Gaussian variable on $l^2$ (Exercise 3.5 from Hairer's lecture notes).












1












$begingroup$


Let's take a sequence ${ a_n }_{ n in mathbb{N} }$ in $l^2$, in other words assume that $sum_{i = 0}^{infty} a_i^2 < infty$.
If ${ xi_n }_{ n in mathbb{N} }$ are i.i.d. $N(0,1)$, we know from Martingale theory that $sum_{i = 0}^{n} a_i xi$ converges a.s. Said that, I'd like to understand better the behavior of the squared sum.



What can we say about the sequence $ S_n = sum_{i = 0}^{n} a_i^2 xi_i^2$?



How can we prove or disprove its a.s. convergence?



The motivation is given from exercise 3.5, page 8, from these lecture notes (http://hairer.org/notes/SPDEs.pdf). If I understood correctly, we are basically asked to prove that $S_n$ converges almost surely, too, but my attempts failed. I tried with other Martingale strategies and direct inequalities, but I suspect the statement to be false and that I misunderstood the exercise. Thanks in advance.






sequences-and-series probability-theory convergence normal-distribution







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let's take a sequence ${ a_n }_{ n in mathbb{N} }$ in $l^2$, in other words assume that $sum_{i = 0}^{infty} a_i^2 < infty$.
    If ${ xi_n }_{ n in mathbb{N} }$ are i.i.d. $N(0,1)$, we know from Martingale theory that $sum_{i = 0}^{n} a_i xi$ converges a.s. Said that, I'd like to understand better the behavior of the squared sum.



    What can we say about the sequence $ S_n = sum_{i = 0}^{n} a_i^2 xi_i^2$?



    How can we prove or disprove its a.s. convergence?



    The motivation is given from exercise 3.5, page 8, from these lecture notes (http://hairer.org/notes/SPDEs.pdf). If I understood correctly, we are basically asked to prove that $S_n$ converges almost surely, too, but my attempts failed. I tried with other Martingale strategies and direct inequalities, but I suspect the statement to be false and that I misunderstood the exercise. Thanks in advance.






    sequences-and-series probability-theory convergence normal-distribution







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let's take a sequence ${ a_n }_{ n in mathbb{N} }$ in $l^2$, in other words assume that $sum_{i = 0}^{infty} a_i^2 < infty$.
      If ${ xi_n }_{ n in mathbb{N} }$ are i.i.d. $N(0,1)$, we know from Martingale theory that $sum_{i = 0}^{n} a_i xi$ converges a.s. Said that, I'd like to understand better the behavior of the squared sum.



      What can we say about the sequence $ S_n = sum_{i = 0}^{n} a_i^2 xi_i^2$?



      How can we prove or disprove its a.s. convergence?



      The motivation is given from exercise 3.5, page 8, from these lecture notes (http://hairer.org/notes/SPDEs.pdf). If I understood correctly, we are basically asked to prove that $S_n$ converges almost surely, too, but my attempts failed. I tried with other Martingale strategies and direct inequalities, but I suspect the statement to be false and that I misunderstood the exercise. Thanks in advance.






      sequences-and-series probability-theory convergence normal-distribution







      share|cite|improve this question











      $endgroup$




      Let's take a sequence ${ a_n }_{ n in mathbb{N} }$ in $l^2$, in other words assume that $sum_{i = 0}^{infty} a_i^2 < infty$.
      If ${ xi_n }_{ n in mathbb{N} }$ are i.i.d. $N(0,1)$, we know from Martingale theory that $sum_{i = 0}^{n} a_i xi$ converges a.s. Said that, I'd like to understand better the behavior of the squared sum.



      What can we say about the sequence $ S_n = sum_{i = 0}^{n} a_i^2 xi_i^2$?



      How can we prove or disprove its a.s. convergence?



      The motivation is given from exercise 3.5, page 8, from these lecture notes (http://hairer.org/notes/SPDEs.pdf). If I understood correctly, we are basically asked to prove that $S_n$ converges almost surely, too, but my attempts failed. I tried with other Martingale strategies and direct inequalities, but I suspect the statement to be false and that I misunderstood the exercise. Thanks in advance.






      sequences-and-series probability-theory convergence normal-distribution




      sequences-and-series probability-theory convergence normal-distribution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 14:56









      Davide Giraudo

      128k17156268




      128k17156268










      asked Dec 14 '18 at 20:32









      user233650user233650

      48629




      48629






















          1 Answer
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          active

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          2












          $begingroup$

          Let $S_n:=sum_{i=0}^na_i^2xi_i^2$. Then for $m,ngeqslant 0$,
          $$
          mathbb Eleftlvert S_{m+n}-S_nrightrvert=sum_{i=n+1}^{n+m}a_i^2mathbb Eleft[xi_i^2right]=sum_{i=n+1}^{n+m}a_i^2
          $$
          and consequently, the sequence $left(S_nright)_{ngeqslant 1}$ is Cauchy in $mathbb L^1$. This sequence converges to a random variable $S$ in probability, and since $left(S_nright)_{ngeqslant 1}$ is non-decreasing, it also converges almost surely.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Hello Davide, thanks for the suggestion. Unfortunately, $L^1$ convergence is not enough for solving the exercise. Rather, I am required to prove a.s. convergence for the full sequence $S_n$.
            $endgroup$
            – user233650
            Dec 15 '18 at 7:13










          • $begingroup$
            I have edited in order to address the almost sure convergence
            $endgroup$
            – Davide Giraudo
            Dec 15 '18 at 9:33










