Editing Weighted arithmetic mean (section)

===Exponentially decreasing weights===
{{see also|Exponentially weighted moving average}}
In the scenario described in the previous section, most frequently the decrease in interaction strength obeys a negative exponential law. If the observations are sampled at equidistant times, then exponential decrease is equivalent to decrease by a constant fraction <math>0<\Delta<1</math> at each time step. Setting <math>w=1-\Delta</math> we can define <math>m</math> normalized weights by
: <math>w_i=\frac {w^{i-1}}{V_1},</math>
where <math>V_1</math> is the sum of the unnormalized weights. In this case <math>V_1</math> is simply
: <math>V_1=\sum_{i=1}^m{w^{i-1}} = \frac {1-w^{m}}{1-w},</math>
approaching <math>V_1=1/(1-w)</math> for large values of <math>m</math>.

The damping constant <math>w</math> must correspond to the actual decrease of interaction strength. If this cannot be determined from theoretical considerations, then the following properties of exponentially decreasing weights are useful in making a suitable choice: at step <math>(1-w)^{-1}</math>, the weight approximately equals <math>{e^{-1}}(1-w)=0.39(1-w)</math>, the tail area the value <math>e^{-1}</math>, the head area <math>{1-e^{-1}}=0.61</math>. The tail area at step <math>n</math> is <math>\le {e^{-n(1-w)}}</math>. Where primarily the closest <math>n</math> observations matter and the effect of the remaining observations can be ignored safely, then choose <math>w</math> such that the tail area is sufficiently small.