1、 n n n元向量
假设
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\begin{split} X&=[X_1,X_2,\cdots,X_i,\cdots ,X_n]^T\\ \mu&= [\mu_1,\mu_2,\cdots,\mu_i,\cdots,\mu_n]^T\\ \sigma&= [\sigma_1,\sigma_2,\cdots,\sigma_i,\cdots,\sigma_n]^T\\ X_i&\sim N(\mu_i,\sigma_i^2)\\ \end{split}
XμσXi=[X1,X2,⋯,Xi,⋯,Xn]T=[μ1,μ2,⋯,μi,⋯,μn]T=[σ1,σ2,⋯,σi,⋯,σn]T∼N(μi,σi2)
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Σ为协方差矩阵。
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\begin{split} \Sigma&=\left[\begin{matrix} Conv(X_1,X_1) & Conv(X_1,X_2) & \cdots & Conv(X_1,X_n) \\ Conv(X_2,X_1) & Conv(X_2,X_2) & \cdots & Conv(X_2,X_n) \\ \vdots & \vdots & \ddots & \vdots \\ Conv(X_n,X_1) & Conv(X_n,X_2) & \cdots & Conv(X_n,X_n) \\ \end{matrix}\right] \end{split}
Σ=
Conv(X1,X1)Conv(X2,X1)⋮Conv(Xn,X1)Conv(X1,X2)Conv(X2,X2)⋮Conv(Xn,X2)⋯⋯⋱⋯Conv(X1,Xn)Conv(X2,Xn)⋮Conv(Xn,Xn)
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\begin{split} \Sigma&=\left[\begin{matrix} \sigma_1^2 & 0 & \cdots & 0 \\ 0 & \sigma_2^2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sigma_n^2 \\ \end{matrix}\right] \end{split}
Σ=
σ120⋮00σ22⋮0⋯⋯⋱⋯00⋮σn2
2 、 n n n元高斯分布
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\begin{split} p(X)&=\frac{1}{(2\pi)^{\frac{n}{2}}\cdot|\Sigma|^{\frac{1}{2}}}\cdot e^{-\frac{(X-\mu)^T\Sigma^{-1}(X-\mu)}{2}} \end{split}
p(X)=(2π)2n⋅∣Σ∣211⋅e−2(X−μ)TΣ−1(X−μ)
其中，
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∣Σ∣为协方差矩阵
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Σ的行列式
3、1元高斯分布
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\begin{split} X&=[X_1]\\ \mu&= [\mu_1]\\ \sigma&= [\sigma_1]\\ X_1&\sim N(\mu_1,\sigma_1^2)\\ \Sigma&=[\sigma_1^2] \end{split}
XμσX1Σ=[X1]=[μ1]=[σ1]∼N(μ1,σ12)=[σ12]
p ( X 1 ) = 1 ( 2 π ) n 2 ⋅ ∣ Σ ∣ 1 2 ⋅ e − ( X − μ ) T Σ − 1 ( X − μ ) 2 = 1 ( 2 π ) 1 2 ⋅ ( σ 1 2 ) 1 2 ⋅ e − ( X 1 − μ 1 ) T ( σ 1 2 ) − 1 ( X 1 − μ 1 ) 2 = 1 2 π ⋅ σ 1 ⋅ e − ( X 1 − μ 1 ) 2 2 ⋅ σ 1 2 \begin{split} p(X_1)&=\frac{1}{(2\pi)^{\frac{n}{2}}\cdot|\Sigma|^{\frac{1}{2}}}\cdot e^{-\frac{(X-\mu)^T\Sigma^{-1}(X-\mu)}{2}} \\ &=\frac{1}{(2\pi)^{\frac{1}{2}}\cdot (\sigma_1^2)^{\frac{1}{2}}}\cdot e^{-\frac{(X_1-\mu_1)^T (\sigma_1^2)^{-1}(X_1-\mu_1)}{2}} \\ &=\frac{1}{\sqrt{2\pi} \cdot \sigma_1}\cdot e^{-\frac{(X_1-\mu_1)^2}{2\cdot \sigma_1^2}} \\ \end{split} p(X1)=(2π)2n⋅∣Σ∣211⋅e−2(X−μ)TΣ−1(X−μ)=(2π)21⋅(σ12)211⋅e−2(X1−μ1)T(σ12)−1(X1−μ1)=2π⋅σ11⋅e−2⋅σ12(X1−μ1)2
2、相互独立的2元高斯分布
