学习理论：单阶段代理损失的(H, R) - 一致界证明

1 导引

我们在上一篇博客《学习理论：预测器-拒绝器多分类弃权学习》中介绍了弃权学习的基本概念和方法，其中包括了下列针对多分类问题的单阶段预测器-拒绝器弃权损失\(L_{\text{abst}}\)：

\[L_{\text{abst}}(h, r, x, y) = \underbrace{\mathbb{I}_{\text{h}(x) \neq y}\mathbb{I}_{r(x) > 0}}_{\text{不弃权}} + \underbrace{c(x) \mathbb{I}_{r(x)\leqslant 0}}_{\text{弃权}}
\]

其中\((x, y)\in \mathcal{X}\times \mathcal{Y}\)（标签\(\mathcal{Y} = \{1, \cdots, n\}\)（\(n\geqslant 2\)）），\((h, r)\in \mathcal{H}\times\mathcal{R}\)为预测器-拒绝器对（\(\mathcal{H}\)和\(\mathcal{R}\)为两个从\(\mathcal{X}\)到\(\mathbb{R}\)的函数构成的函数族），\(\text{h}(x) = \text{arg max}_{y\in \mathcal{Y}} {h(x)}_y\)直接输出实例\(x\)的预测标签。为了简化讨论，在后文中我们假设\(c\in (0, 1)\)为一个常量花费函数。

设\(\mathcal{l}\)为在标签\(\mathcal{Y}\)上定义的0-1多分类损失的代理损失，则我们可以在此基础上进一步定义弃权代理损失\(L\)：

\[L(h, r, x, y) = \mathcal{l}(h, x, y)\phi(-\alpha r(x)) + \psi(c) \phi(\beta r(x))
\]

其中\(\psi\)是非递减函数，\(\phi\)是非递增辅助函数（做为\(z \mapsto \mathbb{I}_{z \leqslant 0}\)的上界），\(\alpha\)、\(\beta\)为正常量。下面，为了简便起见，我们主要对\(\phi(z) = \exp(-z)\)进行分析，尽管相似的分析也可以应用于其它函数\(\phi\)。

在上一篇博客中，我们还提到了单阶段代理损失满足的\((\mathcal{H}, \mathcal{R})\)-一致性界：

定理 1 单阶段代理损失的\((\mathcal{H}, \mathcal{R})\) - 一致性界 假设\(\mathcal{H}\)是对称与完备的。则对\(\alpha=\beta\)，\(\mathcal{l} = \mathcal{l}_{\text{mae}}\)，或者\(\mathcal{l} = \mathcal{l}_{\rho}\)与\(\psi(z) = z\)，或者\(\mathcal{l} = \mathcal{l}_{\rho - \text{hinge}}\)与\(\psi(z) = z\)，有下列\((\mathcal{H}, \mathcal{R})\) - 一致性界对\(h\in \mathcal{H}, r\in \mathcal{R}\)和任意分布成立：

\[R_{L_{\text{abst}}}(h, r) - R_{L_{\text{abst}}}^{*}(\mathcal{H}, \mathcal{R}) + M_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R}) \leqslant \Gamma(R_L(h, r) - R_{L}^{*}(\mathcal{H}, \mathcal{R}) + M_{L}(\mathcal{H}, \mathcal{R}))
\]

其中对\(\mathcal{l} = \mathcal{l}_{\text{mae}}\)取\(\Gamma (z) = \max\{2n\sqrt{z}, nz\}\)；对\(\mathcal{l}=\mathcal{l}_{\rho}\)取\(\Gamma (z) = \max\{2\sqrt{z}, z\}\)；对\(\mathcal{l} = \mathcal{l}_{\rho - \text{hinge}}\)取\(\Gamma (z) = \max\{2\sqrt{nz}, z\}\)。

不过，在上一篇博客中，我们并没有展示单阶段代理损失的\((\mathcal{H}, \mathcal{R})\)-一致性界的详细证明过程，在这片文章里我们来看该如何对该定理进行证明（正好我导师也让我仔细看看这几篇论文中相关的分析部分，并希望我掌握单阶段方法的证明技术）。

2 一些分析的预备概念

我们假设带标签样本\(S=((x_1, y_1), \cdots, (x_m, y_m))\)独立同分布地采自\(p(x, y)\)。则对于目标损失\(L_{\text{abst}}\)和代理损失\(L\)而言，可分别定义\(L_{\text{abst}}\)-期望弃权损失\(R_{L}(h, r)\)（也即目标损失函数的泛化误差）和\(L\)-期望弃权代理损失\(R_{L}(h, r)\)（也即代理损失函数的泛化误差）如下：

\[R_{L_{\text{abst}}}(h, r) = \mathbb{E}_{p(x, y)}\left[L_{\text{abst}}(h, r, x, y)\right], \quad R_{L}(h, r) = \mathbb{E}_{p(x, y)}\left[L(h, r, x, y)\right]
\]

