PRML学习笔记：第2章
\[\Sigma_{a|b}=\Lambda_{aa}^{-1} \tag{2.73}
\]
\[\mu_{a|b}=\mu_a-\Lambda_{aa}^{-1}\Lambda_{ab}(x_b-\mu_b) \tag{2.75}
\]
\[\begin{pmatrix}
\Sigma_{aa} & \Sigma_{ab} \\
\Sigma_{ba} & \Sigma_{bb}
\end{pmatrix}^{-1}
=
\begin{pmatrix}
\Lambda_{aa} & \Lambda_{ab} \\
\Lambda_{ba} & \Lambda_{bb}
\end{pmatrix}
\tag{2.78}
\]
利用公式\((2.76)\)，可得
\[\Lambda_{aa}={\bf M}=({\bf A}-{\bf B}{\bf D}^{-1}{\bf C})^{-1}=(\Sigma_{aa}-\Sigma_{ab}\Sigma_{bb}^{-1}\Sigma_{ba})^{-1} \tag{2.79}
\]
\[{\bf z}=\begin{pmatrix}
{\bf x} \\
{\bf y}
\end{pmatrix}
\tag{2.101}
\]
\[\text{cov}[\bf z]=\bf{R}^{-1}=\begin{pmatrix}
\bf{\Lambda^{-1}} & \bf{\Lambda^{-1}}\bf{A^T} \\
\bf{A\Lambda^{-1}} & \bf{L^{-1}+A\Lambda^{-1}A^T}
\end{pmatrix}
\tag{2.105}
\]
\[\mathbb{E}[\bf z]=\begin{pmatrix}
\bf\mu \\
\bf{A\mu+b}
\end{pmatrix}
\tag{2.108}
\]
对于z，
\[\text{cov}[\bf z]
=\begin{pmatrix}
\Sigma_{xx} & \Sigma_{xy} \\
\Sigma_{yx} & \Sigma_{yy}
\end{pmatrix}
=
\begin{pmatrix}
\Lambda_{xx} & \Lambda_{xy} \\
\Lambda_{yx} & \Lambda_{yy}
\end{pmatrix}^{-1}
=\bf{R}^{-1}
=\begin{pmatrix}
\bf{\Lambda+{\bf A^TLA}} & {\bf -A^TL} \\
{\bf -LA} & {\bf L}
\end{pmatrix}^{-1}
\]
由\((2.73)\)可得：
\[\Sigma_{x|y}=\Lambda_{xx}^{-1}=(\bf{\Lambda+{\bf A^TLA}})^{-1}
\]
\[\color{red}{\mu_{x|y}是如何推导的？}
\]
