Use $\mathbb{E}(X) = \mu$ to prove $\mathbb{E}(xx^T) = \mu \mu^{T} + \Sigma$

Question

Use $\mathbb{E}(X) = \mu$ to prove $\mathbb{E}(x x^T) = \mu \mu^{T} + \Sigma$.

Now, using the results two definitions, show that

$\mathbb{E}[_n x_m] = \mu \mu^{T} + I_{nm} \Sigma$

where $_n$ denotes a data point same from a Gaussian distribution with mean $\mu$ and covariance $\Sigma$, and $I_{nm}$ denotes the (n,m) element of the identity matrix.

Answer 1

$xx^\top$ is a matrix whose $(i,j)$ entry is $x_i x_j$. Using the definition of covariance, we have $$\mathbb{E}[x_i x_j] = \text{Cov}(x_i, x_j) + \mathbb{E}[x_i] \mathbb{E}[x_j] = \Sigma_{i,j} + \mu_i \mu_j.$$ The right-hand side is the $(i,j)$ entry of $\Sigma + \mu\mu^\top$. Thus $E[xx^\top] = \mu\mu^\top + \Sigma$.