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

180 Actuarial Modelling of Claim Counts
On the other hand, T E = e T . Thus, e T x = 0, which implies Ex = 0. Consequently,
This implies for any n ≥ 1 that
I − Px = 0 ⇔ Px = x
x = P n x → x
i.e. x i = ∑ s
j=0 jx j for all i = 0s. Because the right-hand side of these equations
does not depend on i, we have x = ce for some c ∈ . Since we also have
0 = e T x = ce T e = cs + 1 ⇒ c = 0
Thus, I − P + E is invertible. Furthermore, since T I − P = 0, we have
T I − P + E = T E = e T
This proves (4.9).
□
If the number s + 1 of states is small, the matrix I − P + E can easily be inverted. For
larger s + 1, numerical methods have to be used, like the Gaussian elimination algorithm.
Example 4.9 (−1/top scale) We have seen above that p 5
5l = l for l = 0 15,
so that the system needs 5 years to reach stationarity (i.e. the time needed by the best
policyholders starting from level 5 to arrive in level 0). Formula (4.9) gives here
T = 1 1 1 1 1 1
⎛
×
⎜
⎝
2 − exp− 1 1 1 1 exp−
1 − exp− 2 1 1 1 exp−
1 1− exp− 2 1 1 exp−
1 1 1− exp− 2 1 exp−
1 1 1 1− exp− 2 exp−
1 1 1 1 1− exp− 1 + exp−
Applying this formula in the particular case = 01, we get
T 01 = 1 1 1 1 1 1
⎛
×
⎜
⎝
1095163 1 1 1 1 0904837
0095163 2 1 1 1 0904837
1 0095163 2 1 1 0904837
1 1 0095163 2 1 0904837
1 1 1 0095163 2 0904837
1 1 1 1 0095163 1904837
⎞
⎟
⎠
−1
⎞
⎟
⎠
−1