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

Risk Classification 89
2.8 Risk Classification for Portfolio A
So far, we have several competing models for the observed claim frequencies in Portfolio
A. This section purposes to compare these models in order to select the optimal one.
2.8.1 Comparison of Competing models with the Vuong Test
Let us consider two non-nested competing models for the number of claims, with respective
probability mass functions p·x and q·x , where x is a vector of explanatory
variables, and and include the regression parameters as well as some dispersion
coefficients . The corresponding log-likelihoods are
L p =
L q =
n∑
ln pk i x i
i=1
n∑
ln qk i x i
In this regression context, the test statistic proposed by Vuong (1989) is
i=1
T LRNN = L p̂ − L q ̂
√ n
(2.14)
where
2 = 1 n
(
n∑
i=1
ln pk ix i ̂
qk i x i ̂
) 2
−
(
1
n
n∑
i=1
)
ln pk 2
ix i ̂
(2.15)
qk i x i ̂
is the estimate of the variance of the log-likelihood difference. None of the model has to
be true: the test is aimed at selecting the model that is the closer to the true conditional
distribution. The null hypothesis of the test is that the two models are equivalent. Under the
null hypothesis, the test statistic is asymptotically Normally distributed. Rejection in favour
of p happens when T LRNN >c or in favour of q if T LRNN < −c, where c represents the
or 0 1 critical value for some significance level. If T LRNN ≤c then the null hypothesis
is not rejected and the Vuong test cannot discriminate between the two models, given the data.
Now, comparing the three mixed Poisson models with the Vuong test gives:
Negative Binomial against Poisson-Inverse Gaussian
to −0754544 leading to a p-value of 4506 %.
the value of the test statistic is equal
Poisson-LogNormal against Negative Binomial
to −0470894 leading to a p-value of 6378 %.
the value of the test statistic is equal to
Poisson-LogNormal against Poisson-Inverse Gaussian the value of the test statistic is
equal to to −0216702 leading to a p-value of 8284 %.
These results do not enable us to distinguish between the three models which are
therefore statistically equivalent. The Negative Binomial model will be used for the numerical