= Exact Gaussian overflow constraint
{title2=$m+\Phi^{-1}(1-e^{-\gamma})\sqrt v\leq C$}
For a <normal distribution> with mean $m$, positive <variance> $v$, and $\gamma>0$, the target $\mathbb P(X\geq C)\leq e^{-\gamma}$ is equivalent to the displayed constraint, where $\Phi$ is the <standard normal distribution function>. This follows from $\mathbb P(X\geq C)=1-\Phi((C-m)/\sqrt v)$ and strict monotonicity of $\Phi$. Unlike the sufficient <Chernoff bound> margin $\sqrt{2\gamma v}$, this constraint is necessary as well as sufficient. The exact quantile can be negative when $e^{-\gamma}>1/2$.
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