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Weighted arithmetic mean
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==Mathematical definition== Formally, the weighted mean of a non-empty finite [[tuple]] of data <math>\left( x_1, x_2, \dots , x_n \right)</math>, with corresponding non-negative [[weight function|weights]] <math>\left( w_1, w_2, \dots , w_n \right)</math> is :<math>\bar{x} = \frac{ \sum\limits_{i=1}^n w_i x_i}{\sum\limits_{i=1}^n w_i},</math> which expands to: :<math>\bar{x} = \frac{w_1 x_1 + w_2 x_2 + \cdots + w_n x_n}{w_1 + w_2 + \cdots + w_n}.</math> Therefore, data elements with a high weight contribute more to the weighted mean than do elements with a low weight. The weights may not be negative in order for the equation to work{{efn|Technically, negatives may be used if all the values are either zero or negatives. This fills no function however as the weights work as [[absolute values]].}}. Some may be zero, but not all of them (since division by zero is not allowed). The formulas are simplified when the weights are normalized such that they sum up to 1, i.e., <math display=inline>\sum\limits_{i=1}^n {w_i'} = 1</math>. For such normalized weights, the weighted mean is equivalently: :<math>\bar {x} = \sum\limits_{i=1}^n {w_i' x_i}</math>. One can always normalize the weights by making the following transformation on the original weights: :<math>w_i' = \frac{w_i}{\sum\limits_{j=1}^n{w_j}}</math>. The [[ordinary mean]] <math display=inline>\frac {1}{n}\sum\limits_{i=1}^n {x_i}</math> is a special case of the weighted mean where all data have equal weights. If the data elements are [[independent and identically distributed random variables]] with variance <math>\sigma^2</math>, the ''standard error of the weighted mean'', <math>\sigma_{\bar{x}}</math>, can be shown via [[uncertainty propagation]] to be: :<math display="inline"> \sigma_{\bar{x}} = \sigma \sqrt{ \sum\limits_{i=1}^n w_i'^2 } </math> ===Variance-defined weights=== {{main|Inverse-variance weighting}} {{see also|Weighted least squares}} For the weighted mean of a list of data for which each element <math>x_i</math> potentially comes from a different [[probability distribution]] with known [[variance]] <math>\sigma_i^2</math>, all having the same mean, one possible choice for the weights is given by the reciprocal of variance: :<math>w_i = \frac{1}{\sigma_i^2}.</math> The weighted mean in this case is: :<math>\bar{x} = \frac{ \sum_{i=1}^n \left( \dfrac{ x_i}{\sigma_i^{2}} \right)}{\sum_{i=1}^n \dfrac{1}{\sigma_i^{2}}} =\frac{\sum_{i=1}^n\left(x_i\cdot w_i\right)}{\sum_{i=1}^n w_i},</math> and the ''standard error of the weighted mean (with inverse-variance weights)'' is: :<math>\sigma_{\bar{x}} = \sqrt{\frac{ 1 }{\sum_{i=1}^n \sigma_i^{-2}}} =\sqrt{\frac{1}{\sum_{i=1}^n w_i}},</math> Note this reduces to <math> \sigma_{\bar{x}}^2 = \sigma_0^2/n</math> when all <math>\sigma_i = \sigma_0</math>. It is a special case of the general formula in previous section, :<math> \sigma^2_{\bar x} = \sum_{i=1}^n {w_i'^2 \sigma^2_i} = \frac{ \sum_{i=1}^n {\sigma_i^{-4} \sigma^2_i} }{\left(\sum_{i=1}^n \sigma_i^{-2}\right)^2}.</math> The equations above can be combined to obtain: :<math>\bar{x} = \sigma_{\bar{x}}^2 \sum_{i=1}^n \frac{x_i}{\sigma_i^2}.</math> The significance of this choice is that this weighted mean is the [[maximum likelihood estimator]] of the mean of the probability distributions under the assumption that they are independent and [[normally distributed]] with the same mean.
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