Expectation of matrix product

Question

Suppose we have a random matrix $M \in \mathbb{R}^{n\times m}$ such that $\text{E}[M] = 0$ and $\text{E}[M M^\top] = \Sigma$. How does one compute $\text{E}[M^\top M]$?

Answer 1

Assume that $M=\begin{pmatrix}0&a\\0&0\end{pmatrix}$ where $a$ follows a continuous uniform distribution on $[-1,1]$. Then $E(M^TM)=\begin{pmatrix}0&0\\0&1/3\end{pmatrix}$ and $E(MM^T)=\begin{pmatrix}1/3&0\\0&0\end{pmatrix}$ because $E(a^2)=1/2\int_{-1}^1x^2dx=1/3$.

Yet, if the $(m_{i,j})$ are iid and follow the above distribution, then we find when (for example) $n=2,m=3$

$E(MM^T)=I_2,E(M^TM)=(2/3)I_3$.

EDIT. We can generalize (for any $n,m$) as follows. Assume that the $(m_{i,j})$ are iid and follow the same distribution of probability s.t. $E(m_{i,j})=0,E(m_{i,j}^2)=\sigma$.

If $(i,j)\not=(k,l)$, then $E(m_{i,j}m_{k,l})=E(m_{i,j})E(m_{k,l})=0$.

We obtain $E(MM^T)=m\sigma I_n,E(M^TM)=n\sigma I_m$. Thus, if we know $E(MM^T)$, then we can deduce $E(M^TM)$ (without the knowledge of $\sigma$).