除了补充的内容,知识总结部分就只剩下概率统计的第七章和第八章了,分别对应参数估计和假设检验。这两章别的不说,公式和分布类型就特别多和杂,都是采用表格形式来总结和描述,很多东西需要熟记,掌握好对学统计也有一定的帮助

七、参数估计

1. 点估计

(1) 定义

  设$X_1, \, X_2, \, \cdots , \, X_n$为取自总体$X$的样本,若将样本中的某个函数$\hat{\theta}(X_1 , \, X_2, \, \cdots , \, X_n)$作为总体分布中未知参数$\theta$的估计,则称$\hat{\theta}$为$\theta$的点估计。

(2) 估计量与估计值

  设$X_1, \, X_2, \, \cdots , \, X_n$为取自总体$X$的样本,$\theta$是$X$中的未知参数,$x_1, \, x_2, \, \cdots , \, x_n$为相应的一组样本观察值,用统计量$\hat{\theta}(X_1 , \, X_2, \, \cdots , \, X_n)$作为$\theta$的估计时,$\hat{\theta}(X_1 , \, X_2, \, \cdots , \, X_n)$称为$\theta$的估计量,$\hat{\theta}(x_1 , \, x_2, \, \cdots , \, x_n)$称为$\theta$的估计值。

(3) 矩估计法

  矩估计法是用样本矩代替相应的总体矩,样本矩的函数代替相应的总体矩的同一函数,来求未知参数的一种估计方法。

(4) 最大似然估计法

  设总体$X$的密度函数为$f(x; \, \theta_1 , \, \theta_2, \, \cdots, \, \theta_k)$,$\theta_1 , \, \theta_2, \, \cdots, \, \theta_k$为未知参数,则取自$X$的样本$(X_1 , \, X_2, \, \cdots , \, X_n)$的联合分布律或联合密度函数$$L(x_1, \, x_2, \, \cdots , \, x_n; \, \theta_1, \, \theta_2, \, \cdots , \, \theta_k) = \prod_{i=1}^{n}f(x_i; \, \theta_1, \, \theta_2 , \, \cdots , \, \theta_k)$$称为似然函数。若有$\hat{\theta_i} = \hat{\theta_i}(x_1, \, x_2, \, \cdots , \, x_n), \, i = 1, \, 2, \, \cdots , \, k$,使得$$L(x_1, \, x_2 , \, \cdots , \, x_n; \, \hat{\theta_1}, \, \hat{\theta_2}, \, \cdots , \, \hat{\theta_k}) = \max_{\theta_1, \, \theta_2, \, \cdots , \, \theta_k} L(x_1, \, x_2, \, \cdots , \, x_n; \, \theta_1, \, \theta_2 , \, \cdots , \, \theta_k)$$则$\hat{\theta_i}(x_1, \, x_2, \, \cdots , \, x_n)$称为$\theta_i$的最大似然估计值,$\hat{\theta_i}(X_1, \, X_2, \, \cdots , \, X_n)$称为$\theta_i$的最大似然估计量,$i=1, \, 2, \, \cdots , \, k$。

2. 估计量的无偏性、有效性、一致性

(1) 无偏性

  设$\hat{\theta}$为$\theta$的估计量,若$E(\hat{\theta}) = \theta$,则$\hat{\theta}$称为$\theta$的无偏估计。

(2) 有效性

  设$\hat{\theta_1}, \, \hat{\theta_2}$均为$\theta$的无偏估计,如果$D(\hat{\theta_1}) \leqslant D(\hat{\theta_2})$,则称$\hat{\theta_1}$比$\hat{\theta_2}$有效,如果$\theta$的无偏估计$\hat{\theta}$的方差$D(\hat{\theta})$达到最小,则称$\hat{\theta}$为$\theta$的最小方差无偏估计。

(3) 一致性

  设$\hat{\theta}$为$\theta$的估计量,若对$\forall \varepsilon > 0$,有$\lim\limits_{n \to \infty}P \left \{ | \hat{\theta} - \theta | < \varepsilon \right \} = 1$,则称$\hat{\theta}$为$\theta$的一致估计。

