Special case of higher moments of multivariate normal distribution

Question

Consider $X=(X_1,\ldots, X_n)$, $n\in \mathbb N$, such that $X_i \sim \mathcal N (x_i, t)$ independently for $(x_1,\ldots, x_n) \in \mathbb R^n$ and $t\geq 0$. Let $p = 2 q$ for $ q\in \mathbb N$. I am intersted in an expression for $$\mathbb E[\| X \| ^p] = \mathbb E [(X_1^2 +\ldots + X_n^2)^q]$$ which should be a polynomial in $t$, but I do not know how to figure it out as clear expression.

Answer 1

HINT:

All you need to do is to compute the even moments of normal distribution and carefully expand the brackets. Since, $X_i$ are not standard normal then the expressions become very messy even for small $q$. To illustrate what I mean I computed for the first two values of $q$ so that you could get the idea.

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For $q = 1$ you clearly get $$\mathbb{E} \| X \|^2 = nt + \sum_{i=1}^n x_i^2.$$

For $q = 2$ similarly $$\mathbb{E} \| X \|^4 = \sum_{i=1}^n \mathbb{E} X_i^4 + \sum_{i\neq j} \mathbb{E} X_i^2 \cdot \mathbb{E} X_j^2 = 3nt^2 + \sum_{i=1}^n (x_i^4 + 6x_i^2 t) + \sum_{i \neq j}(t + x_i^2)(t + x_j^2).$$

The latter could be written in a canonical form of polynomial as follows $$ \mathbb{E} \| X \|^4 = t^2(n^2 + 2n) + t\left(\sum_{i=1}^n 6x_i^2 + \sum_{i \neq j} (x_i^2 + x_j^2)\right) + \sum_{i=1}^n x_i^4 + \sum_{i \neq j} x_i^2 x_j^2. $$