Ex. 17.6

Recovery of \(\hat{\bm{\Theta}}=\hat{\bm{\Sigma}}^{-1}\) from Algorithm 17.1. Use expression (17.16) to derive the standard partitioned inverse expressions

\[\begin{eqnarray} \theta_{12} &=& -\bb{W}^{-1}_{11}w_{12}\theta_{22}\non\\ \theta_{22} &=&1/(w_{22}-w_{12}^T\bb{W}_{11}^{-1}w_{12}).\non \end{eqnarray}\]

Since \(\hat\beta = \bb{W}_{11}^{-1}w_{12}\), show that \(\hat\theta_{22} = 1/(w_{22}-w_{12}^T\hat\beta)\) and \(\hat\theta_{12} = -\hat\beta\hat\theta_{22}\). Thus \(\hat\theta_{12}\) is a simply rescaling of \(\hat\beta\) by \(-\hat\theta_{22}\).

Soln. 17.6

From (17.16) in the text we have

\[\begin{equation} \label{eq:17-6a} \theta_{12} = -\bb{W}^{-1}_{11}w_{12}\theta_{22}. \end{equation}\]

From (17.15) in the text, we have

\[\begin{equation} w_{12}^T\theta_{12} + w_{22}\theta_{22} = 1.\non \end{equation}\]

Plug \(\eqref{eq:17-6a}\) into it, we get

\[\begin{equation} -w_{12}^T \bb{W}^{-1}_{11}w_{12}\theta_{22} + w_{22}\theta_{22} = 1,\non \end{equation}\]

which gives

\[\begin{equation} \theta_{22} = 1/(w_{22}-w_{12}^T\bb{W}_{11}^{-1}w_{12}).\non \end{equation}\]

The rest follows directly from \(\hat\beta = \bb{W}_11^{-1}w_{12}\).