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.



Expectation of $\left| \frac{\textbf{x}^{H} \textbf{y} }{\| \textbf{x} \|^2} \right|^2$, where $\textbf{x}$ and $\textbf{y}$ are complex Gaussians?
Using the representation in this [answer](https://mathoverflow.net/a/323376/11260), 
$$Z= \frac{|\textbf{x}^{H} \textbf{y} |^2}{ |\textbf{x} |^4} =\frac{\sigma\_y^2}{\sigma\_x^2} \frac{\xi\_{2M}\xi\_{2}}{(\xi\_{2}+\xi\_{2M-2})^2},$$
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?*


