Convergence of “AA” Subsequence of sequence strongly convergent of order $alpha$

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Let $(Omega, mathcal F, mathbb P)$ be a probability space and consider a sequence $X = (X_n)_nin mathbb N$ of integrable random variables, let $X_inftyin L_1$ and $alpha > 0$.



Say that $X$ is convergent of order $alpha$ to $X_infty$ if there exists a $Cgeq 0$ s.t.



$$|X_n- X_infty|_1 leq Cfrac1n^alpha, forall n geq 1.$$



In this case I will write $X to_alpha X_infty$. The motivation for this comes from the Euler-Maruyama method for numerically solving SDE's.



One can prove some nice properties, e.g.



  • $(X)_nin mathbb N to_alpha X$

  • $X to_alpha X_infty$, $n : mathbb N to mathbb N$ monotone $Rightarrow$ $(X_n(k))_k to X_infty$

  • $X to_alpha X_infty$ and $X'to_alpha X_infty$ $Rightarrow Xcap X'to_alpha X_infty$

where $(Xcap X')_2n = X_n$ and $(Xcap X')_2n-1 = X'_n$.



What about the following one:




Is is true that $Xto_alpha X_infty$ and ($X in A $ eventually $Leftrightarrow X'in A$ eventually) $Rightarrow X'to_alpha X_infty$?




Here $Xin A$ eventually if $X_nin A$ for large $n$.



It would be nice to have (*), but I don't see how this could be proven.



I was thinking of $A_k := X - X_infty$, then $Xin A_k$ eventually, hence $X'in A_k$ eventually.
In other words



$$forall kin mathbb N : exists i geq 0 : forall j geq i : |X_j' - X_infty|_1 leq C frac1k^alpha$$



But it doesn't seem like that's enough to conclude that $X'to_alpha X_infty$.




(*) In this case this subsequential structure could be induced by a convergence structure.







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    up vote
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    down vote

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    Let $(Omega, mathcal F, mathbb P)$ be a probability space and consider a sequence $X = (X_n)_nin mathbb N$ of integrable random variables, let $X_inftyin L_1$ and $alpha > 0$.



    Say that $X$ is convergent of order $alpha$ to $X_infty$ if there exists a $Cgeq 0$ s.t.



    $$|X_n- X_infty|_1 leq Cfrac1n^alpha, forall n geq 1.$$



    In this case I will write $X to_alpha X_infty$. The motivation for this comes from the Euler-Maruyama method for numerically solving SDE's.



    One can prove some nice properties, e.g.



    • $(X)_nin mathbb N to_alpha X$

    • $X to_alpha X_infty$, $n : mathbb N to mathbb N$ monotone $Rightarrow$ $(X_n(k))_k to X_infty$

    • $X to_alpha X_infty$ and $X'to_alpha X_infty$ $Rightarrow Xcap X'to_alpha X_infty$

    where $(Xcap X')_2n = X_n$ and $(Xcap X')_2n-1 = X'_n$.



    What about the following one:




    Is is true that $Xto_alpha X_infty$ and ($X in A $ eventually $Leftrightarrow X'in A$ eventually) $Rightarrow X'to_alpha X_infty$?




    Here $Xin A$ eventually if $X_nin A$ for large $n$.



    It would be nice to have (*), but I don't see how this could be proven.



    I was thinking of $A_k := X - X_infty$, then $Xin A_k$ eventually, hence $X'in A_k$ eventually.
    In other words



    $$forall kin mathbb N : exists i geq 0 : forall j geq i : |X_j' - X_infty|_1 leq C frac1k^alpha$$



    But it doesn't seem like that's enough to conclude that $X'to_alpha X_infty$.




    (*) In this case this subsequential structure could be induced by a convergence structure.







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $(Omega, mathcal F, mathbb P)$ be a probability space and consider a sequence $X = (X_n)_nin mathbb N$ of integrable random variables, let $X_inftyin L_1$ and $alpha > 0$.



      Say that $X$ is convergent of order $alpha$ to $X_infty$ if there exists a $Cgeq 0$ s.t.



      $$|X_n- X_infty|_1 leq Cfrac1n^alpha, forall n geq 1.$$



      In this case I will write $X to_alpha X_infty$. The motivation for this comes from the Euler-Maruyama method for numerically solving SDE's.



