Exact Gaussian overflow constraint ($m+{\Phi }^{-1}(1-e^{-\gamma })\sqrt{v}\le C$)

For a normal distribution with mean $m$, positive variance $v$, and $\gamma >0$, the target $\mathbb{P}(X\ge C)\le e^{-\gamma }$ is equivalent to the displayed constraint, where $\Phi$ is the standard normal distribution function. This follows from $\mathbb{P}(X\ge C)=1-\Phi ((C-m)/\sqrt{v})$ 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 $e^{-\gamma }>1/2$.

If a random variable has variance zero and mean $m$, it equals $m$ almost surely. For a target probability strictly between zero and one, $\mathbb{P}(X\ge C)$ meets the target exactly when $m<C$. Substituting zero variance into a positive-variance non-strict quantile constraint incorrectly includes $m=C$, where the overflow probability is one. The distinction comes from the atom at the threshold.