3. 区间估计

置信水平为$1-\alpha$的双侧置信区间表

总体 待估参数 其他参数情况 估计函数与其分布 置信区间
单个正态总体 $\mu$ $\sigma^2$已知 $$\frac{\overline{X} - \mu}{\sigma/ \sqrt{n}} \sim N(0,1)$$ $$(\overline{X} \pm \frac{\sigma}{\sqrt{n}}z_{\frac{\alpha}{2}})$$
$\sigma^2$未知 $$\frac{\overline{X} - \mu}{S/ \sqrt{n}} \sim t(n-1)$$ $$(\overline{X} \pm \frac{S}{\sqrt{n}}t_{\frac{\alpha}{2}}(n-1))$$
$\sigma^2$ $\mu$已知 $$\frac{1}{\sigma^2}\sum_{i=1}^n(X_i- \mu)^2 \sim \chi ^2(n)$$ $$(\frac{\sum_{i=1}^n(X_i- \mu)^2}{\chi_{\frac{\alpha}{2}}^2(n)}, \frac{\sum_{i=1}^n(X_i- \mu)^2}{\chi_{1-\frac{\alpha}{2}}^2(n)})$$
$\mu$未知 $$\frac{(n-1)S^2}{\sigma^2} \sim \chi ^2(n-1)$$ $$(\frac{(n-1)S^2}{\chi_{\frac{\alpha}{2}}^2(n-1)}, \frac{(n-1)S^2}{\chi_{1-\frac{\alpha}{2}}^2(n-1)})$$
两个正态总体 $\mu_1 \, - \, \mu_2$ $\sigma_1^2 , \, \sigma_2^2$已知 $$\frac{(\overline{X}-\overline{Y})-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \sim N(0,1)$$ $$(\overline{X}-\overline{Y} \pm z_{\frac{\alpha}{2}}\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}})$$
$\sigma_1^2 = \sigma_2^2 = \sigma^2$未知 $$\frac{(\overline{X}-\overline{Y})-(\mu_1 - \mu_2)}{S_{\omega }\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \sim t(n_1+n_2-2)$$ $$\begin{align}& ( \overline{X}-\overline{Y} \pm t_{\frac{\alpha}{2}} (n_1+ \\ & n_2-2) S_\omega \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}) \end{align}$$
$\frac{\sigma_1^2}{\sigma_2^2}$ $\mu_1, \, \mu_2$已知 $$\frac{\sum_{i=1}^{n_1}(X_i-\mu_1)^2 / (n_1 \sigma_1^2)}{\sum_{i=1}^{n_2}(Y_i-\mu_2)^2 / (n_2\sigma_2^2)} \sim F(n_1,n_2)$$ $$\begin{align}(\frac{\frac{1}{n_1}\sum_{i=1}^{n_1}(X_i-\mu_1)^2}{\frac{1}{n_2}\sum_{i=1}^{n_2}(Y_i-\mu_2)^2F_{\frac{\alpha}{2}}(n_1,n_2)}&, \\ \frac{\frac{1}{n_1}\sum_{i=1}^{n_1}(X_i-\mu_1)^2}{\frac{1}{n_2}\sum_{i=1}^{n_2}(Y_i-\mu_2)^2F_{1-\frac{\alpha}{2}}(n_1,n_2)})\end{align}$$
$\mu_1, \, \mu_2$未知 $$\frac{S_1^2 / \sigma_1^2}{S_2^2 / \sigma_2^2} \sim F(n_1-1, n_2-1)$$ $$\begin{align}(\frac{S_1^2 }{S_2^2 } \times \frac{1}{F_{\frac{\alpha}{2}}(n_1-1,n_2-1)}&, \\ \frac{S_1^2 }{S_2^2 } \times \frac{1}{F_{1-\frac{\alpha}{2}}(n_1-1,n_2-1)})\end{align}$$

八、假设检验

1. 单个正态总体未知参数假设检验

原假设 备择假设 条件 检验统计量与其分布 拒绝域
$$\begin{align}\mu = \mu_0 & \\ \mu \leqslant \mu_0 & \\ \mu \geqslant \mu_0 \end{align}$$ $$\begin{align}\mu \neq \mu_0 & \\ \mu > \mu_0 & \\ \mu < \mu_0 \end{align}$$ $\sigma^2$已知 $$Z = \frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \sim N(0,1)$$ $$\begin{align} |z| \geqslant z_{\frac{\alpha}{2}} & \\ z \geqslant z_\alpha & \\ z \leqslant - z_\alpha \end{align}$$
$$\begin{align}\mu = \mu_0 & \\ \mu \leqslant \mu_0 & \\ \mu \geqslant \mu_0 \end{align}$$ $$\begin{align}\mu \neq \mu_0 & \\ \mu > \mu_0 & \\ \mu < \mu_0 \end{align}$$ $\sigma^2$未知 $$T=\frac{\overline{X}-\mu_0}{S/ \sqrt{n}} \sim t(n-1)$$ $$\begin{align} |t| \geqslant t_{\frac{\alpha}{2}}(n-1) & \\ t \geqslant t _\alpha (n-1) & \\ t \leqslant - t_\alpha (n-1) \end{align}$$
$$\begin{align}\sigma^2 = \sigma_0^2 & \\ \sigma^2 \leqslant \sigma_0^2 & \\ \sigma^2 \geqslant \sigma_0^2 \end{align}$$ $$\begin{align}\sigma^2 \neq \sigma_0^2 & \\ \sigma^2 > \sigma_0^2 & \\ \sigma^2 < \sigma_0^2 \end{align}$$ $\mu$已知 $$\chi^2 = \frac{1}{\sigma_0^2}\sum_{i=1}^n(X_i- \mu)^2 \sim \chi^2(n)$$ $$\begin{align} \chi^2 \geqslant \chi_{\frac{\alpha}{2}}^2(n) 或& \\ \chi^2 \leqslant \chi_{1-\frac{\alpha}{2}}^2 (n) & \\ \chi^2 \geqslant \chi_\alpha^2(n) & \\ \chi^2 \leqslant \chi_{1-\alpha}^2(n) \end{align}$$
$$\begin{align}\sigma^2 = \sigma_0^2 & \\ \sigma^2 \leqslant \sigma_0^2 & \\ \sigma^2 \geqslant \sigma_0^2 \end{align}$$ $$\begin{align}\sigma^2 \neq \sigma_0^2 & \\ \sigma^2 > \sigma_0^2 & \\ \sigma^2 < \sigma_0^2 \end{align}$$ $\mu$未知 $$\chi^2 = \frac{(n-1)S^2}{\sigma_0^2} \sim \chi^2(n-1)$$ $$\begin{align} \chi^2 \geqslant \chi_{\frac{\alpha}{2}}^2(n-1) 或& \\ \chi^2 \leqslant \chi_{1-\frac{\alpha}{2}}^2 (n-1) & \\ \chi^2 \geqslant \chi_\alpha^2(n-1) & \\ \chi^2 \leqslant \chi_{1-\alpha}^2(n-1) \end{align}$$

