Proof verification: Strong Law of Large Numbers Under Fourth Moment Control












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I'm trying to solve the following exercise.




Show that if $(X_n)_{n =1}^infty$ is a sequence of independent real-valued random variables with mean zero and uniformly bounded fourth moment, $n^{-1}sum_{j=1}^n X_j to 0$ almost surely. (Hint: Think about a fourth moment tail bound.)




Here's what I did.
Without loss, by rescaling $X_i$, we may assume that $mathbf{E} X_i^4 leq 1$. Note that this means $mathbf{E} X_i^2 = |X_i|_2^2 leq |X_i|_4^2 = (mathbf{E} X_i^4)^{1/2} leq 1$. Put $S_n = sum_{j=1}^n X_j$, observe that $mathbf{E}S_n^4 leq n + (n^2 - n) = n^2$, simply because the other cross terms vanish due to independence.



Following the hint, notice that by Markov's inequality,
$$
mathbf{P}(|n^{-1}S_n|geq n^{-1/8}) = mathbf{P}(S_n^4 geq n^{3.5}) leq n^{-3.5} mathbf{E}S_n^4 leq n^{-1.5}.
$$
The upshot of this is that with $A_n = {|n^{-1}S_n| geq n^{-1/8}}$, $mathbf{P}(A_n)$ is summable, and hence $mathbf{P}(A_n~mathrm{i.o.}) = 0$ by Borel-Cantelli Lemma 1.



If $omega$ is such that $|n^{-1}S_n(omega)| geq n^{-1/8}$ for finitely many terms, then $limsup_n |n^{-1}S_n(omega)| = 0$, and hence $n^{-1}S_n(omega) to 0$. But the set of such $omega$ is almost sure (indeed, the complement of $A_n~mathrm{i.o.}$), proving the claim.






probability-theory measure-theory proof-verification







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    $begingroup$


    I'm trying to solve the following exercise.




    Show that if $(X_n)_{n =1}^infty$ is a sequence of independent real-valued random variables with mean zero and uniformly bounded fourth moment, $n^{-1}sum_{j=1}^n X_j to 0$ almost surely. (Hint: Think about a fourth moment tail bound.)




    Here's what I did.
    Without loss, by rescaling $X_i$, we may assume that $mathbf{E} X_i^4 leq 1$. Note that this means $mathbf{E} X_i^2 = |X_i|_2^2 leq |X_i|_4^2 = (mathbf{E} X_i^4)^{1/2} leq 1$. Put $S_n = sum_{j=1}^n X_j$, observe that $mathbf{E}S_n^4 leq n + (n^2 - n) = n^2$, simply because the other cross terms vanish due to independence.



    Following the hint, notice that by Markov's inequality,
    $$
    mathbf{P}(|n^{-1}S_n|geq n^{-1/8}) = mathbf{P}(S_n^4 geq n^{3.5}) leq n^{-3.5} mathbf{E}S_n^4 leq n^{-1.5}.
    $$
    The upshot of this is that with $A_n = {|n^{-1}S_n| geq n^{-1/8}}$, $mathbf{P}(A_n)$ is summable, and hence $mathbf{P}(A_n~mathrm{i.o.}) = 0$ by Borel-Cantelli Lemma 1.



    If $omega$ is such that $|n^{-1}S_n(omega)| geq n^{-1/8}$ for finitely many terms, then $limsup_n |n^{-1}S_n(omega)| = 0$, and hence $n^{-1}S_n(omega) to 0$. But the set of such $omega$ is almost sure (indeed, the complement of $A_n~mathrm{i.o.}$), proving the claim.






    probability-theory measure-theory proof-verification







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I'm trying to solve the following exercise.




      Show that if $(X_n)_{n =1}^infty$ is a sequence of independent real-valued random variables with mean zero and uniformly bounded fourth moment, $n^{-1}sum_{j=1}^n X_j to 0$ almost surely. (Hint: Think about a fourth moment tail bound.)




      Here's what I did.
      Without loss, by rescaling $X_i$, we may assume that $mathbf{E} X_i^4 leq 1$. Note that this means $mathbf{E} X_i^2 = |X_i|_2^2 leq |X_i|_4^2 = (mathbf{E} X_i^4)^{1/2} leq 1$. Put $S_n = sum_{j=1}^n X_j$, observe that $mathbf{E}S_n^4 leq n + (n^2 - n) = n^2$, simply because the other cross terms vanish due to independence.



      Following the hint, notice that by Markov's inequality,
      $$
      mathbf{P}(|n^{-1}S_n|geq n^{-1/8}) = mathbf{P}(S_n^4 geq n^{3.5}) leq n^{-3.5} mathbf{E}S_n^4 leq n^{-1.5}.
      $$
      The upshot of this is that with $A_n = {|n^{-1}S_n| geq n^{-1/8}}$, $mathbf{P}(A_n)$ is summable, and hence $mathbf{P}(A_n~mathrm{i.o.}) = 0$ by Borel-Cantelli Lemma 1.



