= Joint exponential moment in overlap correction
The \b[joint exponential moment in overlap correction] is the <expected value> $\mathbb E[\exp(d-\widehat v/2)]$ that controls the mean of a variance-corrected overlap estimate. It retains the dependence between a logarithmic reading error $d$ and its estimated <variance> $\widehat v$. Separate marginal distributions do not determine this expectation.
For $\widehat\gamma=\gamma e^d$ and $\widetilde\gamma=\widehat\gamma e^{-\widehat v/2}$,
$$
\mathbb E[\widetilde\gamma]/\gamma=\mathbb E[e^{d-\widehat v/2}].
$$
An original multivariate <Gaussian integral> evaluates a useful family exactly. Let $Z\sim N(0,I_n)$, $d=a^{\mathsf T}Z$ and $\widehat v=c+Z^{\mathsf T}BZ$, with $c\ge0$ and $B$ a symmetric <positive semidefinite matrix>. Completing the square gives
$$
\mathbb E[e^{d-\widehat v/2}]
=\frac{e^{-c/2}}{\sqrt{\det(I_n+B)}}
\exp\!\left(\frac12a^{\mathsf T}(I_n+B)^{-1}a\right).
$$
Indeed, 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 $\mathbb E[\widehat v]=c+\operatorname{tr}B$. 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 $B=\operatorname{diag}(1,0)$ uses the same normal component in both the reading and variance estimate, giving $e^{1/4}/\sqrt2\approx0.90794$. Choosing $B=\operatorname{diag}(0,1)$ uses an independent component, giving $e^{1/2}/\sqrt2\approx1.16582$. 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.<Propagation of distributions using a Monte Carlo method by JCGM>[§§5.1–5.2, 6.4.8 and 6.5]{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.
Back to article page
