🎚️参数估计

\( \begin{aligned} &\text{一般地，当总体 }X\text{ 的分布函数 }F(x;\theta_{1},\theta_{2},\dots,\theta_{k})\text{ 的形式已知，}\\ &\text{但其中存在未知参数 }\theta_{1},\theta_{2},\dots,\theta_{k}\text{，甚至分布函数的形式未知时，}\\ &\text{可以利用如下估计参数的方法：设 }(X_{1},X_{2},\dots,X_{n})\text{ 是总体 }X\text{ 的一个样本，}\\ &\text{根据一定的原理，用 }(X_{1},X_{2},\dots,X_{n})\text{ 构造统计量 }\hat{\theta}_{j}=\hat{\theta}_{j}(X_{1},X_{2},\dots,X_{n})，\\ &j=1,2,\dots,k，\text{ 然后再代入样本数据 }(x_{1},x_{2},\dots,x_{n})，\\ &\text{由此对未知参数 }\theta_{j}\;(j=1,2,\dots,k)\text{ 进行估计。}\\[6pt] &\text{这种用 }(X_{1},X_{2},\dots,X_{n})\text{ 构造统计量去估计未知参数的方法称为}\textbf{点估计法}。 \end{aligned} \)

\( \begin{aligned} &\textbf{区间估计中置信区间的定义：}\\ &\text{设总体 }X\text{ 的分布函数为 }F(x;\theta),\;\theta\text{ 为未知参数},\; (X_{1},X_{2},\dots,X_{n})\text{ 是来自总体 }X\text{ 的一个样本}.\\ &\forall\,\alpha\,(0<\alpha<1),\;\text{若存在统计量 }\hat{\theta}_{1}=\hat{\theta}_{1}(X_{1},X_{2},\dots,X_{n})\; \text{和 }\hat{\theta}_{2}=\hat{\theta}_{2}(X_{1},X_{2},\dots,X_{n})\\ &\text{使得 }P\bigl(\hat{\theta}_{1}<\theta<\hat{\theta}_{2}\bigr)=1-\alpha,\\ &\text{则称区间 }\bigl(\hat{\theta}_{1},\hat{\theta}_{2}\bigr)\text{ 是 }\theta\text{ 的置信度为 }1-\alpha\text{ 的置信区间},\\ &\hat{\theta}_{1}\text{ 和 }\hat{\theta}_{2}\text{ 分别称为置信下限与置信上限。} \end{aligned} \)

\( \begin{aligned} &\text{设总体 }X\text{ 的分布函数为 }F\bigl(x;\theta_{1},\theta_{2},\dots,\theta_{k}\bigr)， \text{其中待估计参数为 }\theta_{1},\theta_{2},\dots,\theta_{k}.\\ &\text{假设 }k\text{ 阶原点矩 }\mu_{r}\bigl(\theta_{1},\theta_{2},\dots,\theta_{k}\bigr)=E\bigl(X^{r}\bigr) \text{ 存在，}r=1,2,\dots,k.\\ &\text{由大数定律，可列出矩方程组：}\\ &\qquad \begin{cases} \mu_{1}\bigl(\theta_{1},\theta_{2},\dots,\theta_{k}\bigr)= \dfrac{1}{n}\displaystyle\sum_{i=1}^{n}X_{i},\\[12pt] \mu_{2}\bigl(\theta_{1},\theta_{2},\dots,\theta_{k}\bigr)= \dfrac{1}{n}\displaystyle\sum_{i=1}^{n}X_{i}^{2},\\ \;\;\vdots\\ \mu_{k}\bigl(\theta_{1},\theta_{2},\dots,\theta_{k}\bigr)= \dfrac{1}{n}\displaystyle\sum_{i=1}^{n}X_{i}^{k}. \end{cases}\\[10pt] &\text{若方程组有解（实际上，上述方程组均为近似方程）}\\ &\qquad \hat{\theta}_{1}=\hat{\theta}_{1}(X_{1},X_{2},\dots,X_{n}),\; \hat{\theta}_{2}=\hat{\theta}_{2}(X_{1},X_{2},\dots,X_{n}),\;\dots,\; \hat{\theta}_{k}=\hat{\theta}_{k}(X_{1},X_{2},\dots,X_{n}),\\ &\text{则称 }\hat{\theta}_{1},\hat{\theta}_{2},\dots,\hat{\theta}_{k} \text{ 为参数 }\theta_{1},\theta_{2},\dots,\theta_{k}\text{ 的矩估计量。}\\[6pt] &\text{相应地，其样本值为}\\ &\qquad \hat{\theta}_{1}=\hat{\theta}_{1}(x_{1},x_{2},\dots,x_{n}),\; \hat{\theta}_{2}=\hat{\theta}_{2}(x_{1},x_{2},\dots,x_{n}),\;\dots,\; \hat{\theta}_{k}=\hat{\theta}_{k}(x_{1},x_{2},\dots,x_{n}),\\ &\text{这组数即为参数 }\theta_{1},\theta_{2},\dots,\theta_{k}\text{ 的矩估计值。} \end{aligned} \)

