增加3.10 3.27 3.30 3.32 3.37推导

Chapter3/README.md

@@ -3,7 +3,12 @@
 ### 3.1 基本形式
 ### 3.2 线性回归
 - [3.7](https://github.com/Datawhale18/pumpkin-book/blob/master/Chapter3/chapter3.md)
+- [3.10](https://github.com/Datawhale18/pumpkin-book/blob/master/Chapter3/chapter3.md)
 ### 3.3 对数几率回归
+- [3.27](https://github.com/Datawhale18/pumpkin-book/blob/master/Chapter3/chapter3.md)
+- [3.30](https://github.com/Datawhale18/pumpkin-book/blob/master/Chapter3/chapter3.md)
 ### 3.4 线性判别分析
+- [3.32](https://github.com/Datawhale18/pumpkin-book/blob/master/Chapter3/chapter3.md)
+- [3.37](https://github.com/Datawhale18/pumpkin-book/blob/master/Chapter3/chapter3.md)
 ### 3.5 多分类学习
 ### 3.6 类别不平衡问题

Chapter3/chapter3.md

@@ -32,4 +32,77 @@ $$
 \end{aligned}
 $$
 若令$ \mathbf{X}=(x_1,x_2,...,x_m) $，$\mathbf{X}\_{demean}$为去均值后的$ \mathbf{X} $，$ \mathbf{y}=(y_1,y_2,...,y_m) $，$ \mathbf{y}\_{demean} $为去均值后的$ \mathbf{y} $，其中$ \mathbf{X} $、$ \mathbf{X}\_{demean} $、$ \mathbf{y} $、$ \mathbf{y}\_{demean} $均为m行1列的列向量，代入上式可得：
-$$ w=\cfrac{\mathbf{X}\_{demean}\mathbf{y}\_{demean}^T}{\mathbf{X}\_{demean}\mathbf{X}\_{demean}^T}$$
+$$ w=\cfrac{\mathbf{X}\_{demean}\mathbf{y}\_{demean}^T}{\mathbf{X}\_{demean}\mathbf{X}\_{demean}^T}$$
+### 3.10
+
+$$ \cfrac{\partial E_{\hat{w}}}{\partial \hat{w}}=2\mathbf{X}^T(\mathbf{X}\hat{w}-\mathbf{y}) $$
+
+将$ E_{\hat{w}}=(\mathbf{y}-\mathbf{X}\hat{w})^T(\mathbf{y}-\mathbf{X}\hat{w}) $展开可得：
+$$ E_{\hat{w}}= \mathbf{y}^T\mathbf{y}-\mathbf{y}^T\mathbf{X}\hat{w}-\hat{w}^T\mathbf{X}^T\mathbf{y}+\hat{w}^T\mathbf{X}^T\mathbf{X}\hat{w} $$
+对$ \hat{w} $求导可得：
+$$ \cfrac{\partial E_{\hat{w}}}{\partial \hat{w}}= \cfrac{\partial \mathbf{y}^T\mathbf{y}}{\partial \hat{w}}-\cfrac{\partial \mathbf{y}^T\mathbf{X}\hat{w}}{\partial \hat{w}}-\cfrac{\partial \hat{w}^T\mathbf{X}^T\mathbf{y}}{\partial \hat{w}}+\cfrac{\partial \hat{w}^T\mathbf{X}^T\mathbf{X}\hat{w}}{\partial \hat{w}} $$
+由向量的求导公式可得：
+$$ \cfrac{\partial E_{\hat{w}}}{\partial \hat{w}}= 0-\mathbf{X}^T\mathbf{y}-\mathbf{X}^T\mathbf{y}+(\mathbf{X}^T\mathbf{X}+\mathbf{X}^T\mathbf{X})\hat{w} $$
+$$ \cfrac{\partial E_{\hat{w}}}{\partial \hat{w}}=2\mathbf{X}^T(\mathbf{X}\hat{w}-\mathbf{y}) $$
+
+### 3.27
+
+$$ l(β)=\sum_{i=1}^{m}(-y_iβ^T\hat{\boldsymbol x_i}+\ln(1+e^{β^T\hat{\boldsymbol x_i}})) $$
+
+将式（3.26）代入式（3.25）可得：
+$$ l(β,b)=\sum_{i=1}^{m}\ln(y_ip_1(\boldsymbol{\hat{x_i}};β)+(1-y_i)p_0(\boldsymbol{\hat{x_i}};β)) $$
+其中$ p_1(\boldsymbol{\hat{x_i}};β)=\cfrac{e^{β^T\hat{\boldsymbol x_i}}}{1+e^{β^T\hat{\boldsymbol x_i}}},p_0(\boldsymbol{\hat{x_i}};β)=\cfrac{1}{1+e^{β^T\hat{\boldsymbol x_i}}} $，代入上式可得：
+$$ l(β,b)=\sum_{i=1}^{m}\ln(\cfrac{y_ie^{β^T\hat{\boldsymbol x_i}}+1-y_i}{1+e^{β^T\hat{\boldsymbol x_i}}}) $$
+$$ l(β,b)=\sum_{i=1}^{m}(\ln(y_ie^{β^T\hat{\boldsymbol x_i}}+1-y_i)-\ln(1+e^{β^T\hat{\boldsymbol x_i}})) $$
+又$ y_i $=0或1，则：
+$$ l(β,b) =
