 Bounded errors with prescribed correlation

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.