Etude des marchés d'assurance non-vie à l'aide d'équilibres de ...

tel-00703797, version 2 - 7 Jun 2012
4.3. Asymptotics – the A + B/u rule
Using an integration by part theorem requires that the following limit to exist
fΘ(− log(1 − x))
lim
= lim
x→1 1 − x
t→+∞ fΘ(t)e t .
This requirement is strong and will be satisfied only for light-tailed distributions and a certain
range of parameter values. For example, when Θ is exponentially distributed E(λ), the previous
constraint imposes λ > 1.
Another way to deal with such integral asymptotic is to apply the Pascal formula, assuming
u is an integer. We get
P (X > u) =
u
k=0
u
(1 − q)(−1)
k
k
+∞
0
e −kθ dFΘ(θ).
The integral is (once again) the Laplace transform LΘ of the random variable Θ at k. This
binomial alternating sum requires a special treatment because finding an asymptotic for LΘ(k)
will not help us to derive an asymptotic of the sum. This issue is studied in the next subsection.
4.3.5 Binomial alternating sum for claim tails
In the discrete time framework, the survival function can be expressed as
P (X > u) =
u
k=0
u
(1 − q)(−1)
k
k LΘ(k),
where LΘ denotes the Laplace transform of the random variable Θ.
This integral falls within the framework of the alternating binomial sum defined as
Sn(φ) =
n
k=n0
n
(−1)
k
k φ(k), (4.7)
where 0 ≤ n0 ≤ n, φ is a real function and n ∈ N is large. n0 can be used to exclude first few
points of the sum that would not be defined.
Letting
λ
φ(k) = (1 − q)
α
,
(λ + j) α
in Equation (4.7), we get the distribution function of X when Θ is gamma distributed of
Subsection 4.2.2. Note that, having started with the integral representation +∞
0 P (X =
k|Θ = θ)dFΘ(θ) of P (X = k), we know that the alternating sum is valued on [0, 1]. This is
not immediate without that integral representation.
Let us point out that the probability P (X = k) is a decreasing function of k. This is
not easy to see by using the alternating binomial sum representation (4.7). Here is a simle
proof. To indicate the dependence on the parameter λ, denote by P (X = k)λ. From algebraic
manipulation and using the binomial recurrence equation n n−1 n−1
k = k + k−1 , we get
λ
P (X = k + 1)λ = P (X = k)λ − P (X = k)λ+1
α
.
(λ + 1) α
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