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Inverse-variance weighting combines estimates of the same quantity by giving a more precise estimate a larger weight. Suppose $X_i$ are independent, unbiased estimators of $\theta$, with known variances $0<{\sigma }_i^2<\infty$. For normalized weights,The inverse is taken after squaring the standard uncertainty: twice the standard uncertainty gives one quarter as much weight. The corresponding weighted least squares criterion is ${\sum }_i(X_i-\theta {)}^2/{\sigma }_i^2$; differentiating it with respect to $\theta$ gives the same weighted arithmetic mean.^[ref]

The optimality statement is precise. Any linear unbiased estimator ${\sum }_i{a}_i{X}_i$ requires ${\sum }_i{a}_i=1$. By the Cauchy-Schwarz inequality,Equality holds for the displayed weights. Thus they minimize variance among linear unbiased combinations under these assumptions; a normal distribution is not needed for that proof.

For an original example, combine observed values 10 and 14 with standard uncertainties 1 and 2 in the same units. The normalized weights are $0.8$ and $0.2$, giving estimate $10.8$, variance $0.8$ and standard uncertainty approximately $0.894$. Equal weights give 12 and standard uncertainty $\sqrt{1.25}\approx 1.118$.

Correlated errors change the problem. A contribution common to every reading survives every weighting whose coefficients sum to one; precision limits of weighting correlated repeats derives the resulting floor. More generally, use the full covariance matrix and generalized least squares rather than treating correlated readings as independent.^[ref] Uncertainty in estimated inverse-variance weights is another limitation: variances estimated from the same data are not fixed known quantities, and an accidentally tiny estimated variance does not justify arbitrary confidence.

Uncertainty in estimated inverse-variance weights arises when the variances used in inverse-variance weighting are themselves estimated. A very small estimated variance can give one observation or group a large random weight. The familiar variance formula for fixed, known weights cannot simply be reused with estimated variances treated as exact. NIST specifically warns that weights estimated from few replicates can substantially and unpredictably affect a weighted least squares analysis.^[ref]

Suppose independent groups contain $n_i$ independent normal random variables with common mean $\theta$ and group variance ${\sigma }_i^2$. Let ${\overline{X}}_i$ and $S_i^2$ be the group sample mean and unbiased sample variance, and formFor normal samples, each sample mean is independent of its sample variance. Conditional on the estimated weights, the estimator remains unbiased and hasThus the law of total variance gives unconditional variance equal to the expectation of this expression. The true ${\sigma }_i^2$ still appear; replacing them by the same random $S_i^2$ does not establish the actual conditional variance. Outside the independence assumptions, even conditional unbiasedness can fail.

An original two-group example takes $n_1=n_2=5$ and true ${\sigma }_1^2={\sigma }_2^2=4$, so each group mean has variance $0.8$. If observed sample variances are $0.25$ and $4$, the estimated normalized weights are $16/17$ and $1/17$. For group means $10$ and $14$, the combined estimate is $174/17\approx 10.2353$. The plug-in variance isbut the true conditional variance under the stipulated model isThe corresponding standard deviations are about $0.2169$ and $0.8435$. Equal weights would have variance $0.4$. An accidentally small estimated variance has both overstated precision and moved the combination away from the optimal equal weights in this example.

Reciprocal moments of an estimated variance explain the unstable precision estimate. For a normal sample with $\nu =n-1$, $Z=\nu S^2/{\sigma }^2$ has a chi-squared distribution. Integrating its density givesConsequently $\mathbb{E}[1/S^2]=\nu /[(\nu -2){\sigma }^2]$ for $\nu >2$, while the second moment requires $\nu >4$. The density and its gamma-distribution parameterization are documented in the primary SciPy implementation notes.^[ref] With $n=5$, the inverse sample variance has a finite mean twice the true precision and an infinite variance; the normalized weights remain bounded, so this does not imply their variance is infinite.

More replication or a justified smooth variance model can stabilize weights. NIST describes direct replicate-based weight estimation and variance-function modeling as distinct approaches.^[ref] A proposed validation repeats the complete weight-estimation and combination procedure on simulated or replicated groups, checking actual interval coverage as well as point estimates. A bootstrap must reproduce the weight-estimation stage; holding the fitted weights fixed tests a different procedure.

A weighted centering matrix subtracts the same weighted mean from every component of a vector. Let $a\in {\mathbb{R}}^n$ satisfy $a_i\ge 0$ and $a^{\mathsf{T}}1=1$, where $1$ has every component equal to one. DefineThen $P_{a}x=x-(a^{\mathsf{T}}x)1$ and $a^{\mathsf{T}}{P}_{a}x=0$. The centered vector has zero weighted mean; its unweighted component sum need not vanish.

The projection properties follow directly from normalization:Also $P_{a}1=0$, so its kernel of a linear map is $\mathrm{span}\{1\}$, its range is $\ker a^{\mathsf{T}}$ and its rank is $n-1$. It is an idempotent matrix. For unequal weights it is generally not symmetric: it projects onto the zero-weighted-mean subspace along the constant-vector direction, an oblique projection in ordinary Euclidean geometry.

For an original example, $a=(1/4,3/4{)}^{\mathsf{T}}$ and $x=(4,2{)}^{\mathsf{T}}$ give $a^{\mathsf{T}}x=2.5$ and $P_{a}x=(1.5,-0.5{)}^{\mathsf{T}}$. The weighted mean is zero, while the ordinary mean is $0.5$. Euclidean orthogonal projection of $x$ onto the same line instead gives $(3,-1{)}^{\mathsf{T}}$.

If every $a_i>0$, put $D=\mathrm{diag}(a)$. Then $D{P}_a=P_a^{\mathsf{T}}D=D-a{a}^{\mathsf{T}}$, so this same map is orthogonal for the inner product $⟨u,v{⟩}_D=u^{\mathsf{T}}Dv$. It is the residual-making map for minimizing ${\sum }_i{a}_i(x_i-c{)}^2$ over a fitted constant $c$, the intercept-only case of weighted least squares.^[ref] Zero weights preserve the algebraic projection but make this proposed inner product degenerate. When all weights equal $1/n$, the construction reduces to the usual centering matrix.