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A meta-analysis combines statistical evidence from several studies addressing a common estimand. It may assume a shared effect or use a random-effects meta-analysis to represent between-study variation. Inverse-variance weighted means combine independent effect estimates when their variances are known or estimated; differences in study validity and effect modifiers require substantive assessment, not just weighting.

Let ${\hat{\delta }}_D$ and ${\hat{\delta }}_I$ be independent, approximately unbiased estimates of one consistent treatment effect, with variances $v_D,v_I>0$. The inverse-variance weighted mean givesIndeed, weights $w,1-w$ give variance $w^2{v}_D+(1-w{)}^2{v}_I$; differentiating gives $w=v_I/(v_D+v_I)$, the weight on the direct estimate. Here the three displayed trials give ${\hat{\delta }}_D=-1.33$, $v_D=0.04$, and hence $\hat{\delta }\simeq -1.3203$ with variance $0.03448$. This numerical pooling uses the displayed evidence, not the unavailable data from all twelve trials. A shared study or shared control would introduce covariance, requiring the corresponding correlated-estimator formula; inconsistency would invalidate the common-effect interpretation.

Unbiasedness requires ${\sum }_i{\alpha }_i=1$. Let $S={\sum }_i{\sigma }_i^{-2}$ and minimize $V(\alpha )={\sum }_i{\sigma }_i^2{\alpha }_i^2$. The Lagrange multiplier equations for $V-\lambda ({\sum }_i{\alpha }_i-1)$ are $2{\sigma }_i^2{\alpha }_i=\lambda$, so normalization gives $\lambda =2/S$. HenceThe gradient at this point is $(2/S)1$, whose inner product with every feasible displacement $h$ satisfying $1^{T}h=0$ is zero. The Hessian matrix is $2\mathrm{diag}({\sigma }_1^2,\dots ,{\sigma }_N^2)$, and $h^T{\nabla }^{2}Vh=2{\sum }_i{\sigma }_i^2{h}_i^2>0$ for every nonzero feasible $h$. These are the first- and second-order constrained minimum conditions. Strict convexity also proves global uniqueness of this inverse-variance weighted mean; normal errors are not needed.

Independence makes the variance of an estimator equal toWith ${\alpha }_2=1-{\alpha }_1$, differentiate to obtain $2{\alpha }_1{\sigma }_1^2-2(1-{\alpha }_1){\sigma }_2^2=0$. The second derivative is $2({\sigma }_1^2+{\sigma }_2^2)>0$, so the unique minimum isEquivalently, the inverse-variance weighted mean uses ${\alpha }_i=K{\sigma }_i^{\gamma }$ withWe assume positive measurement variances. A zero-variance observation already supplies the true value exactly.

Equal variances make every weight $1/N$, so the inverse-variance weighted mean reduces to the arithmetic mean. Independence givesThus the variance scales as $N^{-1}$, whereas the uncertainty expressed as a standard deviation scales as $N^{-1/2}$.

For independent study estimates $y_i$ with within-study variances $v_i$, a usual approximate model is $y_i\mid {\delta }_i\sim N({\delta }_i,v_i)$ and ${\delta }_i\sim N(\mu ,{\tau }^2)$. Its marginal distribution is $N(\mu ,v_i+{\tau }^2)$, so for fixed heterogeneity ${\tau }^2$ the inverse-variance weighted mean uses weights $(v_i+{\tau }^2{)}^{-1}$. Here $\mu$ describes the mean effect across comparable studies; ${\tau }^2$ represents between-study variation. Estimation and uncertainty for heterogeneity are essential, especially with few studies.