Given that following two 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 expectation
$$\mathbb{E} \left[ \left| \frac{\textbf{x}^{H} \textbf{y} }{\| \textbf{x} \|^2} \right|^2 \right], $$ where $\mathcal{CN}(.,.)$ is the complex normal random variable.
Using the representation in this answer, $$Z= \frac{|\textbf{x}^{H} \textbf{y} |^2}{ |\textbf{x} |^4} =\frac{\sigma_y^2}{\sigma_x^2} \frac{\xi_{2M}\xi_{2}}{(\xi_{2}+\xi_{2M-2})^2},$$ and integrating over the independent chi-squared variables $\xi_2$, $\xi_{2M}$, and $\xi_{2M-2}$ I find the expectation value $$\Rightarrow\mathbb{E}(Z)=\frac{1}{M-1}(\sigma_y/\sigma_x)^2.$$ More generally, the $p$-th moment is finite for $M>p$, given by $$\mathbb{E}(Z^p)=\frac{(\sigma_y/\sigma_x)^2}{{M-1}\choose{p}}.$$
This answer looks simple enough, I wonder which distribution $P(Z)$ has reciprocals of binomial coefficients as its moments?
To add to the very intriguing accepted answer, here is a quick way to check without doing the difficult bits:
Use the trace trick to factor the mixed expression, $$|x^H y|^2 = x^H y y^H x = \mathrm{tr} \, x^H y y^H x = \mathrm{tr}\, x x^H \, y y^H$$ and move the trace outside to get the expectation $$\mathrm{E}\left[\frac{|x^H y|^2}{\Vert x \Vert^4}\right] = \mathrm{tr} \, E\left[\frac{xx^H}{\Vert x \Vert^4}\right] \Sigma_y\;.$$ Moving the expectation back outside produces $\mathrm{tr} \, xx^H \, \Sigma_y = \sigma_y^2 \, x^H x = \sigma_y^2 \, \Vert x \Vert^2$, since $\Sigma_y = \sigma_y^2 \, I$. The remaining expectation is that of an inverse-chi-squared distribution with $2M$ degrees of freedom, $$E\left[\frac{1}{\Vert x \Vert^2}\right] = \frac{2}{\sigma_x^2} \, \frac{1}{2M-2}$$ so that the final result is indeed $$\frac{\sigma_y^2}{\sigma_x^2} \, \frac{1}{M-1}\;.$$
(This might also be helpful in other situations, since normality is not used until the very end.)