image processing - Are 2nd Order Edge Detectors More Susceptible to Noise?

I am reading about edge detection and I read that 2nd order detectors are more susceptible to noise. Is there a mathematical proof to this ?

Answer

Suppose that the noise is a random vector $X$ with normal zero-mean components of variance $\sigma_i$, mutually independent, then for the linear combination (the $g_i$ being for instance coefficients of a FIR filter): $$Y = \sum_i g_i X_i\,,$$ the variance of $Y$ will be: $$V(Y) =\sum_i g_i^2 \sigma_i^2\,,$$ which boils down to $$\|g\|_2^2 \sigma_2^2\,$$ for a classical stationnary Gaussian noise.

Now start from $g^1=[1,-1]$, and assume you built higher order derivative by self-convolution, for instance $$g^2=g^1 \ast g^1 = [1 ,-2 ,1],$$ $$g^3 = g^1 \ast g^1 \ast g^1 =[1,-3,3,1] ,$$ etc.

You easily get that the coefficients of $g^n$ are $$g^n_k=(-1)^k \binom{n}{k}\,.$$

Hence $\|g^n\|_2^2 = \sum_k \binom{n}{k} ^2 = \binom{2n}{n}$ (a famous identity). For $n=1$, the amplification is $2$, for $n=2$ it will be $6$, and $20$ for the third derivative.

So the noise amplification roughly grows with the derivative order (based on Stirling's formula) as $$\frac{4^n}{\sqrt{\pi n}}\,.$$

So all in all, noise power tends to explode with derivatives.

This does not happen at order $0$, if you consider that traditional low-pass filters having positive coefficients have finite sum, often equal to $1$. Hence, for those $\|g\|_2 \le \|g\|_1=1$. For instance, if one takes standard smoothing "discrete Gaussian" filters of order $m$ with coefficients $$b^m_k=\frac{ \binom{m}{k}}{2^m}\,,$$ then their $\ell_2$ squared norm is: $$\frac{\binom{2m}{m}}{4^m}$$ and one sees that the $4^m$ term cancels in the asymptotics, and you get that, when $m$ grows: $$\|h^m\|_2^2 \approx\frac{1}{\sqrt{\pi m}}\,,$$ and thus tend to reduce the noise power.

Finally, this is not the whole story, since we have not talked about edges, which traditionally are wide-band, and especially hard to detect in high frequencies. A study in the Fourier domains may lead to others aspects, that exceed my time today, but is addressed in @Olli Niemitalo's answer.