= Solution
For an <exponential distribution> with mean $\mu$, the claim <moment-generating function> is $M_I(r)=(1-\mu r)^{-1}$ for $r<1/\mu$. The calculated <adjustment coefficient> must solve
$$
\frac1{1-\mu R_I}=1+(1+\rho)\mu R_I.
$$
Subtract one, divide by the nonzero factor $\mu R_I$, and obtain $(1-\mu R_I)^{-1}=1+\rho$. Thus
$$
\boxed{R_I=\frac{\rho}{(1+\rho)\mu}.}
$$
It belongs to $(0,1/\mu)$ for every $\rho>0$, as required by the transform domain. The Poisson intensity cancels because the premium rate is specified using the same relative loading and mean claim cost.
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