Matching Exponential and Simple Moving Averages by Average Data Age
Summary
The note derives the familiar link between the smoothing factor of an exponential moving average (EMA) and the period of a simple moving average. It defines each observation’s age as the number of periods back from the current point. A simple moving average over N observations gives each observation equal weight, so its average age is (N−1)/2. For an EMA with constant alpha, the geometrically weighted average age is (1−alpha)/alpha.
Equating those average ages yields alpha = 2/(N+1), the stated period-to-smoothing-factor rule. The note attributes this derivation to Brown’s 1963 book on smoothing and forecasting. This is a matching convention based on equal average age; it does not establish that the two averages have identical weights or produce identical outputs, and it does not assess which smoothing choice works best for a trading use case.
Key ideas
- A simple moving average over N observations has an average observation age of (N−1)/2 periods.
- For a constant-alpha EMA, the geometrically weighted observations have average age (1−alpha)/alpha.
- Equating the two average ages gives the conventional smoothing-factor formula alpha = 2/(N+1).
- This equivalence matches average age, not the complete weighting profile of the two averages.
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# Answer by Alex C (score 3, accepted)
# Derivation (or proof) of commonly used formula showing relationship between time and smoothing factor in exponential smoothing
I am using simple exponential smoothing on historical market prices. There is a commonly held view (among market practitioners), that there is a simple relation between the period over which the data is being smoothed - and the smoothing factor (alpha).
The formula is often given (for simple exponential smoothing) as:
```
alpha = 2/(period + 1)
```
Is there any literature out there that proves this relationship?
## Answer by Alex C (score 3, accepted)
https://quant.stackexchange.com/a/33101
The current data point is said to have age 0, the previous has age 1, and so on going backwards.
For a straight N period moving average of the form $\frac{1}{N}(x_t+x_{t-1}+\cdots+x_{t-N+1})$ it is easy to see that the average age of the data is $\frac{N-1}{2}$. Sometimes this is stated in term of "centering": a 3 period moving average is centered on the period $t-1$, i.e the period with age $1=\frac{3-1}{2}$.
A slightly more elaborate calculation shows that for an exponential moving average with constant $\alpha$, the average age of the data is $\frac{1-\alpha}{\alpha}$. (The EMA is of the form $\sum_{k=0}^\infty\alpha(1-\alpha)^k x_{t-k}$ and the average age is $\sum_{k=0}^\infty\alpha(1-\alpha)^k k$ which can be show to converge to $\frac{1-\alpha}{\alpha}$).
To find the EMA most similar to a given MA, we set these two expressions for average age equal, giving the equation $\frac{N-1}{2}=\frac{1-\alpha}{\alpha}$. Solving this for alpha we get $\alpha=\frac{2}{N+1}$ . QED.
This proof was given by Brown in his 1963 book 'Smoothing, Forecasting and Prediction'Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.