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The adjustment coefficient is a positive root of $\lambda (M_X(r)-1)=cr$ in the classical risk model. When the moment-generating function is finite at $R$, the process $\exp (R[{\sum }_{j=1}^{N_t}{X}_j-ct])$ is a continuous-time martingale. It yields the Lundberg inequality and, under the relevant tilted integrability, the Cramér–Lundberg ruin asymptotic.

For NBUE claims in the classical risk model, the equilibrium claim size $Y$ is stochastically bounded by $X$. The positive exponential tail integral formula for moments gives $M_Y(R)\le M_X(R)$. Since $M_Y(R)=1+\theta$ and $M_X(R)=1+(1+\theta )\mu R$, rearrangement proves the displayed bound.

For positive claims with finite variance and $c\le \lambda \mu$, the classical risk model has ruin probability one from any finite capital. At claim times, surplus increments are independent copies of $cT-X$ with mean $c/\lambda -\mu$. Negative mean sends their partial sums to minus infinity by the strong law of large numbers. At zero mean, the increments have finite nonzero variance. For every fixed $K$, the central limit theorem gives limiting probability $1/2$ of a partial sum below $-K$. The probability of unboundedness below is therefore at least $1/2$; as a tail event it has probability zero or one by the Kolmogorov zero-one law, and hence one.

In the classical risk model with positive relative safety loading $\rho$ and adjustment coefficient $R$, tilting the ruin defective renewal equation gives a proper renewal equation. The key renewal theorem yields $e^{Ru}\psi (u)\to \rho /[R{\int }_0^{\infty }x{e}^{Rx}{f}_I(x)dx]$. The constant is positive if the denominator is finite and zero if it is infinite; the claim-size density provides the nonarithmetic hypothesis.

Fix positive claim expected value $\mu$ and relative safety loading $\theta >0$ in a classical risk model. Assume the random claim law has an adjustment coefficient $R_X>0$. By Jensen inequality, $M_X(r)\ge e^{\mu r}$. Thus $e^{\mu R_X}-1-(1+\theta )\mu R_X\le 0$. The expression on the left is strictly convex, has negative derivative at zero and tends to infinity, so its unique positive zero $R_{\mu }$ satisfies $R_X\le R_{\mu }$. If the claim size is not almost surely constant, strict Jensen inequality gives $R_X<R_{\mu }$. Consequently the Lundberg inequality upper bound $e^{-R_{\mu }u}$ for deterministic claims is smaller for $u>0$. Comparing these upper bounds alone does not establish an ordering of actual ultimate ruin 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.

In a classical risk model, an adjustment coefficient $R$ makes $\lambda (M_X(R)-1)-cR=0$. The random-sum transform identity and independent increments show that $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. This nonnegative martingale starts at one. At the ruin stopping time $\tau$, $Z_{\tau }>e^{Ru}$. Applying the optional stopping theorem only to $\tau \wedge t$ yields $\mathbb{P}(\tau \le t)\le e^{-Ru}$; passage to increasing horizons proves the Lundberg inequality without an unjustified expectation identity at an unbounded stopping time.

In a classical risk model, $T_j$ is the time of the $j$th claim. The displayed probability restricts ruin to the first $n$ claims and increases to the ultimate ruin probability. Conditioning on the first arrival and severity gives ${\psi }_{n+1}(u)=\mathbb{E}[{\overline{F}}_X(u+c{T}_1)+{\int }_0^{u+c{T}_1}{\psi }_n(u+c{T}_1-x)f_X(x)dx]$, with ${\psi }_0=0$. For exponential distribution claims of mean $\mu$ and relative safety loading $\rho$, put $d=1+\rho$ and $h=d+1$. Then ${\psi }_1(u)=e^{-u/\mu }/h$ and ${\psi }_2(u)=e^{-u/\mu }(h^{-1}+(u/\mu )h^{-2}+d{h}^{-3})$. The recursion follows because a surviving first claim leaves fresh independent future arrivals and claims. The formulas follow by integrating the rate-$\lambda$ first-arrival density; the claim-arrival rate cancels with $c=d\lambda \mu$.

