Best window length for a moving-average forecast

Question

You forecast a slowly drifting quantity (say, a rolling mean that creeps up by $\delta$ per step) using the average of the last $W$ observations. Each observation carries noise of variance $\sigma^2 = 32$ , and the drift is $\delta = 1$ per step. A window of length $W$ has forecast error

$\text{MSE}(W) \approx \frac{\sigma^2}{W} + \left(\frac{\delta W}{2}\right)^2,$

where the first term is sampling noise and the second is the bias from averaging stale, out-of-date points.

What window length $W$ minimizes the forecast MSE?

Accepted Answer

The idea. A short window uses few points, so it is noisy (high variance) but current. A long window averages away noise but reaches back to stale data that no longer reflects the drifting present (high bias). The best window balances the two. Minimize. Differentiate the MSE with respect to W and set to zero: \frac{d}{dW}\left[\frac{\sigma^2}{W} + \frac{\delta^2 W^2}{4}\right] = -\frac{\sigma^2}{W^2} + \frac{\delta^2 W}{2} = 0. Rearrange: \frac{\delta^2 W^3}{2} = \sigma^2 \implies W^\star = \left(\frac{2\sigma^2}{\delta^2}\right)^{1/3}. Plug in \sigma^2 = 32, \delta = 1: W^\star = \left(\frac{…

Best window length for a moving-average forecast

Your answer

Discussion

Learn the concepts

Related questions