Interestingly, there isn't for Gaussians (see my comment below).
Other simple distributions tend to give biases towards the corners or axis. Perhaps the Gaussian is unique in this regard? I'm not sure.
Other simple distributions tend to give biases towards the corners or axis. Perhaps the Gaussian is unique in this regard? I'm not sure.
I think I found one, but I'm not sure:
Without loss of generality, take f(xi) = k1 * exp(-g(xi)) [1], for some g. Then we need the joint pdf to satisfy f(x1,...,xn) = k2 * exp(-h(R^2)), R=sum(xi^2)^1/2 (the R^2 and h(.) is w.l.g. too). So we get g(x1)+g(x2)=h(x1^2+x2^2). Then assuming the functions g and h analytic we end up needing g(x)= k * x^2, otherwise we get cross terms in the Taylor expansion that can't be cancelled out for all xi. Sounds good?
[1] The function f trivially needs to be symmetric, justifying no loss of generality.
I wanted to make sure I got a proof, since I didn't really find this elsewhere.