= Leave-one-out interlaboratory consensus
{title2=$x_{c,-i}$}
A \b[leave-one-out interlaboratory consensus] estimates a benchmark for laboratory $i$ from the other participants. For a fixed <weighted mean> with positive coefficients $w_j$ and at least two results, define
$$
x_{c,-i}=\frac{\sum_{j\ne i}w_jx_j}{\sum_{j\ne i}w_j},\qquad
D_i=x_i-x_{c,-i}.
$$
The omitted result does not enter its own benchmark. This removes the direct <participant influence on a consensus assigned value> for the stated fixed-weight estimator. It does not establish independence when laboratories share a <systematic error>. The proficiency-testing guide distinguishes participant consensus from expert or independent reference assignment; the leave-one-out construction here is a mathematical comparison, not a claim that the guide requires this scoring procedure.<Selection, Use and Interpretation of Proficiency Testing (PT) Schemes by Brian Brookman and Ian Mann>[§7.2.2 and Annex D.2, methods for establishing assigned values]{ref}
With equal weights and original results $(9,10,11)$, the omitted-result <arithmetic means> are $(10.5,10,9.5)$, giving residuals $(-1.5,0,1.5)$. Residuals from the full <arithmetic mean> are $(-1,0,1)$. In general,
$$
D_i=\frac n{n-1}(x_i-\bar x),\qquad \sum_iD_i=0.
$$
The larger magnitudes reflect a changed comparator rather than a changed laboratory result.
The <joint dependence of leave-one-out consensus residuals> matters when interpreting the collection of scores. Suppose the $X_i$ are <independent random variables> with a common mean and finite common <variance> $\sigma^2$. A result is independent of the <arithmetic mean> of the other results, so
$$
\operatorname{Var}(D_i)=\sigma^2+\frac{\sigma^2}{n-1}
=\frac n{n-1}\sigma^2.
$$
For different participants, however, their leave-one-out benchmarks overlap. <Linearity of covariance> gives
$$
\operatorname{Cov}(D_i,D_j)=-\frac{n\sigma^2}{(n-1)^2},\qquad
\operatorname{Corr}(D_i,D_j)=-\frac1{n-1}\quad(i\ne j).
$$
Thus independent laboratory observations do not give independent residual scores. The zero-sum identity makes their <covariance matrix> a <singular matrix>. For $n=4$ and $\sigma=2$ units, each residual has <standard deviation> $\sqrt{16/3}\approx2.31$ units, and each distinct pair has <correlation coefficient> $-1/3$. No <normal distribution> assumption is needed for these moment identities.
A <leave-one-out inverse-variance weighted consensus> applies an <inverse-variance weighted mean> to the remaining participants when their <standard uncertainties> are known. Under independent unbiased observations with weights $w_j=u_j^{-2}$, put $W=\sum_jw_j$. Then
$$
u^2(x_{c,-i})=\frac1{W-w_i},\qquad
u^2(D_i)=u_i^2+\frac1{W-w_i}.
$$
These are uncertainties under that observation model; the resulting quantity should not automatically be called a <proficiency-test z-score> with a fixed assessment scale. Computing $(\sum_jw_jx_j-w_ix_i)/(W-w_i)$ is algebraically convenient, but direct summation of the remaining terms can avoid cancellation when one participant dominates the total weight.
If a <robust estimator> has weights depending on the full data set, merely deleting one weighted numerator term need not reproduce the estimator fitted without that participant. The estimator must be recomputed according to its specified rule. Finally, a common additive calibration error remains undetectable in these residuals: replacing every $X_i$ by $X_i+B$ leaves all $D_i$ unchanged. <Shared bias in an interlaboratory consensus> therefore remains a separate question from self-influence and the <variance> of the residuals.
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