# A rotation invariant product of probability measures has to be Gaussian or Dirac measure?

I am currently thinking about symmetry of measures and made the following definition (I don't know if there is already any theory in this direction, so in case please let me know in the comments):

Let be $(\Omega,\Sigma,\mu)$ a measure space and $G$ a group (left) operating on $\Omega$. We say that $\mu$ is $G$-invariant, if $\mu(gA)=\mu(A)$ for all $g\in G$ and $A\in\Sigma$.

Now, I'm interested in $\Omega:=\Bbb R^n$, $\Sigma$ the Borel $\sigma$-Algebra of $\Omega$, $G=SO(n)$ the special orthogonal group. (We use the term rotation invariant for $SO(n)$-invariant from now on.) Im wondering, how many measures are rotation invariant, if they are (joint) probability distributions of independent and identically distributed (iid. for short) random variables. I only identified Gaussian and Dirac measures (respectively with expectation zero) with this property, yet. More precisely speaking:

Question: If $\mu:=\nu\otimes\nu\ldots\otimes\nu$ ($n$-times) for a probability (Borel-)measure $\nu$ on $\Bbb R$ is rotation invariant, is it already necessary for $\mu$ to be a Gaussian measure or a Dirac measure (and thus, the same for $\nu$) with expectation zero?

My attempt (which actually inspired my above-mentioned question): Even for two real-valued random variables with Lebesgue-densities this is not easy for me to answer. Let be $X,Y$ real random variables iid. via a continuous Lebesgue-density $f$, meaning the probability measure $\mu$ of $X,Y$ is given by $d\mu=fd\lambda$ for the Lebesgue-measure $\lambda$. Because of independency the Lebesgue-density of the product measure $\mu^2:=\mu\otimes\mu$ is given by $f_{X,Y}(x,y)=f(x)f(y)$. Now, if $\mu^2$ is rotation invariant, $$f(r\cos(\alpha))\cdot f(r\sin(\alpha)) = f(r)\cdot f(0)$$ must hold for all $r\geq 0$ and all $\alpha\in\Bbb R$. Even without restricting $f$ to fulfill $\int fd\lambda=1$, which functions (except the trivial case that $f$ is constant) fulfill this equality at all? Clearly, all scalar multiples of the gaussian density with expectation zero $f(x;\sigma^2)=e^{-\frac12\frac{x^2}{\sigma^2}}$ accomplish that. Is there any other function class?

This is the content of Maxwell's theorem. You can prove it by finding an instance of Cauchy's functional equation satisfied by a function related to the characteristic function. If $\phi(u,v)=E\exp(i(uX+vY))$ then by independence $\phi(u,v)=\phi(u,0)\phi(0,v)$ and by symmetry $\phi(u,v)=g(u^2+v^2)$ for some function $g$. So we have $g(u^2+v^2)=g(u^2)g(v^2)$. Since $\phi$ is a characteristic function, $\phi$ and hence $g$ is continuous, and the function $\log g$ is linear, and so on.