Editing Weighted arithmetic mean (section)

===Vector-valued estimates===
The above generalizes easily to the case of taking the mean of vector-valued estimates. For example, estimates of position on a plane may have less certainty in one direction than another. As in the scalar case, the weighted mean of multiple estimates can provide a [[maximum likelihood]] estimate. We simply replace the variance <math>\sigma^2</math> by the [[covariance matrix]] <math>\mathbf{C}</math> and the [[arithmetic inverse]] by the [[matrix inverse]] (both denoted in the same way, via superscripts); the weight matrix then reads:<ref>{{cite book |last=James |first=Frederick |title=Statistical Methods in Experimental Physics|year=2006|publisher=World Scientific|location=Singapore|isbn=981-270-527-9| edition=2nd|page=324}}</ref>

<math display="block"> \mathbf{W}_i = \mathbf{C}_i^{-1}.</math>

The weighted mean in this case is:
<math display="block">\bar{\mathbf{x}} = \mathbf{C}_{\bar{\mathbf{x}}} \left(\sum_{i=1}^n \mathbf{W}_i \mathbf{x}_i\right),</math>
(where the order of the [[matrix–vector product]] is not [[commutative]]), in terms of the covariance of the weighted mean:
<math display="block">\mathbf{C}_{\bar{\mathbf{x}}} = \left(\sum_{i=1}^n \mathbf{W}_i\right)^{-1},</math>

For example, consider the weighted mean of the point [1 0] with high variance in the second component and [0 1] with high variance in the first component. Then

: <math>\mathbf{x}_1 := \begin{bmatrix}1 & 0\end{bmatrix}^\top, \qquad \mathbf{C}_1 := \begin{bmatrix}1 & 0\\ 0 & 100\end{bmatrix}</math>
: <math>\mathbf{x}_2 := \begin{bmatrix}0 & 1\end{bmatrix}^\top, \qquad \mathbf{C}_2 := \begin{bmatrix}100 & 0\\ 0 & 1\end{bmatrix}</math>

then the weighted mean is:

: <math>
\begin{align}
\bar{\mathbf{x}} & = \left(\mathbf{C}_1^{-1} + \mathbf{C}_2^{-1}\right)^{-1} \left(\mathbf{C}_1^{-1} \mathbf{x}_1 + \mathbf{C}_2^{-1} \mathbf{x}_2\right) \\[5pt]
& =\begin{bmatrix}  0.9901 &0\\ 0& 0.9901\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}0.9901 \\ 0.9901\end{bmatrix}
\end{align}
</math>

which makes sense: the [1 0] estimate is "compliant" in the second component and the [0 1] estimate is compliant in the first component, so the weighted mean is nearly [1 1].