Dependent variance correction in heterodyne normalization

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.

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.

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.

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.