$operatorname{Var}[sum_{i=1}^n (X_i-mu_0)^2]=operatorname{Var}[ sum_{i=1}^n(X_i-overline{X})^2]$?












1












$begingroup$


Consider independent $X_1,ldots, X_nsimmathcal{N}(mu_0,sigma^2)$ with a known $mu_0inmathbb{R}$ and unknown $sigma^2in(0,infty)$. I already know that



$$DeclareMathOperator{Var}{Var}frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2$$



is variance minimising and that



$$EBig[frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2Big]=EBig[frac{1}{n-1} sum_{i=1}^n(X_i-overline{X})^2Big]=sigma^2$$



What I want to show is that



$$frac{1}{n^2}Var[sum_{i=1}^n (X_i-mu_0)^2]=Var[frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2]<Var[frac{1}{n-1}sum_{i=1}^n(X_i-overline{X})^2]=frac{1}{(n-1)^2}Var[sum_{i=1}^n(X_i-overline{X})^2]$$



Since the first moments are the same it would be enough, to know, that



$$EBig[Big(sum_{i=1}^n (X_i-mu_0)^2Big)^2Big]=EBig[Big(sum_{i=1}^n(X_i-overline{X})^2Big)^2Big]$$



Is this true? There might be an easy argument, but I do not find it.



Thanks in advance!






probability probability-theory statistics statistical-inference descriptive-statistics







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Consider independent $X_1,ldots, X_nsimmathcal{N}(mu_0,sigma^2)$ with a known $mu_0inmathbb{R}$ and unknown $sigma^2in(0,infty)$. I already know that



    $$DeclareMathOperator{Var}{Var}frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2$$



    is variance minimising and that



    $$EBig[frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2Big]=EBig[frac{1}{n-1} sum_{i=1}^n(X_i-overline{X})^2Big]=sigma^2$$



    What I want to show is that



    $$frac{1}{n^2}Var[sum_{i=1}^n (X_i-mu_0)^2]=Var[frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2]<Var[frac{1}{n-1}sum_{i=1}^n(X_i-overline{X})^2]=frac{1}{(n-1)^2}Var[sum_{i=1}^n(X_i-overline{X})^2]$$



    Since the first moments are the same it would be enough, to know, that



    $$EBig[Big(sum_{i=1}^n (X_i-mu_0)^2Big)^2Big]=EBig[Big(sum_{i=1}^n(X_i-overline{X})^2Big)^2Big]$$



    Is this true? There might be an easy argument, but I do not find it.



    Thanks in advance!






    probability probability-theory statistics statistical-inference descriptive-statistics







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Consider independent $X_1,ldots, X_nsimmathcal{N}(mu_0,sigma^2)$ with a known $mu_0inmathbb{R}$ and unknown $sigma^2in(0,infty)$. I already know that



      $$DeclareMathOperator{Var}{Var}frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2$$



      is variance minimising and that



      $$EBig[frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2Big]=EBig[frac{1}{n-1} sum_{i=1}^n(X_i-overline{X})^2Big]=sigma^2$$



      What I want to show is that



      $$frac{1}{n^2}Var[sum_{i=1}^n (X_i-mu_0)^2]=Var[frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2]<Var[frac{1}{n-1}sum_{i=1}^n(X_i-overline{X})^2]=frac{1}{(n-1)^2}Var[sum_{i=1}^n(X_i-overline{X})^2]$$



      Since the first moments are the same it would be enough, to know, that



      $$EBig[Big(sum_{i=1}^n (X_i-mu_0)^2Big)^2Big]=EBig[Big(sum_{i=1}^n(X_i-overline{X})^2Big)^2Big]$$



      Is this true? There might be an easy argument, but I do not find it.