设\(R_{{L}^{*}_{\text{abst}}}(\mathcal{H}, \mathcal{R}) = \inf_{h\in \mathcal{H}, r\in \mathcal{R}}R_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R})\)和\(R_{L}^{*}(\mathcal{H}, \mathcal{R}) = \inf_{h\in \mathcal{H}, r\in \mathcal{R}}R_{L}(\mathcal{H}, \mathcal{R})\)分别为\(R_{L_{\text{abst}}}\)和\(R_L\)在\(\mathcal{H}\times \mathcal{R}\)上的下确界。

为了进一步简化后续的分析，我们根据概率的乘法规则将\(R_L(h, r)\)写为：

\[R_{L}(h, r) = \mathbb{E}_{p(x, y)}\left[L(h, r, x, y)\right] = \mathbb{E}_{p(x)}\underbrace{\left[\mathbb{E}_{p(y\mid x)}\left[L(h, r, x, y)\right]\right]}_{\text{conditional risk }C_L}
\]

我们称其中内层的条件期望项为代理损失\(L\)的条件风险（conditional risk）（也称为代理损失\(L\)的pointwise风险），由于在其计算过程中\(y\)取期望取掉了，因此该项只和\(h\)、\(r\)、\(x\)相关，因此我们将其记为\(C_L(h, r, x)\)：

\[C_L(h, r, x) = \mathbb{E}_{p(y\mid x)}\left[L(h, r, x, y)\right] = \sum_{y\in \mathcal{Y}}p(y\mid x)L(h, r, x, y)
\]

我们用\(C^*_L(\mathcal{H}, \mathcal{R}, x) = \inf_{h\in \mathcal{H}, r\in \mathcal{R}} C_L(h, r, x)\)来表示假设类最优（best-in-class） 的\(L\)的条件风险。同理，我们用\(C_{L_{\text{abst}}}\)来表示目标损失\(L_{\text{abst}}\)的条件风险，并用\(C^*_{L_{\text{abst}}}\)来表示假设类最优的\(L_{\text{abst}}\)的条件风险。

根据\(R_{L}^*(h, r)\)和\(C^*_L(\mathcal{H}, \mathcal{R}, x)\)，我们可以表示出最小化能力差距（minimizability gap）：

\[M_L(\mathcal{H}, \mathcal{R}) = R_{L}^*(\mathcal{H}, \mathcal{R}) - \mathbb{E}_{p(x)}\left[C_L^*(\mathcal{H}, \mathcal{R}, x)\right]
\]

\(M_{L_{\text{abst}}}\)的表示同理。

于是，我们可以对要证明的\((\mathcal{H}, \mathcal{R})\)-一致性界进行改写：

\[R_{L_{\text{abst}}}(h, r) - R_{L_{\text{abst}}}^{*}(\mathcal{H}, \mathcal{R}) + M_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R}) \leqslant \Gamma(R_L(h, r) - R_{L}^{*}(\mathcal{H}, \mathcal{R}) + M_{L}(\mathcal{H}, \mathcal{R}))\\
\Rightarrow R_{L_{\text{abst}}}(h, r) - \mathbb{E}_{p(x)}\left[C_{L_{\text{abst}}}^*(\mathcal{H}, \mathcal{R}, x)\right] \leqslant \Gamma\left(R_{L}(h, r) - \mathbb{E}_{p(x)}\left[C_{L}^*(\mathcal{H}, \mathcal{R}, x)\right]\right)
\]

其中\(R_{L_{\text{abst}}}(h, r)\)和\(R_L(h, r)\)分别为\(\mathbb{E}_{p(x)}C_{L_{\text{abst}}}(h, r, x)\)和\(\mathbb{E}_{p(x)}C_{L}(h, r, x)\)，于是上述不等式即为

\[\mathbb{E}_{p(x)}\underbrace{\left[C_{L_{\text{abst}}}(h, r, x) - C_{L_{\text{abst}}}^*(\mathcal{H}, \mathcal{R}, x)\right]}_{\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)} \leqslant \Gamma\left(\mathbb{E}_{p(x)}\underbrace{\left[C_{L}(h, r, x) - C_{L}^*(\mathcal{H}, \mathcal{R}, x)\right]}_{\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)}\right)
\]

我们将上述不等式两边的被取期望的项简记为\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)\)和\(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\)，其中\(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\)被称为校准差距（calibration gap）。由于按定义\(\Gamma(\cdot)\)是凹函数，由Jensen不等式有：

\[\mathbb{E}_{p(x)}\left[\Gamma\left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\right] \leqslant \Gamma\left(\mathbb{E}_{p(x)}\left[\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right]\right)
\]

于是，若我们能证明下述不等式，则原不等式得证：

\[\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)
\]

我们后面将会看到，\((\mathcal{H}, \mathcal{R})\)-一致性界的证明过程中重要的一步即是证明\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)\)能被\(\Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)界定。

3 \(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)\)的表示

我们先来看\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) = C_{L_{\text{abst}}}(h, r, x) - C^*_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R}, x)\)如何表示。根据定义，我们有：

\[\begin{aligned}
C_{L_{\text{abst}}}(h, r, x) &= \sum_{y\in \mathcal{Y}}p(y\mid x)L_{\text{abst}}(h, r, x, y) \\
&= \sum_{y\in \mathcal{Y}}p(y\mid x) \mathbb{I}_{\text{h}(x) \neq y}\mathbb{I}_{r(x) > 0} + c(x) \mathbb{I}_{r(x)\leqslant 0}
\end{aligned}
\]

