , i
k
⊆ {1, . . . , n} and consider 10 for each subset. Summing up and dividing by the number of subsets we see
that E[f X
1
, . . . ,X
k
ϕX
1
, . . . ,X
n
] =E X
f Xi
1
, . . . ,Xi
k
m ϕX
1
, . . . ,X
n
, 11
where m is the number of subsets {i
1
, . . . , i
k
}. We now prove 8. By 11
E[Z
1
ϕX
1
, . . . ,X
n
] = − E[X
1
ϕX
1
, . . . ,X
n
] = − E Y
n
n ϕX
1
, . . . ,X
n
. We now prove 9 for k=2. Considering the subsets
{1, 2} and {2, 1} of {1, . . . , n}, E[Z
2
ϕX
1
, . . . ,X
n
] =E[ −
1 2
Y
2 2
+ Y
1
Y
2
ϕX
1
, . . . ,X
n
] =E[ −
1 2
Y
2 2
+
1 2
Y
2
Y
1
ϕX
1
, . . . ,X
n
] = 0. We now consider 9 for k=3. Then the expectation is the sum of the two terms
E −
1 6
Y
3 3
+ 1
2 Y
1
Y
3 2
ϕX
1
, . . . ,X
n
and E 1
2 Y
2 2
Y
3
− Y
1
Y
2
Y
3
ϕX
1
, . . . ,X
n
In the first term Y
1
can be replaced with
1 3
Y
3
. Hence, it equals zero. In the last term, Y
1
can be replaced with
1 2
Y
2
. Hence, it also equals zero.
A general proof by induction is direct. It is based on 7. In the last member of that relation, the first two terms must be treated jointly; all other terms can be treated separately.
In Theorem 1, F
∗n
is of course the distribution function of the total claim amount X
1
+· · ·+ X
n
in the homogeneous model with fixed N
t
= n. The function I
n
is the n-tuple integral defined by 3. Of course U t, u=1 because ruin
cannot occur if there are no claims. In the general homogeneous model on [0, t], we denote by F
t
the distribution function of the total claim amount S
t
= X
1
+ X
2
+ · · · + X
N
t
in [0, t]. F
t
x = X
n≥0
P N
t
= nF
∗ n
x x ≥ 0.
12
Theorem 1.
a. In the homogeneous model with fixed N
t
= n,
U
n
t ,u =nct
− n
E[I
n
Y
1
, . . . ,Y
n
,ct1X
1
+ · · · + X
n
≤ u+ct].
13
F. De Vylder, M. Goovaerts Insurance: Mathematics and Economics 26 2000 223–238 227
b. In the homogeneous model with fixed N
t
= n and u=0,
U
n
t , 0 =E 1 −
Y
n
ct 1Y
n
≤ ct
= Z
[0,ct]
h 1 −
y ct
i dF
∗ n
y = ct
− 1
Z
[0,ct]
F
∗ n
y dy. 14
c. Prabhu’s formula In the general homogeneous model on [0, t] with u=0, U t , 0 =
Z
[0,ct]
h 1 −
y ct
i dF
t
y = ct
− 1
Z
[0,ct]
F
t
y dy. 15
Proof.
