We know that the conditional variance of a multivariate normal vector $(X,Y)$ is the Schur complement: $$V(X|Y=(y_1,...,y_n))=\Sigma_{XX}-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}$$

I have the intuition that the conditional covariance has the same form but I don't know how to prove it: $$Cov(X_1,X_2|Y=(y_1,...,y_n))=\Sigma_{X_1X_2}-\Sigma_{X_1Y}\Sigma_{YY}^{-1}\Sigma_{YX_2}$$ Where $Cov(X_1,X_2|Y)=E(X_1X_2|Y)-E(X_1|Y)E(X_2|Y)$.

Does someone know how to prove it?

Thank you so much.


1 Answer 1


Since the first equation is true for any $X$ vector, we can define $X=\begin{bmatrix}X_1 \\ X_2\end{bmatrix}$, and $V(X|Y)$ will be $\begin{bmatrix}V(X_1,X_1|Y) & V(X_1,X_2|Y) \\ V(X_2,X_1|Y) & V(X_2,X_2|Y)\end{bmatrix}$. We want the upper right corner of this matrix. Also, we can define $\Sigma_{XX}=\begin{bmatrix}\Sigma_{X_1,X_1} & \Sigma_{X_1,X_2} \\ \Sigma_{X_2,X_1} & \Sigma_{X_2,X_2}\end{bmatrix}$ and $\Sigma_{XY}=\begin{bmatrix}\Sigma_{X_1,Y} \\ \Sigma_{X_2,Y}\end{bmatrix}$, $\Sigma_{YX}=\begin{bmatrix}\Sigma_{X_1,Y} & \Sigma_{X_2,Y}\end{bmatrix}$. Substituting these into the first equation yields: $$\begin{align}V(X|Y) & =\begin{bmatrix}\Sigma_{X_1,X_1} & \Sigma_{X_1,X_2} \\ \Sigma_{X_2,X_1} & \Sigma_{X_2,X_2}\end{bmatrix}-\begin{bmatrix}\Sigma_{X_1,Y} \\ \Sigma_{X_2,Y}\end{bmatrix}\Sigma_{YY}^{-1}\begin{bmatrix}\Sigma_{X_1,Y} & \Sigma_{X_2,Y}\end{bmatrix} \\ & = \begin{bmatrix}\Sigma_{X_1,X_1} & \Sigma_{X_1,X_2} \\ \Sigma_{X_2,X_1} & \Sigma_{X_2,X_2}\end{bmatrix}-\begin{bmatrix}\Sigma_{X_1Y}\Sigma_{YY}^{-1}\Sigma_{YX_1} & \Sigma_{X_1Y}\Sigma_{YY}^{-1}\Sigma_{YX_2} \\ \Sigma_{X_2Y}\Sigma_{YY}^{-1}\Sigma_{YX_1} & \Sigma_{X_2Y}\Sigma_{YY}^{-1}\Sigma_{YX_2}\end{bmatrix}\end{align}$$. The upper right corner is $\Sigma_{X_1,X_2}-\Sigma_{X_1Y}\Sigma_{YY}^{-1}\Sigma_{YX_2}$ just as your wrote.

  • $\begingroup$ Thank you so much gunes. This trick is really helpful. Just one issue that puzzles me: $\Sigma_{XX}-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}$ is a scalar matrix, not a random variable. Therefore, the Schur complement is not equal to the conditional variance $V(X|Y)$ (which is random) but rather to the variance of X conditional to a certain realization of Y $V(X|Y=(y_1,...,y_n))$, isn't it? If so, what is this $(y_1,...,y_n)$? $\endgroup$
    – Dadoo
    Jan 13, 2019 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.