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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.

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.

Replacing a shape-one-half gamma claim law by an exponential distribution with the same mean overestimates the adjustment coefficient. Indeed $R_I-R=\beta [3(1+\rho )-\sqrt{(\rho +9)(\rho +1)}]/[2(1+\rho )]>0$ because the difference of the squared terms is $8\rho (1+\rho )$. The resulting putative Lundberg inequality bound $e^{-R_{I}u}$ is smaller than the valid true-model bound $e^{-Ru}$ for $u>0$; the former is not justified for the true claim law. The true variance is twice the assumed variance, and its exponential tail decays more slowly.

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.

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.

Normalizing the true density givesa gamma distribution of shape $1/2$ and rate $\beta$. Its mean is $1/(2\beta )=\mu$, and the Gamma integral givesPut $x=R/\beta$. Since $\mu \beta =1/2$, the positive adjustment coefficient satisfiesBoth sides are positive in this domain. Squaring, multiplying by $1-x$, and dividing out the zero root $x=0$ yieldsThe discriminant is $(1+\rho {)}^2[(\rho -3{)}^2+16\rho ]=(1+\rho {)}^2(\rho +9)(\rho +1)$. The negative constant term makes one root positive and the other negative. The polynomial is negative at zero and has value four at one, so the positive root lies in $(0,1)$ and is not an extraneous solution introduced by squaring. Therefore the adjustment coefficient for shape-one-half gamma claims is

To compare it with the exponential calculation, write $R_I=2\beta \rho /(1+\rho )$. Their difference issince $9(1+\rho {)}^2-(\rho +9)(\rho +1)=8\rho (1+\rho )>0$. Thus the two proposed Lundberg inequality bounds are ordered asOnly ${\psi }_{\mathrm{true}}(u)\le e^{-Ru}$ is justified for the true claim law. The exponential calculation gives ${\psi }_{\mathrm{exponential}}(u)\le e^{-R_{I}u}$ for a different model, not a valid guaranteed bound on the true ruin probability.

This exponential misspecification of shape-one-half gamma claims is optimistic: the true variance is $1/(2{\beta }^2)=2{\mu }^2$, whereas the assumed exponential variance is ${\mu }^2$. Its tail also decays at rate $\beta$, more slowly than the assumed rate $1/\mu =2\beta$. Equivalently, for $0<r<\beta$, the true transform $(1-2\mu r{)}^{-1/2}$ is larger than $(1-\mu r{)}^{-1}$, since $(1-\mu r{)}^2>1-2\mu r$. This stronger transform growth makes the true adjustment root smaller. Comparing upper bounds alone does not by itself prove an ordering of the two actual ruin probabilities. At $u=0$ both displayed bounds equal one.

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.

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.

In the classical risk model, write the surplus aswhere $N(t)$ is the claim-arrival Poisson process. The ruin time is $\tau =\inf \{t\ge 0:U(t)<0\}$. The ultimate ruin probability isThe positive root $R$ in the question is the adjustment coefficient. The Lundberg inequality and the Cramér–Lundberg approximation are, respectively,The interior adjustment coefficient ruin prefactor is finite here because $R$ lies strictly inside the finite domain of the moment-generating function; the claim-size density supplies the nonarithmetic condition for the asymptotic. Strict convexity makes its denominator positive. The approximation is an asymptotic statement, while the inequality is a bound at every nonnegative initial capital.

The inequality can also be seen directly from the exponential surplus martingale $e^{-RU(t)}$. Its expectation factor per unit time is zero in the exponent because $\lambda (M_X(R)-1)-cR=0$. Apply the optional stopping theorem at $\tau \wedge T$, a bounded time. On $\{\tau \le T\}$, $e^{-RU(\tau )}>1$, so $\mathbb{P}(\tau \le T)\le e^{-Ru}$. Increasing $T$ proves the bound without stopping at an unbounded time.

For any nonnegative claim size, the Tonelli theorem applied to $e^{RX}-1={\int }_0^{X}R{e}^{Rx}dx$ gives the exponential tail identityCombining it with the adjustment equation yieldsThe density $f_Y(x)=(1-F_X(x))/\mu$ integrates to one by the tail integral formula for moments. Thus $Y$ has the equilibrium claim-size distribution, and the preceding formula says $M_Y(R)=1+\theta$.