+\begin{cases} 
+\sum_{i=1}^{m}(-\ln(1+e^{β^T\hat{\boldsymbol x_i}})),  & y_i=0 \\\\
+\sum_{i=1}^{m}(β^T\hat{\boldsymbol x_i}-\ln(1+e^{β^T\hat{\boldsymbol x_i}})), & y_i=1
+\end{cases} $$
+两式综合可得：
+$$ l(β)=\sum_{i=1}^{m}(y_iβ^T\hat{\boldsymbol x_i}-\ln(1+e^{β^T\hat{\boldsymbol x_i}})) $$
+由于此式仍为极大似然估计的似然函数，所以最大化似然函数等价于最小化似然函数的相反数，也即在似然函数前添加负号即可得式（3.27）。
+
+【注】：若式（3.26）中的似然项改写方式为$ p(y_i|\boldsymbol x_i;\boldsymbol w,b)=[p_1(\boldsymbol{\hat{x_i}};β)]^{y_i}[p_0(\boldsymbol{\hat{x_i}};β)]^{1-y_i} $，再将其代入式（3.25）可得：
+$$ l(β)=\sum_{i=1}^{m}(y_i\ln(p_1(\boldsymbol{\hat{x_i}};β))+(1-y_i)\ln(p_0(\boldsymbol{\hat{x_i}};β))) $$
+此式显然更易推导出式（3.27）
+
+### 3.30
+
+$$\frac{\partial l(β)}{\partial β}=-\sum_{i=1}^{m}\hat{\boldsymbol x_i}(y_i-p_1(\hat{\boldsymbol x_i};β))$$
+
+此式可以进行向量化，令$p_1(\hat{\boldsymbol x_i};β)=\hat{y_i}$，代入上式得：
+$$\begin{aligned}
+	\frac{\partial l(β)}{\partial β} &= -\sum_{i=1}^{m}\hat{\boldsymbol x_i}(y_i-\hat{y_i}) \\\\
+	& =\sum_{i=1}^{m}\hat{\boldsymbol x_i}(\hat{y_i}-y_i) \\\\
+	& ={\boldsymbol X^T}(\hat{\boldsymbol y}-\boldsymbol{y}) \\\\
+	& ={\boldsymbol X^T}(p_1(\boldsymbol X;β)-\boldsymbol{y}) \\\\
+\end{aligned}$$
+
+### 3.32
+
+$$J=\cfrac{\boldsymbol w^T(\mu_0-\mu_1)(\mu_0-\mu_1)^T\boldsymbol w}{\boldsymbol w^T(\Sigma_0+\Sigma_1)\boldsymbol w}$$
+
+$$\begin{aligned}
+	J &= \cfrac{\big|\big|\boldsymbol w^T\mu_0-\boldsymbol w^T\mu_1\big|\big|_2^2}{\boldsymbol w^T(\Sigma_0+\Sigma_1)\boldsymbol w} \\\\
+	&= \cfrac{\big|\big|(\boldsymbol w^T\mu_0-\boldsymbol w^T\mu_1)^T\big|\big|_2^2}{\boldsymbol w^T(\Sigma_0+\Sigma_1)\boldsymbol w} \\\\
+	&= \cfrac{\big|\big|(\mu_0-\mu_1)^T\boldsymbol w\big|\big|_2^2}{\boldsymbol w^T(\Sigma_0+\Sigma_1)\boldsymbol w} \\\\
+	&= \cfrac{[(\mu_0-\mu_1)^T\boldsymbol w]^T(\mu_0-\mu_1)^T\boldsymbol w}{\boldsymbol w^T(\Sigma_0+\Sigma_1)\boldsymbol w} \\\\
+	&= \cfrac{\boldsymbol w^T(\mu_0-\mu_1)(\mu_0-\mu_1)^T\boldsymbol w}{\boldsymbol w^T(\Sigma_0+\Sigma_1)\boldsymbol w}
+\end{aligned}$$
+
+### 3.37
+
+$$\boldsymbol S_b\boldsymbol w=\lambda\boldsymbol S_w\boldsymbol w$$
+
+由3.36可列拉格朗日函数：
+$$l(\boldsymbol w)=-\boldsymbol w^T\boldsymbol S_b\boldsymbol w+\lambda(\boldsymbol w^T\boldsymbol S_w\boldsymbol w-1)$$
+对$\boldsymbol w$求偏导可得：
+$$\begin{aligned}
+\cfrac{\partial l(\boldsymbol w)}{\partial \boldsymbol w} &= -\cfrac{\partial(\boldsymbol w^T\boldsymbol S_b\boldsymbol w)}{\partial \boldsymbol w}+\lambda \cfrac{(\boldsymbol w^T\boldsymbol S_w\boldsymbol w-1)}{\partial \boldsymbol w} \\\\
+	&= -(\boldsymbol S_b+\boldsymbol S_b^T)\boldsymbol w+\lambda(\boldsymbol S_w+\boldsymbol S_w^T)\boldsymbol w
+\end{aligned}$$
+又$\boldsymbol S_b=\boldsymbol S_b^T,\boldsymbol S_w=\boldsymbol S_w^T$，则：
+$$\cfrac{\partial l(\boldsymbol w)}{\partial \boldsymbol w} = -2\boldsymbol S_b\boldsymbol w+2\lambda\boldsymbol S_w\boldsymbol w$$
+令导函数等于0即可得式3.37。