Let $A$ and $B$ be real-valued, centered and independent Gaussian random variables such that \begin{equation} Cov(A,A)=Cov(B,B):=C \end{equation} Since they are independent, it is clear that \begin{equation} Cov(A,B)=Cov(B,A)=0. \end{equation}
Now, for a fixed even $n \in \mathbb{N}$, let $X_1, \cdots, X_n$ be any given random variables such that \begin{equation} \text{each } X_i \text{ is either } A \text{ or } B \end{equation} for all $i=1, \cdots, n$.
Then, I wonder if I can express the their moment \begin{equation} E[X_1 \cdots X_n] \end{equation} as \begin{equation} \prod_{i=1}^n C_i \end{equation} where $C_i = C_1, \cdots, C_n$ are some prescribed values. Moreover, are these $C_i$'s expressible in terms of the variance $C$ above?
Wikipedia article https://en.wikipedia.org/wiki/Multivariate_normal_distribution provides a way to compute the higher moments in general cases. However, I cannot figure out myself if the moments factorize when the covariance matrix is proportional to the identity.
Could anyone please help me?
The expectation satisfies $$E[X_1\cdots X_n]=E[A^k\cdot B^{n-k}]=E[A^k]\cdot E[B^{n-k}]$$ where k is the number of $X_i$ that are equal to $A$.
Since $A$ and $B$ are centered, their moments are equal to their central moments. The central moments of a Gaussian random variable can be expressed in terms of the standard deviation by the following formulae, \begin{align} E[(X-E[X])^{n}]&=\sigma^{n}(n-1)!!&\text{for n even}\\\\ E[(X-E[X])^{n}]&=0&\text{for n odd} \end{align} This gives us that $$E[A^k]\cdot E[B^{n-k}]=\bigr((k-1)!!\cdot C^{k/2}\bigl)\cdot \bigr((n-k-1)!!C^{(n-k)/2}\bigl)=(k-1)!!\cdot (n-k-1)!! \cdot C^{n/2}$$ if $k$ and $n-k$ are both even, and $$E[A^k]\cdot E[B^{n-k}]=0$$ if either $k$ or $n-k$ are odd.
Since $n$ is even, the requirement reduces to simply that $k$ be even, and we have \begin{align} E[X_1\cdots X_n]&=E[A^k]\cdot E[B^{n-k}]=(k-1)!!\cdot (n-k-1)!!\cdot C^{n/2} &\text{for k even}\\ E[X_1\cdots X_n]&=E[A^k]\cdot E[B^{n-k}]=0&\text{for k odd} \end{align}