高斯判别分析

高斯判别分析（Gaussian discriminative analysis）属于概率生成式模型，并不是直接计算p(y|x)的概率，而是基于bayes,比较p(y=1|x)和p(y=0|x)的大小，从而确定分类
贝叶斯公式：
p ( y ∣ x ) = p ( x ∣ y ) p ( y ) p ( x ) p(y|x)=\frac {p(x|y)p(y)}{p(x)} p(y∣x)=p(x)p(x∣y)p(y)​
p(x)项和p(y)没有关系，所以可以去掉，原式可以写为基于联合概率建模,形式
$argmaxp(y|x)=argmaxp(x|y)p(y)=argmaxp(x,y)$
这里$p(y)是先验概率，p(y|x)是后验概率，p(x|y)是似然函数$

假定：
$p(y)\sim B(1,p)$
p ( x ∣ y = 1 ) ∼ N ( μ 1 , σ ) p(x|y=1) \ \ \thicksim N(\mu_1,\sigma) p(x∣y=1)  ∼N(μ1​,σ)
p ( x ∣ y = 0 ) ∼ N ( μ 2 , σ ) p(x|y=0) \ \ \thicksim N(\mu_2,\sigma) p(x∣y=0)  ∼N(μ2​,σ)
令$y=1$
则有：
p ( y ) = ρ y ( 1 − ρ ) 1 − y p(y)=\rho^y(1-\rho)^{1-y} p(y)=ρy(1−ρ)1−y
p ( x ∣ y ) = N ( μ 1 , σ ) y N ( μ 2 , σ ) 1 − y p(x|y) = N(\mu_1,\sigma)^yN(\mu_2,\sigma)^{1-y} p(x∣y)=N(μ1​,σ)yN(μ2​,σ)1−y
建立似然函数有
$L(\theta )=\log \prod P(x|y)p(y)$
$=\sum \log P(x|y)p(y)$
= ∑ log ⁡ ( N ( μ 1 , σ ) y N ( μ 2 , σ ) 1 − y ρ y ( 1 − ρ ) 1 − y ) \quad =\sum \log ( N(\mu_1,\sigma)^yN(\mu_2,\sigma)^{1-y}\rho^y(1-\rho)^{1-y}) =∑log(N(μ1​,σ)yN(μ2​,σ)1−yρy(1−ρ)1−y)
= ∑ log ⁡ ( N ( μ 1 , σ ) y N ( μ 2 , σ ) 1 − y ) + log ⁡ ( ρ y ( 1 − ρ ) 1 − y ) \quad =\sum \log ( N(\mu_1,\sigma)^yN(\mu_2,\sigma)^{1-y})+\log(\rho^y(1-\rho)^{1-y}) =∑log(N(μ1​,σ)yN(μ2​,σ)1−y)+log(ρy(1−ρ)1−y)
= ∑ log ⁡ ( N ( μ 1 , σ ) y ) + log ⁡ ( N ( μ 2 , σ ) 1 − y ) + log ⁡ ( ρ y ( 1 − ρ ) 1 − y ) \quad =\sum \log ( N(\mu_1,\sigma)^y)+\log(N(\mu_2,\sigma)^{1-y})+\log(\rho^y(1-\rho)^{1-y}) =∑log(N(μ1​,σ)y)+log(N(μ2​,σ)1−y)+log(ρy(1−ρ)1−y)
所以 θ = ( μ 1 , μ 2 , σ , ρ ) \theta=(\mu_1,\mu_2,\sigma,\rho) θ=(μ1​,μ2​,σ,ρ)
最后求解 θ ^ = a r g m a x L ( θ ) \hat \theta = argmaxL(\theta) θ^=argmaxL(θ)
1：求$\rho$
∂ L ( θ ) ∂ ρ = d ∑ ( log ⁡ ρ y + log ⁡ ( 1 − ρ ) 1 − y ) \frac{\partial L(\theta)}{\partial \rho}=d\sum( \log \rho^y+\log(1-\rho)^{1-y}) ∂ρ∂L(θ)​=d∑(logρy+log(1−ρ)1−y)
= ∑ ( y 1 ρ + ( 1 − y ) 1 1 − ρ ( − 1 ) ) = 0 \quad =\sum(y \frac{1}{\rho}+(1-y) \frac{1}{1-\rho}(-1)) =0 =∑(yρ1​+(1−y)1−ρ1​(−1))=0
$=\sum (y(1-\rho )-(1-y)\rho )=0$
$=\sum (\rho -y\rho -y+y\rho )=0$
$=\sum (\rho -y)=0$
所以有
$\sum y=\sum \rho$
因为：
$\sum =N$
y=1的个数有
$\sum y=N1$
y=0的个数有
$\sum (1-y)=N2$
$N1+N2=N$
所以
$N1=N\rho$
ρ = N 1 N \rho = \frac{N_1}{N} ρ=NN1​​

