[TOC]

1. 異常検知と変化検知(井出剛 杉山将)

1.1. 証明補足

1.1.1. p49(4.12)

$\frac{\partial \Psi \left(A\right)}{\partial A}=\frac{1}{N}XC{X}^T$

Proof. $$\begin{array}{cc}& \frac{\partial \Psi \left(A\right)}{\partial A}\\ =& \frac{1}{\partial A}(\frac{1}{N}\sum _{n=1}^N\sum _{j\in N^{\left(n\right)}}((1-\mu )d_A^2(x^{\left(n\right)},x^{\left(j\right)})\\ & +\mu \sum _{l=1}^{N}I[y^{\left(l\right)}\ne y^{\left(n\right)}][1+d_A^2(x^{\left(n\right)},x^{\left(j\right)})-d_A^2(x^{\left(n\right)},x^{\left(l\right)}){]}_{+}))\end{array}$$ Since $$\sum _{l=1}^N[y^{\left(l\right)}\ne y^{\left(n\right)}][1+d_A^2(x^{\left(n\right)},x^{\left(j\right)})-d_A^2(x^{\left(n\right)},x^{\left(l\right)}){]}_{+}=\sum _{l\in N_{n,j}}1+d_A^2(x^{\left(n\right)},x^{\left(j\right)})-d_A^2(x^{\left(n\right)},x^{\left(l\right)})$$ and $$\begin{array}{cc}& \frac{1}{\partial A}{d}_A^2(x^{\left(n\right)},x^{\left(j\right)})\\ =& \frac{1}{\partial A}(x^{\left(n\right)}-x^{\left(j\right)}{)}^{T}A(x^{\left(n\right)}-x^{\left(j\right)})\\ =& (x^{\left(n\right)}-x^{\left(j\right)})(x^{\left(n\right)}-x^{\left(j\right)}{)}^T(\because \frac{\partial }{\partial X}{a}^{T}Xb=a{b}^T)\\ =& [x^{\left(1\right)},\dots ,x^{\left(N\right)}](e_n-e_j)(e_n-e_j{)}^T[x^{\left(1\right)},\dots ,x^{\left(N\right)}{]}^T\\ =& X{C}^{(n,j)}{X}^T\end{array}$$ proof of $\frac{\partial }{\partial X}{a}^{T}Xb=a{b}^T$ is here

Then, $$\begin{array}{cc}& =\frac{1}{N}X(\sum _{n=1}^N\sum _{j\in N^{\left(n\right)}}((1-\mu )C^{(n,j)}+\mu \sum _{l\in N_{n,j}}{C}^{(n,j)}-C^{(n,l)}))X^T\\ & =\frac{1}{N}XC{X}^T\end{array}$$

1.1.2. 多変量正規分布のベイズ公式p97

平均$\mu$, 分散共分散行列$\Sigma$の多次元正規分布を$N(x\mid \mu ,\Sigma )$と表す. つまり$N$次元確率ベクトル$X$が$N(x\mid \mu ,\Sigma )$に従うとき$X$の確率密度関数は $$f\left(x\right)=\frac{|\Sigma {|}^{-1/2}}{(2\pi {)}^{n/2}}\exp (-\frac{1}{2}(x-\mu {)}^T{\Sigma }^{-1}(x-\mu ))$$

である。ここで$n$次元確率ベクトル$X$と$m$次元確率ベクトル$Y$が以下を満たすと仮定する。 $$\begin{array}{cc}p(y\mid x)& =N(y\mid Ax+b,L^{-1})\\ p\left(x\right)& =N(x\mid \mu ,{\Lambda }^{-1})\end{array}$$

ただし、$A\in R^{m\times n}$,$b\in R^m$, $L^{-1}\in R^{m\times m}$, $\mu \in R^{n\times n},{\Lambda }^{-1}\in R^{n\times n}$

この時,以下が成り立つ $$\begin{array}{cc}p(x\mid y)& =N(x\mid M\{\left(A^{T}L\right(y-b)+\Lambda \mu \},M)\\ p\left(y\right)& =N(y\mid A\mu +b,L^{-1}+A{\Lambda }^{-1}{A}^T)\end{array}$$

