Conditional independence from a partitioned covariance

Topic: Graphical models · more questions on this topic
‹ A three-variable chain How many graphs? ›

Using the conditional normal, verify directly for the covariance matrix of the three-variable chain that \operatorname{Cov}(X_1,X_3\mid X_2)=0 but that \operatorname{Cov}(X_1,X_3)\neq 0.

Worked answer (click to show)

Take A=\{1,3\} and B=\{2\}. Then \Sigma_{AB}=(0.8,0.8)^\top, \Sigma_{BB}=1, and

\Sigma_{A\mid B} =\begin{pmatrix} 1 & 0.64 \\ 0.64 & 1 \end{pmatrix} -\begin{pmatrix} 0.8 \\ 0.8 \end{pmatrix} \begin{pmatrix} 0.8 & 0.8 \end{pmatrix} =\begin{pmatrix} 1-0.64 & 0.64-0.64 \\ 0.64-0.64 & 1-0.64 \end{pmatrix} =\begin{pmatrix} 0.36 & 0 \\ 0 & 0.36 \end{pmatrix}.

The conditional covariance is 0, hence conditional independence for a normal vector, and the conditional variances are 0.36, the innovation variance. The unconditional covariance is \operatorname{Cov}(X_1,X_3)=0.64\neq 0.