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tel-00703797, version 2 - 7 Jun 2012
The Error function From page 164 of Olver et al. (2010), we have
erfc(z) ∼ e−z2
z √ π
N−1
k=0
k 1.3.5 . . . (2k − 1)
(−1)
(2z2 ) k + RN(z)
4.6. Appendix
for z → +∞ and | arg(z)| < 3π/4 We can deduce the asymptotic of the error function at −∞
or +∞ using erf(x) = 1 − erfc(x), erf(−x) = −erf(x) and erfc(−x) = 2 − erfc(x). We get
erf(x) ∼ +∞ 1 − e−x2
x √ π , erfc(x) ∼ +∞
4.6.2 For the continuous time model
The special case of the Lévy distribution
e−x2 x √ π , erf(x) ∼ e−x2
−1 +
−∞ x √ π , erfc(x) ∼ e−x2
2 −
−∞ x √ π .
As for the incomplete gamma function, the function Γ(., .; .) satisfies a recurrence on the
a parameter,
Γ(a + 1, x; b) = aΓ(a, x; b) + bΓ(a − 1, x; b) + x a e −x−b/x ,
see Theorem 2.2 of Chaudry and Zubair (2002). Thus we deduce
Γ(−3/2, x; b) = 1
Γ(1/2, x; b) + 1/2Γ(−1/2, x; b) − x
b
−1/2 e −x−b/x
.
As reported in Theorem 2.6 and Corollary 2.7 of Chaudry and Zubair (2002), Γ(a, x; b) has a
simpler expresssion in terms of the error function when a = 1/2, −1/2, . . . ,
and
Therefore, we have
Γ(−3/2, x; b) =
with
It yields
√
π
Γ(1/2, x; b) =
2
√
π
Γ(−1/2, x; b) =
2 √
b
√ π
2b
Γ(−3/2, θ0u; α 2 u/4) = 2√ π
α 2 u
e 2√ b erfc
−e 2√ b erfc
x +
x +
√
b
x
√
b
x
+ e −2√ b erfc
+ e −2√ b erfc
x −
√
b
x
√
b
x − .
x
1 − 1
2 √
e
b
2√
b
erfc (d+) + 1 + 1
2 √
e
b
−2√b erfc (d−) − 2
√ e
πx −x−b/x
,
d+ = √ x +
√
b
x and d− = √ √
b
x −
x .
1 − 1
α √
e
u
α√
u
erfc (d+) + 1 + 1
α √
u
e −α√ u erfc (d−)
− 2
√ e
πuθ0
−uθ0−α2 /(4θ0)
,
207