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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.

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.

Condition on the claim count distribution. If $\mu =\mathbb{E}{X}_1$ and $v=\mathrm{Var}(X_1)$, then independence of the count and the independent and identically distributed random variables givesThe empty sum is zero. The law of total expectation and law of total variance therefore prove the random sum of independent claims formulasFor the variance calculation assume the count and severity have finite second moments; the expectation identity needs the corresponding integrability. Nonnegative claims also permit the expectation identity with extended nonnegative values.

Represent each policy payment as $S_i=B_i{Y}_i$, where $B_i$ is a Bernoulli random variable of parameter $q_i$, independent of its severity. Equivalently, $Y_i$ specifies the payment distribution conditional on a claim. ThenIndependence of the policies makes their variances add. PutThe heterogeneous individual claims model consequently has

For the Poisson approximation, define the severity mixture distributionLet ${\phi }_i(t)=\mathbb{E}[e^{it{Y}_i}]$ be the individual characteristic functions. Conditioning on each Poisson count and then multiplying transforms of the independent policy payments givesThis is Poisson superposition of insurance portfolios. By the uniqueness theorem for characteristic functions,with count and severities independent. Its compound Poisson distribution has distribution functionwhere $F^{\ast k}$ is the convolution distribution of $k$ independent severities and $F^{\ast 0}$ is the distribution concentrated at zero. Characteristic functions avoid assuming that severity moment-generating functions exist. Since $\mathbb{E}Z=M/Q$ and $\mathbb{E}{Z}^2=A/Q$, the random-sum formulas give

Use $T_{\mathrm{B}}$ to distinguish the second, compound-binomial approximation from the Poisson one. It has $K_{\mathrm{B}}\sim \mathrm{Bin}(n,p)$ with $p=Q/n$ and the same severity law $F$. Thus $\mathbb{E}{K}_{\mathrm{B}}=Q$ and $\mathrm{Var}(K_{\mathrm{B}})=Q(1-p)$. Applying the random sum of independent claims identities again givesHence

Both approximations match the exact expected value. The Cauchy-Schwarz inequality gives $M^2\le nC$, proving the variance comparison for compound portfolio approximations:The excess variances are respectively $C-M^2/n$ and $C$. The compound-binomial model is therefore closer in variance; equality with the exact variance occurs when all $q_i{\mu }_i$ are equal. This variance comparison is not a general ordering of approximation errors or tail probabilities.

The Poisson replacement is useful when all claim probabilities are small: it replaces the individual restriction of at most one claim by an unrestricted Poisson count, and the variance error is of second order in those probabilities. The compound-binomial replacement retains a count bounded by $n$, but pools both frequencies and severity laws and can lose policy-specific information. If all $q_i$ and all $F_i$ agree, it is exactly the original portfolio law. In a heterogeneous portfolio, agreement of means, or even variances, does not make either approximate law exact. Neither approximation is guaranteed accurate merely by having many policies.

If $a+b=0$, then $p_1=(a+b)p_0=0$. Applying the count recurrence successively gives $p_n=0$ for every $n\ge 1$. Normalization forcesThus the claim count distribution is a point mass at zero, and the aggregate claims model also has $S=0$ almost surely. In particular the recursion starts at $g_0=1$ and gives no positive-loss probabilities.

With $a=0$ and $b>0$, the recurrence is $p_n=(b/n)p_{n-1}$. Iteration gives $p_n=p_0{b}^n/n!$. Since ${\sum }_{n\ge 0}{b}^n/n!=e^b$, normalization givesHence the claim count distribution is a Poisson distribution of mean $b$, and the Panjer recursion begins at $g_0=e^{-b}$.