标签：AC 方差 limits sum rac 样本 应用 bsp 证明
（1）如果$DX$存在，则$E{{S}^{2}}=DX,EM_{2}^{*}=\frac{n-1}{n}DX$;
（2）对任意实数$\mu $，有
$\sum\limits_{i=1}^{n}{({{X}_{i}}}-\overline{X}{{)}^{2}}\le \sum\limits_{i=1}^{n}{({{X}_{i}}}-\mu {{)}^{2}}$.
证明 :
（1） $E{{S}^{2}}=E\frac{1}{n-1}({{\sum\limits_{i=1}^{n}{X_{i}^{2}-n\overline{X}}}^{2}})=\frac{1}{n-1}(\sum\limits_{i=1}^{n}{EX_{i}^{2}-nE{{\overline{X}}^{2}}^{{}})}$
$=\frac{n}{n-1}(E{{X}^{2}}-E{{\overline{X}}^{2}})=\frac{n}{n-1}(DX+{{(EX)}^{2}}-D\overline{X}-{{(E\overline{X})}^{2}})$
$=\frac{n}{n-1}(DX+{{(EX)}^{2}}-\frac{DX}{n}-{{(EX)}^{2}})=DX$
(2) ${{\sum\limits_{i=1}^{n}{({{X}_{i}}-\overline{X})}}^{2}}={{\sum\limits_{i=1}^{n}{(({{X}_{i}}-\mu )+(\mu -\overline{X}))}}^{2}}$
$=\sum\limits_{i=1}^{n}{{{({{X}_{i}}-\mu )}^{2}}}+n{{(\mu -\overline{X})}^{2}}+2(\mu -\overline{X})\sum\limits_{i=1}^{n}{({{X}_{i}}-\mu )}$
$=\sum\limits_{i=1}^{n}{{{({{X}_{i}}-\mu )}^{2}}}+n{{(\mu -\overline{x})}^{2}}-2(\mu -\overline{X})(n\mu -\sum\limits_{i=1}^{n}{{{X}_{i}}})$
$=\sum\limits_{i=1}^{n}{{{({{X}_{i}}-\mu )}^{2}}}-n{{(\mu -\overline{X})}^{2}}\le \sum\limits_{i=1}^{n}{{{({{X}_{i}}-\mu )}^{2}}}$
标签：AC 方差 limits sum rac 样本 应用 bsp 证明
原文地址：https://www.cnblogs.com/wf-strongteam/p/9043052.html
