$$
\begin{gathered}
X \sim N(\mu,\Sigma)=\frac{1}{(2\pi)^{\frac{p}{2}}|\Sigma|^{\frac{1}{2}}}\text{exp}\left(- \frac{1}{2}(x-\mu)^{T}\Sigma^{-1}(x-\mu)\right)\
x \in \mathbb{R}^{p},r.v.\
\end{gathered}
$$
已知
$$
\begin{gathered}
x=\begin{pmatrix}
x_{a} \ x_{b}
\end{pmatrix},\mu=\begin{pmatrix}
\mu_{a} \ \mu_{b}
\end{pmatrix},\Sigma=\begin{pmatrix}
\Sigma_{aa} & \Sigma_{ab} \ \Sigma_{ba} & \Sigma_{bb}
\end{pmatrix}\
x_{a}为m \times 1,x_{b}为 n \times 1,m+n=p
\end{gathered}
$$
求$P(x_{a}),P(x_{b}|x_{a})$，求得后可以由对称性得到$P(x_{b}),P(x_{a}|x_{b})$
定理：
已知
$$X \sim N(\mu,\Sigma),x \in \mathbb{R}^{p},y=Ax+B,y \in \mathbb{R}^{p}$$
则有
$$y \sim N(A \mu+B,A \Sigma A^{T})$$
先求$x_{a}$的分布
$$
\begin{aligned}
x_{a}&=\underbrace{\begin{pmatrix}I_{m} & O_{n}\end{pmatrix}}_{A}\underbrace{\begin{pmatrix}
x_{a} \ x_{b}
\end{pmatrix}}_{x}\
E(x_{a})&=\begin{pmatrix}I_{m} & O\end{pmatrix}\begin{pmatrix}
\mu_{a} \ \mu_{b}
\end{pmatrix}=\mu_{a}\
\text{Var}(x_{a})&=\begin{pmatrix}
I_{m} & O
\end{pmatrix}\begin{pmatrix}
\Sigma_{aa} & \Sigma_{ab} \ \Sigma_{ba} & \Sigma_{bb}
\end{pmatrix}\begin{pmatrix}
I_{m} \ O
\end{pmatrix}=\Sigma_{aa}
\end{aligned}
$$
因此$x_{a}\sim N(\mu_{a},\Sigma_{aa})$
再求$x_{b}|x_{a}$的分布，令
$$
\left{\begin{aligned}&x_{b \cdot a}=x_{b}-\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}\&\mu_{b \cdot a}=\mu_{b}-\Sigma_{ba}\Sigma_{aa}^{-1}\mu_{a}\&\Sigma_{bb \cdot a}=\Sigma_{bb}-\Sigma_{ba}\Sigma_{aa}^{-1}\Sigma_{ab}\end{aligned}\right.
$$
有
$$
\begin{aligned}
x_{b \cdot a}&=\underbrace{\begin{pmatrix}- \Sigma_{ba}\Sigma_{aa}^{-1} & I_{n}
\end{pmatrix}}_{A}\underbrace{\begin{pmatrix}
x_{a} \ x_{b}
\end{pmatrix}}_{x}\
E(x_{b \cdot a})&=\begin{pmatrix}- \Sigma_{ba}\Sigma_{aa}^{-1} & I_{n}
\end{pmatrix}\begin{pmatrix}
\mu_{a} \ \mu_{b}
\end{pmatrix}=\mu_{b}-\Sigma_{ba}\Sigma_{aa}^{-1}\mu_{a}=\mu_{b \cdot a}\
\text{Var}(x_{b \cdot a})&=\begin{pmatrix}- \Sigma_{ba}\Sigma_{aa}^{-1} & I_{n}
\end{pmatrix}\begin{pmatrix}
\Sigma_{aa} & \Sigma_{ab} \ \Sigma_{ba} & \Sigma_{bb}
\end{pmatrix}\begin{pmatrix}
-\Sigma_{aa}^{-1}\Sigma_{ba}^{T} \ I_{n}
\end{pmatrix}\
&=\Sigma_{bb}-\Sigma_{ba}\Sigma_{aa}^{-1}\Sigma_{ab}=\Sigma_{bb \cdot a}
\end{aligned}
$$
因此$x_{b \cdot a}\sim N(\mu_{b \cdot a},\Sigma_{bb \cdot a})$
这里要求$x_{b}|x_{a}$，即
$$
\begin{aligned}
x_{b \cdot a}&=x_{b}-\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}\
