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The joint dependence of leave-one-out consensus residuals arises because different laboratories' benchmarks reuse many of the same observations. Excluding each result from its own leave-one-out interlaboratory consensus does not make the complete collection of residuals independent.

Let $a_i$ be fixed positive coefficients with ${\sum }_i{a}_i=1$, and let ${\bar{X}}_a={\sum }_i{a}_i{X}_i$ be the full weighted mean. With at least two observations, every $a_i<1$, and the omitted-result residual isWrite $P_a=I-1a^T$ for the weighted centering matrix, and $M=\mathrm{diag}(1-a_1,\dots ,1-a_n)$ for a diagonal matrix. ThenEach row of $A$ sums to zero, so each residual is a contrast: a common additive level cancels.

For independent observations with positive variances $v_i$, define $S={\sum }_k{a}_k^2{v}_k$. Expanding linearity of covariance givesEqual coefficients and equal variances reduce these expressions toThe familiar negative covariance belongs to this model. Arbitrary fixed coefficients can produce a different sign.

An original three-laboratory counterexample sets $v_1=v_2=v_3=1$ and $a=(0.05,0.05,0.90)$. Then $S=0.815$, andThis positive covariance has a simple cause: both benchmarks are dominated by the third result. Increasing $X_3$ pushes both $D_1$ and $D_2$ downward. The coefficients are stipulated to expose the dependence; with equal observation variances, they are not inverse-variance coefficients.

For positive inverse-variance weights $a_i=v_i^{-1}/W$, where $W={\sum }_k{v}_k^{-1}$, one has $a_i{v}_i=1/W$ and $S=1/W$. The off-diagonal expression becomesThus the leave-one-out inverse-variance weighted consensus has a special sign pattern under its independent known-variance assumptions.

The coefficients also impose a deterministic constraint,Consequently the residual covariance matrix is a singular matrix. Under the independent-observation model with every $v_i>0$, its matrix rank is $n-1$: the weighted centering matrix removes the common-level direction and the invertible diagonal matrix $M$ preserves that rank. For equal coefficients the constraint is simply ${\sum }_i{D}_i=0$.

These dependencies matter when evaluating several laboratories together. Products of single-residual probabilities generally do not describe joint events. A specified joint model or a set of appropriately chosen contrasts can represent the available information without treating redundant residuals as independent observations.

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.