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












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

          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021375%2foperatornamevar-sum-i-1n-x-i-mu-02-operatornamevar-sum-i-1n%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021375%2foperatornamevar-sum-i-1n-x-i-mu-02-operatornamevar-sum-i-1n%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          How to change which sound is reproduced for terminal bell?

          Image
          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
          Read more

          Can I use Tabulator js library in my java Spring + Thymeleaf project?

          Image
          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
          Read more

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
          Read more