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Given that the random variables $\textbf{x} \sim \mathcal{CN}(\textbf{0}_{M},\sigma_{x}^{2}\textbf{I}_{M})$ and $\textbf{y} \sim \mathcal{CN}(\textbf{0}_{M},\sigma_{y}^{2}\textbf{I}_{M})$ are independent, what would be the p.d.f. of

$$Z = \left| \frac{\textbf{x}^{H} \textbf{y} }{\| \textbf{x} \|^2} \right|^2, $$ where $\mathcal{CN}(.,.)$ is the complex normal random variable.

Which kind of variable change could be applied here?

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    $\begingroup$ Eigendecomposition? $\endgroup$ – LeechLattice Feb 16 '19 at 12:59
  • $\begingroup$ @Bullet51, could you explain how to to that, please? $\endgroup$ – Felipe Augusto de Figueiredo Feb 16 '19 at 17:10
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You can use the rotational invariance of the Gaussian distribution to take $\mathbf{y}=2^{-1/2}\sigma_y(\sqrt \xi_{2M},0,0,\ldots,0)$, with $\xi_{2M}$ distributed independently of $\mathbf{x}$ according to a chi-squared distribution with $2M$ degrees of freedom. Then $$Z=\tfrac{1}{2}\sigma_y^2 \xi_{2M}|z|^2\;\;\text{with}\;\; z=\frac{x_1}{\sum_{n=1}^M |x_n|^2}.$$ This can be further reduced to $$Z=\frac{\sigma_y^2}{\sigma_x^2} \frac{\xi_{2M}\xi_{2}}{(\xi_{2}+\xi_{2M-2})^2},$$ with $\xi_2$ and $\xi_{2M-2}$ independently chi-squared distributed with $2$, respectively, $2M-2$ degrees of freedom.

I don't think the distribution of this rational function of three independent chi-squared distributions has a closed form expression. For $M\gg 1$ one has simply $$Z\rightarrow \frac{\sigma_y^2}{2M\sigma_x^2}\xi_2,$$ so $Z$ has for large $M$ a chi-squared distribution with 2 degrees of freedom.

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  • $\begingroup$ Thanks for your answer. Do you think it is possible to find a closed-form expression for the moments of $Z$? $\endgroup$ – Felipe Augusto de Figueiredo Feb 16 '19 at 14:03
  • $\begingroup$ Please, could you explain how you arrived at the limit when $M \gg 1$? $\endgroup$ – Felipe Augusto de Figueiredo Feb 16 '19 at 17:39
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    $\begingroup$ for large $M$ the sum of the square of $2M$ normally distributed independent variables self-averages to $2M$, so $\xi_{2M}$ in the numerator and $\xi_2+\xi_{2M-2}$ in the denominator can both be replaced by $2M$. $\endgroup$ – Carlo Beenakker Feb 16 '19 at 18:10
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    $\begingroup$ for the moments of $Z$, see mathoverflow.net/a/323393/11260 $\endgroup$ – Carlo Beenakker Feb 16 '19 at 18:51
  • $\begingroup$ Dear Carlo, do you think you could help me with this problem? mathoverflow.net/questions/337341/… thanks! $\endgroup$ – Felipe Augusto de Figueiredo Aug 3 '19 at 10:46

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