Back out the drift from the chosen window

Question

A team forecasts a drifting quantity with a moving average and uses window $W = 5$ , which they claim is MSE-optimal. The observation noise is $\sigma^2 = 250$ . The optimal window is $W^\star = (2\sigma^2/\delta^2)^{1/3}$ , where $\delta$ is the drift per step.

What drift $\delta$ does their choice imply?

Accepted Answer

The idea. The optimal window is set by the balance between noise and drift. If the window and the noise are known, the implied drift is pinned down. Invert the formula. Cube both sides of W^\star = (2\sigma^2/\delta^2)^{1/3}: (W^\star)^3 = \frac{2\sigma^2}{\delta^2} \implies \delta^2 = \frac{2\sigma^2}{(W^\star)^3}. Plug in W^\star = 5, \sigma^2 = 250: \delta^2 = \frac{2 	imes 250}{125} = \frac{500}{125} = 4 \implies \delta = 2. So a 5-point window is only justified if the quantity drifts by about 2 per step. If the true drift is faster than that, the 5-point window is too long: it leans on …

Back out the drift from the chosen window

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