In the deterministic case, for two matrices $A$ and $B$ with appropriate matrices, we know that $$tr((A^{T}B)^{2})\leq tr(A^{T}A)tr(B^{T}B)$$ which is the trace form of Cauchy-Schwarz-Inequality (CSI).

However, when we wish to minimize some tracking error in closed control loop for stochastic systems, we are confused about the fact that, whether the following statement is true or not. If correct, how to prove it? In such a case, $A$ and $B$ are random matrices. $$E\{tr((A^{T}B)^{2})\}\leq E\{tr(A^{T}A)tr(B^{T}B)\}$$ or $$E\{tr((A^{T}B)^{2})\}\leq E\{tr(A^{T}A)\}E\{tr(B^{T}B)\}$$

Rigorously speaking, for stochastic matrices, if the first inequality (deterministic CSI) is established 'almost surely', then we can get the second one.

allmatrices (without needing to refer to any probability), so why would one need any specific a.s. argument for stochastic matrices? $\endgroup$ – Suvrit Sep 21 '18 at 11:57