Given the following two R.V.s


$$z\_1 = \frac{x\_1}{|x\_1|^2 + |x\_2|^2 + \cdots + |x\_M|^2}$$


and


$$z\_2 = \frac{x\_2}{|x\_1|^2 + |x\_2|^2 + \cdots + |x\_M|^2}$$


where $x\_i \sim \mathcal{CN}(0,a), \forall i$ and $a > 0$. 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.


Based on these results [(1)](https://mathoverflow.net/questions/322246/correlation-between-r-v-s-following-a-distribution-that-is-the-ration-between-c) and [(2)](https://mathoverflow.net/questions/290092/distribution-of-ratio-between-complex-gaussian-and-chi-square-r-v-s) and on the observation that for $>5$ the real and imaginary parts of $z\_i \forall i$ are normally distributed with mean equal to $0,$ can we say that $z\_{1}$ and $z\_2$ are independent?



Independence of r.v.'s following a distribution that is the ratio between complex Gaussian and Chi-square r.v.'s
*The answer is "no":* 


Take $M>2$ and evaluate


$$a\,\mathbb{E}[|z\_1|^2]=a\,\mathbb{E}[|z\_2|^2]=\frac{1}{2M(M-1)},$$
$$a^2\,\mathbb{E}[|z\_1|^2|z\_2|^2]=\frac{1}{4M(M+1)(M-1)(M-2)},$$
$$\Rightarrow a^2\,\mathbb{E}[|z\_1|^2|z\_2|^2]-a^2\,\mathbb{E}[|z\_1|^2]\mathbb{E}[|z\_2|^2]=\frac{1}{2M^6}+{\cal O}(M^{-7}),$$
so $z\_1$ and $z\_2$ are correlated no matter how large $M$.


