Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 39 / 1

Use the usual aggregate claims model assumption that the claim count $N$ is independent of the sequence of severities. Conditional on $N=n$, the probability generating function of the sum is the product of the $n$ severity generating functions. The empty sum has value zero. Thus the law of total expectation gives the random-sum transform identityfor $|z|<1$, with the usual continuous interpretation at $z=1$.

For a claim count distribution in the Panjer claim-count class, multiply the count recurrence by $n{z}^{n-1}$ and sum. With $m=n-1$,Consequently $(1-az)G_N^{\prime }=(a+b)G_N$, orThe chain rule applied to the composition then gives

For $r\ge 1$, compare the coefficients of $z^{r-1}$. The two products giveMove the $j=0$ term to the left and combine the remaining coefficients, since $a(r-j)+(a+b)j=ar+bj$. The Panjer recursion with zero severities is thereforeFor the usual nondegenerate proper Panjer laws, $a<1$, so $1-a{f}_0>0$. If a degenerate count $N\equiv 0$ is represented by redundant constants making this denominator zero, the aggregate is identically zero and must be handled directly rather than dividing by zero.

Because severities are nonnegative, the aggregate is zero exactly when every included severity is zero. Count-severity independence gives the initializationIn particular $g_0$ generally exceeds $p_0$ when zero severities are possible.

The count-severity independence is necessary and is implicit, rather than explicitly stated, in the printed random-sum setup. For a concrete counterexample without it, take independent Bernoulli-$1/2$ variables $X_j$ and put $N=X_1$. Then $S=X_1$, so $G_S(z)=(1+z)/2$, whereas $G_N(G_X(z))=(3+z)/4$. Independence of the $X_j$ among themselves alone cannot justify the requested formula.

Within the approximation that a factory has either zero or one accident, let $N$ be the accident count. The factories' independence gives $N\sim \mathrm{Binomial}(m,p)$. Given an accident, its cluster of claims has a Poisson distribution of mean $\lambda$, independently across factories. Its probability generating function is $G_X(z)=e^{\lambda (z-1)}$. Thus this is a binomial accident portfolio with Poisson clusters, withA factory can contribute no claims either because it has no accident or because its accident's Poisson cluster is empty. Hence the no-claim probability differs from the no-accident probability $q^m$.

For $0\le n\le m$, the accident probability mass function is $p_n=\left(\genfrac{}{}{0ex}{}{m}{n}\right)p^n{q}^{m-n}$. For $1\le n\le m$,Therefore its Panjer claim-count class parameters areAt $n=m+1$ the multiplier is zero, and afterwards both successive masses are zero, so the recurrence holds for all $n\ge 1$. Finally, $f_0=e^{-\lambda }$ and $G_N(z)=(q+pz{)}^m$ reproduce the same initialization $g_0=G_N(f_0)$.