In three dimensions (to keep the notation reasonably simple) you need a distribution with density f such that f(x)f(y) f(z) is a function of x^2 + y^2 + z^2 (i. e. the distance from the origin). It looks like the Gaussian (up to scaling) is the only one.

Yes, the fact that i.i.d. Gaussians give a rotationally symmetric joint distribution is a defining property of the Gaussian.

Ah I suspected as much. Can you provide proof?

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.

Perhaps coincidentally, someone just asked about this at math.stackexchange: http://math.stackexchange.com/questions/1255637/joint-pdf-of...

Yes, that was me :)

I wanted to make sure I got a proof, since I didn't really find this elsewhere.

You have to assume that the one-dimensional marginals are Gaussian for otherwise the statement is not correct.

Hmm I'm not sure I get your point. I'm trying to prove that the joint pdf of N iid RV's is isotropic if and only if the RV's are gaussian. If I assume the pdfs are gaussian in the first place the proof isn't valid?

Oh, I see. I thought you were trying to prove something different.