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Boundary-sensitive Gaussian overlap inversion describes the loss of stable radius information when a symmetric log-overlap ratio approaches its finite-model limit. A central reading can remain formally invertible while small measurement uncertainty reaches values for which no finite radius ratio exists.

For known $h=\log q>0$, define $D=D_h$, $u=\log \rho$ andThe inversion of a symmetric Gaussian overlap log-ratio requires $|D|<4h$. At the upper limit $D\to 4h$, the radius ratio tends to zero; at the lower limit it tends to infinity.

An original expansion near the upper limit puts $ϵ=4h-D>0$. Expanding the hyperbolic tangent givesThus the approximate relative radius uncertainty is $u(\rho )/\rho \simeq u(D)/(2ϵ)$ when linearization is still justified. The distance to the boundary, rather than the magnitude of $D$ alone, controls this sensitivity.

For stipulated $q=2$ and a central log-ratio only 0.01 below $4\log 2$, the asymptotic radius ratio is about 0.0365. A standard uncertainty 0.02 in $D$ is twice the boundary distance, so a small symmetric error bar on the inferred radius is not credible. In a hypothetical Gaussian readout model centered there, roughly 31% of readings fall beyond the finite-inversion limit. This is a statistical illustration, not an observed receiver distribution.

Interval propagation for bounded overlap data shows the asymmetry directly. A stated interval in $D$ that reaches the upper bound maps to radius ratios arbitrarily close to zero; it does not yield a symmetric finite relative interval. An interval wholly inside the domain maps monotonically, with its endpoint order reversed.

Discarding out-of-range Monte Carlo draws would condition on a selected subset and change the distribution being propagated. Instead, examine the forward measurement model, admissible parameter range and calibration uncertainty, or construct an explicitly stated inference using the measured data. JCGM 101 separates input-distribution assignment from numerical propagation and states conditions for valid Monte Carlo use.^[ref]

Finally, proximity to the mathematical boundary is not the only possible problem. A change in shared normalization, polarization or wavefront curvature can violate the one-radius model. NIST's receiver apparatus illustrates the need for optical matching and separate electrical corrections.^[ref] Numerical clipping cannot replace those physical checks.

Dependent variance correction in heterodyne normalization occurs when an overlap reading and the estimated logarithmic variance used to correct it share random information. Correcting the reading is then a joint nonlinear calculation; multiplying separately averaged correction factors can give the wrong answer, even when the variance estimate is unbiased.

Write $\hat{\gamma }=\gamma e^d$ and $\tilde{\gamma }=\hat{\gamma }{e}^{-\hat{v}/2}$, with $d$ a centered normal random variable of variance $v$. The exact mean ratio isThe joint exponential moment in overlap correction cannot generally be replaced by $\mathbb{E}[e^d]\mathbb{E}[e^{-\hat{v}/2}]$. That factorization requires independence. JCGM 101 treats a measurement function of jointly distributed inputs and distinguishes choosing that joint model from propagating it.^[ref]

An original counterexample takes $\hat{v}=d^2$. This estimate is nonnegative and has expected value $v$, because the log-error mean is assumed known to be zero. It is completely dependent on the reading. It is a mathematical stress test, not a recommended way to estimate receiver noise from one observation.

Completing the square in a Gaussian integral givesSetting $a=t=1$ therefore yieldsAt stipulated $v=1$, the ratio is $e^{1/4}/\sqrt{2}\approx 0.90794$. An unbiased variance estimate has left a downward mean shift. The independence-based upward-shift argument in uncertain variance in heterodyne lognormal bias correction does not apply.

For comparison, let $d^{\prime }$ be an independent copy and use $\hat{v}=(d^{\prime }{)}^2$. The variance estimate has exactly the same marginal distribution, but the mean ratio becomes $e^{v/2}/\sqrt{1+v}$, approximately 1.16582 at $v=1$. Changing only the dependence reverses the direction of the residual bias. Both are illustrative probability models, not measured optical noise levels.

In a physical analysis, record which calibration records, fitted residuals and normalization readings are reused. A Monte Carlo method must reproduce those shared random inputs rather than independently drawing a new variance estimate wherever the formula uses it. An independent calibration sample simplifies some dependencies but can introduce drift between calibration and measurement. Neither strategy replaces checks on heterodyne RF-to-DC gain transfer check or stable optical alignment; the probability model describes residual uncertainty after those measurement paths have been specified.

