Given the following two R.V.s


$$z\_{1} = \frac{x\_{1}}{|x\_{1}|^2 + |x\_{2}|^2 + ... + |x\_{M}|^2}$$


and


$$z\_{2} = \frac{x\_{2}}{|x\_{1}|^2 + |x\_{2}|^2 + ... + |x\_{M}|^2}$$


where $x\_{i} \sim \mathcal{CN}(0,a), \forall i$ and $a > 1$. As can be seen, the denominator follows a Chi-square distribution with $2M$ degrees of freedom as $x\_{i}$ are i.i.d. R.V.s.


Is it possible to calculate 


$$\mathbb{E} \{ z\_{1} z\_{2}^{\*}\},$$


and show that the variables are correlated or not? Not that $\*$ is the conjugate.



Correlation between r.v.'s following a distribution that is the ration between complex Gaussian and Chi-square r.v.'s
First here, by the Cauchy--Schwarz inequality, 
$$E|z\_1z^\*\_2|=E|z\_1|\,|z^\*\_2|\le\sqrt{E|z\_1|^2E|z^\*\_2|^2}=E|z\_1|^2<\infty,$$
by [Addition in response to the modification of the OP's original question](https://mathoverflow.net/questions/290092/distribution-of-ratio-between-complex-gaussian-and-chi-square-r-v-s). 
So, $Ez\_1z^\*\_2$ exists and is finite. 
Therefore and because the joint distribution of the pair $(-z\_1,z^\*\_2)$ is the same as that of $(z\_1,z^\*\_2)$, we conclude that 
$$Ez\_1z^\*\_2=E(-z\_1)z^\*\_2=0.$$


**Detail sdded in response to the OP's comment:** The $x\_i$'s are iid and, for each $i$, the distribution of $-x\_i$ is the same as that of $x\_i$. So, the joint distribution of $(-x\_1,x\_2,\dots,x\_M)$ is the same as that of $(x\_1,x\_2,\dots,x\_M)$. Also, $|-x\_1|=|x\_1|$. So, the random pair
\begin{multline\*}
 (z\_1,z^\*\_2)=g(x\_1,x\_2,\dots,x\_M):= \\ 
 \Big(\frac{x\_{1}}{|x\_{1}|^2 + |x\_{2}|^2 + \dots + |x\_{M}|^2},
 \frac{x\_{2}}{|x\_{1}|^2 + |x\_{2}|^2 + \dots + |x\_{M}|^2}\Big) 
\end{multline\*} 
equals 
\begin{multline\*}
 (-z\_1,z^\*\_2)= 
 \Big(\frac{-x\_{1}}{|-x\_{1}|^2 + |x\_{2}|^2 + \dots + |x\_{M}|^2},
 \frac{x\_{2}}{|x\_{1}|^2 + |x\_{2}|^2 + \dots + |x\_{M}|^2}\Big) \\ 
 =g(-x\_1,x\_2,\dots,x\_M)
\end{multline\*} 
in distribution, 
as was stated above. 


