Given a vector of correlated Gaussian random variables $\vec{X}=\left(X_1, ..., X_k\right)$ that are normally distributed $\vec{X} \sim \mathcal{N}\left(\vec{\mu}, \Sigma\right)$ with arbitrary $\Sigma$ and $\vec{\mu}$ values what's the variance (i.e. the second central moment) of the random variable $\lVert X \lVert_2$?
From length of Gaussian Random Vector we know that the general probability density function of $\sqrt{Q(\vec{X})} = \sqrt{X^TAX}$ does not follow any known laws.
However I'm only looking for the variance. From the post we know that when analyzing the quadratic form (see linked post) $$ Q(\vec{X})=\sum_{i=1}^k\lambda_i\left(U_i+b_i\right)^2=\sum_{i=1}^k\lambda_i\left(U_i^2 + 2b_iU_i + b_i^2\right) $$ we obtain a superposition of Chi-Squared-Distributions and zero-mean Gaussian distributions since $\vec{U}$ is multivariate normal with identity covariance matrix and expectation zero. Using these rules we obtain:
- $var\left(U_i^2\right) = 2$ (Chi-Squared)
- $var\left(U_i\right) = 1$ (uniform Gaussian)
- $var\left(2b_iU_i\right) = 4b_i^2$
- $var\left(b_i^2\right) = 0$ (constant)
- $cov\left(U_i^2, 2b_iU_i\right) = E\left[2b_iU_i^3\right] = 2b_iE\left[U_i^3\right] = 0$
- $cov\left(U_i^2, b_i^2\right) = 0$
- $var\left(U_i^2 + 2b_iU_i + b_i^2\right) = 2 + 4b_i^2$, and hence
$$var\left(Q(\vec{X})\right) = \sum_{i=1}^k\lambda_i^2(2 + 4b_i^2)$$
So the only step missing would to find $var\left(\sqrt{Q(\vec{X})}\right)$ given $var\left(Q(\vec{X})\right)$. Unfortunately it doesn't appear that simple according to Mean and variance of $\sqrt{X}$ given mean and variance of X.
So how to move on with this? Will the definition of the central moment be helpful? Assuming we are able to come up with the pdf of $\sqrt{Q(\vec{X})}$ I'm afraid this might only yield ugly integral terms that are impossible to solve or even approximate in the general case...
edit: fixed the variance of Chi-Squared to 2
