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.



Expectation of ratio between product of gaussian r.v.'s and generalized gamma r.v
First, by the Cauchy-Schwarz inequality, we have 
$$\mathbb{E} [ | \textbf{z}\_1^{H} \textbf{z}\_2 | ] \le\sqrt{\mathbb{E}[|\textbf{z}\_1|^2] \mathbb{E}[|\textbf{z}\_2|^2]} = \mathbb{E}[|\textbf{z}\_1|^2]<\infty,$$


hence, $\mathbb{E} [ \textbf{z}\_1^{H} \textbf{z}\_2 ]$ exists and is finite. 
Therefore and because the joint distribution of the pair $(-\textbf{z}\_1^H,\textbf{z}\_2)$ is the same as that of $(\textbf{z}\_1^H, \textbf{z}\_2)$, we conclude that 
$$\mathbb{E}[\textbf{z}\_1^{H} \textbf{z}\_2] = \mathbb{E} [ (-\textbf{z}\_1)^H \textbf{z}\_2 ] = 0.$$


If $\textbf{z}\_{i} \forall i$ is a vector composed of i.i.d. random variables following a symmetric distribution around 0, then
$$\mathbb{E}[\textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})^{H} \textbf{h}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})] = 0,$$
where $\textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})$ and $ \textbf{h}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})$ are functions of the random variables $\textbf{z}\_{i} \forall i$, which present the same symmetric distribution around 0.


**Proof**


First, by using the Cauchy-Schwarz inequality, we find that
$$\mathbb{E} [ | \textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})^{H} \textbf{h}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M}) | ] \le\sqrt{\mathbb{E}[|\textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})|^2] \mathbb{E}[|\textbf{h}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})|^2]} = \mathbb{E}[|\textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})|^2]<\infty,$$
and thus, $\mathbb{E}[\textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})^{H} \textbf{h}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})]$ exists and if finite. Next, as $\textbf{z}\_{i} \forall i$ is a vector composed of i.i.d. random variables following a symmetric distribution around 0, then, the distribution $f\_{\textbf{z}\_{i}}(\textbf{z}\_{i})$ is the same as $f\_{\textbf{z}\_{i}}(-\textbf{z}\_{i})$. Therefore, the joint distribution $f\_{\textbf{z}\_{1},\textbf{z}\_{2},\ldots,\textbf{z}\_{M}}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})$ is the same as that of $f\_{\textbf{z}\_{1},\textbf{z}\_{2},\ldots,\textbf{z}\_{M}}(-\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})$. We also know that the distribution of $\textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})$ is equal to the distribution of $\textbf{g}(-\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})$ and that the distribution of $\textbf{h}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})$ is equal to the distribution of $\textbf{h}(-\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})$. Therefore, we conclude that $\mathbb{E}[\textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})^{H} \textbf{h}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})] = \mathbb{E} [(-\textbf{g}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M}))^{H} \textbf{h}(\textbf{z}\_{1}, \textbf{z}\_{2}, \ldots, \textbf{z}\_{M})] ] = 0$.


