Variance of Hadamard Product

Question

Multivariate vectors $\textbf{X} \sim N(0, \textbf{A})$ and $\textbf{Y} \sim N(0, \textbf{B})$

Now I want to show that

$Var(\textbf{X} \odot \textbf{Y}) = \textbf{A} \odot \textbf{B}$ (Matrix) (Hadamard product)

What I have done

Simplified $Var(\textbf{X} \odot \textbf{Y}) = E(\textbf{X} \odot \textbf{Y} \cdot \textbf{X} \odot \textbf{Y}) - E(\textbf{X} \odot \textbf{Y})E(\textbf{X} \odot \textbf{Y})$

How can I proceed from here?

Thanks in advance!

Answer 1

If ${\bf U}$ is a random column vector with zero mean vector, then the $i,j$ element of the covariance matrix of $\bf U$ is $$(\operatorname{Var}({\bf U}))_{i,j}=E(U_i U_j).\tag1$$ Now suppose $\bf X$ and $\bf Y$ are independent mean zero random vectors with covariance matrices $\bf A$ and $\bf B$ respectively. Define the Hadamard product ${\bf X}\odot{\bf Y}$ by $$({\bf X}\odot{\bf Y})_i:=X_iY_i.\tag2 $$ Check that ${\bf X}\odot{\bf Y}$ has zero mean vector, so apply (1) with ${\bf U}:={\bf X}\odot{\bf Y}$ to find its covariance matrix: $$\begin{align} (\operatorname{Var}({\bf X}\odot{\bf Y}))_{i,j}&\stackrel{(1)}=E({\bf X}\odot{\bf Y})_i ({\bf X}\odot{\bf Y})_j\\ &\stackrel{(2)}=E(X_iY_i)(X_jY_j) =E(X_iX_j)(Y_iY_j)\\&\stackrel{(3)}=E(X_iX_j)E(Y_iY_j)\\&\stackrel{(1)}=A_{i,j}B_{i,j}=({\bf A}\odot {\bf B})_{i,j} \end{align} $$ Step (3) uses independence of $\bf X$ and $\bf Y$. The final equality is the definition of Hadamard product for matrices.