Eigenmode alignment in overlap variance correction describes how shared calibration directions change the mean of a variance-corrected overlap, even when the marginal distributions of the reading and estimated variance are fixed. In the quadratic Gaussian overlap correction model, let $Z$ have the standard multivariate normal distribution, $d=a^{\mathsf{T}}Z$ and $\hat{v}=c+Z^{\mathsf{T}}BZ$, with symmetric positive semidefinite matrix $B$ and $c\ge 0$. The corrected estimate is $\tilde{\gamma }=\gamma \exp (d-\hat{v}/2)$.

Write $B=Q\mathrm{diag}({\lambda }_k)Q^{\mathsf{T}}$ with orthogonal matrix $Q$, and let $\alpha =Q^{\mathsf{T}}a$. Completing the square in the Gaussian density givesThe eigenvalues determine the variance estimator's distribution. Its eigenvectors determine which of those directions also enters $d$. Rotating $B$ relative to a fixed $a$ preserves those eigenvalues but changes the last sum.

An original rank-one example takes $\Vert a\Vert =1$, $B=0.2u{u}^{\mathsf{T}}$ with unit vector $u$, and $c=0.8$. If $\theta$ is the angle between $a$ and $u$, direct inversion of $I+B$ givesParallel alignment gives a mean ratio about $0.9282$; perpendicular alignment gives about $1.0089$. Both estimated variances have mean one and distribution $0.8+0.2Z_1^2$. Unbiased estimation of the logarithmic variance therefore does not itself make the corrected arithmetic overlap unbiased.

A Rayleigh quotient bounds every orientation when $a\ne 0$:The endpoints occur when $a$ lies in the respective extreme eigenspaces. Substituting these bounds into the logarithmic mean formula gives a deterministic range over all alignments with the stated eigenvalues.

There is an instructive dependence trap. Symmetry gives $\mathrm{cov}(d,\hat{v})=0$, while the Gaussian fourth moments giveZero linear covariance between the signed residual and the variance estimate therefore does not establish that they are independent random variables. Squaring the residual exposes shared directions.

For an actual calibration, the joint input model must describe which records are reused. JCGM 101 separates assigning input distributions from propagating them and presents the multivariate normal distribution as one possible joint assignment.^[ref] The examples and orientation bounds above are derived probability models, not measured heterodyne performance.

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.

The feasible-set boundary in overlap inversion separates joint measurements admitting a finite positive Gaussian radius ratio from measurements incompatible with that forward model. For radius-setting parameter $h>0$ and log-ratio $D$, the admissible region isAt its upper boundary the inferred radius ratio tends to zero; at its lower boundary it becomes arbitrarily large. The joint radius-setting and log-ratio bounds must therefore be checked against the sloping joint boundary, not against a fixed limit evaluated only at the central $h$.

An original ellipsoid calculation gives an exact test for a common joint-set model. Let $x=(D,h{)}^{\mathsf{T}}$ lie in $(x-\mu {)}^{\mathsf{T}}{\Sigma }^{-1}(x-\mu )\le k^2$, with $\Sigma$ positive definite. For any vector $l$, the maximum of $l^{\mathsf{T}}x$ isTo see this, write $x=\mu +{\Sigma }^{1/2}y$, $\Vert y\Vert \le k$, and apply the Cauchy-Schwarz inequality; equality is attained along the corresponding transformed direction.

Use $l_{+}=(1,-4{)}^{\mathsf{T}}$ and $l_{-}=(-1,-4{)}^{\mathsf{T}}$. The entire ellipse is strictly admissible if both displayed maxima are negative and its smallest $h$ is positive. This ellipsoidal admissibility test for overlap data checks every point of the stated joint set. It does not require sampling a coarse grid that might miss the closest approach to the boundary.

For stipulated means $(D_0,h_0)=(2.5,0.7)$, standard-scale parameters $s_D=0.1$, $s_h=0.03$ and $k=2$, the central distance $4h_0-D_0$ is 0.3. With zero covariance, the upper-boundary excursion isso the ellipse crosses that boundary. With correlation $+0.8$, the covariance is 0.0024 and the excursion becomesThat ellipse stays below the upper boundary; the lower boundary and positive-$h$ condition are also satisfied. Correlation changes the geometry even though both marginal uncertainty scales are unchanged.

