Sixth and eighth moments of a complex Gaussian random vector

Question

Considering the random vector $\textbf{h} = [h_{1}, h_{2}, \ldots, h_{M}]^{T} \sim \mathcal{CN}(\textbf{0}_{M},d\textbf{I}_{M \times M})$, what would be the expectation of

$\mathbb{E}[\| \textbf{h} \|^{6}]$ and $\mathbb{E}[\| \textbf{h} \|^{8}]$ ?

Answer 1

Note: the following treatment assumes $d$ generalises the variance in a univariate $N(0,\,\sigma^2)$ distribution. I'm also not an expert on the complex Gaussian distribution, so I'll "flatten" the variable to a $2M$-dimensional real multivariate Gaussian. (This allows us to drop the $||$ signs around even powers.) There might be some powers-of-$2$ errors, but hopefully you can fix the sketch below.

Let $a=\frac{1}{2d}$, so unitarity is the condition $\int d^{2M}\mathbf{h}\exp -ah^2=(\pi/a)^{M}$. Differentiating both sides $k$ times with respect to $a$, and dividing by the original equation, gives $$\mathbf{E}h^{2k}=(M)^{(k)} a^{-k}=(M)^{(k)} (2d)^{k},$$where $(x)^{(k)}:=\prod_{i=0}^{k-1}(x+i)$ is the rising Pochhammer symbol. This technique for evaluating even moments is well-known for the case $M=1/2$ [sic], say with $d=1$ for brevity. For example, if $H\sim N(0,\,1)$ then $$\mathbb{E}H^2=\frac{1}{2}2=1,\,\mathbb{E}H^4=\frac{1}{2}\frac{3}{2}2^2=3,$$and similarly $\mathbb{E}H^6=15,\,\mathbb{E}H^8=105$. In theory, the $6$th & $8$th moments are respectively $8d^3 M(M+1)(M+2),\,16d^4 M(M+1)(M+2)(M+3)$.