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Normalizing the true density givesa gamma distribution of shape $1/2$ and rate $\beta$. Its mean is $1/(2\beta )=\mu$, and the Gamma integral givesPut $x=R/\beta$. Since $\mu \beta =1/2$, the positive adjustment coefficient satisfiesBoth sides are positive in this domain. Squaring, multiplying by $1-x$, and dividing out the zero root $x=0$ yieldsThe discriminant is $(1+\rho {)}^2[(\rho -3{)}^2+16\rho ]=(1+\rho {)}^2(\rho +9)(\rho +1)$. The negative constant term makes one root positive and the other negative. The polynomial is negative at zero and has value four at one, so the positive root lies in $(0,1)$ and is not an extraneous solution introduced by squaring. Therefore the adjustment coefficient for shape-one-half gamma claims is

To compare it with the exponential calculation, write $R_I=2\beta \rho /(1+\rho )$. Their difference issince $9(1+\rho {)}^2-(\rho +9)(\rho +1)=8\rho (1+\rho )>0$. Thus the two proposed Lundberg inequality bounds are ordered asOnly ${\psi }_{\mathrm{true}}(u)\le e^{-Ru}$ is justified for the true claim law. The exponential calculation gives ${\psi }_{\mathrm{exponential}}(u)\le e^{-R_{I}u}$ for a different model, not a valid guaranteed bound on the true ruin probability.

This exponential misspecification of shape-one-half gamma claims is optimistic: the true variance is $1/(2{\beta }^2)=2{\mu }^2$, whereas the assumed exponential variance is ${\mu }^2$. Its tail also decays at rate $\beta$, more slowly than the assumed rate $1/\mu =2\beta$. Equivalently, for $0<r<\beta$, the true transform $(1-2\mu r{)}^{-1/2}$ is larger than $(1-\mu r{)}^{-1}$, since $(1-\mu r{)}^2>1-2\mu r$. This stronger transform growth makes the true adjustment root smaller. Comparing upper bounds alone does not by itself prove an ordering of the two actual ruin probabilities. At $u=0$ both displayed bounds equal one.