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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 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 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.

Consider the secant slopeThe moment-generating function is continuous on the interior of its finite domain and has right derivative $M^{\prime }(0)=\mu$. Thus $H(r)\to \mu$ as $r\downarrow 0$. For every fixed positive claim amount $x$, the function $(e^{rx}-1)/r$ is strictly increasing in $r>0$: its derivative has numerator $e^{rx}(rx-1)+1>0$, since this numerator starts at zero and has derivative $rx{e}^{rx}$ as a function of $rx$. Taking expected values preserves the strict inequality. Therefore $H$ is continuous and strictly increasing. Equivalently, the strictly convex transform $M$ has strictly increasing secant slopes from the origin.

If $r_{\infty }<\infty$, the assumed blow-up of $M$ gives $H(r)\to \infty$. If $r_{\infty }=\infty$, choose $a>0$ with $d=\mathbb{P}(X_1\ge a)>0$. Such an $a$ exists because the claims are positive. Then $M(r)\ge d{e}^{ar}$, so again $H(r)\to \infty$. This exponential lower bound is needed at an infinite endpoint: mere divergence of $M$ would not, by itself, establish divergence of $M(r)/r$.

The target $(1+\theta )\mu$ strictly exceeds the limiting slope $\mu$. The intermediate value theorem and strict monotonicity therefore giveMultiplying by $R$ gives the defining adjustment coefficient equation. The zero root of the undivided equation is excluded. This is the secant-slope existence criterion for an adjustment coefficient.