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Let $P(z)$, $F(z)$ and $G(z)$ be the probability generating functions of $N$, one severity, and $S$, respectively. Positivity of the severities gives $F(0)=0$ and $g_0=p_0$; the random-sum transform identity gives $G=P\circ F$. The count recurrence in the Panjer claim-count class impliesUse the chain rule and substitute $F(z)$ to obtainEquating the coefficient of $z^{r-1}$ givesThe factor multiplying $f_j{g}_{r-j}$ after rearrangement is $a(r-j)+(a+b)j=ar+bj$. Therefore the Panjer recursion isThese power-series calculations are valid inside the convergence discs of the probability generating functions, and hence justify the coefficient identities without assumptions about positive exponential moments.

For a Poisson distribution count, $p_n/p_{n-1}=\lambda /n$, so $a=0$ and $b=\lambda$. The aggregate recursion becomesTo obtain a moment recursion, write ${\mu }_j=\mathbb{E}{X}^j$, $m_j=\mathbb{E}{S}^j$, and $m_0=1$. For nonnegative measurable $h$, conditioning on $N=n$ and using exchangeability of the severities giveswhere $X$ on the right is independent of $S$. The second equality uses $n{p}_n=\lambda p_{n-1}$, the Poisson size-bias identity. Choosing $h(s)=s^{k-1}$, applying the binomial theorem, and using independence proves the raw moment recursion for a compound Poisson distribution:For any particular $k$, a finite severity $k$th moment suffices: $({\sum }_{i=1}^n{X}_i{)}^k\le n^{k-1}{\sum }_{i=1}^n{X}_i^k$, and the Poisson distribution has finite moments of every order. No positive moment-generating function domain is required.

The first three iterations giveSubtracting the appropriate powers of the expected value yieldsIn particular, the aggregate variance and third central moment involve the raw second and third severity moments, rather than their centered counterparts.