= Leave-one-out inverse-variance weighted consensus
{title2=$\widehat\theta_{-i}$}
A \b[leave-one-out inverse-variance weighted consensus] applies an <inverse-variance weighted mean> to all laboratories except the one being assessed. Suppose their results $X_j$ are <independent random variables> with common <expected value> $\theta$ and known positive <variances> $u_j^2$. For laboratory $i$, define
$$
W_{-i}=\sum_{j\ne i}u_j^{-2},\qquad
\widehat\theta_{-i}=\frac{\sum_{j\ne i}X_ju_j^{-2}}{W_{-i}}.
$$
Under this model it is the <best linear unbiased estimator> of $\theta$ from the remaining results. The omitted observation does not determine its own <leave-one-out interlaboratory consensus> value.
The minimum-variance property follows directly from the <Cauchy-Schwarz inequality>. For any fixed coefficients $c_j$ satisfying $\sum_{j\ne i}c_j=1$, the <linear unbiased estimator> $T=\sum_{j\ne i}c_jX_j$ has
$$
1=\left(\sum_{j\ne i}(c_ju_j)u_j^{-1}\right)^2
\leq\left(\sum_{j\ne i}c_j^2u_j^2\right)W_{-i}
=\operatorname{Var}(T)W_{-i}.
$$
Equality holds precisely for $c_j=u_j^{-2}/W_{-i}$. Therefore
$$
\boxed{\operatorname{Var}(\widehat\theta_{-i})=W_{-i}^{-1}}.
$$
This is a comparison among <linear unbiased estimators> under the stated independent-error model; it requires no <normal distribution> assumption.
For an original example, assess a laboratory reporting $x_i=12$ with <standard uncertainty> $u_i=0.5$. The other laboratories report $(10,12,8)$ with respective <standard uncertainties> $(1,2,4)$, all in the same units. Their total precision is $W_{-i}=1+1/4+1/16=21/16$, giving
$$
\widehat\theta_{-i}=\frac{10+12/4+8/16}{21/16}=\frac{72}{7}\simeq10.286,
\qquad u(\widehat\theta_{-i})=\sqrt{\frac{16}{21}}\simeq0.873.
$$
The held-out difference is $D_i=12/7\simeq1.714$, with
$$
u(D_i)=\sqrt{u_i^2+W_{-i}^{-1}}=\sqrt{\frac{85}{84}}\simeq1.006.
$$
The denominator includes the benchmark's uncertainty even though $X_i$ was excluded. This uncertainty-normalized difference and a <proficiency-test z-score> based on a predetermined assessment scale answer different questions.
<Shared bias in an interlaboratory consensus> changes the uncertainty model. Suppose instead $X_j=\theta+B+\varepsilon_j$, where the common reference contribution $B$ has <variance> $u_B^2$ and the mutually independent residuals, independent also of $B$, have <variances> $s_j^2$. Coefficients summing to one preserve $B$ in the benchmark. Its <variance> is $u_B^2+\sum c_j^2s_j^2$, so the minimizing coefficients use $s_j^{-2}$, and
$$
u^2(\widehat\theta_{-i})=u_B^2+\left(\sum_{j\ne i}s_j^{-2}\right)^{-1}.
$$
Using the full marginal uncertainties $\sqrt{u_B^2+s_j^2}$ as if they were independent changes the coefficients and loses the shared-reference structure. The <propagation of uncertainty> retains the <covariance> of dependent inputs.<Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results by Barry N. Taylor and Chris E. Kuyatt>[Appendix A, equation A-3, covariance terms in the measurement model]{ref}
If the held-out laboratory shares the same $B$, that contribution cancels in $D_i$, while it remains in the benchmark's absolute <measurement uncertainty>. Precision of a difference and knowledge of an absolute value are distinct properties. Estimated, result-dependent uncertainties or additional correlations require analysis of the actual estimator; the fixed known-variance proof does not establish optimality for those cases.
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