approximate inversion, are justified by theoretical arguments. 2.6 Application: optimal reinsurance An insurer is looking for an optimal reinsurance for a portfolio consisting of 20000 one-year life insurance policies that are grouped as follows: Insured amount by Number of policies 10000 2 3 5000 5000 36 2. The individual risk model The probability of dying within one year is 0.01 for each insured, and the policies are independent. The insurer wants to optimize the probability of being able to meet his financial obligations by choosing the best retention, which is the maximum payment per policy. The remaining part of a claim is paid by the reinsurer. For example, if the retention is 1.6 and someone with insured amount 2 dies, then the insurer pays 1.6, the reinsurer pays 0.4. After collecting the premiums, the insurer holds a capital B from which he has to pay the claims and the reinsurance premium This premium is assumed to be 120% of the net premium. First, we set the retention equal to 2. From the point of view of the insurer, the policies are then distributed as follows: Insured amount by Number of policies n 1½ 10000 10000 The expected value and the variance of the insurer's total claim amount S are equal to ES +2 = 10000 × 1 × 0.01+10000×2×0.01 = 300, Vars=b(1-4)+2(1-42) =10000x 1 x 0.01 x 0.99+10000x4x0.01 x 0.99-495. (2.74) By applying the CLT, we get for the probability that the costs S plus the reinsurance premium 1.2x0.01 x 5000 x 1-60 exceed the available capital B [S-ES Pris+60>B-Pr B-360] √495 (B-360) 495 (2.75) We leave it to the reader to determine this same probability for retentions between 2 and 3, as well as to determine which retention for a given B leads to the largest probability of survival. See the exercises with this section. 2.7 Exercises Section 2.2