Zero-truncated claim-count distribution ($G_{N\mid N>0}(z)=\frac{G_N(z)-G_N(0)}{1-G_N(0)}$)

For a nonnegative integer claim count $N$ with $p=\mathbb{P}(N=0)<1$, the conditional law given $N>0$ has masses $\mathbb{P}(N=n)/(1-p)$ for $n\ge 1$. Its probability generating function is $(G_N(z)-p)/(1-p)$. Truncation at zero is meaningful only when the positive-count event has positive probability.

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 $p_1=\lambda /(e^{\lambda }-1)$.