= Solution
In the <classical risk model>, the insurer's surplus is
$$
U(t)=u+ct-C(t),\qquad C(t)=\sum_{j=1}^{N(t)}X_j,
$$
where $u\ge0$ 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> $\theta$ means
$$
\boxed{c=(1+\theta)\lambda\mu,\qquad\theta>0,}
$$
so premium income exceeds the expected claim cost per unit time. The ruin <stopping time> is $\tau=\inf\{t\ge0:U(t)<0\}$, 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> is
$$
\boxed{\psi(u)\le e^{-Ru},\qquad u\ge0.}
$$
Here is a proof, including the stopping argument. By conditioning on the <Poisson distribution> of $N(t)$, the <random-sum transform identity> gives
$$
\mathbb E e^{rC(t)}=\exp\big(\lambda t(M(r)-1)\big).
$$
The <adjustment coefficient> equation is $\lambda(M(R)-1)=cR$. Consequently the <exponential surplus martingale>
$$
Z_t=\exp\big(R(C(t)-ct)\big),\qquad Z_0=1,
$$
has <expected value> one at every finite time. More strongly, the claim process has <independent increments>, so for $s\le t$,
$$
\begin{aligned}
\mathbb E[Z_t\mid\mathcal F_s]
&=Z_s\,\mathbb E\exp\big(R(C(t)-C(s)-c(t-s))\big)\\
&=Z_s\exp\big((t-s)(\lambda(M(R)-1)-cR)\big)=Z_s.
\end{aligned}
$$
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,
$$
1=\mathbb E Z_{\tau\wedge T}
\ge \mathbb E\big[Z_\tau\mathbf1_{\{\tau\le T\}}\big]
\ge e^{Ru}\mathbb P(\tau\le T).
$$
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 gives
$$
\frac{\mu r}{1-\mu r}=(1+\theta)\mu r.
$$
Discard the zero root and divide by $\mu r$. The unique positive solution is
$$
\boxed{R=\frac{\theta}{(1+\theta)\mu}},
$$
which 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 $h(0)=0$, $h'(0)=-\theta\mu<0$, $h''(r)=\mu^2e^{\mu r}>0$, 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 $e^x<1/(1-x)$. Hence
$$
h(R)=e^{\mu R}-1-(1+\theta)\mu R
<\frac1{1-\mu R}-1-(1+\theta)\mu R=0.
$$
It follows that
$$
\boxed{R<R_\mu,\qquad e^{-R_\mu u}<e^{-Ru}\quad(u>0).}
$$
Thus \b[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.
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