You are likely used to computing the arithmetic mean (average) of a set of numbers. For $N$ values $(x_1,x_2,x_3,\dots,x_N)$, the mean is
| $\overline x = \dfrac{1}{N} \displaystyle \sum_{i=1}^N x_i$ | (1) |
Suppose, however, that we do not want each value to contribute equally. Instead, we may want to weight some values more (or less) heavily in the mean than others since we have more (or less) confidence in their value. If each value has an associated uncertainty $\sigma$ , – so that we have $(x_1 \pm\sigma_1,x_2 \pm\sigma_2,x_3 \pm\sigma_3,\dots,x_N\pm\sigma_N)$ – an appropriate weight might be $w_i = 1/\sigma_i^2$. Such a choice means that having a large error bar weights that point less than if it had a small error bar. Now, when we take the average, we sum over all weighted values $w_ix_i$ and divide by an appropriate normalization such that
| $\overline x = \dfrac{ \sum_{i=1}^N w_ix_i}{\sum_{i=1}^N w_i} = \dfrac{ \sum_{i=1}^N x_i/\sigma^2_i}{\sum_{i=1}^N 1/\sigma^2_i}$ | (2) |
Notice that when all the values have equal weight, ($w_i = 1$ for all $i$ ), we recover Eq. 1. As each of our values had an uncertainty, we want to propagate that uncertainty through the average. Crunching the numbers, we find that the uncertainty in the mean is given by
| $\sigma^2_{\overline{x}} = \dfrac{1}{\sum_{i=1}^N w_i}= \dfrac{1}{\sum_{i=1}^N 1/\sigma_i^2}$. | (3) |