2：求 μ 1 \mu_1 μ1​
∂ L ( θ ) ∂ μ 1 = d ∑ log ⁡ ( N ( μ 1 , σ ) y ) \frac{\partial L(\theta)}{\partial \mu_1}=d\sum \log ( N(\mu_1,\sigma)^y) ∂μ1​∂L(θ)​=d∑log(N(μ1​,σ)y)
定义：
∑ = [ σ 1 2 0 ⋯ 0 0 σ 2 2 ⋯ 0 ⋮ ⋯ ⋯ ⋮ 0 0 ⋯ σ n 2 ] ∑_{}^{} = \left[ \begin{matrix} σ_{1}^2&0&\cdots&0\\ 0&σ_{2}^2&\cdots&0\\ \vdots&\cdots&\cdots&\vdots\\ 0&0&\cdots&σ_{n}^2 \end{matrix}\right] ∑​=⎣⎢⎢⎢⎡​σ12​0⋮0​0σ22​⋯0​⋯⋯⋯⋯​00⋮σn2​​⎦⎥⎥⎥⎤​
$\sum$代表协方差矩阵， i行j列的元素值表示不同元素的协方差

因为现在变量之间是相互独立的，所以只有对角线上 (i = j)存在非0元素，其他地方都等于0，且元素与它本身的协方差就等于方差
∑是一个对角阵，根据对角矩阵的性质，它的逆矩阵表示为：
( ∑ ) − 1 = [ 1 σ 1 2 0 ⋯ 0 0 1 σ 2 2 ⋯ 0 ⋮ ⋯ ⋯ ⋮ 0 0 ⋯ 1 σ n 2 ] (∑_{}^{})^{-1} = \left[ \begin{matrix} \frac{1}{σ_{1}^2}&0&\cdots&0\\ 0&\frac{1}{σ_{2}^2}&\cdots&0\\ \vdots&\cdots&\cdots&\vdots\\ 0&0&\cdots&\frac{1}{σ_{n}^2} \end{matrix}\right] (∑​)−1=⎣⎢⎢⎢⎢⎡​σ12​1​0⋮0​0σ22​1​⋯0​⋯⋯⋯⋯​00⋮σn2​1​​⎦⎥⎥⎥⎥⎤​
对角矩阵的行列式 = 对角元素的乘积
σ z = ∣ ∑ ∣ 1 2 = σ 1 σ 2 . . . . . σ n σ_{z}= \left|∑_{}^{}\right|^\frac{1}{2} =σ_{1}σ_{2}.....σ_{n} σz​=∣∑​∣21​=σ1​σ2​.....σn​

展开有
∂ L ( θ ) ∂ μ 1 = d ∑ y log ⁡ ( 1 2 π ) n ∣ ∑ ∣ 1 2 e x p ( − 1 2 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) ) \frac{\partial L(\theta)}{\partial \mu_1}=d\sum y\log (\frac{1}{\sqrt{2\pi})^n|\sum|^{\frac{1}{2}}}exp(-\frac{1}{2}(x-\mu_1)^T\sum^{-1}(x-\mu_1)) ∂μ1​∂L(θ)​=d∑ylog(2π ​)n∣∑∣21​1​exp(−21​(x−μ1​)T∑−1(x−μ1​))
∂ L ( θ ) ∂ μ 1 = d ∑ y log ⁡ ( 1 2 π ) n ∣ ∑ ∣ 1 2 ) − y 1 2 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) \frac{\partial L(\theta)}{\partial \mu_1}=d\sum y\log (\frac{1}{\sqrt{2\pi})^n|\sum|^{\frac{1}{2}}})-y\frac{1}{2}(x-\mu_1)^T\sum^{-1}(x-\mu_1) ∂μ1​∂L(θ)​=d∑ylog(2π ​)n∣∑∣21​1​)−y21​(x−μ1​)T∑−1(x−μ1​)
这里的第一个$\sum$是求和符号

