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Let $\Lambda ={\sum }_{i=1}^m{\lambda }_i$. For $\Lambda >0$, the resulting compound Poisson distribution hasThus the merged claim size probability distribution is a rate-weighted mixture distribution, not an equally weighted mixture distribution unless the rates agree.

To prove this without any moment-generating function assumption, let ${\chi }_i(t)$ be the characteristic function of a claim of type $i$. Conditioning on its Poisson distribution count givesIndependence of the aggregate risks therefore givesThe weighted characteristic function inside is precisely that of the displayed mixture distribution. This is the transform of a sum of independent claims with independent count $\mathrm{Pois}(\Lambda )$, proving the claim by uniqueness of characteristic functions. This is Poisson superposition of insurance portfolios. If $\Lambda =0$, all risks make no claims almost surely and $S=0$; the severity law is then irrelevant.

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.

Write $\lambda ={\sum }_i{\lambda }_i$, and suppose first that $\lambda >0$. Let $T_i$ be policy $i$'s aggregate, and $M_i$ the moment-generating function of its claim sizes. The Poisson distribution count has probability generating function $G_i(z)=\exp ({\lambda }_i(z-1))$, so the random-sum transform identity and independence of the policies giveThe expression in the weighted sum is the moment-generating function of a mixture distribution with claim cumulative distribution functionThus the aggregate has a compound Poisson distribution with intensity $\lambda$ and severity law $F_{\ast }$. This identification remains valid without positive exponential moments, using the same calculation for Laplace transforms of nonnegative random variables and their uniqueness. It is the Poisson superposition of insurance portfolios.

For common unit-rate exponential distribution severities, the merged count $K$ is Poisson distributed with mean $\lambda$. Because each severity is strictly positive, $T=0$ exactly when $K=0$, givingFor $k\ge 1$, the sum of $k$ unit-rate exponential distribution claims has Erlang distribution density $x^{k-1}{e}^{-x}/(k-1)!$. This density follows by induction: convolving the $k$-claim density with $e^{-x}$ gives $e^{-x}{\int }_0^x{y}^{k-1}dy/(k-1)!=e^{-x}{x}^k/k!$. Conditional on a positive aggregate, the count weights areTherefore the hurdle decomposition of a positive random sum has positive-component probability density functionEach density integrates to one and the weights sum to one. Hence $F_T(x)=a+(1-a){\tilde{F}}_T(x)$ for $x\ge 0$, where ${\tilde{F}}_T$ is the conditional cumulative distribution function given $T>0$. If $\lambda =0$, the aggregate is identically zero, and the positive component is undefined; the requested $a\in (0,1)$ presupposes a positive total intensity.

For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

First derive the survival renewal equation for a classical risk model. Before the first claim, capital rises deterministically. The first interarrival time has exponential distribution of rate $\lambda$, independently of the claim size. Conditioning on its time and size, and using the Markov property after that claim, givesA claim larger than the capital available then causes immediate ruin and contributes zero. There is almost surely a first claim because $\lambda >0$.

Put $g(v)={\int }_0^v\phi (v-x)f(x)dx$. Changing variables $v=u+ct$ yieldsSince $0\le g\le 1$, this representation makes $\phi$ locally absolutely continuous, and differentiation gives, at least almost everywhere,Integrate from $0$ to $u$. The integrands are nonnegative, so Tonelli theorem permits interchange in the double integral. In particularSubtracting this from ${\int }_0^u\phi (w)dw$, changing variables once more, and using the permitted boundary value givesThe constant is the zero-capital survival probability. Positive relative safety loading means $c>\lambda \mu$, so it is positive. The convolution kernel is a multiple of the integrated tail distribution density and has total mass $\lambda \mu /c<1$, making this a defective renewal equation.

For the individual portfolios define ${\rho }_i={\lambda }_i{\mu }_i/c_i$. Under the positive-loading convention of the question, $0<{\rho }_i<1$. At zero capital, portfolio $i$ survives with probability $1-{\rho }_i$. The independence of random variables of the entire portfolio processes makes their ultimate survival events independent. ConsequentlyThe exponential form is not needed for this zero-capital probability under positive loading; only the claim mean enters. For completeness, the company description does not separately repeat positive loading for every portfolio. If nonpositive loadings are allowed, certain ruin with nonpositive loading and finite claim variance instead givesHere the positive part sets a nonpositive factor to zero. To justify the additional case, observe capital at claim times: its independent increments are $c_{i}T-X$, with mean $c_i/{\lambda }_i-{\mu }_i$. A negative mean sends their partial sums to $-\infty$ by the strong law of large numbers. At zero mean, these increments have finite nonzero variance. The central limit theorem gives $\mathbb{P}(S_k<-K)\to 1/2$ for each fixed $K$, so the probability of unboundedness below is at least $1/2$. That event is unchanged by altering finitely many increments and hence is a tail event; the Kolmogorov zero-one law makes its probability one. Ruin therefore occurs almost surely also at zero loading.

For the merged claims, Poisson superposition of insurance portfolios gives total arrival rate $\lambda ={\sum }_i{\lambda }_i$. Each arrival is from portfolio $i$ with probability ${\lambda }_i/\lambda$, independently of other arrival labels. Thus its claim-size mixture distribution has densityThe weights sum to one, so this density integrates to one. An independent transform verification uses the aggregate for one accounting period. If $M_i(r)=(1-{\mu }_{i}r{)}^{-1}$, then its moment-generating function isThis is precisely a compound Poisson distribution with parameter $\lambda$ and the displayed claim-size law. The transform identity can safely be read at $r\le 0$; positive arguments must be below the relevant poles.

Premium incomes and initial capitals add. Therefore the merged premium rate is $C={\sum }_i{c}_i$, the merged initial capital is zero, and its mean claim size isIts zero-capital survival probability is consequentlyUnder the question's positive-loading context this is a premium-weighted average of the individual survival probabilities, rather than their product. If arbitrary individual loadings are admitted, recompute the loading of the merged portfolio; its finite-variance mixture law gives the complete formulaAn individually nonpositive loading does not force merged ruin when aggregate premiums still exceed aggregate expected claims. The survival probability under risk pooling is at least their product because a product of numbers in $[0,1]$ is at most each factor. Pooling allows one portfolio's surplus to cover another's deficit, so merged survival does not require every original portfolio to remain solvent separately.