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A 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, defineThe 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.^[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,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, soFor different participants, however, their leave-one-out benchmarks overlap. Linearity of covariance givesThus 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 $\sqrt{16/3}\approx 2.31$ 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 $W={\sum }_j{w}_j$. ThenThese 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 $({\sum }_j{w}_j{x}_j-w_i{x}_i)/(W-w_i)$ 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.

Two random variables have negative covariance when $\mathbb{E}[(X-\mathbb{E}X)(Y-\mathbb{E}Y)]<0$, assuming finite second moments. Opposite deviations from their respective expected values then outweigh matching deviations on average. For example, if $Y=c-X$ for a constant $c$, the bilinear identities of covariance give $\mathrm{Cov}(X,Y)=-\mathrm{Var}(X)$, which is strictly negative when $X$ has positive variance.^[ref]

Sampling without replacement gives a useful example in which the negative sign comes from a shared finite resource. Choose a simple random sample $S$ of exactly $k$ distinct members from a statistical population of size $N$, with $0<k<N$. For each member define the inclusion indicator $I_i=1_{\{i\in S\}}$. Its inclusion probability is $p=k/N$. For different members $i,j$, the joint inclusion probability isSubtracting the product of the expected values givesThe Pearson correlation coefficient follows because each inclusion indicator has variance $p(1-p)$. Once member $i$ is included, only $k-1$ places remain for the other $N-1$ members, reducing member $j$'s conditional probability of inclusion below $k/N$.

For an original exact example, draw two members from four. The six unordered sample pairs are equiprobable. For $I_1,I_2$, the outcomes $(0,0)$ and $(1,1)$ each have probability $1/6$, while $(0,1)$ and $(1,0)$ each have probability $1/3$. Hence $\mathrm{Cov}(I_1,I_2)=1/6-1/4=-1/12$, and their Pearson correlation coefficient is $-1/3$. Matching outcomes remain possible despite the negative covariance.

A fixed total alone does not force every pair among three or more random variables to have negative covariance. If ${\sum }_j{X}_j=c$, thenThis constrains the sum of the off-diagonal entries in a row of the covariance matrix. It does not fix each entry's sign. For a counterexample, let $Z$ be equally likely to be zero or one, and set $X_1=Z$, $X_2=Z$, $X_3=2-2Z$. Their total is always two, yet $\mathrm{Cov}(X_1,X_2)=\mathrm{Var}(Z)=1/4>0$.

In the simple random sample, exchangeability makes the off-diagonal covariances equal, so the fixed-total identity does determine each of them: $(N-1)\mathrm{Cov}(I_i,I_j)=-p(1-p)$. The same distinction between a linear constraint and individual pair signs matters for the joint dependence of leave-one-out consensus residuals.