= 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> $1/2$. Set
$$
X=u_XZ,\qquad Y=u_Y\left(\rho Z+\sqrt{1-\rho^2}\,W\right).
$$
Both <expected values> vanish. Since $\mathbb E[Z^2]=\mathbb E[W^2]=1$ and $\mathbb E[ZW]=0$, direct expansion gives
$$
\operatorname{Var}(X)=u_X^2,\qquad\operatorname{Var}(Y)=u_Y^2,\qquad\operatorname{Cov}(X,Y)=\rho u_Xu_Y.
$$
There are at most four distinct states, each obtained from an equally likely sign pair. Also $|X|\le u_X$ and $|Y|\le u_Y(|\rho|+\sqrt{1-\rho^2})\le\sqrt2u_Y$. 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 $\Sigma=Q_r\Lambda_rQ_r^{\mathsf T}$ 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 satisfies
$$
\mathbb E[E]=0,\qquad\operatorname{Cov}(E)=L\mathbb E[SS^{\mathsf T}]L^{\mathsf T}=LL^{\mathsf T}=\Sigma.
$$
The <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 condition
$$
a^{\mathsf T}\mu-\lVert L^{\mathsf T}a\rVert_1\ge b,
$$
where $\lVert\cdot\rVert_1$ 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 $\theta+\delta\theta\ge0.4866411\,\mathrm V>0$, while the minimum amplitude-minus-threshold is
$$
0.5-|0.02-0.009|-0.01\sqrt{0.19}=0.4846411\ldots\,\mathrm V>0.
$$
All 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. Let
$$
L=\begin{pmatrix}1&0\\0.8&0.6\end{pmatrix},\qquad O=\frac1{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}.
$$
For independent fair signs $S$, both $LS$ and $LOS$ have <covariance matrix> $\left(\begin{smallmatrix}1&0.8\\0.8&1\end{smallmatrix}\right)$ because $O$ is an <orthogonal matrix>. Yet the first coordinate of $LS$ is always $\pm1$, whereas that of $LOS$ is $0$ with <probability> $1/2$ and each of $\pm\sqrt2$ with <probability> $1/4$. Hence $\mathbb P(|X|>1.2)$ is respectively \b[zero] and \b[one half], despite identical <expected values> and <covariance matrices>. A nonlinear risk calculation needs more than those moments.
\Image[/bounded-correlated-errors.png]
{title=Two bounded four-state error models with identical covariance and different tail probabilities. All points are constructed, not measured data.}
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