Aggregate recursion for zero-truncated Panjer counts ($g_k=p_1{f}_k+{\sum }_{j=1}^{k-1}(a+bj/k)f_j{g}_{k-j}$)

Let $N\ge 1$ have probability mass function $p_n$ satisfying $p_n=(a+b/n)p_{n-1}$ for $n\ge 2$. 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 $(1-az)P^{\prime }(z)=(a+b)P(z)+p_1$. Since $G=P\circ F$, the chain rule gives $(1-aF)G^{\prime }=((a+b)G+p_1)F^{\prime }$. Comparing coefficients of $z^{k-1}$ gives the recursion. The forcing term $p_1{f}_k$ distinguishes this algorithm from the usual Panjer recursion. For a zero-truncated Poisson distribution with parameter $\lambda$, take $a=0$, $b=\lambda$ and $p_1=\lambda /(e^{\lambda }-1)$.