 Ellipsoidal admissibility test for overlap data

An ellipsoidal admissibility test for overlap data checks whether every point of a specified joint calibration ellipsoid permits finite Gaussian radius inversion. For the inversion of a symmetric Gaussian overlap log-ratio, write $x=(D,h{)}^{\mathsf{T}}$ with $D$ the measured logarithmic ratio and $h$ the positive radius-setting parameter. The admissible set is $D-4h<0$ and $-D-4h<0$. Together these strict inequalities already imply $h>0$.

Suppose the joint set isFor nonsingular $\Sigma$ this is equivalently $(x-\mu {)}^{\mathsf{T}}{\Sigma }^{-1}(x-\mu )\le k^2$. The parameterized form also accommodates a singular covariance matrix or a lower-dimensional deterministic set. Its support function in direction $l$ isIndeed, $l^{\mathsf{T}}Ay=(A^{\mathsf{T}}l{)}^{\mathsf{T}}y$, so the Cauchy-Schwarz inequality gives the bound, attained by a parallel $y$ when $A^{\mathsf{T}}l\ne 0$.

Let $\mu =(D_0,h_0{)}^{\mathsf{T}}$, marginal scales $s_D,s_h$, and off-diagonal entry $c_{Dh}$. Substituting $l_{+}=(1,-4{)}^{\mathsf{T}}$ and $l_{-}=(-1,-4{)}^{\mathsf{T}}$ gives two exact tests:Both must hold. Equality means that the closed joint set touches a boundary where no finite positive radius ratio exists; testing only the center would miss that point.

For an original example, take $D_0=2.4$, $h_0=0.7$, $s_D=0.1$, $s_h=0.03$ and $k=2$. Zero covariance gives upper-boundary excursion $2\sqrt{0.0244}\approx 0.3124$, smaller than the central margin $0.4$. The lower-boundary test also passes. With correlation coefficient $-0.8$, however, $c_{Dh}=-0.0024$ and the upper excursion becomes $2\sqrt{0.0436}\approx 0.4176$: the joint set crosses the forbidden region despite identical marginal scales.

The offending point need not be found by grid search. When $l^{\mathsf{T}}\Sigma l>0$, the maximizing point isFor the correlated example and $l=l_{+}$, this gives approximately $(D,h)=(2.5877,0.6425)$, with $D-4h\approx 0.0176>0$. It is a concrete incompatible calibration pair.

The ellipsoid here is a stated joint set. Interpreting it as a confidence region requires a separate sampling or uncertainty model; choosing $k=2$ alone supplies no universal coverage probability. JCGM 101 distinguishes assignment of joint input distributions from propagation and summaries.^[ref] If the set fails, report the incompatible region and reconsider the input model or measurement. Clipping it before inversion changes the set being propagated.