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Why is the smoothing coefficient of the EMA (exponential moving average) calculated as:

$${\displaystyle \alpha =2/(N+1)}?$$

Brown R.G, on page 107 of "Smoothing, forecasting and prediction of discrete time series (1963)" goes about giving an explanation using the following folowing derivation for what he calls the average age of the data set:

(1) $$avg=0\alpha+1\alpha\beta+2\alpha\beta^2+...$$ (2) $$avg=\alpha\sum_{k=0}^{\infty}k\beta^k$$ (3) $$avg=\frac{\beta}{\alpha}$$

He then goes about saying that if want the exponential average age of the data set (3) to be the same as the simple (N-datapoints) moving average you just need to solve the equation:

$${\displaystyle \beta/\alpha =(N-1)/2}?$$

Or in plain english:

The exponential average age of the data = simple N average age of the data.

It's all fine but I cannot bridge the gap between equations 2 and 3, how did he solve the power serie into that fraction?

Note:

"The average age is the age of each piece of data used in the average, weighted as the data of that age would be weighted. In the exponential smoothing process, the weight given data k periods ago is is:

$$\alpha\beta^k, \beta=(1-\alpha)$$

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Bridging the gap between equation 2 and 3:

Let us use the fact that the infinite series $\sum_{k=0}^\infty k \cdot \beta^k$ has a closed-form solution given by (see this post for proof):

$$\sum_{k=0}^\infty k \cdot \beta^k = \frac{\beta}{(1-\beta)^2},$$

when $|\beta| = |1-\alpha|<1$ or $\alpha < 1$, for the infinite series to converge to the above result.


Now, let us calculate the average in a few more steps:

\begin{align} \bar{k} &= \alpha \sum_{k=0}^\infty k \cdot \beta^k \\ &= \alpha \cdot \frac{\beta}{(1-\beta)^2} \\ &= \alpha \cdot \frac{\beta}{\left(1-(1-\alpha)\right)^2}\\ &= \alpha \cdot \frac{\beta}{\alpha^2}\\ &= \frac{\beta}{\alpha} \end{align}

I hope this clarifies some things.

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    $\begingroup$ That's perfect. Thank you! $\endgroup$
    – MBdr
    Nov 30 '21 at 0:58

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