首先要知道加法模型和指数损失函数。

加法模型

f ( x ) = ∑ m = 1 M α m G m ( x ) f(x) = \sum\limits_{m=1}^{M}\alpha_mG_{m}(x) f(x)=m=1MαmGm(x)

加法模型是一个加和模型,每一列训练一个分类器 G m ( x ) ​ G_{m}(x)​ Gm(x),并且基于这个分类器的误差,得到这个分类器的权重 α m ​ \alpha_m​ αm

指数损失函数

L ( y , f ( x ) ) = exp ⁡ [ − y f ( x ) ] L(y,f(x)) = \exp[-yf(x)] L(y,f(x))=exp[yf(x)]

对于分类模型而言,上述损失函数,在分类正确的时候,指数部分为负数;在分类错误的时候,指数部分为正数,符合损失函数的意义。

由加法模型的定义:
f ( x ) = ∑ m = 1 M α m G m ( x ) f(x) = \sum\limits_{m=1}^{M}\alpha_mG_{m}(x) f(x)=m=1MαmGm(x)
得到迭代公式:
f m ( x ) = f m − 1 ( x ) + α m G m ( x ) f_m(x) = f_{m-1}(x) + \alpha_mG_{m}(x) fm(x)=fm1(x)+αmGm(x)
此时认为函数 f m − 1 ( x ) f_{m-1}(x) fm1(x) 是已知的,每一轮迭代求的是分类器 G m ( x ) G_{m}(x) Gm(x) 和这个分类器的权重 α m \alpha_m αm。将上式代入损失函数得:
L ( y , f ( x ) ) = ∑ i = 1 N exp ⁡ [ − y i ( f m − 1 ( x ) + α m G m ( x ) ) ] L(y,f(x)) = \sum_{i=1}^{N}\exp[-y_i(f_{m-1}(x) + \alpha_mG_{m}(x))] L(y,f(x))=i=1Nexp[yi(fm1(x)+αmGm(x))]
把指数加法因子展开,变成指数的乘积,得到:
L ( y , f ( x ) ) = ∑ i = 1 N [ exp ⁡ [ − y i ( f m − 1 ( x ) ] [ exp ⁡ ( − y i α m G m ( x ) ) ] ] L(y,f(x)) = \sum_{i=1}^{N}[\exp[-y_i(f_{m-1}(x)][\exp(-y_i\alpha_mG_{m}(x))]] L(y,f(x))=i=1N[exp[yi(fm1(x)][exp(yiαmGm(x))]]
由于 y i y_i yi f m − 1 ( x ) f_{m-1}(x) fm1(x) 已知,可以令 w ‾ m i = exp ⁡ [ − y i f m − 1 ( x i ) ] \overline w_{mi} =\exp[-y_if_{m-1}(x_i)] wmi=exp[yifm1(xi)],于是
L ( y , f ( x ) ) = ∑ i = 1 N w ‾ m i exp ⁡ ( − y i α m G m ( x ) ) L(y,f(x)) = \sum_{i=1}^{N}\overline w_{mi} \exp(-y_i\alpha_mG_{m}(x)) L(y,f(x))=i=1Nwmiexp(yiαmGm(x))
于是分类器 G m ( x ) ​ G_{m}(x)​ Gm(x) 和这个分类器的权重 α m ​ \alpha_m​ αm 可以表示成:
( α m , G m ( x ) ) = a r g    m i n    ⎵ α , G ∑ i = 1 N w ‾ m i exp ⁡ ( − y i α m G m ( x ) ) (\alpha_m,G_{m}(x)) = \underbrace{arg\;min\;}_{\alpha,G} \sum_{i=1}^{N}\overline w_{mi} \exp(-y_i\alpha_mG_{m}(x)) (αm,Gm(x))=α,G argmini=1Nwmiexp(yiαmGm(x))
先求 G m ( x ) G_{m}(x) Gm(x),看上面的式子,分类器的权重 α m \alpha_m αm 可以认为是一个确定的数, G m ( x ) G_{m}(x) Gm(x) 是使得分错的(带权重的)样本里损失函数最小的那个,可以写成:
G m ∗ ( x ) = a r g    m i n    ⎵ G ∑ i = 1 N w ‾ m i I ( y i ≠ G ( x i ) ) G_{m}^*(x) = \underbrace{arg\;min\;}_{G} \sum_{i=1}^{N}\overline w_{mi} I(y_i \neq G(x_i)) Gm(x)=G argmini=1NwmiI(yi̸=G(xi))
注意:重点理解上面两个式子的等价性。得到 G m ∗ ( x ) G_{m}^{*}(x) Gm(x) 以后,再求 α m ∗ \alpha_m^* αm。还是看损失函数,可以写成:
L ( y , f ( x ) ) = ∑ i = 1 N w ‾ m i exp ⁡ ( − y i α m G m ( x ) ) = ∑ y i = G m ( x i ) w ‾ m i e − α + ∑ y i ≠ G m ( x i ) w ‾ m i e α = ∑ y i = G m ( x i ) w ‾ m i e − α + ∑ y i ≠ G m ( x i ) w ‾ m i e − α − ∑ y i ≠ G m ( x i ) w ‾ m i e − α + ∑ y i ≠ G m ( x i ) w ‾ m i e α = e − α ∑ i = 1 N w ‾ m i + ( e α − e − α ) ∑ y i ≠ G m ( x i ) w ‾ m i = e − α ∑ i = 1 N w ‾ m i + ( e α − e − α ) ∑ i = 1 N w ‾ m i I ( y i ≠ G m ( x i ) ) \begin {aligned} L(y,f(x)) &= \sum_{i=1}^{N}\overline