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Point-Point Distance--1-D

Given a unit Line Segment $[0, 1]$, pick two points at random on it. Call the first point $x_1$ and the second point $x_2$. Find the distribution of distances $d$ between points. The probability of the points being a (Positive) distance $d$ apart (i.e., without regard to ordering) is given by

$\displaystyle P(d)$ $\textstyle =$ $\displaystyle {\int_0^1\int_0^1 \delta(d-\vert x_2-x_1\vert)\,dx_1\,dx_2\over \int_0^1\int_0^1 dx_1\,dx_2}$  
  $\textstyle =$ $\displaystyle (1-d)[H(1-d)-H(d-1)+H(d)-H(-d)]$  
  $\textstyle =$ $\displaystyle \left\{\begin{array}{ll} 2(1-d) & \mbox{for $0\leq d\leq 1$}\\  0 & \mbox{otherwise,}\end{array}\right.$ (1)

where $\delta$ is the Dirac Delta Function and $H$ is the Heaviside Step Function. The Moments are then
$\displaystyle \mu'_m$ $\textstyle =$ $\displaystyle \int_0^1 d^m P(d)\,dd = 2\int_0^1 d^m(1-d)\,dd$  
  $\textstyle =$ $\displaystyle 2\left[{{d^{m+1}\over m+1}-{d^{m+2}\over m+2}}\right]_0^1$  
  $\textstyle =$ $\displaystyle 2\left({{1\over m+1}-{1\over m+2}}\right)=2\left[{(m+2)-(m+1)\over (m+1)(m+2)}\right]$  
  $\textstyle =$ $\displaystyle {2\over (m+1)(m+2)}$  
  $\textstyle =$ $\displaystyle \left\{\begin{array}{ll} {1\over (n+1)(2n+1)} & \mbox{for $m=2n$}\\  {1\over (n+1)(2n+3)} & \mbox{for $m=2n+1$,}\end{array}\right.$ (2)

giving Moments about 0
$\displaystyle \mu'_1$ $\textstyle =$ $\displaystyle {\textstyle{1\over 3}}$ (3)
$\displaystyle \mu'_2$ $\textstyle =$ $\displaystyle {\textstyle{1\over 6}}$ (4)
$\displaystyle \mu'_3$ $\textstyle =$ $\displaystyle {\textstyle{1\over 10}}$ (5)
$\displaystyle \mu'_4$ $\textstyle =$ $\displaystyle {\textstyle{1\over 15}}.$ (6)

The Moments can also be computed directly without explicit knowledge of the distribution

