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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.

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.