Thursday, 22 August 2013

Mean of moving average

Mean of moving average

Define an interval $[0,\dots,n]$ and let $A_i=1$ for $k$ distinct values
of $i$ and $A_i = 0$ otherwise.
Now, for an EMA we have the basic formula: $$\mathrm{EMA} = S_i = S_{i-1}
+ \alpha * (A_i - S_{i-1})$$ Where $\alpha$ is a weighting factor $(0 <
\alpha < 1)$.
Is it true that the average of $S$ over the interval $[0,\dots,n]$ simply
equals $\alpha k/n$?

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