= Solution
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> by
$$
\tau_u=\inf\{t\ge0:U(t)<0\},\qquad\psi(u)=\mathbb P(\tau_u<\infty),\quad u\ge0.
$$
The <relative safety loading> is $\rho=c/(\lambda\mu)-1>0$, so $c=(1+\rho)\lambda\mu$. The <adjustment coefficient> is the nonzero positive solution
$$
\boxed{\lambda\{M(R)-1\}=cR.}
$$
Existence 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 $\varphi=1-\psi$ in the <survival renewal equation for a classical risk model> yields
$$
\psi(u)=\frac{\lambda}{c}\left(\mu-\int_0^u\overline F(x)\,dx\right)+\frac{\lambda}{c}\int_0^u\psi(u-x)\overline F(x)\,dx.
$$
By the <tail integral formula for moments>, $\mu=\int_0^\infty\overline F(x)\,dx$. Therefore this is the <defective renewal equation>
$$
\psi(u)=\frac\lambda c\int_u^\infty\overline F(x)\,dx+\frac\lambda c\int_0^u\psi(u-x)\overline F(x)\,dx.
$$
The original kernel has mass $\lambda\mu/c=1/(1+\rho)<1$. For the <exponential tilt of the ruin renewal kernel>, set
$$
k_R(x)=\frac\lambda c e^{Rx}\overline F(x),\qquad
h_R(u)=\frac\lambda c e^{Ru}\int_u^\infty\overline F(x)\,dx.
$$
Multiplying by $e^{Ru}$ gives the required proper <renewal equation>
$$
\boxed{Z(u)=h_R(u)+\int_0^u Z(u-x)k_R(x)\,dx,\qquad Z(u)=e^{Ru}\psi(u).}
$$
To verify that this is a probability renewal kernel, <Tonelli theorem> gives, for $r>0$ in the finite-transform domain,
$$
\int_0^\infty e^{rx}\overline F(x)\,dx=\mathbb E\int_0^X e^{rx}\,dx=\frac{M(r)-1}{r}.
$$
The <adjustment coefficient> equation consequently implies $\int_0^\infty k_R(x)\,dx=1$. Its <expected value> is
$$
m_R=\int_0^\infty xk_R(x)\,dx
=\frac\lambda c\frac{RM'(R)-(M(R)-1)}{R^2}
=\frac{\lambda M'(R)-c}{cR}.
$$
It is finite because $R$ is interior to the finite-transform domain and is positive because the strict convex crossing has derivative $\lambda M'(R)-c>0$.
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 $z=h+z*K$ satisfies $z(u)\to m^{-1}\int_0^\infty h$. It has <renewal representation> $z=h*\sum_{j\ge0}K^{*j}$; 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:
$$
\begin{aligned}
\int_0^\infty h_R(u)\,du
&=\frac\lambda c\int_0^\infty\overline F(x)\int_0^x e^{Ru}\,du\,dx\\
&=\frac\lambda{cR}\left(\frac{M(R)-1}{R}-\mu\right)
=\frac{c-\lambda\mu}{cR}.
\end{aligned}
$$
All hypotheses of the <key renewal theorem> are now checked, so the <interior adjustment coefficient ruin prefactor> is
$$
\boxed{\lim_{u\to\infty}e^{Ru}\psi(u)=\frac{c-\lambda\mu}{\lambda M'(R)-c}
=\frac{\rho\mu}{M'(R)-(1+\rho)\mu}.}
$$
This proves the requested <Cramér–Lundberg ruin asymptotic>, including its constant.
For the two-component <hyperexponential distribution>, conditioning on the chosen exponential component gives
$$
\overline F(x)=\frac12 e^{-x}+\frac12 e^{-x/2},\quad x\ge0,
\qquad
\boxed{\mu=\frac32.}
$$
Its <moment-generating function> and derivative are
$$
M(r)=\frac1{2(1-r)}+\frac1{2(1-2r)},\qquad
M'(r)=\frac1{2(1-r)^2}+\frac1{(1-2r)^2},\quad r<\frac12.
$$
The <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> in
$$
\boxed{0<R<\frac12.}
$$
The adjustment equation, divided by $R$, becomes
$$
\frac1{2(1-R)}+\frac1{1-2R}=\frac32(1+\rho).
$$
Using this identity to subtract $(1+\rho)\mu$ from $M'(R)$ gives
$$
M'(R)-\frac32(1+\rho)=\frac{R}{2(1-R)^2}+\frac{2R}{(1-2R)^2}.
$$
Thus the asymptotic constant in terms of $\rho$ and $R$ is
$$
\boxed{\lim_{u\to\infty}e^{Ru}\psi(u)
=\frac{3\rho}{R\{(1-R)^{-2}+4(1-2R)^{-2}\}}.}
$$
The 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>.
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