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

tel-00703797, version 2 - 7 Jun 2012
2.2. A one-period model
where λ j⋆ ∈ R 3 are Lagrange multipliers, see, e.g., Facchinei and Kanzow (2009). In the last
part of equation (2.7), gj(x ⋆ j )T λ j⋆ = 0 is the complementarity equation implying that the l
constraint gl j is either active (gl j (x⋆j ) = 0) or inactive (gl j (x⋆j ) > 0), but λj⋆
l = 0.
We suppose that x⋆ j ∈]x, x[. Hence, λj⋆ 2 = λj⋆ 3 = 0. There are two cases: either the solvency
constraint g1 j is active or not. Let us assume the solvency constraint is inactive. Insurer j’s
premium equilibrium verifies ∇xjOj(x ⋆ ) = 0, i.e.
nj
n
x
1 − 2βj
⋆ j
mj(x⋆ ) + βj
πj
+ βj
mj(x⋆
= 0. (2.8)
)
Let x j y be the premium vector with the j component being y, i.e. x j y = (x1, . . . , xj−1, y, xj+1, . . . , xI).
We denote by z a parameter of interest and define the function F as
F j x(z, y) = ∂Oj
(x
∂xj
j y, z),
where the objective function depends (also) on the interest parameter z. Equation (2.8) can
be rewritten as F j
x⋆(z, x⋆j ) = 0.
By the continuous differentiability of F with respect to z and y and the fact that F j x(z, y) =
0 has at least one solution (z0, y0), we can invoke the implicit function theorem, see Appendix
2.6.1. So there exists a function ϕ defined in a neighborhood of (z0, y0) such that F j x(z, ϕ(z)) =
0 and ϕ(z0) = y0. Furthermore, if ∂F j x
∂y (z0, y0) = 0, the derivative of ϕ is given by
In our case, we have
∂F j x
ϕ ′ (z) = −
∂F j x
∂z
∂F j x
∂y
(z, y)
(z, y)
y=ϕ(z)
∂y (z, y) = ∂2Oj (x j y, z) = −2αj < 0.
nmj(x)
∂x 2 j
As a consequence, the sign of ϕ ′ is simply
Let us consider z = πj. We have
sign ϕ ′ (z) = sign
.
nj
∂F j x
(z, ϕ(z))
∂z
∂F j x njβj
(z, y) = > 0.
∂z nmj(x)
Thus, the function πj ↦→ x ⋆ j (πj) is increasing.
Let z be the sensitivity coefficient βj. We have
∂F j
x nj
(z, y) = −2βj
∂z n
y
mj(x)
πj
+ 1 + .
mj(x)
.
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