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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.

A continuous martingale is a continuous-time martingale whose paths are continuous outside one null event. Continuity does not remove the integrability or conditional-expectation requirements in the martingale definition.

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.

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.

As in the intended independent-mark construction, take the entire sequence $(X_j)$ independent of the Poisson process, not merely independent of one fixed count. Then $Y$ is a Compound Poisson process. For deterministic $s<t$, its increment $Y_t-Y_s$ is independent of ${\mathcal{F}}_s=\sigma (Y_u:0\le u\le s)$. To justify this carefully, first condition on $N_s=k$. The past of $Y$ uses only the counting history up to $s$ and marks among $X_1,\dots ,X_k$. The future count has independent rate-$\lambda$ Poisson process increments, and future marks $X_{k+1},X_{k+2},\dots$ are fresh with the common law. Their joint conditional law is the same for every $k$ and every past history, so the increment is independent even of this larger marked history, hence of ${\mathcal{F}}_s$. Marks equal to zero do not cause a problem: ${\mathcal{F}}_s$ may then reveal less than the full counting history.

By part (b), with Poisson parameter $\lambda (t-s)$,The proposed process is adapted to ${\mathcal{F}}_t$, nonnegative, andIn particular, no finite expectation of a mark is required. Also $Z_s>0$, andis independent of ${\mathcal{F}}_s$ and has mean one. Therefore the exponential Laplace martingale of a compound Poisson process satisfiesThis proves every condition for a continuous-time martingale in the requested filtration. When $q=0$, $\varphi (0)=1$ and the same conclusion reduces to the constant martingale $Z_t=1$.

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 $c_{\mathrm{prem}}$.

Let $(M_t{)}_{t\ge 0}$ be a right-continuous continuous-time martingale with ${\sup }_t\mathbb{E}|M_t|<\infty$. Restricting it to the nonnegative rational times gives a countable martingale. The argument of part (b), applied on successively finer rational grids, shows that it has a finite almost-sure limit as the rational time tends to infinity. Right-continuity and the upcrossing characterization prevent the values at arbitrary times from having a different limit. Thus $M_t$ converges almost surely to a finite integrable random variable as $t\to \infty$, which is the Continuous-time martingale convergence theorem.