Is the expected value of determinant times random matrix proportional to expected value of random matrix.

Question

Let $\mathbf{X}$ be a random matrix. Is it true that $$ E[\det(\mathbf{X})^h \mathbf{X} ] \propto E[\mathbf{X}], $$ where $h$ is a scalar and $\det(.)$ denotes the determinant?

Answer 1

This is unfortunately not true. Consider the following counterexample for $h=1$:

\begin{align*} X(\omega_{1}) &= \begin{pmatrix} 2 & \ 1 \\ 0 & \ 2 \end{pmatrix} \\ X(\omega_{2}) &= \begin{pmatrix} 1 & -1 \\ 0 & \ 1\end{pmatrix} \end{align*} with $P(\omega_{1}) = P(\omega_{2}) = \frac{1}{2}$. Then we have $$ \mathbb{E}[X] = \begin{pmatrix} \frac{3}{2} & 0 \\ 0 & \frac{3}{2} \end{pmatrix}.$$

On the other hand, you can see that $\det(X)(\omega_{1}) = 4$ and $\det(X)(\omega_{2}) = 1$, so \begin{align*} [\det(X) \cdot X](\omega_{1}) &= \begin{pmatrix} 8 & \ 4 \\ 0 & \ 8 \end{pmatrix} \\ [\det(X) \cdot X](\omega_{2}) &= \begin{pmatrix} 1 & -1 \\ 0 & \ 1\end{pmatrix} \end{align*}

As such, we have $$ \mathbb{E}[\det(X) \cdot X] = \begin{pmatrix} \frac{9}{2} & \frac{3}{2} \\ 0 & \frac{9}{2}\end{pmatrix}. $$