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The monetary amount of an individual insurance claim. Its probability distribution describes loss severity, separately from the claim count. In an independent random sum of independent claims, severity mean $\mu$ and expected count combine to give aggregate mean $\mathbb{E}N\mu$.

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.

Mix the conditional Poisson distribution over the rate-$\nu$ exponential distribution intensity. The Gamma integral giveswhere $p=\nu /(1+\nu )$ and $q=1/(1+\nu )$. Thus the annual claim count has the zero-based geometric distribution, with probabilities $\mathbb{P}(N=0)=p$, $\mathbb{P}(N=1)=pq$ and $\mathbb{P}(N\ge 2)=q^2$.

For the intended Markov chain calculation, take these annual counts to be independent. Order the states by discounts $0,\alpha ,\beta$. From either of the first two states, a positive count returns the policy to zero, while a zero count raises it by one level. From the top, two or more claims send it to zero, exactly one sends it to $\alpha$, and no claim keeps it at $\beta$. Hence the transition matrix isWrite its stationary distribution as $({\pi }_0,{\pi }_{\alpha },{\pi }_{\beta })$. The final two balance equations give $q{\pi }_{\beta }=p{\pi }_{\alpha }$ and ${\pi }_{\alpha }=p{\pi }_0+pq{\pi }_{\beta }$. It follows that ${\pi }_{\alpha }=p{\pi }_0/[q(1+p)]$ and ${\pi }_{\beta }=p^2{\pi }_0/[q^2(1+p)]$. Normalizing yields the three-level discount equilibrium with geometric annual counts:The Markov chain is irreducible for $0<p,q<1$ and has a self-loop, so it is aperiodic and this equilibrium is the limiting distribution. Weighting the three premiums $c,c(1-\alpha ),c(1-\beta )$ givesThere is a necessary qualification to the temporal model: the one-year mixture alone does not specify independence between years. A persistent claim intensity can destroy the discount Markov property. If one fixed $\Lambda$ is drawn for the policyholder and annual counts are conditionally independent given it, then histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the top after three years but give different next-year no-claim probabilities. The first history is three zero counts, so its posterior intensity has exponential rate $\nu +3$ and next no-claim probability $(\nu +3)/(\nu +4)$. The second is a positive count followed by two zero counts; its posterior density is proportional to $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$, giving next no-claim probability $(\nu +2)/(\nu +4)$. Thus the displayed matrix and premium require independent annual marginal sampling, for example a fresh intensity each year. They are not valid for a persistent unobserved policy intensity without conditioning on that intensity.

Under the persistent-intensity interpretation, one can instead condition on $\lambda$. Put $a_0=e^{-\lambda }$ and $b_1=\lambda e^{-\lambda }$. The conditional transition matrix has rows $(1-a_0,a_0,0)$, $(1-a_0,0,a_0)$ and $(1-a_0-b_1,b_1,a_0)$. Its stationary distribution isThe corresponding population limiting expected value of the premium is $c{\int }_0^{\infty }[1-\alpha {\pi }_{\alpha }(\lambda )-\beta {\pi }_{\beta }(\lambda )]\nu e^{-\nu \lambda }d\lambda$. This gives the correct alternative when the intensity persists, and makes the extra assumption behind the requested marginal matrix explicit.

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, defining the empty sum to be zero. Independence gives $\mathbb{E}[S\mid N=k]=k\mathbb{E}{X}_1$. By the law of total expectation,For nonnegative claims this also follows from Tonelli theorem; for signed claims finite $\mathbb{E}N$ and $\mathbb{E}|X_1|$ ensure integrability. Conditional independence likewise givesThis random-sum transform identity holds wherever the composed series is finite, or as an extended nonnegative expectation. It is not permission to use an algebraic continuation outside the convergence domain.

For the zero-based geometric distribution, summing the geometric series givesThe exponential distribution of mean $\mu$ has $M_{X_1}(t)=(1-\mu t{)}^{-1}$ for $t<1/\mu$. ThereforeThe stricter bound $t<p/\mu$ is necessary: the geometric series requires $q{M}_{X_1}(t)<1$. At and above $p/\mu$ that series diverges; at and above $1/\mu$ even one claim has an infinite transform. Thus the displayed rational expression represents the moment-generating function only on its stated domain.

Since all claims are strictly positive, $S=0$ exactly when $N=0$, so $\mathbb{P}(S=0)=p$. Splitting the transform givesUniqueness of the moment-generating function identifies a mixture distribution with a zero atom of weight $p$ and an exponential positive component of weight $q$ and rate $p/\mu$. In particular,The density integrates to $q$, not one. One can also verify it directly: conditional on $N=k\ge 1$, the sum has Erlang distribution density $x^{k-1}{e}^{-x/\mu }/[{\mu }^k(k-1)!]$. Multiplying by $p{q}^k$ and summing over $k$ gives $(pq/\mu )e^{-x/\mu }{e}^{qx/\mu }$, the same density. This is the zero-based geometric sum of exponential variables.

Under excess of loss reinsurance, the ceded amount for each claim is $(X_j-M{)}_{+}$, and the monthly reduction is their random sum. The tail integral formula for moments givesConditioning on $N$ again yieldsThe retained payment is ${\sum }_{j=1}^N\min (X_j,M)$; it is generally not $\min (S,M)$.

Under aggregate stop loss reinsurance, the reduction is $(S-\tilde{M}{)}_{+}$. The aggregate tail is $\mathbb{P}(S>x)=q{e}^{-px/\mu }$ for $x\ge 0$, soThe prefactor is positive. Equality with the per-claim recovery is equivalent to $p\tilde{M}=M$, giving the unique reinsurance retentionThis is the relation for equal expected recoveries from excess of loss and stop loss; since $p<1$, the aggregate reinsurance retention exceeds the individual reinsurance retention. These are reductions in claim payouts, before any reinsurance premium is deducted.

Let a fixed policy intensity have prior density $3{\theta }^{-4}$ on $\theta >1$, and let $n$ annual claim counts be conditionally independent with Poisson distribution mean $\theta$. The expected process variance is $3/2$ and the variance of hypothetical means is $3/4$, so the Bühlmann credibility factor is $n/(n+2)$. A total count $k$ gives the displayed credibility premium. By contrast, the Bayesian posterior density is proportional to ${\theta }^{k-4}{e}^{-n\theta }$ on $\theta >1$. Its mean is the ratio ${\int }_1^{\infty }{\theta }^{k-3}{e}^{-n\theta }d\theta /{\int }_1^{\infty }{\theta }^{k-4}{e}^{-n\theta }d\theta$. For $n=2,k=5$, that Bayesian credibility estimate is $5/3$, whereas the affine estimate is $2$.

Suppose annual claim counts are independent with geometric distribution mass $p{q}^j$, $j\ge 0$. In a no claims discount system with discounts $0,\alpha ,\beta$, a claim returns either of the first two levels to zero; at the top, one claim sends the level to $\alpha$ and two or more send it to zero. The transition matrix is $\left(\begin{array}{ccc}q& p& 0\\ q& 0& p\\ q^2& pq& p\end{array}\right)$. Solving $\pi P=\pi$ gives the displayed stationary distribution. With full premium $c$, the stationary expected value of the premium is $c[1-(\alpha pq+\beta p^2)/(1-p^{2}q)]$. Independent annual sampling is essential; a persistent random intensity does not justify this marginal transition matrix.