= Solution
Use $c_{\rm prem}$ for the premium income rate, reserving $c$ later for the smaller exponential decay rate. In the <classical risk model>, the surplus is
$$
U_t=u+c_{\rm prem}t-\sum_{j=1}^{N_t}X_j,
\qquad c_{\rm prem}=(1+\rho)\lambda\mu.
$$
Here $\rho$ is the <relative safety loading>. The aggregate claims form a <Compound Poisson process>. Define $L_t=\sum_{j=1}^{N_t}X_j-c_{\rm prem}t$, so ruin occurs when $L_t>u$. By <independent increments> and the <exponential formula for a marked Poisson sum>,
$$
\mathbb E[e^{R(L_t-L_s)}]
=\exp\{(t-s)[\lambda(M(R)-1)-c_{\rm prem}R]\}=1.
$$
Thus $Z_t=e^{RL_t}$ 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 $\{\tau\leq T\}$, $L_\tau>u$, so
$$
1=\mathbb E[Z_{\tau\wedge T}]
\geq e^{Ru}\mathbb P(\tau\leq T).
$$
Letting $T$ increase proves \b[the Lundberg inequality and the unscaled limit]:
$$
\boxed{\psi(u)\leq e^{-Ru},\qquad
\lim_{u\to\infty}\psi(u)=0.}
$$
For the precise asymptotic, put
$$
h=\frac{\lambda\mu}{c_{\rm prem}}=\frac1{1+\rho},\quad
z(u)=e^{Ru}\psi(u),\quad
k(x)=h e^{Rx}f_I(x),\quad
g(u)=h e^{Ru}\int_u^\infty f_I(x)\,dx.
$$
The given exponential integral identity makes $k$ a probability density. Multiplying the given <defective renewal equation> by $e^{Ru}$ turns it into the ordinary <renewal equation>
$$
z(u)=g(u)+\int_0^u z(u-x)k(x)\,dx.
$$
For 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> gives
$$
\int_0^\infty g(u)\,du
=h\int_0^\infty f_I(x)\frac{e^{Rx}-1}{R}\,dx
=\frac{1-h}{R}.
$$
Furthermore
$$
g(u)=\int_u^\infty e^{-R(x-u)}k(x)\,dx,\qquad
g'(u)=Rg(u)-k(u)\quad\hbox{almost everywhere}.
$$
Thus $g$ is continuous and integrable, and $\int_0^\infty|g'(u)|\,du\leq R\int g+1<\infty$. On a mesh of width $\delta$, the difference between its upper and lower sums is at most $\delta\int|g'|$; its upper sum is at most $\int g+\delta\int|g'|$. This proves <direct Riemann integrability> rather than assuming it. Also $0\leq z(u)\leq1$ by the <Lundberg inequality>, so the solution is locally bounded. Its <renewal representation> is $z=g*U_k$, where $U_k=\sum_{j\geq0}K^{*j}$ 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 $J=\int_0^\infty xe^{Rx}f_I(x)\,dx$, the tilted interarrival <expected value> is $m_k=hJ$. The <key renewal theorem> gives \b[the <Cramér–Lundberg ruin asymptotic>]
$$
\boxed{\lim_{u\to\infty}e^{Ru}\psi(u)
=\frac{1-h}{RhJ}
=\frac{\rho}{R\displaystyle\int_0^\infty xe^{Rx}f_I(x)\,dx}=A.}
$$
If $J=\infty$, 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:
$$
h=\psi(0)=a+b,\qquad
\boxed{\rho=\frac{1-a-b}{a+b}}.
$$
One can identify the <adjustment coefficient> without silently assuming $A>0$. For $0<r<c$, set
$$
P(r)=\int_0^\infty e^{ru}\psi(u)\,du
=\frac{a}{c-r}+\frac{b}{d-r}.
$$
It 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 gives
$$
P(r)=\frac h r[F_I(r)-1]+hP(r)F_I(r),
\qquad
hF_I(r)=\frac{h+rP(r)}{1+rP(r)}.
$$
As $r\uparrow c$, $P(r)\to\infty$ because $a>0$. By <monotone convergence theorem>, $hF_I(c)=1$. 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 \b[the remaining constants]
$$
\boxed{R=c,\qquad A=a,\qquad \rho=\frac{1-a-b}{a+b}.}
$$
In 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 $c_{\rm prem}$.
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