I am taking a class related to image processing and we were taught about Gaussian Filters that are related to the following Gaussian Function:
$$G(u,v) = \frac{1}{2\pi\sigma^2}e^{-\frac{u^2 + v^2}{2\sigma^2}}$$
We were told in class that if apply a convolution with standard deviation $\sigma$ twice with a Gaussian filter it is equivalent to applying a Gaussian filter with standard deviation of $\sqrt{2}\sigma$. I want to actually be able to prove this but I cannot for the life of me do this, I have found resources related to Fourier analysis but I am not familiar with Fourier analysis and we are not expected to know Fourier analysis. I tried (probably naively) just multiplying two Gaussian functions with each other and then setting $\sigma' = \sqrt{2}\sigma$ to see if I can get an equivalence and I cannot get it to work. Any help would be greatly appreciated!
The important thing to remember is that you are dealing with multiple convolutions gaussian filter. If you have any probability background this is no different to multiple draws from a gaussian distribution and thus the same laws apply.
Theorem: The normal distribution is stable. Specifically, suppose that X has the normal distribution with mean μ∈ℝ and variance $\sigma^2$∈(0,∞). If (X1,X2,…,Xn) are independent copies of X, then X1+X2+⋯+Xn has the same distribution as (n−$\sqrt{n}$)μ+$\sqrt{n}$X, namely normal with mean nμ and variance n$\sigma^2$.
A proof of this is under Theorem 27 of this link.
Now the the standard deviation is the square root of the variance and thus $\sqrt{n}\sigma$.