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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.

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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. 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.

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  • $\begingroup$ Dear, thanks. I don't understand the inequality, is this some known one? Could you please, give further details on that? $\endgroup$ – Felipe Augusto de Figueiredo Feb 2 at 9:44
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    $\begingroup$ The only inequality in the above answer is an instance of the Cauchy--Schwarz inequality, which is now made explicit in the answer. $\endgroup$ – Iosif Pinelis Feb 3 at 1:49
  • $\begingroup$ I've accepted your answer as the correct one as I have simulated the correlation and it is zero as you pointed out, however, I fail to see how you conclude that the joint distribution of the pair (−𝑧1,𝑧∗2) is the same as that of (𝑧1,𝑧∗2) once the joint pdf is unknown. Is it possible to conclude that without knowing the joint pdf of (𝑧1,𝑧∗2)? Thanks! $\endgroup$ – Felipe Augusto de Figueiredo Feb 3 at 11:17
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    $\begingroup$ @FelipeAugustodeFigueiredo : I have added the detail you requested. $\endgroup$ – Iosif Pinelis Feb 3 at 15:34
  • $\begingroup$ Thanks, it is clear now. $\endgroup$ – Felipe Augusto de Figueiredo Feb 3 at 17:23

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