In a classical risk model, before the first claim at time $t$, available capital is $u+ct$. A claim of size $x\le u+ct$ leaves future survival probability $\phi (u+ct-x)$ by the Markov property. Integrating over the independent first-arrival exponential distribution and claim density gives $\phi (u)={\int }_0^{\infty }\lambda e^{-\lambda t}{\int }_0^{u+ct}\phi (u+ct-x)f(x)dxdt$.

For a classical risk model with adjustment coefficient $R>0$, the ultimate ruin probability from capital $u\ge 0$ satisfies $\psi (u)\le e^{-Ru}$. Stop the exponential continuous-time martingale at ruin or a finite horizon, bound its value on the ruin event, and then increase the horizon.

Write the surplus process in the classical risk model as $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with positive independent claims. Condition on the first claim time, which has exponential distribution of rate $\lambda$. If that claim occurs at $t$, the available capital is $u+ct$ and survival requires its size not to exceed that capital. The independent future process then givesPut $A(v)={\int }_0^v\varphi (v-x)f(x)dx$ and change variables $v=u+ct$. We obtainThe convolution of bounded $\varphi$ with the integrable density is continuous; differentiation of this expression is therefore justified. It proves the survival version of the ruin integro-differential equation:

The density is a hyperexponential distribution with mixture weights $3/4,1/4$ and exponential rates $4,2$. Its total integral is one and its mean isThe relative safety loading means $c=(1+\rho )\lambda {\mu }_X$. Substitution of $\rho =3/5$ gives $c=\lambda /2$ and $\lambda /c=2$.

Set $A_a(u)={\int }_0^u\varphi (u-x)e^{-ax}dx$. Differentiating in its equivalent form ${\int }_0^u\varphi (v)e^{-a(u-v)}dv$ gives $A_a^{\prime }=\varphi -a{A}_a$. The survival equation becomesApply the constant-coefficient operator $(D+4)(D+2)$, where $D=d/du$. Since $(D+4)A_4=\varphi$ and $(D+2)A_2=\varphi$,Expanding and cancelling yields $\boxed{{\varphi }^{\prime \prime \prime }+4{\varphi }^{\prime \prime }+3{\varphi }^{\prime }=0}$. Thus $\varphi (u)=A+B{e}^{-u}+C{e}^{-3u}$.

Positive loading also fixes the limiting boundary value $\varphi (u)\to 1$. Indeed the aggregate-claims process minus premium income tends to minus infinity almost surely by the strong law of large numbers. Its supremum over all time is consequently finite almost surely, since paths have finitely many jumps on bounded intervals. The probability that this supremum exceeds $u$ tends to zero. Hence $A=1$. At zero, the given value is $\varphi (0)=3/8$, and the integral equation supplies the extra condition ${\varphi }^{\prime }(0)=2\varphi (0)=3/4$. Thereforeso $B=-9/16$ and $C=-1/16$. The nonruin probability isThis is the two-exponential-mixture survival probability. It increases from $3/8$ to $1$ and satisfies the original integral equation, not just the differentiated equation; the zero-capital derivative condition prevents introducing extraneous solutions.

Write the surplus in the classical risk model as $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with premium rate $c=(1+\rho )\lambda \mu$. The adjustment coefficient is the positive solution ofFor the exponential distribution of mean $\mu$, $M_X(r)=(1-\mu r{)}^{-1}$ for $r<1/\mu$. Cancelling the nonzero root $R$ in its equation gives $\lambda \mu /(1-\mu R)=c$, and henceIt is strictly between zero and $1/\mu$, as required by the transform domain.

The ultimate ruin probability is $\psi (u)=\mathbb{P}({\inf }_{t\ge 0}{U}_t<0)$. The Lundberg inequality statesRuin can occur only at a claim arrival. Thus the event defining the finite-claim ruin probability ${\psi }_n(u)$ is contained in the ultimate ruin event, giving ${\psi }_n(u)\le \psi (u)\le e^{-Ru}$ for every $n\ge 1$ and $u>0$.

