Adjustment coefficient for shape-one-half gamma claims ($R=\beta \frac{\rho -3+\sqrt{(\rho +9)(\rho +1)}}{2(1+\rho )},\beta =1/(2\mu )$)

The gamma distribution of shape one half and rate $\beta$ has moment-generating function $(1-r/\beta {)}^{-1/2}$. With mean $\mu =1/(2\beta )$ and positive relative safety loading $\rho$, write $x=R/\beta$. Dividing out the zero root after squaring the adjustment equation gives $(1+\rho {)}^2{x}^2-(1+\rho )(\rho -3)x-4\rho =0$. Its unique positive root lies between zero and one, so it belongs to the transform domain and gives the display.

Replacing a shape-one-half gamma claim law by an exponential distribution with the same mean overestimates the adjustment coefficient. Indeed $R_I-R=\beta [3(1+\rho )-\sqrt{(\rho +9)(\rho +1)}]/[2(1+\rho )]>0$ because the difference of the squared terms is $8\rho (1+\rho )$. The resulting putative Lundberg inequality bound $e^{-R_{I}u}$ is smaller than the valid true-model bound $e^{-Ru}$ for $u>0$; the former is not justified for the true claim law. The true variance is twice the assumed variance, and its exponential tail decays more slowly.