第一项和 μ 1 \mu_1 μ1​无关，所以也就是
− 1 2 d μ ∑ y ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) = 0 - \frac{1}{2}d_\mu\sum y(x-\mu_1)^T\sum^{-1}(x-\mu_1) =0 −21​dμ​∑y(x−μ1​)T∑−1(x−μ1​)=0
− 1 2 d μ ∑ y ( x T ∑ − 1 − μ 1 T ∑ − 1 ) ( x − μ 1 ) = 0 - \frac{1}{2}d_\mu\sum y(x^T\sum^{-1}-\mu_1^T\sum^{-1})(x-\mu_1) =0 −21​dμ​∑y(xT∑−1−μ1T​∑−1)(x−μ1​)=0
− 1 2 d μ ∑ y ( x T ∑ − 1 x − x T ∑ − 1 μ 1 − μ 1 T ∑ − 1 x + μ 1 T ∑ − 1 μ 1 ) = 0 - \frac{1}{2}d_\mu\sum y(x^T\sum^{-1}x-x^T\sum^{-1}\mu_1-\mu_1^T\sum^{-1}x+\mu_1^T\sum^{-1}\mu_1)=0 −21​dμ​∑y(xT∑−1x−xT∑−1μ1​−μ1T​∑−1x+μ1T​∑−1μ1​)=0
− 1 2 d μ ∑ y ( x T ∑ − 1 x − x T ∑ − 1 μ 1 − μ 1 T ∑ − 1 x + μ 1 T ∑ − 1 μ 1 ) = 0 - \frac{1}{2}d_\mu\sum y(x^T\sum^{-1}x-x^T\sum^{-1}\mu_1-\mu_1^T\sum^{-1}x+\mu_1^T\sum^{-1}\mu_1)=0 −21​dμ​∑y(xT∑−1x−xT∑−1μ1​−μ1T​∑−1x+μ1T​∑−1μ1​)=0
也就是
− 1 2 ∑ y ( − 2 x T ∑ − 1 + 2 ∑ − 1 μ 1 ） = 0 - \frac{1}{2}\sum y(-2x^T\sum^{-1}+2\sum^{-1}\mu_1）=0 −21​∑y(−2xT∑−1+2∑−1μ1​）=0
∑ y ( x T ∑ − 1 − ∑ − 1 μ 1 ） = 0 \sum y(x^T\sum^{-1}-\sum^{-1}\mu_1）=0 ∑y(xT∑−1−∑−1μ1​）=0
∑ y ( x − μ 1 ） = 0 \sum y(x-\mu_1）=0 ∑y(x−μ1​）=0
∑ x y = ∑ y μ 1 \sum xy=\sum y\mu_1 ∑xy=∑yμ1​
μ 1 = ∑ x y ∑ y = ∑ x y N 1 \mu_1 =\frac{\sum xy}{\sum y} =\frac{\sum xy}{N1} μ1​=∑y∑xy​=N1∑xy​

