L(P,w)≡−H(P)+w0(1−∑yP(y∣x))+∑i=1nwi(Ep‾(fi)−EP(fi))=∑x,yP~(x)P(y∣x)logP(y∣x)+w0(1−∑yP(y∣x))+∑i=1nwi(∑x,yP~(x,y)fi(x,y)−∑x,yP~(x)P(y∣x)fi(x,y))\begin{aligned} L(P, w) & \equiv-H(P)+w_{0}\left(1-\sum_{y} P(y | x)\right)+\sum_{i=1}^{n} w_{i}\left(E_{\overline{p}}\left(f_{i}\right)-E_{P}\left(f_{i}\right)\right) \\=& {\color{red}\sum_{x, y} \tilde{P}(x) P(y | x) \log P(y | x) } +{\color{blue} w_{0}\left(1-\sum_{y} P(y | x)\right)} \\ &+\sum_{i=1}^{n} w_{i}\left(\sum_{x, y} \tilde{P}(x, y) f_{i}(x, y)-\sum_{x, y} \tilde{P}(x) P(y | x) f_{i}(x, y)\right) \end{aligned}L(P,w)=≡−H(P)+w0(1−y∑P(y∣x))+i=1∑nwi(Ep(fi)−EP(fi))x,y∑P~(x)P(y∣x)logP(y∣x)+w0(1−y∑P(y∣x))+i=1∑nwi(x,y∑P~(x,y)fi(x,y)−x,y∑P~(x)P(y∣x)fi(x,y))
L(P,w)L(P, w)L(P,w)对P(y|x)求导 假如是对P(y1∣x1)P(y_1|x_1)P(y1∣x1)求导,如下
∂L(P,W)∂P(y1∣x1)=P~(x1)(logP(y1∣x1)+1)−w0+∑i=1nwi∑x,yP~(x)P(y∣x)fi(x,y)\frac{\partial L(P,W)}{\partial P(y_1|x_1)}={\color{red}\tilde {P}(x_1)(logP(y_1|x_1)+1)}{\color{blue}-w_0}+\sum_{i=1}^nw_i\sum_{x,y}\tilde P(x)P(y|x)f_i(x,y)∂P(y1∣x1)∂L(P,W)=P~(x1)(logP(y1∣x1)+1)−w0+∑i=1nwi∑x,yP~(x)P(y∣x)fi(x,y)
其中红色部分是因为只有x=x1,y=y1x=x_1,y=y_1x=x1,y=y1那一项才含有P(y1∣x1)P(y_1|x_1)P(y1∣x1)
应用∑xP~(x)=1\sum_x\tilde P(x)=1∑xP~(x)=1
=P~(x1)(logP(y1∣x1)+1)+∑xP~(x)w0+∑x,yP~(x)∑i=1nwiP(y∣x)fi(x,y)=\tilde{P}(x_1)(logP(y_1|x_1)+1) +\sum_x\tilde P(x)w_0+\sum_{x,y}\tilde P(x)\sum_{i=1}^nw_iP(y|x)f_i(x,y)=P~(x1)(logP(y1∣x1)+1)+∑xP~(x)w0+∑x,yP~(x)∑i=1nwiP(y∣x)fi(x,y)
问题是最后怎么推出
∑x,yP~(x)(logP(y∣x)+1−w0−∑i=1nwifi(x,y))\sum_{x, y} \tilde{P}(x)\left(\log P(y | x)+1-w_{0}-\sum_{i=1}^{n} w_{i} f_{i}(x, y)\right)x,y∑P~(x)(logP(y∣x)+1−w0−i=1∑nwifi(x,y))
的???