When is the distribution of two iid random variables spherically symmetric?

In this post we show that the joint distribution of two iid random variables is spherically symmetric iff the marginal distribution is Gaussian.

Proof: If the marginal distribution is Gaussian, the joint distribution is clearly spherical. Below we show the converse: if the joint distribution is spherical, the marginal distribution is Gaussian.

Let the marginal distribution be $g(x)$, so $p(x,y) = g(x)g(y)$. Spherical means the gradient of the joint distribution is proportional to $(x,y)$. That is $$\nabla p(x,y) = \begin{bmatrix} g'(x) g(y) \\ g(x) g'(y)\end{bmatrix} = c(x,y) \begin{bmatrix}x \\ y \end{bmatrix}.$$

To eliminate the constant of proportionality $c(x,y)$, we divide the first term in the gradient by the second, to get $$ {g'(x) g(y) \over g(x) g'(y)} = {x \over y}.$$ Recognizing $g'(x)/g(x) = d\ln g(x)/dx$, we get $$ {d \ln g(x)/ dx \over d \ln g(y)/dy } = {x \over y}.$$

This motivates expressing $g(x) = e^{\phi(x)}$, for some unknown $\phi(x)$. Then $d \ln g(x)/dx = \phi'(x),$ and we get $${\phi'(x) \over \phi'(y)} = {x \over y}.$$

We then set $y$ to 1, and get $$ {d \phi(x) \over dx} = \phi'(1) x \implies d\phi(x) = \phi'(1) x dx \implies \phi(x) = {\phi'(1) \over 2}x^2 + c,$$ which then implies that $$ g(x) = e^{\phi(x)} \propto e^{\phi'(1) {x^2 \over 2}},$$ and the proof is complete.

$$\blacksquare$$