\( \begin{aligned} &\text{求最大似然估计的一般步骤如下：}\\ &\textbf{(1)}\;\text{写出似然函数 }L(\theta_{1},\theta_{2},\dots,\theta_{k})\text{ 或对数似然函数 }\ln L(\theta_{1},\theta_{2},\dots,\theta_{k});\\ &\textbf{(2)}\;\text{写出似然方程组}\\ &\qquad \frac{\partial L(\theta_{1},\theta_{2},\dots,\theta_{k})}{\partial \theta_{i}} =0,\;i=1,2,\dots,k\\ &\text{或}\\ &\qquad \frac{\partial\,\ln L(\theta_{1},\theta_{2},\dots,\theta_{k})}{\partial \theta_{i}} =0,\;i=1,2,\dots,k;\\ &\textbf{(3)}\;\text{求解上述方程组得到 }\widehat{\theta}_{i}=\widehat{\theta}_{i}(x_{1},x_{2},\dots,x_{n}),\\ &\qquad \text{于是 }\widehat{\theta}_{i}=\widehat{\theta}_{i}(x_{1},x_{2},\dots,x_{n}) \text{ 称为参数 }\theta_{i}\,(i=1,2,\dots,k)\text{ 的最大似然估计值},\\ &\qquad \text{相应的统计量 }\widehat{\theta}_{i}(X_{1},X_{2},\dots,X_{n}) \text{ 称为参数 }\theta_{i}\,(i=1,2,\dots,k)\text{ 的最大似然估计量}. \end{aligned} \)

\( \begin{aligned} &\text{设参数 }\theta\text{ 的估计量为 }\hat{\theta}=\hat{\theta}(X_{1},X_{2},\dots,X_{n}),\;\text{若}\;E(\hat{\theta})=\theta,\\ &\text{则称 }\hat{\theta}\text{ 是 }\theta\text{ 的无偏估计量；反之若 }E(\hat{\theta})\neq\theta,\;\text{则称 }\varepsilon=E(\hat{\theta})-\theta\text{ 为估计量 }\hat{\theta}\text{ 的偏差。} \end{aligned} \)

\( \begin{aligned} &\text{设 }\hat{\theta}_{1}=\hat{\theta}_{1}(X_{1},X_{2},\dots,X_{n})\text{ 和 }\hat{\theta}_{2}=\hat{\theta}_{2}(X_{1},X_{2},\dots,X_{n})\text{ 均为参数 }\theta\text{ 的无偏估计量},\\ &\text{若 }D(\hat{\theta}_{1})\lt D(\hat{\theta}_{2}),\;\text{则称 }\hat{\theta}_{1}\text{ 比 }\hat{\theta}_{2}\text{ 有效。} \end{aligned} \)

\( \begin{aligned} &\text{设 }\hat{\theta}_{n}=\hat{\theta}_{n}(X_{1},X_{2},\dots,X_{n})\text{ 是参数 }\theta\text{ 的估计量，若随机变量序列 }\{\hat{\theta}_{n}\}\text{ 依概率收敛于 }\theta,\\ &\text{即 }\forall\,\varepsilon>0,\;\lim_{n\to+\infty}P\bigl(\,|\hat{\theta}_{n}-\theta|<\varepsilon\,\bigr)=1 \;\text{或}\;\lim_{n\to+\infty}P\bigl(\,|\hat{\theta}_{n}-\theta|\ge\varepsilon\,\bigr)=0,\\ &\text{则称 }\hat{\theta}_{n}\text{ 是 }\theta\text{ 的一致估计量（或相合估计量）。} \end{aligned} \)

