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

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

Multiplying on both sides <strong>of</strong> the equality by n−1<br />

X uu i/u i X u and summing over n = 1<br />

to +, weget<br />

u i<br />

<br />

<br />

<br />

<br />

u S u = a X uu i <br />

i u S u + a + bu i <br />

i u X u S u<br />

i<br />

Comparing identical powers <strong>of</strong> u on both sides <strong>of</strong> the equality gives:<br />

s∑<br />

s∑<br />

s i gs = a fxs i − x i gs − x + a + b fxx i gs − x<br />

x=0<br />

⇔ s i gs = as i f0gs +<br />

⇔ gs =<br />

1<br />

1 − af0<br />

x≠0<br />

x=0<br />

s∑<br />

fxgs − x ( )<br />

as i + bx i<br />

x≠0<br />

(<br />

s∑<br />

fxgs − x a + b x )<br />

i<br />

<br />

s i<br />

which ends the pro<strong>of</strong>.<br />

□<br />

The Panjer formula in dimension 1 immediately follows from Property 9.1 by putting<br />

k = 1.<br />

Multivariate De Pril Algorithm<br />

Another interesting problem is to compute the multivariate convolution <strong>of</strong> a random vector.<br />

This is actually the result we will need in the following sections. Let X i = X i1 X ik T ,<br />

i = 1 2n, be independent and identically distributed realizations <strong>of</strong> the random vector<br />

X = X 1 X k T . We want to obtain the distribution <strong>of</strong> the random vector<br />

( ) n∑ T<br />

n∑<br />

S = S 1 S k T = X i1 X ik (9.6)<br />

i=1<br />

i=1<br />

The probabiliy mass function <strong>of</strong> S is the n-fold convolution f ⋆n x. The multivariate version<br />

<strong>of</strong> De Pril’s formula provides a recursive formula to derive the distribution <strong>of</strong> S.<br />

Property 9.2 Let X i = X i1 X ik T be independent and identically distributed<br />

realizations <strong>of</strong> the random vector X defined on the nonnegative integers and with probability<br />

function such that f0>0. Then, the following recursion holds :<br />

f ⋆n 0 = f n 0<br />

f ⋆n s = 1 ( )<br />

s∑ n + 1<br />

x<br />

f0 s i − 1 f ⋆n s − xfx<br />

i<br />

s i ≥ 1 i= 1k<br />

x≠0<br />

Pro<strong>of</strong> Let us introduce an auxilliary random vector W with probability mass function<br />

h0 = 0 and<br />

hx =<br />

fx<br />

1 − f0 x > 0

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