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

Bonus-Malus Scales 179
does not depend on the starting class. This means that the nth power P n of the one-step
transition matrix P converges to a matrix with all the same rows T , that is
⎛
lim
n→+ Pn = = ⎜
⎝
T
T
T
exactly as we saw in the introductory examples.
Let us now explain how to compute the l s. Taking the limit in both sides of (4.5)
for n →+, we see that the vector is the unique probabilistic solution to the system
of linear equations
j =
⎞
⎟
⎠
s∑
l p lj j ∈ 0s (4.7)
l=0
In matrix notation, (4.7) can be written as
{ T = T P
T e = 1
(4.8)
where e is a column vector of 1s. This means that is the left eigenvector u 0 of P
encountered above. Thus we see that if the initial distribution in the scale is , then the
probability distribution remains equal to .
4.4.2 Rolski–Schmidli–Schmidt–Teugels Formula
Let E be the s +1×s +1 matrix all of whose entries are 1, i.e. consisting of s +1 column
vectors e. Then, the following result provides a direct method to get .
Property 4.2 Assume that the stochastic matrix P is regular. Then the matrix I −P+
E is invertible and the solution of (4.8) is given by
T = e T I − P + E −1 (4.9)
Proof
Let us first check that I − P + E is invertible. We must show that
I − P + Ex = 0 ⇒ x = 0
From (4.8), we have T I − P = 0. Thus,
I − P + Ex = 0 ⇒ 0 = T I − P + Ex = 0 + T Ex
⇔ T Ex = 0