Talk:Inverse-variance weighting: Difference between revisions

Statistics Stub‑class Mid‑importance

This article is within the scope of WikiProject Statistics, a collaborative effort to improve the coverage of statistics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.StatisticsWikipedia:WikiProject StatisticsTemplate:WikiProject StatisticsStatistics articles Stub This article has been rated as Stub-class on Wikipedia's content assessment scale. Mid This article has been rated as Mid-importance on the importance scale.

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

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

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

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

Finding ${\displaystyle x}$ that mini/maximizes ${\displaystyle {\mathcal {L}}(x)}$ should give the "best" estimator of the weighted-mean of the ${\displaystyle x_{i}}$ values, taking the uncertainties into account:

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

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

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

Decomposing the variance from the formula, we get:

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