X ∼ N ( μ , Σ ) = 1 ( 2 π ) p 2 ∣ Σ ∣ 1 2 exp ( − 1 2 ( x − μ ) T Σ − 1 ( x − μ ) ) x ∈ R p , r . v . \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} X∼N(μ,Σ)=(2π)2p∣Σ∣211exp(−21(x−μ)TΣ−1(x−μ))x∈Rp,r.v.
已知
x = ( x a x b ) , μ = ( μ a μ b ) , Σ = ( Σ a a Σ a b Σ b a Σ b b ) x a 为 m × 1 , x b 为 n × 1 , m + n = p \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} x=(xaxb),μ=(μaμb),Σ=(ΣaaΣbaΣabΣbb)xa为m×1,xb为n×1,m+n=p
求 P ( x a ) , P ( x b ∣ x a ) P(x_{a}),P(x_{b}|x_{a}) P(xa),P(xb∣xa),求得后可以由对称性得到 P ( x b ) , P ( x a ∣ x b ) P(x_{b}),P(x_{a}|x_{b}) P(xb),P(xa∣xb)
定理:
已知
X ∼ N ( μ , Σ ) , x ∈ R p , y = A x + B , y ∈ R p X \sim N(\mu,\Sigma),x \in \mathbb{R}^{p},y=Ax+B,y \in \mathbb{R}^{p} X∼N(μ,Σ),x∈Rp,y=Ax+B,y∈Rp
则有
y ∼ N ( A μ + B , A Σ A T ) y \sim N(A \mu+B,A \Sigma A^{T}) y∼N(Aμ+B,AΣAT)
先求 x a x_{a} xa的分布
x a = ( I m O n ) ⏟ A ( x a x b ) ⏟ x E ( x a ) = ( I m O ) ( μ a μ b ) = μ a Var ( x a ) = ( I m O ) ( Σ a a Σ a b Σ b a Σ b b ) ( I m O ) = Σ a a \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} xaE(xa)Var(xa)=A
(ImOn)x
(xaxb)=(ImO)(μaμb