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@@ -1,48 +1,64 @@
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-# 5.1 神经元模型
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-# 5.2 感知机与多层网络
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-# 5.3 误差逆传播算法
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-对于训练例$(x_k,y_k)$, 假定神经网络输出为 $\hat{y}_k = (\hat{y}_1^k , \hat{y}_2^k,...,\hat{y}_l^k)$
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-$$ \hat{y}_j^k = f(\beta_j - \theta_j) $$ (5.3)
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-设网络在 $(x_k,y_k)$的均方误差为
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- $$ E_k = \frac{1}{2} \sum^l_{j=1}(\hat{y}_j^k -y_j^k)^2$$ (5.4)
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-
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-> 这里 1/2是为了求导方便消去系数,只是个规定
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-
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-设 学习率为 $\eta$, 以目标的负梯度方向对参数进行调整的含义是:
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-目标函数(均方误差 $E_k$)在 $w_{hj}$处的负梯度对参数进行调整:
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-$$ \Delta w_{hj} = - \eta\frac{\partial E_k}{\partial w_{hj}}$$ (5.6)
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-
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-根据链式法则:
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-$$ \frac{\partial E_k}{\partial w_{hj}} = \frac{\partial E_k}{\partial \hat{y}_j^k}\frac{\partial \hat{y}_j^k}{\partial \beta_j}\frac{\partial \beta_j}{\partial w_{hj}}$$ (5.7)
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-
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-根据 $\beta_j$ 的定义,
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-$$ \beta_j = \sum^q_{h=1} w_{hj}b_h$$
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-
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-有
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-$$\frac{\partial \beta_j}{\partial w_{hj}} = b_h$$ (5.8)
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-图 5.2 的Sigmoid 函数为
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-$$ sigmoid(x) = \frac{1}{1+e^{-x}}$$
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+### 5.2
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+$$\Delta w_i = \eta(y-\hat{y})x_i$$
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+[推导]:此处感知机的模型为:
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+$$y=f(\sum_{i} w_i x_i - \theta)$$
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+将$\theta$看成哑结点后,模型可化简为:
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+$$y=f(\sum_{i} w_i x_i)=f(\boldsymbol w^T \boldsymbol x)$$
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+其中$f$为阶跃函数。<br>根据《统计学习方法》§2可知,假设误分类点集合为$M$,$\boldsymbol x_i \in M$为误分类点,$\boldsymbol x_i$的真实标签为$y_i$,模型的预测值为$\hat{y_i}$,对于误分类点$\boldsymbol x_i$来说,此时$\boldsymbol w^T \boldsymbol x_i \gt 0,\hat{y_i}=1,y_i=0$或$\boldsymbol w^T \boldsymbol x_i \lt 0,\hat{y_i}=0,y_i=1$,综合考虑两种情形可得:
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+$$(\hat{y_i}-y_i)\boldsymbol w \boldsymbol x_i>0$$
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+所以可以推得损失函数为:
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+$$L(\boldsymbol w)=\sum_{\boldsymbol x_i \in M} (\hat{y_i}-y_i)\boldsymbol w \boldsymbol x_i$$
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+损失函数的梯度为:
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+$$\nabla_w L(\boldsymbol w)=\sum_{\boldsymbol x_i \in M} (\hat{y_i}-y_i)\boldsymbol x_i$$
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+随机选取一个误分类点$(\boldsymbol x_i,y_i)$,对$\boldsymbol w$进行更新:
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+$$\boldsymbol w \leftarrow \boldsymbol w-\eta(\hat{y_i}-y_i)\boldsymbol x_i=\boldsymbol w+\eta(y_i-\hat{y_i})\boldsymbol x_i$$
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+显然式5.2为$\boldsymbol w$的第$i$个分量$w_i$的变化情况
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+### 5.12
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+$$\Delta \theta_j = -\eta g_j$$
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+[推导]:因为
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+$$\Delta \theta_j = -\eta \cfrac{\partial E_k}{\partial \theta_j}$$
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+又
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+$$
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+\begin{aligned}
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+\cfrac{\partial E_k}{\partial \theta_j} &= \cfrac{\partial E_k}{\partial \hat{y_j^k}} \cdot\cfrac{\partial \hat{y_j^k}}{\partial \theta_j} \\\\
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+&= (\hat{y_j^k}-y_j^k) \cdot f’(\beta_j-\theta_j) \cdot (-1) \\\\
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+&= -(\hat{y_j^k}-y_j^k)f’(\beta_j-\theta_j) \\\\
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+&= g_j
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+\end{aligned}
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+$$
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所以
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-$$ f' = \frac{e^{-x}}{(1+e^{-x})^2} = f(x)(1-f(x))$$ (5.9)
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-根据 5.3和 5.4
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-$$ - \frac{\partial E_k}{\partial \hat{y}_j^k}\frac{\partial \hat{y}_j^k}{\partial \beta_j} = - (\hat{y}_j^k - y_j^k)f'(\beta_j - \theta_j) = - (\hat{y}_j^k - y_j^k)[\hat{y}_j^k(1-\hat{y}_j^k)] = \hat{y}_j^k(1-\hat{y}_j^k)(y_j^k - \hat{y}_j^k) = g_j$$
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-(5.10)
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->$g_j$ 是设出来的变量
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-
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-将 (5.10)和(5.8)带入(5.7)再带入5.6,可以得到BP算法中关于$w_{hj}$的更新公式
