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Given \begin{equation}\label{eq:definition_of_z} \begin{split} \textbf{Z} = \left[\begin{array}{cccc} {z}_{11} & {z}_{12} & \cdots & {z}_{1P} \\ {z}_{21} & {z}_{22} & \cdots & {z}_{2P} \\ {z}_{31} & {z}_{32} & \cdots & {z}_{3P} \\ \vdots & \vdots & \ddots & \vdots \\ {z}_{M1} & {z}_{M2} & \cdots & {z}_{MP} \\ \end{array} \right] = \left[\begin{array}{cccc} \textbf{z}_{1} & \textbf{z}_{2} & \cdots & \textbf{z}_{P} \\ \end{array} \right] \end{split} \end{equation}

where the $z_{mp}$ are i.i.d. $\forall \ m,p$ with distribution $z_{mp} \sim \mathcal{CN}(0,c)$ and $\textbf{z}_{i}$ represents each of the columns of $\textbf{Z} \sim \mathcal{CN}(\textbf{0}_{M \times P},\text{cM}\textbf{I}_{P})$.

Other necessary definition is given by

\begin{equation}\label{eq:definition_of_z_z} \textbf{Z}^{H} \textbf{Z} = \left[\begin{array}{c} \textbf{z}_{1}^{H} \\ \textbf{z}_{2}^{H} \\ \vdots \\ \textbf{z}_{P}^{H} \\ \end{array} \right] \left[\begin{array}{cccc} \textbf{z}_{1} & \textbf{z}_{2} & \cdots & \textbf{z}_{P} \\ \end{array} \right] = \left[\begin{array}{cccc} \textbf{z}_{1}^{H}\textbf{z}_{1} & \textbf{z}_{1}^{H} \textbf{z}_{2} & \cdots & \textbf{z}_{1}^{H} \textbf{z}_{P} \\ \textbf{z}_{2}^{H} \textbf{z}_{1} & \textbf{z}_{2}^{H} \textbf{z}_{2} & \cdots & \textbf{z}_{2}^{H} \textbf{z}_{P} \\ \vdots & \vdots & \ddots & \vdots \\ \textbf{z}_{P}^{H} \textbf{z}_{1} & \textbf{z}_{P}^{H} \textbf{z}_{2} & \cdots & \textbf{z}_{P}^{H} \textbf{z}_{P} \\ \end{array} \right] \end{equation}

Next we define the following ratio:

\begin{equation}\label{eq:channel_matrix} \frac{ \textbf{Z}^{H} \textbf{Z} }{ \text{Tr} \left( \textbf{Z}^{H} \textbf{Z} \right)^{2} } = \frac{ \textbf{Z}^{H} \textbf{Z} }{ \left( \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P} \right)^{2} } \end{equation}

where $\text{Tr}$ is the Trace operator.

It is important to note that the elements of the main diagonal of $\textbf{Z}^{H} \textbf{Z}$, namely $\textbf{z}_{1}^{H}\textbf{z}_{1}, \ \textbf{z}_{2}^{H}\textbf{z}_{2}, \ \cdots, \ \textbf{z}_{P}^{H}\textbf{z}_{P} \sim \Gamma(M,2c)$.

I'd like to prove that the following expectation is equal to 0.

\begin{equation}\label{eq:4} \mathbb{E} \left\lbrace \frac{\textbf{z}_{i}^{H}\textbf{z}_{j}}{\left( \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P} \right)^{2} } \right\rbrace = 0, \ \text{when} \ i \neq j. \end{equation}

I know for a fact that the above statement is true due to some Matlab simulations I ran but I'd like to prove that mathematically.

It's also worth mentioning that $\textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P}$ results in a scalar random variable.

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  • $\begingroup$ I'm a little confused by the notation; are the $z_{ij}$ iid? $\endgroup$ – Dougal Jan 28 '17 at 0:38
  • $\begingroup$ @Dougal, I've just fixed the notations and added more information and yes, $z_{mp}$ are i.i.d. $\forall \ m,p$ with distribution $z_{mp} \sim \mathcal{CN}(0,c)$. I hope this answers your question. $\endgroup$ – Felipe Augusto de Figueiredo Jan 28 '17 at 9:57

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