ただし、$M:=(A^{T}LA+\Lambda {)}^{-1}$

(proof) $x$と$y$の同時分布 $z:=\left(\begin{array}{c}x\\ y\end{array}\right)$を定義すると, $p\left(z\right)=p\left(x\right)p(y\mid x)$なので, $$\begin{array}{cc}\ln p\left(z\right)& =\ln p\left(x\right)+\ln p(y\mid x)\\ & =-\frac{1}{2}(x-\mu {)}^T\Lambda (x-\mu )-\frac{1}{2}(y-Ax-b{)}^{T}L(y-Ax-b)+\text{const}\\ & =-\frac{1}{2}\{(\text{2nd order of}x,y)+-2(x^T,y^T)\left(\begin{array}{c}\Lambda \mu -A^{T}Lb\\ Lb\end{array}\right)+\text{const}\}\\ & \because (L^{-1}{)}^T=L^{-1}\Rightarrow L^T=L\end{array}$$ $x,y$の2次の項に対しては $$\begin{array}{cc}\text{2nd order of}x,y& =x^T(\Lambda +A^{T}LA)x+y^{T}Ly-y^{T}LAx-x^T{A}^{T}Ly\\ & =(x^T,y^T)\left(\begin{array}{cc}\Lambda +A^{T}LA& -A^{T}L\\ -LA& L\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ & =z^{T}Rz\end{array}$$ ゆえに, $$\begin{array}{c}\ln p\left(z\right)=-\frac{1}{2}\{z^{T}Rz-2z^T\left(\begin{array}{c}\Lambda \mu -A^{T}Lb\\ Lb\end{array}\right)\}+\text{const}\end{array}$$ また、シューアの補行列より $$\begin{array}{c}{R}^{-1}={\left(\begin{array}{cc}\Lambda +A^{T}LA& -A^{T}L\\ -LA& L\end{array}\right)}^{-1}=\left(\begin{array}{cc}{\Lambda }^{-1}& {\Lambda }^{-1}{A}^T\\ A{\Lambda }^{-1}& L^{-1}+A{\Lambda }^{-1}{A}^T\end{array}\right)\end{array}$$ であるので, $$\begin{array}{cc}p\left(z\right)& =\text{const}\cdot \exp \{-\frac{1}{2}(z^{T}Rz-2z^{T}c)\}\\ & \text{where}c:=\left(\begin{array}{c}\Lambda \mu -A^{T}Lb\\ Lb\end{array}\right)\\ & =\text{const}\cdot \exp \{-\frac{1}{2}(z-{\mu }_z{)}^{T}R(z-{\mu }_z)\}\\ & \text{wehere}c=R{\mu }_z\iff {\mu }_z=R^{-1}c\end{array}$$ $\int dzf\left(z\right)=1$より $$f\left(z\right)=\frac{|R^{-1}{|}^{-1/2}}{(2\pi {)}^{2N/2}}\exp (-\frac{1}{2}(z-{\mu }_z{)}^{T}R(z-{\mu }_z))$$ また, $${\mu }_z=R^{-1}c=\left(\begin{array}{cc}{\Lambda }^{-1}& {\Lambda }^{-1}{A}^T\\ A{\Lambda }^{-1}& L^{-1}+A{\Lambda }^{-1}{A}^T\end{array}\right)\left(\begin{array}{c}\Lambda \mu -A^{T}Lb\\ Lb\end{array}\right)=\left(\begin{array}{c}\mu \\ A\mu +b\end{array}\right)$$ 多変数正規分布の分割公式を$p\left(y\right)$に対して用いると $$\begin{array}{cc}p\left(y\right)& =(p\left(x_b\right)=N(x_b\mid {\mu }_b,{\Sigma }_{bb}))\\ & =N(y\mid A\mu +b,L^{-1}+A{\Lambda }^{-1}{A}^T)\end{array}$$ 同様に $$\begin{array}{c}p(x\mid y)=N({\mu }_{x\mid y},{\Sigma }_{x\mid y})\end{array}$$ に関しても多変数正規分布の分割公式を用いると, $$\begin{array}{cc}{\mu }_{x\mid y}& =(={\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b))\\ & =\mu -(A^{T}LA+\Lambda {)}^{-1}\cdot (-A^{T}L\left)\right(y-A\mu -b)\\ & =(A^{T}LA+\Lambda {)}^{-1}(A^{T}L(y-b)-A^{T}LA\mu )+\mu \\ & =(A^{T}LA+\Lambda {)}^{-1}(A^{T}L(y-b)+\Lambda \mu )-(A^{T}LA+\Lambda {)}^{-1}(\Lambda \mu +A^{T}LA\mu )+\mu \\ & =(A^{T}LA+\Lambda {)}^{-1}(A^{T}L(y-b)+\Lambda \mu )\\ & \\ {\Sigma }_{x\mid y}& =({\Sigma }_{a\mid b}={\Lambda }_{aa}^{-1})\\ & =(\Lambda +A^{T}LA{)}^{-1}\end{array}$$ 参考文献

