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
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'
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