Talk:Inverse-variance weighting

Confusing symbols in the Multivariate case section

The notation used in the section on the multivariate case is quite confusing, in the that the $\Sigma$ is used to indicate both a sum and a covariance matrix. Additionally, the symbol $Var$ is used to denote a covariance matrix, whereas in the rest of the article is used to mean variance.

I have boldly edited the equations to use the more common symbol $\mathbf {C}$ for covariance matrices. The older formulae are retained below.

$\Sigma _{i}$ of the individual estimates $x_{i}$:

${\hat {x}}=\left(\sum _{i}\Sigma _{i}^{-1}\right)^{-1}\sum _{i}\Sigma _{i}^{-1}x_{i}$

$Var({\hat {x}})=\left(\sum _{i}\Sigma _{i}^{-1}\right)^{-1}$ — Preceding unsigned comment added by Glopk (talk • contribs) 16:11, 8 September 2023 (UTC)

Derivation from maximum likelihood?

Let there be a set of $n$ measurements $x_{i}$, each with uncertainty $\sigma _{i}$, of a variable $X$. A "gaussian" probability distribution function of $x$ with respect to each measurment is:

$p_{i}(x)\propto {\frac {1}{\sigma _{i}}}\mathrm {e} ^{\frac {\left(x-x_{i}\right)^{2}}{2\sigma _{i}^{2}}}$

The log-likelihood of $X=x$ given the measurements, (${\mathcal {L}}$ could be multiplied with -1, doesn't matter):

${\mathcal {L}}(x)=-\sum _{i=1}^{n}\log(\sigma _{i})+\sum _{i=1}^{n}{\frac {\left(x-x_{i}\right)^{2}}{2\sigma _{i}^{2}}}$

Finding $x$ that maximizes likelihood should give the "best" estimator of the weighted-mean of the $x_{i}$ values, taking the uncertainties into account:

${\frac {\partial {\mathcal {L}}(x)}{\partial x}}=\sum _{i=1}^{n}{\frac {x-x_{i}}{\sigma _{i}^{2}}}=0$

So then, from the above the "best" ${\hat {x}}$ is:

${\hat {x}}={\frac {\sum _{i=1}^{n}{\frac {x_{i}}{\sigma _{i}^{2}}}}{\sum _{i=1}^{n}{\frac {1}{\sigma _{i}^{2}}}}}$

Decomposing the variance of ${\hat {x}}$, we get:

$\mathrm {V} [{\hat {x}}]=\mathrm {V} \left[{\frac {\sum _{i=1}^{n}{\frac {x_{i}}{\sigma _{i}^{2}}}}{\sum _{i=1}^{n}{\frac {1}{\sigma _{i}^{2}}}}}\right]=\left(\sum _{i=1}^{n}{\frac {1}{\sigma _{i}^{2}}}\right)^{-2}\left(\sum _{i=1}^{n}\mathrm {V} \left[\sum _{i=1}^{n}{\frac {x_{i}}{\sigma _{i}^{2}}}\right]\right)=\left(\sum _{i=1}^{n}{\frac {1}{\sigma _{i}^{2}}}\right)^{-2}\cdot \sum _{i=1}^{n}{\frac {1}{\sigma _{i}^{4}}}\mathrm {V} \left[x_{i}\right]=$ $=\left(\sum _{i=1}^{n}{\frac {1}{\sigma _{i}^{2}}}\right)^{-2}\cdot \sum _{i=1}^{n}{\frac {1}{\sigma _{i}^{2}}}=\left(\sum _{i=1}^{n}{\frac {1}{\sigma _{i}^{2}}}\right)^{-1}$ Blakut (talk) 08:52, 14 June 2023 (UTC)