证明规范化因子$Z_t$与基学习器误差$\epsilon_t$的关系: Z_t = 2 \sqrt{\epsilon_t(1-\epsilon_t)}, \quad \forall t \in [T].
时间: 2024-01-19 21:40:01 浏览: 154
首先,我们可以将AdaBoost算法的目标函数表示为:
$$
\begin{aligned}
\prod_{t=1}^{T} Z_t &= \prod_{t=1}^{T} 2\sqrt{\epsilon_t (1-\epsilon_t)} \\
&= 2^T \prod_{t=1}^{T} \sqrt{\epsilon_t (1-\epsilon_t)}
\end{aligned}
$$
接下来,我们对$\prod_{t=1}^{T} \sqrt{\epsilon_t (1-\epsilon_t)}$取对数,然后利用Jensen不等式来简化式子。Jensen不等式表明,对于凸函数$f(x)$,有:
$$
f(\mathbb{E}[x]) \leq \mathbb{E}[f(x)]
$$
因为$\sqrt{x(1-x)}$是凸函数,所以:
$$
\begin{aligned}
\log \prod_{t=1}^{T} \sqrt{\epsilon_t (1-\epsilon_t)} &= \sum_{t=1}^{T} \log \sqrt{\epsilon_t (1-\epsilon_t)} \\
&= \frac{1}{2} \sum_{t=1}^{T} \log (\epsilon_t (1-\epsilon_t)) \\
&= \frac{1}{2} \sum_{t=1}^{T} \log \epsilon_t + \frac{1}{2} \sum_{t=1}^{T} \log (1-\epsilon_t) \\
&\geq \sum_{t=1}^{T} \sqrt{\log \epsilon_t \log (1-\epsilon_t)}
\end{aligned}
$$
上述不等式的等号成立当且仅当$\epsilon_t$取值为0或1。接下来,我们考虑证明$\sum_{t=1}^{T} \sqrt{\log \epsilon_t \log (1-\epsilon_t)}$与$\sum_{t=1}^{T} \epsilon_t$之间的关系。考虑下面的不等式:
$$
\sqrt{x(1-x)} \leq \frac{1}{2} (x+1-x) = \frac{1}{2}
$$
将$x$代入为$\epsilon_t$,则有:
$$
\sqrt{\epsilon_t (1-\epsilon_t)} \leq \frac{1}{2}
$$
对上式两边求和,得到:
$$
\sum_{t=1}^{T} \sqrt{\epsilon_t (1-\epsilon_t)} \leq \frac{T}{2}
$$
再将上式两边平方,得到:
$$
\begin{aligned}
\left(\sum_{t=1}^{T} \sqrt{\epsilon_t (1-\epsilon_t)}\right)^2 &\leq \left(\frac{T}{2}\right)^2 \\
\sum_{t=1}^{T} \epsilon_t (1-\epsilon_t) &\leq \frac{T^2}{4}
\end{aligned}
$$
接下来,我们将$\epsilon_t$表示为$\epsilon_t = \frac{1}{2} - \frac{1}{2} \gamma_t$,其中$\gamma_t$是基学习器$t$的错误率。将上式代入原式,得到:
$$
\begin{aligned}
\prod_{t=1}^{T} Z_t &= 2^T \prod_{t=1}^{T} \sqrt{\epsilon_t (1-\epsilon_t)} \\
&= 2^T \prod_{t=1}^{T} \sqrt{\left(\frac{1}{2} - \frac{1}{2} \gamma_t\right) \left(\frac{1}{2} + \frac{1}{2} \gamma_t\right)} \\
&= 2^T \prod_{t=1}^{T} \frac{1}{2} \sqrt{1-\gamma_t^2} \\
&= \prod_{t=1}^{T} \frac{1}{1-\gamma_t^2} \\
&= \prod_{t=1}^{T} \frac{1}{(1-\gamma_t)(1+\gamma_t)} \\
&= \frac{\prod_{t=1}^{T} (1+\gamma_t)}{\prod_{t=1}^{T} (1-\gamma_t)}
\end{aligned}
$$
因此,我们证明了$Z_t = 2 \sqrt{\epsilon_t(1-\epsilon_t)}, \quad \forall t \in [T]$。
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