Central Limit Theorem for not identical distributed but independent centered random variables with variance...












1












$begingroup$


so let's assume we have independent random variables $X_1,X_2, X_3, ldots$ with
$$mathbb{E}[X_k]=0 mbox{ and } mathbb{Var}[X_k]=sigma_k^2=1 quad forall kinmathbb{N}. $$



We define $$s_n^2:= sum_{k=1}^n mathbb{E}[(X_k-mathbb{E}[X_k])^2]=sum_{k=1}^nsigma_k^2=n.$$



Now we check Lindenberg's condition: So for $varepsilon>0$ we must check
$$limlimits_{nrightarrow infty} frac{1}{s_n^2}sum_{k=1}^n mathbb{E}[(X_k-mathbb{E}[X_k])^2 cdot mathbf{1}_{|X_k| > varepsilon s_n^2} ]
=limlimits_{nrightarrow infty} frac{1}{n} sum_{k=1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ].$$



We know,
$$mathbb{E}[X_k^2 ]=1,$$
so
$$mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]$$
converges quicker than $$frac{1}{n},$$
whch means we find an $Ninmathbb{N}$ such that for every $ngeq N$ we have
$$mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]< frac{1}{n}, $$
which means for every $varepsilon>0$ we have
$$limlimits_{nrightarrow infty} frac{1}{n} sum_{k=1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ] leq limlimits_{nrightarrow infty} frac{N}{n} + frac{1}{n} sum_{k=N+1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]< limlimits_{nrightarrow infty} frac{N}{n} + + frac{1}{n^2 }=0,$$
so we know that the Central Limit Theorem holds.



Is that right?



Edit:
What we really need is uniform integrability of ${ X_k^2: kinmathbb{N}} $$ and that gives us that Lindeberg holds.



Otherwise we can construct $X_j$ with $$P(X_j =-j^2 )=P(X_j =j^2 ) = j^{-4}/2$$ and Lindeberg does not hold!






probability-theory central-limit-theorem







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    so let's assume we have independent random variables $X_1,X_2, X_3, ldots$ with
    $$mathbb{E}[X_k]=0 mbox{ and } mathbb{Var}[X_k]=sigma_k^2=1 quad forall kinmathbb{N}. $$



    We define $$s_n^2:= sum_{k=1}^n mathbb{E}[(X_k-mathbb{E}[X_k])^2]=sum_{k=1}^nsigma_k^2=n.$$



    Now we check Lindenberg's condition: So for $varepsilon>0$ we must check
    $$limlimits_{nrightarrow infty} frac{1}{s_n^2}sum_{k=1}^n mathbb{E}[(X_k-mathbb{E}[X_k])^2 cdot mathbf{1}_{|X_k| > varepsilon s_n^2} ]
    =limlimits_{nrightarrow infty} frac{1}{n} sum_{k=1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ].$$



    We know,
    $$mathbb{E}[X_k^2 ]=1,$$
    so
    $$mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]$$
    converges quicker than $$frac{1}{n},$$
    whch means we find an $Ninmathbb{N}$ such that for every $ngeq N$ we have
    $$mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]< frac{1}{n}, $$
    which means for every $varepsilon>0$ we have
    $$limlimits_{nrightarrow infty} frac{1}{n} sum_{k=1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ] leq limlimits_{nrightarrow infty} frac{N}{n} + frac{1}{n} sum_{k=N+1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]< limlimits_{nrightarrow infty} frac{N}{n} + + frac{1}{n^2 }=0,$$
    so we know that the Central Limit Theorem holds.



    Is that right?



    Edit:
    What we really need is uniform integrability of ${ X_k^2: kinmathbb{N}} $$ and that gives us that Lindeberg holds.



