= Solution
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> implies
$$
P'(z)=\sum_{n\ge1}np_nz^{n-1}
=\sum_{n\ge1}(an+b)p_{n-1}z^{n-1}
=azP'(z)+(a+b)P(z).
$$
Use the <chain rule> and substitute $F(z)$ to obtain
$$
(1-aF(z))G'(z)=(a+b)F'(z)G(z).
$$
Equating the coefficient of $z^{r-1}$ gives
$$
r g_r-a\sum_{j=1}^r(r-j)f_jg_{r-j}
=(a+b)\sum_{j=1}^rj f_jg_{r-j}.
$$
The factor multiplying $f_jg_{r-j}$ after rearrangement is $a(r-j)+(a+b)j=ar+bj$. Therefore the <Panjer recursion> is
$$
\boxed{g_0=p_0,\qquad g_r=\sum_{j=1}^r\left(a+\frac{bj}{r}\right)f_jg_{r-j}\quad(r\ge1).}
$$
These 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 becomes
$$
\boxed{g_0=e^{-\lambda},\qquad g_r=\frac\lambda r\sum_{j=1}^rj f_jg_{r-j}.}
$$
To 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 gives
$$
\begin{aligned}
\mathbb E[S h(S)]
&=\sum_{n\ge1}p_n n\,\mathbb E\left[X_1h\left(X_1+\sum_{j=2}^nX_j\right)\right]\\
&=\lambda\sum_{m\ge0}p_m\,\mathbb E\left[Xh\left(X+\sum_{j=1}^mX_j\right)\right]
=\lambda\mathbb E[Xh(X+S)],
\end{aligned}
$$
where $X$ on the right is independent of $S$. The second equality uses $np_n=\lambda p_{n-1}$, 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>:
$$
\boxed{m_k=\lambda\sum_{j=0}^{k-1}\binom{k-1}{j}\mu_{j+1}m_{k-1-j},\qquad k\ge1.}
$$
For any particular $k$, a finite severity $k$th <moment> suffices: $(\sum_{i=1}^nX_i)^k\le n^{k-1}\sum_{i=1}^nX_i^k$, and the <Poisson distribution> has finite <moments> of every order. No positive <moment-generating function> domain is required.
The first three iterations give
$$
m_1=\lambda\mu_1,\qquad
m_2=\lambda\mu_2+\lambda^2\mu_1^2,\qquad
m_3=\lambda\mu_3+3\lambda^2\mu_1\mu_2+\lambda^3\mu_1^3.
$$
Subtracting the appropriate powers of the <expected value> yields
$$
\boxed{\mathbb ES=\lambda\mu_1,\qquad
\operatorname{Var}(S)=\lambda\mu_2,\qquad
\mathbb E[(S-\mathbb ES)^3]=\lambda\mu_3.}
$$
In particular, the aggregate <variance> and third <central moment> involve the raw second and third severity <moments>, rather than their centered counterparts.
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