For the explicit calculations, take $\mu =1$, put $d=1+\rho =c/\lambda$ and $h=d+1=2+\rho$, and let $T$ be the first arrival time. It has rate-$\lambda$ exponential distribution, independent of the severity. Ruin on the first claim is the event $X_1>u+cT$, soFor ruin by the second claim, condition on whether the first claim ruins the insurer or leaves capital $u+cT-x$. The independent future arrivals and severities have the same law as a fresh classical risk model. ThereforeThe two terms in the first line refer to disjoint ruin events, so there is no double counting.

For a direct check of the Lundberg inequality, now $R=1-1/d$. The ratio ${\psi }_1(u)/e^{-Ru}=h^{-1}{e}^{-u/d}$ is less than one. For the second finite-claim ruin probability, put $A=h^{-1}+d{h}^{-3}$ and $B=h^{-2}$. Its ratio is $e^{-u/d}(A+Bu)$. Sincethe derivative of this ratio is $e^{-u/d}[B-(A+Bu)/d]<0$ for $u\ge 0$, and its value at zero is $A<1$. This verifies both required bounds directly from the calculated probabilities.

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.

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.

For the Poisson process of claims and relative safety loading $\rho$, the premium rate in the classical risk model is $c=(1+\rho )\lambda \mu$. Dividing the adjustment coefficient equation $\lambda (M(r)-1)=cr$ by $\lambda$ explains why the arrival rate cancels. Interpret $r_{\infty }$ as the upper endpoint of the finite moment-generating function domain, so $M(r)<\infty$ for $0\le r<r_{\infty }$ and its stated divergence occurs at that endpoint.

Set $F(r)=M(r)-1-(1+\rho )\mu r$. ThenDifferentiation is valid inside the finite-transform domain, and its positive neighbourhood gives finite moments of every order. Thus $F$ is strictly convex and negative just to the right of zero. If $r_{\infty }<\infty$, divergence of $M(r)$ makes $F(r)\to \infty$ at the endpoint. If $r_{\infty }=\infty$, choose $a>0$ with $\mathbb{P}(X_1\ge a)>0$; then $M(r)\ge \mathbb{P}(X_1\ge a)e^{ar}$, which eventually exceeds every linear function. Again $F(r)\to \infty$. The intermediate value theorem gives a positive zero.

There is at most one positive zero: strict convexity and $F(0)=0$ make the secant slope $F(r)/r$ strictly increasing for $r>0$. It starts at $-\rho \mu$ and crosses zero exactly once. HenceThis proves the secant-slope existence criterion for an adjustment coefficient in the present setting.

Put $m_j=\mathbb{E}{X}_1^j$. For $u>0$, the positive remainder of the exponential series gives $e^u>1+u+u^2/2$. Apply it to $u=R{X}_1$ and take expectations:Dividing by $R>0$ yieldsKeeping the cubic term similarly givesThe polynomial $g(r)=m_3{r}^2+3m_{2}r-6\rho \mu$ is strictly increasing for $r\ge 0$, is negative at zero and tends to infinity. Let $r_2$ be its unique positive zero. The last inequality says $g(R)<0$, soFinally, $3m_2{r}_1=6\rho \mu$, whence $g(r_1)=m_3{r}_1^2>0$. Monotonicity gives $r_2<r_1$. Thus the requested upper bounds can in fact be sharpened to the strict ordering $0<R<r_2<r_1$. The bounds themselves need not lie in the finite-transform domain; the argument only evaluates the actual transform at $R$ and polynomial moments elsewhere. These are polynomial moment bounds for the adjustment coefficient.

For rate-$\alpha$ exponential distribution claims,The positive root satisfies $1/(\alpha -R)=(1+\rho )/\alpha$ after dividing out $R$, soThe cubic-bound equation reduces to $r_2^2+\alpha r_2-\rho {\alpha }^2=0$, selecting the displayed positive root. In particular $R<\alpha$, as required by the transform domain, for every $\rho >0$.

In the classical risk model, the insurer's surplus iswhere $u\ge 0$ is initial capital, $N(t)$ is a Poisson process of rate $\lambda >0$, and the claim sizes are positive independent and identically distributed random variables with expected value $\mu$, independent of the claim-arrival process. Premium flows in at constant rate $c$. A positive relative safety loading $\theta$ meansso premium income exceeds the expected claim cost per unit time. The ruin stopping time is $\tau =\inf \{t\ge 0:U(t)<0\}$, with $\tau =\infty$ if ruin never occurs, and the ultimate ruin probability is $\psi (u)=\mathbb{P}(\tau <\infty )$.

