For a distribution $p_{\mathbf{X}}$, $p_{\mathbf{X}}(\mathbf{x}) = p(x_1) \cdots p(x_d)$ if $p(\mathbf{x}) = q(r)$?

Question

For a distribution $p_{\mathbf{X}}$ of a random vector $\mathbf{X}$, assume $p_{\mathbf{X}}(\mathbf{x}) = p_{\mathbf{X}}(\mathbf{y})$ for any $\mathbf{x}, \mathbf{y}$ such that $\lVert \mathbf{x}\rVert =\lVert \mathbf{y}\rVert$. Then is it true that $$ p_{\mathbf{X}}(\mathbf{x}) = p(x_1) \cdots p(x_d)? $$ In other words, if the distribution $p_{\mathbf{X}}$ is symmetric along all directions, then are all of the coordinate functions $X_1,\dots,X_d$ (mutually) independent?

Answer 1

Edit: After further clarifications

Assuming we don't have independence of the components of the vector then functional dependence on the radial component only will not be enough to claim it will factor as independent marginals, consider the bivariate Cauchy distribution $$ p(x,y) = \frac{1}{2\pi} \frac{c}{(c^2 + x^2 + y^2 )^{3/2}}, c > 0 $$ then you can show that $$ p(Y | X = x) = \frac{(c^2 + x^2)}{2(c^2 + x^2 + y^2 )^{3/2}}. $$

So using the bivariate Cauchy example we have seen that even though the density function depends only on the magnitude $\| \textbf{x} \|$ of it's components it does not factor into a product of independent marginals.

So while the answer to your question as posed is a negative there is an asymptotic approximation that may be of some interest, although note that it is only a statement about the projection of the original distribution onto the unit sphere $S^{n-1} \subset \mathbb{R}^n$ and so while it says these projections are asymptotically iid Gaussian it of course does not say the original distribution in the ambient space $\mathbb{R}^n$ factors as a product.

Let $\textbf{X} = (X_1,\ldots,X_n)$ be such that the scaled random variable $\textbf{Y}=(Y_1,\ldots,Y_n)$ with components $$ Y_k = \frac{X_k}{\| \textbf{X} \|} $$ is distributed uniformly on the unit sphere. That is we take a random variable such that the density is a function of the norm $\| \textbf{x} \|$ only and project it to a density defined over the unit sphere. Also let $\Phi(z)$ denote the distribution of the standard normal. Then if $\textbf{Y}_n$ is a sequence of observations of these scaled random variables you have that as $n \rightarrow \infty$ then $$ \mathbb{P}( n^{1/2} Y_{n1} \leq y_1, \ldots , n^{1/2}Y_{nn} \leq y_{n} ) \xrightarrow{\mathscr{D}} \prod_{i=1}^{n} \Phi(y_i), $$ where by $Y_{nk}$ we mean the $k$th component of the $n$th random vector $\textbf{Y}$. So that in the limit and appropriately scaled the observation of such a random variable is approximately the i.i.d. product Gaussian case.

Original answer, assuming independence It is a classical result, (and can even be extended to infinite sequences) that if the distribution is rotatable, i.e. the probability remains invariant under rotations, by which I mean that if $x = (x_1,\ldots,x_n)$ and $y = (y_1,\ldots,y_n)$ are two points such that $x \neq y$ but $$ \sqrt{\sum_i x_i^2} = \sqrt{\sum_i y_i^2}, $$ then we have that $p(x) = p(y)$. So that as a function the probability density function is in only dependent on the radius of it's argument then we have

Theorem [Rotatability and independence ] Let $X_1,\ldots,X_n$ be independent random variables with $n \geq 2$. Then $(X_1,\ldots,X_n)$ is rotatable if and only if the $X_i$ are i.i.d centered Gaussians.