# Convexity of a Certain Set of Covariance Matrices

Hello,

My question is about a certain set of matrices being convex or not. I'll start with some preliminaries in order to define myself properly. Let $X_1,U,X_2$ be three zero-mean Gaussian random vectors (RVs) of dimension $N\times 1$, that admit the Markov relation $X_1-U-X_2$. Let us use the notations: $\Sigma_i=\mathbb{E}[X_iX_i^T]$, for $i=1,2$, $\Sigma_U=\mathbb{E}[UU^T]$, $\Sigma_{iU}=\mathbb{E}[X_iU^T]$, for $i=1,2$ and finally $\Sigma_{12}=\mathbb{E}[X_1X_2^T]$. The Markov relation is equivalent to the fact that the auto- and cross- covariance matrices of the RV satisfy: $\Sigma_{12}=\Sigma_{1U}\Sigma_U^{-1}\Sigma_{U2}$.

Let us define the matrix:

$\Sigma=\left( \begin{array}{ccc} \Sigma_1 & \Sigma_{1U} & A\\\ \Sigma_{1U}^T & \Sigma_U & \Sigma_{2U}^T\\\ A^T & \Sigma_{2U} & \Sigma_{2}\end{array}\right)$.

where $A=\Sigma_{1U}\Sigma_{U}^{-1}\Sigma_{U2}$. My question regards the set of all legitimate matrices $\Sigma$, is this set convex? How can one check this?

Obviously not. The data $\Sigma_{U,1,2,1U, U2}$ are independent apart from inequalities that leaves room for an open set. If the set was convex, $A$ would be a linear function of these data, but it is not. A natural question is: what is the convex hull of the set of such matrices? – Denis Serre Jan 19 '13 at 22:24
I'm sorry, but I didn't understand your answer fully. When you say that the data are independent apart from some inequalities do you mean the following: $\Sigma_i-\Sigma_{iU}\Sigma_U^{-1}\Sigma_{Ui}\geq 0$, and the non-negativety constraints, i.e., $\Sigma_i\geq 0$ and $\Sigma_U\geq 0$, where $i=1,2$. Are there any additional relations between the data matrices? What do you mean by "leaves room for an open set"? And finally, why $A$ must be linear if the set is to be convex? There are convex sets that are not linear. I'd appreciate if you could clarify this for me. Thank you in advance! – Ziv Goldfeld Jan 20 '13 at 10:03