1.2. Appendix

1.2.1. 2次形式の微分

Let $X$ be a $N\times N$ matrix, and $a,b$ are $N$-dimensional vectors. Then, $\frac{\partial }{\partial X}{a}^{T}Xb=a{b}^T$where $a^T$ is the transpose of $a$

(Proof)

$(A{)}_{i,j}$ denotes the $(i,j)$ element of matrix $A$. $$\begin{array}{cc}(\frac{\partial }{\partial X}{a}^{T}Xb{)}_{ij}& =\frac{\partial }{\partial x_{i,j}}{a}^{T}Xb\\ & =\frac{\partial }{\partial x_{i,j}}\sum _{k=1}^N\left(\sum _{l=1}^N{a}_k{x}_{k,l}\right)b_l\\ & =a_i{b}_j\\ & =(a{b}^T{)}_{i,j}\end{array}$$

1.2.2. シューアの補行列

シューアの補行列(Schur complement matrix) Suppose A, B, C, D are respectivesly $p\times p,p\times q,q\times p$ and $q\times q$ matrices, and $D$ is invertible. If $M=A-B{D}^{-1}C$ is invertible, then

$$\begin{array}{c}{\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)}^{-1}=\left(\begin{array}{cc}{M}^{-1}& -M^{-1}B{D}^{-1}\\ -D^{-1}C{M}^{-1}& D^{-1}+D^{-1}C{M}^{-1}B{D}^{-1}\end{array}\right)\end{array}$$

Similarly, if $A$ and $R=D-C{A}^{-1}B$ are invertible

$$\begin{array}{c}{\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)}^{-1}=\left(\begin{array}{cc}{A}^{-1}+A^{-1}B{R}^{-1}C{A}^{-1}& -A^{-1}B{R}^{-1}\\ -R^{-1}C{A}^{-1}& R^{-1}\end{array}\right)\end{array}$$

Woodbury matrix identity If we compare partitioned matrices, we get below results.

$$\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& =A^{-1}+A^{-1}B(D-C{A}^{-1}B{)}^{-1}C{A}^{-1}\\ (D-C{A}^{-1}B{)}^{-1}& =D^{-1}+D^{-1}C(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\\ (A-B{D}^{-1}C{)}^{-1}B{D}^{-1}& =A^{-1}B(D-C{A}^{-1}B{)}^{-1}\\ (D-C{A}^{-1}B{)}^{-1}C{A}^{-1}& =D^{-1}C(A-B{D}^{-1}C{)}^{-1}\end{array}$$

(proof) If we use LDU decomposition, we have shown that $$\begin{array}{c}\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)=\left(\begin{array}{cc}{I}_p& B{D}^{-1}\\ 0& I_q\end{array}\right)\left(\begin{array}{cc}A-B{D}^{-1}C& 0\\ 0& D\end{array}\right)\left(\begin{array}{cc}{I}_p& 0\\ D^{-1}C& I_q\end{array}\right)\end{array}$$ Generally, it is true that $det\left(AB\right)=det\left(A\right)det\left(B\right)$, so

$$\begin{array}{c}\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)=det(A-B{D}^{-1}C)det\left(D\right)\ne 0\end{array}$$ Therefore, $$\begin{array}{cc}{\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)}^{-1}& =\left(\begin{array}{cc}{I}_p& 0\\ -D^{-1}C& I_q\end{array}\right)\left(\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& 0\\ 0& D^{-1}\end{array}\right)\left(\begin{array}{cc}{I}_p& -B{D}^{-1}\\ 0& I_q\end{array}\right)\\ & ={\left(\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& -(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\\ -D^{-1}C(A-B{D}^{-1}C{)}^{-1}& D^{-1}+D^{-1}C(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\end{array}\right)}^{-1}\\ & =\left(\begin{array}{cc}{M}^{-1}& -M^{-1}B{D}^{-1}\\ -D^{-1}C{M}^{-1}& D^{-1}+D^{-1}C{M}^{-1}B{D}^{-1}\end{array}\right)\end{array}$$ Similarly, $$\begin{array}{c}\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)=\left(\begin{array}{cc}I& 0\\ C{A}^{-1}& I\end{array}\right)\left(\begin{array}{cc}A& 0\\ 0& D-C{A}^{-1}B\end{array}\right)\left(\begin{array}{cc}I& A^{-1}B\\ 0& I\end{array}\right)\end{array}$$ Reference detail of Schur Complement