      Thanks in advance!






      probability probability-theory statistics statistical-inference descriptive-statistics







      share|cite|improve this question











      $endgroup$




      Consider independent $X_1,ldots, X_nsimmathcal{N}(mu_0,sigma^2)$ with a known $mu_0inmathbb{R}$ and unknown $sigma^2in(0,infty)$. I already know that



      $$DeclareMathOperator{Var}{Var}frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2$$



      is variance minimising and that



      $$EBig[frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2Big]=EBig[frac{1}{n-1} sum_{i=1}^n(X_i-overline{X})^2Big]=sigma^2$$



      What I want to show is that



      $$frac{1}{n^2}Var[sum_{i=1}^n (X_i-mu_0)^2]=Var[frac{1}{n}sum_{i=1}^n (X_i-mu_0)^2]<Var[frac{1}{n-1}sum_{i=1}^n(X_i-overline{X})^2]=frac{1}{(n-1)^2}Var[sum_{i=1}^n(X_i-overline{X})^2]$$



      Since the first moments are the same it would be enough, to know, that



      $$EBig[Big(sum_{i=1}^n (X_i-mu_0)^2Big)^2Big]=EBig[Big(sum_{i=1}^n(X_i-overline{X})^2Big)^2Big]$$



      Is this true? There might be an easy argument, but I do not find it.



      Thanks in advance!






      probability probability-theory statistics statistical-inference descriptive-statistics




      probability probability-theory statistics statistical-inference descriptive-statistics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 14:39









      J.G.

      26.6k22642




      26.6k22642










      asked Dec 1 '18 at 13:59









      user408858user408858

      482213




      482213






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          Recall that if $X_1,X_2,...,X_n$ iid $N(mu, sigma ^ 2)$, then
          $$ DeclareMathOperator{Var}{Var}
          frac{sum ( X_i - mu ) ^ 2}{ sigma ^ 2} sim chi^2_n, quad frac{sum ( X_i - bar{X}_n ) ^ 2}{ sigma ^ 2} sim chi^2_{n-1},
          $$
          and
          $$
          Var(chi^2_n)=2n, quad Var(chi^2_{n-1})=2(n-1).
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Where $overline{X}_n=frac{1}{n}sum_{i=1}^nX_i$?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:10










          • $begingroup$
            @user408858 Right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:10










          • $begingroup$
            So I get $Var[sum_{i=1}^n (X_i-mu_0)^2]= 2nsigma^4$ and $Var[ sum_{i=1}^n(X_i-overline{X})^2]=2(n-1)sigma^4$, right?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:13












          • $begingroup$
            Yup, that is right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:18











          Your Answer





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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Recall that if $X_1,X_2,...,X_n$ iid $N(mu, sigma ^ 2)$, then
          $$ DeclareMathOperator{Var}{Var}
          frac{sum ( X_i - mu ) ^ 2}{ sigma ^ 2} sim chi^2_n, quad frac{sum ( X_i - bar{X}_n ) ^ 2}{ sigma ^ 2} sim chi^2_{n-1},
          $$
          and
          $$
          Var(chi^2_n)=2n, quad Var(chi^2_{n-1})=2(n-1).
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Where $overline{X}_n=frac{1}{n}sum_{i=1}^nX_i$?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:10










          • $begingroup$
            @user408858 Right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:10










          • $begingroup$
            So I get $Var[sum_{i=1}^n (X_i-mu_0)^2]= 2nsigma^4$ and $Var[ sum_{i=1}^n(X_i-overline{X})^2]=2(n-1)sigma^4$, right?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:13












          • $begingroup$
            Yup, that is right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:18
















          1












          $begingroup$

          Recall that if $X_1,X_2,...,X_n$ iid $N(mu, sigma ^ 2)$, then
          $$ DeclareMathOperator{Var}{Var}
          frac{sum ( X_i - mu ) ^ 2}{ sigma ^ 2} sim chi^2_n, quad frac{sum ( X_i - bar{X}_n ) ^ 2}{ sigma ^ 2} sim chi^2_{n-1},
          $$
          and
          $$
          Var(chi^2_n)=2n, quad Var(chi^2_{n-1})=2(n-1).
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Where $overline{X}_n=frac{1}{n}sum_{i=1}^nX_i$?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:10










          • $begingroup$
            @user408858 Right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:10










          • $begingroup$
            So I get $Var[sum_{i=1}^n (X_i-mu_0)^2]= 2nsigma^4$ and $Var[ sum_{i=1}^n(X_i-overline{X})^2]=2(n-1)sigma^4$, right?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:13












