# How Does Gaussian Blur Affect Image Variance

I am trying to find a way to analytically estimate how the noise (or variance in pixel values) will be affected for a zero-frequency image of random noise after applying a Gaussian blur.

In more detail, say I have an image of pixels with an average $μ$ and random noise of variance $σ_{0}^{2}$. The image has no features in it, but is simply that normal noise. I now apply a Gaussian blur (or Gaussian filter) with a kernel of standard deviation $σ_{G}$. I would like a way to analytically estimate (and I know it will have to be an approximation) what the variance of the pixel values in the filtered image will be ($σ_{f}^{2}$).

Wikipedia has an equation that clearly states what I need. However, they do not have any references and I will like to have a stronger reference for it. I have been unable to find this anywhere else. Any ideas where I'll be able to find this? Does anyone know if that is the correct equation? And maybe how to derive it?

In Gaussian blur the value of each output pixel is calculated as a weighted sum of all input pixels: $$\text{out}(x, y) = \sum^\infty_{j = -\infty} \sum^\infty_{i = -\infty} \frac{1}{{2\pi \sigma_G^2}} e^{-\frac{i^2 + j^2}{2 \sigma_G^2}}\text{in}(x+i,y+j).$$ We want to calculate the variance $\text{Var}[\text{out}(x, y)] = \sigma^2_f$ based on the variances $\text{Var}[\text{in}(x + i, y + i)] = \sigma^2_0$. We require that $\text{in}(x + i, y + i)$ are independent, which you should mention in your question. Based on the identity (variance of linear combination): $$Var\left[\sum_i c_i X_i\right] = \sum_i c^2_i Var[X_i],$$ where $c_i$ are constants and $X_i$ are independent random variables, $$\sigma^2_f = \sigma^2_0\sum^\infty_{j = -\infty} \sum^\infty_{i = -\infty} \left(\frac{1}{{2\pi \sigma_G^2}} e^{-\frac{i^2 + j^2}{2 \sigma_G^2}}\right)^2.$$ For a large $\sigma_G$, the squared Gaussian is rather smooth and the sum can be approximated by an integral as: $$\sigma^2_f \approx \sigma^2_0\int^\infty_{-\infty} \int^\infty_{-\infty} \left(\frac{1}{{2\pi \sigma_G^2}} e^{-\frac{i^2 + j^2}{2 \sigma_G^2}}\right)^2\,\text{d}i\,\text{d}j = \frac{\sigma^2_0}{4\pi \sigma_G^2}\\ \implies \sigma_f \approx \frac{\sigma_0}{2\sigma_G\sqrt{\pi}},$$ which is the same result as in Wikipedia. You can calculate the integral using Wolfram Alpha.