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For positive integer severities with probability mass function $f_j$ and independent count in the Panjer claim-count class, the aggregate probability mass function starts at $g_0=p_0$ and satisfies the displayed recurrence. If $P,F,G$ are the count, severity and aggregate probability generating functions, respectively, then $G=P\circ F$ and $(1-aF)G^{\prime }=(a+b)F^{\prime }G$. Equating the coefficient of $z^{r-1}$ gives $r{g}_r={\sum }_{j=1}^r[a(r-j)+(a+b)j]f_j{g}_{r-j}$. The positivity assumption excludes zero severities; otherwise the initial value and recursion denominator change.

For a claim count distribution in the Panjer claim-count class and independent nonnegative integer severities with mass $f_j$, the probability generating function composition is $G_S=G_N\circ G_X$. Combining it with $(1-az)G_N^{\prime }=(a+b)G_N$ gives $(1-a{G}_X)G_S^{\prime }=(a+b)G_S{G}_X^{\prime }$. Equating coefficients yields the displayed recursion. Zero severities change both the initial mass and the denominator. For nondegenerate proper Panjer count laws $a<1$, so the denominator is positive. An identically zero count is handled directly if its redundant parameter choice makes the denominator vanish.

Within the approximation that a factory has either zero or one accident, let $N$ be the accident count. The factories' independence gives $N\sim \mathrm{Binomial}(m,p)$. Given an accident, its cluster of claims has a Poisson distribution of mean $\lambda$, independently across factories. Its probability generating function is $G_X(z)=e^{\lambda (z-1)}$. Thus this is a binomial accident portfolio with Poisson clusters, withA factory can contribute no claims either because it has no accident or because its accident's Poisson cluster is empty. Hence the no-claim probability differs from the no-accident probability $q^m$.

For $0\le n\le m$, the accident probability mass function is $p_n=\left(\genfrac{}{}{0ex}{}{m}{n}\right)p^n{q}^{m-n}$. For $1\le n\le m$,Therefore its Panjer claim-count class parameters areAt $n=m+1$ the multiplier is zero, and afterwards both successive masses are zero, so the recurrence holds for all $n\ge 1$. Finally, $f_0=e^{-\lambda }$ and $G_N(z)=(q+pz{)}^m$ reproduce the same initialization $g_0=G_N(f_0)$.

Use the usual aggregate claims model assumption that the claim count $N$ is independent of the sequence of severities. Conditional on $N=n$, the probability generating function of the sum is the product of the $n$ severity generating functions. The empty sum has value zero. Thus the law of total expectation gives the random-sum transform identityfor $|z|<1$, with the usual continuous interpretation at $z=1$.

For a claim count distribution in the Panjer claim-count class, multiply the count recurrence by $n{z}^{n-1}$ and sum. With $m=n-1$,Consequently $(1-az)G_N^{\prime }=(a+b)G_N$, orThe chain rule applied to the composition then gives

For $r\ge 1$, compare the coefficients of $z^{r-1}$. The two products giveMove the $j=0$ term to the left and combine the remaining coefficients, since $a(r-j)+(a+b)j=ar+bj$. The Panjer recursion with zero severities is thereforeFor the usual nondegenerate proper Panjer laws, $a<1$, so $1-a{f}_0>0$. If a degenerate count $N\equiv 0$ is represented by redundant constants making this denominator zero, the aggregate is identically zero and must be handled directly rather than dividing by zero.

Because severities are nonnegative, the aggregate is zero exactly when every included severity is zero. Count-severity independence gives the initializationIn particular $g_0$ generally exceeds $p_0$ when zero severities are possible.

The count-severity independence is necessary and is implicit, rather than explicitly stated, in the printed random-sum setup. For a concrete counterexample without it, take independent Bernoulli-$1/2$ variables $X_j$ and put $N=X_1$. Then $S=X_1$, so $G_S(z)=(1+z)/2$, whereas $G_N(G_X(z))=(3+z)/4$. Independence of the $X_j$ among themselves alone cannot justify the requested formula.

