Let $A\in\mathbb{R}^{n\times m}$ be a random matrix that each of its rows or columns is a Gaussian vector with iid components. More formally we have $A_{i,\cdot}\sim\mathcal{N}(0,I_m),A_{\cdot,j}\sim\mathcal{N}(0,I_n)$ for all $i\in[n],j\in[m]$, and furthermore, $\mathrm{law}(A)$ has support almost everywhere. Is $\mathrm{law}(A)$ a multivariate Gaussian?
Example ($m=n=2$): let $A=\left[\begin{matrix}a& b\\ c & d\end{matrix}\right]$. Assuming that $(a,b),(c,d),(a,c),(b,d)\sim\mathrm{N}(0,I_2)$, and $\mathrm{support}(\mathop{law}(A))$ is non-zero almost everywhere. Is $\mathop{law}(a,b,c,d)$ a multivariate Gaussian?
Possible generalization: if each row and column follow a multivariate p-Stable distribution, namely Cauchy for $p=1$, would the matrix follow a joint p-Stable distribution?
Best Answer
No. Answer to a previous version of the problem:
Let $X,Y\sim N(0,1)$ iid, let $S=\pm1$ be independent of $X,Y$ and let $$A=\pmatrix{X&Y\\Y&S X}.$$ Note that $SX\sim N(0,1)$, and $(X,Y)\sim N(0,I_2)$, $(Y,SX)\sim N(0,I_2)$ but $(X,SX)\nsim N(0,I_2)$.
With the current, changed version of the problem: There are rvs $X,Z$ which are marginally $N(0,1)$ but not jointly Gaussian, whose joint density function is positive everywhere. (For example, let $f(x,z)=3\phi(x,z)/2$ in the 1st and 3d quadrants and $f(x,z)=\phi(x,z)/2$ in the 2nd and 4th quadrants, where $\phi$ is the density of $N(0,I_2)$. Now construct $$A=\pmatrix{X&Y_1\\Y_2&Z},$$ where $Y_1,Y_2$ are independent $N(0,1)$ rv.s