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

Condition first on the count in the random sum of independent claims. For $N=k$, independence of the summands and independence from $N$ giveThe empty product is one when $k=0$. Taking expected values proves the random-sum transform identity:For positive claim sizes this composition is always finite for $u\le 0$, and is an identity of Laplace transforms of nonnegative random variables after writing $u=-s$. For positive $u$, it holds wherever the composed series is finite; no positive exponential moment is implied merely by positivity of the summands.

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.

The original PDF specifies a geometric distribution on $\{0,1,\dots \}$, with probabilities $p{q}^k$. Its probability generating function is $p/(1-qz)$. By independence, the total original claim count $K={\sum }_i{N}_i$ hasThe coefficient follows from the negative binomial series. Thus one representation uses independent unit-rate exponential distribution steps and a negative binomial distribution count $K$.

There is a second useful random sum of independent claims representation. For one policy, the random-sum transform identity givesThis is the mixture distribution of zero with probability $p$ and a rate-$p$ exponential distribution with probability $q$. Consequentlywith the binomial distribution count independent of the independent steps. Equality follows by multiplying the moment-generating functions. This is a negative-binomial sum of exponential claims.

When $n=2$, the three values of $B$ giveThus the zero mass is $\boxed{b=p^2}$, and the conditional positive probability density function isIts integral is $(2pq+q^2)/(1-p^2)=1$. Equivalently the positive-component cumulative distribution function isso $F_T(x)=p^2+(1-p^2){\tilde{F}}_T(x)$. For $x<0$, $F_T(x)=0$.