Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 38 / 1 / a

Write $\lambda ={\sum }_i{\lambda }_i$, and suppose first that $\lambda >0$. Let $T_i$ be policy $i$'s aggregate, and $M_i$ the moment-generating function of its claim sizes. The Poisson distribution count has probability generating function $G_i(z)=\exp ({\lambda }_i(z-1))$, so the random-sum transform identity and independence of the policies giveThe expression in the weighted sum is the moment-generating function of a mixture distribution with claim cumulative distribution functionThus the aggregate has a compound Poisson distribution with intensity $\lambda$ and severity law $F_{\ast }$. This identification remains valid without positive exponential moments, using the same calculation for Laplace transforms of nonnegative random variables and their uniqueness. It is the Poisson superposition of insurance portfolios.

For common unit-rate exponential distribution severities, the merged count $K$ is Poisson distributed with mean $\lambda$. Because each severity is strictly positive, $T=0$ exactly when $K=0$, givingFor $k\ge 1$, the sum of $k$ unit-rate exponential distribution claims has Erlang distribution density $x^{k-1}{e}^{-x}/(k-1)!$. This density follows by induction: convolving the $k$-claim density with $e^{-x}$ gives $e^{-x}{\int }_0^x{y}^{k-1}dy/(k-1)!=e^{-x}{x}^k/k!$. Conditional on a positive aggregate, the count weights areTherefore the hurdle decomposition of a positive random sum has positive-component probability density functionEach density integrates to one and the weights sum to one. Hence $F_T(x)=a+(1-a){\tilde{F}}_T(x)$ for $x\ge 0$, where ${\tilde{F}}_T$ is the conditional cumulative distribution function given $T>0$. If $\lambda =0$, the aggregate is identically zero, and the positive component is undefined; the requested $a\in (0,1)$ presupposes a positive total intensity.