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-$$ \Delta w_{hj} = \eta g_j b_h$$ (5.11)
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-
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-同理有:
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-
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-$$\Delta\theta_j = -\eta g_j $$ (5.12)
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-
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-> 注意这里是把 $\theta_j$ 看做为输入为 -1,系数或者说权重为 $\theta$ 的w, 和$\Delta w_{hj}$ 具有等价的地位, 取 $b_h = -1$ 带入(5.11)得到的
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-$$\Delta v_{ih} = -\eta \frac{\partial E_k}{\partial v_{ih}} $$
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-$$\frac{\partial E_k}{\partial v_{ih}} = \frac{E_k}{\partial{b_h}}\frac{\partial b_h}{\partial \alpha_h}\frac{\partial \alpha_h}{\partial x_i} = e_hx_i$$
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-
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-所以
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-$$\Delta v_{ih} = -\eta e_h x_i $$ (5.13)
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-这里的
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- $$e_h = \frac{E_k}{\partial{b_h}}\frac{\partial b_h}{\partial \alpha_h} = -\sum_{j=1}^l \frac{\partial E_k}{\partial \beta_j}\frac{\partial \beta_j}{\partial b_h}f'(\alpha - \gamma_h) = \sum_{j=1}^lw_{hj}g_jf'(\alpha_h - \gamma_h) = b_h(1-b_h)\sum_{j=1}^lw_{hj}g_j$$
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-同(5.12)理得到
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-$$\Delta \gamma_h = -\eta e_h$$ (5.14)
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+$$\Delta \theta_j = -\eta \cfrac{\partial E_k}{\partial \theta_j}=-\eta g_j$$
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+### 5.13
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+$$\Delta v_{ih} = \eta e_h x_i$$
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+[推导]:因为
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+$$\Delta v_{ih} = -\eta \cfrac{\partial E_k}{\partial v_{ih}}$$
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+又
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+$$
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+\begin{aligned}
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+\cfrac{\partial E_k}{\partial v_{ih}} &= \sum_{j=1}^{l} \cfrac{\partial E_k}{\partial \hat{y_j^k}} \cdot \cfrac{\partial \hat{y_j^k}}{\partial \beta_j} \cdot \cfrac{\partial \beta_j}{\partial b_h} \cdot \cfrac{\partial b_h}{\partial \alpha_h} \cdot \cfrac{\partial \alpha_h}{\partial v_{ih}} \\\\
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+&= \sum_{j=1}^{l} \cfrac{\partial E_k}{\partial \hat{y_j^k}} \cdot \cfrac{\partial \hat{y_j^k}}{\partial \beta_j} \cdot \cfrac{\partial \beta_j}{\partial b_h} \cdot \cfrac{\partial b_h}{\partial \alpha_h} \cdot x_i \\\\
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+&= \sum_{j=1}^{l} \cfrac{\partial E_k}{\partial \hat{y_j^k}} \cdot \cfrac{\partial \hat{y_j^k}}{\partial \beta_j} \cdot \cfrac{\partial \beta_j}{\partial b_h} \cdot f’(\alpha_h-\gamma_h) \cdot x_i \\\\
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+&= \sum_{j=1}^{l} \cfrac{\partial E_k}{\partial \hat{y_j^k}} \cdot \cfrac{\partial \hat{y_j^k}}{\partial \beta_j} \cdot w_{hj} \cdot f’(\alpha_h-\gamma_h) \cdot x_i \\\\
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+&= \sum_{j=1}^{l} (-g_j) \cdot w_{hj} \cdot f’(\alpha_h-\gamma_h) \cdot x_i \\\\
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+&= -f’(\alpha_h-\gamma_h) \cdot \sum_{j=1}^{l} g_j \cdot w_{hj} \cdot x_i\\\\
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+&= -b_h(1-b_h) \cdot \sum_{j=1}^{l} g_j \cdot w_{hj} \cdot x_i \\\\
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+&= -e_h \cdot x_i
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+\end{aligned}
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+$$
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+所以
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+$$\Delta v_{ih} = -\eta \cdot -e_h \cdot x_i=\eta e_h x_i$$
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+### 5.14
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+$$\Delta \gamma_h= -\eta e_h$$
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+[推导]:因为
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+$$\Delta \gamma_h = -\eta \cfrac{\partial E_k}{\partial \gamma_h}$$
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+又
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+$$
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+\begin{aligned}
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+\cfrac{\partial E_k}{\partial \gamma_h} &= \sum_{j=1}^{l} \cfrac{\partial E_k}{\partial \hat{y_j^k}} \cdot \cfrac{\partial \hat{y_j^k}}{\partial \beta_j} \cdot \cfrac{\partial \beta_j}{\partial b_h} \cdot \cfrac{\partial b_h}{\partial \gamma_h} \\\\
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+&= \sum_{j=1}^{l} \cfrac{\partial E_k}{\partial \hat{y_j^k}} \cdot \cfrac{\partial \hat{y_j^k}}{\partial \beta_j} \cdot \cfrac{\partial \beta_j}{\partial b_h} \cdot f’(\alpha_h-\gamma_h) \cdot (-1) \\\\
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+&= -\sum_{j=1}^{l} \cfrac{\partial E_k}{\partial \hat{y_j^k}} \cdot \cfrac{\partial \hat{y_j^k}}{\partial \beta_j} \cdot w_{hj} \cdot f’(\alpha_h-\gamma_h)\\\\
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+&=e_h
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+\end{aligned}
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+$$
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+所以
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+$$\Delta \gamma_h= -\eta e_h$$
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