= Aggregate recursion for zero-truncated Panjer counts
{title2=$g_k=p_1f_k+\sum_{j=1}^{k-1}(a+bj/k)f_jg_{k-j}$}
Let $N\ge1$ have <probability mass function> $p_n$ satisfying $p_n=(a+b/n)p_{n-1}$ for $n\ge2$. Let independent positive integer claim sizes have <probability mass function> $f_j$, independently of $N$. The aggregate <probability mass function> $g_k$ satisfies the displayed recursion, with $g_0=0$. To prove it, write $P,F,G$ for the count, severity and aggregate <probability generating functions>. The count recurrence gives $(1-az)P'(z)=(a+b)P(z)+p_1$. Since $G=P\circ F$, the <chain rule> gives $(1-aF)G'=((a+b)G+p_1)F'$. Comparing coefficients of $z^{k-1}$ gives the recursion. The forcing term $p_1f_k$ distinguishes this algorithm from the usual <Panjer recursion>. For a <zero-truncated Poisson distribution> with parameter $\lambda$, take $a=0$, $b=\lambda$ and $p_1=\lambda/(e^\lambda-1)$.
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