由于是关于\(y\)的条件期望，上式最后一行中只需要对\(\mathbb{I}_{\text{h}(x) \neq y}\)进行加权求和即可。为了进一步对\(C_{L_{\text{abst}}}(h, r, x)\)进行表示，我们需要对\(r(x)\)的正负情况进行分类讨论：

1. \(r(x) > 0\)：此时\(C_{L_{\text{abst}}}(h, r, x) = \sum_{y\in \mathcal{Y}}p(y\mid x) \mathbb{I}_{\text{h}(x) \neq y} = 1 - p(\text{h}(x)\mid x)\)。
2. \(r(x) \leqslant 0\)：此时\(C_{L_{\text{abst}}}(h, r, x) = c\)。

接下来我们来看\(C^*_{L_{\text{abst}}}\)如何表示。我们假设拒绝函数集\(\mathcal{R}\)是完备的（也即对任意\(x\in \mathcal{X}, \{r(x): r\in \mathcal{R}\} = \mathbb{R}\)），那么\(\mathcal{R}\)也是弃权正规的（也即使得对任意\(x\in \mathcal{X}\)，存在\(r_1, r_2\in \mathcal{R}\)满足\(r_1(x) > 0\)与\(r_2(x) \leqslant 0\)）。于是我们有

\[\begin{aligned}
C^*_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R}, x) &= \inf_{h\in \mathcal{H}, r\in \mathcal{R}}C_{L_{\text{abst}}}(h, r, x)\\
& = \min \left\{\min_{h\in \mathcal{H}}\left(1 - p\left( \text{h}(x)\mid x\right)\right), c\right\}\\
& = 1 - \max\left\{\max_{h\in \mathcal{H}}p\left(\text{h}(x)\mid x\right), 1 - c\right\}
\end{aligned}
\]

我们假设\(\mathcal{H}\)是对称的且完备的（具体定义参见博客《学习理论：预测器-拒绝器多分类弃权学习》），则我们有\(\{\text{h}(x): h\in \mathcal{H}\} = \mathcal{Y}\)，于是

\[C^*_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R}, x) = 1 - \max\left\{\max_{y\in \mathcal{Y}}p\left(y\mid x\right), 1 - c\right\}
\]

为了进一步对\(C^*_{L_{\text{abst}}}(\mathcal{H}, \mathcal{y}, x)\)进行表示，我们需要对\(\max_{y\in \mathcal{Y}}p(y\mid x)\)和\((1 - c)\)的大小比较情况进行分类讨论：

1. \(\max_{y\in \mathcal{Y}}p(y\mid x) > 1 - c\)：此时\(C^*_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R}, x) = 1 - \max_{y\in \mathcal{Y}}p(y\mid x)\)。
2. \(\max_{y\in \mathcal{Y}}p(y\mid x) \leqslant 1 - c\)：此时\(C^*_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R}, x) = c\)。

于是，我们有：

\[\begin{aligned}
\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) &= C_{L_{\text{abst}}}(h, r, x) - C^*_{L_{\text{abst}}}(\mathcal{H}, \mathcal{R}, x) \\
& = \left\{\begin{aligned}
&\max_{y\in \mathcal{Y}}p(y\mid x) - p(\text{h}(x)\mid x)\quad &\text{if } \max_{y\in \mathcal{Y}} p(y\mid x) > (1 - c)，r(x) > 0 \\
&1 - c - p(\text{h}(x)\mid x) \quad &\text{if } \max_{y\in \mathcal{Y}} p(y\mid x) \leqslant (1 - c)，r(x) > 0 \\
&0 \quad &\text{if } \max_{y\in \mathcal{Y}} p(y\mid x) \leqslant (1 - c)，r(x) \leqslant 0 \\
&\max_{y\in \mathcal{Y}}p(y\mid x) - 1 + c \quad &\text{if } \max_{y\in \mathcal{Y}} p(y\mid x) > (1 - c)，r(x) \leqslant 0 \\
\end{aligned}\right.
\end{aligned}
\]

4 \(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\)的表示

4.1 分类讨论的准备

接下来我们来看\(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) = C_L(h, r, x) - C^*_L(\mathcal{H}, \mathcal{R}, x)\)如何表示。根据定义，若\(\alpha = \beta\)，\(\phi(z) = \exp(-z)\)，我们有：

\[\begin{aligned}
C_L(h, r, x) &= \sum_{y\in \mathcal{Y}}p(y\mid x)L(h, r, x, y) \\
&= \sum_{y\in \mathcal{Y}}p(y\mid x) \mathcal{l}(h, x, y)e^{\alpha r(x)} + \psi(c) e^{-\alpha r(x)}
\end{aligned}
\]

由于是关于\(y\)的条件期望，上式最后一行中只需要对\(\mathcal{l}(h, x, y)\)进行加权求和即可。在后文中我们将会针对下列三种不同的\(\mathcal{l}\)函数以及\(\psi(z)\)的选择情况来分别对\(C_L(h, r, x)\)进行讨论：

