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For $X\sim N(\mu 1,C)$ with a known positive-definite matrix $C$, the maximum-likelihood estimator of $\mu$ is $\hat{\mu }=(1^T{C}^{-1}X)/(1^T{C}^{-1}1)$. It is unbiased with variance $(1^T{C}^{-1}1{)}^{-1}$, attaining the Cramér-Rao bound. It is also the best linear unbiased estimator. Pairwise admissible correlation coefficients alone do not ensure that $C$ is a valid covariance matrix; the entire matrix must be symmetric and a positive semidefinite matrix. Inverse-based formulas require positive definiteness.

Let $Y=X\beta +\epsilon$ with a fixed full column rank design matrix, $\mathbb{E}\epsilon =0$ and $\mathrm{Cov}(\epsilon )={\sigma }^{2}I$. Then ordinary least squares is a best linear unbiased estimator: every response-linear unbiased estimator has at least as much variance in every coefficient direction. A normal distribution is unnecessary.^[ref]

For a proof put $A=(X^{\mathsf{T}}X{)}^{-1}{X}^{\mathsf{T}}$. A competing estimator $CY$ is unbiased for every $\beta$ exactly when $CX=I$. Write $C=A+D$; then $DX=0$, so the cross terms vanish andThis positive semidefinite matrix inequality proves the simultaneous directional comparison. The scope is linear unbiased estimators, not all conceivable estimators: a biased ridge regression estimate may improve mean squared error for some signals. With correlated or unequal-variance errors, the corresponding known-covariance result uses generalized least squares.

For independent unbiased estimators $X_i$ of a common mean with known positive variances $v_i$, the linear unbiased estimator with smallest variance has weights $w_i=v_i^{-1}/{\sum }_j{v}_j^{-1}$ and variance $({\sum }_j{v}_j^{-1}{)}^{-1}$. Unbiasedness requires ${\sum }_i{w}_i=1$. Minimizing ${\sum }_i{v}_i{w}_i^2$ by a Lagrange multiplier gives the weights, and the strictly positive diagonal Hessian matrix proves a unique global minimum. This conclusion requires no normality assumption.

A leave-one-out inverse-variance weighted consensus applies an inverse-variance weighted mean to all laboratories except the one being assessed. Suppose their results $X_j$ are independent random variables with common expected value $\theta$ and known positive variances $u_j^2$. For laboratory $i$, defineUnder this model it is the best linear unbiased estimator of $\theta$ from the remaining results. The omitted observation does not determine its own leave-one-out interlaboratory consensus value.

The minimum-variance property follows directly from the Cauchy-Schwarz inequality. For any fixed coefficients $c_j$ satisfying ${\sum }_{j\ne i}{c}_j=1$, the linear unbiased estimator $T={\sum }_{j\ne i}{c}_j{X}_j$ hasEquality holds precisely for $c_j=u_j^{-2}/W_{-i}$. ThereforeThis is a comparison among linear unbiased estimators under the stated independent-error model; it requires no normal distribution assumption.

For an original example, assess a laboratory reporting $x_i=12$ with standard uncertainty $u_i=0.5$. The other laboratories report $(10,12,8)$ with respective standard uncertainties $(1,2,4)$, all in the same units. Their total precision is $W_{-i}=1+1/4+1/16=21/16$, givingThe held-out difference is $D_i=12/7\simeq 1.714$, withThe denominator includes the benchmark's uncertainty even though $X_i$ was excluded. This uncertainty-normalized difference and a proficiency-test z-score based on a predetermined assessment scale answer different questions.

Shared bias in an interlaboratory consensus changes the uncertainty model. Suppose instead $X_j=\theta +B+{\epsilon }_j$, where the common reference contribution $B$ has variance $u_B^2$ and the mutually independent residuals, independent also of $B$, have variances $s_j^2$. Coefficients summing to one preserve $B$ in the benchmark. Its variance is $u_B^2+\sum c_j^2{s}_j^2$, so the minimizing coefficients use $s_j^{-2}$, andUsing the full marginal uncertainties $\sqrt{u_B^2+s_j^2}$ as if they were independent changes the coefficients and loses the shared-reference structure. The propagation of uncertainty retains the covariance of dependent inputs.^[ref]

If the held-out laboratory shares the same $B$, that contribution cancels in $D_i$, while it remains in the benchmark's absolute measurement uncertainty. Precision of a difference and knowledge of an absolute value are distinct properties. Estimated, result-dependent uncertainties or additional correlations require analysis of the actual estimator; the fixed known-variance proof does not establish optimality for those cases.

Linearity of expectation gives ${\mathbb{E}}_{\mu }[\hat{\mu }]=({\alpha }_1+{\alpha }_2)\mu$. Thus a linear unbiased estimator for every value of the distance modulus satisfies $\boxed{{\alpha }_1+{\alpha }_2=1}$. Requiring unbiasedness at the single value $\mu =0$ would not impose this constraint; an unbiased estimator must work throughout the parameter space.

For this model, the Gauss-Markov theorem states that $\hat{\beta }$ is the unique minimum-variance linear unbiased estimator of $\beta$: ifis unbiased for every $\beta$, then $\mathrm{var}(\tilde{\beta })\ge \mathrm{var}(\hat{\beta })$.

Indeed,so unbiasedness is equivalent to ${\sum }_i{a}_i{x}_i=1$. WriteThe constraint becomes ${\sum }_i{c}_i{x}_i=0$. Since the errors are independent with common variance ${\sigma }^2$,Equality holds exactly when every $c_i=0$, which gives $a_i=x_i/S_{xx}$ and $\tilde{\beta }=\hat{\beta }$.