w_{mi} \exp(-y_i\alpha_mG_{m}(x)) \\ &=\sum_{y_i = G_m(x_i)}\overline w_{mi} e^{-\alpha} + \sum_{y_i \neq G_m(x_i)}\overline w_{mi} e^{\alpha} \\ &=\sum_{y_i = G_m(x_i)}\overline w_{mi} e^{-\alpha} + \sum_{y_i \neq G_m(x_i)}\overline w_{mi} e^{-\alpha} - \sum_{y_i \neq G_m(x_i)}\overline w_{mi} e^{-\alpha} + \sum_{y_i \neq G_m(x_i)}\overline w_{mi} e^{\alpha} \\ &= e^{-\alpha}\sum_{i=1}^{N}\overline w_{mi} + (e^{\alpha} - e^{-\alpha})\sum_{y_i \neq G_m(x_i)}\overline w_{mi} \\ &= e^{-\alpha}\sum_{i=1}^{N}\overline w_{mi} + (e^{\alpha} - e^{-\alpha})\sum_{i=1}^N \overline w_{mi} I(y_i \neq G_m(x_i)) \end {aligned} L(y,f(x))=i=1Nwmiexp(yiαmGm(x))=yi=Gm(xi)wmieα+yi̸=Gm(xi)wmieα=yi=Gm(xi)wmieα+yi̸=Gm(xi)wmieαyi̸=Gm(xi)wmieα+yi̸=Gm(xi)wmieα=eαi=1Nwmi+(eαeα)yi̸=Gm(xi)wmi=eαi=1Nwmi+(eαeα)i=1NwmiI(yi̸=Gm(xi))
把上式对 α \alpha α 求导,再令导函数为 0 0 0,得:
− e − α ∑ i = 1 N w ‾ m i + ( e α + e − α ) ∑ i = 1 N w ‾ m i I ( y i ≠ G m ( x i ) ) = 0 -e^{-\alpha}\sum_{i=1}^{N}\overline w_{mi} + (e^{\alpha} + e^{-\alpha})\sum_{i=1}^N \overline w_{mi} I(y_i \neq G_m(x_i)) = 0 eαi=1Nwmi+(eα+eα)i=1NwmiI(yi̸=Gm(xi))=0
上式两边同时除以 ∑ i = 1 N w ‾ m i ​ \sum_{i=1}^{N}\overline w_{mi}​ i=1Nwmi,得:
− e − α + ( e α + e − α ) ∑ i = 1 N w ‾ m i I ( y i ≠ G m ( x i ) ) ∑ i = 1 N w ‾ m i = 0 -e^{-\alpha} + (e^{\alpha} + e^{-\alpha})\cfrac{\sum_{i=1}^N \overline w_{mi} I(y_i \neq G_m(x_i))}{\sum_{i=1}^{N}\overline w_{mi}} = 0 eα+(eα+eα)i=1Nwmii=1NwmiI(yi̸=Gm(xi))=0
∑ i = 1 N w ‾ m i I ( y i ≠ G m ( x i ) ) ∑ i = 1 N w ‾ m i = e m ​ \cfrac{\sum_{i=1}^N \overline w_{mi} I(y_i \neq G_m(x_i))}{\sum_{i=1}^{N}\overline w_{mi}} = e_m​ i=1Nwmii=1NwmiI(yi̸=Gm(xi))=em,则有:
− e − α + ( e α + e − α ) e m = 0 ( e α + e − α ) e m = e − α ( e 2 α + 1 ) e m = 1 e 2 α + 1 = 1 e m e 2 α = 1 e m − 1 = 1 − e m e m 2 α = log ⁡ 1 − e m e m α = 1 2 log ⁡ 1 − e m e m \begin {aligned} -e^{-\alpha} + (e^{\alpha} + e^{-\alpha})e_m &= 0 \\ (e^{\alpha} + e^{-\alpha})e_m &= e^{-\alpha}\\ (e^{2\alpha} + 1)e_m &= 1 \\ e^{2\alpha} + 1 &= \cfrac{1}{e_m} \\ e^{2\alpha} &= \cfrac{1}{e_m} -1 = \cfrac{1 - e_m}{e_m} \\ 2\alpha &= \log\cfrac{1 - e_m}{e_m} \\ \alpha &= \cfrac{1}{2}\log\cfrac{1 - e_m}{e_m} \end {aligned} eα+(eα+eα)em(eα+eα)em(e2α+1)eme2α+1e2α2αα=0=eα=1=em1=em11=em1em=logem1em=21logem1em
于是得到使得损失函数最小的
α m ∗ = 1 2 log ⁡ 1 − e m e m \alpha_m^* = \cfrac{1}{2}\log\cfrac{1 - e_m}{e_m} αm=21logem1em
这里
e m = ∑ i = 1 N w ‾ m i I ( y i ≠ G m ( x i ) ) ∑ i = 1 N w ‾ m i = ∑ i = 1 N w m i I ( y i ≠ G m ( x i ) ) e_m =\cfrac{\sum_{i=1}^N \overline w_{mi} I(y_i \neq G_m(x_i))}{\sum_{i=1}^{N}\overline w_{mi}} =\sum_{i=1}^N w_{mi} I(y_i \neq G_m(x_i)) em=i=1Nwmii=1NwmiI(yi̸=Gm(xi))=i=1NwmiI(yi̸=Gm(xi))

