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Merge pull request #43 from Ye980226/master

8.16
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1 gewijzigde bestanden met toevoegingen van 49 en 0 verwijderingen
  1. 49 0
      docs/chapter8/chapter8.md

+ 49 - 0
docs/chapter8/chapter8.md

@@ -61,3 +61,52 @@ $$
 \\ & =\underset{y \in\{-1,1\}}{\arg \max } P(f(x)=y | \boldsymbol{x})
 \\ & =\underset{y \in\{-1,1\}}{\arg \max } P(f(x)=y | \boldsymbol{x})
 \end{aligned}
 \end{aligned}
 $$
 $$
+
+## 8.16
+由式(8.13)可知
+$$
+\begin{aligned} \ell_{\exp }\left(H_{t-1}+h_{t} | \mathcal{D}\right) & \simeq \mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\left(1-f(\boldsymbol{x}) h_{t}(\boldsymbol{x})+\frac{f^{2}(\boldsymbol{x}) h_{t}^{2}(\boldsymbol{x})}{2}\right)\right] \\ &=\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\left(1-f(\boldsymbol{x}) h_{t}(\boldsymbol{x})+\frac{1}{2}\right)\right] \quad \text { . } \end{aligned}
+$$
+
+求得的$h_t$使得损失函数最小,所以得出(8.14)
+$$
+\begin{aligned}
+h_{t}(\boldsymbol{x})&=\underset{h}{\arg \min } \ell_{\exp }\left(H_{t-1}+h | \mathcal{D}\right)
+\\&
+\begin{array}{l}{ =\arg \min _{h} \mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\left(1-f(\boldsymbol{x}) h(\boldsymbol{x})+\frac{1}{2}\right)\right]} \\ {=\arg \max _{h} \mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})} f(\boldsymbol{x}) h(\boldsymbol{x})\right]} \\ {=\underset{\boldsymbol{h}}{\arg \max } \mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[\frac{e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}}{\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\right.}} f(\boldsymbol{x}) h(\boldsymbol{x})\right]}\end{array} \end{aligned}
+$$
+
+最后一行式子里$\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\right]$因为是基于学习器$H_{t-1}$的指数损失函数的期望,所以是大于0的常数,所以这里不影响损失函数最小时,参数h的值。
+
+式(8.15)
+$$
+\mathcal{D}_{t}(\boldsymbol{x})=\frac{\mathcal{D}(\boldsymbol{x}) e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}}{\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\right]}
+$$
+
+这里的$\mathcal{D}$是x的概率分布
+
+## 8.16
+$$
+\begin{aligned} h_{t}(\boldsymbol{x}) &=\underset{h}{\arg \max } \mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[\frac{e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}}{\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\right]} f(\boldsymbol{x}) h(\boldsymbol{x})\right] \\ &=\underset{\boldsymbol{h}}{\arg \max } \mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}_{t}}[f(\boldsymbol{x}) h(\boldsymbol{x})] \end{aligned}
+$$
+
+假设x的概率分布是f(x)
+(注:本书中概率分布全都是$\mathcal{D(x)}$)
+
+$$
+\mathbb{E(g(x))}=\sum_{i=1}^{|D|}f(x)g(x)
+$$
+故可得
+
+$$
+\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H(\boldsymbol{x})}\right]=\sum_{i=1}^{|D|} \mathcal{D}\left(\boldsymbol{x}_{i}\right) e^{-f\left(\boldsymbol{x}_{i}\right) H\left(\boldsymbol{x}_{i}\right)}
+$$
+由式(8.15)可知
+$$
+\mathcal{D}_{t}\left(\boldsymbol{x}_{i}\right)=\mathcal{D}\left(\boldsymbol{x}_{i}\right) \frac{e^{-f\left(\boldsymbol{x}_{i}\right) H_{t-1}\left(\boldsymbol{x}_{i}\right)}}{\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\right]}
+$$
+
+所以式(8.16)可以表示为
+$$
+\begin{aligned} & \mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[\frac{e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}}{\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x})}\right]} f(\boldsymbol{x}) h(\boldsymbol{x})\right] \\=& \sum_{i=1}^{|D|} \mathcal{D}\left(\boldsymbol{x}_{i}\right) \frac{e^{-f\left(\boldsymbol{x}_{i}\right) H_{t-1}\left(\boldsymbol{x}_{i}\right)}}{\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H_{t-1}(\boldsymbol{x}) }]  \right.}f(x_i)h(x_i) \\=& \sum_{i=1}^{|D|} \mathcal{D}_{t}\left(\boldsymbol{x}_{i}\right) f\left(\boldsymbol{x}_{i}\right) h\left(\boldsymbol{x}_{i}\right) \\=& \mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}_{t}}[f(\boldsymbol{x}) h(\boldsymbol{x})] \end{aligned}
+$$