Assume the stated positive adjustment coefficient $R$ exists and that the claim moment-generating function is finite at $R$. The Lundberg inequality isHere is a proof, including the stopping argument. By conditioning on the Poisson distribution of $N(t)$, the random-sum transform identity givesThe adjustment coefficient equation is $\lambda (M(R)-1)=cR$. Consequently the exponential surplus martingalehas expected value one at every finite time. More strongly, the claim process has independent increments, so for $s\le t$,Thus $Z$ is a nonnegative continuous-time martingale for the filtration generated by arrivals and claim sizes. For a fixed finite horizon $T$, the optional stopping theorem at the bounded stopping time $\tau \wedge T$ gives $\mathbb{E}{Z}_{\tau \wedge T}=1$. This bounded use is legitimate: the stopped values belong to the closed martingale on $[0,T]$ given by conditional expectations of the integrable variable $Z_T$.

On $\{\tau \le T\}$, the surplus is negative, and therefore $C(\tau )-c\tau =u-U(\tau )>u$. By nonnegativity on the complementary event,The ruin events increase to $\{\tau <\infty \}$ as $T$ increases, so continuity of probability gives the claimed Lundberg inequality. No expectation identity at the unbounded stopping time $\tau$ is needed. Positive loading by itself does not guarantee a positive adjustment coefficient for every claim law; the existence assumption is used here.

For the exponential distribution of claim sizes, $M(r)=1/(1-\mu r)$ for $r<1/\mu$. Substitution in the adjustment coefficient equation givesDiscard the zero root and divide by $\mu r$. The unique positive solution iswhich lies strictly inside the finite-transform domain.

For deterministic claim sizes, the claim moment-generating function is $e^{\mu r}$. Define $h(r)=e^{\mu r}-1-(1+\theta )\mu r$. It satisfies $h(0)=0$, $h^{\prime }(0)=-\theta \mu <0$, $h^{\prime \prime }(r)={\mu }^2{e}^{\mu r}>0$, and $h(r)\to \infty$ as $r\to \infty$. Thus its strict convexity gives exactly one positive zero $R_{\mu }$, with $h(r)<0$ for $0<r<R_{\mu }$.

Put $x=\mu R=\theta /(1+\theta )\in (0,1)$. Since $-\log (1-x)>x$, we have $e^x<1/(1-x)$. HenceIt follows thatThus the deterministic-claim Lundberg inequality provides a smaller upper bound at positive capital; at zero capital both bounds equal one. This is a special case of deterministic claims maximize the adjustment coefficient at fixed mean and loading: fixing the mean and loading, variability lowers the exponential decay coefficient relative to deterministic claims. The ordering of these upper bounds alone is not a proof that the actual ultimate ruin probabilities are ordered.

The claim law is an Erlang distribution of shape two and rate $2/\beta$. Direct integration, using ${\int }_0^{\infty }x{e}^{-ax}dx=a^{-2}$ for $a>0$, gives its moment-generating functionDifferentiating at zero gives $\mathbb{E}X=\beta$. The positive relative safety loading condition for the classical risk model is thereforeThe adjustment coefficient is the positive root in the finite-transform domain ofSet $z=\beta R/2$ and $k=2c/(\lambda \beta )>2$. For $R>0$, division by $z$ reduces the equation toThe two roots are $z=(2k-1\pm \sqrt{1+4k})/(2k)$. The minus root lies strictly between zero and one: positivity follows from $(2k-1{)}^2-(1+4k)=4k(k-2)>0$, and the numerator is smaller than $2k$. The plus root exceeds one and is outside the moment-generating function's domain. Consequently the adjustment coefficient for shape-two Erlang claims isThe zero root of the original equation is not an adjustment coefficient. The increasing secant slope $(M_X(r)-1)/r$ on $(0,2/\beta )$ also establishes uniqueness of the positive root.