If a set intersects the forbidden region, report the incompatibility and examine calibration, model adequacy and the meaning of the input bounds. Clipping points and applying the inverse to the survivors changes the propagated set or probability distribution. JCGM 101 separates input assignment, propagation and output summaries and gives conditions for valid Monte Carlo propagation.^[ref] An illustrative ellipse is a deterministic set here; interpreting it as a confidence region requires an additional statistical model.

Interval propagation for bounded overlap data maps an allowed measurement interval through the nonlinear inversion of a symmetric Gaussian overlap log-ratio. It exposes asymmetry and finite-model limits that a central value with a symmetric standard uncertainty can hide. A deterministic input bound and a statistical confidence interval have different meanings, even if the endpoint calculation is the same.

For known $h=\log q>0$, defineA finite positive radius ratio requires $-4h<D<4h$. Within this range $\rho$ decreases monotonically with $D$. Thus an input interval $[D_L,D_U]$ lying strictly inside that range maps exactly toThe endpoint order reverses. This is direct monotone propagation, without a small-uncertainty approximation.

An original example sets $q=2$ and $D\in [1.4,1.6]$. The inferred radius ratio lies approximately between 0.4739 and 0.5306. The endpoints should be computed from the full inverse, rather than by forcing equal absolute distances around the central inferred radius. Such asymmetry becomes much stronger near either finite-model boundary.

If an interval reaches $D=4h$ from below, it permits radius ratios arbitrarily close to zero. Reaching $D=-4h$ permits arbitrarily large radius ratios. Boundary values themselves correspond only to limiting ratios, not finite positive solutions. If the interval extends beyond the model range, some allowed observations have no solution in this forward model. Intersecting the interval with the admissible range can describe its compatible portion, but it does not justify discarding the incompatible observations or retaining the original claimed coverage without further analysis.

Uncertainty in $h$ changes the boundary too. For a rectangular set with $0<D_L\le D_U<4h_L$ and $h\in [h_L,h_U]$, $\rho$ increases with $h$. Consequently the extreme compatible radius ratios areFor negative $D$, the dependence on $h$ reverses; a sign-spanning interval requires checking that dependence instead of blindly using the positive-case formula. A correlated joint uncertainty region in $(D,h)$ need not fill the rectangle, so independent interval bounds can overstate the feasible range.

Joint radius-setting and log-ratio bounds therefore belong with the experimental calibration, not only the final arithmetic. JCGM 101 separates assigning input distributions from propagation, defines output coverage intervals and states conditions for its Monte Carlo method.^[ref] A correctly propagated deterministic set is not automatically a 95% confidence interval; any probability statement requires the input model and its calibration. Optical alignment and state-gain checks remain necessary because an interval calculation cannot repair an incorrect forward model.

The joint exponential moment in overlap correction is the expected value $\mathbb{E}[\exp (d-\hat{v}/2)]$ that controls the mean of a variance-corrected overlap estimate. It retains the dependence between a logarithmic reading error $d$ and its estimated variance $\hat{v}$. Separate marginal distributions do not determine this expectation.

For $\hat{\gamma }=\gamma e^d$ and $\tilde{\gamma }=\hat{\gamma }{e}^{-\hat{v}/2}$,An original multivariate Gaussian integral evaluates a useful family exactly. Let $Z\sim N(0,I_n)$, $d=a^{\mathsf{T}}Z$ and $\hat{v}=c+Z^{\mathsf{T}}BZ$, with $c\ge 0$ and $B$ a symmetric positive semidefinite matrix. Completing the square givesIndeed, combining the Gaussian density with the correction changes its precision matrix to $I_n+B$ and shifts its center to $(I_n+B{)}^{-1}a$. Integration gives the determinant factor. Positive semidefiniteness keeps the variance estimate nonnegative and the integral finite. This is a quadratic Gaussian overlap correction model, not a receiver specification.

The log-error variance is $a^{\mathsf{T}}a$, while $\mathbb{E}[\hat{v}]=c+\mathrm{tr}B$. Equality of those quantities makes the variance estimate unbiased; it does not make the corrected overlap unbiased.

