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

148 Actuarial Modelling of Claim Counts
Hence, the value of c is does not depend on s and the same weight is given to all the past
annual claim numbers. Now, inserting the value of c is that we just obtained in (3.14) finally
gives
which in turn yields.
c i0 =
c is =
iTi +1
1 + V i ∑ T i
t=1 it
iTi +1V i
1 + V i ∑ T i
t=1
it
The expected claim frequency for year T i + 1 given past claims history is
̂N iTi +1 = iTi +1
1 + V i N i•
1 + V i i•
(3.16)
where N i• and i• have been defined in (3.1). Note that ̂N iTi +1 is the best linear predictor
of each of the true unknown means iTi +1 i , of the Bayesian credibility premium
EN iTi +1N i1 N iTi and of the number of claims N iTi +1 for year T i + 1.
The linear predictor for year T i + 1 thus appears as the product of the a priori
expected claim frequency, iTi +1, times an approximation of the theoretical correction
E i N i1 N iTi . This approximation possesses a particularly simple interpretation since
it entails a malus when N i• > i• , that is, if the policyholder reported more claims than
expected a priori.
Remark 3.2 Note that (3.16) agrees with the result obtained in the Poisson-Gamma case.
This is because the Bayesian credibility premium is linear in the past observations in the
Poisson-Gamma case. The term exact credibility is used to describe the situation where
the linear credibility premium equals the Bayesian one. Intuitively speaking, using linear
credibility formulas in the mixed Poisson model boils down to approximating the mixing
distribution with the Gamma distribution (or, equivalently, the distribution of the claim
numbers with the Negative Binomial one).
Remark 3.3 Note that ̂N iTi +1 could have been obtained by a direct application of the
Bühlmann–Straub formula. Let us consider a sequence of random variables X 1 X 2 X 3
such that, given a random variable , the X t s are independent, with finite first and second
moments
Now, define M 2 and 2 as
= EX t and EX t = = E
E [ VX t ] = 2
w t
and V [ EX t ] = M 2