PumpkinBook/docs/chapter3/chapter3.md

3.7

$$ w=\cfrac{\sum_{i=1}^{m}y_i(xi-\bar{x})}{\sum{i=1}^{m}xi^2-\cfrac{1}{m}(\sum{i=1}^{m}x_i)^2} $$

[推导]：令式（3.5）等于0： $$ 0 = w\sum_{i=1}^{m}xi^2-\sum{i=1}^{m}(y_i-b)xi $$ $$ w\sum{i=1}^{m}xi^2 = \sum{i=1}^{m}y_ixi-\sum{i=1}^{m}bxi $$ 由于令式（3.6）等于0可得$ b=\cfrac{1}{m}\sum{i=1}^{m}(y_i-wxi) $，又$ \cfrac{1}{m}\sum{i=1}^{m}yi=\bar{y} $，$ \cfrac{1}{m}\sum{i=1}^{m}x_i=\bar{x} $，则$ b=\bar{y}-w\bar{x} $，代入上式可得： $$ \begin{aligned}

w\sum_{i=1}^{m}x_i^2 & = \sum_{i=1}^{m}y_ix_i-\sum_{i=1}^{m}(\bar{y}-w\bar{x})x_i \\
w\sum_{i=1}^{m}x_i^2 & = \sum_{i=1}^{m}y_ix_i-\bar{y}\sum_{i=1}^{m}x_i+w\bar{x}\sum_{i=1}^{m}x_i \\
w(\sum_{i=1}^{m}x_i^2-\bar{x}\sum_{i=1}^{m}x_i) & = \sum_{i=1}^{m}y_ix_i-\bar{y}\sum_{i=1}^{m}x_i \\
w & = \cfrac{\sum_{i=1}^{m}y_ix_i-\bar{y}\sum_{i=1}^{m}x_i}{\sum_{i=1}^{m}x_i^2-\bar{x}\sum_{i=1}^{m}x_i}

\end{aligned} $$ 又$ \bar{y}\sum_{i=1}^{m}xi=\cfrac{1}{m}\sum{i=1}^{m}yi\sum{i=1}^{m}xi=\bar{x}\sum{i=1}^{m}yi $，$ \bar{x}\sum{i=1}^{m}xi=\cfrac{1}{m}\sum{i=1}^{m}xi\sum{i=1}^{m}xi=\cfrac{1}{m}(\sum{i=1}^{m}xi)^2 $，代入上式即可得式（3.7）： $$ w=\cfrac{\sum{i=1}^{m}y_i(xi-\bar{x})}{\sum{i=1}^{m}xi^2-\cfrac{1}{m}(\sum{i=1}^{m}x_i)^2} $$

【注】：式（3.7）还可以进一步化简为能用向量表达的形式，将$ \cfrac{1}{m}(\sum_{i=1}^{m}xi)^2=\bar{x}\sum{i=1}^{m}x_i $代入分母可得： $$ \begin{aligned}

 w & = \cfrac{\sum_{i=1}^{m}y_i(x_i-\bar{x})}{\sum_{i=1}^{m}x_i^2-\bar{x}\sum_{i=1}^{m}x_i} \\
 & = \cfrac{\sum_{i=1}^{m}(y_ix_i-y_i\bar{x})}{\sum_{i=1}^{m}(x_i^2-x_i\bar{x})}

\end{aligned} $$ 又$ \bar{y}\sum_{i=1}^{m}xi=\bar{x}\sum{i=1}^{m}yi=\sum{i=1}^{m}\bar{y}xi=\sum{i=1}^{m}\bar{x}yi=m\bar{x}\bar{y}=\sum{i=1}^{m}\bar{x}\bar{y} $，则上式可化为： $$ \begin{aligned}

w & = \cfrac{\sum_{i=1}^{m}(y_ix_i-y_i\bar{x}-x_i\bar{y}+\bar{x}\bar{y})}{\sum_{i=1}^{m}(x_i^2-x_i\bar{x}-x_i\bar{x}+\bar{x}^2)} \\
& = \cfrac{\sum_{i=1}^{m}(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^{m}(x_i-\bar{x})^2} 

