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Write $P(z)=\mathbb{E}[z^N]$, $F(z)=\mathbb{E}[z^{X_1}]$ and $G(z)=\mathbb{E}[z^{S_N}]$ for the three probability generating functions. Since both the claim count and every claim size are positive, $P(0)=F(0)=G(0)=0$. Conditioning on the count and using independence gives the random-sum transform identityThe crucial point is that the count recurrence begins at $n=2$. Multiplying it by $n{z}^{n-1}$ and summing, we obtainConsequently $(1-az)P^{\prime }(z)=(a+b)P(z)+p_1$. The chain rule applied to the aggregate probability generating function therefore givesThese identities hold inside the unit disk, or as identities of formal power series. Since $F(0)=0$, every coefficient of the composition depends on only finitely many count probabilities.

Compare the coefficient of $z^{k-1}$. The product $F{G}^{\prime }$ contributes ${\sum }_{j=1}^{k-1}(k-j)f_j{g}_{k-j}$, and $F^{\prime }G$ contributes ${\sum }_{j=1}^{k-1}j{f}_j{g}_{k-j}$. HenceThe required aggregate recursion for zero-truncated Panjer counts is initialized by $g_0=0$ and $g_1=p_1{f}_1$, and for every $k\ge 1$ it isThe empty sum at $k=1$ is zero. Every term on the right is known or has a smaller aggregate index. Omitting $p_1{f}_k$ would incorrectly apply the usual Panjer recursion with a zero initial value, producing zero for every aggregate probability.

For the zero-truncated Poisson distribution, $p_n/p_{n-1}=\lambda /n$ for $n\ge 2$, so $a=0$, $b=\lambda$ and $p_1=\lambda /(e^{\lambda }-1)$, with $\lambda >0$. ThusAs a direct probability generating function check, $P(z)=(e^{\lambda z}-1)/(e^{\lambda }-1)$ and therefore $G(z)=(e^{\lambda F(z)}-1)/(e^{\lambda }-1)$; differentiation reproduces this recursion.