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\begin{split} X&=[X_1,X_2]^T\\ \mu&= [\mu_1,\mu_2]^T\\ \sigma&= [\sigma_1,\sigma_2]^T\\ X_i&\sim N(\mu_i,\sigma_i^2)\\ \Sigma&=\left[\begin{matrix} \sigma_1^2 & 0 \\ 0 & \sigma_2^2 \\ \end{matrix}\right] \end{split}
XμσXiΣ=[X1,X2]T=[μ1,μ2]T=[σ1,σ2]T∼N(μi,σi2)=[σ1200σ22]
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\begin{split} p(X)&=p([X_1,X_2]^T) \\ &=\frac{1}{(2\pi)^{\frac{n}{2}}\cdot|\Sigma|^{\frac{1}{2}}}\cdot e^{-\frac{(X-\mu)^T\Sigma^{-1}(X-\mu)}{2}} \\ &=\frac{1}{(2\pi)^{\frac{2}{2}}\cdot \left|\begin{matrix} \sigma_1^2 & 0 \\ 0 & \sigma_2^2 \\ \end{matrix}\right|^{\frac{1}{2}}}\cdot e^{-\frac{\Bigg(\left[\begin{matrix} X_1 \\ X_2 \\ \end{matrix}\right]-\left[\begin{matrix} \mu_1 \\ \mu_2 \\ \end{matrix}\right]\Bigg)^T\left[\begin{matrix} \sigma_1^2 & 0 \\ 0 & \sigma_2^2 \\ \end{matrix}\right]^{-1}\Bigg(\left[\begin{matrix} X_1 \\ X_2 \\ \end{matrix}\right]-\left[\begin{matrix} \mu_1 \\ \mu_2 \\ \end{matrix}\right]\Bigg)}{2}} \\ &=\frac{1}{2\pi\cdot \sigma_1\cdot \sigma_2}\cdot e^{-\frac{\left[\begin{matrix} X_1 -\mu_1\\ X_2 -\mu_2 \\ \end{matrix}\right]^T\left[\begin{matrix} \frac{1}{\sigma_1^2} & 0 \\ 0 & \frac{1}{\sigma_2^2} \\ \end{matrix}\right] \left[\begin{matrix} X_1-\mu_1 \\ X_2-\mu_2 \\ \end{matrix}\right]}{2}} \\ &=\frac{1}{2\pi\cdot \sigma_1\cdot \sigma_2}\cdot e^{-\frac{\left[\begin{matrix} X_1 -\mu_1, X_2 -\mu_2 \\ \end{matrix}\right]\left[\begin{matrix} \frac{1}{\sigma_1^2} & 0 \\ 0 & \frac{1}{\sigma_2^2} \\ \end{matrix}\right] \left[\begin{matrix} X_1-\mu_1 \\ X_2-\mu_2 \\ \end{matrix}\right]}{2}} \\ &=\frac{1}{2\pi\cdot \sigma_1\cdot \sigma_2}\cdot e^{-\frac{\left[\begin{matrix} \frac{X_1 -\mu_1}{\sigma_1^2}, \frac{X_2 -\mu_2}{\sigma_2^2} \\ \end{matrix}\right] \left[\begin{matrix} X_1-\mu_1 \\ X_2-\mu_2 \\ \end{matrix}\right]}{2}} \\ &=\frac{1}{2\pi\cdot \sigma_1\cdot \sigma_2}\cdot e^{-\frac{\frac{(X_1 -\mu_1)^2}{\sigma_1^2}-\frac{(X_2 -\mu_2)^2}{\sigma_2^2}}{2}} \\ &=\frac{1}{\sqrt{2\pi}\cdot \sigma_1}\cdot e^{-\frac{(X_1 -\mu_1)^2}{2\sigma_1^2}} \cdot \frac{1}{\sqrt{2\pi}\cdot \sigma_2}\cdot e^{-\frac{(X_2 -\mu_2)^2}{2\sigma_2^2}} \end{split}
p(X)=p([X1,X2]T)=(2π)2n⋅∣Σ∣211⋅e−2(X−μ)TΣ−1(X−μ)=(2π)22⋅
σ1200σ22
211⋅e−2([X1X2]−[μ1μ2])T[σ1200σ22]−1([X1X2]−[μ1μ2])=2π⋅σ1⋅σ21⋅e−2[X1−μ1X2−μ2]T[σ12100σ221][X1−μ1X2−μ2]=2π⋅σ1⋅σ21⋅e−2[X1−μ1,X2−μ2][σ12100σ221][X1−μ1X2−μ2]=2π⋅σ1⋅σ21⋅e−2[σ12X1−μ1,σ22X2−μ2][X1−μ1X2−μ2]=2π⋅σ1⋅σ21⋅e−2σ12(X1−μ1)2−σ22(X2−μ2)2=2π⋅σ11⋅e−2σ12(X1−μ1)2⋅2π⋅σ21⋅e−2σ22(X2−μ2)2
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