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The covariance measures the average product of two centered deviations. For real square-integrable random variables $X$ and $Y$,Deviations with the same sign contribute positively; deviations with opposite signs contribute negatively. Their magnitudes and probabilities determine the net result. Taking $Y=X$ recovers the variance. The definitions and algebraic properties are presented in Orloff and Bloom, MIT 18.05, Covariance and Correlation, sections 2–2.1.

For a worked example, let $X,Z$ be independent random variables, each equally likely to be $-1$, $0$ or $1$, and put $Y=2X+Z$. All nine pairs $(X,Z)$ have probability $1/9$. The expected values are zero, while linearity of covariance and the vanishing covariance of independent random variables giveThe shared $X$ term makes large $X$ accompany large $Y$ on average. Replacing $2X$ by $-2X$ reverses the covariance to $-4/3$ without changing either variance.

The covariance is useful whenever deviations are added or subtracted. Expanding the square givesFor the worked model, these are $20/3$ and $4/3$. Positive covariance increases variation in the sum while reducing variation in the difference: the same shared fluctuation reinforces one combination and partly cancels in the other. The identity applies beyond independent random variables.

The size of a covariance depends on the measurement scale. For constants $a,b,c,d$, linearity of covariance gives $\mathrm{Cov}(aX+b,cY+d)=ac\mathrm{Cov}(X,Y)$. Adding offsets has no effect; rescaling either measurement rescales the covariance. The Pearson correlation coefficient divides by both standard deviations to describe the standardized linear relationship.

A raw laser pulse energy estimate and its baseline correction can contain the same readout error. Let $Y$ be the raw gate integral, $B$ the estimated integrated background and $\hat{U}=Y-B$ the corrected result. Linearity of covariance gives the exact identityFor this linear measurement model, the law of propagation of uncertainty is exact whenever the input variances exist. Its general form includes the sensitivity coefficients and covariance terms.^[ref] Positive covariance reduces the variance of this difference; it would instead increase the variance of a sum.

An original shared-offset model makes the mechanism explicit. For fixed gate duration $T$, writewhere $c$ is a common additive power-readout offset with standard uncertainty ${\sigma }_c$, while ${\epsilon }_Y$ and ${\epsilon }_B$ are independent zero-mean energy errors, independent also of $c$. Then $\mathrm{Cov}(Y,B)=T^2{\sigma }_c^2$, and shared-offset cancellation in a baseline-corrected pulse integral gives $\hat{U}=U+{\epsilon }_Y-{\epsilon }_B$. The shared offset contributes no uncertainty to this difference. It need not be small or separately known.

Choose $T=10\mu \mathrm{s}$, ${\sigma }_c=0.20\mathrm{mW}$, $u({\epsilon }_Y)=0.30\mathrm{nJ}$ and $u({\epsilon }_B)=0.40\mathrm{nJ}$. The shared energy contribution is $2.0\mathrm{nJ}$ in each reading. Thus $u^2(Y)=4.09{\mathrm{nJ}}^2$, $u^2(B)=4.16{\mathrm{nJ}}^2$ and $\mathrm{Cov}(Y,B)=4.00{\mathrm{nJ}}^2$, yielding $u(\hat{U})=0.50\mathrm{nJ}$. Incorrectly discarding the covariance gives $\sqrt{8.25}\approx 2.87\mathrm{nJ}$.

Cancellation weakens if the offset changes between acquisitions. Replace $c$ by two offsets with the same standard uncertainty ${\sigma }_c$ and correlation coefficient $\rho$. Under the same independent residual-error assumptions,At $\rho =0.90$, the chosen example gives $u(\hat{U})=\sqrt{1.05}\approx 1.025\mathrm{nJ}$. A baseline taken far away in time cannot simply be assigned the covariance of an unchanged offset.

A sampled implementation has the same structure. If a vector of readings $\mathbf{x}$ has covariance matrix $C$, and fixed integration weights give $Y={\mathbf{q}}^T\mathbf{x}$ and $B={\mathbf{w}}^T\mathbf{x}$, thenA constant offset cancels when the weights sum to the same gate duration: ${\mathbf{q}}^{T}1={\mathbf{w}}^{T}1=T$. This does not justify using pulse-contaminated readings as background. Matched blank sequences, collected with the actual timing, help characterize the needed noise correlations.

The historical NIST TC-24 apparatus used a thermocouple-based absorbing plate and averaged baseline measurements before and after laser injection to compensate background and thermal drift.^[ref] The offset model above is an independent illustration, not a reconstruction of that instrument's uncertainty budget.

For real random vectors $\mathbf{a}\in {\mathbb{R}}^p$ and $\mathbf{b}\in {\mathbb{R}}^q$ with finite second moments, their cross-covariance matrix collects the covariances between the two sets of coordinates:It has size $p\times q$, and exchanging the two vectors transposes it: $C_{ba}=C_{ab}^T$. Unlike a covariance matrix of a vector with itself, a cross-covariance matrix need not be square or symmetric.

For an original construction, let centered independent random variables $X,Z$ have variances1and4, and putLinearity of covariance givesThe second entry contains a negative contribution from $Z$, although the total is positive. Also $\mathrm{Var}(b)=13$ andThis joint covariance matrix is positive semidefinite. That restriction couples its diagonal and off-diagonal blocks; an arbitrary rectangular matrix cannot be combined with arbitrary within-vector covariances to produce a valid joint model.

For fixed real matrices $A,B$ and constant vectors $c,d$, direct substitution into the definition givesThis identity propagates shared input fluctuations through two different linear readouts, for example when assessing shared calibration error.

From $n$ independent identically distributed paired observations, the usual unbiased estimate isThe pairing must be preserved: independently shuffling the second set changes the estimated association. NIST's multivariate tutorial gives the coordinate sample-covariance formula and its assembly into a covariance matrix; the rectangular formula above is the block joining the two chosen coordinate groups.^[ref]

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.

For any finite list $t_1,\dots ,t_k\in [0,1]$, the random vector $(U_{t_1},\dots ,U_{t_k},B_1)$ has a multivariate normal distribution, since it is a linear image of a vector of Brownian motion coordinates. The Brownian covariance kernel and linearity of covariance giveThe independence of uncorrelated jointly normal variables implies that $(U_{t_1},\dots ,U_{t_k})$ is independent of $B_1$. Extending from finite-coordinate cylinder events by independence extended from generating pi-systems provesThus the Brownian bridge is independent of its endpoint. Its covariance kernel is $\mathrm{Cov}(U_s,U_t)=\min (s,t)-st$, confirming that $U$ is the standard Brownian bridge. This is the Brownian bridge independence from its endpoint property.

For distinct sibling supernovae $s$ and $s^{\prime }$ in the same galaxy, the shared fluctuation is the only common random term. Independence and linearity of covariance therefore giveTheir correlation coefficient isThis is the intraclass correlation coefficient of the galaxy-level random intercept model.

Here $\hat{\alpha }=\alpha +n^{-1}{\sum }_i{ϵ}_i$. By linearity of covariance and independence,Thus the two estimators are uncorrelated random variables.