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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.

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.