Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 39 2 Solution Created 2026-10-03 Updated 2026-10-14
Let and write for the standard normal density, with . The change of variables converts the density integral intoComplete the square using . Integration gives the truncated lognormal momentHere is the standard normal distribution function; take and . The identity holds for every real , including the full-interval limits, because the shifted Gaussian integral remains finite.
The limited excess of loss reinsurance payment is . It is zero below , equals for , and equals above . Since the lognormal distribution has a continuous density,Put , , and . Applying the moment identity with and gives the conditional lognormal layer recoveryThe denominator is , and payments outside that event are zero. The three numerator terms are respectively the truncated first moment, the attachment subtracted from partial payments, and the capped payments on claims above the layer. This proves the conditional expectation including the limit on each reinsurer payment.
For claims inflation by , set . Its log-mean is and its log-standard-deviation stays . The attachment and payment limit stay fixed. Therefore the requested new conditional mean isExplicitly, in the preceding fraction replace by and replace by . Equivalently, claims inflation of a fixed reinsurance layer givesbecause and the new conditioning event is . Merely multiplying the original conditional mean by would omit both of these threshold changes.