For a concrete two-dimensional example, take $a=(1,0{)}^{\mathsf{T}}$ and $c=0$. Choosing $B=\mathrm{diag}(1,0)$ uses the same normal component in both the reading and variance estimate, giving $e^{1/4}/\sqrt{2}\approx 0.90794$. Choosing $B=\mathrm{diag}(0,1)$ uses an independent component, giving $e^{1/2}/\sqrt{2}\approx 1.16582$. Both variance estimates have the same chi-squared distribution and mean one. Their different joint relationships to the reading reverse the direction of the remaining bias.

In a measurement implementation, preserve which calibration residuals and normalization records are shared. JCGM 101 distinguishes joint input assignment from numerical propagation and discusses multivariate Gaussian inputs and prior uncertainty calculations.^[ref] A Monte Carlo method can reproduce the integral for a specified joint model; independently resampling shared records would test another model. Compare simulated means with this exact family before interpreting numerical convergence as evidence about a physical receiver.

Joint radius-setting and log-ratio bounds retain the allowed relationship between the radius-setting parameter $h$ and measured log-ratio $D$ when inferring an optical radius ratio. Separate marginal intervals form a rectangle that may contain combinations excluded by shared calibration information. Propagating that rectangle can therefore exaggerate the feasible radius range.

For $h>0$ and $|D|<4h$, defineFor a stated joint set $\mathcal{F}$, the compatible ratios are $\{\rho (D,h):(D,h)\in \mathcal{F},|D|<4h\}$. This set definition precedes any approximation. A deterministic set is not automatically a statistical confidence region; a probability interpretation needs the corresponding input model.

An original exactly linked calibration example has $0.6\le h\le 0.8$ and $D=2h$. The marginal intervals are $D\in [1.2,1.6]$ and $h\in [0.6,0.8]$, but their extreme values cannot be chosen independently. Writing $t=\tanh (h/2)$ and using $\tanh h=2t/(1+t^2)$ givesThus the joint model gives approximately $0.5216\le \rho \le 0.5447$. Treating the two marginal intervals as an independent rectangle instead allows approximately $0.4139\le \rho \le 0.6246$. The wider result contains parameter combinations that violate $D=2h$. All values describe a stipulated calibration relation, not observed optical data.

For a positive-$D$ rectangle wholly inside the inversion range, monotonicity places the minimum ratio at $(D_U,h_L)$ and maximum at $(D_L,h_U)$. That shortcut is valid for the rectangle, but those corners may be absent from the actual joint set. With nonlinear dependencies, optimize over the stated joint set or parameterize it by the genuinely independent calibration inputs.

The feasible-set boundary in overlap inversion also matters. If part of a joint set crosses $|D|=4h$, some allowed observations have no finite positive solution in the assumed model. Report that incompatibility and examine the measurement model; silently discarding those points changes the set or probability distribution. JCGM 101 distinguishes input assignment, propagation and output summaries, and specifies conditions for its Monte Carlo approach.^[ref]

Experimentally, retain common calibration records when deriving both $h$ and $D$. A shared scale or fitted parameter should be varied consistently through both calculations. Independent draws from their marginals erase that information. Conversely, imposing a convenient correlation without physical calibration evidence can create an unjustifiably narrow range. Interval propagation for bounded overlap data is a numerical step conditional on this joint experimental model.

A quadratic Gaussian overlap correction models an estimated logarithmic variance as a nonnegative quadratic function of Gaussian calibration residuals. It gives an exact mean for a corrected overlap while retaining the dependence between the reading and the variance estimate. The model is useful for testing uncertainty propagation software and understanding which shared residuals matter.

Let $Z\sim N(0,I_n)$, $d=a^{\mathsf{T}}Z$ and $\hat{v}=c+Z^{\mathsf{T}}BZ$, with $B$ a symmetric positive semidefinite matrix and $c\ge 0$. For $\tilde{\gamma }=\gamma \exp (d-\hat{v}/2)$, the joint exponential moment in overlap correction givesThe eigenvalues of $B$ describe how strongly each calibration direction enters the variance estimate, while the components of $a$ describe how those same directions enter the reading. Eigenmode alignment in overlap variance correction therefore affects the bias even when the variance estimate's marginal distribution is unchanged.