      If $omega$ is such that $|n^{-1}S_n(omega)| geq n^{-1/8}$ for finitely many terms, then $limsup_n |n^{-1}S_n(omega)| = 0$, and hence $n^{-1}S_n(omega) to 0$. But the set of such $omega$ is almost sure (indeed, the complement of $A_n~mathrm{i.o.}$), proving the claim.






      probability-theory measure-theory proof-verification







      share|cite|improve this question











      $endgroup$




      I'm trying to solve the following exercise.




      Show that if $(X_n)_{n =1}^infty$ is a sequence of independent real-valued random variables with mean zero and uniformly bounded fourth moment, $n^{-1}sum_{j=1}^n X_j to 0$ almost surely. (Hint: Think about a fourth moment tail bound.)




      Here's what I did.
      Without loss, by rescaling $X_i$, we may assume that $mathbf{E} X_i^4 leq 1$. Note that this means $mathbf{E} X_i^2 = |X_i|_2^2 leq |X_i|_4^2 = (mathbf{E} X_i^4)^{1/2} leq 1$. Put $S_n = sum_{j=1}^n X_j$, observe that $mathbf{E}S_n^4 leq n + (n^2 - n) = n^2$, simply because the other cross terms vanish due to independence.



      Following the hint, notice that by Markov's inequality,
      $$
      mathbf{P}(|n^{-1}S_n|geq n^{-1/8}) = mathbf{P}(S_n^4 geq n^{3.5}) leq n^{-3.5} mathbf{E}S_n^4 leq n^{-1.5}.
      $$
      The upshot of this is that with $A_n = {|n^{-1}S_n| geq n^{-1/8}}$, $mathbf{P}(A_n)$ is summable, and hence $mathbf{P}(A_n~mathrm{i.o.}) = 0$ by Borel-Cantelli Lemma 1.



      If $omega$ is such that $|n^{-1}S_n(omega)| geq n^{-1/8}$ for finitely many terms, then $limsup_n |n^{-1}S_n(omega)| = 0$, and hence $n^{-1}S_n(omega) to 0$. But the set of such $omega$ is almost sure (indeed, the complement of $A_n~mathrm{i.o.}$), proving the claim.






      probability-theory measure-theory proof-verification




      probability-theory measure-theory proof-verification






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      edited Dec 12 '18 at 0:47







      Drew Brady

















      asked Dec 11 '18 at 23:25









      Drew BradyDrew Brady

      719315




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          This is almost correct, except for the crucial part: it is not necessarily true that the expected value of $S_n^4$ will not be greater than $n$. You fail to take into account the $sim n^2$ terms $mathbb{E}[X_i^2X_j^2]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, @Mindlack, good point. I think, however since $umapsto u^4$ is convex, we have that $(1/n sum_{j=1}^n X_j(omega))^4 leq 1/n sum_{j=1}^n X_j(omega)^4$, in which case we actually have that $mathbf{E} (S_n/n)^4 leq 1$.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:22












          • $begingroup$
            Ah, but this bound is not strong enough to carry forth the argument.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:25










          • $begingroup$
            @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 0:26












          • $begingroup$
            Thanks for the edits. Can you take a look at my revised argument. Tell me if you think the details look right now. I forgot that $L^p$ is monotone on probability spaces!
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:48












          • $begingroup$
            The edited solution works now!
            $endgroup$
            – Mindlack
            Dec 12 '18 at 12:00











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          $begingroup$

          This is almost correct, except for the crucial part: it is not necessarily true that the expected value of $S_n^4$ will not be greater than $n$. You fail to take into account the $sim n^2$ terms $mathbb{E}[X_i^2X_j^2]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, @Mindlack, good point. I think, however since $umapsto u^4$ is convex, we have that $(1/n sum_{j=1}^n X_j(omega))^4 leq 1/n sum_{j=1}^n X_j(omega)^4$, in which case we actually have that $mathbf{E} (S_n/n)^4 leq 1$.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:22












          • $begingroup$
            Ah, but this bound is not strong enough to carry forth the argument.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:25










          • $begingroup$
            @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 0:26












          • $begingroup$
            Thanks for the edits. Can you take a look at my revised argument. Tell me if you think the details look right now. I forgot that $L^p$ is monotone on probability spaces!
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:48












          • $begingroup$
            The edited solution works now!
            $endgroup$
            – Mindlack
            Dec 12 '18 at 12:00
















          2












          $begingroup$

          This is almost correct, except for the crucial part: it is not necessarily true that the expected value of $S_n^4$ will not be greater than $n$. You fail to take into account the $sim n^2$ terms $mathbb{E}[X_i^2X_j^2]$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, @Mindlack, good point. I think, however since $umapsto u^4$ is convex, we have that $(1/n sum_{j=1}^n X_j(omega))^4 leq 1/n sum_{j=1}^n X_j(omega)^4$, in which case we actually have that $mathbf{E} (S_n/n)^4 leq 1$.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:22












