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

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82 <strong>Actuarial</strong> <strong>Modelling</strong> <strong>of</strong> <strong>Claim</strong> <strong>Counts</strong><br />

0.8<br />

Overdispersion <strong>of</strong> the data<br />

Empirical variances<br />

0.6<br />

0.4<br />

0.2<br />

0.0<br />

0.0 0.1 0.2 0.3 0.4<br />

Empirical means<br />

Figure 2.9 Mean–Variance pairs for the risk classes, Portfolio A.<br />

A quadratic curve (without intercept) has been fitted to the mean–variance couples by<br />

weighted least-squares (the weights being the exposures <strong>of</strong> the risk classes). The high value<br />

<strong>of</strong> the R 2 coefficient (86.41 %) supports the quality <strong>of</strong> the fit. This shows that equation (2.11)<br />

is supported by the data, and provides empirical evidence for a mixed Poisson model.<br />

2.4.6 Testing for Overdispersion<br />

The graphical test <strong>of</strong> the previous section is an easy way <strong>of</strong> detecting overdispersion. But<br />

many statistical tests for the overdispersion assumption have been developed in the literature.<br />

Testing for overdispersion can be done by testing for the Poisson distribution against a mixed<br />

Poisson model. One problem with standard specification tests (such as likelihood ratio tests)<br />

occurs when the null hypothesis is on the boundary <strong>of</strong> the parameter space, as explained in<br />

Chapter 1. When a parameter is bounded by the H 0 hypothesis, the estimate is also bounded<br />

and the asymptotic Normality <strong>of</strong> the maximum likelihood estimator no longer holds under<br />

H 0 . Consequently, a correction must be made.<br />

Alternatively, testing for overdispersion can be based on the variance function. According<br />

to (2.11), the variance function <strong>of</strong> a heterogeneity model is <strong>of</strong> the form:<br />

VN i = i + 2 i (2.12)<br />

with = V i being the variance <strong>of</strong> the random effect. Therefore, we have to test the null<br />

hypothesis H 0 : = 0 against H 1 : >0. The following score statistics can be used to test the<br />

Poisson distribution against heterogeneity models with a variance function <strong>of</strong> the form (2.12):<br />

T 1 =<br />

∑ n<br />

i=1<br />

(k i −̂ i 2 − k i<br />

)<br />

√<br />

2 ∑ n<br />

i=1̂ 2 i

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