1. \(\mathcal{l} = \mathcal{l}_{\text{mae}}\)，\(\psi(z) = z\)；
2. \(\mathcal{l} = \mathcal{l}_{\rho}\)，\(\psi(z) = z\)；
3. \(\mathcal{l} = \mathcal{l}_{\rho-\text{hinge}}\)，\(\psi(z) = nz\)。

注 这三种不同\(\mathcal{l}\)的定义参见博客《学习理论：预测器-拒绝器多分类弃权学习》），我在这里把它们的定义贴一下：

• 平均绝对误差损失：\(\mathcal{l}_{\text{mae}}(h, x, y) = 1 - \frac{e^{{h(x)}_y}}{\sum_{y^{\prime}\in \mathcal{Y}}e^{{h(x)}_{y^{\prime}}}}\)；
• 约束\(\rho\)-合页损失：\(\mathcal{l}_{\rho-\text{hinge}}(h, x, y) = \sum_{y^{\prime}\neq y}\phi_{\rho-\text{hinge}}(-{h(x)}_{y^{\prime}}), \rho > 0\)，其中\(\phi_{\rho-\text{hinge}}(z) = \max\{0, 1 - \frac{z}{\rho}\}\)为\(\rho\)-合页损失，且约束条件\(\sum_{y\in \mathcal{Y}}{h(x)}_y=0\)。
• \(\rho\)-间隔损失：\(\mathcal{l}_{\rho}(h, x, y) = \phi_{\rho}({\rho_h (x, y)})\)，其中\(\rho_{h}(x, y) = h(x)_y - \max_{y^{\prime} \neq y}h(x)_{y^{\prime}}\)是置信度间隔，\(\phi_{\rho}(z) = \min\{\max\{0, 1 - \frac{z}{\rho}\}, 1\}, \rho > 0\)为\(\rho\)-间隔损失。

4.2 \(\mathcal{l} = \mathcal{l}_{\text{mae}}\)，\(\psi(z) = z\)

在这种情况下\(C_L(h, r, x)\)可以表示为：

\[\begin{aligned}
C_L(h, r, x) &= \sum_{y\in \mathcal{Y}}p(y\mid x) \underbrace{\left(1 - \frac{e^{{h(x)}_y}}{\sum_{y^{\prime}\in \mathcal{Y}}e^{{h(x)}_{y^{\prime}}}}\right)}_{\mathcal{l}_{\text{mae}}}e^{\alpha r(x)} + c e^{-\alpha r(x)} \\
&= \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)}
\end{aligned}
\]

其中\(s_h(x, y) = \frac{e^{{h(x)}_y}}{\sum_{y^{\prime}\in \mathcal{Y}}e^{{h(x)}_{y^{\prime}}}}\)。

于是

\[\begin{aligned}
C_L^*(\mathcal{H}, \mathcal{R}, x) &= \inf_{h\in \mathcal{H}, r\in\mathcal{R}} \left\{\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)}\right\} \\
&= \inf_{r\in\mathcal{R}} \left\{\inf_{h\in \mathcal{H}}\left\{\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)\right\}e^{\alpha r(x)} + c e^{-\alpha r(x)}\right\}
\end{aligned}
\]

由于假设了\(\mathcal{H}\)是对称的与完备的，我们有

\[\begin{aligned}
&\inf_{h\in \mathcal{H}}\left\{\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y\right))\right\} \\
&= 1 - \sup_{h\in \mathcal{H}}\sum_{y\in \mathcal{Y}}p(y\mid x)s_h(x, y) \\
&= 1 - \max_{y\in \mathcal{Y}}p(y\mid x)\quad \left(s_h(x, y)\in (0, 1)\right)
\end{aligned}
\]

注 实际上，对任意\(h\in \mathcal{H}\)，有：

\[\begin{aligned}
&\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right) - \left(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right)\right) \\
&= \max_{y\in \mathcal{Y}} p(y\mid x) - \sum_{y\in \mathcal{Y}}p(y\mid x)s_h(x, y) \\
&= \max_{y\in \mathcal{Y}} p(y\mid x) - \left(p\left(\text{h}(x)\mid x\right)s_h\left(x, \text{h}(x)\right) + \sum_{y\neq \text{h}(x)}p(y\mid x)s_h(x, y)\right) \\
&\geqslant \max_{y\in \mathcal{Y}} p(y\mid x) - \left(p\left(\text{h}(x)\mid x\right)s_h\left(x, \text{h}(x)\right) + \max_{y\in \mathcal{Y}}p(y\mid x)\left(1 - s_h\left(x, \text{h}(x)\right)\right)\right) \\
&= s_h\left(x, \text{h}(x)\right)\left(\max_{y\in \mathcal{Y}}p(y\mid x) - p\left(\text{h}(x)\mid x\right)\right) \\
&\geqslant \frac{1}{n} \left(\max_{y\in \mathcal{Y}}p(y\mid x) - p\left(\text{h}(x)\mid x\right)\right)
\end{aligned}
\]

