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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.

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    $\begingroup$ the first conjectured inequality is just the integration of your initial inequality (CSI) $\endgroup$
    – Fedor Petrov
    Commented Sep 21, 2018 at 10:54
  • $\begingroup$ Rigorously speaking, for stochastic matrices, the first inequality (deterministic CSI) should be established 'almost surely', then we can get the second one. Furthermore, I am wondering that how to give and prove such an inequality if it exists. $\endgroup$
    – Dude-Ray
    Commented Sep 21, 2018 at 11:17
  • $\begingroup$ @Dude-Ray the deterministic inequality holds for all matrices (without needing to refer to any probability), so why would one need any specific a.s. argument for stochastic matrices? $\endgroup$
    – Suvrit
    Commented Sep 21, 2018 at 11:57
  • $\begingroup$ @Suvrit, in the stochastic scenario, $A$ and $B$ are random matrices rather than deterministic ones. $\endgroup$
    – Dude-Ray
    Commented Sep 21, 2018 at 12:23
  • $\begingroup$ You're asking for a certain inequality to hold almost surely. But because you have the deterministic inequality, the one that you want to hold almost surely will actually hold surely. $\endgroup$
    – user114263
    Commented Sep 21, 2018 at 13:21

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As Fedor said in the comments, the first inequality is correct by integrating the deterministic inequality over the probability space.

The second inequality, however, has no chance to be correct: Take $A=B$ to be $1\times 1$ matrices. Then the inequality reads $$ \mathbf{E}\,A^4\le (\mathbf{E}\,A^2)(\mathbf{E}\,A^2),$$ which cannot be correct. As an example, if $A$ is a standard real Gaussian, the lhs. is $3$, while the rhs. is $1$.

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