[GeeTwo]

Originally Posted by Richard Wallace

So the distribution of lengths is Rayleigh. Neat example.

What would the distribution look like if the line segments were contained in a cube (3D) boundary? That one may be easier to visualize using a Monte Carlo simulation.

Originally Posted by Caleb Sykes

Rayleigh distributions can't have upper bounds though, right? This distribution will clearly have an upper bound at sqrt(2).

Similar, but not quite a Rayleigh Distribution.

Originally Posted by Rayleigh distribution

The Rayleigh distribution, named for William Strutt, Lord Rayleigh, is the distribution of the magnitude of a two-dimensional random vector whose coordinates are independent, identically distributed, mean 0 normal variables.

After going to signed x ̅ = x₁-x₂ and y ̅ = y₁-y₂ components, but before collapsing the four quadrants, this problem is to evaluate the mean of the magnitude of a two-dimensional random vector whose coordinates are independent, identcially distributed, mean 0 triangular variables. That is, x ̅ and y ̅ are defined by the triangle function, max(0,1-abs(x)), rather than the normal distribution which would mean one proportional to e^-ax2 (aka the Gaussian "bell curve"). As Caleb notes, this distribution, like the normal distribution, has no upper bound (an infinitely long tail).

If the working space were a cube, the small end of the distribution would start out with zero slope and a parabolic ramp of 4πr². It would peak at a somewhat higher length (WAG of 0.6), have a similar change of curvature at length 1, and have an upper bound at √3.

Spoiler for Did you notice?:

For those who noticed, yes, the 2πr in the square case is the length of a circle of radius r, and 4πr² is the area of a sphere of radius r in the cubic case. In the near-zero-radius limit, the size of the locus of points r units away from any given point is the population density.