          • $begingroup$
            Yup, that is right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:18














          1












          1








          1





          $begingroup$

          Recall that if $X_1,X_2,...,X_n$ iid $N(mu, sigma ^ 2)$, then
          $$ DeclareMathOperator{Var}{Var}
          frac{sum ( X_i - mu ) ^ 2}{ sigma ^ 2} sim chi^2_n, quad frac{sum ( X_i - bar{X}_n ) ^ 2}{ sigma ^ 2} sim chi^2_{n-1},
          $$
          and
          $$
          Var(chi^2_n)=2n, quad Var(chi^2_{n-1})=2(n-1).
          $$






          share|cite|improve this answer











          $endgroup$



          Recall that if $X_1,X_2,...,X_n$ iid $N(mu, sigma ^ 2)$, then
          $$ DeclareMathOperator{Var}{Var}
          frac{sum ( X_i - mu ) ^ 2}{ sigma ^ 2} sim chi^2_n, quad frac{sum ( X_i - bar{X}_n ) ^ 2}{ sigma ^ 2} sim chi^2_{n-1},
          $$
          and
          $$
          Var(chi^2_n)=2n, quad Var(chi^2_{n-1})=2(n-1).
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 14:30









          Bernard

          121k740116




          121k740116










          answered Dec 1 '18 at 14:08









          V. VancakV. Vancak

          11.1k2926




          11.1k2926












          • $begingroup$
            Where $overline{X}_n=frac{1}{n}sum_{i=1}^nX_i$?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:10










          • $begingroup$
            @user408858 Right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:10










          • $begingroup$
            So I get $Var[sum_{i=1}^n (X_i-mu_0)^2]= 2nsigma^4$ and $Var[ sum_{i=1}^n(X_i-overline{X})^2]=2(n-1)sigma^4$, right?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:13












          • $begingroup$
            Yup, that is right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:18


















          • $begingroup$
            Where $overline{X}_n=frac{1}{n}sum_{i=1}^nX_i$?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:10










          • $begingroup$
            @user408858 Right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:10










          • $begingroup$
            So I get $Var[sum_{i=1}^n (X_i-mu_0)^2]= 2nsigma^4$ and $Var[ sum_{i=1}^n(X_i-overline{X})^2]=2(n-1)sigma^4$, right?
            $endgroup$
            – user408858
            Dec 1 '18 at 14:13












          • $begingroup$
            Yup, that is right
            $endgroup$
            – V. Vancak
            Dec 1 '18 at 14:18
















          $begingroup$
          Where $overline{X}_n=frac{1}{n}sum_{i=1}^nX_i$?
          $endgroup$
          – user408858
          Dec 1 '18 at 14:10




          $begingroup$
          Where $overline{X}_n=frac{1}{n}sum_{i=1}^nX_i$?
          $endgroup$
          – user408858
          Dec 1 '18 at 14:10












          $begingroup$
          @user408858 Right
          $endgroup$
          – V. Vancak
          Dec 1 '18 at 14:10




          $begingroup$
          @user408858 Right
          $endgroup$
          – V. Vancak
          Dec 1 '18 at 14:10












          $begingroup$
          So I get $Var[sum_{i=1}^n (X_i-mu_0)^2]= 2nsigma^4$ and $Var[ sum_{i=1}^n(X_i-overline{X})^2]=2(n-1)sigma^4$, right?
          $endgroup$
          – user408858
          Dec 1 '18 at 14:13






          $begingroup$
          So I get $Var[sum_{i=1}^n (X_i-mu_0)^2]= 2nsigma^4$ and $Var[ sum_{i=1}^n(X_i-overline{X})^2]=2(n-1)sigma^4$, right?
          $endgroup$
          – user408858
          Dec 1 '18 at 14:13














          $begingroup$
          Yup, that is right
          $endgroup$
          – V. Vancak
          Dec 1 '18 at 14:18




          $begingroup$
          Yup, that is right
          $endgroup$
          – V. Vancak
          Dec 1 '18 at 14:18


















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