We know that the conditional variance of a multivariate normal vector $(X,Y)$ is equal to the Schur complement: $$V(X|Y)=\Sigma_{XX}-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}$$
However, $\Sigma_{XX}-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}$ is a matrix of scalars, whereas $V(X|Y)$ is a matrix of random variables function of Y. Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y, $V(X|Y=(y_1,...,y_n))$, is it correct? If so, who is this $(y_1,...,y_n)$?
Thank you very much.
You are right when you said:
Regard the vector $$Z = \begin{bmatrix} X \\ Y \end{bmatrix} \sim N( \underbrace{\begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix}}_{\mu}; \underbrace{\begin{bmatrix} \Sigma_{XX} & \Sigma_{XY} \\ \Sigma_{XY} & \Sigma_{YY} \end{bmatrix}}_{\Sigma} )$$ All we did is partition a given Normal vector into $2$ subvectors. Now, if you want to find $V(X|Y)$, it turns out to be (using Schurs' complement), nothing other than the equation you have provided.