Under the given NBUE condition, the stochastic order comparison is $1-F_Y(x)\le 1-F_X(x)$ at every $x\ge 0$. Apply the same exponential tail identity to both variables to obtainConsequentlyand rearrangement proves the adjustment-coefficient lower bound for NBUE claims:This uses the specified distributional comparison, not an unwarranted assumption that an arbitrary equilibrium claim law is always smaller than its original law.

For the final example, $f_X(x)=x{e}^{-x}$ is the shape-two, rate-one gamma distribution, with $\mu =2$, survival function ${\overline{F}}_X(x)=(1+x)e^{-x}$, and moment-generating function $M_X(r)=(1-r{)}^{-2}$ for $r<1$. With $\theta =1/10$, the positive adjustment root satisfiesDividing out the zero root and simplifying gives $11R^2-17R+1=0$. Only the smaller root is in the finite-transform interval $(0,1)$, soThe equilibrium density and survival function areThus ${\overline{F}}_Y(x)\le (1+x)e^{-x}={\overline{F}}_X(x)$ for every $x\ge 0$, proving that this claim law is new better than used in expectation. Finally the lower bound equals $1/22$. Our exact root is larger because $16>7\sqrt{5}$, equivalently $256>245$. Henceverifying the requested bound for the example.

For a unit-rate exponential distribution, $\mu =1$ and $M_X(r)=1/(1-r)$ for $r<1$. The adjustment coefficient equation isIts nonzero root isIt lies in $(0,1)$ for $\theta >0$, so the moment-generating function is finite there and the Lundberg inequality applies.

The severity is an equal-weight mixture distribution of rate-$2$ and rate-$2/3$ exponential distributions. Its expected value isThe moment-generating function, finite exactly for $r<2/3$, isCanceling the nonzero $r$ in the adjustment coefficient equation and multiplying by the positive denominator on $0<r<2/3$ givesThus the required polynomial isSince $P(0)=4\theta >0$, $P(2/3)=-2<0$, and the leading coefficient is positive, one root lies in $(0,2/3)$ and the other exceeds $2/3$. Only the smaller root is in the moment-generating function domain. ConsequentlyThe rationalized expression avoids subtraction of nearly equal numbers when $\theta$ is small.

The erroneously chosen same-mean exponential distribution gives $R_{\exp }=\theta /(1+\theta )$. Direct substitution shows $P(R_{\exp })=-\theta <0$, so $R_{\exp }$ lies strictly between the two polynomial roots, and in particular $R_{\mathrm{mix}}<R_{\exp }$. Thereforewhile both bounds equal one at $u=0$. The exponential misspecification produces a smaller claimed upper bound. The Lundberg inequality certifies $e^{-R_{\mathrm{mix}}u}$ for the actual mixture; it does not certify the smaller expression computed from the wrong severity law.

Let $C_t$ be the cumulative claim amount and $U_t=u+ct-C_t$ the surplus in the classical risk model. The supplied adjustment coefficient satisfies $\lambda (M_X(R)-1)=cR$. The Compound Poisson process has independent increments, and the random-sum transform identity givesIt follows that $Z_t=\exp (R(C_t-ct))$ is a nonnegative martingale with $Z_0=1$: conditional on the past, the multiplicative increment has expected value one. This is the exponential surplus martingale.

Let $\tau =\inf \{t\ge 0:U_t<0\}$ be the ruin stopping time. Surplus decreases only at claim arrivals, so on $\{\tau \le t\}$ the deficit at ruin gives $C_{\tau }-c\tau >u$ and $Z_{\tau }>e^{Ru}$. The optional stopping theorem at the bounded stopping time $\tau \wedge t$ yieldsAs $t$ increases without bound, the events $\{\tau \le t\}$ increase to eventual ruin. Hence the ultimate ruin probability satisfies the Lundberg inequalityThe proof uses only bounded-time stopping; it does not presume that the martingale can be stopped with equality of expected values at an unbounded ruin time.

For the original exponential distribution, cancel the nonzero root into obtainIt lies strictly below the transform pole $1/\mu$.