\[ \begin{array}{c} \boxed{\text{单个正态总体}}\\[4pt] \begin{array}{|c|c|c|} \hline \text{待估参数} & \text{已知 / 未知量及枢轴量} & 1-\alpha\;\text{置信区间} \\ \hline \mu & \sigma^{2}\;\text{已知},\; U=\dfrac{\bar X-\mu}{\sigma/\sqrt n}\sim N(0,1) & \Bigl(\bar X-u_{\alpha/2}\;\dfrac{\sigma}{\sqrt n},\; \bar X+u_{\alpha/2}\;\dfrac{\sigma}{\sqrt n}\Bigr) \\ \hline \mu & \sigma^{2}\;\text{未知},\; T=\dfrac{\bar X-\mu}{S/\sqrt n}\sim t(n-1) & \Bigl(\bar X-t_{\alpha/2}(n-1)\;\dfrac{S}{\sqrt n},\; \bar X+t_{\alpha/2}(n-1)\;\dfrac{S}{\sqrt n}\Bigr) \\ \hline \sigma^{2} & \mu\;\text{未知},\; \chi^{2}=\dfrac{(n-1)S^{2}}{\sigma^{2}}\sim\chi^{2}(n-1) & \Bigl(\dfrac{(n-1)S^{2}}{\chi^{2}_{\alpha/2}(n-1)},\; \dfrac{(n-1)S^{2}}{\chi^{2}_{1-\alpha/2}(n-1)}\Bigr) \\ \hline \sigma^{2} & \mu\;\text{已知},\; \chi^{2}=\dfrac{1}{\sigma^{2}}\displaystyle\sum_{i=1}^{n}(X_{i}-\mu)^{2}\sim\chi^{2}(n) & \Bigl(\dfrac{\displaystyle\sum_{i=1}^{n}(X_{i}-\mu)^{2}}{\chi^{2}_{\alpha/2}(n)},\; \dfrac{\displaystyle\sum_{i=1}^{n}(X_{i}-\mu)^{2}}{\chi^{2}_{1-\alpha/2}(n)}\Bigr) \\ \hline \end{array} \\[10pt] \boxed{\text{两个正态总体}}\\[4pt] \begin{array}{|c|c|c|} \hline \text{待估参数} & \text{已知 / 未知量及枢轴量} & 1-\alpha\;\text{置信区间} \\ \hline \mu_{1}-\mu_{2} & \begin{array}{l} \sigma_{1}^{2},\sigma_{2}^{2}\text{ 均未知且不一定相等},\;n_{1}=n_{2}=n\\ \text{配对样本：}\;Z_{i}=X_{i}-Y_{i},\\ \bar Z=\dfrac1n\sum_{i=1}^{n}Z_{i},\; S_{Z}^{2}=\dfrac1{n-1}\sum_{i=1}^{n}(Z_{i}-\bar Z)^{2}\\ T=\dfrac{\bar Z-(\mu_{1}-\mu_{2})}{S_{Z}/\sqrt n}\sim t(n-1) \end{array} & \Bigl(\bar Z-t_{\alpha/2}(n-1)\;\dfrac{S_{Z}}{\sqrt n},\; \bar Z+t_{\alpha/2}(n-1)\;\dfrac{S_{Z}}{\sqrt n}\Bigr) \\ \hline \mu_{1}-\mu_{2} & \begin{array}{l} \sigma_{1}^{2},\sigma_{2}^{2}\text{ 均未知},\;n_{1},n_{2}\gg50\\ S_{1}^{2},S_{2}^{2}\text{ 为样本方差}\\ U\approx\dfrac{\bar X-\bar Y-(\mu_{1}-\mu_{2})} {\sqrt{\dfrac{S_{1}^{2}}{n_{1}}+\dfrac{S_{2}^{2}}{n_{2}}}} \sim N(0,1) \end{array} & \Bigl(\bar X-\bar Y-u_{\alpha/2}\,\sqrt{\dfrac{S_{1}^{2}}{n_{1}}+\dfrac{S_{2}^{2}}{n_{2}}},\; \bar X-\bar Y+u_{\alpha/2}\,\sqrt{\dfrac{S_{1}^{2}}{n_{1}}+\dfrac{S_{2}^{2}}{n_{2}}}\Bigr) \\ \hline \dfrac{\sigma_{1}^{2}}{\sigma_{2}^{2}} & \begin{array}{l} \text{独立样本，}\sigma_{1}^{2},\sigma_{2}^{2}\text{ 均未知}\\ F=\dfrac{S_{1}^{2}/\sigma_{1}^{2}}{S_{2}^{2}/\sigma_{2}^{2}} =\dfrac{S_{1}^{2}}{S_{2}^{2}}\;\dfrac{\sigma_{2}^{2}}{\sigma_{1}^{2}} \sim F(n_{1}-1,n_{2}-1) \end{array} & \Bigl(\dfrac{S_{1}^{2}/S_{2}^{2}}{F_{\alpha/2}(n_{1}-1,n_{2}-1)},\; \dfrac{S_{1}^{2}/S_{2}^{2}}{F_{1-\alpha/2}(n_{1}-1,n_{2}-1)}\Bigr) \\ \hline \end{array} \end{array} \]

\( \begin{aligned} &\text{设总体 }X\text{ 的分布任意，}\;(X_{1},X_{2},\dots,X_{n})\text{ 为来自该总体的一组样本}.\\ &\text{欲对未知参数 }\mu=E(X)\text{ 进行区间估计}.\;\text{由中心极限定理，当 }n\text{ 充分大时}\\ &\qquad U\;=\;\frac{\bar X-\mu}{S/\sqrt n}\;\dot\sim\;N(0,1).\\[6pt] &\text{对于给定 }\alpha\;(0\lt \alpha\lt 1),\;\text{有}\\ &\qquad P\bigl(|U|\lt u_{\alpha/2}\bigr)\approx1-\alpha,\\ &\text{即}\\ &\qquad P\Bigl(\bar X-u_{\alpha/2}\,\tfrac{S}{\sqrt n}\lt \mu \lt \bar X+u_{\alpha/2}\,\tfrac{S}{\sqrt n}\Bigr)\approx1-\alpha.\\[8pt] &\text{因此，}\mu\text{ 的近似 }1-\alpha\text{ 置信区间为}\\ &\qquad \Bigl(\bar X-u_{\alpha/2}\,\tfrac{S}{\sqrt n},\;\bar X+u_{\alpha/2}\,\tfrac{S}{\sqrt n}\Bigr).\\[8pt] &\text{一般要求 }n\gt 50\text{ 以保证近似精度；随着 }n\text{ 增大，近似效果更佳}. \end{aligned} \)