例 6.3.4

似然函数及其对数分别为

$$L(\mu, \sigma^2) = \prod_{i=1}^{n}\left(\frac{1}{\sqrt{2\pi}\sigma}\exp\left\{-\frac{(x_i-\mu)^2}{2\sigma^2}\right\}\right) = (2\pi\sigma^2)^{-n/2}\exp\left\{-\frac{1}{2\sigma^2}\sum_{i=1}^{n}(x_i-\mu)^2\right\},$$ $$\ln L(\mu, \sigma^2) = -\frac{1}{2\sigma^2}\sum_{i=1}^{n}(x_i-\mu)^2 - \frac{n}{2}\ln\sigma^2 - \frac{n}{2}\ln(2\pi),$$

将 $\ln L(\mu, \sigma^2)$ 分别关于两个分量求偏导并令其为 0 即得到似然方程组

$$\left.\frac{\partial \ln L(\mu, \sigma^2)}{\partial \mu}\right|_{(\hat{\mu},\hat{\sigma}^2)} = \frac{1}{\hat{\sigma}^2}\sum_{i=1}^{n}(x_i - \hat{\mu}) = 0, \tag{6.3.6}$$ $$\left.\frac{\partial \ln L(\mu, \sigma^2)}{\partial \sigma^2}\right|_{(\hat{\mu},\hat{\sigma}^2)} = \frac{1}{2\hat{\sigma}^4}\sum_{i=1}^{n}(x_i - \hat{\mu})^2 - \frac{n}{2\hat{\sigma}^2} = 0. \tag{6.3.7}$$

解此方程组，由 (6.3.6) 式可得 $\mu$ 的最大似然估计为

$$\hat{\mu} = \frac{1}{n}\sum_{i=1}^{n}x_i = \bar{x},$$

将之代入 (6.3.7) 式给出 $\sigma^2$ 的最大似然估计

$$\hat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 = s_n^2,$$

利用二阶导数矩阵的非正定性可以说明上述估计使得似然函数取极大值。