概率论与数理统计

数理统计总结

2019-06-13 21:14 CST
2019-06-13 23:59 CST
CC BY-NC 4.0

注意:样本方差:

$$S^2 = \dfrac{1}{n-1} \sum_{i=1}^n(X_i - \bar{X})^2.$$

参数估计方法:

  1. 矩估计
  2. 极大似然估计(对数、求导)

三大正态总体的分布:

  1. $\chi^2$分布:$X \sim N(0, 1)$ $$\chi^2_n = \sum_{i=1}^n X_i^2 \sim \chi^2(n).$$ 性质:自由度可加,以及 $$E(\chi^2) = n, \ \ D(\chi^2) = 2n.$$
  2. $t$分布:$X \sim N(0, 1), Y \sim \chi^2(n)$ $$T = \dfrac{X}{\sqrt{Y/n}} \sim t(n).$$ 性质:密度函数是偶函数。
  3. $F$分布:$U \sim \chi^2(n_1), V \sim \chi^2(n_2)$且相互独立, $$F = \dfrac{U/n_1}{V/n_2} \sim F(n_1, n_2).$$ 性质:
    • 若$F \sim F(n_1, n_2)$,则$\dfrac{1}{F} \sim F(n_2, n_1)$。
    • 若$T \sim t(n)$,则$T^2 \sim F(1, n)$。

五个重要的性质:

  1. $\bar{X} \sim N(\mu, \dfrac{\sigma^2}{n}), \dfrac{\bar{X} - \mu}{\sigma/\sqrt{n}} \sim N(0, 1).$ 当$\sigma^2$已知时,$\mu$的置信度为$1 - \alpha$的区间为 $$\left( \bar{X} - u_{\alpha / 2} \dfrac{\sigma}{\sqrt{n}}, \bar{X} + u_{\alpha / 2} \dfrac{\sigma}{\sqrt{n}} \right).$$ 当任意分布的$n$充分大时,根据中心极限定理,有 $$\dfrac{\sum\limits_{i=1}^n X_i - n\mu}{\sqrt{n} \sigma} \longrightarrow N(0, 1),$$ 由此可得均值$\mu$的置信度为$1 - \alpha$的置信区间为 $$\left( \bar{X} - u_{\alpha/2}\dfrac{S}{\sqrt{n}}, \bar{X} + u_{\alpha/2}\dfrac{S}{\sqrt{n}} \right).$$ $H_0: \mu = \mu_0$的拒绝域为 $$W = \left\lbrace \vert U \vert = \vert \dfrac{\bar{X}-\mu_0}{\sigma/\sqrt{n}} \vert \geqslant u_{\alpha/2} \right\rbrace.$$

  2. $T = \dfrac{\bar{X} - \mu}{S/\sqrt{n}} \sim t(n-1).$ 当$\sigma^2$未知时,$\mu$的置信度为$1 - \alpha$的区间为 $$\left( \bar{X} - t_{\alpha / 2} (n-1) \dfrac{S}{\sqrt{n}}, \bar{X} + t_{\alpha / 2} (n-1) \dfrac{S}{\sqrt{n}} \right).$$ $H_0: \mu = \mu_0$的拒绝域为 $$W = \left\lbrace \vert T \vert = \vert \dfrac{\bar{X} - \mu_0}{S / \sqrt{n}} \vert \geqslant t_{\alpha/2}(n-1) \right\rbrace.$$

  3. $\dfrac{(n-1)S^2}{\sigma^2} \sim \chi^2(n-1).$ 当$\mu$未知时,$\sigma^2$的置信度为$1 - \alpha$的区间为 $$\left( \dfrac{(n-1)S^2}{\chi^2_{\alpha / 2}(n-1)}, \dfrac{(n-1)S^2}{\chi^2_{1 - \alpha / 2}(n-1)} \right).$$ $H_0: \sigma^2 = \sigma^2_0$的拒绝域为 $$W = \left\lbrace \chi^2 \leqslant \chi^2_{1-\alpha/2}(n-1) \right\rbrace \cup \left\lbrace \chi^2 \geqslant \chi^2_{\alpha/2}(n-1) \right\rbrace.$$

  4. $F = \dfrac{S_1^2\sigma_2^2}{S_2^2\sigma_1^2} \sim F(n_1 - 1, n_2 - 1).$ 若$\mu_1, \mu_2$已知,$\sigma^2_1 / \sigma^2_2$的置信度为$1 - \alpha$的置信区间为 $$\left( \dfrac{S^2_1}{S^2_2} \cdot \dfrac{1}{F_{\alpha/2}(n_1 - 1, n_2 - 1)}, \dfrac{S^2_1}{S^2_2} \cdot \dfrac{1}{F_{1-\alpha/2}(n_1 - 1, n_2 - 1)} \right).$$ $H_0: \sigma^2_1 = \sigma^2_2$的拒绝域为 $$W = \left\lbrace F \leqslant F_{1-\alpha/2}(n_1-1, n_2-1) \right\rbrace \cup \left\lbrace F \geqslant F_{\alpha/2}(n_1-1, n_2-1) \right\rbrace.$$

  5. $U = \dfrac{(\bar{X} - \bar{Y}) - (\mu_1 - \mu_2)}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}} \sim N(0, 1).$ 若$\sigma^2_1, \sigma^2_2$已知,$\mu_1 - \mu_2$的置信度为$1 - \alpha$的置信区间为 $$\left( \bar{X} - \bar{Y} - u_{\alpha / 2} \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}}, \bar{X} - \bar{Y} + u_{\alpha / 2} \sqrt{\dfrac{\sigma^2_1}{n_1} + \dfrac{\sigma^2_2}{n_2}} \right).$$ $H_0: \mu_1 = \mu_2$的拒绝域为 $$W = \left\lbrace \vert U \vert \geqslant u_{\alpha/2} \right\rbrace.$$