 Joint exponential moment in overlap correction

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.