Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say you have 2 iid random variables $x,y\sim U[0,1]^k$, i.e. the uniform distribution over the k-dimensional unit cube. What's the expected value of the Euclidean distance between them when they have been normalized by the maximum distance possible, i.e. $\sqrt k$?

For $k=1$, I worked out that this is 1/3. For $k=100$, monte carlo simulations tell me it's a little over 0.4.

I tried to work out the math but $$\frac{1}{\sqrt{k}}\int_{S_x} \int_{S_y} \sqrt{(x-y)^T (x-y)}\;dx\;dy$$ where $S_x,S_y=[0,1]^k$ for general $k$ is beyond me.

share|cite|improve this question
up vote 5 down vote accepted

The asymptotics is $1/\sqrt6=0.40824829$.

To see this, consider i.i.d. random variables $X_i$ and $Y_i$ uniform on $[0,1]$ and write the quantity to be computed as $I_k=\mathrm E\left(\sqrt{Z_k}\right)$ with $$ Z_k=\frac1k\sum_{i=1}^k(X_i-Y_i)^2. $$ By the strong law of large numbers for i.i.d. bounded random variables, when $k\to\infty$, $Z_k\to z$ almost surely and in every $L^p$, with $z=\mathrm E(Z_1)$. In particular, $I_k\to \sqrt{z}$. Numerically, $$ z=\iint_{[0,1]^2}(x-y)^2\mathrm{d}x\mathrm{d}y=2\int_0^1x^2\mathrm{d}x-2\left(\int_0^1x\mathrm{d}x\right)^2=2\frac13-2\left(\frac12\right)^2=\frac16. $$

Edit (This answers a different question asked by the OP in the comments.)

Consider the maximum of $n\gg1$ independent copies of $kZ_k$ with $k\gg1$ and call $M_{n,k}$ its square root. A heuristics to estimate the typical behaviour of $M_{n,k}$ is as follows.

By the central limit theorem (and in a loose sense), $Z_k\approx z+N\sqrt{v/k}$ where $v$ is the variance of $Z_1$ and $N$ is a standard gaussian random variable. In particular, for every given nonnegative $s$, $$ \mathrm P\left(Z_k\ge z+s\right)\approx\mathrm P\left(N^2\ge ks^2/v\right). $$ Furthermore, the typical size of $M_{n,k}^2$ is $z+s$ where $s$ solves $\mathrm P(Z_k\ge z+s)\approx1/n$. Choose $q(n)$ such that $\mathrm P(N\ge q(n))=1/n$, that is, $q(n)$ is a so-called quantile of the standard gaussian distribution. Then, the typical size of $M_{n,k}^2$ is $k(z+s)$ where $s$ solves $ks^2/v=q(n)^2$. Finally, $$ M_{n,k}\approx \sqrt{kz+q(n)\sqrt{kv}}. $$ Numerically, $z=1/6$, $v=7/180$, and you are interested in $k=1'000$. For $n=10'000$, $q(n)=3.719$ yields a typical size $M_{n,k}\approx13.78$ and for $n=100'000$, $q(n)=4.265$ hence $M_{n,k}\approx13.90$ (these should be compared to the values you observed).

To make rigorous such estimates and to understand why, in a way, $M_{n,k}$ concentrates around the typical value we computed above, see here.

share|cite|improve this answer
Thanks! This makes perfect sense. One question about monte carlo experiments: there's a fairly big difference in the expected value when I fix the max distance possible to the theoretical $\sqrt(k)$ and when I take the actual max that's been observed from the samples. The latter actually gives the sample expectation to be about 0.8ish. I know that observing $\sqrt{k}$ from samples is near impossible. But will the monte carlo estimate converge almost surely to $1/\sqrt{6}$ for large enough $n$? – JasonMond Sep 13 '11 at 15:35
Wait, you simulate (pairs of) points in the unit cube and not their distances, do you? So I am not sure of the reason why you have to fix a max distance. – Did Sep 13 '11 at 16:39
Let me elaborate: if $X$ and $Y$ are independent and uniform on $[0,1]$, neither $|X-Y|$ nor $(X-Y)^2$ are uniform on $[0,1]$. So if you simulate either one of these directly as uniform (instead of simulating $X$ uniform then $Y$ uniform then combining them), the result has little to do with $I_k$. – Did Sep 13 '11 at 17:07
I'm simulating as you describe it (in R: for(i in 1:10000) {v <- nrm(runif(1000),runif(1000)); if (v > mx) mx <- v;} where nrm is euclidian distance and mx stores the maximum). But with 10000 samples, the maximum distance I get with $k=1000$ is 13.8. With 100000 samples, it ekes out a bit more with 14.0. Theoretically, the maximum distance possible is $\sqrt{1000}\approx 31.6$ and with an infinite number of samples, I guess I will get very close to this value. And the expected value of pairwise distance would be $1/\sqrt{6}$. But not with the sample maximum. – JasonMond Sep 14 '11 at 2:01
@JasonMond: What is your concern with the maximum? And, but the way, a much more efficient way to generate your sample in $R$ is to do: v<-replicate(10000,sqrt(sum((runif(1000)-runif(1000))^2)). – cardinal Sep 14 '11 at 9:56

This probably isn't anything nice for general $k$. You're asking for $$ E \sqrt{ \sum_{i=1}^k (X_i-Y_i)^2/k } $$ where $X_i$ and $Y_i$ are independent uniform(0,1). However, let $Z_i = (X_i-Y_i)^2$. Now you're looking for $$ E \sqrt{ (\sum_{i=1}^k Z_k)/k }$$ and let's think about what happens when $k$ gets large. When $k$ is large, by the law of large numbers $(\sum_{i=1}^k Z_k)/k$ will be concentrated around $E(Z_k)$. Therefore as $k$ gets large I expect that the answer approaches $\sqrt{E(Z_k)}$, which is $\sqrt{1/6}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.