What is the distribution of the distance between 2 random points in a cube?

Question

Assume 3 random variables $x,y,z \sim U(0,1)$, which we consider to be a 3D coordinates in a cartesian box.

Assume we define now $k(x_1,y_1,z_1,x_2,y_2,z_2) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$

What's the distribution of $k$?

My try is:

Consider 2 random variables $x_1, x_2 \sim U(0,1)$ then define $p=x_1 - x_2$.

Clearly $p \sim U(-1,1)$, therefore $\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2} \sim U(0,\sqrt{3})$.

Your $p$ is not $\sim U(-1,1)$; it is triangularly distributed. The correct answer is below! — Parcly Taxel, Oct 25 '17 at 03:39

score 4 · Accepted Answer · answered Oct 25 '17 at 03:37

4

MathWorld calls this cube line picking and gives a closed form for the distribution due to Mathai et al. $$P(k=d)=\begin{cases} -d^2((d-8)d^2+\pi(6d-4))&0\le d\le 1\\ 2d((d^2-8\sqrt{d^2-1}+3)d^2-4\sqrt{d^2-1}+12d^2\sec^{-1}d+\pi(3-4d)-1/2)&1\le d\le\sqrt2\\ d((1+d^2)(6\pi+8\sqrt{d^2-2}-5-d^2)-16d\csc^{-1}\sqrt{2-2/d^2}+16d\tan^{-1}d\sqrt{d^2-2}-24(d^2+1)\tan^{-1}\sqrt{d^2-2})&\sqrt2\le d\le\sqrt3 \end{cases}$$

answered Oct 25 '17 at 03:37

Parcly Taxel

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Is this only for the unit cube though? If not, does a known extension exist? – Teg Louis Jul 06 '23 at 13:36

What is the distribution of the distance between 2 random points in a cube?

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