$$\nonumber
    \mathbf{J}=\mathbf{I}_{\frac{n_T}{2}}\otimes\left[%
    \begin{array}{cc}
      0 & 1 \\
      -1 & 0 \\
    \end{array}%
    \right].$$

2.  $
        \mathbb{E}\left[\mathbf{H}\mathbf{H}_e^H\right]=\mathbb{E}\left[\mathbf{H}\mathbf{J}^T\mathbf{H}^T\right]=\mathbf{0}
        $.

The proof is straightforward and uninformative and thus it is omitted.

Eq.  can be rewritten as $$\begin{aligned}
\nonumber
I_{sA} & =\frac{1}{2}\log_2\Bigg(\det\left(\mathbf{I}_{n_T}
+\frac{\rho}{n_T}(\mathbf{H})^H\mathbf{H}\right) \times\\
&\det\left(\mathbf{I}_{n_T}+\frac{\rho}{n_T}\mathbf{H}'_e\left(\mathbf{I}_{n_T}
+\frac{\rho}{n_T}(\mathbf{H})^H\mathbf{H}\right)^{-1}(\mathbf{H}'_e)^H\right) \Bigg) \nonumber\\
&= \frac{1}{2}\log_2\det\left(\mathbf{I}_{n_T} +\frac{\rho}{n_T}(\mathbf{H})^H\mathbf{H}\right) +\frac{1}{2}\;\times \nonumber \\
 &  \log_2\det\Big(\mathbf{I}_{n_T} + \frac{\rho}{n_T}\mathbf{H}'_e\left(\mathbf{I}_{n_T}
+\frac{\rho}{n_T}(\mathbf{H})^H\mathbf{H}\right)^{-1}(\mathbf{H}'_e)^H\Big)\;.\nonumber\end{aligned}$$ Since $\mathbf{H}'_e\left(\mathbf{I}_{n_T} +\frac{\rho}{n_T}(\mathbf{H})^H\mathbf{H}\right)^{-1}(\mathbf{H}'_e)^H$ is a positive semidefinite matrix, it follows immediately that the rate achieved with the stacked Alamouti is lower bounded by $$\begin{aligned}
\nonumber
I_{sA} \geq \frac{1}{2}\log_2\det\left(\mathbf{I}_{n_T} +\frac{\rho}{n_T}(\mathbf{H})^H\mathbf{H}\right),\end{aligned}$$ which is half the capacity of a MIMO system with $n_T$ transmit and $n_R$ receive antennas.

Another lower bound is obtained for the case $n_T\leq n_R $ by applying Minkowski’s determinant inequality [@HornJohnson p.482] ($\det(\mathbf{A}+\mathbf{B}) \geq
  (\det(\mathbf{A})^{\frac{1}{n}}+\det(\mathbf{B})^{\frac{1}{n}})^{n}$, $\mathbf{A}\succ0, \mathbf{B}\succeq0$) to  $$\begin{aligned}
R_{sA}& =\mathbb{E}\left[\frac{1}{2}\log_2\det\left(\mathbf{I}+\frac{\rho}{n_T}(\mathbf{H}')^H\mathbf{H}'\right)\right]
\nonumber \\ & \geq
\frac{n_T}{2}\mathbb{E}\left[\log_2\left(1+\rho\det\left(\frac{1}{n_T}(\mathbf{H}')^H\mathbf{H}'\right)^{\frac{1}{n_T}}\right)\right]\nonumber
\\
 & =\frac{n_T}{2}\mathbb{E}\left[\log_2\left(1+\rho\det\left(\frac{1}{n_T}(\mathbf{H}^H \mathbf{H}+
 \mathbf{H}_e^H\mathbf{H}_e)\right)^{\frac{1}{n_T}}\right)
 \right].\nonumber
 \end{aligned}$$ Applying again Minkowski’s determinant inequality results in $$\begin{aligned}
 R_{sA} \geq
\frac{n_T}{2}\mathbb{E}\Bigg[\log_2\Big(1+\rho\det\left(\frac{1}{n_T}\mathbf{H}^H\mathbf{H}\right) ^{1/n_T} \nonumber \\
+\rho\det\left(\frac{1}{n_T}\mathbf{H}_e^H\mathbf{H}_e\right)^{\frac{1}{n_T}}\Big)\Bigg].\nonumber\end{aligned}$$ Since $\mathbf{H}_e$ is obtained simply by conjugating and exchanging some elements of the actual matrix $\mathbf{H}$, it can be shown that the eigenvalues of $(\mathbf{H}_e)^H(\mathbf{H}_e)$ are the same as the eigenvalues of $\mathbf{H}^H(\mathbf{H})$. Therefore, the lower bound is equal to $$\begin{aligned}
 R_{sA}\geq
\frac{n_T}{2}\mathbb{E}\left[\log_2\left(1+\rho\exp\ln\left(2\det\left(\frac{1}{n_T}\mathbf{H}^H\mathbf{H}\right)^{\frac{1}{n_T}}
\right)\right)\right].\nonumber\end{aligned}$$ Since $\log_2(1+ce^x)$ 