With the extra expenses, the insurer's payment per claim is $Y=X+A$, where $A$ has exponential distribution of expected value $\mu /2$ and is independent of $X$. The convolution of independent random variables gives a hypoexponential distribution withKeeping the relative safety loading fixed means using the new expected payment: the premium rate becomes $c_{\mathrm{new}}=\lambda (1+\theta )(3\mu /2)$. It does not mean keeping the old premium rate fixed. The adjustment coefficient with independent claim expenses therefore solvesSet $t=\mu R$, cancel $t>0$, and simplify:The quadratic is positive at $t=0$ and equals $-4$ at $t=1$. Its leading coefficient is positive, so the smaller root is in $(0,1)$ and the larger root exceeds $1$. Only the smaller root lies in the finite moment-generating function domain. HenceFor $\theta =1$,Thus the new adjustment coefficient is about $9.72\%$ smaller, even though the premium rate has been increased to retain the same relative safety loading. The Lundberg inequality consequently has a slower exponential decay rate as a function of capital.

Use $c_{\mathrm{prem}}$ for the premium income rate, reserving $c$ later for the smaller exponential decay rate. In the classical risk model, the surplus isHere $\rho$ is the relative safety loading. The aggregate claims form a Compound Poisson process. Define $L_t={\sum }_{j=1}^{N_t}{X}_j-c_{\mathrm{prem}}t$, so ruin occurs when $L_t>u$. By independent increments and the exponential formula for a marked Poisson sum,Thus $Z_t=e^{R{L}_t}$ is a nonnegative continuous-time martingale with $Z_0=1$, because the adjustment coefficient makes the exponent vanish.

Let $\tau =\inf \{t:U_t<0\}$. Apply the optional stopping theorem at the bounded stopping time $\tau \wedge T$. On $\{\tau \le T\}$, $L_{\tau }>u$, soLetting $T$ increase proves the Lundberg inequality and the unscaled limit:

For the precise asymptotic, putThe given exponential integral identity makes $k$ a probability density. Multiplying the given defective renewal equation by $e^{Ru}$ turns it into the ordinary renewal equationFor clarity, the version of the key renewal theorem used here is: if the interarrival law is nonarithmetic, has mean $m_k\in (0,\infty )$, and $g$ is directly Riemann integrable, the locally bounded solution of this renewal equation satisfies $z(u)\to m_k^{-1}{\int }_0^{\infty }g(v)dv$. The infinite-mean version gives zero for nonnegative directly Riemann integrable $g$.

All the hypotheses can be checked here. The density $k$ gives a nonarithmetic distribution. The Tonelli theorem givesFurthermoreThus $g$ is continuous and integrable, and ${\int }_0^{\infty }|g^{\prime }(u)|du\le R\int g+1<\infty$. On a mesh of width $\delta$, the difference between its upper and lower sums is at most $\delta \int |g^{\prime }|$; its upper sum is at most $\int g+\delta \int |g^{\prime }|$. This proves direct Riemann integrability rather than assuming it. Also $0\le z(u)\le 1$ by the Lundberg inequality, so the solution is locally bounded. Its renewal representation is $z=g\ast U_k$, where $U_k={\sum }_{j\ge 0}{K}^{\ast j}$ and $K(dx)=k(x)dx$; the residual after iteration tends to zero on compact intervals because sums of positive interarrivals tend to infinity.

Writing $J={\int }_0^{\infty }x{e}^{Rx}{f}_I(x)dx$, the tilted interarrival expected value is $m_k=hJ$. The key renewal theorem gives the Cramér–Lundberg ruin asymptoticIf $J=\infty$, the same formula is interpreted as $A=0$. A positive finite asymptotic constant requires $J<\infty$; this extra integrability is not explicitly stated in the paper.

For the final two-exponential case, evaluate the defective renewal equation at zero:One can identify the adjustment coefficient without silently assuming $A>0$. For $0<r<c$, setIt is finite and positive. Integrating the nonnegative terms of the defective renewal equation, using the Tonelli theorem, first shows that $F_I(r)={\int }_0^{\infty }{e}^{rx}{f}_I(x)dx$ is finite and then givesAs $r\uparrow c$, $P(r)\to \infty$ because $a>0$. By monotone convergence theorem, $h{F}_I(c)=1$. The integrated tail distribution in the classical risk model has density $f_I(x)=\mathbb{P}(X_1>x)/\mu$, so $F_I(r)=[M(r)-1]/(\mu r)$ by the tail integral formula for moments. Hence $c$ solves the adjustment equation, and its stipulated uniqueness implies $R=c$. Finally the displayed form of $\psi$ gives the remaining constantsIn particular the decay exponent $d$ and the coefficient $b$ do not affect $R$ or $A$. The $c$ in these final answers is the printed decay rate, not $c_{\mathrm{prem}}$.