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

tel-00703797, version 2 - 7 Jun 2012
In our case ∗ , we have
and
F j x(z, y) = nj
n lgj
j (xj y)[1 − Sj(x j y)(y − πj)],
∂F j x
∂z (z, y) = ∂2Oj (x
∂z∂xj
j y, z), and ∂F j x
∂y (z, y) = ∂2Oj The first-order derivative is given by
2.3. Refinements of the one-period model
∂x 2 j
(x j y, z).
∂F j x
(z, y) = 2nj
∂y n lgj j (xjy)Sj(x j y) (y − πj)Sj(x j y) − 1 − nj
n lgj j (xjy)(y−πj)
Using F j x(z, ϕ(z)) = 0 whatever z represents, it simplifies to
∂F j x
(z, ϕ(z)) = −nj
∂y n lgj j (xj ϕ(z)
)(ϕ(z) − πj)
Let z now be the insurer’s break-even premium z = πj. We have
Thus, the derivative of ϕ is
l=j
∂F j x
∂z (z, y) = nj lg j
j (xj y)Sj(x j y).
ϕ ′ (z) =
(ϕ(z) − z)
Sj
x j
ϕ(z)
f
l=j
′ j1 (ϕ(z), xl) 2 lg l j
By definition, F j x(z, ϕ(z)) = 0 is equivalent to
1 = Sj x j
ϕ(z) (ϕ(z) − z).
l=j
f ′ j1(ϕ(z), xl) 2 lg l j(x j
ϕ(z) ).
x j
.
ϕ(z)
f ′ j1(y, xl) 2 lg l j(x j y).
Thus ϕ(z) − z > 0. We conclude that ϕ ′ (z) > 0, i.e. the function πj ↦→ x ⋆ j (πj) is increasing.
Let z be the intercept lapse parameter z = µj. By differentiating the lapse probability, we
have
∂ lg j
j
∂z (xjy) = − lg j
j (xjy)
lg l j(x j y) and ∂ lgkj ∂z (xj
y)
We get
l=j
j=k
= − lg k j (x j y)
lg l j(x j y) + lg k j (x j y).
∂F j x
∂z (z, y) = −nj lg j
j (xj y)(1 − lg j
j (xj y)) 1 − Sj(x j y)(y − πj) − nj lg j
j (xj y) 2 Sj(x j y).
Note the first term when y = ϕ(z) since F j x(z, ϕ(z)) = 0. We finally obtain
ϕ ′
Sj x
(x) = −
j
ϕ(z) lg j
j x j
ϕ(z)
(ϕ(z) − z)
f ′ j1 (ϕ(z), xl) 2 lg l
j x j
.
ϕ(z)
∗. To simplify, we do not stress the dependence of lg k
j and Sj on f.
l=j
l=j
111