# What is the fourth moment of a Euclidean Norm?

Let $$X=\lVert M^\top p\rVert_2$$, where $$M$$ is an $$n\times n$$ non-random matrix and $$p\sim N(0,I_{n\times n})$$ is an $$n\times 1$$ vector, and$$\lVert \cdot\rVert_2$$ is the Euclidean norm.

Using some linear algebra, we can determine that $$$$\mathbb{E}[X^2]=tr(MM^\top),$$$$ where $$tr(.)$$ is the trace operator.

Now suppose instead we are interested in the 4th moment - i.e., $$$$\mathbb{E}[X^4],$$$$ How can this be obtained? For instance, is there an extension of Isserlis' Theorem which is appropriate in this setting?

• – whuber
Oct 3, 2023 at 15:31

\begin{align} E[X^4] &= E[X^2X^2] \\ &= \text{Cov}(X^2,X^2) + E[X^2]^2 \\ &= \text{Var}(X^2) + E[X^2]^2 \end{align} Note that $$X^2 = \| M^T p \|_2^2 = p^T M M^T p$$, which follows a generalized chi-squared distribution when $$p\sim N(0,I_{n\times n})$$. Then, \begin{align} E[X^4] &= \text{Var}(X^2) + E[X^2]^2 \\ &= 2\text{Tr}\left[MM^TMM^T\right] + \text{Tr}\left[MM^T\right]^2 \end{align}