求$\sum$
矩阵的迹相关定理：
t r ( A ) = ∑ A i i tr(A)=\sum A_{ii} tr(A)=∑Aii​
$tr(AB)=tr(BA)$
$tr(ABC)=tr(CBA)$
∂ t r ( A B ) ∂ A = B T \frac{\partial tr(AB)}{\partial A}=B^T ∂A∂tr(AB)​=BT
|A|表示矩阵A的行列式
∂ ∣ A ∣ ∂ A = ∣ A ∣ . A − 1 \frac{\partial |A|}{\partial A}=|A|.A^{-1} ∂A∂∣A∣​=∣A∣.A−1
如果a∈实数，则有tr(a)=a
令：
C 1 = { x i ∣ y = 1 ; x i ∈ 1... n } C1=\{x_i |y=1;x_i \in 1...n\} C1={xi​∣y=1;xi​∈1...n}
C 2 = { x i ∣ y = 0 ; x i ∈ 1... n } C2=\{x_i |y=0;x_i \in 1...n\} C2={xi​∣y=0;xi​∈1...n}
$|C1|=N1$
$|C2|=N2$
$N1+N2=N$
原函数对$\sum$ 求偏导有
∂ J ( θ ) ∂ ∑ = d ( ∑ x i ∈ C 1 log ⁡ ( N ( μ 1 , ∑ ) + ∑ x i ∈ C 2 log ⁡ ( N ( μ 2 , ∑ ) ) = 0 \frac{\partial J(\theta)}{\partial \sum}=d(\displaystyle \sum_{x_i \in C1}\log ( N(\mu_1,\sum) +\displaystyle \sum_{x_i \in C2}\log ( N(\mu_2,\sum)) =0 ∂∑∂J(θ)​=d(xi​∈C1∑​log(N(μ1​,∑)+xi​∈C2∑​log(N(μ2​,∑))=0
令：
f ( μ 1 ) = ∑ x i ∈ C 1 log ⁡ ( N ( μ 1 , ∑ ) f(\mu_1) =\displaystyle \sum_{x_i \in C1}\log ( N(\mu_1,\sum) f(μ1​)=xi​∈C1∑​log(N(μ1​,∑)
f ( μ 1 ) = ∑ x i ∈ C 1 log ⁡ ( 1 2 π ) n ∣ ∑ ∣ 1 2 e x p ( − 1 2 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) ) f(\mu_1) = \sum_{x_i \in C1}\log ( \frac{1}{\sqrt{2\pi})^n|\sum|^{\frac{1}{2}}}exp(-\frac{1}{2}(x-\mu_1)^T \sum^{-1}(x-\mu_1)) f(μ1​)=∑xi​∈C1​log(2π ​)n∣∑∣21​1​exp(−21​(x−μ1​)T∑−1(x−μ1​))
f ( μ 1 ) = ∑ x i ∈ C 1 log ⁡ 1 2 π ) n ∣ ∑ ∣ 1 2 − 1 2 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) f(\mu_1) = \sum_{x_i \in C1}\log \frac{1}{\sqrt{2\pi})^n|\sum|^{\frac{1}{2}}}-\frac{1}{2}(x-\mu_1)^T\sum^{-1}(x-\mu_1) f(μ1​)=∑xi​∈C1​log2π ​)n∣∑∣21​1​−21​(x−μ1​)T∑−1(x−μ1​)
f ( μ 1 ) = ∑ x i ∈ C 1 log ⁡ 1 2 π ) n − 1 2 log ⁡ ∣ ∑ ∣ − 1 2 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) f(\mu_1) =\sum_{x_i \in C1}\log \frac{1}{\sqrt{2\pi})^n}-{\frac{1}{2}}\log |\sum|-\frac{1}{2}(x-\mu_1)^T\sum^{-1}(x-\mu_1) f(μ1​)=∑xi​∈C1​log2π ​)n1​−21​log∣∑∣−21​(x−μ1​)T∑−1(x−μ1​)
把求和符号带人有
f ( μ 1 ) = ∑ x i ∈ C 1 log ⁡ 1 2 π ) n − 1 2 ∑ x i ∈ C 1 log ⁡ ∣ ∑ ∣ − 1 2 ∑ x i ∈ C 1 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) f(\mu_1) =\sum_{x_i \in C1}\log \frac{1}{\sqrt{2\pi})^n}-{\frac{1}{2}} \sum_{x_i \in C1}\log |\sum|-\frac{1}{2}\sum_{x_i \in C1}(x-\mu_1)^T\sum^{-1}(x-\mu_1) f(μ1​)=∑xi​∈C1​log2π ​)n1​−21​∑xi​∈C1​log∣∑∣−21​∑xi​∈C1​(x−μ1​)T∑−1(x−μ1​)

∑ x i ∈ C 1 log ⁡ 1 2 π ) n 和 ∑ 无 关 ， 记 作 常 识 C 3 \displaystyle \sum_{x_i \in C1}\log \frac{1}{\sqrt{2\pi})^n} 和\sum无关，记作常识C3 xi​∈C1∑​log2π ​)n1​和∑无关，记作常识C3

