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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.

In the classical risk model write $U(t)=u+ct-{\sum }_{j=1}^{N(t)}{X}_j$, where the Poisson process $N(t)$ has rate $\lambda$ and is independent of the claim sizes. Define the ruin time and ultimate ultimate ruin probability byThe relative safety loading is $\rho =c/(\lambda \mu )-1>0$, so $c=(1+\rho )\lambda \mu$. The adjustment coefficient is the nonzero positive solutionExistence and uniqueness follow from the secant-slope existence criterion for an adjustment coefficient. Explicitly, $\lambda (M(r)-1)-cr$ has derivative $\lambda \mu -c<0$ at zero and is strictly convex for positive claims. The assumed divergence of $M$ makes it cross zero once at a positive $R<r_{\infty }$. When $r_{\infty }=\infty$, an exponential lower bound from any positive tail event shows that $M(r)/r\to \infty$. In particular $R$ is inside the finite-transform domain, not at its endpoint.

Put $\overline{F}=1-F$. Replacing $\phi =1-\psi$ in the survival renewal equation for a classical risk model yieldsBy the tail integral formula for moments, $\mu ={\int }_0^{\infty }\overline{F}(x)dx$. Therefore this is the defective renewal equationThe original kernel has mass $\lambda \mu /c=1/(1+\rho )<1$. For the exponential tilt of the ruin renewal kernel, setMultiplying by $e^{Ru}$ gives the required proper renewal equationTo verify that this is a probability renewal kernel, Tonelli theorem gives, for $r>0$ in the finite-transform domain,The adjustment coefficient equation consequently implies ${\int }_0^{\infty }{k}_R(x)dx=1$. Its expected value isIt is finite because $R$ is interior to the finite-transform domain and is positive because the strict convex crossing has derivative $\lambda M^{\prime }(R)-c>0$.

We quote the key renewal theorem in the following form: for a nonarithmetic distribution $K$ of positive increments with finite positive mean $m$, and a directly Riemann integrable nonnegative function $h$, the locally bounded solution of $z=h+z\ast K$ satisfies $z(u)\to m^{-1}{\int }_0^{\infty }h$. It has renewal representation $z=h\ast {\sum }_{j\ge 0}{K}^{\ast j}$; this follows by iterating the equation, since the probability that arbitrarily many positive increments have sum at most a fixed $u$ tends to zero.

Here $K_R(dx)=k_R(x)dx$ is absolutely continuous, and hence nonarithmetic distribution, even if the original claim law has atoms. The function $h_R$ is continuous. Choose $\delta >0$ with $M(R+\delta )<\infty$. The Markov inequality gives $\overline{F}(x)\le M(R+\delta )e^{-(R+\delta )x}$, and hence $h_R(u)\le C_{\delta }{e}^{-\delta u}$. Continuity on compact intervals and this exponential bound make the upper Riemann sums finite with uniformly vanishing tails, proving direct Riemann integrability.

A further application of Tonelli theorem evaluates the forcing integral:All hypotheses of the key renewal theorem are now checked, so the interior adjustment coefficient ruin prefactor isThis proves the requested Cramér–Lundberg ruin asymptotic, including its constant.

For the two-component hyperexponential distribution, conditioning on the chosen exponential component givesIts moment-generating function and derivative areThe moment-generating function diverges at $1/2$, while the derivative of the adjustment equation at zero is negative. Strict convexity therefore places its unique positive adjustment coefficient inThe adjustment equation, divided by $R$, becomesUsing this identity to subtract $(1+\rho )\mu$ from $M^{\prime }(R)$ givesThus the asymptotic constant in terms of $\rho$ and $R$ isThe denominator is positive on the identified domain. If desired, $R$ is the smaller root of $6(1+\rho )R^2-(5+9\rho )R+3\rho =0$; the other algebraic root lies outside the positive finite-transform interval and is not an adjustment coefficient.

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.