p ( x ∣ m , Σ ) = 1 ( 2 π ) D ∣ Σ ∣ e − 1 2 ( x − m ) ⊤ Σ − 1 ( x − m ) p(x|m,\Sigma) = \frac{1}{\sqrt{(2\pi)^{D}|\Sigma|}}e^{-\frac{1}{2}(x-m)^\top \Sigma^{-1}(x-m)} p(x∣m,Σ)=(2π)D∣Σ∣ ​1​e−21​(x−m)⊤Σ−1(x−m) 或者 p ( x ∣ m , Σ ) = ( 2 π ) − D / 2 ∣ Σ ∣ − 1 / 2 exp ⁡ ( − 1 2 ( x − m ) ⊤ Σ − 1 ( x − m ) ) p(x|m,\Sigma) = (2\pi)^{-D/2}|\Sigma|^{-1/2}\exp \left(-\frac{1}{2}(x-m)^\top \Sigma^{-1}(x-m)\right) p(x∣m,Σ)=(2π)−D/2∣Σ∣−1/2exp(−21​(x−m)⊤Σ−1(x−m)) 其中 x , m ∈ R D , Σ ∈ R D × D x,m \in R^{D}, \Sigma \in R^{D\times D} x,m∈RD,Σ∈RD×D.

记为 x ∼ N ( m , Σ ) x\sim N(m,\Sigma) x∼N(m,Σ).

假设 x , y x,y x,y 为联合高斯随机变量： [ x y ] ∼ N ( [ μ x μ y ] , [ A C C ⊤ D ] ) = N ( [ μ x μ y ] , [ A ^ C ^ C ^ ⊤ D ^ ] − 1 ) \left[ \begin{array}{c} x\\ y \end{array} \right] \sim N\left( \left[ \begin{array}{c} \mu_x\\ \mu_y \end{array} \right], \left[ \begin{array}{c} A & C\\ C^\top & D \end{array} \right] \right)= N\left( \left[ \begin{array}{c} \mu_x\\ \mu_y \end{array} \right], \left[ \begin{array}{c} \hat{A} & \hat{C}\\ \hat{C}^\top & \hat{D} \end{array} \right]^{-1} \right) [xy​]∼N([μx​μy​​],[AC⊤​CD​])=N([μx​μy​​],[A^C^⊤​C^D^​]−1) 则 x ∼ N ( μ x , A ) x \sim N(\mu_x, A) x∼N(μx​,A)

假设 x , y x,y x,y 为联合高斯随机变量： [ x y ] ∼ N ( [ μ x μ y ] , [ A C C ⊤ D ] ) = N ( [ μ x μ y ] , [ A ^ C ^ C ^ ⊤ D ^ ] − 1 ) \left[ \begin{array}{c} x\\ y \end{array} \right] \sim N\left( \left[ \begin{array}{c} \mu_x\\ \mu_y \end{array} \right], \left[ \begin{array}{c} A & C\\ C^\top & D \end{array} \right] \right)= N\left( \left[ \begin{array}{c} \mu_x\\ \mu_y \end{array} \right], \left[ \begin{array}{c} \hat{A} & \hat{C}\\ \hat{C}^\top & \hat{D} \end{array} \right]^{-1} \right) [xy​]∼N([μx​μy​​],[AC⊤​CD​])=N([μx​μy​​],[A^C^⊤​C^D^​]−1) 则 x ∣ y ∼ N ( μ x + C B − 1 ( y − μ y ) , A − C B − 1 C ⊤ ) x| y \sim N(\mu_x + CB^{-1}(y-\mu_y), A - CB^{-1}C^\top) x∣y∼N(μx​+CB−1(y−μy​),A−CB−1C⊤)

两个高斯分布的乘积为未归一化的高斯分布： N ( x ∣ a , A ) N ( y ∣ b , B ) = Z − 1 N ( x ∣ c , C ) N(x|a,A)N(y|b,B) = Z^{-1}N(x|c,C) N(x∣a,A)N(y∣b,B)=Z−1N(x∣c,C) 其中 c = C ( A − 1 a + B − 1 b ) C = ( A − 1 + B − 1 ) − 1 Z = ( 2 π ) − D / 2 ∣ A + B ∣ − 1 / 2 exp ⁡ ( − 1 2 ( a − b ) ⊤ ( A + B ) − 1 ( a − b ) ) c = C(A^{-1}a + B^{-1}b) \\ C = (A^{-1} + B^{-1})^{-1} \\ Z = (2\pi)^{-D/2}|A+B|^{-1/2}\exp \left(-\frac{1}{2}(a-b)^\top (A+B)^{-1}(a-b)\right) c=C(A−1a+B−1b)C=(A−1+B−1)−1Z=(2π)−D/2∣A+B∣−1/2exp(−21​(a−b)⊤(A+B)−1(a−b))