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

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 $x<0$, $F_T(x)=0$.