# 训练误差分析

adaboost的训练误差为：

$$
ERR = \frac{\text{分类错误的样本数}}{\text{总样本数}} = \frac{\sum\_{i=1}^NI(G(x\_i) \neq y\_i)}{N}
$$

adaboost算法最终分类器的误差界为：

$$
\begin{aligned}
ERR = \frac{1}{N}\sum\_{i=1}^NI(G(x\_i) \neq y\_i)   &&{1}
\le \frac{1}{N}\sum\_{i=1}^N\exp(-y\_if(x\_i))    &&{2}
\= \prod\_mZ\_m    &&{3}
\= \prod\_{m=1}^M(2\sqrt {e\_m(1-e\_m)})    &&{4}
\= \prod\_{m=1}^M(2\sqrt {1-4\gamma\_m^2})    &&{5}
\le \exp(-2\sum\_{m=1}^M\gamma\_m^2)   &&{6}
\end{aligned}
$$

说明： （1）：ERR的定义\
（2）:

$$
I(G(x\_i) \neq y\_i) =  I(G(x\_i) \neq y\_i) +  0
$$

当$$G(x\_i) \neq y\_i$$时，

$$
y\_if(x\_i)\lt 0 \Rightarrow \exp(-y\_if(x\_i))\gt 1 = I(G(x\_i) \neq y\_i)
$$

当$$G(x\_i) = y\_i$$时，

$$
y\_if(x\_i)\gt 0 \Rightarrow \exp(-y\_if(x\_i))\gt 1 = I(G(x\_i) \neq y\_i) \gt 0
$$

等式得证\
（3）：

$$
\begin{aligned}
\frac{1}{N}\sum\_{i}^N\exp(-y\_if(x\_i)) \\
\= \frac{1}{N}\sum\_{i}^N\exp(-\sum\_{m}^Ma\_my\_iG\_m(x\_i)), \text {公式6} \\
\=  \sum\_i w\_{1i}\sum\_{i}^N\exp(-\sum\_{m}^Ma\_my\_iG\_m(x\_i)), \text {公式1} \\
\=  Z\_1\sum\_i w\_{2i}\sum\_{2}^N\exp(-\sum\_{m}^Ma\_my\_iG\_m(x\_i)), \text {公式8.4} \\
\= \prod\_{m=1}^MZ\_m
\end{aligned}
$$

（4）：

$$
\begin{aligned}
Z\_m = \sum\_{i=1}^Nw\_{mi}\exp (-a\_my\_iG\_m(x\_i))  \\
\= \sum\_{y\_i=G\_m(x\_i)}w\_{mi}e^{-a\_m} + \sum\_{y\_i\neq G\_m(x\_i)}w\_{mi}e^{a\_m}, y\_iG\_m(x\_i)\text{代表对样本i是否分类正确}  \\
\= (1-e\_m)\exp(-a\_m) + e\_m\exp(a\_m), \text{公式3}  \\
\= (1-e\_m)\sqrt{\frac{e\_m}{1-e\_m}} + (e\_m)\sqrt{\frac{1-e\_m}{e\_m}}, t\text{公式4}  \\
\= 2\sqrt{(1-e\_m)e\_m}
\end{aligned}
$$

（5）：令$$\gamma = \frac{1}{2}-e\_m$$\
（6）：【？】泰勒公式


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://windmising.gitbook.io/lihang-tongjixuexifangfa/adaboost/2.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.