Inverse-variance weight
= Inverse-variance weight
{title2=$w_i=v_i^{-1}$}
A weight $w_i=1/v_i$ gives more weight to a more precise estimate. In <random-effects meta-analysis> the <variance> of the corresponding <marginal distribution> is $v_i+\tau^2$, giving weight $(v_i+\tau^2)^{-1}$. A percentage weight divides the individual weight by the sum of all weights.