\end{aligned} $$ 若令$ \boldsymbol{x}=(x_1,x_2,...,xm) $，$ \boldsymbol{x}{d} $为去均值后的$ \boldsymbol{x} $，$ \boldsymbol{y}=(y_1,y_2,...,ym) $，$ \boldsymbol{y}{d} $为去均值后的$ \boldsymbol{y} $，其中$ \boldsymbol{x} $、$ \boldsymbol{x}{d} $、$ \boldsymbol{y} $、$ \boldsymbol{y}{d} $均为m行1列的列向量，代入上式可得： $$ w=\cfrac{\boldsymbol{y}{d}^T\boldsymbol{x}{d}}{\boldsymbol{x}d^T\boldsymbol{x}{d}}$$

3.10

$$ \cfrac{\partial E_{\hat{\boldsymbol w}}}{\partial \hat{\boldsymbol w}}=2\mathbf{X}^T(\mathbf{X}\hat{\boldsymbol w}-\mathbf{y}) $$

[推导]：将$ E{\hat{\boldsymbol w}}=(\mathbf{y}-\mathbf{X}\hat{\boldsymbol w})^T(\mathbf{y}-\mathbf{X}\hat{\boldsymbol w}) $展开可得： $$ E{\hat{\boldsymbol w}}= \mathbf{y}^T\mathbf{y}-\mathbf{y}^T\mathbf{X}\hat{\boldsymbol w}-\hat{\boldsymbol w}^T\mathbf{X}^T\mathbf{y}+\hat{\boldsymbol w}^T\mathbf{X}^T\mathbf{X}\hat{\boldsymbol w} $$ 对$ \hat{\boldsymbol w} $求导可得： $$ \cfrac{\partial E{\hat{\boldsymbol w}}}{\partial \hat{\boldsymbol w}}= \cfrac{\partial \mathbf{y}^T\mathbf{y}}{\partial \hat{\boldsymbol w}}-\cfrac{\partial \mathbf{y}^T\mathbf{X}\hat{\boldsymbol w}}{\partial \hat{\boldsymbol w}}-\cfrac{\partial \hat{\boldsymbol w}^T\mathbf{X}^T\mathbf{y}}{\partial \hat{\boldsymbol w}}+\cfrac{\partial \hat{\boldsymbol w}^T\mathbf{X}^T\mathbf{X}\hat{\boldsymbol w}}{\partial \hat{\boldsymbol w}} $$ 由向量的求导公式可得： $$ \cfrac{\partial E{\hat{\boldsymbol w}}}{\partial \hat{\boldsymbol w}}= 0-\mathbf{X}^T\mathbf{y}-\mathbf{X}^T\mathbf{y}+(\mathbf{X}^T\mathbf{X}+\mathbf{X}^T\mathbf{X})\hat{\boldsymbol w} $$ $$ \cfrac{\partial E_{\hat{\boldsymbol w}}}{\partial \hat{\boldsymbol w}}=2\mathbf{X}^T(\mathbf{X}\hat{\boldsymbol w}-\mathbf{y}) $$

3.27

$$ l(\beta)=\sum_{i=1}^{m}(-y_i\beta^T\hat{\boldsymbol x}_i+\ln(1+e^{\beta^T\hat{\boldsymbol x}_i})) $$

