正态分布
\[f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^\frac{(x-\mu)^2}{2\sigma^2}\]
正太分布的推导
这里介绍高斯的推导方法，采用误差
\[\begin{align*} L(\theta) &= L(\theta;x_1,\cdots,x_n)=f(e_1)\cdots f(e_n) \\ & = f(x_1-\theta)\cdots f(x_n-\theta) \end{align*}\]
\(\frac{d \log L(\theta)}{d \theta} = 0\)
\(\sum_{i=1}^n \frac{f'(x_i-\theta)}{f(x_i-\theta)} = 0\)
令 \(g(x) = \frac{f'(x)}{f(x)}\)
\(\sum_{i=1}^n g(x_i-\theta) = 0\)
\[\begin{equation} \sum_{i=1}^n g(x_i-\bar{x}) = 0 \quad (1) \end{equation}\]
- 式中取 \(n=2\), 有
\(g(x_1-\bar{x}) + g(x_2-\bar{x}) = 0\)
由于此时有 \(x_1-\bar{x} = -(x_2-\bar{x})\)