样本方差的性质

(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

                     $sumlimits_{i=1}^{n}{({{X}_{i}}}-overline{X}{{)}^{2}}le sumlimits_{i=1}^{n}{({{X}_{i}}}-mu {{)}^{2}}$

证明 :

(1)           $E{{S}^{2}}=Efrac{1}{n-1}({{sumlimits_{i=1}^{n}{X_{i}^{2}-noverline{X}}}^{2}})=frac{1}{n-1}(sumlimits_{i=1}^{n}{EX_{i}^{2}-nE{{overline{X}}^{2}}^{{}})}$

                          $=frac{n}{n-1}(E{{X}^{2}}-E{{overline{X}}^{2}})=frac{n}{n-1}(DX+{{(EX)}^{2}}-Doverline{X}-{{(Eoverline{X})}^{2}})$

                          $=frac{n}{n-1}(DX+{{(EX)}^{2}}-frac{DX}{n}-{{(EX)}^{2}})=DX]$

(2)            ${{sumlimits_{i=1}^{n}{({{X}_{i}}-overline{X})}}^{2}}={{sumlimits_{i=1}^{n}{(({{X}_{i}}-mu )+(mu -overline{X}))}}^{2}}$

                                      $=sumlimits_{i=1}^{n}{{{({{X}_{i}}-mu )}^{2}}}+n{{(mu -overline{X})}^{2}}+2(mu -overline{X})sumlimits_{i=1}^{n}{({{X}_{i}}-mu )}$

                                      $=sumlimits_{i=1}^{n}{{{({{X}_{i}}-mu )}^{2}}}+n{{(mu -overline{x})}^{2}}-2(mu -overline{X})(nmu -sumlimits_{i=1}^{n}{{{X}_{i}}})]$

                                      $=sumlimits_{i=1}^{n}{{{({{X}_{i}}-mu )}^{2}}}-n{{(mu -overline{X})}^{2}}le sumlimits_{i=1}^{n}{{{({{X}_{i}}-mu )}^{2}}}$

原文地址:https://www.cnblogs.com/wf-strongteam/p/9042334.html