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：

$$\Sigma_{x|y}=\Lambda_{xx}^{-1}=(\bf{\Lambda+{\bf A^TLA}})^{-1}$$

$$\color{red}{\mu_{x|y}是如何推导的？}$$