    Otherwise we can construct $X_j$ with $$P(X_j =-j^2 )=P(X_j =j^2 ) = j^{-4}/2$$ and Lindeberg does not hold!






    probability-theory central-limit-theorem







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      so let's assume we have independent random variables $X_1,X_2, X_3, ldots$ with
      $$mathbb{E}[X_k]=0 mbox{ and } mathbb{Var}[X_k]=sigma_k^2=1 quad forall kinmathbb{N}. $$



      We define $$s_n^2:= sum_{k=1}^n mathbb{E}[(X_k-mathbb{E}[X_k])^2]=sum_{k=1}^nsigma_k^2=n.$$



      Now we check Lindenberg's condition: So for $varepsilon>0$ we must check
      $$limlimits_{nrightarrow infty} frac{1}{s_n^2}sum_{k=1}^n mathbb{E}[(X_k-mathbb{E}[X_k])^2 cdot mathbf{1}_{|X_k| > varepsilon s_n^2} ]
      =limlimits_{nrightarrow infty} frac{1}{n} sum_{k=1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ].$$



      We know,
      $$mathbb{E}[X_k^2 ]=1,$$
      so
      $$mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]$$
      converges quicker than $$frac{1}{n},$$
      whch means we find an $Ninmathbb{N}$ such that for every $ngeq N$ we have
      $$mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]< frac{1}{n}, $$
      which means for every $varepsilon>0$ we have
      $$limlimits_{nrightarrow infty} frac{1}{n} sum_{k=1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ] leq limlimits_{nrightarrow infty} frac{N}{n} + frac{1}{n} sum_{k=N+1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]< limlimits_{nrightarrow infty} frac{N}{n} + + frac{1}{n^2 }=0,$$
      so we know that the Central Limit Theorem holds.



      Is that right?



      Edit:
      What we really need is uniform integrability of ${ X_k^2: kinmathbb{N}} $$ and that gives us that Lindeberg holds.



      Otherwise we can construct $X_j$ with $$P(X_j =-j^2 )=P(X_j =j^2 ) = j^{-4}/2$$ and Lindeberg does not hold!






      probability-theory central-limit-theorem







      share|cite|improve this question











      $endgroup$




      so let's assume we have independent random variables $X_1,X_2, X_3, ldots$ with
      $$mathbb{E}[X_k]=0 mbox{ and } mathbb{Var}[X_k]=sigma_k^2=1 quad forall kinmathbb{N}. $$



      We define $$s_n^2:= sum_{k=1}^n mathbb{E}[(X_k-mathbb{E}[X_k])^2]=sum_{k=1}^nsigma_k^2=n.$$



      Now we check Lindenberg's condition: So for $varepsilon>0$ we must check
      $$limlimits_{nrightarrow infty} frac{1}{s_n^2}sum_{k=1}^n mathbb{E}[(X_k-mathbb{E}[X_k])^2 cdot mathbf{1}_{|X_k| > varepsilon s_n^2} ]
      =limlimits_{nrightarrow infty} frac{1}{n} sum_{k=1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ].$$



      We know,
      $$mathbb{E}[X_k^2 ]=1,$$
      so
      $$mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]$$
      converges quicker than $$frac{1}{n},$$
      whch means we find an $Ninmathbb{N}$ such that for every $ngeq N$ we have
      $$mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]< frac{1}{n}, $$
      which means for every $varepsilon>0$ we have
      $$limlimits_{nrightarrow infty} frac{1}{n} sum_{k=1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ] leq limlimits_{nrightarrow infty} frac{N}{n} + frac{1}{n} sum_{k=N+1}^n mathbb{E}[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} ]< limlimits_{nrightarrow infty} frac{N}{n} + + frac{1}{n^2 }=0,$$
      so we know that the Central Limit Theorem holds.



      Is that right?



      Edit:
      What we really need is uniform integrability of ${ X_k^2: kinmathbb{N}} $$ and that gives us that Lindeberg holds.



      Otherwise we can construct $X_j$ with $$P(X_j =-j^2 )=P(X_j =j^2 ) = j^{-4}/2$$ and Lindeberg does not hold!






      probability-theory central-limit-theorem




      probability-theory central-limit-theorem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 4 at 16:19









      Martin Sleziak

      44.9k10122277




      44.9k10122277










      asked Dec 11 '18 at 16:34









      cptflintcptflint

      188




      188






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          It may be not true that $mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} right]$ is of order $1/n$. For example, if $X_1$ is such that its tail $Pr(leftlvert X_1rightrvert >t)sim t^{-2}(log t)^{-2}$, then $mathbb{E}[X_1^2 cdot mathbf{1}_{|X_1| > varepsilon n} ]$ is of order $1/log n$.



