Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 34 / 1 / c

Conditioning on the intensity in the Poisson mixture givesThus $N$ has the negative binomial distribution with two successes and success probability $p$, counting failures; explicitly $\mathbb{P}(N=j)=(j+1)p^2{q}^j$ for $j\ge 0$. The probability generating function is finite for real $z<1/q$.

The claim-size moment-generating function is $M_X(t)=(1-\mu t{)}^{-1}$. Substituting into the aggregate moment-generating function and using $p+q=1$ givesPut $L(t)=p/(p-\mu t)$, the moment-generating function of an exponential distribution with rate $p/\mu$. The identitythen yieldsThe gamma-mixed Poisson aggregate with exponential claims has three nonnegative mixture weights summing to one. By uniqueness of the moment-generating function near zero, the aggregate distribution isHere ${\delta }_0$ is the Dirac measure at zero. In particular $\mathbb{P}(S=0)=p^2$, consistently with $\mathbb{P}(N=0)$. The positive components have respective expected values $\mu /p$ and $2\mu /p$; their mixture distribution accounts for the possibility of no aggregate payout.