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

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          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





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          1 Answer
          1






          active

          oldest

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          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


















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