= Solution
For nontrivial positive <claim sizes> in the <classical risk model> and $\lambda>0$, the <adjustment coefficient> is the unique positive solution
$$
\boxed{\lambda(M_X(R)-1)=cR,\qquad 0<R<r_\infty.}
$$
Indeed, $h(r)=\lambda(M_X(r)-1)-cr$ has $h(0)=0$, $h'(0)=\lambda\mu-c<0$, and is <strictly convex>. It becomes positive near the right endpoint of the <moment-generating function> domain. For a finite endpoint this follows from the assumed divergence; for an infinite endpoint, $\mathbb P(X\ge\varepsilon)>0$ for some $\varepsilon>0$ gives an exponential lower bound on $M_X(r)$, which eventually exceeds any linear function. <Convexity> gives existence and uniqueness of the positive crossing and $\lambda M_X'(R)-c>0$.
Put $q=\lambda\mu/c$ and $\overline F=1-F$. Substituting $\varphi=1-\psi$ into the supplied <survival renewal equation for a classical risk model>, and using $\int_0^\infty\overline F(x)dx=\mu$, yields
$$
\psi(u)=\frac{\lambda}{c}\int_u^\infty\overline F(x)\,dx
+\frac{\lambda}{c}\int_0^u\psi(u-x)\overline F(x)\,dx.
$$
Multiplying by $e^{Ru}$ gives the proper renewal-type equation
$$
\boxed{Z(u)=g_R(u)+\int_0^u Z(u-x)k_R(x)\,dx,}
\qquad
k_R(x)=\frac{\lambda}{c}e^{Rx}\overline F(x),\quad
g_R(u)=\frac{\lambda}{c}e^{Ru}\int_u^\infty\overline F(x)\,dx.
$$
The <exponential tilt of the ruin renewal kernel> turns a defective kernel into a <probability density function>, because the tail integration identity gives
$$
\int_0^\infty k_R(x)dx
=\frac{\lambda}{c}\frac{M_X(R)-1}{R}=1.
$$
This identity follows also by integrating $e^{RX}-1=R\int_0^X e^{Rx}dx$ and applying <Tonelli theorem>. The tilted <probability density function> defines a <nonarithmetic distribution>, even if the <claim size> <probability distribution> itself has atoms.
Differentiate $(M_X(R)-1)/R$ to calculate its <mean>:
$$
m_R=\int_0^\infty xk_R(x)dx
=\frac{\lambda}{c}\frac{RM_X'(R)-M_X(R)+1}{R^2}
=\frac{\lambda M_X'(R)-c}{cR}.
$$
It is finite and positive because $R$ is inside the finite <moment-generating function> domain. Next, reversing the order of integration gives
$$
\begin{aligned}
\int_0^\infty g_R(u)du
&=\frac{\lambda}{c}\int_0^\infty\overline F(x)\int_0^x e^{Ru}du\,dx\\
&=\frac{\lambda}{cR}\left(\frac{M_X(R)-1}{R}-\mu\right)
=\frac{c-\lambda\mu}{cR}.
\end{aligned}
$$
The forcing function is <directly Riemann integrable>. Choose $\varepsilon>0$ with $M_X(R+\varepsilon)<\infty$. <Markov's inequality> bounds $\overline F(x)$ by $M_X(R+\varepsilon)e^{-(R+\varepsilon)x}$, so $g_R(u)$ is bounded by a constant times $e^{-\varepsilon u}$. It is continuous; on finite intervals continuity controls its <Riemann sums>, and the exponential envelope controls the tails. These are sufficient conditions for <direct Riemann integrability>.
Iterating the <renewal equation> by <convolution> gives $Z=g_R*U_R$, where $U_R=\sum_{n\ge0}k_R^{*n}$ is the <renewal measure>. The remainder after $n$ iterations is bounded on each fixed interval by a constant times the chance that a sum of $n$ positive interarrivals remains in that interval, which tends to zero. The <key renewal theorem>, applicable to this <nonarithmetic distribution> of finite <mean> and the directly integrable forcing, now gives the <interior adjustment coefficient ruin prefactor>:
$$
\boxed{A=\lim_{u\to\infty}e^{Ru}\psi(u)
=\frac{\int_0^\infty g_R(u)du}{m_R}
=\frac{c-\lambda\mu}{\lambda M_X'(R)-c}.}
$$
Both numerator and denominator are positive, so $0<A<\infty$.
For the unit-rate shape-two <Erlang distribution>, $\mu=2$, $M_X(r)=(1-r)^{-2}$ and $M_X'(r)=2(1-r)^{-3}$. Write $\delta=\lambda/c$, so positive loading requires $0<\delta<1/2$. Dividing the <adjustment coefficient> equation by its nonzero root gives $\delta(2-R)=(1-R)^2$. Thus
$$
\boxed{R=1-\frac{\delta+\sqrt{\delta^2+4\delta}}2,\qquad
A=\frac{1-2\delta}{2\delta(1-R)^{-3}-1}
=\frac{(1-R)(3-2R)}{3-R}.}
$$
The last simplification uses $\delta=(1-R)^2/(2-R)$; it is the <Erlang shape-two ruin prefactor>. Its value depends on the premium-to-arrival ratio, which is not specified numerically.
For the final exponential representation, evaluating at zero in the <survival renewal equation for a classical risk model> gives \b[$\lambda\mu/c=\psi(0)=a+b$]. If the slower term is present, necessarily $a>0$ and $e^u\psi(u)\to a$. Comparing with the finite positive asymptotic constant just proved yields
$$
\boxed{R=1,\qquad A=a,\qquad\lambda\mu/c=a+b\quad(a>0).}
$$
A smaller <adjustment coefficient> would make the limit zero and a larger one would make it infinite. Here the <leading exponential term determines a ruin adjustment coefficient>. The coefficient $b$ need not be assumed positive to identify the leading term.
The wording does not specify that $a$ is nonzero. If $a=0$ and $b>0$, the only surviving term instead gives \b[$R=6$, $A=b$ and $\lambda\mu/c=b$]. This case is realizable: claims with an <exponential distribution> of rate $\beta=6+\lambda/c$ have <ultimate ruin probability> $(\lambda/(c\beta))e^{-6u}$. Negative $a$ would make the <ultimate ruin probability> negative for large $u$, and both coefficients zero are incompatible with a nontrivial positive claim rate and <mean>. These observations specify the coefficient qualification needed for the usual $R=1$ answer.
Back to article page
