Theorem: Nuisance cancellation and robust margin

Under (eq:Y),

If $|r_{h,i}|\le {ϵ}_{h,i}$, put $E_h={\sum }_i|w_i|{ϵ}_{h,i}$. The contrast intervals are

For a fixed alternative they are disjoint exactly when

with remaining gap

For an alternative family, a common oriented certificate instead requires one sign $\sigma \in \{-1,1\}$ and

An infimum of absolute values does not establish a common oriented threshold when the sign varies.

Proof. Multiply (eq:Y) by the weights and sum. The two nuisance terms vanish. The residual magnitude is bounded by the weighted sum of its component bounds. Two centred closed intervals with radii $E_0,E_1$ and centre distance $|\lambda t{\mathcal{B}}_{r_{\mathrm{c}}}|$ are disjoint precisely under the displayed strict inequality. The family statement applies the same orientation to every alternative.