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

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

The maximum likelihood estimator for and a solve<br />

(<br />

)<br />

<br />

n∑<br />

La= a + k<br />

˜x i k i − i<br />

i = 0<br />

a + i<br />

i=1<br />

Note that these equations are similar to the ones obtained in the Poisson case except that i<br />

is now replaced with i a + k i /a + i .<br />

Remark 2.4 Let us give an intuitive explanation for the ratio a + k i /a + i involved<br />

in the Negative Binomial likelihood equations. The joint probability density function <strong>of</strong><br />

N i i equals<br />

( ) ki<br />

i i 1<br />

exp− i i <br />

k i ! a aa a−1<br />

i<br />

exp−a i <br />

∝ exp − i i k i+a−1<br />

i exp−a i <br />

The probability density function <strong>of</strong> i given N i = k i is<br />

exp − i a + i a+k i−1<br />

i<br />

∫ +<br />

0<br />

exp − a + i a+k i−1<br />

d<br />

= exp − i a + i a+k a + <br />

i−1 i a+k i<br />

i<br />

a + k i <br />

so that i given N i = k i follows the am a + k i a+ i distribution. Therefore,<br />

E i N i = k i = a + k i<br />

a + i<br />

and the maximum likelihood estimators in the Negative Binomial regression model solve<br />

n∑<br />

i=1<br />

)<br />

x i<br />

(k i − i E i N i = k i = 0<br />

Compared to the Poisson likelihood equations, the predicted expected claim number i is<br />

replaced with its update i E i N i = k i based on the information contained in the number<br />

k i <strong>of</strong> claims filed by policyholder i.<br />

As already shown in Section 2.3.7, it is possible to solve the Negative Binomial likelihood<br />

equations with the help <strong>of</strong> the Newton–Raphson iterative procedure. Starting values for the<br />

Newton–Raphson iterative procedure are usually obtained as follows: The Poisson maximum<br />

likelihood estimator ̂ is known to be consistent, so that we keep it as a reasonable starting<br />

value. We still have to find an initial estimate for = V i = 1/a. So, we first compute<br />

the variance <strong>of</strong> N i which is given by<br />

VN i = EN i + ( d i expscore i ) 2

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