这个结论我们会在后面的证明中多次用到。该结论的一个推论是如果分类器\(h^*\)为贝叶斯最优分类器（也即\(p(\text{h}^*(x)\mid x) = \max_{y\in \mathcal{Y}} p(y\mid x)\)），则\(\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right) - \left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right) \geqslant 0\)，可直观地将其理解为\(\mathbb{E}_{p(y\mid x)}\left[\mathcal{l}_{\text{mae}}\right]\)更可能接近其下确界。

于是

\[C_L^*(\mathcal{H}, \mathcal{R}, x) = \inf_{r\in\mathcal{R}} \left\{\left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)}\right\}
\]

记上式中需要求极值的部分为泛函\(F(r)\)，则其泛函导数为

\[\frac{\delta F}{\delta r(x)} = \alpha \left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)e^{\alpha r(x)} - c\alpha e^{-\alpha r(x)}
\]

令\(\frac{\delta F}{\delta p(x)} = 0\)（对\(\forall x\in \mathcal{X}\)），解得\(r^*(x) = -\frac{1}{2\alpha}\log \left(\frac{1 - \max_{y\in \mathcal{Y}}p(y\mid x)}{c}\right)\)。将其代入\(F(r)\)可得：

\[C_L^*(\mathcal{H}, \mathcal{R}, x) = 2\sqrt{c(1 - \max_{y\in \mathcal{Y}}p(y\mid x))}
\]

于是

\[\begin{aligned}
\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) &= C_L(h, r, x) - C^*_L(\mathcal{H}, \mathcal{R}, x) \\
&= \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} - 2\sqrt{c(1 - \max_{y\in \mathcal{Y}}p(y\mid x))}
\end{aligned}
\]

为了构建\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)\)和\(\Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)的不等式关系，接下来我们将会采用第3节中类似的做法，针对\(\max_{y\in \mathcal{Y}} p(y\mid x)\)与\(1 - c\)的大小比较情况与\(r(x)\)的正负情况来对\(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\)进行分类讨论：

1. \(\max_{y\in \mathcal{Y}} p(y\mid x) > (1 - c)\)，\(r(x) > 0\)：
   
   此时

   \[\begin{aligned}
   \Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) &= \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} - 2\sqrt{c\left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)} \\
   & \geqslant \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} - \left(c + \underbrace{\left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)}_{<c}\right) \\
   & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (\text{AM-GM inequality}) \\
   &\geqslant \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} - ce^{-\alpha r(x)} - \left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)e^{\alpha r(x)} \\
   &\geqslant \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right) - \left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)\\
   &\geqslant \frac{1}{n} \left(\max_{y\in \mathcal{Y}}p(y\mid x) - p\left(\text{h}(x)\mid x\right)\right) \\
   &= \frac{1}{n} \Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)
   \end{aligned}
   \]

   （其中\(\text{AM-GM inequality}\)为算术-几何平均值不等式）
   
   取\(\Gamma (z) = nz\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)得证。

2. \(\max_{y\in \mathcal{Y}} p(y\mid x) \leqslant (1 - c)\)，\(r(x) > 0\)：

   \[ \begin{aligned}
   \Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) &= \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} - 2\sqrt{c\left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)} \\
   & \geqslant \underbrace{\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h\left(x, y\right)\right)}_{\geqslant c}e^{\alpha r(x)} + c e^{-\alpha r(x)} - 2\sqrt{c\left(\sum_{y\in \mathcal{Y}}p(y\mid x)\left(1 - s_h(x, y)\right)\right)} \\
   & \geqslant \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right) + c - 2\sqrt{c\left(\sum_{y\in \mathcal{Y}}p(y\mid x)\left(1 - s_h(x, y)\right)\right)} \\
   &= \left(\sqrt{\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)} - \sqrt{c}\right)^2 \\
   &= \left(\frac{\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right) - c}{\sqrt{\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)} + \sqrt{c}}\right)^2 \\
   &\geqslant \left(\frac{\sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right) - \left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right) + \left(1 - \max_{y\in \mathcal{Y}}p(y\mid x) - c\right)}{2}\right)^2 \\
   &\geqslant \left(\frac{\frac{1}{n} \left(\max_{y\in \mathcal{Y}}p(y\mid x) - p\left(\text{h}(x)\mid x\right)\right) + \frac{1}{n}\left(1 - \max_{y\in \mathcal{Y}}p(y\mid x) - c\right)}{2}\right)^2 \\
   &= \frac{1}{4n^2}\left(1 - c - p\left(\text{h}(x)\mid x\right)\right)^2 \\
   &= \frac{\Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)^2}{4n^2}
   \end{aligned}
   \]

   取\(\Gamma (z) = 2n\sqrt{z}\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)得证。

3. \(\max_{y\in \mathcal{Y}} p(y\mid x) \leqslant (1 - c)\)，\(r(x) \leqslant 0\)：
   
   由于此时\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) = 0\)，因此\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma\left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)对任意\(\Gamma \geqslant 0\)成立。