          • $begingroup$
            Ah, but this bound is not strong enough to carry forth the argument.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:25










          • $begingroup$
            @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 0:26












          • $begingroup$
            Thanks for the edits. Can you take a look at my revised argument. Tell me if you think the details look right now. I forgot that $L^p$ is monotone on probability spaces!
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:48












          • $begingroup$
            The edited solution works now!
            $endgroup$
            – Mindlack
            Dec 12 '18 at 12:00














          2












          2








          2





          $begingroup$

          This is almost correct, except for the crucial part: it is not necessarily true that the expected value of $S_n^4$ will not be greater than $n$. You fail to take into account the $sim n^2$ terms $mathbb{E}[X_i^2X_j^2]$.






          share|cite|improve this answer









          $endgroup$



          This is almost correct, except for the crucial part: it is not necessarily true that the expected value of $S_n^4$ will not be greater than $n$. You fail to take into account the $sim n^2$ terms $mathbb{E}[X_i^2X_j^2]$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 23:48









          MindlackMindlack

          3,87018




          3,87018












          • $begingroup$
            Ah, @Mindlack, good point. I think, however since $umapsto u^4$ is convex, we have that $(1/n sum_{j=1}^n X_j(omega))^4 leq 1/n sum_{j=1}^n X_j(omega)^4$, in which case we actually have that $mathbf{E} (S_n/n)^4 leq 1$.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:22












          • $begingroup$
            Ah, but this bound is not strong enough to carry forth the argument.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:25










          • $begingroup$
            @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 0:26












          • $begingroup$
            Thanks for the edits. Can you take a look at my revised argument. Tell me if you think the details look right now. I forgot that $L^p$ is monotone on probability spaces!
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:48












          • $begingroup$
            The edited solution works now!
            $endgroup$
            – Mindlack
            Dec 12 '18 at 12:00


















          • $begingroup$
            Ah, @Mindlack, good point. I think, however since $umapsto u^4$ is convex, we have that $(1/n sum_{j=1}^n X_j(omega))^4 leq 1/n sum_{j=1}^n X_j(omega)^4$, in which case we actually have that $mathbf{E} (S_n/n)^4 leq 1$.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:22












          • $begingroup$
            Ah, but this bound is not strong enough to carry forth the argument.
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:25










          • $begingroup$
            @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$.
            $endgroup$
            – Mindlack
            Dec 12 '18 at 0:26












          • $begingroup$
            Thanks for the edits. Can you take a look at my revised argument. Tell me if you think the details look right now. I forgot that $L^p$ is monotone on probability spaces!
            $endgroup$
            – Drew Brady
            Dec 12 '18 at 0:48












          • $begingroup$
            The edited solution works now!
            $endgroup$
            – Mindlack
            Dec 12 '18 at 12:00
















          $begingroup$
          Ah, @Mindlack, good point. I think, however since $umapsto u^4$ is convex, we have that $(1/n sum_{j=1}^n X_j(omega))^4 leq 1/n sum_{j=1}^n X_j(omega)^4$, in which case we actually have that $mathbf{E} (S_n/n)^4 leq 1$.
          $endgroup$
          – Drew Brady
          Dec 12 '18 at 0:22






          $begingroup$
          Ah, @Mindlack, good point. I think, however since $umapsto u^4$ is convex, we have that $(1/n sum_{j=1}^n X_j(omega))^4 leq 1/n sum_{j=1}^n X_j(omega)^4$, in which case we actually have that $mathbf{E} (S_n/n)^4 leq 1$.
          $endgroup$
          – Drew Brady
          Dec 12 '18 at 0:22














          $begingroup$
          Ah, but this bound is not strong enough to carry forth the argument.
          $endgroup$
          – Drew Brady
          Dec 12 '18 at 0:25




          $begingroup$
          Ah, but this bound is not strong enough to carry forth the argument.
          $endgroup$
          – Drew Brady
          Dec 12 '18 at 0:25












          $begingroup$
          @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$.
          $endgroup$
          – Mindlack
          Dec 12 '18 at 0:26






          $begingroup$
          @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$.
          $endgroup$
          – Mindlack
          Dec 12 '18 at 0:26














          $begingroup$
          Thanks for the edits. Can you take a look at my revised argument. Tell me if you think the details look right now. I forgot that $L^p$ is monotone on probability spaces!
          $endgroup$
          – Drew Brady
          Dec 12 '18 at 0:48






          $begingroup$
          Thanks for the edits. Can you take a look at my revised argument. Tell me if you think the details look right now. I forgot that $L^p$ is monotone on probability spaces!
          $endgroup$
          – Drew Brady
          Dec 12 '18 at 0:48














          $begingroup$
          The edited solution works now!
          $endgroup$
          – Mindlack
          Dec 12 '18 at 12:00




          $begingroup$
          The edited solution works now!
          $endgroup$
          – Mindlack
          Dec 12 '18 at 12:00


















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