Now set $\beta =1$. Under quota share reinsurance, the retained payment is $Y_{\alpha }=\alpha X$. For $0<\alpha \le 1$, the change of variables $x=y/\alpha$ yieldsHere the second gamma parameter is the rate. In particular, $\mathbb{E}{Y}_{\alpha }=\alpha$ and $M_{Y_{\alpha }}(r)=(1-\alpha r/2{)}^{-2}$. At full cession $\alpha =0$ the retained payment instead has a point mass at zero.

The reinsurer's expected annual payments are $\lambda (1-\alpha )$, so its loaded premium is $(1+{\theta }_R)\lambda (1-\alpha )$. Subtract it from the original premium income to obtain the direct insurer's net premium rateIts expected retained claim outflow is $\lambda \alpha$. Thusby the stipulated lower bound on $\alpha$. In particular $c_{\alpha }>0$, so the retained portfolio satisfies the net profitability condition. Its relative safety loading is ${\theta }_R-({\theta }_R-\theta )/\alpha$.

Apply the same adjustment coefficient for shape-two Erlang claims formula with scale $\alpha$ and net premium $c_{\alpha }$. WritinggivesEquivalently it is the positive solution, with $R_{\alpha }<2/\alpha$, of

For fixed positive initial capital, a larger adjustment coefficient reduces the exponential upper bound ${\psi }_{\alpha }(u)\le e^{-R_{\alpha }u}$ from the Lundberg inequality. It also makes the large-capital decay exponent in the Cramér–Lundberg ruin asymptotic larger. These are the reasons for maximizing $R_{\alpha }$ as a ruin-risk criterion; the coefficient alone need not order exact finite-capital ruin probabilities for arbitrary different models.

If ${\theta }_R=\theta$, then $c_{\alpha }=(1+\theta )\lambda \alpha$ and $k_{\alpha }=2(1+\theta )$ is constant. The equal-loading quota share and vanishing retention relation becomesThe numerator is positive for $\theta >0$, so $R_{\alpha }$ decreases strictly with $\alpha$ and tends to infinity as $\alpha \downarrow 0$. There is no maximizing retention in the stipulated interval $0<\alpha \le 1$; the best limit is full cession. If the endpoint $\alpha =0$ is admitted as an idealized contract, all claims are transferred and the net premium income is zero, leaving the direct insurer's capital constant and its ultimate ruin probability zero. At that endpoint the coefficient equation is identically zero, so it no longer defines a unique finite adjustment coefficient. Calling $\alpha =0$ optimal requires this explicit extension of the original admissible set.

The surplus in the classical risk model is $U(t)=u+ct-S(t)$. With ruin defined as a strictly negative surplus, its ultimate ruin probability isLet ${\overline{F}}_X(v)={\int }_v^{\infty }{f}_X(x)dx$. The first arrival time has exponential distribution of rate $\lambda$, independent of the first claim. Before that claim no ruin occurs, because the premium rate is positive. Given its arrival at $t$ and size $x$, available capital just before it is $u+ct$. If $x>u+ct$ ruin is immediate; otherwise the independent increments and fresh claim sizes make subsequent ruin probability $\psi (u+ct-x)$. Thus first-claim conditioning givesThis is the ruin counterpart of the first-claim decomposition for survival probability. DefineThe convolution term is continuous: extend the bounded function $\psi$ by zero to negative arguments and use continuity of translations in $L^1$ for the integrable density $f_X$. The tail function is also continuous. Changing variables to $v=u+ct$ yieldsDifferentiating this exponentially weighted integral establishes differentiability rather than presuming it. The fundamental theorem of calculus givesReversing the integration variable in the convolution therefore proves the ruin integro-differential equation:At zero the derivative is the right derivative; the same representation supplies it.

For the exponential claim law, introduceDifferentiating the explicit convolution gives $A^{\prime }=\alpha \psi -\alpha A$. Henceusing the proved integral equation. Differentiating that equation once more givesso the required ordinary differential equation isPositive relative safety loading means $c>\lambda /\alpha$, so $\kappa =\alpha -\lambda /c>0$. The general solution is $a+b{e}^{-\kappa u}$.