2. 两个正态总体假设检验

原假设 备择假设 条件 检验统计量与其分布 拒绝域
$$\begin{align}\mu = \mu_0 & \\ \mu \leqslant \mu_0 & \\ \mu \geqslant \mu_0 \end{align}$$ $$\begin{align}\mu \neq \mu_0 & \\ \mu > \mu_0 & \\ \mu < \mu_0 \end{align}$$ $\sigma^2$已知 $$Z = \frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \sim N(0,1)$$ $$\begin{align} |z| \geqslant z_{\frac{\alpha}{2}} & \\ z \geqslant z_\alpha & \\ z \leqslant - z_\alpha \end{align}$$
$$\begin{align}\mu = \mu_0 & \\ \mu \leqslant \mu_0 & \\ \mu \geqslant \mu_0 \end{align}$$ $$\begin{align}\mu \neq \mu_0 & \\ \mu > \mu_0 & \\ \mu < \mu_0 \end{align}$$ $\sigma^2$未知 $$T=\frac{\overline{X}-\mu_0}{S/ \sqrt{n}} \sim t(n-1)$$ $$\begin{align} |t| \geqslant t_{\frac{\alpha}{2}}(n_1+n_2-2) & \\ t \geqslant t _\alpha (n_1+n_2-2) & \\ t \leqslant - t_\alpha (n_1+n_2-2) \end{align}$$
$$\begin{align}\sigma^2 = \sigma_0^2 & \\ \sigma^2 \leqslant \sigma_0^2 & \\ \sigma^2 \geqslant \sigma_0^2 \end{align}$$ $$\begin{align}\sigma^2 \neq \sigma_0^2 & \\ \sigma^2 > \sigma_0^2 & \\ \sigma^2 < \sigma_0^2 \end{align}$$ $\mu$已知 $$\chi^2 = \frac{1}{\sigma_0^2}\sum_{i=1}^n(X_i- \mu)^2 \sim \chi^2(n)$$ $$\begin{align} F \geqslant F_{\frac{\alpha}{2}}(n_1,n_2) 或& \\ F \leqslant F _{1-\frac{\alpha}{2}} (n_1,n_2) & \\ F \leqslant F_{1-\alpha} (n_1,n_2) & \\ F \geqslant F_\alpha (n_1, n_2) \end{align}$$
$$\begin{align}\sigma^2 = \sigma_0^2 & \\ \sigma^2 \leqslant \sigma_0^2 & \\ \sigma^2 \geqslant \sigma_0^2 \end{align}$$ $$\begin{align}\sigma^2 \neq \sigma_0^2 & \\ \sigma^2 > \sigma_0^2 & \\ \sigma^2 < \sigma_0^2 \end{align}$$ $\mu$未知 $$\chi^2 = \frac{(n-1)S^2}{\sigma_0^2} \sim \chi^2(n-1)$$ $$\begin{align} F \geqslant F_{\frac{\alpha}{2}}(n_1-1,n_2-1) 或& \\ F \leqslant F _{1-\frac{\alpha}{2}} (n_1-1,n_2-1) & \\ F \leqslant F_{1-\alpha} (n_1-1,n_2-1) & \\ F \geqslant F_\alpha (n_1-1, n_2-1) \end{align}$$