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

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<strong>Credibility</strong> Models for <strong>Claim</strong> <strong>Counts</strong> 137<br />

determination <strong>of</strong> conditions on 0 b 1 2 s−1 so that s 0 b is not void, or<br />

the obtention <strong>of</strong> the elements in s 0 b with the minimal number <strong>of</strong> support points.<br />

These elements possess some extremal properties and will be used here in connection with<br />

credibility formulas. Henceforth, we tacitly assume that 0b and 1 2 s−1 are such<br />

that the associated moment space s 0 b is not void and is not a singleton (i.e.,<br />

s 0 b consists <strong>of</strong> at least two distinct distribution functions).<br />

De Vylder’s Moment Problem<br />

De Vylder (1996, Section 8.3) investigated the following problem: within s 0 b ,<br />

determine the random variables X s<br />

min<br />

and Xs max such that the inequalities<br />

EX s<br />

min s ≤ EX s ≤ EX s<br />

max s hold for all X ∈ s 0 b (3.8)<br />

Explicit solutions to (3.8) are available for s up to five. The supports <strong>of</strong> the extremal<br />

distributions are given in Tables 3.5–3.6.<br />

As shown in Denuit, De Vylder & Lefèvre (1999), the random variables X s<br />

X s<br />

min and<br />

max involved in (3.8) give bounds on EX for every function that is s − 2 times<br />

differentiable, with a convex s − 2th derivative (such functions are called s-convex; see<br />

Roberts & Varberg (1973) for details). In that context, X s<br />

min<br />

is called the s-convex<br />

minimum, and Xmax s is called the s-convex maximum.<br />

Table 3.5<br />

s = 1to5.<br />

Probability distribution <strong>of</strong> X s<br />

min ∈ s0 b achieving the lower bound in (3.8) for<br />

s Support points Probability masses<br />

1 0 1<br />

2 1 1<br />

3 0<br />

2 − 2 1<br />

2<br />

2<br />

1<br />

2 1<br />

2<br />

4 r + = 3 − 1 2 + √ 3 − 1 2 2 − 4 2 − 2 1 1 3 − 2 2 <br />

2 2 − 2 1 <br />

r − = 3 − 1 2 − √ 3 − 1 2 2 − 4 2 − 2 1 1 3 − 2 2 <br />

2 2 − 2 1 <br />

1 − r −<br />

r + − r −<br />

1 − 1 − r −<br />

r + − r −<br />

5 0 1− q + − q −<br />

t + = 1 4 − 2 3 + √ 1 4 − 2 3 2 − 4 1 3 − 2 2 2 4 − 2 3 <br />

2 1 3 − 2 2 q + = 2 − t − 1<br />

t + t + − t − <br />

t − = 1 4 − 2 3 − √ 1 4 − 2 3 2 − 4 1 3 − 2 2 2 4 − 2 3 <br />

2 1 3 − 2 2 q − = 2 − t + 1<br />

t − t − − t +

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