Dot product of normal random vector with covariance matrix in between is chi squared distributed

Question

I wish to prove that if $Y \sim N(0, \Omega)$, then $Y^T \Omega^{-1}Y \sim \chi^2_{k}$ where $k$ is the rank of $\Omega$. Any hint on how this can be done?

Answer 1

First, the dimension of $\Omega$ should be $(k\times k)$. Only in this case it is invertible. Next look at https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix#Real_symmetric_matrices

Covariance matrix $\Omega$ can be decomposed as $$ \Omega = Q\Lambda Q^T $$ where $Q$ is an orthogonal matrix, and $\Lambda$ is a diagonal matrix with positive elements on diagonal. Next, let $\sqrt{\Lambda}$ is a diagonal matrix which diagonal elements are square roots of the diagonal elements of $\Lambda$. Then $$ \Omega=Q\sqrt{\Lambda}\sqrt{\Lambda}Q^T, \quad \Omega^{-1}=Q(\sqrt{\Lambda})^{-1}(\sqrt{\Lambda})^{-1}Q^T. $$ Then $$ Y^T\Omega^{-1}Y = Y^TQ(\sqrt{\Lambda})^{-1}(\sqrt{\Lambda})^{-1}Q^T Y = \big((\sqrt{\Lambda})^{-1}Q^T Y\big)^T \big((\sqrt{\Lambda})^{-1}Q^T Y\big) $$ This is the sum of squares of coordinates of the vector $(\sqrt{\Lambda})^{-1}Q^T Y$. Look at this vector. Since it is a linear transformation of normal vector $Y$, it has multivariate normal distribution too. It's covariance matrix is $$ \text{Var}((\sqrt{\Lambda})^{-1}Q^T Y) = (\sqrt{\Lambda})^{-1}Q^T \underbrace{\text{Var}(Y)}_{=\Omega} \big((\sqrt{\Lambda})^{-1}Q^T \big)^T $$ $$ =(\sqrt{\Lambda})^{-1}Q^T\times \underbrace{Q\sqrt{\Lambda}\sqrt{\Lambda}Q^T}_{=\Omega} \times Q (\sqrt{\Lambda})^{-1} = I $$ where $I$ is $(k\times k)$ identity matrix. So the vector $(\sqrt{\Lambda})^{-1}Q^T Y$ has multivariate standard normal distribution and sum of squares of its coordinates has $\chi^2_k$ distribution.