I was reading a paper and saw this is mentioned there, but I cannot figure out how this can analytically be proven?


Simply put: take an independent identically distributed Gaussian noise (one observation in blue, left) and its gradient (in green, right). The amplitude is roughly multiplied by the norm of the gradient filter (here $h=[1\,-1]$), which is $\sqrt{2}$, which you can see from the picture.

Noise and Gradient

Their average energy $E$ thus differ by a factor of $|h\|^2=\sqrt{2}^2 = 2$. As said by Olli, a similar question was answered in 2nd order edge detectors more susceptible to noise?. The part "can be magnified twice" is important, since this does not happen for all noises or all difference filters.

More generally, a random signal $x(t)$ passing through a time-invariant filter with frequency response $H(f)$ produces a signal $y(t)$ with PSD:

$$ S_y(f) = S_x(f) |H(f)|^2\,. $$

Since the Fourier domain is energy-preserving, you get that

$$\|H\|^2\ =|h\|^2=2\,.$$

This is a reason why people have developed noise-robust differentiation, often combined with smoothing (eg Savitzky-Golay filters). See examples at Differentiation.

Note that in the continuous case, you should take care of non integrable processes, see for instance Variance of white Gaussian noise.


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