− 1 2 ∑ x i ∈ C 1 log ⁡ ∣ ∑ ∣ = − 1 2 N 1 log ⁡ ∣ ∑ ∣ -{\frac{1}{2}}\displaystyle \sum_{x_i \in C1}\log |\sum|=-\frac{1}{2}N1\log |\sum| −21​xi​∈C1∑​log∣∑∣=−21​N1log∣∑∣

由于 ( x − μ 1 ) T (x-\mu_1)^T (x−μ1​)T是(1xn)维
∑ − 1 \sum^{-1} ∑−1是pxp 维
( x − μ 1 ) (x-\mu_1) (x−μ1​)是px1维
所以 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) (x-\mu_1)^T\sum^{-1}(x-\mu_1) (x−μ1​)T∑−1(x−μ1​)结果是实数
也就可以表示为
( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) = t r ( ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) ) (x-\mu_1)^T\sum^{-1}(x-\mu_1)=tr((x-\mu_1)^T\sum^{-1}(x-\mu_1)) (x−μ1​)T∑−1(x−μ1​)=tr((x−μ1​)T∑−1(x−μ1​))
= t r ( ( x − μ 1 ) T ( x − μ 1 ) ∑ − 1 ) =tr((x-\mu_1)^T(x-\mu_1)\sum^{-1}) =tr((x−μ1​)T(x−μ1​)∑−1)
∑ x i ∈ C 1 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) = ∑ x i ∈ C 1 t r ( ( x − μ 1 ) T ( x − μ 1 ) ∑ − 1 ) \sum_{x_i \in C1}(x-\mu_1)^T\sum^{-1}(x-\mu_1)= \sum_{x_i \in C1}tr((x-\mu_1)^T(x-\mu_1)\sum^{-1}) ∑xi​∈C1​(x−μ1​)T∑−1(x−μ1​)=∑xi​∈C1​tr((x−μ1​)T(x−μ1​)∑−1)
∑ x i ∈ C 1 ( x − μ 1 ) T ∑ − 1 ( x − μ 1 ) = t r ( ∑ x i ∈ C 1 ( x − μ 1 ) T ( x − μ 1 ) ∑ − 1 ) \sum_{x_i \in C1}(x-\mu_1)^T\sum^{-1}(x-\mu_1)=tr( \sum_{x_i \in C1}(x-\mu_1)^T(x-\mu_1)\sum^{-1}) ∑xi​∈C1​(x−μ1​)T∑−1(x−μ1​)=tr(∑xi​∈C1​(x−μ1​)T(x−μ1​)∑−1)
因为有：方差矩阵
S 1 = 1 N 1 ( ∑ x i ∈ C 1 ( x − μ 1 ) T ( x − μ 1 ) ) S1=\frac{1}{N1}(\displaystyle \sum_{x_i \in C1}(x-\mu_1)^T(x-\mu_1)) S1=N11​(xi​∈C1∑​(x−μ1​)T(x−μ1​))
所以
∑ x i ∈ C 1 ( x − μ 1 ) T ∣ ∑ ∣ − 1 ( x − μ 1 ) = N 1 t r ( S 1 ∑ − 1 ) \sum_{x_i \in C1}(x-\mu_1)^T|\sum|^{-1}(x-\mu_1)=N1tr(S1\sum^{-1}) ∑xi​∈C1​(x−μ1​)T∣∑∣−1(x−μ1​)=N1tr(S1∑−1)

