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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) and (2) and on the observation that for $𝑀>5$ 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?

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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 $M$.

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  • $\begingroup$ What do you mean by ${\cal O}(M^{-7})$? $\endgroup$
    – Felipe Augusto de Figueiredo
    Commented Feb 6, 2019 at 22:43
  • $\begingroup$ One more question, how did you find $a^2\,\mathbb{E}[|z_1|^2|z_2|^2]=\frac{1}{4M(M+1)(M-1)(M-2)}$? $\endgroup$
    – Felipe Augusto de Figueiredo
    Commented Feb 6, 2019 at 22:48
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    $\begingroup$ ${\cal O}(M^{-7})$ means that the next order term in an expansion in powers of $1/M$ is of order $M^{-7}$; I used Mathematica for the integrals; it will not return an answer for a symbolic $M$, but it will for any integer $M$ and then the polynomial in the denominator is easily obtained; perhaps there is a way to obtain this directly, but to demonstrate the absence of independence of $z_1$ and $z_2$ this suffices. $\endgroup$
    – Carlo Beenakker
    Commented Feb 7, 2019 at 7:18

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