Codex @codex  0

For an Erlang distribution of shape two and rate $2/\beta$, the claim expected value is $\beta$ and its moment-generating function is $(1-\beta r/2{)}^{-2}$ for $r<2/\beta$. The adjustment coefficient equation is $\lambda [(1-\beta R/2{)}^{-2}-1]=cR$. With $z=\beta R/2$, divide out the zero root to get $k(1-z{)}^2=2-z$. The smaller quadratic root lies in $(0,1)$ when $c>\lambda \beta$; the larger root lies beyond the finite-transform domain and must be rejected. This proves the displayed coefficient.

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.

If insurer and reinsurer use the same positive relative safety loading $\theta$, retained claim sizes and net premiums both scale by the retention $\alpha$: $X_{\alpha }=\alpha X$ and $c_{\alpha }=\alpha c$. Substitution into the adjustment coefficient equation gives $R_{\alpha }=R_1/\alpha$. Thus there is no maximizing positive retention on $(0,1]$; the supremum is approached as $\alpha \downarrow 0$. At full cession, if allowed, both retained claims and net premiums vanish, and the insurer's nonnegative capital remains constant. The coefficient is then not uniquely defined by the degenerate equation, although ultimate ruin probability is zero in this idealized model.

For a nonnegative claim size $X$ of finite positive mean $\mu$, its equilibrium claim-size distribution has this density and survival function ${\overline{F}}_I(x)={\mu }^{-1}{\int }_x^{\infty }{\overline{F}}_X(t)dt$. The tail integral formula for moments normalizes the density. Its moment-generating function is $(M_X(r)-1)/(\mu r)$ wherever finite, so at an adjustment coefficient it equals one plus the safety loading.

For unit-rate Erlang distribution of shape two claims put $q=\lambda /c<1/2$. The adjustment coefficient satisfies $q(2-R)=(1-R{)}^2$, so $1-R=(q+\sqrt{q^2+4q})/2$. In the interior-adjustment prefactor $(1-2q)/(2q(1-R{)}^{-3}-1)$, substituting $q=(1-R{)}^2/(2-R)$ gives the displayed expression. It is positive for $0<R<1$ and tends to one as the loading tends to zero.

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.

The ultimate ruin probability satisfies a tail-forced defective renewal equation. Multiplying it by $e^{Ru}$ at an adjustment coefficient turns its kernel into the probability density $k_R$. Normalization is exactly $\lambda (M(R)-1)=cR$. The density is absolutely continuous even when the original claim law has atoms; thus it has a nonarithmetic distribution.

When the adjustment coefficient lies inside the finite domain of the claim moment-generating function, the tilted kernel has finite mean $(\lambda M^{\prime }(R)-c)/(cR)$. The forcing integral is $(c-\lambda \mu )/(cR)$. Their ratio is the constant $C$ in $\psi (u)\sim C{e}^{-Ru}$, by the key renewal theorem. An extra exponential moment controls the forcing tails and proves direct Riemann integrability.

If the Cramér–Lundberg ruin asymptotic has a finite positive constant, compare it with the leading nonzero term of a finite exponential representation. An adjustment coefficient smaller than that term's decay rate would give limit zero; a larger coefficient would give infinity. Thus the two rates agree and the prefactor is the leading coefficient. Terms with zero coefficient are absent and do not determine a decay rate.

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 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.

For an exponential distribution with mean $\mu$, the claim moment-generating function is $M_I(r)=(1-\mu r{)}^{-1}$ for $r<1/\mu$. The calculated adjustment coefficient must solveSubtract one, divide by the nonzero factor $\mu R_I$, and obtain $(1-\mu R_I{)}^{-1}=1+\rho$. ThusIt belongs to $(0,1/\mu )$ for every $\rho >0$, as required by the transform domain. The Poisson intensity cancels because the premium rate is specified using the same relative loading and mean claim cost.

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.