How to derive $E(XX^T)$?

Question

I've seen the identity of $$X \sim \mathcal{N}(m, \Sigma) ~ \Rightarrow ~ E[xx^T] = \Sigma+mm^T $$ in the matrix cookbook. How does one derive this term? (Also, how does it look like for any random distribution $X$?)

Answer 1

$$\begin{align}\Sigma&=\Bbb{E}\left[(X-m)(X-m)^{\top}\right]\\&=\Bbb{E}\left[XX^{\top}\right]-2\cdot\Bbb{E}\left[Xm^{\top}\right]+\Bbb{E}\left[mm^{\top}\right]~~~~~~~~~\left(\text{since}~\Bbb{E}\left[Xm^{\top}\right]=\Bbb{E}\left[mX^{\top}\right]\right)\\&=\Bbb{E}\left[XX^{\top}\right]-2\cdot\Bbb{E}\left[X\right]m^{\top}+mm^{\top}\\&=\Bbb{E}\left[XX^{\top}\right]-mm^{\top}\end{align}$$