      One can prove some nice properties, e.g.



      • $(X)_nin mathbb N to_alpha X$

      • $X to_alpha X_infty$, $n : mathbb N to mathbb N$ monotone $Rightarrow$ $(X_n(k))_k to X_infty$

      • $X to_alpha X_infty$ and $X'to_alpha X_infty$ $Rightarrow Xcap X'to_alpha X_infty$

      where $(Xcap X')_2n = X_n$ and $(Xcap X')_2n-1 = X'_n$.



      What about the following one:




      Is is true that $Xto_alpha X_infty$ and ($X in A $ eventually $Leftrightarrow X'in A$ eventually) $Rightarrow X'to_alpha X_infty$?




      Here $Xin A$ eventually if $X_nin A$ for large $n$.



      It would be nice to have (*), but I don't see how this could be proven.



      I was thinking of $A_k := X - X_infty$, then $Xin A_k$ eventually, hence $X'in A_k$ eventually.
      In other words



      $$forall kin mathbb N : exists i geq 0 : forall j geq i : |X_j' - X_infty|_1 leq C frac1k^alpha$$



      But it doesn't seem like that's enough to conclude that $X'to_alpha X_infty$.




      (*) In this case this subsequential structure could be induced by a convergence structure.







      share|cite|improve this question












      Let $(Omega, mathcal F, mathbb P)$ be a probability space and consider a sequence $X = (X_n)_nin mathbb N$ of integrable random variables, let $X_inftyin L_1$ and $alpha > 0$.



      Say that $X$ is convergent of order $alpha$ to $X_infty$ if there exists a $Cgeq 0$ s.t.



      $$|X_n- X_infty|_1 leq Cfrac1n^alpha, forall n geq 1.$$



      In this case I will write $X to_alpha X_infty$. The motivation for this comes from the Euler-Maruyama method for numerically solving SDE's.



      One can prove some nice properties, e.g.



      • $(X)_nin mathbb N to_alpha X$

      • $X to_alpha X_infty$, $n : mathbb N to mathbb N$ monotone $Rightarrow$ $(X_n(k))_k to X_infty$

      • $X to_alpha X_infty$ and $X'to_alpha X_infty$ $Rightarrow Xcap X'to_alpha X_infty$

      where $(Xcap X')_2n = X_n$ and $(Xcap X')_2n-1 = X'_n$.



      What about the following one:




      Is is true that $Xto_alpha X_infty$ and ($X in A $ eventually $Leftrightarrow X'in A$ eventually) $Rightarrow X'to_alpha X_infty$?




      Here $Xin A$ eventually if $X_nin A$ for large $n$.



      It would be nice to have (*), but I don't see how this could be proven.



      I was thinking of $A_k := X - X_infty$, then $Xin A_k$ eventually, hence $X'in A_k$ eventually.
      In other words



      $$forall kin mathbb N : exists i geq 0 : forall j geq i : |X_j' - X_infty|_1 leq C frac1k^alpha$$



      But it doesn't seem like that's enough to conclude that $X'to_alpha X_infty$.




      (*) In this case this subsequential structure could be induced by a convergence structure.









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 28 at 17:19









      Stefan Perko

      8,31611641




      8,31611641




















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          I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
          $$
          |X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
          $$
          so you can only say $X'to_alpha/2X_infty.$






          share|cite|improve this answer




















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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
            $$
            |X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
            $$
            so you can only say $X'to_alpha/2X_infty.$






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
              $$
              |X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
              $$
              so you can only say $X'to_alpha/2X_infty.$






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
                $$
                |X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
                $$
                so you can only say $X'to_alpha/2X_infty.$






                share|cite|improve this answer












                I think the following map be a counterexample. For any sequence $X$, define $X'$ by $X'_n = X_sqrtn$, where the index is rounded down. Certainly $X'$ has the same eventuality set as $X$. However,
                $$
                |X'_n-X_infty|_1=|X_sqrtn-X_infty|_1le C/(sqrtn)^alpha=C/n^alpha/2
                $$
                so you can only say $X'to_alpha/2X_infty.$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 28 at 19:19









                Mike Earnest

                17.5k11749




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