Distribution of the sample variance of values from a multivariate normal distribution

What is the sampling distribution of the variance of a collection of variables that follow a multivariate normal distribution? Specifically, assume that the $n-$dimensional vector $\boldsymbol \sim \mathcal(\boldsymbol<\mu>,\boldsymbol)$, where $\boldsymbol<\mu>$ and $\boldsymbol$ are known. Denote the sample mean by $\bar = \sum_^n x_i/n$. What is the distribution of $s^2_x \equiv \sum_^n (x_i - \bar)^2/(n-1)$? I know that the sample variance of a collection of independent and identically distributed normal variables follows a chi-squared distribution, but have been unable to find an extension to the case of correlated normal variables. I have posed this question more generally, but I am specifically interested in the case of exchangeable variables $x_i$ which marginally have the same variance but are positively correlated with each other. I have simulated the problem with various variance and correlation parameters and suspect that the sample variance is chi-squared in this instance as well, but would like a reliable reference for this result if true. It seems that a transformation of a multivariate normal distribution would be useful here. As far as I understand, we can write $\boldsymbol = \boldsymbol <\mu>+ \boldsymbol\boldsymbol$, where $\boldsymbol$ is an $n-$dimensional vector of independent and identically distributed unit normal variables and $\boldsymbol$ is the Cholesky decomposition of $\boldsymbol$ such that $\boldsymbol\boldsymbol' = \boldsymbol$.

• variance
• multivariate-normal-distribution
• chi-squared-distribution

$\begingroup$ Exchangeability implies they all have equal means, so you might as well assume $\mu=0.$ That considerably simplifies the problem. $\endgroup$

$\begingroup$ I agree, in the exchangeable case we can set $\boldsymbol <\mu>= 0$ without loss of generality. I figured it would be best to pose the question in its more general form for future people searching for answers. $\endgroup$