An original eigenvector calculation writes $B=Q\mathrm{diag}({\lambda }_k)Q^{\mathsf{T}}$ and $\alpha =Q^{\mathsf{T}}a$. Taking logarithms givesThis form avoids taking a determinant and a potentially large exponential separately. It also displays which calibration directions overlap the readout direction. Unbiased variance estimation requires $c+{\sum }_k{\lambda }_k=\Vert a{\Vert }^2$; it does not force the displayed logarithm to vanish.

For a stipulated two-direction example, take $a=(1,0{)}^{\mathsf{T}}$, $c=0.9$ and one nonzero eigenvalue 0.1. If that eigenvalue lies along the reading direction, $B=\mathrm{diag}(0.1,0)$ and the mean ratio is about 0.9578. Rotating it into the independent direction, $B=\mathrm{diag}(0,0.1)$, gives about 1.00235. Both estimates have the form $0.9+0.1Z_1^2$ in distribution, and both have mean one. Only their joint relationship to $d$ changes.

A small-$B$ expansion, with $c=\Vert a{\Vert }^2-\mathrm{tr}B\ge 0$, makes the distinction sharper:The shared-direction term can appear at first order; the independent-direction contribution begins differently. All examples are probability calculations, not measured receiver performance.

In an actual analysis, build $a$ and $B$ from the stated measurement model and reused calibration records. JCGM 101 distinguishes input-distribution assignment from propagation and provides multivariate Gaussian inputs as one particular joint model.^[ref] Agreement of a Monte Carlo method with the exact formula checks implementation conditional on those inputs. It does not establish that Gaussian residuals or a quadratic variance estimator describe the physical experiment.

Thermal-product dependence beyond covariance is the influence of higher joint moments on uncertainty in a nonlinear thermal expansion correction. Two input models can have identical means, variances and covariance yet predict different uncertainty in a product. A thermal-product input-distribution check must therefore examine how the temperature offset and expansion-coefficient error occur together.

Let $a={\mu }_a+A$ and $\theta ={\mu }_{\theta }+T$, where the centered variables have expected values zero, variances $s_a^2,s_{\theta }^2$ and covariance $c=\mathbb{E}[AT]$. Direct expansion givesThe third joint moments and the mixed fourth moment of thermal inputs are extra information. For centered inputs, only the last two terms remain. A first-order uncertainty propagation rule at zero input means would miss the product entirely; the GUM explicitly requires higher-order treatment when nonlinearity matters.^[ref]

An original counterexample keeps even the marginal shapes fixed. Let $Z$ have a standard normal distribution and let $S$ independently take values $+1$ and $-1$ with equal probabilities. Define $A=s_{a}Z$ and $T=s_{\theta }SZ$. Both have centered normal distributions, the quoted variances and zero covariance. They are not independent: their absolute standardized values are identical.

Their product is $s_a{s}_{\theta }S{Z}^2$. Its expected value is zero, while its variance is $3s_a^2{s}_{\theta }^2$, because $\mathbb{E}[Z^4]=3$. Independent centered normal inputs with the same individual variances instead give product variance $s_a^2{s}_{\theta }^2$. The joint model, rather than a marginal histogram or correlation coefficient alone, determines the difference.

For an illustrative length correction $d=La\theta$ with zero means, $L=100\mathrm{mm}$, $s_a=10^{-6}{\mathrm{K}}^{-1}$ and $s_{\theta }=0.1\mathrm{K}$, these models give standard deviations $\sqrt{3}\times 10\approx 17.32$ nm and 10 nm. This constructed dependence is a mathematical test case, not a proposed universal thermometer model.

In a physical gauge block study, track paired temperature and coefficient information, shared thermometer corrections and any procedure that estimates one input using the other. NIST's thermal correction explicitly combines block length, expansion coefficient and temperature difference.^[ref] Preserve that dependence when generating joint draws. JCGM 101 distinguishes model formulation from propagation and gives a multivariate Gaussian input model as a specific choice, not a consequence of normal marginals alone.^[ref] More simulation trials reduce numerical error conditional on the chosen model; they cannot identify unmeasured joint moments.