4. \(\max_{y\in \mathcal{Y}} p(y\mid x) > (1 - c)\)，\(r(x) \leqslant 0\)：

   \[ \begin{aligned}
   \Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) &= \sum_{y\in \mathcal{Y}}p(y\mid x) \left(1 - s_h(x, y)\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} - 2\sqrt{c\left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)} \\
   &\geqslant \left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)\underbrace{e^{\alpha r(x)}}_{\leqslant 1} + c \underbrace{e^{-\alpha r(x)}}_{\geqslant 1} - 2\sqrt{c\left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)} \\
   &\geqslant 1 - \max_{y\in \mathcal{Y}}p(y\mid x) + c - 2\sqrt{c\left(1 - \max_{y\in \mathcal{Y}}p(y\mid x)\right)} \\
   &= \left(\sqrt{1 - \max_{y\in \mathcal{Y}}p(y\mid x)} - \sqrt{c}\right)^2 \\
   &= \left(\frac{1 - \max_{y\in \mathcal{Y}}p(y\mid x) - c}{\sqrt{1 - \max_{y\in \mathcal{Y}}p(y\mid x)} + \sqrt{c}}\right)^2 \\
   &\geqslant \left(\frac{\max_{y\in \mathcal{Y}}p(y\mid x) - 1 + c}{2}\right)^2 \\
   &= \frac{\Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)^2}{4}
   \end{aligned}
   \]

   取\(\Gamma (z) = 2\sqrt{z}\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)得证。

综上所述，若取\(\Gamma(z) = \max\{\Gamma_1(z), \Gamma_2(z), \Gamma_3(z)\} = \max\{2n\sqrt{z}, nz\}\)，则恒有\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)。于是\(\mathcal{l} = \mathcal{l}_{\text{mae}}\)，\(\psi(z) = z\)时单阶段代理损失的\((\mathcal{H}, \mathcal{R})\)-一致性界得证。

4.3 \(\mathcal{l} = \mathcal{l}_{\rho}\)，\(\psi(z) = z\)

在这种情况下\(C_L(h, r, x)\)可以表示为：

\[\begin{aligned}
C_L(h, r, x) &= \sum_{y\in \mathcal{Y}}p(y\mid x) \underbrace{\min\left\{\max\left\{0, 1 - \frac{\rho_h(x, y)}{\rho}\right\}, 1\right\}}_{\mathcal{l}_{\rho}}e^{\alpha r(x)} + c e^{-\alpha r(x)} \\
&= \left(1 - \sum_{y\in \mathcal{Y}} p(y\mid x)\max\left\{\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\}, 0\right\}\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} \\
&= \left(1 - \sum_{y\in \mathcal{Y}} p(y\mid x)\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\}\right)e^{\alpha r(x)} + c e^{-\alpha r(x)}
\end{aligned}
\]

其中\(\rho_h(x, y) = h(x)_y - \max_{y^{\prime}\neq y}h(x)_{y^{\prime}}\)为间隔。

由于假设了\(\mathcal{H}\)是对称的与完备的，我们有

\[\begin{aligned}
&\inf_{h\in \mathcal{H}}\left\{1 - \sum_{y\in \mathcal{Y}} p (y\mid x)\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\}\right\} \\
&= 1 - \sup_{h\in \mathcal{H}}\sum_{y\in \mathcal{Y}}p(y\mid x)\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\} \\
&= 1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right)\quad (\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\}\in [0, 1])
\end{aligned}
\]

注 实际上，对任意\(h\in \mathcal{H}\)，有：

\[\begin{aligned}
&\left(1 - \sum_{y\in \mathcal{Y}} p (y\mid x)\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\}\right) - \left(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right)\right) \\
&= \max_{y\in \mathcal{Y}} p(y\mid x) - \sum_{y\in \mathcal{Y}} p (y\mid x)\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\} \\
&= \max_{y\in \mathcal{Y}} p(y\mid x) - \min \left\{1, \frac{\rho_h\left(x, \text{h}(x)\right)}{\rho}\right\}p\left(\text{h}(x)\mid x\right) \\
&\geqslant \max_{y\in \mathcal{Y}}p\left(y\mid x\right) - p\left(\text{h}(x)\mid x\right)
\end{aligned}
\]

和之前\(\mathcal{l}_{\text{mae}}\)的证明类似，这个结论我们会在后面的证明中多次用到。

于是和之前\(\mathcal{l}_{\text{mae}}\)类似，我们有

\[C_L^*(\mathcal{H}, \mathcal{R}, x) = 2\sqrt{c(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right))}
\]

于是

\[\begin{aligned}
\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) &= C_L(h, r, x) - C^*_L(\mathcal{H}, \mathcal{R}, x) \\
&= \left(1 - \sum_{y\in \mathcal{Y}} p (y\mid x)\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\}\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} - 2\sqrt{c(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right))}
\end{aligned}
\]

为了构建\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)\)和\(\Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)的不等式关系，接下来我们将会采用\(\mathcal{l}_{\text{mae}}\)的证明中类似的做法，针对\(\max_{y\in \mathcal{Y}} p(y\mid x)\)与\(1 - c\)的大小比较情况与\(r(x)\)的正负情况来对\(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\)进行分类讨论：

