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The number of insurance claims in a specified exposure period. Its probability generating function combines with the moment-generating function of an independent claim size through the random-sum transform identity. A Poisson distribution, geometric distribution, or negative binomial distribution provides a possible count model, with different dispersion and zero-claim probabilities.

The expected value premium principle charges the expected value of the covered loss plus a proportional margin. For a nonnegative random variable $L$ with finite expected value, and a fixed dimensionless loading $\theta \ge 0$, it setsHere $\theta =0$ gives the expected claim cost alone; $\theta >0$ gives a positive margin when $\mathbb{E}[L]>0$. The loading is a fraction of the expected value, not of the largest possible claim size. Specify the covered period and use a consistent monetary valuation date. The rule itself does not determine $\theta$. ^[ref].

Consider an explicitly toy one-year policy: it pays £1,000 with probability $p=0.1$, and zero otherwise. Write $L=1000B$, where $B$ has the Bernoulli distribution with parameter $p$. The expected value is $1000(0.1)=\pounds 100$. A loading $\theta =0.2$ adds £20, making the premium £120. Neither the £100 expected value nor the £120 premium is a possible realized claim: the actual payment is £0 or £1,000. The first diagram separates these quantities. All numerical examples and both diagrams here are exact consequences of invented assumptions, not measured insurance data.

Why pay £120 for a loss whose expected value is only £100? Conditional on the insurer meeting its obligation, the policyholder replaces an uncertain £1,000 bill with a known £120 payment. The insurer can combine many such policies, so a few claims are funded by premiums from the whole pool. Winkel develops this motivation and the importance of independence in §15.2 of ^[ref].

To see the pooling mechanism precisely, suppose there are $n$ toy policies with mutually independent Bernoulli random variables $B_1,\dots ,B_n$, each of parameter $0.1$. Every claim size is £1,000, with at most one claim per policy. The aggregate claims model isUsing linearity of expectation and addition of variances for independent random variables, the expected value and standard deviation of this binomial distribution aggregate areThus 100 policies have expected aggregate claims £10,000 and aggregate standard deviation £3,000; 10,000 policies have expected aggregate claims £1,000,000 and aggregate standard deviation £30,000. Total uncertainty grows, but much more slowly than the expected value. Per policy, $S_n/n$ has expected value £100 and standard deviation $\pounds 300/\sqrt{n}$. Equivalently, the aggregate coefficient of variation, its standard deviation divided by its positive expected value, isIt falls from $300\%$ for one policy to $30\%$ for 100 policies and $3\%$ for 10,000. This is the useful beauty of independent pooling: an erratic individual cost becomes a comparatively stable average cost. The law of large numbers describes the corresponding convergence of the average toward its expected value.

At the same fixed loading $\theta =0.2$, total premium is $\Pi (S_n)=120n$, with expected surplus before other outgoings $120n-100n=20n$. The loading margin divided by the aggregate standard deviation is $20n/(300\sqrt{n})=\sqrt{n}/15$. This ratio rises with $n$ under our independence assumption; it is not a specified loss probability. For $n=100$, premiums total £12,000, but 13 claims cost £13,000. The one-year probability that claims exceed collected premiums is exactlyA positive loading therefore leaves about a $19.8\%$ chance of a one-year claims deficit in this toy pool. A claims deficit need not mean insolvency if capital is available. This calculation assumes premiums are collected before claims and ignores expenses and investment income; it is not an ultimate ruin probability.

The expected value premium principle also has a revealing limitation. With a common fixed $\theta$, linearity of expectation gives $\Pi (L_1+L_2)=\Pi (L_1)+\Pi (L_2)$ even without independence. Pooling therefore does not automatically reduce the premium per policy under this rule, and two losses with the same expected value receive the same premium regardless of their variance or large-loss probabilities. Any decision to change the loading after pooling requires information beyond the formula.

