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Let$\begin{bmatrix} \mathbf X\\ \mathbf Y \end{bmatrix}\sim N( \begin{bmatrix} \mathbf \mu_X\\ \mathbf \mu_Y \end{bmatrix}, \begin{bmatrix} \Sigma_{XX} & \Sigma_{XY}\\ \Sigma_{YX} & \Sigma_{YY} \end{bmatrix})$. If $\Sigma_{XX}$ is full rank, it is easy to show that (we can find that the covariance between $\mathbf Y - \Sigma_{YX}\Sigma_{XX}^{-1}\mathbf X$ and $\mathbf X$ is zero) $$ \mathbf Y|\mathbf X \sim N(\mu_Y+\Sigma_{YX}\Sigma^{-1}_{XX}(\mathbf X-\mu_X),\Sigma_{YY}-\Sigma_{YX}\Sigma^{-1}_{XX}\Sigma_{XY}) $$ But if $\Sigma_{XX}$ is not full rank, what is the distribution of $\mathbf Y|\mathbf X$? (Here, the inverse of $\Sigma_{XX}$ does not exist, so the above method can not be used.)

Four
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  • The same result holds with $\Sigma^{-1}{XX}$ replaced by a generalized inverse $\Sigma^{-}{XX}$. – StubbornAtom Apr 22 '21 at 12:52
  • How to obtain the result? If $\Sigma_{XX}^{-1}$ is replaced by $\Sigma_{XX}^-$, then the covariance between $\mathbf Y - \Sigma_{YX}\Sigma_{XX}^{-}\mathbf X$ and $\mathbf X$ is not zero. I don't know obtain the result. – Four Apr 23 '21 at 03:24
  • They are still uncorrelated. See https://stats.stackexchange.com/a/270934/119261, https://math.stackexchange.com/q/3595992/321264. – StubbornAtom Apr 23 '21 at 14:49

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