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Let $\mathbf x=[x_1, ... ,x_K]^T$, $x\sim\mathcal C\mathcal N(\mathbf 0,\sigma_x^2\mathbf I)$, I'm interested in calculating $E\bigl[\|\mathbf x\|^8\bigr]$ in terms of $\sigma_x^2$ and $K$. I'm wondering which way would be best, using the multinomial theorem or to express the problem as follows: \begin{align} E\bigl[\|\mathbf{x}\|^8\bigr]&=E[\left(\sum_{i=1}^K |x_i|^2\right)^4 \cr &=E\biggl[\sum_{i=1}^K\sum_{j=1}^K\sum_{k=1}^K\sum_{l=1}^K|x_i|^2|x_j|^2|x_k|^2|x_l|^2\biggr]\cr &=E\biggl[\sum_{i=1}^K|x_i|^2|x_i|^2|x_i|^2|x_i|^2\biggr]+E\biggl[\sum_{i=1}^K\sum_{j=1,j\ne i}^K|x_i|^2|x_i|^2|x_j|^2|x_j|^2\biggr]+\dotsm\\ \end{align} I know both ways are tedious, is there another better way to calculate the expectation?

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The easiest way I know would be to utilize the MGF of a $\chi^2.$


Note that $$Y_K = \frac{2}{\sigma_x^2}\sum_{i=1}^K |x_i|^2 \sim \chi^2(2K)$$

Therefore, $$E[\|\mathbf{x}\|^8] = E\left[\left(\sum_{i=1}^K|x_i|^2\right)^4\right] = \frac{\sigma_x^8}{2^4}E[Y_K^4]$$

Since the $MGF$ of a $\chi^2(2K)$ is given by $$(1-2t)^{-K} = \sum_{n=0}^\infty \binom{-K}{n}(-2t)^n = \sum_{n=0}^\infty 2^n(K)_n\frac{t^n}{n!}$$ we can conclude $$\frac{\sigma_x^8}{2^4}E[Y_K^4] = \frac{\sigma_x^8}{2^4}\cdot2^4(K)_4 = \sigma_x^8\cdot K(K+1)(K+2)(K+3).$$


Notation: $(x)_k = x(x+1)\cdots(x+k-1)$ is the Pochhammer Symbol.

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  • $\begingroup$ That was awesome. Thank you. $\endgroup$ – Aymen Kareem Dec 27 '19 at 18:42

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