1、 n n n元向量
假設 n n n元隨機變量 X X X
X = [ X 1 , X 2 , ? , X i , ? , X n ] T μ = [ μ 1 , μ 2 , ? , μ i , ? , μ n ] T σ = [ σ 1 , σ 2 , ? , σ i , ? , σ n ] T X i ~ N ( μ i , σ i 2 ) \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?)?
Σ \Sigma Σ為協方差矩陣。
Σ = [ C o n v ( X 1 , X 1 ) C o n v ( X 1 , X 2 ) ? C o n v ( X 1 , X n ) C o n v ( X 2 , X 1 ) C o n v ( X 2 , X 2 ) ? C o n v ( X 2 , X n ) ? ? ? ? C o n v ( X n , X 1 ) C o n v ( X n , X 2 ) ? C o n v ( X n , X n ) ] \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?)? ??
當 X 1 , X 2 , ? , X i , ? , X n X_1,X_2,\cdots,X_i,\cdots ,X_n X1?,X2?,?,Xi?,?,Xn?之間相互獨立時,有
Σ = [ σ 1 2 0 ? 0 0 σ 2 2 ? 0 ? ? ? ? 0 0 ? σ n 2 ] \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?= ?σ12?0?0?0σ22??0??????00?σn2?? ??
2 、 n n n元高斯分布
p ( X ) = 1 ( 2 π ) n 2 ? ∣ Σ ∣ 1 2 ? e ? ( X ? μ ) T Σ ? 1 ( X ? μ ) 2 \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∣21?1??e?2(X?μ)TΣ?1(X?μ)??
其中, ∣ Σ ∣ |\Sigma| ∣Σ∣為協方差矩陣 Σ \Sigma Σ的行列式
3、1元高斯分布
此時
X = [ X 1 ] μ = [ μ 1 ] σ = [ σ 1 ] X 1 ~ N ( μ 1 , σ 1 2 ) Σ = [ σ 1 2 ] \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} XμσX1?Σ?=[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∣21?1??e?2(X?μ)TΣ?1(X?μ)?=(2π)21??(σ12?)21?1??e?2(X1??μ1?)T(σ12?)?1(X1??μ1?)?=2π??σ1?1??e?2?σ12?(X1??μ1?)2??
2、相互獨立的2元高斯分布
此時
X = [ X 1 , X 2 ] T μ = [ μ 1 , μ 2 ] T σ = [ σ 1 , σ 2 ] T X i ~ N ( μ i , σ i 2 ) Σ = [ σ 1 2 0 0 σ 2 2 ] \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} XμσXi?Σ?=[X1?,X2?]T=[μ1?,μ2?]T=[σ1?,σ2?]T~N(μi?,σi2?)=[σ12?0?0σ22??]?
p ( X ) = p ( [ X 1 , X 2 ] T ) = 1 ( 2 π ) n 2 ? ∣ Σ ∣ 1 2 ? e ? ( X ? μ ) T Σ ? 1 ( X ? μ ) 2 = 1 ( 2 π ) 2 2 ? ∣ σ 1 2 0 0 σ 2 2 ∣ 1 2 ? e ? ( [ X 1 X 2 ] ? [ μ 1 μ 2 ] ) T [ σ 1 2 0 0 σ 2 2 ] ? 1 ( [ X 1 X 2 ] ? [ μ 1 μ 2 ] ) 2 = 1 2 π ? σ 1 ? σ 2 ? e ? [ X 1 ? μ 1 X 2 ? μ 2 ] T [ 1 σ 1 2 0 0 1 σ 2 2 ] [ X 1 ? μ 1 X 2 ? μ 2 ] 2 = 1 2 π ? σ 1 ? σ 2 ? e ? [ X 1 ? μ 1 , X 2 ? μ 2 ] [ 1 σ 1 2 0 0 1 σ 2 2 ] [ X 1 ? μ 1 X 2 ? μ 2 ] 2 = 1 2 π ? σ 1 ? σ 2 ? e ? [ X 1 ? μ 1 σ 1 2 , X 2 ? μ 2 σ 2 2 ] [ X 1 ? μ 1 X 2 ? μ 2 ] 2 = 1 2 π ? σ 1 ? σ 2 ? e ? ( X 1 ? μ 1 ) 2 σ 1 2 ? ( X 2 ? μ 2 ) 2 σ 2 2 2 = 1 2 π ? σ 1 ? e ? ( X 1 ? μ 1 ) 2 2 σ 1 2 ? 1 2 π ? σ 2 ? e ? ( X 2 ? μ 2 ) 2 2 σ 2 2 \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∣21?1??e?2(X?μ)TΣ?1(X?μ)?=(2π)22?? ?σ12?0?0σ22?? ?21?1??e?2([X1?X2??]?[μ1?μ2??])T[σ12?0?0σ22??]?1([X1?X2??]?[μ1?μ2??])?=2π?σ1??σ2?1??e?2[X1??μ1?X2??μ2??]T[σ12?1?0?0σ22?1??][X1??μ1?X2??μ2??]?=2π?σ1??σ2?1??e?2[X1??μ1?,X2??μ2??][σ12?1?0?0σ22?1??][X1??μ1?X2??μ2??]?=2π?σ1??σ2?1??e?2[σ12?X1??μ1??,σ22?X2??μ2???][X1??μ1?X2??μ2??]?=2π?σ1??σ2?1??e?2σ12?(X1??μ1?)2??σ22?(X2??μ2?)2??=2π??σ1?1??e?2σ12?(X1??μ1?)2??2π??σ2?1??e?2σ22?(X2??μ2?)2??