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Let $\mathbf{U}$ and $\mathbf{V}$ be random unitary matrices independent of random input vector $\mathbf{x}$. Moreover, $\mathbf{z}$ be random iid complex Gaussian vector with zero mean and identity covaiance matrix. Output random vector $\mathbf{y}$ is equal to \begin{align} \mathbf{y}_{m\times 1}=\mathbf{U}_{m\times m}\mathbf{D}_{m\times n}\mathbf{V}_{n\times n}\mathbf{x}_{n\times 1}+\mathbf{z}_{m\times 1}, \end{align} where $m<n$ and $\mathbf{D}=[\mathbf{I}_m,\mathbf{0}_{m\times (n-m)}]$ in which $\mathbf{I}_m$ is the identity matrix of size $m$.

Then, what is the following mutual information maximization \begin{align} \max_{p(\mathbf{x}): \mathbb{E}[\mathbf{x}^\mathrm{H}\mathbf{x}]=1} I(\mathbf{x};\mathbf{y}), \end{align} where $(.)^\mathrm{H}$ is the Hermitian operator.

My Guess is that the compex Gaussian distribution is the maximizing distribution because it is unitary invariant.

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  • $\begingroup$ does the maximum exist? I would think $I$ can become arbitrarily large by choosing $\mathbf{x}=(x_0,0,0,\ldots 0,0)$ with $E[x_0^2]=1$ and then giving $x_0^2$ an arbitrarily large variance. $\endgroup$ May 17, 2021 at 8:45
  • $\begingroup$ I do not understand your comment. Variance of $x_0$ is $1$ as you write. $\endgroup$
    – Math_Y
    May 17, 2021 at 9:47
  • $\begingroup$ Why $var(x_0^2)$ is important? This maximization is known for example for simple case $\mathbf{y}=\mathbf{x}+\mathbf{z}$. $\endgroup$
    – Math_Y
    May 17, 2021 at 10:14

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