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Actuarial statistics uses probability theory and statistical inference to quantify insurance claims, premiums, and the uncertainty of future losses. Aggregate claims model describes portfolios, reinsurance redistributes losses, and the Bühlmann model combines experience with population information.

Per-claim reinsurance under which the insurer retains each claim up to a level $M$ and the reinsurer pays its positive part $(x-M{)}_{+}$. The annual retained amount is a sum of capped claims, so it is generally different from the cap of the annual sum in aggregate stop loss reinsurance.

A reinsurance layer with attachment $M$ and payment limit $A$ pays nothing below $M$, the excess between $M$ and $M+A$, and $A$ above that. Its expected recovery is ${\int }_M^{M+A}\mathbb{P}(X>x)dx$, by the tail integral formula for moments. For continuous claims, conditioning on penetration of the layer divides this expectation by $\mathbb{P}(X>M)$.

Under quota share reinsurance, the direct insurer retains a fixed fraction $\alpha$ of each claim and the reinsurer pays the other fraction. Thus their annual claim totals are $\alpha S$ and $(1-\alpha )S$. Under excess of loss reinsurance with retention $d$, the direct insurer pays $\min (X_j,d)$ on each individual claim and the reinsurer pays $(X_j-d{)}_{+}$. Under stop loss reinsurance with annual retention $D$, the direct insurer pays $\min (S,D)$ and the reinsurer pays $(S-D{)}_{+}$. The last two contracts differ because one caps each claim and the other caps the aggregate.

Conditioning on the annual Poisson distribution count gives the compound Poisson distribution transform:This equality is understood on the domain where the moment-generating function is finite. By the expected value premium principle, annual premium income is $c=(1+\theta )\lambda \mu$. Without reinsurance, final capital is $W_0=u+c-S$, and therefore

For the quota-share contract, the reinsurer's expected claim cost is $(1-\alpha )\lambda \mu$, so its annual premium isThe direct insurer keeps its original premium income, pays this reinsurance premium and retains only $\alpha S$. ThusandThe reinsurer's loading is applied to the ceded expected loss, not to the direct insurer's original premium.

Maximizing expected exponential utility $\mathbb{E}[-e^{-\beta W}]$ is equivalent to minimizing the exponent $H(\alpha )=\log \mathbb{E}[e^{-\beta W}]$. For exponential distribution claim sizes of mean $\mu$,so on the finite-utility domain $\beta \mu \alpha <1$,For a nontrivial positive arrival rate, differentiation givesThe unique unconstrained stationary retention is thereforeIt is positive and lies below $1/(\beta \mu )$. If ${\alpha }_{\ast }<1$, it is the unique optimal retention in the printed range $0<\alpha <1$. If ${\alpha }_{\ast }\ge 1$, expected utility increases throughout that open interval and its supremum is approached as $\alpha \uparrow 1$; there is no attained maximum in the strictly open range. Allowing the usual no-reinsurance endpoint gives the compact formulaRetentions with $\beta \mu \alpha \ge 1$ have infinite exponential loss moment and expected utility $-\infty$, so they cannot displace the finite stationary solution. The stationary point and full-retention endpoint just described stay within the finite-transform domain whenever they are selected.

For the requested comparisons,Higher reinsurer loading increases optimal retention, up to full retention: reinsurance is more expensive. The direct insurer's loading $\theta$ does not affect the optimizer. It adds only the constant $-\beta \lambda \mu \theta$ to $H$, reflecting the translation property of constant absolute risk aversion. These observations give the quota-share retention under exponential utility rule; $u$ and the positive arrival rate also drop out of the stationary equation. In the degenerate case $\lambda =0$, no claims or premiums occur, $W=u$, and every retained fraction has the same expected utility.

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.

A threshold limiting the insurer's retained payment under a reinsurance contract. Under excess of loss reinsurance it applies to each claim size; under aggregate stop loss reinsurance it applies to the total loss over the stated period. These contracts therefore cap different quantities even when their thresholds have the same numerical value.