a. Taking the distribution of the random vector T
1
, . . . , T
n
, X
1
, . . . , X
n
into account see Section 1, the nonruin probability U
n
t, u equals U
n
t ,u =nt
− n
Z · · ·
Z
D
dt
1
· · · dt
n
dF x
1
· · · dF x
n
, where the integration domain D is the subset of R
2n
of points t
1
, . . . , t
n
, x
1
, . . . , x
n
satisfying the relations x
1
≥ 0, . . . ,x
n
≥ 0,
0 ≤ t
1
≤ · · · ≤ t
n
≤ t
x
1
≤ u+ct
1
, x
1
+ x
2
≤ u+ct
2
, . . . ,x
1
+ x
2
+ · · · + x
n
≤ u+ct
n
. Instead of t
1
, . . . , t
n
, let us take the new integration variables t
1
= ct
1
, . . . , t
n
= ct
n
. Then U
n
t ,u =nct
− n
Z · · ·
Z
1
dτ
1
· · · dτ
n
dF x
1
· · · dF x
n
, where the integration domain 1 is the subset of R
2n
of points t
1
, . . . , t
n
, x
1
, . . . , x
n
satisfying the relations x
1
≥ 0, . . . ,x
n
≥ 0,
0 ≤ τ
1
≤ · · · ≤ τ
n
≤ ct,
x
1
≤ u+τ
1
, x
1
+ x
2
≤ u+τ
2
, . . . ,x
1
+ x
2
+ · · · + x
n
≤ u+τ
n
. These relations are equivalent to the relations
x
1
≥ 0, . . . ,x
n
≥ 0,
x
1
+ · · · + x
n
≤ u+ct ,
16 x
1
+ · · · + x
n−1
+ x
n
− u
+
≤ τ
n
≤ ct ,
x
1
+ · · · + x
n−1
− u
+
≤ τ
n−1
≤ τ
n
, · · ·
· · · · · ·
· · · · · ·
· · · · · ·
x
1
+ x
2
− u
+
≤ τ
2
≤ τ
3
, x
1
− u
+
≤ τ
1
≤ τ
2
. With the notation y
k
= x
1
+ · · · + x
k
− u
+
, these relations are 16 and y
n
≤ τ
n
≤ ct,
y
n−1
≤ τ
n−1
≤ τ
n
, . . . ,y
2
≤ τ
2
≤ τ
3
, y
1
≤ τ
1
≤ τ
2
. 17
Then, by Fubini, U
n
t ,u =nct
− n
Z · · ·
Z
1
′
Z · · ·
Z
1
′′
dτ
1
· · · dτ
n
dF x
1
· · · dF x
n
, 18
where the integration domain 1
′
is the subset of R
n
of points x
1
, . . . , x
n
satisfying 16 and where for fixed x
1
, . . . , x
n
, the integration domain 1
′′
is the subset of R
n
of points t
1
, . . . , t
n
satisfying 17. The n-tuple integral in square brackets of 18 is I
n
y
1
, . . . , y
n
,ct defined by the relation 3. Incorporating the integration domain 1
′
in the integrand by the use of the indicator function 1., it is clear that 18 is the same as 13.
228 F. De Vylder, M. Goovaerts Insurance: Mathematics and Economics 26 2000 223–238
b. Let us now assume that u=0. By Lemmas 1 and 2, I
n
Y
1
, . . . , Y
n
, ct can be replaced with ct
n
n −
Y
n
nct
n−1
n − 1 =
ct
n
n
− 1
1 − Y
n
ct in 13. This proves the first relation 14. The second relation 14 is obvious. Then the last relation 14 results
from an integration by parts. c. 15 results from 1, 12 and 14.
General expectations such as E[ϕX
1
, . . . , X
n
] are almost impossible to evaluate numerically, fastly and precisely enough, if n is large. In particular, 13 is not useful for the practical evaluation of U
n
t, u. On the contrary, expectations such as E[ϕX
1
+ · · · + X
n
] can be calculated. Indeed, they are reduced to single integrals
E[ϕX
1
+ · · · + X
n
] = Z
I
ϕy dF
∗ n
y. In practice I is a bounded interval and F the discretized, i.e. F is hold back as a long vector in the computer
program. Then F
∗n
can be calculated iteratively and only two successive long vectors F
∗k
and F
∗ k +1
k=1,2, . . . , n must be stored simultaneously in the process at each stage.
If the discretized F has ν components, then the direct calculation of E[ϕX
1
, . . . ,X
n
] = Z
· · · Z
D
ϕx
1
, . . . ,x
n
dF x
1
· · · dF x
n
may need the evaluation of ν
n
terms, to be summed up. In the case of U
n
t, u the evaluation of ϕx
1
, . . . , x
n
goes via relations such as 4. All this must be done for values n=0, 1, 2, . . . , n
and then Ut, u should result from 1. In some cases ν and n
may be pretty large, say ν=1000 and n =
100. Clearly this rudimentary procedure is hopeless, even with the fastest computers available today.
3. Homogeneous model with equalized claim amounts
