How the best window scales with noise

Question

The MSE-optimal moving-average window for a drifting quantity is $W^\star = (2\sigma^2/\delta^2)^{1/3}$ , where $\sigma^2$ is observation noise and $\delta$ is drift per step. Suppose the observation noise $\sigma^2$ suddenly increases by a factor of 8, with the drift unchanged.

By what factor should the optimal window length change?

Accepted Answer

The idea. The window depends on the noise only through a cube root, so multiplying the noise scales the window by the cube root of that same factor. You do not need the actual numbers. Apply the scaling. Since W^\star \propto (\sigma^2)^{1/3}, multiplying \sigma^2 by 8 multiplies W^\star by 8^{1/3} = 2. The window should double. Even though the noise grew eightfold, the optimal window only doubles, because the cube root heavily dampens the response. This is why practical smoothing windows are fairly insensitive to noise estimates: you can misjudge \sigma^2 by a large factor and be off on the …

How the best window scales with noise

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