A thermal-product input-distribution check tests whether the probability model used for a nonlinear thermal correction describes the available temperature and material information. In $d=La\theta$, with coefficient difference $a$ and temperature offset $\theta$, the uncertainty depends on their joint behaviour, not just on two separately quoted standard uncertainties.

Start with the physical inputs: which thermometer calibration is shared, how coefficient estimates were obtained, and whether either quantity was inferred using the other. Translate those relationships into a joint distribution or an explicit dependence model. JCGM 101 distinguishes assigning input distributions from propagating them through a measurement model; numerical sampling cannot supply missing physical information.^[ref]

An original counterexample shows why zero covariance is insufficient for the independent-input formula. Let $Z$ be standard normal and defineBoth inputs have zero mean, with variances ${\sigma }_a^2$ and ${\sigma }_{\theta }^2$, and their covariance is zero because the required Gaussian odd moments vanish. They are nevertheless dependent: $\theta$ is determined by $a$.

Their product has zero mean and varianceusing Gaussian moments $\mathbb{E}[Z^2]=1$, $\mathbb{E}[Z^4]=3$, $\mathbb{E}[Z^6]=15$. Independence would instead give ${\sigma }_a^2{\sigma }_{\theta }^2$. With $L=100\mathrm{mm}$, ${\sigma }_a=10^{-6}{\mathrm{K}}^{-1}$ and ${\sigma }_{\theta }=0.1\mathrm{K}$, the resulting standard deviation is about 22.36 nm rather than 10 nm. This is a deliberately constructed probability example, not a claim about typical thermometers.

Thermal-product dependence beyond covariance matters because higher moments enter the product variance. A multivariate normal distribution is a particular joint model whose moments are determined by its means and covariance; arbitrary inputs sharing those same summaries need not behave identically.

Check units, support, marginal shapes and dependence before simulation. Compare exact or limiting cases with a Monte Carlo method, and retain a sensitivity analysis for plausible alternative input models. The GUM's first-order rule requires higher-order treatment when nonlinearity matters.^[ref] A small simulation error measures numerical precision conditional on the model, not proof that the model is physically correct.

Uncertain variance in heterodyne lognormal bias correction matters when a measured overlap is corrected using an estimated, rather than known, logarithmic variance. The known-variance correction is a nonlinear function of that estimate, so replacing the variance by an unbiased estimate does not automatically preserve unbiasedness of the corrected overlap.

Suppose $\hat{\gamma }=\gamma e^d$, with $d\sim N(0,v)$, and use $\tilde{\gamma }=\hat{\gamma }{e}^{-\hat{v}/2}$. If the variance estimate is independent of the overlap readout,Even if $\mathbb{E}[\hat{v}]=v$, Jensen's inequality gives $\mathbb{E}[e^{-\hat{v}/2}]\ge e^{-v/2}$. Consequently the mean corrected estimate is at least $\gamma$ under these independence and unbiased-variance assumptions. Equality requires a nonrandom variance estimate in the ordinary finite-variance case.

An original exact example uses a variance estimate from an independent normal calibration sample, with $\nu \hat{v}/v$ following a chi-squared distribution with $\nu$ degrees of freedom. Its Laplace transform givesThe logarithm of this factor is $v/2-(\nu /2)\log (1+v/\nu )$, which is positive for $v>0$. With stipulated $v=1$ and $\nu =2$, the factor is $e^{1/2}/1.5\approx 1.09915$: about a 9.91% remaining mean shift. This intentionally broad statistical example is not a typical receiver specification.

For fixed $v$ and large $\nu$, the logarithm is $v^2/(4\nu )+O({\nu }^{-2})$. More independent calibration information reduces this effect. Merely taking more overlap readings while retaining the same uncertain variance estimate does not remove its influence.

Dependent variance correction in heterodyne normalization requires the joint quantity $\mathbb{E}[e^{d-\hat{v}/2}]$ instead of the product of expectations above. A variance inferred from the same readout cannot simply be declared independent. Propagate the joint calibration and readout model, with a stated target summary. JCGM 101 treats input distributions from prior uncertainty calculations and numerical propagation explicitly.^[ref]

The optical gain-matching condition for heterodyne normalization still concerns physical calibration paths. This calculation addresses the statistical correction of residual log-errors; it does not imply that an imperfect RF or DC calibration can be repaired solely by subtracting a fitted variance.