How to prove that $E[||\mathbf x||^4]=K(K+1)$? where $K$ is the length of $\mathbf x$

Question

Let $\mathbf x=[x_1, ... ,x_K]^T$, $x\sim\mathcal C\mathcal N(\mathbf 0,\sigma_x^2\mathbf I)$, I have worked out the value of $E[||\mathbf x||^4]$ by assuming values for the vector $\mathbf x$ and found it was $K(K+1)$ when $\sigma_x^2=1$. Can I prove it in the same manner as the answer in Expectation of $x^4$. Also, what is the above expectation when $\sigma_x^2\ne 1$? What is $E[||\mathbf x||^8]$?

Answer 1

Let $y_i=\operatorname{re}(x_i)$ and $z_i=\operatorname{im}(x_i)$. Then, for $\sigma_x=1$, \begin{align} \mathsf{E}|\mathbf{x}|^4&=\mathsf{E}\left(\sum_{i=1}^K (y_i^2+z_i^2)\right)^2 \\ &=\sum_{i=1}^K\mathsf{E}(y_i^2+z_i^2)^2+\sum_{i=1}^K\sum_{j\ne i}\mathsf{E}(y_i^2+z_i^2)(y_j^2+z_j^2) \\ &=\sum_{i=1}^K(\mathsf{E}y_i^4+\mathsf{E}z_i^4+2\mathsf{E}y_i^2\mathsf{E}z_i^2)+\sum_{i=1}^K\sum_{j\ne i}\mathsf{E}(y_i^2+z_i^2)\mathsf{E}(y_j^2+z_j^2) \\ &=\sum_{i=1}^K 8\cdot2^{-2}+\sum_{i=1}^K\sum_{j\ne i}4 \cdot 2^{-2}=K(K+1). \end{align}

For $\sigma_x\ne 1$, note that $\mathbf{x}/\sigma_x\sim \mathcal{CN}(0,I)$.