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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!

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  • $\begingroup$ Hint: Fourier transform of convolution. $\endgroup$ – Lenar Hoyt Nov 1 '15 at 13:54
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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 $\sqrt{n}\sigma$.

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