Let $\mathbf x=[x_1, ... ,x_K]^T$, $x\sim\mathcal C\mathcal N(\mathbf 0,\sigma_x^2\mathbf I)$, I believe that the distribution of $||\mathbf x||^2=\mathbf x^{\dagger}\mathbf x$ is Erlang (Please see attached photo from "Performance analysis of cloud radio access networks with distributed multiple antenna remote radio heads"). Is there anything known about the distribution of $\left (\mathbf x^{\dagger}\mathbf x\right )^2$? If not, does it converge in distribution? How can we know which distribution? If yes, can we prove it?
This is what I can say for the moment. We have from the discussion:
$V \equiv x^+x \sim \sum_{i=1}^N e_i$
where $e_i$ are independent and exponentially distributed $e_i \sim Exp(\lambda)$.
Than we have $V^2=(x^+x)^2 \sim \sum_{i,j}e_ie_j$. This is the variable we are interested in. Let's try some asymptotics.
If we call $S_N= V/N$ by the central limit theorem for $N$ large:
$(S_N-\frac{1}{\lambda})\sqrt{N} \sim \mathcal{N}(0,\frac{1}{\lambda^2})$ [1]
applying the delta method with $g(x)=x^2$:
$(S^2_N-\frac{1}{\lambda^2})\sqrt{N} \sim \mathcal{N}(0,\frac{4}{\lambda^4})$
and $V^2=S_N^2N^2$ so that:
$(\frac{V^2}{N^2}-\frac{1}{\lambda^2})\sqrt{N} \sim \mathcal{N}(0,\frac{4}{\lambda^4})$ [2]
and informally :
$V^2 \sim \mathcal{N}(\frac{N^2}{\lambda^2},\frac{4}{\lambda^4}N^3)$
All convergences are intended in distribution.