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For independent rate-$\lambda$ exponential distribution claims and a zero-based negative binomial distribution count of shape $k$ and parameter $p$, the random-sum transform identity gives $M_S(t)=[p(\lambda -t)/(p\lambda -t){]}^k$. This equals the transform of a compound binomial distribution with count $\mathrm{Bin}(k,1-p)$ and rate-$p\lambda$ claim sizes. Conditional on its positive count $n$, the total has Erlang distribution, whose survival function is the inner exponential-polynomial sum. Conditioning yields the displayed finite mixture distribution formula for $x>0$. It replaces an infinite mixture and includes the zero-total atom $p^k$ through the omitted $n=0$ term.

Condition on the claim count. The empty random sum of independent claims is zero, and independence gives $\mathbb{E}[e^{tS}\mid N=n]=M_X(t{)}^n$. Therefore the random-sum transform identity iswherever the composition is finite. This follows directly from the law of total expectation, rather than replacing the random count by its mean.

Put $q=1-p$. For the specified negative binomial distribution, the negative binomial series gives $G_N(z)=[p/(1-qz){]}^k$. A rate-$\lambda$ exponential distribution has moment-generating function $M_X(t)=\lambda /(\lambda -t)$. HenceThe last expression is exactly the moment-generating function of a compound binomial distribution with independent count and severitiesThe moment-generating function exists in a neighborhood of zero, so its uniqueness identifies the distributions. In particular, the new claim sizes have mean $1/(p\lambda )$, not $1/\lambda$, and both aggregate distributions have an atom $p^k$ at zero. This is the negative-binomial sum of exponential claims representation with the severity rate explicitly retained.

Conditional on $N=n\ge 1$, the aggregate has Erlang distribution with density ${\lambda }^n{s}^{n-1}{e}^{-\lambda s}/(n-1)!$ on $s>0$. Repeated integration by parts of its tail givesIndeed, after substituting $v=\lambda s$, the integral ${\int }_{\lambda x}^{\infty }{e}^{-v}{v}^{n-1}dv/(n-1)!$ reduces recursively to the corresponding integral with exponent $n-2$, terminating at $e^{-\lambda x}$. Applying the law of total probability yields the original infinite mixture distribution tailApplying the same argument to the compound binomial distribution gives instead the finite Erlang-mixture tail for a negative-binomial exponential aggregate:The two tails are equal. The binomial representation replaces an infinite sum by a finite exact calculation, eliminating the need to choose a claim-count truncation and control its omitted tail. The inner sums can be evaluated recursively, using $a_{j+1}=a_j(p\lambda x)/(j+1)$ for their exponential-weighted terms.

For the binomial distribution, $G_N(z)=(1-p+pz{)}^n$. Thus the compound binomial distribution hasSet $h(t)=1-p+p{M}_{X_1}(t)$. Since $h(0)=1$ and $h^{(j)}(0)=p{m}_j$, differentiating $n\log h$ three times givesThe first two cumulants, for comparison, are $np{m}_1$ and $n(p{m}_2-p^2{m}_1^2)$.

Choose deterministic claim sizes $X_i=d>0$. Then $S=dN$, so $m_j=d^j$ andThis is strictly negative for $p>1/2$ and $n\ge 1$. A concrete negative-skewness example is $n=1$, $p=3/4$, $d=1$, for whichThere is no contradiction with positive claims: bounded Bernoulli-type totals can have more mass near their upper value and a longer left tail.

Put $Q={\sum }_i{q}_i$, $M={\sum }_i{q}_i{\mu }_i$, and $A={\sum }_i{q}_i\mathbb{E}{Y}_i^2$. A compound Poisson distribution with count mean $Q$ and severity law ${\sum }_i(q_i/Q)F_i$ has variance $V_{\mathrm{P}}=A$. A compound binomial distribution with $n$ trials, probability $Q/n$, and that same severity law has variance $V_{\mathrm{B}}=A-M^2/n$. The individual-policy variance is $V=A-{\sum }_i(q_i{\mu }_i{)}^2$. The Cauchy-Schwarz inequality gives ${\sum }_i(q_i{\mu }_i{)}^2\ge M^2/n$, proving the ordering. Equality of the first two variances requires constant $q_i{\mu }_i$; it need not imply equality of the distributions. If all probabilities and severity laws are identical, the binomial compound approximation is exactly the individual model.