Dependence can remove the stability benefit. For arbitrary square-integrable losses, the aggregate variance includes their covariances:In the extreme toy common-shock model, every policy has $L_i=1000B$ for the same Bernoulli random variable $B$. Each policy still has expected value £100, so total premium is still $120n$, but $S_n=1000nB$ has standard deviation $300n$ and coefficient of variation $3$ for every $n$. With probability $0.1$, all policies claim together and losses exceed premiums by $880n$. The second diagram compares this model with independent pooling. Real shared events can invalidate independence; Winkel uses flood claims as an example in §15.2 of Actuarial Science II. The common-shock curve is an illustrative limiting model, not a fit to flood data.

The earlier $n^{-1/2}$ relative-variability calculation needs finite variance and the stated model of independent and identically distributed random variables. A finite expected value alone suffices to define the premium but does not imply finite variance. The rule makes no promise of solvency: losses, available capital and the timing of payments must be modelled separately. In the classical risk model, claim arrival rate $\lambda$ and mean claim size $\mu$ give premium rate $c=(1+\theta )\lambda \mu$; relative safety loading is then $\theta$. That model tracks capital through time and asks for ultimate ruin probability, a different question from calculating an expected margin. ^[ref]. Under quota share reinsurance, the reinsurer can apply the expected value premium principle to its ceded fraction $(1-\alpha )S$ using its own loading, which need not equal the direct insurer's loading.

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.

In the classical risk model, the surplus with initial capital $u\ge 0$ iswhere $N(t)$ is a Poisson process of rate $\lambda$, and the nonnegative independent claim sizes, of mean $\mu >0$, are independent of that process. Positive relative safety loading $\rho$ makes expected premium income exceed expected claim outflow. With ruin time $T=\inf \{t\ge 0:U(t)<0\}$, the ultimate ruin probability isThe adjustment coefficient is the positive root $R$ in the finite domain of the claim moment-generating function satisfyingThe zero root is excluded by definition.

Under the stated transform-divergence assumption this positive root exists and is unique. Indeed $F(r)=M_X(r)-1-(1+\rho )\mu r$ has $F(0)=0$, derivative $F^{\prime }(0)=-\rho \mu <0$, and strictly positive second derivative on its positive domain. At a finite endpoint the diverging transform makes $F$ positive eventually. At an infinite endpoint, choose $a>0$ with $\mathbb{P}(X\ge a)>0$; the bound $M_X(r)\ge \mathbb{P}(X\ge a)e^{ar}$ makes it dominate the linear term. Strict convexity then gives exactly one positive crossing. This is the secant-slope existence criterion for an adjustment coefficient.

The Lundberg inequality isFor a short justification, the compound-Poisson transform and the defining equation make $e^{-RU(t)}$ a nonnegative martingale starting at $e^{-Ru}$. Stop it at $T\wedge t$, a bounded stopping time. On $\{T\le t\}$ its value is greater than one. The optional stopping theorem therefore gives $\mathbb{P}(T\le t)\le e^{-Ru}$; let $t$ increase to infinity. No expectation identity at an unbounded ruin time is required.

Define $N_i={\sum }_{k=1}^N{1}_{\{X_k=a_i\}}$. Grouping equal claim sizes gives the pathwise identityConditional on $N=n$, the $n$ independent claims receive category labels with probabilities $p_i$. There are $n!/(n_1!\cdots n_m!)$ assignments producing prescribed category counts, and every assignment has probability ${\prod }_i{p}_i^{n_i}$. Thus the conditional probability distribution is a multinomial distribution:when the nonnegative counts sum to $n$, and is zero otherwise.

For $0\le z_i\le 1$, condition first on $N$ and use the probability generating function of a Poisson distribution:This factorization is the joint probability generating function of independent Poisson distribution variables. HenceA category with $p_i=0$ has zero count almost surely. The categorical thinning of a Poisson claim count produces independence after averaging over the random total, even though the counts are constrained to sum to $n$ conditional on that total.

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.

For nontrivial positive claim sizes in the classical risk model and $\lambda >0$, the adjustment coefficient is the unique positive solutionIndeed, $h(r)=\lambda (M_X(r)-1)-cr$ has $h(0)=0$, $h^{\prime }(0)=\lambda \mu -c<0$, and is strictly convex. It becomes positive near the right endpoint of the moment-generating function domain. For a finite endpoint this follows from the assumed divergence; for an infinite endpoint, $\mathbb{P}(X\ge \epsilon )>0$ for some $\epsilon >0$ gives an exponential lower bound on $M_X(r)$, which eventually exceeds any linear function. Convexity gives existence and uniqueness of the positive crossing and $\lambda M_X^{\prime }(R)-c>0$.