$\displaystyle \mu'_1$ $\textstyle =$ $\displaystyle {\int_0^1 \int_0^1 \vert x_2-x_1\vert\,dx_1\,dx_2 \over \int_0^1 \int_0^1 dx_1\,dx_2}$  
  $\textstyle =$ $\displaystyle \int_0^1 \int_0^1 \vert x_2-x_1\vert\,dx_1\,dx_2$  
  $\textstyle =$ $\displaystyle {\int_0^1 \int_0^1\atop\scriptstyle x_2-x_1>0} (x_2-x_1)\,dx_1\,dx_2+ {\int_0^1 \int_0^1\atop\scriptstyle x_2-x_1<0} (x_1-x_2)\,dx_1\,dx_2$  
  $\textstyle =$ $\displaystyle \int_0^1 \int_{x_1}^1 (x_2-x_1)\,dx_1\,dx_2+ \int_0^1 \int_0^{x_1} (x_2-x_1)\,dx_1\,dx_2$  
  $\textstyle =$ $\displaystyle \int_0^1 \left[{{\textstyle{1\over 2}}{x_2}^2-x_1x_2}\right]_{x_1...
...dx_1+\int_0^1 \left[{x_1x_2-{\textstyle{1\over 2}}{x_2}^2}\right]^{x_1}_0\,dx_1$  
  $\textstyle =$ $\displaystyle \int_0^1 \left[{({\textstyle{1\over 2}}-x_1)-({\textstyle{1\over ...
..._1+ \int_0^1 \left[{({x_1}^2-{\textstyle{1\over 2}}{x_1}^2)-(0-0)}\right]\,dx_1$  
  $\textstyle =$ $\displaystyle \int_0^1 ({\textstyle{1\over 2}}-x_1+{x_1}^2)\,dx_1 = [{\textstyle{1\over 2}}x_1-{\textstyle{1\over 2}}{x_1}^2+{\textstyle{1\over 3}}{x_1}^3]^1_0$  
  $\textstyle =$ $\displaystyle ({\textstyle{1\over 2}}-{\textstyle{1\over 2}}+{\textstyle{1\over 3}})-(0-0+0) = {\textstyle{1\over 3}}$ (7)
$\displaystyle \mu'_2$ $\textstyle =$ $\displaystyle \int_0^1\int_0^1 (\vert x_2-x_1\vert)^2\,dx_2\,dx_1$  
  $\textstyle =$ $\displaystyle \int_0^1\int_0^1 (x_2-x_1)^2\,dx_1\,dx_2$  
  $\textstyle =$ $\displaystyle \int_0^1\int_0^1 ({x_2}^2-2x_1x_2+{x_1}^2)\,dx_1\,dx_2$  
  $\textstyle =$ $\displaystyle \int_0^1 [{\textstyle{1\over 3}}{x_2}^3-x_1{x_2}^2+{x_1}^2x_2]^1_0\,dx_1$  
  $\textstyle =$ $\displaystyle \int_0^1 ({\textstyle{1\over 3}}-x_1+{x_1}^2)\,dx_1 = [{\textstyle{1\over 3}}{x_1}^3-{\textstyle{1\over 2}}{x_1}^2+{\textstyle{1\over 3}}x_1]^1_0$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 3}}-{\textstyle{1\over 2}}+{\textstyle{1\over 3}} = {\textstyle{1\over 6}}.$ (8)

The Moments about the Mean are therefore
$\displaystyle \mu_2$ $\textstyle =$ $\displaystyle \mu_2'-{\mu_1'}^2={\textstyle{1\over 6}}-({\textstyle{1\over 3}})^2={\textstyle{1\over 18}}$ (9)
$\displaystyle \mu_3$ $\textstyle =$ $\displaystyle \mu_3'-3\mu_2'\mu_1'+2(\mu_1')^3={\textstyle{1\over 135}}$ (10)
$\displaystyle \mu_4$ $\textstyle =$ $\displaystyle \mu_4'-4\mu_3'\mu_1'+6\mu_2'(\mu_1')^2-3(\mu_1')^4={\textstyle{1\over 135}},$ (11)

so the Mean, Variance, Skewness, and Kurtosis are
$\displaystyle \mu$ $\textstyle =$ $\displaystyle \mu_1'={\textstyle{1\over 3}}$ (12)
$\displaystyle \sigma^2$ $\textstyle =$ $\displaystyle \mu_2={\textstyle{1\over 18}}$ (13)
$\displaystyle \gamma_1$ $\textstyle =$ $\displaystyle {\mu_3\over\sigma^3}={\textstyle{2\over 5}}\sqrt{2}$ (14)
$\displaystyle \gamma_2$ $\textstyle =$ $\displaystyle {\mu_4\over\sigma^4}-3=-{\textstyle{3\over 5}}.$ (15)

The probability distribution of the distance between two points randomly picked on a Line Segment is germane to the problem of determining the access time of computer hard drives. In fact, the average access time for a hard drive is precisely the time required to seek across 1/3 of the tracks (Benedict 1995).

See also Point-Point Distance--2-D, Point-Point Distance--3-D, Point-Quadratic Distance, Tetrahedron Inscribing, Triangle Inscribing in a Circle


Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 930-931, 1985.

Benedict, B. Using Norton Utilities for the Macintosh. Indianapolis, IN: Que, pp. B-8-B-9, 1995.

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© 1996-9 Eric W. Weisstein