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A covariance matrix describes second moments, but does not specify a complete probability distribution. A bounded model can realize a prescribed covariance matrix exactly while keeping every possible error finite. This is useful for checking a nonlinear measurement model without accidentally generating physically impossible inputs; it is a constructed model, not evidence that an actual instrument has that distribution.

For two errors with desired standard deviations $u_X,u_Y$ and correlation coefficient $\rho \in [-1,1]$, let $Z,W$ be independent random variables with the Rademacher distribution: each takes $-1$ and $+1$ with probability $1/2$. SetBoth expected values vanish. Since $\mathbb{E}[Z^2]=\mathbb{E}[W^2]=1$ and $\mathbb{E}[ZW]=0$, direct expansion givesThere are at most four distinct states, each obtained from an equally likely sign pair. Also $|X|\le u_X$ and $|Y|\le u_Y(|\rho |+\sqrt{1-{\rho }^2})\le \sqrt{2}{u}_Y$. At $|\rho |=1$ the second sign has no effect, leaving only two distinct states. If either standard deviation is zero, the corresponding error is zero and its correlation coefficient is undefined; the covariance construction still makes sense.

More generally, a finite-support realization of a covariance matrix exists for every real symmetric positive semidefinite matrix $\Sigma$. By the spectral theorem, write $\Sigma =Q_r{\Lambda }_r{Q}_r^{\mathsf{T}}$ using its $r$ positive eigenvalues and corresponding orthonormal eigenvectors, and put $L=Q_r{\Lambda }_r^{1/2}$. Let $S$ have $r$ independent Rademacher distribution coordinates. The random vector $E=LS$ then satisfiesThe identity matrix appears because the signs have unit variance and zero pairwise covariance. There are at most $2^r$ distinct states, with component bounds $|E_i|\le {\sum }_j|L_{ij}|\le \sqrt{r}\sqrt{{\Sigma }_{ii}}$ by the Cauchy-Schwarz inequality. When $r=0$, take $E=0$. Thus singular covariance matrices are allowed; a nonsingular Cholesky decomposition is not required.

This full sign model also makes robust linear constraints for bounded covariance models exact. To require $a^{\mathsf{T}}(\mu +LS)\ge b$ for every sign state, choose each sign to minimize its contribution. The result is the necessary and sufficient conditionwhere $\Vert \cdot {\Vert }_1$ is the L1 norm. This is a small robust optimization calculation. A physical domain must be checked in the chosen units; clipping inadmissible samples afterwards generally changes both the probability distribution and its covariance matrix.

For example, in correlated threshold and amplitude errors in RC timing, take nominal amplitude $H=1\mathrm{V}$, threshold $\theta =0.5\mathrm{V}$, $u_H=0.02\mathrm{V}$, $u_{\theta }=0.01\mathrm{V}$ and $\rho =0.9$. Use $\delta H=0.02Z$ and $\delta \theta =0.01(0.9Z+\sqrt{0.19}W)$. Every state has $\theta +\delta \theta \ge 0.4866411\mathrm{V}>0$, while the minimum amplitude-minus-threshold isAll four states therefore remain within the RC threshold-crossing time domain. Enumerating them supplies an exact check for this particular bounded error model; it does not establish the instrument's actual tail probabilities.

The nonuniqueness of bounded uncertainty laws with a fixed covariance is visible even in two dimensions. LetFor independent fair signs $S$, both $LS$ and $LOS$ have covariance matrix $\left(\begin{array}{cc}1& 0.8\\ 0.8& 1\end{array}\right)$ because $O$ is an orthogonal matrix. Yet the first coordinate of $LS$ is always $\pm 1$, whereas that of $LOS$ is $0$ with probability $1/2$ and each of $\pm \sqrt{2}$ with probability $1/4$. Hence $\mathbb{P}(|X|>1.2)$ is respectively zero and one half, despite identical expected values and covariance matrices. A nonlinear risk calculation needs more than those moments.

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 $\mathrm{Cov}(E)=L{L}^{\mathsf{T}}$. For nominal input $\mu$, require $a^{\mathsf{T}}(\mu +E)\ge b$ for every sign state. Writing $c=L^{\mathsf{T}}a$, direct minimization givesEach sign can independently oppose its coefficient, including when some coefficients vanish. Thus the exact necessary and sufficient condition is $a^{\mathsf{T}}\mu -\Vert L^{\mathsf{T}}a{\Vert }_1\ge b$, 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.^[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 $\left(\begin{array}{cc}1& 0.6\\ 0.6& 1\end{array}\right){\mathrm{V}}^2$. To keep the actual first input at least as large as the second, let the nominal difference be $d={\mu }_1-{\mu }_2$. ThenA 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 $|E_1|\le 1\mathrm{V}$, $|E_2|\le 1.4\mathrm{V}$ would demand $d\ge 2.4\mathrm{V}$, 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.