[推导]：将式（3.26）代入式（3.25）可得： $$ l(\beta)=\sum_{i=1}^{m}\ln\left(y_ip_1(\hat{\boldsymbol x}_i;\beta)+(1-y_i)p_0(\hat{\boldsymbol x}_i;\beta)\right) $$ 其中$ p_1(\hat{\boldsymbol x}_i;\beta)=\cfrac{e^{\beta^T\hat{\boldsymbol x}_i}}{1+e^{\beta^T\hat{\boldsymbol x}_i}},p_0(\hat{\boldsymbol x}_i;\beta)=\cfrac{1}{1+e^{\beta^T\hat{\boldsymbol x}i}} $，代入上式可得： $$\begin{aligned} l(\beta)&=\sum{i=1}^{m}\ln\left(\cfrac{y_ie^{\beta^T\hat{\boldsymbol x}_i}+1-y_i}{1+e^{\beta^T\hat{\boldsymbol x}i}}\right) \ &=\sum{i=1}^{m}\left(\ln(y_ie^{\beta^T\hat{\boldsymbol x}_i}+1-y_i)-\ln(1+e^{\beta^T\hat{\boldsymbol x}_i})\right) \end{aligned}$$ 由于$ yi $=0或1，则： $$ l(\beta) = \begin{cases} \sum{i=1}^{m}(-\ln(1+e^{\beta^T\hat{\boldsymbol x}_i})), & yi=0 \ \sum{i=1}^{m}(\beta^T\hat{\boldsymbol x}_i-\ln(1+e^{\beta^T\hat{\boldsymbol x}_i})), & yi=1 \end{cases} $$ 两式综合可得： $$ l(\beta)=\sum{i=1}^{m}\left(y_i\beta^T\hat{\boldsymbol x}_i-\ln(1+e^{\beta^T\hat{\boldsymbol x}_i})\right) $$ 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(\hat{\boldsymbol x}_i;\beta)]^{y_i}[p_0(\hat{\boldsymbol x}_i;\beta)]^{1-yi} $，再将其代入式（3.25）可得： $$ l(\beta)=\sum{i=1}^{m}\left(y_i\ln(p_1(\hat{\boldsymbol x}_i;\beta))+(1-y_i)\ln(p_0(\hat{\boldsymbol x}_i;\beta))\right) $$ 此式显然更易推导出式（3.27）

3.30

$$\frac{\partial l(\beta)}{\partial \beta}=-\sum_{i=1}^{m}\hat{\boldsymbol x}_i(y_i-p_1(\hat{\boldsymbol x}_i;\beta))$$

[解析]：此式可以进行向量化，令$p_1(\hat{\boldsymbol x}_i;\beta)=\hat{y}_i$，代入上式得： $$\begin{aligned}

\frac{\partial l(\beta)}{\partial \beta} &= -\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;\beta)-\boldsymbol{y}) \\

\end{aligned}$$

3.32

$$J=\cfrac{\boldsymbol w^T(\mu_0-\mu_1)(\mu_0-\mu_1)^T\boldsymbol w}{\boldsymbol w^T(\Sigma_0+\Sigma_1)\boldsymbol w}$$

