Let $\mathbf{Z}$ be a $m\times n$ matrix with zero-mean unit-variance i.i.d complex Gaussian entries. What is the maximum value of mutual information $I(\mathbf{X}_1,\mathbf{X}_2;\mathbf{Y})$ such that \begin{align} \mathbf{Y}=\mathbf{X}_1^\mathrm{H}\mathbf{X}_2+\mathbf{X}_1^\mathrm{H}\mathbf{Z}, \end{align} where $\mathbf{X}_1$ and $\mathbf{X}_2$ are two $m\times n$ arbitrary random matrices and $(.)^\mathrm{H}$ denotes hermitian of matrix. Moreover, the column norm of $\mathbf{X}_i$'s are less than or equal to one.
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1$\begingroup$ I believe that in your setup, the MI is unbounded. To see why, let $\mathbf{X}_1 = \mathbf{I}$ with probability 1, so $\mathbf{Y}=\mathbf{X_2} + \mathbf{Z}$. We then have $I(\mathbf{X_1},\mathbf{X_2};\mathbf{Y})=I(\mathbf{X_2};\mathbf{Y})=h(\mathbf{Y}) - h(\mathbf{Y}\vert \mathbf{X_2}) \ge h(\mathbf{X}_2) - h(\mathbf{Z})$. Now the differential entropy $ h(\mathbf{Z})$ is a finite quantity, while the differential entropy $h(\mathbf{X}_2)$ can be arbitrarily large. Hence, the MI can be arbitrarily large. $\endgroup$– ArtemyCommented Apr 13, 2021 at 4:14
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$\begingroup$ You are right. We have a condition that $\mathbf{X}_2$ is bounded. I append it to question. $\endgroup$– Math_YCommented Apr 13, 2021 at 10:29
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