Finding an isotropic joint density from a marginal I'm trying to find out weither it is possible or not to recover an isotropic bivariate pdf from one of its marginal pdf. 
By isotropic, I mean that the density only depends on the radius when expressed in polar coordinates 
$$ p(x,y) = 2 \pi \, r \, p(r), \quad r=\sqrt{x^2+y^2}.$$
So if I have $$p(x) = \int_{-\infty}^{\infty} p(x,y) \, dy,$$ can I recover $p(x, y)$ ? My intuition says that there should be a unic isotropic bivariate pdf $p(x,y)$ that produces the marginal density $p(x)$ so it should be possible, but I have been unable to find a proof so far.  
 A: Here's a solution under an extra assumption. Assume that the isotropic density can be written as a finite sum of Gaussian distributions, i.e.
$$p(x,y) = f(\sqrt{x^2+y^2}) = \sum_k c_k e^{-a_k (x^2+y^2)},$$
where all $a_k > 0$ are distinct. Then the marginal may be written
$$p(x) = \int p(x,y) dy = \sum_k \frac{\sqrt{\pi}}{\sqrt{a_k}} c_k e^{-a_k (x^2+y^2)}.$$
The RHS is a linear combination of a finite number of linearly independent functions, so $p(x)$ determines the coefficients. Hence, under the extra assumption that $p(x,y)$ is a finite sum of Gaussians in $r$, the marginal determines the joint distribution.
(Remark: I think the extra assumption can probably be weakened a lot, but I am not sure about the technicalities. For example, the finite sum could be replaced by an infinite sum or an integral $p(x,y) = \int c(a) e^{-a (x^2+y^2)} da$. Especially with the integral representation, I'm not sure about the uniqueness of the resulting representation of $p(x)$... So I just leave this with the extra assumption above.)
