Positive definiteness of block matrices of covariance matrix It is known that a proper covariance matrix is symmetric and at least positive semi definite. I wanted to know whether those properties extends to its  block matrices.

Referring the section from Wikipedia above,  $\Sigma$ is symmetric and at least positive semi definite. Do those properties carry over to $K_{xx}, K_{xy}, K_{yy}$ ?
 A: Lemma. If $K$ is a matrix that is the dispersion matrix of any random vector, then $K$ is positive semidefinite.
Proof. Suppose $K$ is the dispersion matrix of the vector $x,$ which we may assumed centred. Then $K = E(x x^\intercal)$ by definition. If $a$ is any constant vector, $a^\intercal K a = E(a^\intercal x x^\intercal a)$ and since $a^\intercal x$ is a real number, the random variable $a^\intercal x x^\intercal a = (a^\intercal x)^2 \geq 0$ and $a^\intercal K a \geq 0.$ QED
If a random vector $x$ is now split as $x^\intercal = (x_1^\intercal, x_2^\intercal),$ then $xx^\intercal = \begin{bmatrix}x_1x_1^\intercal &x_1x_2^\intercal\\x_2x_1^\intercal &x_2x_2^\intercal\end{bmatrix}$ and after taking expectation, we see that the main diagonal blocks are dispersion matrices of random vectors, thus positive semidefinite. Off-diagonal blocks need not be squared, thus asking for positiveness is nonsensical in this case, but even if they were squared, you will get that those blocks are of the form $K_{xy} = E(xy^\intercal)$ and the argument in the lemma above simply does not carry over except under special circumstances (i.e. off-diagonal blocks will rarely have any special property).
