# p.d.f. of $\left| \frac{\textbf{x}^{H} \textbf{y} }{\| \textbf{x} \|^2} \right|^2$, where $\textbf{x}$ and $\textbf{y}$ are complex Gaussians?

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?

• Eigendecomposition? – LeechLattice Feb 16 '19 at 12:59
• @Bullet51, could you explain how to to that, please? – Felipe Augusto de Figueiredo Feb 16 '19 at 17:10

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.
• Thanks for your answer. Do you think it is possible to find a closed-form expression for the moments of $Z$? – Felipe Augusto de Figueiredo Feb 16 '19 at 14:03
• Please, could you explain how you arrived at the limit when $M \gg 1$? – Felipe Augusto de Figueiredo Feb 16 '19 at 17:39
• 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 $2M$. – Carlo Beenakker Feb 16 '19 at 18:10
• for the moments of $Z$, see mathoverflow.net/a/323393/11260 – Carlo Beenakker Feb 16 '19 at 18:51