1. \(\max_{y\in \mathcal{Y}} p(y\mid x) > (1 - c)\)，\(r(x) > 0\)：
   
   此时

   \[\begin{aligned}
   \Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) &= \left(1 - \sum_{y\in \mathcal{Y}} p (y\mid x)\min\left\{1, \frac{\rho_h(x, y)}{\rho}\right\}\right)e^{\alpha r(x)} + c e^{-\alpha r(x)} - 2\sqrt{c\left(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right)\right)} \\
   &\geqslant \frac{1}{4}\left(1 - c - p\left(\text{h}\left(x\right)\mid x\right)\right)^2 \\
   &= \frac{\Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)^2}{4}
   \end{aligned}
   \]

   （由于证明步骤与\(\mathcal{l}_{\text{mae}}\)类似，这里对证明步骤进行了一些精简，下面同理）
   
   取\(\Gamma_2 (z) = 2\sqrt{z}\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)得证。

2. \(\max_{y\in \mathcal{Y}} p(y\mid x) \leqslant (1 - c)\)，\(r(x) > 0\)：

   \[\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) \geqslant \max_{y\in \mathcal{Y}}p(y\mid x) - p(\text{h}(x)\mid x) = \Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)
   \]

   取\(\Gamma_1 (z) = z\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)得证。

3. \(\max_{y\in \mathcal{Y}} p(y\mid x) \leqslant (1 - c)\)，\(r(x) \leqslant 0\)：
   
   由于此时\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) = 0\)，因此\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)对任意\(\Gamma \geqslant 0\)成立。

4. \(\max_{y\in \mathcal{Y}} p(y\mid x) > (1 - c)\)，\(r(x) \leqslant 0\)：

   \[\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) \geqslant \left(\frac{\max_{y\in \mathcal{Y}}p(y\mid x) - 1 + c}{2}\right)^2 = \frac{\Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)^2}{4}
   \]

   取\(\Gamma_3 (z) = 2\sqrt{z}\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)得证。

综上所述，若取\(\Gamma(z) = \max\{\Gamma_1(z), \Gamma_2(z), \Gamma_3(z)\} = \max\{2\sqrt{z}, z\}\)，则恒有\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)。于是\(\mathcal{l} = \mathcal{l}_{\rho}\)，\(\psi(z) = z\)时单阶段代理损失的\((\mathcal{H}, \mathcal{R})\)-一致性界得证。

4.4 \(\mathcal{l} = \mathcal{l}_{\rho-\text{hinge}}\)，\(\psi(z) = nz\)

在这种情况下\(C_L(h, r, x)\)可以表示为：

\[\begin{aligned}
C_L(h, r, x) &= \sum_{y\in \mathcal{Y}}p(y\mid x) \underbrace{\sum_{y^{\prime} \neq y}\max\left\{0, 1 + \frac{h(x)_{y^{\prime}}}{\rho}\right\}}_{\mathcal{l}_{\rho}-\text{hinge}}e^{\alpha r(x)} + nce^{-\alpha r(x)} \\
&= \sum_{y\in \mathcal{Y}}\left(1 - p(y\mid x)\right)\max\left\{0, 1 + \frac{h(x)_y}{\rho}\right\}e^{\alpha r(x)} + nce^{-\alpha r(x)} \\
\end{aligned}
\]

由于假设了\(\mathcal{H}\)是对称的与完备的，我们有

\[\begin{aligned}
&\inf_{h\in \mathcal{H}}\left\{\sum_{y\in \mathcal{Y}}\left(1 - p(y\mid x)\right)\max\left\{0, 1 + \frac{h(x)_y}{\rho}\right\}\right\} \\
&= n - \sup_{h\in \mathcal{H}}\sum_{y\in \mathcal{Y}}p(y\mid x)\max\left\{0, 1 + \frac{h(x)_y}{\rho}\right\} \\
&= n\left(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right)\right)
\end{aligned}
\]

注 实际上，若取\(h_{\rho}\)使得\(h_{\rho}(x)_y = \left\{\begin{aligned}
&h(x)_y\quad &\text{if } y\notin \left\{y_{\max}, \text{h}(x)\right\} \\
&-\rho \quad &\text{if } y = \text{h}(x) \\
&h\left(x\right)_{y_{\text{max}}} + h\left(x\right)_{\text{h}(x)} + \rho \quad &\text{if } y = y_{\text{max}} \\
\end{aligned}\right.\)满足约束\(\sum_{y\in \mathcal{Y}}h_{\rho}(y\mid x)=0\)，其中\(y_{\max} = \text{arg max}_{y\in \mathcal{Y}}p(y\mid x)\)，则对任意\(h\in \mathcal{H}\)有：

\[\begin{aligned}
&\sum_{y\in \mathcal{Y}}\left(1 - p(y\mid x)\right)\max\left\{0, 1 + \frac{h(x)_y}{\rho}\right\} - n\left(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right)\right) \\
&\geqslant \sum_{y\in \mathcal{Y}}\left(1 - p(y\mid x)\right)\min\left\{n, \max\left\{0, 1 + \frac{h(x)_y}{\rho}\right\}\right\} - n\left(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right)\right) \\
&\geqslant \sum_{y\in \mathcal{Y}}\left(1 - p(y\mid x)\right)\min\left\{n, \max\left\{0, 1 + \frac{h(x)_y}{\rho}\right\}\right\} \\
&\quad - \sum_{y\in \mathcal{Y}}\left(1 - p(y\mid x)\right)\min\left\{n, \max\left\{0, 1 + \frac{h_{\rho}(x)_y}{\rho}\right\}\right\} \\
&= \left(p(y_{\text{max}}\mid x) - p(\text{h}(x)\mid x)\right)\min\left\{n, 1 + \frac{h(x)_{\text{h}(x)}}{\rho}\right\} \\
&\geqslant \max_{y\in \mathcal{Y}}p\left(y\mid x\right) - p\left(\text{h}\left(x\right)\mid x\right)
\end{aligned}
\]