          Note that it would suffices to have uniform integrability, that is,
          $$
          lim_{Rto +infty}sup_{kgeqslant 1}mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > R} right]=0.
          $$



          Otherwise, the are counter-examples.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think we have uniform integrability, because $sigma_k^2=1$, but we haven't unform integrability of the family $X_k^2$, right?
            $endgroup$
            – cptflint
            Dec 11 '18 at 23:24










          • $begingroup$
            We indeed have integrability of each $X_k$ but in general not uniform integrability, unless you have other assumptions.
            $endgroup$
            – Davide Giraudo
            Dec 12 '18 at 9:44










          • $begingroup$
            I think we need $exists p>2, C>0: left[ forall k inN: EW[|X_k|^p ] leq C right]$, right? Then I have unfirom integrability of $X_k^2$, which means that we can use Lindeberg.
            $endgroup$
            – cptflint
            Dec 12 '18 at 9:58












          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%2f3035474%2fcentral-limit-theorem-for-not-identical-distributed-but-independent-centered-ran%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












          $begingroup$

          It may be not true that $mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} right]$ is of order $1/n$. For example, if $X_1$ is such that its tail $Pr(leftlvert X_1rightrvert >t)sim t^{-2}(log t)^{-2}$, then $mathbb{E}[X_1^2 cdot mathbf{1}_{|X_1| > varepsilon n} ]$ is of order $1/log n$.



          Note that it would suffices to have uniform integrability, that is,
          $$
          lim_{Rto +infty}sup_{kgeqslant 1}mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > R} right]=0.
          $$



          Otherwise, the are counter-examples.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think we have uniform integrability, because $sigma_k^2=1$, but we haven't unform integrability of the family $X_k^2$, right?
            $endgroup$
            – cptflint
            Dec 11 '18 at 23:24










          • $begingroup$
            We indeed have integrability of each $X_k$ but in general not uniform integrability, unless you have other assumptions.
            $endgroup$
            – Davide Giraudo
            Dec 12 '18 at 9:44










          • $begingroup$
            I think we need $exists p>2, C>0: left[ forall k inN: EW[|X_k|^p ] leq C right]$, right? Then I have unfirom integrability of $X_k^2$, which means that we can use Lindeberg.
            $endgroup$
            – cptflint
            Dec 12 '18 at 9:58
















          1












          $begingroup$

          It may be not true that $mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} right]$ is of order $1/n$. For example, if $X_1$ is such that its tail $Pr(leftlvert X_1rightrvert >t)sim t^{-2}(log t)^{-2}$, then $mathbb{E}[X_1^2 cdot mathbf{1}_{|X_1| > varepsilon n} ]$ is of order $1/log n$.



          Note that it would suffices to have uniform integrability, that is,
          $$
          lim_{Rto +infty}sup_{kgeqslant 1}mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > R} right]=0.
          $$



          Otherwise, the are counter-examples.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think we have uniform integrability, because $sigma_k^2=1$, but we haven't unform integrability of the family $X_k^2$, right?
            $endgroup$
            – cptflint
            Dec 11 '18 at 23:24










          • $begingroup$
            We indeed have integrability of each $X_k$ but in general not uniform integrability, unless you have other assumptions.
            $endgroup$
            – Davide Giraudo
            Dec 12 '18 at 9:44










          • $begingroup$
            I think we need $exists p>2, C>0: left[ forall k inN: EW[|X_k|^p ] leq C right]$, right? Then I have unfirom integrability of $X_k^2$, which means that we can use Lindeberg.
            $endgroup$
            – cptflint
            Dec 12 '18 at 9:58














          1












          1








          1





          $begingroup$

          It may be not true that $mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} right]$ is of order $1/n$. For example, if $X_1$ is such that its tail $Pr(leftlvert X_1rightrvert >t)sim t^{-2}(log t)^{-2}$, then $mathbb{E}[X_1^2 cdot mathbf{1}_{|X_1| > varepsilon n} ]$ is of order $1/log n$.



