Pairwise distance distribution for point clouds (normal distribution) I have a point cloud (2D for now) of $N$ normally distributed points (with a certain $\sigma$).
My first question would be how the pairwise distance distribution looks (just by chance I discovered a normal distribution folded by the error function does the job well in fitting the distance distribution, but I have problems in quantifying the parameter like normalization factor).
The second question (probably easy if the first part is solved): I now have a second point cloud nearby (distance $d$ to the first cloud, also normal distribution, same $\sigma$, $N_2$ number of points). How does that change the distribution (that now of course has an additional peak).
Any input is appreciated. Thanks a lot (this is my first post - I am sorry if I posted this incorrectly)!
Below you find links to images of the distribution for an example population (the fit in the second picture is just two gauss folded with an error function - as mentioned: this fits nicely but I am not sure about the parameter relation).
point cloud: http://i.imgur.com/2d9ir6P.png
pdist histogram: http://i.imgur.com/jPhdlvL.png
 A: For your first question:
Let $X_1$ and $X_2$ be two points. they are normally distributed, so their difference is normally distributed:
$$ X_1 - X_2 \sim N(0,2\sigma^2) $$
You are asking a question about the norm of that difference. A quick google gives us that the norm of a Gaussian random-variable is a Rayleigh distribution and that the square norm is Gamma (in case you would need that)
For your second question, you now have two cases for your distance: are you measuring the distance between two points in the same cloud or in different clouds ? your distribution of norms is thus a mixture, with each component corresponding to one of those cases.
The first component (when the two points come from the same distribution) is the Rayleigh distribution of question 1. The second component is the norm of a Gaussian which is not centered at 0. I'm sure somebody has solved that before and that you can find it
A: Since I am not able to edit my comment ... here is the solution again, nicely formatted:
For two point clouds spaced $d$ with individual sigma $c$ and a total of $N$ events:
We get a total of $N_D=(N-1)\cdot N/2$ distances. As mentioned by Guillaume the total PDF is a sum of distances inside one cluster - represented by the Rayleigh distribution (with $\sigma=\sqrt{2}\cdot c$) and the distances between the two cluster by the Rice distribution with the distance $d$ as the first parameter. So we get:
$P=N_D \cdot P_{rayleigh}(x,\sqrt{2}\cdot c) + N_D \cdot P_{rice}(x,d,\sqrt{2}\cdot c)$
If the events are not equally distributed you need an additional scaling factor.
Note that the rayleigh distribution is a special case of the rice distribution for $d=0$.
