# 估计量的评选标准

1，无偏性： $$E(\hat{\theta})=\theta$$。

$E(\bar{x})=E\left(\frac{1}{n}\sum_{i=1}^nx_i\right)=\frac{1}{n}\sum_{i=1}^nE(x_i)=\frac{1}{n}\cdot n\mu=\mu$

\begin{align*}\sum_{i=1}^n(x_i-\bar{x})^2&=\sum_{i=1}^n(x_i^2-2x_i\bar{x}+\bar{x}^2)\\ &=\sum_{i=1}^nx_i^2-2\bar{x}\sum_{i=1}^nx_i+n\bar{x}^2\\ &=\sum_{i=1}^nx_i^2-2\cdot\frac{1}{n}\sum_{i=1}^nx_i\cdot\sum_{i=1}^nx_i+n\bar{x}^2\\ &=\sum_{i=1}^nx_i^2-2n\frac{1}{n}\sum_{i=1}^nx_i\cdot\frac{1}{n}\sum_{i=1}^nx_i+n\bar{x}^2\\ &=\frac{1}{n}\sum_{i=1}^nx_i^2-2n\bar{x}^2+n\bar{x}^2\\ &=\frac{1}{n}\sum_{i=1}^nx_i^2-n\bar{x}^2\end{align*}

$S^2=\frac{1}{n-1}\left(\sum_{i=1}^nx_i^2-n\bar{x}^2\right)$

\begin{align*}E(S^2)&=E\left\{\frac{1}{n-1}\left(\sum_{i=1}^nx_i^2-n\bar{x}^2\right)\right\}\\ &=\frac{1}{n-1}\left(\sum_{i=1}^nE(x_i^2)-nE(\bar{x}^2)\right)\\ &=\frac{1}{n-1}\left(\sum_{i=1}^n(\sigma^2+\mu^2)-n(D(\bar{x})+E(\bar{x}^2))\right)\\&=\frac{1}{n-1}\left(\sum_{i=1}^n(\sigma^2+\mu^2)-n(D(\bar{x})+E(\bar{x}^2))\right)\\&=\frac{1}{n-1}\left(n\sigma^2+n\mu^2-n(\frac{\sigma^2}{n}+\mu^2\right)\\ &=\frac{1}{n-1}(n-1)\sigma^2=\sigma^2\end{align*}

2，有效性：若 $$D(\hat{\theta_1})\le D(\hat{\theta_2})$$，称 $$\hat{\theta_1}$$ 比 $$\hat{\theta_2}$$ 更有效。

\begin{align*}E(\hat{\theta_2})&=E\left(\frac{1}{2}x_1+\frac{1}{3}x_2+\frac{1}{6}x_3)\right)\\ &=\frac{1}{2}E(x_1)+\frac{1}{3}E(x_2)+\frac{1}{6}E(x_3))\\&=\frac{1}{2}\mu+\frac{1}{3}\mu+\frac{1}{6}\mu)=\mu\end{align*}

\begin{align*}D(\hat{\theta_1})&=D\left(\frac{1}{3}(x_1+x_2+x_3)\right)\\ &=\frac{1}{9}(D(x_1)+D(x_2)+D(x_3))\\ &=\frac{1}{9}(\sigma^2+\sigma^2+\sigma^2)=\frac{1}{3}\sigma^2\end{align*}

\begin{align*}D(\hat{\theta_1})&=D\left(\frac{1}{2}x_1+\frac{1}{3}x_2+\frac{1}{6}x_3)\right)\\ &=\frac{1}{4}D(x_1)+\frac{1}{9}D(x_2)+\frac{1}{36}D(x_3)\\ &=\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{36}\right)\sigma^2=\frac{7}{18}\sigma^2\end{align*}

3，相合性（一致性）：若对任意的 $$\epsilon>0$$，

$\lim_{n\to \infty}P\{|\hat{\theta}-\theta|<\epsilon\}=1$

\begin{align*}P\{|\bar{x}-E(\bar{x})|<\epsilon\}\ge 1-\frac{D(\bar{x})}{\epsilon^2}\end{align*}

\begin{align*}P\{|\bar{x}-\mu)|<\epsilon\}&\ge 1-\frac{D(\bar{x})}{\epsilon^2}\\ &=1-\frac{1}{\epsilon}\cdot\frac{\sigma^2}{n}\\ &\to 1,\quad n\to\infty\end{align*}