1.2.3. 多変数正規分布の分割公式

$N$次元正規分布の確率密度関数を

$$N(x\mid \mu ,\Sigma )=\frac{|\Sigma {|}^{-1/2}}{(2\pi {)}^{N/2}}\exp (-\frac{1}{2}(x-\mu {)}^T{\Sigma }^{-1}(x-\mu ))$$

と表すことにする.このとき,$x$が次のように分割されたとする.

$$x=\left(\begin{array}{c}{x}_a\\ x_b\end{array}\right)$$

これに対応して, $\mu ,\Sigma$と$\Lambda :={\Sigma }^{-1}$が次のように分割されているとする.

$$\mu =\left(\begin{array}{c}{\mu }_a\\ {\mu }_b\end{array}\right)\Sigma =\left(\begin{array}{cc}{\Sigma }_{aa}& {\Sigma }_{ab}\\ {\Sigma }_{ba}& {\Sigma }_{bb}\end{array}\right)\Lambda =\left(\begin{array}{cc}{\Lambda }_{aa}& {\Lambda }_{ab}\\ {\Lambda }_{ba}& {\Lambda }_{bb}\end{array}\right)$$

このとき, $p\left(x_a\right),p(x_a\mid x_b)$は以下に従う.

$$\begin{array}{c}p\left(x_b\right)=N(x_b\mid {\mu }_b,{\Sigma }_{bb})=N(x_b\mid {\mu }_b,[{\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab}{]}^{-1})\end{array}$$

$$\begin{array}{c}p\left(x_a|x_b\right)=N(x_a\mid {\mu }_{a\mid b},{\Sigma }_{a\mid b})\end{array}$$

ただし,

$$\begin{array}{c}{\mu }_{a\mid b}={\mu }_a+{\Sigma }_{ab}{\Sigma }_{bb}^{-1}(x_b-{\mu }_b)={\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b)\end{array}$$

$$\begin{array}{c}{\Sigma }_{a\mid b}={\Sigma }_{aa}-{\Sigma }_{ab}{\Sigma }_{bb}^{-1}{\Sigma }_{ba}={\Lambda }_{aa}^{-1}\end{array}$$

(Proof)

