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  • 样本方差性质及其应用

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

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  • 原文地址:https://www.cnblogs.com/wf-strongteam/p/9043052.html
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