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Let $f(x)$ be a continuous probability distribution in the plane. It is obvious that if $X$ and $X'$ are two independent random samples from $f$, then $\mathbf{E}(\|X - X'\|) \leq 2 \mathbf{E}(\|X\|)$ by the triangle inequality. Can this upper bound be made tighter if we assume that $f$ is rotationally symmetric about the origin., i.e. $f(x) = g(\|x\|)$ for some function $g$?

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3 Answers 3

up vote 5 down vote accepted

Using the pareto distribution $f(x) = \frac{\alpha}{x^{\alpha+1}}$ ($x > 1$) , the ratio $\frac{E(||X-X'||)}{E(||X||)}$ approaches a $2$ as $\alpha$ tends to 1.

To find such a distribution, consider that all else equal, you want to maximize the difference $||X||-||X'||$ since the angle between the two is independent. This means that you want a distribution that extends to infinity as flatly as possible.

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Simpler than a Pareto distribution is a Bernoulli distribution with $p \to 0.$ –  Douglas Zare Aug 31 '12 at 21:25
    
So you'd get a mixture between a circle of radius R and a Dirac at 0. $E(||X||)$ would be $(1-p)R$ and $E(||X−X′||)$ would be $p(1−p)R+(1−p)^2 R 4/\pi$ thus $E(||X−X′||)/E(||X||)=p+(1−p)4/\pi$... Am I missing something? –  Arthur B Aug 31 '12 at 21:54
    
By convention, the parameter $p$ is the chance that the Bernoulli random variable is $1$, not $0.$ The expected magnitude of $X$ is $p$. $E\|X-X'\| = 2p(1-p) + p^2 \frac 4 \pi.$ The ratio goes to $2$ as $p \to 0.$ –  Douglas Zare Sep 2 '12 at 8:11
    
Ahh, I forgot the factor of $2$ in $2p(1−p)$... yes, that is much simpler. My first idea was to go with Bernoulli,s but I made the same stupid mistake and thought it didn't work. –  Arthur B Sep 4 '12 at 13:39

The conditional expectation $$\eqalign{E[ \|X - X'\| | \|X\| = r, \|X'\| = s] &= \frac{1}{2\pi} \int_0^{2\pi} \sqrt{r^2 + s^2 - 2 r s \cos \theta}\ d\theta \cr &= \frac{2(r+s)}{\pi} EllipticE(2 \sqrt{rs}/(r+s))\cr}$$ where EllipticE is Maple's version of the complete elliptic integral of the second kind. Note that $0 < 2 \sqrt{rs}/(r+s) \le 1$, with $1$ occurring for $r=s$. On the interval $[0,1]$, $1 \le EllipticE(x) \le \pi/2$. Thus $$ \frac{4}{\pi} E[\|X\|] = \frac{2}{\pi} E[\|X\|+\|X'\|] \le E[\|X - X'\|] \le E[\|X\|+\|X'\|] = 2 E[\|X\|] $$

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4  
These are sharp as constants. If $\|X\|$ is constant, then $E\|X-X'\| = \frac{4}{\pi}E\|X\|.$ If $P(\|X\|/\|X'\| \in (1/r,r)$ is close to $0$, which is achievable as Arthur B points out, then $E\|X-X'\|$ is close to $2 E\|X\|.$ –  Douglas Zare Aug 29 '12 at 20:37
    
Thanks! It doesn't appear that I can accept both answers, but I would if I could =/ –  Charlie Aug 29 '12 at 22:46

Under the assumption that $f$ is radial, the angle between $X$ and $X'$ is uniformly distributed in $[0,\pi]$ (since the signed angle that each of $X$ and $X'$ makes with the $x$-axis is uniformly distributed by rotational invariance, hence so is their difference). Thus the distribution of $\|X-X'\|$ is the same as the distribution of $\| (r,0)-s e^{i\alpha}\|$ where $\alpha$ is an angle chosen uniformly at random, and $r, s$ are two independent samples of the distribution $g$ (on $[0,\infty)$). Its expectation thus equals (after a little calculation) $$ \pi^{-1}\int_0^\pi \int \int \sqrt{r^2+s^2-2rs\cos(\alpha)} g(r) g(s) dr ds d\alpha, $$ or $$ \pi^{-1}\int_0^\pi \int \int \sqrt{(r+s)^2-2rs(1+\cos(\alpha))} g(r) g(s) dr ds d\alpha. $$ I suppose you could get some bound in terms of $\mathbb{E}(\|X\|)$ from here which is better than $2\mathbb{E}(\|X\|)$, but there is a much more pleasant expression for the expectation of $\mathbb{E}\|X-X'\|^2$: as before, this is $$ \pi^{-1}\int_0^\pi \int \int r^2+s^2-2rs\cos(\alpha) g(r) g(s) dr ds d\alpha, $$ which can be readily evaluated to $2\mathbb{E}(\|X\|^2)-4\mathbb{E}(\|X\|^2)/\pi$.

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