x_{b}&=x_{b \cdot a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}\
x_{b}|x_{a}&=(x_{b \cdot a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a})|x_{a}\
x_{b}|x_{a}&=x_{b \cdot a}|x_{a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}|x_{a}\
x_{b}|x_{a}&=x_{b \cdot a}|x_{a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}
\end{aligned}
$$
这里如果有$x_{b \cdot a}|x_{a}=x_{b \cdot a}$，就可以有
$$
x_{b}|x_{a}=x_{b \cdot a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}
$$
若$X \sim N(\mu,\Sigma)$，则$Mx \bot Nx \Leftrightarrow M \Sigma N^{T}=0$
证明：
因为$x \sim N(\mu,\Sigma)$，有$Mx \sim N(M \mu,M \Sigma M^{T}),Nx \sim N(N \mu,N \Sigma N^{T})$
$$\begin{aligned} \text{cov}(Mx,Nx)&=E[(Mx-M \mu)(Nx-N \mu)^{T}]\&=M \cdot E[(x-\mu)(x-\mu)^{T}]\cdot N^{T}\&=M \Sigma N^{T}\end{aligned}$$
又因为$Mx \bot Nx$且均为高斯分布，则有$\text{Cov}(Mx ,Nx)=M \Sigma N^{T}=0$
$$
\begin{aligned}
\Sigma&=\begin{pmatrix}
\Sigma_{aa} & \Sigma_{ab} \ \Sigma_{ba} & \Sigma_{bb}
\end{pmatrix}\
x_{b \cdot a}&=x_{b}-\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}=\underbrace{\begin{pmatrix}
-\Sigma_{ba}\Sigma_{aa}^{-1} & I
\end{pmatrix}}_{M}\begin{pmatrix}
x_{a} \ x_{b}
\end{pmatrix}\
x_{a}&=\underbrace{\begin{pmatrix}
I & O
\end{pmatrix}}_{N}\begin{pmatrix}
x_{a} \ x_{b}
\end{pmatrix}\
M \Sigma N^{T}&=\begin{pmatrix}
-\Sigma_{ba}\Sigma_{aa}^{-1} & I
\end{pmatrix}\begin{pmatrix}
\Sigma_{aa} & \Sigma_{ab} \ \Sigma_{ba} & \Sigma_{bb}
\end{pmatrix}\begin{pmatrix}
I & O
\end{pmatrix}=0
\end{aligned}
$$
因此$x_{b \cdot a}\bot x_{a}\Rightarrow x_{b \cdot a}|x_{a}=x_{b \cdot a }$，就有
$$
x_{b}|x_{a}=x_{b \cdot a}|x_{a}+\Sigma_{ba}\Sigma_{aa}^{-1}|x_{a}=x_{b \cdot a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}
$$
因此
$$
\begin{aligned}
E(x_{b}|x_{a})&=E(x_{b \cdot a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a})\
&=\mu_{b \cdot a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}\
&=\mu_{b}-\Sigma_{ba}\Sigma_{aa}^{-1}\mu_{a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a}\
\text{Var}(x_{b}|x_{a})&=\text{Var}(x_{b \cdot a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a})\
&=\text{Var}(x_{b \cdot a})\
&=\Sigma_{bb \cdot a}
\end{aligned}
$$
因此$x_{b}|x_{a} \sim N(\mu_{b}-\Sigma_{ba}\Sigma_{aa}^{-1}\mu_{a}+\Sigma_{ba}\Sigma_{aa}^{-1}x_{a},\Sigma_{bb}-\Sigma_{ba}\Sigma_{aa}^{-1}\Sigma_{ab})$
