= Dependent variance correction in heterodyne normalization
\b[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 $\widehat\gamma=\gamma e^d$ and $\widetilde\gamma=\widehat\gamma e^{-\widehat v/2}$, with $d$ a centered <normal random variable> of variance $v$. The exact mean ratio is
$$
\frac{\mathbb E[\widetilde\gamma]}\gamma
=\mathbb E\!\left[e^{d-\widehat v/2}\right].
$$
The <joint exponential moment in overlap correction> cannot generally be replaced by $\mathbb E[e^d]\mathbb E[e^{-\widehat v/2}]$. That factorization requires <independence>. JCGM 101 treats a measurement function of jointly distributed inputs and distinguishes choosing that joint model from propagating it.<Propagation of distributions using a Monte Carlo method by JCGM>[§§5.1–5.2, 6.4.8 and 6.5]{ref}
An original counterexample takes $\widehat v=d^2$. 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> gives
$$
\mathbb E[e^{td-ad^2/2}]
=\frac{\exp\!\left(t^2v/[2(1+av)]\right)}{\sqrt{1+av}},\qquad a\ge0.
$$
Setting $a=t=1$ therefore yields
$$
\frac{\mathbb E[\widetilde\gamma]}\gamma
=\frac{\exp\!\left(v/[2(1+v)]\right)}{\sqrt{1+v}}.
$$
At stipulated $v=1$, the ratio is $e^{1/4}/\sqrt2\approx0.90794$. 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 $d'$ be an independent copy and use $\widehat v=(d')^2$. 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.
Back to article page
