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For positive claim sizes, a random sum of independent claims is zero exactly when its count is zero. Its law is a mixture distribution of a zero atom of mass $p=\mathbb{P}(N=0)$ and, with weight $1-p$, a random sum whose count has the zero-truncated claim-count distribution. An independent Bernoulli random variable multiplying that positive component gives the same law. This is distinct from arbitrarily inserting additional zeros into an otherwise unchanged count law.

The moment-generating function of a mixture distribution is the mixture of its component transforms, soFor the aggregate, positivity of every claim implies $S=0$ exactly when $N=0$. ChooseIf $q>0$, take $\tilde{N}$ to have the zero-truncated claim-count distribution, namely the conditional law of $N$ given $N>0$. ThenChoose this count independently of a fresh independent claim-size sequence and put $Z={\sum }_{i=1}^{\tilde{N}}{X}_i$. Its probability distribution is that of $S$ conditional on being positive. Thus the hurdle decomposition of a positive random sum givesand an independent Bernoulli random variable $B$ of success probability $q$ realizes $S\stackrel{d}{=}BZ$. This establishes the distributional representation, including that $Z$ is itself a positive random sum of independent claims. If $p=1$, the aggregate is identically zero; set $q=0$ and choose any positive $Z$, for example one claim. Conditioning the count on positivity is then unnecessary and would be undefined.

For the specified geometric distribution on the nonnegative integers,An exponential distribution of expected value $\mu$ has transform $M_X(r)=(1-\mu r{)}^{-1}$. Substitution yieldsHence $Z$ is exponential with rate $1/(2\mu )$ and expected value $2\mu$. This is the geometric sum of exponential variables with a rescaling of the claim mean. Identification can also use the uniqueness theorem for Laplace transforms of nonnegative random variables by taking $r\le 0$.

The resulting distribution function isIts jump of size $1/2$ at zero is important: the aggregate law is not a purely continuous exponential distribution.

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.

Conditioning on the intensity in the Poisson mixture givesThus $N$ has the negative binomial distribution with two successes and success probability $p$, counting failures; explicitly $\mathbb{P}(N=j)=(j+1)p^2{q}^j$ for $j\ge 0$. The probability generating function is finite for real $z<1/q$.

The claim-size moment-generating function is $M_X(t)=(1-\mu t{)}^{-1}$. Substituting into the aggregate moment-generating function and using $p+q=1$ givesPut $L(t)=p/(p-\mu t)$, the moment-generating function of an exponential distribution with rate $p/\mu$. The identitythen yieldsThe gamma-mixed Poisson aggregate with exponential claims has three nonnegative mixture weights summing to one. By uniqueness of the moment-generating function near zero, the aggregate distribution isHere ${\delta }_0$ is the Dirac measure at zero. In particular $\mathbb{P}(S=0)=p^2$, consistently with $\mathbb{P}(N=0)$. The positive components have respective expected values $\mu /p$ and $2\mu /p$; their mixture distribution accounts for the possibility of no aggregate payout.

Independent claim Poisson processes of rates ${\lambda }_i$ merge into a Poisson process of rate $\lambda ={\sum }_i{\lambda }_i$, by the Superposition theorem for Poisson point processes. The merged claim law is a mixture distribution of the individual claim laws with weights ${\lambda }_i/\lambda$. Equivalently, multiplication of the individual compound-Poisson transforms produces $\exp (\lambda [{\sum }_i({\lambda }_i/\lambda )M_i(r)-1])$.