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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 $c_{\mathrm{prem}}$.

For positive claims with finite nonzero expected value $\mu$, the function $(M_X(r)-1)/r$ is continuous and strictly increasing on the positive finite-transform domain, starting at $\mu$. If $M_X$ diverges at a finite upper endpoint, or is finite for all positive arguments, the secant slope tends to infinity. In the latter case use $M_X(r)\ge \mathbb{P}(X\ge a)e^{ar}$ for some $a>0$ of positive tail probability. Every target $(1+\theta )\mu$ with positive relative safety loading therefore has exactly one positive root.

In the classical risk model with density $3e^{-4x}+e^{-2x}/2$ and relative safety loading $3/5$, the claim mean is $5/16$ and $\lambda /c=2$. The convolution terms $A_a(u)={\int }_0^u\varphi (u-x)e^{-ax}dx$ satisfy $A_a^{\prime }=\varphi -a{A}_a$. Applying $(D+4)(D+2)$ to ${\varphi }^{\prime }=2\varphi -6A_4-A_2$ yields ${\varphi }^{(3)}+4{\varphi }^{\prime \prime }+3{\varphi }^{\prime }=0$. Its solution is fixed by $\varphi (\infty )=1$, $\varphi (0)=3/8$ and ${\varphi }^{\prime }(0)=3/4$, giving the displayed probability. The derivative condition comes from the original integral equation and excludes spurious solutions of the higher-order equation.

For rate-$\alpha$ exponential distribution claims and $c>\lambda /\alpha$, the ruin integro-differential equation reduces to ${\psi }^{\prime \prime }+(\alpha -\lambda /c){\psi }^{\prime }=0$. Differentiate the convolution and use its equation $A^{\prime }=\alpha \psi -\alpha A$ to eliminate it. Positive relative safety loading implies $\psi (u)\to 0$ as $u\to \infty$, and the zero-capital survival probability gives $\psi (0)=\lambda /(\alpha c)$. These two conditions determine the displayed solution. The adjustment coefficient is $\alpha -\lambda /c$.

In a classical risk model with positive relative safety loading, zero initial capital has ultimate survival probability $1-\lambda \mu /c$, where $\lambda$ is the claim-arrival rate, $\mu$ the claim expected value and $c$ the premium rate. It depends on the claim law through its mean alone. It is not the event of never receiving a claim: premium accumulates between arrivals.