Distribution sum of squared correlated normal random variables

Question

Assume we have $n$ non-iid standard normal random variables $X_i$. I'm interested in the distribution of $Z=\sum X_i^2$. It is clear to me, that the sum of independent $n$ squared standard normal variables will follow a chi-squared distribution with n degrees of freedom. However, as said above, we assume the $X_i$ to be pairwise correlated. To be more precise:

Assume we have $n$ non-iid standard normal random variables $X_i (i=1,...n)$ with mean vector $\vec{\mu}= \vec 0$ and covariance matrix $\Sigma$. Is it possible to calculate the distibution of $Z$ in dependence of $\sigma_{ij}$? Or $\rho_{ij}$ (as pairwise correlation coefficient)?

Answer 1

The distribution of $Z$ will depend on the full joint distribution of $X = (X_1,X_2,...,X_n)^T$. If you assume for example that $X$ has a multivariate normal distribution with covariance matrix $\Sigma$, then by performing a unitary transformation that diagonalizes the covariance matrix you can express $Z$ as a weighted sum of independent $\chi^2_1$ random variables :

$$ Z = X^TX = \lambda_1 u_1^2 + \lambda_2 u_2^2 + ... + \lambda_n u_n^2$$ $$ u_i^2 \overset{\mathrm{iid}}{\sim} \chi^2_1$$

where $\lambda_i$ are the eigenvalues of $\Sigma$.

For $n=2$ you can find a closed form for the distribution of $Z$, but not for the general case (see e.g. here). You can however find the moments of $Z$, for example:

$$ Var(Z) = \sum_i \lambda_i^2 Var(u_i^2) = 2 \sum_i \lambda_i^2$$