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fix 8.6
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archwalker committed May 28, 2020
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Expand Up @@ -84,12 +84,12 @@ $$
$$

[解析]:由公式(8.5)中对于符号$\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}[\cdot]$的解释可知

$$
\begin{aligned}
\ell_{\exp }(H | \mathcal{D}) &=\mathbb{E}_{\boldsymbol{x} \sim \mathcal{D}}\left[e^{-f(\boldsymbol{x}) H(\boldsymbol{x})}\right] \\
&=\sum_{\boldsymbol{x} \in D} \mathcal{D}(\boldsymbol{x}) e^{-f(\boldsymbol{x}) H(\boldsymbol{x})} \\
&=\sum_{i=1}^{|D|} \mathcal{D}\left(\boldsymbol{x}_{i}\right)\left(e^{-H\left(\boldsymbol{x}_{i}\right)} P\left(f\left(\boldsymbol{x}_{i}\right)=1 | \boldsymbol{x}_{i}\right)+e^{H\left(\boldsymbol{x}_{i}\right)} P\left(f\left(\boldsymbol{x}_{i}\right)=-1 | \boldsymbol{x}_{i}\right)\right)
&=\sum_{i=1}^{|D|} \mathcal{D}\left(\boldsymbol{x}_{i}\right)\left(e^{-H\left(\boldsymbol{x}_{i}\right)} \mathbb{I}\left(f\left(\boldsymbol{x}_{i}\right)=1\right)+e^{H\left(\boldsymbol{x}_{i}\right)} \mathbb{I}\left(f\left(\boldsymbol{x}_{i}\right)=-1\right)\right)\\
&=e^{-H\left(\boldsymbol{x}_{i}\right)} P\left(f\left(\boldsymbol{x}_{i}\right)=1 | \boldsymbol{x}_{i}\right)+e^{H\left(\boldsymbol{x}_{i}\right)} P\left(f\left(\boldsymbol{x}_{i}\right)=-1 | \boldsymbol{x}_{i}\right)
\end{aligned}
$$

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