和之前\(\mathcal{l}_{mae}\)、\(\mathcal{l}_{\rho}\)的证明类似，这个结论我们会在后面的证明中多次用到。

于是和之前\(\mathcal{l}_{mae}\)、\(\mathcal{l}_{\rho}\)类似，我们有

\[C_L^*(\mathcal{H}, \mathcal{R}, x) = 2\sqrt{n^2c(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right))}
\]

于是

\[\begin{aligned}
\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) &= C_L(h, r, x) - C^*_L(\mathcal{H}, \mathcal{R}, x) \\
&= \sum_{y\in \mathcal{Y}}\left(1 - p(y\mid x)\right)\max\left\{0, 1 + \frac{h(x)_y}{\rho}\right\}e^{\alpha r(x)} + c e^{-\alpha r(x)} - 2\sqrt{c(1 - \max_{y\in \mathcal{Y}}p\left(y\mid x\right))}
\end{aligned}
\]

为了构建\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)\)和\(\Gamma \left(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\right)\)的不等式关系，接下来我们将会采用\(\mathcal{l}_{\text{mae}}\)、\(\mathcal{l}_{\rho}\)的证明中类似的做法，针对\(\max_{y\in \mathcal{Y}} p(y\mid x)\)与\(1 - c\)的大小比较情况与\(r(x)\)的正负情况来对\(\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)\)进行分类讨论：

1. \(\max_{y\in \mathcal{Y}} p(y\mid x) > (1 - c)\)，\(r(x) > 0\)：
   
   此时

   \[\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x)
   \geqslant \max_{y\in \mathcal{Y}}p(y\mid x) - p(\text{h}(x)\mid x) = \Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)^2
   \]

   取\(\Gamma_1 (z) = z\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)得证。

2. \(\max_{y\in \mathcal{Y}} p(y\mid x) \leqslant (1 - c)\)，\(r(x) > 0\)：

   \[\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) \geqslant \frac{1}{4n}\left(1 - c - p\left(\text{h}\left(x\right)\mid x\right)\right)^2 = \frac{\Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)^2}{4n}
   \]

   取\(\Gamma_1 (z) = 2\sqrt{nz}\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)得证。

3. \(\max_{y\in \mathcal{Y}} p(y\mid x) \leqslant (1 - c)\)，\(r(x) \leqslant 0\)：
   
   由于此时\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) = 0\)，因此\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)对任意\(\Gamma \geqslant 0\)成立。

4. \(\max_{y\in \mathcal{Y}} p(y\mid x) > (1 - c)\)，\(r(x) \leqslant 0\)：

   \[\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x) \geqslant n\left(\frac{\max_{y\in \mathcal{Y}}p(y\mid x) - 1 + c}{2}\right)^2 = \frac{n\Delta C_{\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x)^2}{4}
   \]

   取\(\Gamma_3 (z) = 2\sqrt{z/n}\)，于是\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)得证。

综上所述，若取\(\Gamma(z) = \max\{\Gamma_1(z), \Gamma_2(z), \Gamma_3(z)\} = \max\{2\sqrt{nz}, z\}\)，则恒有\(\Delta C_{L_\text{abst}, \mathcal{H}, \mathcal{R}}(h, r, x) \leqslant \Gamma (\Delta C_{L, \mathcal{H}, \mathcal{R}}(h, r, x))\)。于是\(\mathcal{l} = \mathcal{l}_{\rho-\text{hinge}}\)，\(\psi(z) = nz\)时单阶段代理损失的\((\mathcal{H}, \mathcal{R})\)-一致性界得证。

参考

• [1] Mao A, Mohri M, Zhong Y. Predictor-rejector multi-class abstention: Theoretical analysis and algorithms[C]//International Conference on Algorithmic Learning Theory. PMLR, 2024: 822-867.
• [2] Cortes C, DeSalvo G, Mohri M. Boosting with abstention[J]. Advances in Neural Information Processing Systems, 2016, 29.
• [3] Ni C, Charoenphakdee N, Honda J, et al. On the calibration of multiclass classification with rejection[J]. Advances in Neural Information Processing Systems, 2019, 32.
• [4] Han Bao: Learning Theory Bridges Loss Functions
• [5] Crammer K, Singer Y. On the algorithmic implementation of multiclass kernel-based vector machines[J]. Journal of machine learning research, 2001, 2(Dec): 265-292.
• [6] Awasthi P, Mao A, Mohri M, et al. Multi-Class $ H $-Consistency Bounds[J]. Advances in neural information processing systems, 2022, 35: 782-795.