权重更新公式

那么权重如何更新呢?已知:
f m ( x ) = f m − 1 ( x ) + α m G m ( x ) f_m(x) = f_{m-1}(x) + \alpha_mG_{m}(x) fm(x)=fm1(x)+αmGm(x)
和我们之前的定义:
w ‾ m i = exp ⁡ [ − y i f m − 1 ( x i ) ] \overline w_{mi} =\exp[-y_if_{m-1}(x_i)] wmi=exp[yifm1(xi)]
于是就有:
w ‾ m + 1 , i = exp ⁡ [ − y i f m ( x i ) ] = exp ⁡ [ − y i [ f m − 1 ( x i ) + α m G m ( x i ) ] ] = exp ⁡ [ − y i f m − 1 ( x i ) ] exp ⁡ [ − y i α m G m ( x i ) ] = w ‾ m , i exp ⁡ [ − y i α m G m ( x i ) ] \begin {aligned} \overline w_{m+1,i} &=\exp[-y_if_{m}(x_i)] \\ &=\exp[-y_i[f_{m-1}(x_i) + \alpha_mG_{m}(x_i)]]\\ &=\exp[-y_if_{m-1}(x_i)]\exp[-y_i\alpha_mG_{m}(x_i)]\\ &=\overline w_{m,i}\exp[-y_i\alpha_mG_{m}(x_i)] \end {aligned} wm+1,i=exp[yifm(xi)]=exp[yi[fm1(xi)+αmGm(xi)]]=exp[yifm1(xi)]exp[yiαmGm(xi)]=wm,iexp[yiαmGm(xi)]