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Let random matrix $\mathbf{X} \in \mathbb{C^{\mathrm{m} \times \mathrm{n}}}$ and random vector $\mathbf{y} \in \mathbb{C^{\mathrm{m} \times 1}}$ are unknown distributed, but their covariance and correlation are known, $\mathbf{C}_{\mathbf{X}}$, $\mathbf{C}_{\mathbf{y}}$, and $\mathbf{C}_{\mathbf{Xy}}$. The question is whether there is any bound for the following expression, (like Cauchy-Schwarz or Jensen's)

$?\le \mathbb{E}[(\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} \mathbf{X}^{\mathrm{H}} \mathbf{y}] \le ?$

I have to add that each element in matrix $\mathbf{X}$ or $\mathbf{y}$ is the output of a sign function, or in other words

Let $x : = \mathbf{X}_{i,j}, \ \forall i=1,\ldots m, \ \forall j=1,\ldots n$ , $\ y : = \mathbf{y}_{l}, \ \forall l=1,\ldots m $
then $x = sgn(z), \ y = sgn(t)$ where both $z$ and $t$ are gaussian random variables, and $sgn(.)$ means sign function. Also, I have to mention $\le$ here means an element-wise inequality since the corresponding expectation is an $\mathrm{n} \times 1$ vector.

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  • $\begingroup$ Does $(\cdot) ^{\mathrm H}$ mean "Hermitian adjoint"? $\endgroup$
    – Alex M.
    Commented Feb 20, 2023 at 12:22
  • $\begingroup$ @ Alex M: It means conjugate transpose of a matrix. $\endgroup$
    – A. R.
    Commented Feb 20, 2023 at 14:44

1 Answer 1

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I wonder maybe this could solve the problem, but maybe is wrong.

Let $f_1(\mathbf{X}^{\mathrm{H}}\mathbf{X}) = (\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} $ and as we know it's a convex function. Also $f_2(\mathbf{X},\mathbf{y}) = \mathbf{X}^{\mathrm{H}} \mathbf{y}$ is a convex function. Then

$ f_1(\mathbb{E}[\mathbf{X}^{\mathrm{H}}\mathbf{X}]) f_2(\mathbb{E}[\mathbf{X},\mathbf{y}]) \le \mathbb{E}[f_1(\mathbf{X}^{\mathrm{H}}\mathbf{X})f_2(\mathbf{X},\mathbf{y})] = \mathbb{E}[(\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} \mathbf{X}^{\mathrm{H}} \mathbf{y}]$

so we have $ (\mathbb{E}[\mathbf{X}^{\mathrm{H}}\mathbf{X}])^{-1} \mathbb{E}[\mathbf{X}^{\mathrm{H}}\mathbf{y}] \le \mathbb{E}[(\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1} \mathbf{X}^{\mathrm{H}} \mathbf{y}]$

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  • $\begingroup$ It is not clear how one could define a function by $f_1(\mathbf{X}^{\mathrm{H}}\mathbf{X}) = (\mathbf{X}^{\mathrm{H}} \mathbf{X})^{-1}$. What does this mean? Who is $f_1 (x)$ if $x$ is a real number? Do you mean that $f_1 (x) = \frac 1 x$ for $x>0$? $\endgroup$
    – Alex M.
    Commented Feb 20, 2023 at 12:20
  • $\begingroup$ @ Alex M: Yes $f_1(\mathbf{Z}) = (\mathbf{Z})^{-1}$ . $\endgroup$
    – A. R.
    Commented Feb 20, 2023 at 13:07

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