f ( μ 1 ) = − 1 2 ( C 3 + N 1 log ⁡ ∣ ∑ ∣ + N 1 t r ( S 1 ∑ − 1 ) f(\mu_1) =-\frac{1}{2}(C3+N1\log |\sum|+N1tr(S1\sum^{-1}) f(μ1​)=−21​(C3+N1log∣∑∣+N1tr(S1∑−1)
同理：
f ( μ 2 ) = − 1 2 ( C 4 + N 2 log ⁡ ∣ ∑ ∣ + N 2 t r ( S 2 ∑ − 1 ) f(\mu_2) =-\frac{1}{2}(C4+N2\log |\sum|+N2tr(S2\sum^{-1}) f(μ2​)=−21​(C4+N2log∣∑∣+N2tr(S2∑−1)
对原函数求导可以写为
∂ J ( θ ) ∂ ∑ = d ( f ( μ 1 ) + f ( μ 2 ) ) = 0 \frac{\partial J(\theta)}{\partial \sum}=d(f(\mu_1)+f(\mu_2)) =0 ∂∑∂J(θ)​=d(f(μ1​)+f(μ2​))=0
∂ J ( θ ) ∂ ∑ = d ( − 1 2 ( N 1 log ⁡ ∣ ∑ ∣ + N 1 t r ( S 1 ∑ − 1 ) − 1 2 ( N 2 log ⁡ ∣ ∑ ∣ + N 2 t r ( S 2 ∑ − 1 ) ) = 0 \frac{\partial J(\theta)}{\partial \sum}=d(-\frac{1}{2}(N1\log |\sum|+N1tr(S1\sum^{-1})-\frac{1}{2}(N2\log |\sum|+N2tr(S2\sum^{-1})) =0 ∂∑∂J(θ)​=d(−21​(N1log∣∑∣+N1tr(S1∑−1)−21​(N2log∣∑∣+N2tr(S2∑−1))=0
∂ J ( θ ) ∂ ∑ = d ( − 1 2 ( N log ⁡ ∣ ∑ ∣ + N 1 t r ( S 1 ∑ − 1 ) + N 2 t r ( S 2 ∑ − 1 ) ) ) = 0 \frac{\partial J(\theta)}{\partial \sum}=d(-\frac{1}{2}(N\log |\sum|+N1tr(S1\sum^{-1})+N2tr(S2\sum^{-1}))) =0 ∂∑∂J(θ)​=d(−21​(Nlog∣∑∣+N1tr(S1∑−1)+N2tr(S2∑−1)))=0
∂ J ( θ ) ∂ ∑ = − 1 2 ( N 1 ∣ ∑ ∣ ∣ ∑ ∣ ∑ − 1 + N 1 t r ( ∑ − 1 S 1 ) + N 2 t r ( ∑ − 1 S 2 ) ) = 0 \frac{\partial J(\theta)}{\partial \sum}=-\frac{1}{2}(N\frac{1}{|\sum|} |\sum|\sum^{-1}+N1tr(\sum^{-1}S1)+N2tr(\sum^{-1}S2)) =0 ∂∑∂J(θ)​=−21​(N∣∑∣1​∣∑∣∑−1+N1tr(∑−1S1)+N2tr(∑−1S2))=0
∂ J ( θ ) ∂ ∑ = − 1 2 ( N ∑ − 1 + N 1 t r ( ∑ − 1 S 1 ) + N 2 t r ( ∑ − 1 S 2 ) ) = 0 \frac{\partial J(\theta)}{\partial \sum}=-\frac{1}{2}(N\sum^{-1}+N1tr(\sum^{-1}S1)+N2tr(\sum^{-1}S2)) =0 ∂∑∂J(θ)​=−21​(N∑−1+N1tr(∑−1S1)+N2tr(∑−1S2))=0
∂ J ( θ ) ∂ ∑ = − 1 2 ( N ∑ − 1 − N 1 S 1 T ∑ − 2 − N 2 S 2 T ∑ − 2 ) = 0 \frac{\partial J(\theta)}{\partial \sum}=-\frac{1}{2}(N\sum^{-1}-N1S1^T\sum^{-2}-N2S2^T\sum^{-2})=0 ∂∑∂J(θ)​=−21​(N∑−1−N1S1T∑−2−N2S2T∑−2)=0
两边乘以 ∑ 2 \sum^{2} ∑2有
N ∑ = N 1 S 1 T + N 2 S 2 T N\sum =N1S1^T+N2S2^T N∑=N1S1T+N2S2T
∑ = N 1 S 1 T + N 2 S 2 T N \sum =\frac{N1S1^T+N2S2^T}{N} ∑=NN1S1T+N2S2T​
由于方差矩阵的对称型，所以可写为
∑ = N 1 S 1 + N 2 S 2 N \sum =\frac{N1S1+N2S2}{N} ∑=NN1S1+N2S2​

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