$\newcommand{\cov}{\operatorname{cov}}\newcommand{\corr}{\operatorname{corr}}$I have two dependent multivariate random variables, or random vectors, $X$ and $Y$, their respective covariance matrices $\Sigma_X = \cov(X,X)$, $\Sigma_Y = \cov(Y,Y)$ and a non-zero correlation matrix $P_{XY} = \corr(X, Y)$.
I would like to compute the covariance matrix of the sum of the two random vectors:
\begin{align} Z {}={}& X + Y \\ \Sigma_Z {}={}& \Sigma_X + \Sigma_Y + 2 \cov(X, Y) \end{align}
Intuitively, I would assume the cross-covariance can be computed as follows:
$$\cov(X, Y) = P_{XY} \sqrt{\Sigma_X} \sqrt{\Sigma_Y}$$
But I can see that this matrix is not necessarily symmetric, and therefore the resulting $\Sigma_Z$ wouldn't be either. Am I wrong? Is it at all possible to compute $\Sigma_Z$ from the information I have? If not, is there some way to approximate it?
EDIT: I think it isn't correct to write:
$$ \Sigma_Z = \Sigma_X + \Sigma_Y + 2 \cov(X, Y) $$
as $\cov(X,Y) \ne \cov(Y,X)$, but instead $\cov(X,Y) = \cov(Y,X)^T$.
Rewriting the equation for $\Sigma_Z$ to:
\begin{align} \Sigma_Z {}={}& \Sigma_X + \Sigma_Y + \cov(X, Y) + \cov(Y, X) \\ {}={}& \Sigma_X + \Sigma_Y + \cov(X, Y) + \cov(X, Y)^T \end{align}
I can now prove that indeed $\Sigma_Z$ is guaranteed to be symmetric. As antkam points out in the chat below, it isn't necessary that $\cov(X, Y) + \cov(X, Y)^T$ be positive semi-definite for $\Sigma_Z$ to be positive semi-definite.