まず, 以下を示す $$\begin{array}{c}{\Lambda }_{ab}^T={\Lambda }_{ba},{\Lambda }_{aa}^T={\Lambda }_{aa},{\Lambda }_{bb}^T={\Lambda }_{bb}\end{array}$$ $\Sigma$は正定値対称行列より、任意の首座行列式が正であるので、${\Sigma }_{aa}$に逆行列が存在する。シューアの補行列(Schur complement matrixを用いると, $$\begin{array}{cc}{\Lambda }_{ab}& =-{\Sigma }_{aa}^{-1}{\Sigma }_{ab}{R}^{-1}\\ R^T& =({\Sigma }_{bb}-{\Sigma }_{ba}{\Sigma }_{aa}^{-1}{\Sigma }_{ab}{)}^T\\ & ={\Sigma }_{bb}-{\Sigma }_{ba}{\Sigma }_{aa}^{-1}{\Sigma }_{ab}(\because {\Sigma }^T=\Sigma )\\ & =R\\ {\Lambda }_{ab}^T& =-R^{-1}{\Sigma }_{ba}{\Sigma }_{aa}^{-1}\\ & ={\Lambda }_{ba}\end{array}$$ $\Lambda$にシューアの補行列を当てはめると, 公式中において $$\begin{array}{c}{A}^T=A\\ D^T=D\\ B^T=C\end{array}$$ である。 よって,　公式中において $$\begin{array}{c}{R}^T=(D-C{A}^{-1}B{)}^T=D-C{A}^{-1}B=R\end{array}$$ より,${\Lambda }_{bb}^T={\Lambda }_{bb}$ また、公式中において $$\begin{array}{c}(-A^{-1}B{R}^{-1}{)}^T=-R^{-1}C{A}^{-1}\end{array}$$ であるので、${\Lambda }_{aa}^T={\Lambda }_{aa}$ よって、(11)は示された. 続いて本題に戻る.(11)を用いると $$\begin{array}{cc}& (x-\mu {)}^T{\Sigma }^{-1}(x-\mu )\\ =& {\left(\begin{array}{c}{y}_a\\ y_b\end{array}\right)}^T\left(\begin{array}{cc}{\Lambda }_{aa}& {\Lambda }_{ab}\\ {\Lambda }_{ba}& {\Lambda }_{bb}\end{array}\right)\left(\begin{array}{c}{y}_a\\ y_b\end{array}\right)\\ & \text{where}{y}_a=x_a-{\mu }_a,y_b=x_b-{\mu }_b\\ =& {y_a}^T{\Lambda }_{aa}{y}_a+2{y_a}^T{\Lambda }_{ab}{y}_b+{y_b}^T{\Lambda }_{bb}{y}_b\\ =& (y_a+{\Lambda }_{aa}^{-1}{\Lambda }_{ab}{y}_b{)}^T{\Lambda }_{aa}(y_a+{\Lambda }_{aa}^{-1}{\Lambda }_{ab}{y}_b)+{y_b}^T({\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab})x_b\\ =& (x_a-({\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b){)}^T{\Lambda }_{aa}(x_a-({\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b))\\ & +{y_b}^T({\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab})y_b\end{array}$$ であり, $$\begin{array}{cc}p\left(x_b\right)=& \int d{x}_{a}p\left(x\right)\\ =& \text{const}\cdot \exp \{-\frac{1}{2}{y_b}^T({\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab})y_b\}\\ & +\int d{x}_a\exp \{-\frac{1}{2}(x_a-({\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b){)}^T{\Lambda }_{aa}(x_a-({\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b))\}\\ =& \text{const}\cdot \exp \{-\frac{1}{2}(x_b-{\mu }_b{)}^T({\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab}\left)\right(x_b-{\mu }_b)\}\end{array}$$ $\int d{x}_{b}p\left(x_b\right)=1$であるので, $$p\left(x_b\right)=N(x_b\mid {\mu }_b,[{\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab}{]}^{-1})$$ また, $\Sigma$は正定値対称行列より,$\Lambda$も正定値対称行列であることを踏まえシューアの補行列を$\Lambda$に対して用いると, $$\begin{array}{c}{\Sigma }_{bb}=({\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab}{)}^{-1}\end{array}$$ が示せる.また、 $$\begin{array}{c}p(x_a\mid x_b)=\frac{p\left(x\right)}{p\left(x_b\right)}\end{array}$$ であるので, $$\begin{array}{cc}\ln p(x_a\mid x_b)=& \ln p\left(x\right)-\ln p\left(x_b\right)\\ =& -\frac{1}{2}\{(x_a-({\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b){)}^T{\Lambda }_{aa}(x_a-({\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b))\\ & +{y_b}^T({\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab})y_b\}\\ & +\frac{1}{2}\{{y_b}^T({\Lambda }_{bb}-{\Lambda }_{ba}{\Lambda }_{aa}^{-1}{\Lambda }_{ab})y_b\}\\ & +\text{const}\\ =& -\frac{1}{2}\{(x_a-({\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b){)}^T{\Lambda }_{aa}(x_a-({\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b))\}+\text{const}\end{array}$$ ゆえに, $\int d{x}_{a}p(x_a\mid x_b)=1$であることを考慮すると, $$\begin{array}{cc}{\mu }_{a\mid b}& ={\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b)\\ {\Sigma }_{a\mid b}& ={\Lambda }_{aa}^{-1}\end{array}$$ ただし, $$\begin{array}{cc}{\mu }_{a\mid b}& ={\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b)\\ {\Sigma }_{a\mid b}& ={\Lambda }_{aa}^{-1}\end{array}$$ シューアの補行列を$\Sigma$に対して用いると, p $$\begin{array}{cc}{\Lambda }_{aa}^{-1}& =(M=A-B{D}^{-1}C)\\ & ={\Sigma }_{aa}-{\Sigma }_{ab}{\Sigma }_{bb}^{-1}{\Sigma }_{ba}\end{array}$$

$$\begin{array}{cc}{\Lambda }_{aa}^{-1}{\Lambda }_{ab}& =(M\cdot (-M)B{D}^{-1}=-B{D}^{-1})\\ & =-{\Sigma }_{ab}{\Sigma }_{bb}^{-1}\end{array}$$

ただし式変形で公式に対応するものは()でくくった. よって、${\mu }_{a\mid b},{\Sigma }_{a\mid b}$は以下のように変形出来る. $$\begin{array}{cc}{\mu }_{a\mid b}& ={\mu }_a-{\Lambda }_{aa}^{-1}{\Lambda }_{ab}(x_b-{\mu }_b)\\ & ={\mu }_a+{\Sigma }_{ab}{\Sigma }_{bb}^{-1}(x_b-{\mu }_b)\\ {\Sigma }_{a\mid b}& ={\Lambda }_{aa}^{-1}\\ & ={\Sigma }_{aa}-{\Sigma }_{ab}{\Sigma }_{bb}^{-1}{\Sigma }_{ba}\end{array}$$