Distribution of sum of dependent Gaussians

Question

If we know $X \sim \mathcal N(\mu, \Sigma)$, is there a closed-form expression for the distribution of $(X-\mu)^\top \Lambda^{-1} (X - \mu)$? I know there is one if $\Lambda = \Sigma$, but if we just assume $\Lambda$ to be any arbitrary positive definite matrix, can we still obtain some distribution? Maybe a gamma or some chi-squared distribution? Any help would be appreciated!

Answer 1

$X - \mu = \Sigma^{1/2} Z$ where $Z \sim N(0, I_n)$. So $$(X-\mu)^\top \Lambda^{-1} (X-\mu) = Z^\top \Sigma^{1/2} \Lambda^{-1} \Sigma^{1/2} Z = \|\Lambda^{-1/2} \Sigma^{1/2} Z\|^2.$$

So it suffices to understand the distribution of $\|AZ\|^2$ where $A$ is positive definite and $Z \sim N(0, I_n)$, since you can take $A=\Lambda^{-1/2} \Sigma^{1/2}$ above for your specific question.

Because $A$ is positive definite, it can be written as $A=UDU^\top$ where $U$ is orthogonal and $D$ is diagonal with positive entries. Then $\|AZ\|^2 = \|UDU^\top Z\|^2 = \|D \tilde{Z}\|^2$ where $\tilde{Z} := U^\top Z$ is also $N(0, I_n)$ due to rotational invariance of the standard multivariate Gaussian distribution. So your random variable is a weighted sum of squares of independent Gaussians, i.e. $\sum_{i=1}^n d_{ii}^2 \tilde{Z}_i^2$. This is a generalization of the chi-squared distribution, but I don't think it has a name.