Actuarial Modelling of Claim Counts Risk Classification, Credibility ...

340 Actuarial Modelling of Claim Counts
It can be verified that the first order conditions are as follows:
⎧
2
⎪⎨
100 1 + 1 + 1 − + 010 1 + 1 + 1 −
= 100
2 1 + 1 + 1 − + 010
2 1 + 1 + 1 −
⎪⎩
2 001 1 + 1 + 1 − = 001
2 1 + 1 + 1 −
where
xyz a b c =
x y z
s x t y u z s t u ∣
∣∣s=at=bu=c
xyz
2 a b c = x y z
s x t y u z s2 t 2 u 2
∣
∣
s=at=bu=c
for x y z ∈ 0 1. Again, numerical procedures are needed to find the solution of this
optimization problem.
9.3.3 Financial Equilibrium
Analyzing the financial equilibrium of the system now amounts to checking whether
r t t
N 1• N 2• I 12• t is equal to 1 with the optimal values t and t . To this end, we
need the joint distribution of the random vector N 1• N 2• I 12• . The joint probability mass
function of this vector is given in the following result that extends Property 9.3 in the present
setting.
Property 9.4
For fixed , the following recursive formulas
g ⋆t x y z = f ⋆t x y t − z for 0 ≤ z ≤ t x y ≥ 0 and x + y>0
g ⋆t 0 0 0 = e −t
fx y 0 = e − x+y 1 − q x q y
x!y!
for x y ≥ 0 and x + y>0
f0 0 1 = e −
( )
x∑ y∑ t + 1
g ⋆t x y z = e
− 1 g ⋆t x − u y − v z − 1gu v 1
u=0 v=0
z
for 1 ≤ z ≤ t x y ≥ 0 and x + y>z− 1
hold true with the convention that the defined functions take the value 0 where they have
not been defined.
Proof
It is trivial that for t = 1 we have
fx 0z = e −1+ x z
x!z!
f0 1 0 = e −1+
xz>0