How can I show that the expected pairwise square euclidean distance between points in $X$ is $Θ(d)$?

Where $X$ is a $(x_1,...x_n)$ of points generated uniformly at random in the unit, d is d-dimensional cube , $x=(x(1),...x(d))$ the generic point has its -th component $x(i)$ chosen uniformly at random in$ [0,1] $independently of other components and points.

$\Theta(d)$ represent the largest possible distance is d.

I try to reconduct this problem to Bertrand Paradox but i dont think is right. Maybe I that show that $E(||x−y||2)=Θ(d)$ , because is a hint but i dont know how.

i m following this path: https://stats.stackexchange.com/questions/22488/probability-that-uniformly-random-points-in-a-rectangle-have-euclidean-distance

but is different to my point.


  • 1
    $\begingroup$ Try solving the case when $d=1$ first. Then use the linearity of expectation. $\endgroup$ – John L. Nov 30 '18 at 17:05

Let $\vec{x},\vec{y}$ be two random $d$-dimensional vectors chosen uniformly and independently from $[0,1]^d$. That is, $x_1,\ldots,x_d,y_1,\ldots,y_d$ are all uniform random samples of the uniform distribution over $[0,1]$. Then $$ \mathbb{E}[\|\vec{x}-\vec{y}\|^2] = \mathbb{E}\left[\sum_{i=1}^d (x_i-y_i)^2\right] = \sum_{i=1}^d \mathbb{E}[(x_i-y_i)]^2 = d \operatorname*{\mathbb{E}}_{x,y \sim [0,1]} [(x-y)^2]. $$ Let $C = \mathbb{E}[(x-y)^2]$. Then the expected squared distance of two points in $[0,1]^d$ is $Cd$.

It is not hard to calculate $C$ explicitly: $$ C = \mathbb{E}[((x-1/2) - (y-1/2))^2] = \mathbb{E}[(x-1/2)^2] + 2\mathbb{E}[(x-1/2)(y-1/2)] + \mathbb{E}[(y-1/2)^2] = \\ 2\mathbb{E}[(x-1/2)^2] + 2\mathbb{E}[x-1/2] \mathbb{E}[y-1/2] = 2\mathbb{E}[(x-1/2)^2] = \\ 2\int_0^1 (x-1/2)^2 \, dx = 2\int_0^1 x^2-x+\frac{1}{4} \, dx = 2\left(\frac{1}{3} - \frac{1}{2} + \frac{1}{4}\right) = \frac{1}{6}. $$

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  • $\begingroup$ Why in C calculation, the expectation became (x-1/2)(y-1/2) ? You take middle point in [0,1] ? thanks for you answer $\endgroup$ – theantomc Dec 1 '18 at 11:15
  • $\begingroup$ Here $1/2$ is the expectation of $x$. $\endgroup$ – Yuval Filmus Dec 1 '18 at 15:53
  • $\begingroup$ you simple apply $\int_{0}^{1} x = 1/2$. And after we apply the linearity on the expectation? Because i don't caught why after we simplify the expression for reach the $2E(x-1/2)^2$ $\endgroup$ – theantomc Dec 1 '18 at 16:10
  • $\begingroup$ Give it a few more hours. $\endgroup$ – Yuval Filmus Dec 1 '18 at 16:13
  • $\begingroup$ see this on "uniform distribution" For a random variable following this distribution, the expected value is then m1 = (a + b)/2 en.wikipedia.org/wiki/Uniform_distribution_(continuous) $\endgroup$ – vzn Dec 2 '18 at 17:37

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