Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 40 3 Solution Created 2026-10-03 Updated 2026-10-07
Use the classical risk model surplus , with , and write for its ultimate survival probability. The first-claim decomposition for survival probability follows by conditioning on the first arrival time of the Poisson process. That time has exponential distribution of rate . A claim at time leaves capital if , after which the Markov property restarts the same risk model. ConsequentlyPut and change variable :The convolution is continuous since is bounded and is integrable. Differentiating proves the survival integro-differential equation
For the specified claim law, expand its probability density function as . This is an equal-weight mixture distribution of exponential distributions with rates and . In the convolution, set to obtainSince , the requested coefficients are
For the exponential-mixture differential equation for survival probability, define , and . Differentiating the integrals yieldsApply to the last equation, where . The first two equations eliminate , givingThusThe characteristic polynomial factors asThe ordinary differential equation therefore has solution .
The claim expected value is . The relative safety loading is , and the given zero-capital survival probability is . The limit at infinity sets . A third condition comes from the original survival integro-differential equation: its convolution vanishes at zero, so . Thereforewhich gives and . The final survival probability and ultimate ruin probability areBoth exponential coefficients of are positive. Hence increases from to one, and decreases from to zero. The initial slope is essential: the two stated boundary values alone do not determine all three constants of the eliminated differential equation.