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 $g$?
Using the pareto distribution $f(x) = \frac{\alpha}{x^{\alpha+1}}$ ($x > 1$) , the ratio $\frac{E(XX')}{E(X)}$ approaches a $2$ as $\alpha$ tends to 1.
To find such a distribution, consider that all else equal, you want to maximize the difference $XX'$ since the angle between the two is independent. This means that you want a distribution that extends to infinity as flatly as possible.

$\begingroup$ Simpler than a Pareto distribution is a Bernoulli distribution with $p \to 0.$ $\endgroup$ – Douglas Zare Aug 31 '12 at 21:25

$\begingroup$ So you'd get a mixture between a circle of radius R and a Dirac at 0. $E(X)$ would be $(1p)R$ 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? $\endgroup$ – Arthur B Aug 31 '12 at 21:54

$\begingroup$ By convention, the parameter $p$ is the chance that the Bernoulli random variable is $1$, not $0.$ The expected magnitude of $X$ is $p$. $E\XX'\ = 2p(1p) + p^2 \frac 4 \pi.$ The ratio goes to $2$ as $p \to 0.$ $\endgroup$ – Douglas Zare Sep 2 '12 at 8:11

$\begingroup$ 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. $\endgroup$ – 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\] $$

4$\begingroup$ These are sharp as constants. If $\X\$ is constant, then $E\XX'\ = \frac{4}{\pi}E\X\.$ If $P(\X\/\X'\ \in (1/r,r)$ is close to $0$, which is achievable as Arthur B points out, then $E\XX'\$ is close to $2 E\X\.$ $\endgroup$ – Douglas Zare Aug 29 '12 at 20:37

$\begingroup$ Thanks! It doesn't appear that I can accept both answers, but I would if I could =/ $\endgroup$ – 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 $\XX'\$ 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) $$ \pi^{1}\int_0^\pi \int \int \sqrt{r^2+s^22rs\cos(\alpha)} g(r) g(s) dr ds d\alpha, $$ or $$ \pi^{1}\int_0^\pi \int \int \sqrt{(r+s)^22rs(1+\cos(\alpha))} g(r) g(s) dr ds d\alpha. $$ 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 $\mathbb{E}\XX'\^2$: as before, this is $$ \pi^{1}\int_0^\pi \int \int r^2+s^22rs\cos(\alpha) g(r) g(s) dr ds d\alpha, $$ which can be readily evaluated to $2\mathbb{E}(\X\^2)4\mathbb{E}(\X\^2)/\pi$.