Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 43 / 1 / Solution

Let $P(z)={\sum }_{n\ge 0}{p}_n{z}^n$, $F(z)={\sum }_{j\ge 1}{f}_j{z}^j$, and $G(z)={\sum }_{k\ge 0}{g}_k{z}^k$ be the count, claim-size and aggregate probability generating functions. Independence in the aggregate claims model givesThe Panjer claim-count class recurrence can be written $n{p}_n=(an+b)p_{n-1}$. Multiply by $z^{n-1}$ and sum to obtainUsing the chain rule in $G=P\circ F$ then yieldsThe coefficient of $z^{k-1}$ on the left is $k{g}_k-a{\sum }_{j=1}^k{f}_j(k-j)g_{k-j}$, where the $j=k$ term is zero. On the right it is $(a+b){\sum }_{j=1}^{k}j{f}_j{g}_{k-j}$. Rearranging gives the Panjer recursionThe initial value is $p_0$ because all claim sizes are positive: zero total loss occurs exactly when $N=0$. Only already computed aggregate probabilities occur on the right. The count normalization determines $p_0$ from $a,b$ as well: with $w_0=1$ and $w_n={\prod }_{j=1}^n(a+b/j)$, we have $p_n=p_0{w}_n$ and $p_0=({\sum }_{n\ge 0}{w}_n{)}^{-1}$. The assumed valid count law makes this sum finite.

Finally $p_0$ must be positive. If it were zero, the recurrence would make every $p_n$ zero, contradicting normalization. At $n=1$ the recurrence therefore givesThis proves the sign constraint without assuming any particular count family. The three parts below also provide explicit initial probabilities for the aggregate recursion.