What is the standard deviation of a gaussian filter? I was referring to this thesis, High dimensional gaussian filtering for computational photography (or pdf), for understanding gaussian filters.
However, I didn't understand one part. The gaussian filter is given by
$$
\hat{v_i} = \sum_{j}\exp(-|p_i-p_j|^2 / 2)\,v_j
$$
If $p_i$ and $p_j$ are position vectors then it is simply the Gaussian blur. Now I didn't understand if we divide $p_i$ and $p_j$ by a certain constant, the standard deviation of the filter is changed. I didn't get how it occurs. Also I didn't get what you mean by the standard deviation of a filter?
 A: This question actually has (next to) nothing to do with "standard deviation" in the sense understood in statistics.  A more precise term for what is being changed is the "radius" or "half-width" of the filter.  Whatever it is called--standard deviation, radius, whatever--it always refers to the typical distance over which the kernel (here, $e^{-|p_i-p_j|^2/2}$) has appreciable magnitude.
When you uniformly rescale the $p_i$, you are effectively changing the units of distance measurement.  (For instance, dividing 12 feet by 3 converts it to 4 yards.)  It's the same distance but it is expressed with a different number.  But because the numeric value of this distance has changed, the values of the kernel change too.
It helps to know that most kernels (and the Gaussian is typical) attain their highest values at 0 (i.e., when $p_i = p_j$) and decrease from there.  Thus, decreasing the numeric value in the kernel's argument (which is never negative) increases the kernel's value.  This means that positions that once were so far apart to contribute essentially nothing to the kernel might now start contributing something.  In short, shrinking the numbers increases the apparent "reach" of the kernel itself: its radius (aka "standard deviation") extends proportionately.
