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 $g$?



Expected distance between two points in the plane
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.


