Inferring symmetry of a distribution from its marginals

Question

Let $X=[X_1,\ldots,X_n]$ be a continuous random vector of size $n$ with density function $f_X(x_1,\ldots,x_n)$. If all the marginals \begin{align*} \int \ldots \int f_X(x_1,\ldots,x_n)\, dx_2dx_3\ldots dx_n \\ \int \ldots \int f_X(x_1,\ldots,x_n)\, dx_1dx_3\ldots dx_n \\ \hspace{-3cm} \vdots \\ \int \ldots \int f_X(x_1,\ldots,x_n)\, dx_1dx_2\ldots dx_{n-1} \end{align*} are symmetric, then is the original density function $f_X(x)$ also symmetric?

Answer 1

I think the density need not be symmetric. Since the problem has nothing to do with absolute continuity as far as I see, I will argue without densities below.

1. Let $(\cdot,\cdot)$ denote the inner product in $\mathbb{R}^n$. If we could show that for every $b\in\mathbb{R}^n$, the linear combination $(b,X)$ is a symmetric random variable, then we could argue that $$E\exp\{i(b,X)\}=E\exp\{-i(b,X)\}=E\exp\{i(b,-X)\}$$ so that $X$ and $-X$ would have the same characteristic function and thus the same distribution, so that $P(X\in A) = P(-X\in A)$ for every Borel set $A\in\mathbb{R}^n$.
2. It is sufficient to give a counterexample for $n=2$. Let $Y,Y_1,Y_2,Y_3$ be independent identically distributed random variables. Define $X=(X_1,X_2)=(Y_1-Y_2,Y_2-Y_3)$. By construction, $X_1$ and $X_2$ have a symmetric distribution. For $X$ to have the same distribution as $-X$, their characteristic functions $\varphi_X$ and $\varphi_{-X}$ need to be the same. So we need $$\varphi_X(b) = \varphi_{-X}(b)=\varphi_X(-b)$$ for all $b\in\mathbb{R}^2$. Plugging in, we see that this amounts to $$\varphi_Y(b_1)\varphi_Y(b_2-b_1)\varphi_Y(-b_2)=\varphi_Y(-b_1)\varphi_Y(b_1-b_2)\varphi_Y(b_2)\,.$$ For general $Y$, this is not true for all $b=(b_1,b_2)$.