Gaussian Blur - Standard Deviation, Radius and Kernel Size
What's the relationship between sigma and radius? I've read that sigma is equivalent to radius, I don't see how sigma is expressed in pixels. Or is "radius" just a name for sigma, not related to pixels?


There are three things at play here. The variance, ($\sigma^2$), the radius, and the number of pixels. Since this is a 2-dimensional gaussian function, it makes sense to talk of the covariance matrix $\boldsymbol{\Sigma}$ instead. Be that as it may however, those three concepts are weakly related. 

First of all, the 2-D gaussian is given by the equation:

$$
g({\bf z}) = \frac{1}{\sqrt{(2 \pi)^2 |\boldsymbol{\Sigma}|}} e^{-\frac{1}{2} ({\bf z}-\boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} \ ({\bf z}-\boldsymbol{\mu})}
$$

Where ${\bf z}$ is a column vector containing the $x$ and $y$ coordinate in your image. So, ${\bf z} = \begin{bmatrix} x \\ y\end{bmatrix}$, and $\boldsymbol{\mu}$ is a column vector codifying the mean of your gaussian function, in the $x$ and $y$ directions $\boldsymbol{\mu} = \begin{bmatrix} \mu_x \\ \mu_y\end{bmatrix}$.

Example: 

Now, let us say that we set the covariance matrix $\boldsymbol{\Sigma} = \begin{bmatrix} 1   &amp; 0 \\ 0 &amp; 1\end{bmatrix}$, and $\boldsymbol{\mu} = \begin{bmatrix} 0 \\ 0\end{bmatrix}$. I will also set the number of pixels to be $100$ x $100$. Furthermore, my 'grid', where I evaluate this PDF, is going to be going from $-10$ to $10$, in both $x$ and $y$. This means I have a grid resolution of $\frac{10 - (-10)}{100} = 0.2$. But this is completely arbitrary. With those settings, I will get the probability density function image on the left. Now, if I change the 'variance', (really, the covariance), such that $\boldsymbol{\Sigma} = \begin{bmatrix} 9   &amp; 0 \\ 0 &amp; 9\end{bmatrix}$ and keep everything else the same, I get the image on the right. 



The number of pixels are still the same for both, $100$ x $100$, but we changed the variance. Suppose instead we do the same experiment, but use $20$ x $20$ pixels instead, but I still ran from $-10$ to $10$. Then, my grid has a resolution of $\frac{10-(-10)}{20} = 1$. If I use the same covariances as before, I get this:



These are how you must understand the interplay between those variables. If you would like the code, I can post that here as well.


  How do I choose sigma?


The choice of the variance/covariance-matrix of your gaussian filter is extremely application dependent. There is no 'right' answer. That is like asking what bandwidth should one choose for a filter. Again, it depends on your application. Typically, you want to choose a gaussian filter such that you are nulling out a considerable amount of high frequency components in your image. One thing you can do to get a good measure, is compute the 2D DFT of your image, and overlay its co-efficients with your 2D gaussian image. This will tell you what co-efficients are being heavily penalized. 

For example, if your gaussian image has a covariance so wide that it is encompassing many high frequency coefficients of your image, then you need to make its covariance elements smaller.