Cramér–Lundberg ruin asymptotic ($e^{Ru}\psi (u)\to \rho /(R\int x{e}^{Rx}{f}_I(x)dx)$)

In the classical risk model with positive relative safety loading $\rho$ and adjustment coefficient $R$, tilting the ruin defective renewal equation gives a proper renewal equation. The key renewal theorem yields $e^{Ru}\psi (u)\to \rho /[R{\int }_0^{\infty }x{e}^{Rx}{f}_I(x)dx]$. The constant is positive if the denominator is finite and zero if it is infinite; the claim-size density provides the nonarithmetic hypothesis.

If the Cramér–Lundberg ruin asymptotic has a finite positive constant, compare it with the leading nonzero term of a finite exponential representation. An adjustment coefficient smaller than that term's decay rate would give limit zero; a larger coefficient would give infinity. Thus the two rates agree and the prefactor is the leading coefficient. Terms with zero coefficient are absent and do not determine a decay rate.

When the adjustment coefficient lies inside the finite domain of the claim moment-generating function, the tilted kernel has finite mean $(\lambda M^{\prime }(R)-c)/(cR)$. The forcing integral is $(c-\lambda \mu )/(cR)$. Their ratio is the constant $C$ in $\psi (u)\sim C{e}^{-Ru}$, by the key renewal theorem. An extra exponential moment controls the forcing tails and proves direct Riemann integrability.

For unit-rate Erlang distribution of shape two claims put $q=\lambda /c<1/2$. The adjustment coefficient satisfies $q(2-R)=(1-R{)}^2$, so $1-R=(q+\sqrt{q^2+4q})/2$. In the interior-adjustment prefactor $(1-2q)/(2q(1-R{)}^{-3}-1)$, substituting $q=(1-R{)}^2/(2-R)$ gives the displayed expression. It is positive for $0<R<1$ and tends to one as the loading tends to zero.