# Independence of r.v.'s following a distribution that is the ratio between complex Gaussian and Chi-square r.v.'s

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 $$đť‘€>5$$ 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?

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 $$M$$.
• What do you mean by ${\cal O}(M^{-7})$? – Felipe Augusto de Figueiredo Feb 6 '19 at 22:43
• One more question, how did you find $a^2\,\mathbb{E}[|z_1|^2|z_2|^2]=\frac{1}{4M(M+1)(M-1)(M-2)}$? – Felipe Augusto de Figueiredo Feb 6 '19 at 22:48
• ${\cal O}(M^{-7})$ 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. – Carlo Beenakker Feb 7 '19 at 7:18