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

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

LogNormal Distribution<br />

Recall that a random variable X is Normally distributed with mean and variance 2 ,<br />

denoted as X ∼ or 2 , if its distribution function is<br />

where<br />

( x − <br />

)<br />

Fx = <br />

<br />

x = 1 √<br />

2<br />

∫ x<br />

−<br />

exp−y 2 /2 dy (1.44)<br />

Now, a random variable X is LogNormally distributed with parameters and (notation<br />

X ∼ or 2 )iflnX is Normally distributed with mean and variance 2 , that is, if<br />

its probability density function is given by<br />

fx =<br />

(<br />

1<br />

√ exp − 1<br />

)<br />

2x 2 ln x − 2 2 <br />

If X ∼ or 2 , then its mean is<br />

)<br />

EX = exp<br />

( + 2<br />

<br />

2<br />

and its variance<br />

VX = exp ( 2 + 2)( exp 2 − 1 ) <br />

x>0<br />

Poisson-LogNormal Distribution<br />

Taking =− 2 /2 (to ensure that E = 1), the probability density function <strong>of</strong> ∼<br />

or− 2 /2 2 is<br />

(<br />

1<br />

f =<br />

√ 2 exp − ln + )<br />

2 /2 2<br />

>0 (1.45)<br />

2 2<br />

The probability mass function <strong>of</strong> the Poisson-LogNormal distribution is given by<br />

PrN = k = 1<br />

√ d k<br />

2 k!<br />

∫ <br />

Coming back to (1.28) and (1.29), we easily see that<br />

0<br />

exp−d k−1 exp<br />

(− ln + )<br />

2 /2 2<br />

d<br />

2 2<br />

EN = and VN = + 2( exp 2 − 1 ) <br />

It can be shown that / √ V = 2 + exp 2 in (1.31) for the Poisson-LogNormal<br />

distribution. Therefore the skewness <strong>of</strong> a Poisson-LogNormal distribution exceeds the<br />

skewness <strong>of</strong> the Poisson-Inverse Gaussian distribution having the same mean and the same<br />

variance.

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