= Solution
Consider the secant slope
$$
H(r)=\frac{M(r)-1}{r},\qquad 0<r<r_\infty.
$$
The <moment-generating function> is continuous on the interior of its finite domain and has right derivative $M'(0)=\mu$. Thus $H(r)\to\mu$ as $r\downarrow0$. For every fixed positive claim amount $x$, the function $(e^{rx}-1)/r$ is strictly increasing in $r>0$: its derivative has numerator $e^{rx}(rx-1)+1>0$, since this numerator starts at zero and has derivative $rx e^{rx}$ as a function of $rx$. Taking <expected values> preserves the strict inequality. Therefore $H$ is continuous and strictly increasing. Equivalently, the <strictly convex> transform $M$ has strictly increasing secant slopes from the origin.
If $r_\infty<\infty$, the assumed blow-up of $M$ gives $H(r)\to\infty$. If $r_\infty=\infty$, choose $a>0$ with $d=\mathbb P(X_1\ge a)>0$. Such an $a$ exists because the claims are positive. Then $M(r)\ge de^{ar}$, so again $H(r)\to\infty$. This exponential lower bound is needed at an infinite endpoint: mere divergence of $M$ would not, by itself, establish divergence of $M(r)/r$.
The target $(1+\theta)\mu$ strictly exceeds the limiting slope $\mu$. The <intermediate value theorem> and strict monotonicity therefore give
$$
\boxed{\exists!\ R\in(0,r_\infty):\ H(R)=(1+\theta)\mu.}
$$
Multiplying by $R$ gives the defining <adjustment coefficient> equation. The zero root of the undivided equation is excluded. This is the <secant-slope existence criterion for an adjustment coefficient>.
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