Let $P(z)={\sum }_{n\ge 0}{p}_n{z}^n$, $F(z)={\sum }_{j\ge 1}{f}_j{z}^j$, and $G(z)={\sum }_{k\ge 0}{g}_k{z}^k$ be the count, claim-size and aggregate probability generating functions. Independence in the aggregate claims model givesThe Panjer claim-count class recurrence can be written $n{p}_n=(an+b)p_{n-1}$. Multiply by $z^{n-1}$ and sum to obtainUsing the chain rule in $G=P\circ F$ then yieldsThe coefficient of $z^{k-1}$ on the left is $k{g}_k-a{\sum }_{j=1}^k{f}_j(k-j)g_{k-j}$, where the $j=k$ term is zero. On the right it is $(a+b){\sum }_{j=1}^{k}j{f}_j{g}_{k-j}$. Rearranging gives the Panjer recursionThe initial value is $p_0$ because all claim sizes are positive: zero total loss occurs exactly when $N=0$. Only already computed aggregate probabilities occur on the right. The count normalization determines $p_0$ from $a,b$ as well: with $w_0=1$ and $w_n={\prod }_{j=1}^n(a+b/j)$, we have $p_n=p_0{w}_n$ and $p_0=({\sum }_{n\ge 0}{w}_n{)}^{-1}$. The assumed valid count law makes this sum finite.

Finally $p_0$ must be positive. If it were zero, the recurrence would make every $p_n$ zero, contradicting normalization. At $n=1$ the recurrence therefore givesThis proves the sign constraint without assuming any particular count family. The three parts below also provide explicit initial probabilities for the aggregate recursion.

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.

Let $G(z)={\sum }_{n\ge 0}{p}_n{z}^n$ be the probability generating function. For $|z|<1$ the series can be differentiated term by term: $\sum n{p}_n|z{|}^{n-1}$ is finite, even without a finite claim-count expected value. Multiply the Panjer claim-count class recurrence by $n$ and put $m=n-1$. ThenHenceThis argument is valid throughout the open unit disk. Values at boundary points may be obtained by a limit when the required derivatives exist; one need not assume $\mathbb{E}N<\infty$ to establish the identity.

The exponential distribution with rate $\beta >0$ has moment-generating function $\beta /(\beta -t)$ for $t<\beta$. Shifting it by $\alpha$ multiplies the moment-generating function by $e^{\alpha t}$. Therefore the shifted-exponential Poisson mixture has probability generating functionThe probability generating function converges and is analytic for $|z|<\beta +1$, and in particular is finite for real $0\le z<\beta +1$.

Write $p=\beta /(\beta +1)$ and $q=1/(\beta +1)$, so $p+q=1$. The factor $e^{\alpha (z-1)}$ is the probability generating function of $V\sim \mathrm{Poisson}(\alpha )$, while $p/(1-qz)$ is that of a geometric distribution on $\{0,1,\dots \}$, with $\mathbb{P}(W=j)=p{q}^j$. Choose $V$ and $W$ independently. The product rule for the probability generating function of a sum of independent counts then proves $N\stackrel{d}{=}V+W$. The support-zero convention for $W$ is essential here.

The convolution of independent random variables gives the finite sumTaking the logarithmic derivative of the probability generating function givesCompare coefficients of $z^{n-1}$, for $n\ge 2$. The left side is $(\beta +1)n{p}_n-(n-1)p_{n-1}$, and the right side is $\{1+\alpha (\beta +1)\}{p}_{n-1}-\alpha p_{n-2}$. Hence the shifted-exponential Poisson count recursion isThe constant and linear coefficients supply the starting valuesThese two values determine every subsequent probability by the recursion. The negative second term does not mean that the distribution has negative probabilities: the convolution formula exhibits every $p_n$ as a sum of nonnegative quantities.

At $\alpha =0$ the independent Poisson distribution component is identically zero and the count has a geometric distribution. Both the finite sum and the recursion reduce to $p_n=q{p}_{n-1}$. Consequently the Panjer claim-count class parameters areAlthough the shifted model initially has positive $\alpha$, this zero-shift boundary case is well defined.