[推导]： $$\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

$$\boldsymbol S_b\boldsymbol w=\lambda\boldsymbol S_w\boldsymbol w$$

[推导]：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。

3.43

$$\begin{aligned} \boldsymbol S_b &= \boldsymbol S_t - \boldsymbol Sw \ &= \sum{i=1}^N m_i(\boldsymbol\mu_i-\boldsymbol\mu)(\boldsymbol\mu_i-\boldsymbol\mu)^T \end{aligned}$$ [推导]：由式3.40、3.41、3.42可得： $$\begin{aligned} \boldsymbol S_b &= \boldsymbol S_t - \boldsymbol Sw \ &= \sum{i=1}^m(\boldsymbol x_i-\boldsymbol\mu)(\boldsymbol xi-\boldsymbol\mu)^T-\sum{i=1}^N\sum_{\boldsymbol x\in X_i}(\boldsymbol x-\boldsymbol\mu_i)(\boldsymbol x-\boldsymbol\mui)^T \ &= \sum{i=1}^N\left(\sum_{\boldsymbol x\in X_i}\left((\boldsymbol x-\boldsymbol\mu)(\boldsymbol x-\boldsymbol\mu)^T-(\boldsymbol x-\boldsymbol\mu_i)(\boldsymbol x-\boldsymbol\mui)^T\right)\right) \ &= \sum{i=1}^N\left(\sum_{\boldsymbol x\in X_i}\left((\boldsymbol x-\boldsymbol\mu)(\boldsymbol x^T-\boldsymbol\mu^T)-(\boldsymbol x-\boldsymbol\mu_i)(\boldsymbol x^T-\boldsymbol\mui^T)\right)\right) \ &= \sum{i=1}^N\left(\sum_{\boldsymbol x\in X_i}\left(\boldsymbol x\boldsymbol x^T - \boldsymbol x\boldsymbol\mu^T-\boldsymbol\mu\boldsymbol x^T+\boldsymbol\mu\boldsymbol\mu^T-\boldsymbol x\boldsymbol x^T+\boldsymbol x\boldsymbol\mu_i^T+\boldsymbol\mu_i\boldsymbol x^T-\boldsymbol\mu_i\boldsymbol\mui^T\right)\right) \ &= \sum{i=1}^N\left(\sum_{\boldsymbol x\in X_i}\left(- \boldsymbol x\boldsymbol\mu^T-\boldsymbol\mu\boldsymbol x^T+\boldsymbol\mu\boldsymbol\mu^T+\boldsymbol x\boldsymbol\mu_i^T+\boldsymbol\mu_i\boldsymbol x^T-\boldsymbol\mu_i\boldsymbol\mui^T\right)\right) \ &= \sum{i=1}^N\left(-\sum_{\boldsymbol x\in Xi}\boldsymbol x\boldsymbol\mu^T-\sum{\boldsymbol x\in Xi}\boldsymbol\mu\boldsymbol x^T+\sum{\boldsymbol x\in Xi}\boldsymbol\mu\boldsymbol\mu^T+\sum{\boldsymbol x\in X_i}\boldsymbol x\boldsymbol\mui^T+\sum{\boldsymbol x\in X_i}\boldsymbol\mui\boldsymbol x^T-\sum{\boldsymbol x\in X_i}\boldsymbol\mu_i\boldsymbol\mui^T\right) \ &= \sum{i=1}^N\left(-m_i\boldsymbol\mu_i\boldsymbol\mu^T-m_i\boldsymbol\mu\boldsymbol\mu_i^T+m_i\boldsymbol\mu\boldsymbol\mu^T+m_i\boldsymbol\mu_i\boldsymbol\mu_i^T+m_i\boldsymbol\mu_i\boldsymbol\mu_i^T-m_i\boldsymbol\mu_i\boldsymbol\mui^T\right) \ &= \sum{i=1}^N\left(-m_i\boldsymbol\mu_i\boldsymbol\mu^T-m_i\boldsymbol\mu\boldsymbol\mu_i^T+m_i\boldsymbol\mu\boldsymbol\mu^T+m_i\boldsymbol\mu_i\boldsymbol\mui^T\right) \ &= \sum{i=1}^Nm_i\left(-\boldsymbol\mu_i\boldsymbol\mu^T-\boldsymbol\mu\boldsymbol\mu_i^T+\boldsymbol\mu\boldsymbol\mu^T+\boldsymbol\mu_i\boldsymbol\mui^T\right) \ &= \sum{i=1}^N m_i(\boldsymbol\mu_i-\boldsymbol\mu)(\boldsymbol\mu_i-\boldsymbol\mu)^T \end{aligned}$$

3.44

$$\max\limits_{\mathbf{W}}\cfrac{ tr(\mathbf{W}^T\boldsymbol S_b \mathbf{W})}{tr(\mathbf{W}^T\boldsymbol S_w \mathbf{W})}$$ [解析]：此式是式3.35的推广形式，证明如下： 设$\mathbf{W}=[\boldsymbol w_1,\boldsymbol w_2,...,\boldsymbol wi,...,\boldsymbol w{N-1}]$，其中$\boldsymbol w_i$为$d$行1列的列向量，则： $$\left{ \begin{aligned} tr(\mathbf{W}^T\boldsymbol Sb \mathbf{W})&=\sum{i=1}^{N-1}\boldsymbol w_i^T\boldsymbol S_b \boldsymbol w_i \ tr(\mathbf{W}^T\boldsymbol Sw \mathbf{W})&=\sum{i=1}^{N-1}\boldsymbol w_i^T\boldsymbol S_w \boldsymbol wi \end{aligned} \right.$$ 所以式3.44可变形为： $$\max\limits{\mathbf{W}}\cfrac{ \sum_{i=1}^{N-1}\boldsymbol w_i^T\boldsymbol S_b \boldsymbol wi}{\sum{i=1}^{N-1}\boldsymbol w_i^T\boldsymbol S_w \boldsymbol w_i}$$ 对比式3.35易知上式即为式3.35的推广形式。