= Robust linear constraints for bounded covariance models
<Robust linear constraints for bounded covariance models> require a linear inequality to hold for every error state admitted by the selected bounded model. A <covariance matrix> fixes second moments; the robust guarantee additionally depends on the <support of a probability distribution> or other specified uncertainty set. This distinction matters when a physical limit must never be crossed within the model.
Let $E=LS$ as in <finite-support realization of a covariance matrix>, where all coordinates of $S$ are independent fair signs. Then $\operatorname{Cov}(E)=LL^{\mathsf T}$. For nominal input $\mu$, require $a^{\mathsf T}(\mu+E)\ge b$ for every sign state. Writing $c=L^{\mathsf T}a$, direct minimization gives
$$
\min_{s_j\in\{-1,1\}}a^{\mathsf T}(\mu+Ls)
=a^{\mathsf T}\mu-\sum_j|c_j|.
$$
Each sign can independently oppose its coefficient, including when some coefficients vanish. Thus the exact necessary and sufficient condition is $a^{\mathsf T}\mu-\lVert L^{\mathsf T}a\rVert_1\ge b$, using the <L1 norm>. The same extremum holds over the full box $s_j\in[-1,1]$ because a linear function attains an extreme at a vertex. This is a <support function> calculation in <robust optimization>; the general worst-case formulation and its dependence on the uncertainty set are illustrated in the primary optimization text.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[§4.4.2, printed p. 157, robust linear programming, equation (4.37); the text's illustrated uncertainty set is an ellipsoid]{ref}
For an original two-input example in volts, let $E_1=(1\,\mathrm V)Z$ and $E_2=(1\,\mathrm V)(0.6Z+0.8W)$, with independent fair signs $Z,W$. The <covariance matrix> is $\left(\begin{smallmatrix}1&0.6\\0.6&1\end{smallmatrix}\right)\,\mathrm V^2$. To keep the actual first input at least as large as the second, let the nominal difference be $d=\mu_1-\mu_2$. Then
$$
(\mu_1+E_1)-(\mu_2+E_2)=d+(0.4Z-0.8W)\,\mathrm V,
\qquad\min=d-1.2\,\mathrm V.
$$
A nominal difference of $1.0\,\mathrm V$ fails in the state $Z=-1,W=1$, despite being positive nominally. A difference of $1.3\,\mathrm V$ leaves a guaranteed $0.1\,\mathrm V$ margin under this model. Using only the separate bounds $|E_1|\le1\,\mathrm V$, $|E_2|\le1.4\,\mathrm V$ would demand $d\ge2.4\,\mathrm V$, because that larger box admits joint states the sign model forbids.
The <standard deviation> of $E_1-E_2$ is instead $\sqrt{0.8}\,\mathrm V$. It is not the hard bound $1.2\,\mathrm V$. A different bounded law with the same <covariance matrix> can change the hard bound, as <nonuniqueness of bounded uncertainty laws with a fixed covariance> explains. For several linear inequalities, apply the support calculation to each row. This verifies an assumed model's feasible states; it does not establish that the chosen support contains every error of an actual instrument.
Back to article page
