Let $\mathbf x=[x_1, ... ,x_K]^T$, $x_k\sim\mathcal C\mathcal N(\mathbf 0,\sigma_x^2\mathbf I)$, $\mathbf y=[y_1, ... ,y_K]^T$, $y_k\sim\mathcal C\mathcal N(\mathbf 0,\sigma_y^2\mathbf I)$, and suppose $z=\mathbf x^T\mathbf y^*\mathbf x^H\mathbf x$, where T, *, and H are for transpose, conjugate, and hermitian transpose respectively. I'm interested in calculating the mean of $z$ and $E[|z|^2]$. As it's evident, the mean of $z$ is zero, so we can calculate $E[|z|^2]=\sigma_z^2$ as follows. Since $z$ is Gaussiane with zero mean and variance of $\sigma_z^2=M^2\sigma_x^4\sigma_y^2$, then $g=\frac{2}{\sigma_z^2}|z|^2\sim \chi^2(2)$. Moreover, $E[|z|^2]=\frac{\sigma_z^2}{2}E[g]=\sigma_z^2=M^2\sigma_x^4\sigma_y^2$.
The above seems correct to me but when simulating the equation in Matlab I don't get the same result. Can anyone tell me where I was wrong? I guess $E[z]\ne{0}$. If it's not zero then what is it?
Write $(\quad)^{\dagger}$ for the conjugate transpose operation. Then $$z=(\mathbf{y}^{\dagger}\mathbf{x})(\mathbf{x}^{\dagger}\mathbf{x})$$ indeed satisfies $$\mathrm{E}[z]=0$$ because of reflection symmetry of the expectation of $\mathbf{y}^{\dagger}$. However, $$\begin{split}\mathrm{E}[z^*z]&=\mathrm{E}[(\mathbf{x}^{\dagger}\mathbf{x})(\mathbf{x}^{\dagger}\mathbf{y})(\mathbf{y}^{\dagger}\mathbf{x})(\mathbf{x}^{\dagger}\mathbf{x})]\\ &=\mathrm{E}[(\mathbf{x}^{\dagger}\mathbf{x})(\mathbf{x}^{\dagger}\mathrm{E}[\mathbf{y}\mathbf{y}^{\dagger}]\mathbf{x})(\mathbf{x}^{\dagger}\mathbf{x})] \\ &=\mathrm{E}[(\mathbf{x}^{\dagger}\mathbf{x})(\mathbf{x}^{\dagger}(\sigma_y^2\mathsf{I})\mathbf{x})(\mathbf{x}^{\dagger}\mathbf{x})]\\ &=\sigma_y^2 \mathrm{E}[(\mathbf{x}^{\dagger}\mathbf{x})^3]\\ &=K(K+1)(K+2)\sigma_x^6\sigma_y^2\text{.} \end{split}$$