Almost sure convergence of a martingale sum












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Consider the senquence of iid r.v. $(Y_k)_{kgeq1}$ such that $mathbb{P}(Y_k=1)=mathbb{P}(Y_k=-1)=frac{1}{2}$ and then consider the process $X=(X_k)_{ngeq1}$ such that $X_n=sum_{k=1}^nfrac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?






probability martingales almost-everywhere







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    1












    $begingroup$


    Consider the senquence of iid r.v. $(Y_k)_{kgeq1}$ such that $mathbb{P}(Y_k=1)=mathbb{P}(Y_k=-1)=frac{1}{2}$ and then consider the process $X=(X_k)_{ngeq1}$ such that $X_n=sum_{k=1}^nfrac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?






    probability martingales almost-everywhere







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Consider the senquence of iid r.v. $(Y_k)_{kgeq1}$ such that $mathbb{P}(Y_k=1)=mathbb{P}(Y_k=-1)=frac{1}{2}$ and then consider the process $X=(X_k)_{ngeq1}$ such that $X_n=sum_{k=1}^nfrac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?






      probability martingales almost-everywhere







      share|cite|improve this question











      $endgroup$




      Consider the senquence of iid r.v. $(Y_k)_{kgeq1}$ such that $mathbb{P}(Y_k=1)=mathbb{P}(Y_k=-1)=frac{1}{2}$ and then consider the process $X=(X_k)_{ngeq1}$ such that $X_n=sum_{k=1}^nfrac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?






      probability martingales almost-everywhere




      probability martingales almost-everywhere






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 12:21







      StabiloBoss

















      asked Jan 7 at 12:12









      StabiloBossStabiloBoss

      1016




      1016






















          3 Answers
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          One can see that
          $$
          mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
          $$
          hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.






          share|cite|improve this answer









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            1












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            $EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Sorry I forgot to divide se term in the sum please check the edited question
              $endgroup$
              – StabiloBoss
              Jan 7 at 12:22










            • $begingroup$
              @StabiloBoss I have answered the new version of your question also.
              $endgroup$
              – Kavi Rama Murthy
              Jan 7 at 12:28



















            0












            $begingroup$

            For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.



            Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.






            share|cite|improve this answer











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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              One can see that
              $$
              mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
              $$
              hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                One can see that
                $$
                mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
                $$
                hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  One can see that
                  $$
                  mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
                  $$
                  hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.






                  share|cite|improve this answer









                  $endgroup$



                  One can see that
                  $$
                  mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
                  $$
                  hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 7 at 12:28









                  Davide GiraudoDavide Giraudo

                  128k17156268




                  128k17156268























                      1












                      $begingroup$

                      $EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Sorry I forgot to divide se term in the sum please check the edited question
                        $endgroup$
                        – StabiloBoss
                        Jan 7 at 12:22










                      • $begingroup$
                        @StabiloBoss I have answered the new version of your question also.
                        $endgroup$
                        – Kavi Rama Murthy
                        Jan 7 at 12:28
















                      1












                      $begingroup$

                      $EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Sorry I forgot to divide se term in the sum please check the edited question
                        $endgroup$
                        – StabiloBoss
                        Jan 7 at 12:22










                      • $begingroup$
                        @StabiloBoss I have answered the new version of your question also.
                        $endgroup$
                        – Kavi Rama Murthy
                        Jan 7 at 12:28














                      1












                      1








                      1





                      $begingroup$

                      $EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.






                      share|cite|improve this answer











                      $endgroup$



                      $EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 7 at 12:33

























                      answered Jan 7 at 12:15









                      Kavi Rama MurthyKavi Rama Murthy

                      74.6k53270




                      74.6k53270












                      • $begingroup$
                        Sorry I forgot to divide se term in the sum please check the edited question
                        $endgroup$
                        – StabiloBoss
                        Jan 7 at 12:22










                      • $begingroup$
                        @StabiloBoss I have answered the new version of your question also.
                        $endgroup$
                        – Kavi Rama Murthy
                        Jan 7 at 12:28


















                      • $begingroup$
                        Sorry I forgot to divide se term in the sum please check the edited question
                        $endgroup$
                        – StabiloBoss
                        Jan 7 at 12:22










                      • $begingroup$
                        @StabiloBoss I have answered the new version of your question also.
                        $endgroup$
                        – Kavi Rama Murthy
                        Jan 7 at 12:28
















                      $begingroup$
                      Sorry I forgot to divide se term in the sum please check the edited question
                      $endgroup$
                      – StabiloBoss
                      Jan 7 at 12:22




                      $begingroup$
                      Sorry I forgot to divide se term in the sum please check the edited question
                      $endgroup$
                      – StabiloBoss
                      Jan 7 at 12:22












                      $begingroup$
                      @StabiloBoss I have answered the new version of your question also.
                      $endgroup$
                      – Kavi Rama Murthy
                      Jan 7 at 12:28




                      $begingroup$
                      @StabiloBoss I have answered the new version of your question also.
                      $endgroup$
                      – Kavi Rama Murthy
                      Jan 7 at 12:28











                      0












                      $begingroup$

                      For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.



                      Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.



                        Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.



                          Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.






                          share|cite|improve this answer











                          $endgroup$



                          For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.



                          Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 7 at 12:29

























                          answered Jan 7 at 12:19









                          MindlackMindlack

                          4,910211




                          4,910211






























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