Put $q=\lambda \mu /c$ and $\overline{F}=1-F$. Substituting $\phi =1-\psi$ into the supplied survival renewal equation for a classical risk model, and using ${\int }_0^{\infty }\overline{F}(x)dx=\mu$, yieldsMultiplying by $e^{Ru}$ gives the proper renewal-type equationThe exponential tilt of the ruin renewal kernel turns a defective kernel into a probability density function, because the tail integration identity givesThis identity follows also by integrating $e^{RX}-1=R{\int }_0^X{e}^{Rx}dx$ and applying Tonelli theorem. The tilted probability density function defines a nonarithmetic distribution, even if the claim size probability distribution itself has atoms.

Differentiate $(M_X(R)-1)/R$ to calculate its mean:It is finite and positive because $R$ is inside the finite moment-generating function domain. Next, reversing the order of integration givesThe forcing function is directly Riemann integrable. Choose $\epsilon >0$ with $M_X(R+\epsilon )<\infty$. Markov's inequality bounds $\overline{F}(x)$ by $M_X(R+\epsilon )e^{-(R+\epsilon )x}$, so $g_R(u)$ is bounded by a constant times $e^{-\epsilon u}$. It is continuous; on finite intervals continuity controls its Riemann sums, and the exponential envelope controls the tails. These are sufficient conditions for direct Riemann integrability.

Iterating the renewal equation by convolution gives $Z=g_R\ast U_R$, where $U_R={\sum }_{n\ge 0}{k}_R^{\ast n}$ is the renewal measure. The remainder after $n$ iterations is bounded on each fixed interval by a constant times the chance that a sum of $n$ positive interarrivals remains in that interval, which tends to zero. The key renewal theorem, applicable to this nonarithmetic distribution of finite mean and the directly integrable forcing, now gives the interior adjustment coefficient ruin prefactor:Both numerator and denominator are positive, so $0<A<\infty$.

For the unit-rate shape-two Erlang distribution, $\mu =2$, $M_X(r)=(1-r{)}^{-2}$ and $M_X^{\prime }(r)=2(1-r{)}^{-3}$. Write $\delta =\lambda /c$, so positive loading requires $0<\delta <1/2$. Dividing the adjustment coefficient equation by its nonzero root gives $\delta (2-R)=(1-R{)}^2$. ThusThe last simplification uses $\delta =(1-R{)}^2/(2-R)$; it is the Erlang shape-two ruin prefactor. Its value depends on the premium-to-arrival ratio, which is not specified numerically.

For the final exponential representation, evaluating at zero in the survival renewal equation for a classical risk model gives $\lambda \mu /c=\psi (0)=a+b$. If the slower term is present, necessarily $a>0$ and $e^u\psi (u)\to a$. Comparing with the finite positive asymptotic constant just proved yieldsA smaller adjustment coefficient would make the limit zero and a larger one would make it infinite. Here the leading exponential term determines a ruin adjustment coefficient. The coefficient $b$ need not be assumed positive to identify the leading term.

The wording does not specify that $a$ is nonzero. If $a=0$ and $b>0$, the only surviving term instead gives $R=6$, $A=b$ and $\lambda \mu /c=b$. This case is realizable: claims with an exponential distribution of rate $\beta =6+\lambda /c$ have ultimate ruin probability $(\lambda /(c\beta ))e^{-6u}$. Negative $a$ would make the ultimate ruin probability negative for large $u$, and both coefficients zero are incompatible with a nontrivial positive claim rate and mean. These observations specify the coefficient qualification needed for the usual $R=1$ answer.

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.

If zero-sized claims are allowed, a compound Poisson distribution count parameter need not be unique. Poisson thinning of zero claims gives the displayed positive-claim count parameter and the conditional claim size probability distribution given $X>0$, without changing the aggregate. If no positive claims occur, the aggregate is identically zero. This matters when representing a finite weighted sum of Poisson distribution counts whose coefficients include zero.