          Note that it would suffices to have uniform integrability, that is,
          $$
          lim_{Rto +infty}sup_{kgeqslant 1}mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > R} right]=0.
          $$



          Otherwise, the are counter-examples.






          share|cite|improve this answer











          $endgroup$



          It may be not true that $mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > varepsilon n} right]$ is of order $1/n$. For example, if $X_1$ is such that its tail $Pr(leftlvert X_1rightrvert >t)sim t^{-2}(log t)^{-2}$, then $mathbb{E}[X_1^2 cdot mathbf{1}_{|X_1| > varepsilon n} ]$ is of order $1/log n$.



          Note that it would suffices to have uniform integrability, that is,
          $$
          lim_{Rto +infty}sup_{kgeqslant 1}mathbb{E}left[X_k^2 cdot mathbf{1}_{|X_k| > R} right]=0.
          $$



          Otherwise, the are counter-examples.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 11:08

























          answered Dec 11 '18 at 19:22









          Davide GiraudoDavide Giraudo

          128k17155268




          128k17155268












          • $begingroup$
            I think we have uniform integrability, because $sigma_k^2=1$, but we haven't unform integrability of the family $X_k^2$, right?
            $endgroup$
            – cptflint
            Dec 11 '18 at 23:24










          • $begingroup$
            We indeed have integrability of each $X_k$ but in general not uniform integrability, unless you have other assumptions.
            $endgroup$
            – Davide Giraudo
            Dec 12 '18 at 9:44










          • $begingroup$
            I think we need $exists p>2, C>0: left[ forall k inN: EW[|X_k|^p ] leq C right]$, right? Then I have unfirom integrability of $X_k^2$, which means that we can use Lindeberg.
            $endgroup$
            – cptflint
            Dec 12 '18 at 9:58


















          • $begingroup$
            I think we have uniform integrability, because $sigma_k^2=1$, but we haven't unform integrability of the family $X_k^2$, right?
            $endgroup$
            – cptflint
            Dec 11 '18 at 23:24










          • $begingroup$
            We indeed have integrability of each $X_k$ but in general not uniform integrability, unless you have other assumptions.
            $endgroup$
            – Davide Giraudo
            Dec 12 '18 at 9:44










          • $begingroup$
            I think we need $exists p>2, C>0: left[ forall k inN: EW[|X_k|^p ] leq C right]$, right? Then I have unfirom integrability of $X_k^2$, which means that we can use Lindeberg.
            $endgroup$
            – cptflint
            Dec 12 '18 at 9:58
















          $begingroup$
          I think we have uniform integrability, because $sigma_k^2=1$, but we haven't unform integrability of the family $X_k^2$, right?
          $endgroup$
          – cptflint
          Dec 11 '18 at 23:24




          $begingroup$
          I think we have uniform integrability, because $sigma_k^2=1$, but we haven't unform integrability of the family $X_k^2$, right?
          $endgroup$
          – cptflint
          Dec 11 '18 at 23:24












          $begingroup$
          We indeed have integrability of each $X_k$ but in general not uniform integrability, unless you have other assumptions.
          $endgroup$
          – Davide Giraudo
          Dec 12 '18 at 9:44




          $begingroup$
          We indeed have integrability of each $X_k$ but in general not uniform integrability, unless you have other assumptions.
          $endgroup$
          – Davide Giraudo
          Dec 12 '18 at 9:44












          $begingroup$
          I think we need $exists p>2, C>0: left[ forall k inN: EW[|X_k|^p ] leq C right]$, right? Then I have unfirom integrability of $X_k^2$, which means that we can use Lindeberg.
          $endgroup$
          – cptflint
          Dec 12 '18 at 9:58




          $begingroup$
          I think we need $exists p>2, C>0: left[ forall k inN: EW[|X_k|^p ] leq C right]$, right? Then I have unfirom integrability of $X_k^2$, which means that we can use Lindeberg.
          $endgroup$
          – cptflint
          Dec 12 '18 at 9:58


















          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%2f3035474%2fcentral-limit-theorem-for-not-identical-distributed-but-independent-centered-ran%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

          Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

          Image
          2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
          Read more

          ComboBox Display Member on multiple fields

          Image
          2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
          Read more

          Is it possible to collect Nectar points via Trainline?

          Image
          2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
          Read more