          • $begingroup$
            Monotonicity! Yes, that was the piece I was missing. Thanks so much and have a nice weekend!
            $endgroup$
            – user233650
            Dec 15 '18 at 14:54












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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          2












          $begingroup$

          Let $S_n:=sum_{i=0}^na_i^2xi_i^2$. Then for $m,ngeqslant 0$,
          $$
          mathbb Eleftlvert S_{m+n}-S_nrightrvert=sum_{i=n+1}^{n+m}a_i^2mathbb Eleft[xi_i^2right]=sum_{i=n+1}^{n+m}a_i^2
          $$
          and consequently, the sequence $left(S_nright)_{ngeqslant 1}$ is Cauchy in $mathbb L^1$. This sequence converges to a random variable $S$ in probability, and since $left(S_nright)_{ngeqslant 1}$ is non-decreasing, it also converges almost surely.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Hello Davide, thanks for the suggestion. Unfortunately, $L^1$ convergence is not enough for solving the exercise. Rather, I am required to prove a.s. convergence for the full sequence $S_n$.
            $endgroup$
            – user233650
            Dec 15 '18 at 7:13










          • $begingroup$
            I have edited in order to address the almost sure convergence
            $endgroup$
            – Davide Giraudo
            Dec 15 '18 at 9:33










          • $begingroup$
            Monotonicity! Yes, that was the piece I was missing. Thanks so much and have a nice weekend!
            $endgroup$
            – user233650
            Dec 15 '18 at 14:54
















          2












          $begingroup$

          Let $S_n:=sum_{i=0}^na_i^2xi_i^2$. Then for $m,ngeqslant 0$,
          $$
          mathbb Eleftlvert S_{m+n}-S_nrightrvert=sum_{i=n+1}^{n+m}a_i^2mathbb Eleft[xi_i^2right]=sum_{i=n+1}^{n+m}a_i^2
          $$
          and consequently, the sequence $left(S_nright)_{ngeqslant 1}$ is Cauchy in $mathbb L^1$. This sequence converges to a random variable $S$ in probability, and since $left(S_nright)_{ngeqslant 1}$ is non-decreasing, it also converges almost surely.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Hello Davide, thanks for the suggestion. Unfortunately, $L^1$ convergence is not enough for solving the exercise. Rather, I am required to prove a.s. convergence for the full sequence $S_n$.
            $endgroup$
            – user233650
            Dec 15 '18 at 7:13










          • $begingroup$
            I have edited in order to address the almost sure convergence
            $endgroup$
            – Davide Giraudo
            Dec 15 '18 at 9:33










          • $begingroup$
            Monotonicity! Yes, that was the piece I was missing. Thanks so much and have a nice weekend!
            $endgroup$
            – user233650
            Dec 15 '18 at 14:54














          2












          2








          2





          $begingroup$

          Let $S_n:=sum_{i=0}^na_i^2xi_i^2$. Then for $m,ngeqslant 0$,
          $$
          mathbb Eleftlvert S_{m+n}-S_nrightrvert=sum_{i=n+1}^{n+m}a_i^2mathbb Eleft[xi_i^2right]=sum_{i=n+1}^{n+m}a_i^2
          $$
          and consequently, the sequence $left(S_nright)_{ngeqslant 1}$ is Cauchy in $mathbb L^1$. This sequence converges to a random variable $S$ in probability, and since $left(S_nright)_{ngeqslant 1}$ is non-decreasing, it also converges almost surely.






          share|cite|improve this answer











          $endgroup$



          Let $S_n:=sum_{i=0}^na_i^2xi_i^2$. Then for $m,ngeqslant 0$,
          $$
          mathbb Eleftlvert S_{m+n}-S_nrightrvert=sum_{i=n+1}^{n+m}a_i^2mathbb Eleft[xi_i^2right]=sum_{i=n+1}^{n+m}a_i^2
          $$
          and consequently, the sequence $left(S_nright)_{ngeqslant 1}$ is Cauchy in $mathbb L^1$. This sequence converges to a random variable $S$ in probability, and since $left(S_nright)_{ngeqslant 1}$ is non-decreasing, it also converges almost surely.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 '18 at 9:33

























          answered Dec 14 '18 at 21:26









          Davide GiraudoDavide Giraudo

          128k17156268




          128k17156268












          • $begingroup$
            Hello Davide, thanks for the suggestion. Unfortunately, $L^1$ convergence is not enough for solving the exercise. Rather, I am required to prove a.s. convergence for the full sequence $S_n$.
            $endgroup$
            – user233650
            Dec 15 '18 at 7:13










          • $begingroup$
            I have edited in order to address the almost sure convergence
            $endgroup$
            – Davide Giraudo
            Dec 15 '18 at 9:33










          • $begingroup$
            Monotonicity! Yes, that was the piece I was missing. Thanks so much and have a nice weekend!
            $endgroup$
            – user233650
            Dec 15 '18 at 14:54


















          • $begingroup$
            Hello Davide, thanks for the suggestion. Unfortunately, $L^1$ convergence is not enough for solving the exercise. Rather, I am required to prove a.s. convergence for the full sequence $S_n$.
            $endgroup$
            – user233650
            Dec 15 '18 at 7:13










          • $begingroup$
            I have edited in order to address the almost sure convergence
            $endgroup$
            – Davide Giraudo
            Dec 15 '18 at 9:33










          • $begingroup$
            Monotonicity! Yes, that was the piece I was missing. Thanks so much and have a nice weekend!
            $endgroup$
            – user233650
            Dec 15 '18 at 14:54
















          $begingroup$
          Hello Davide, thanks for the suggestion. Unfortunately, $L^1$ convergence is not enough for solving the exercise. Rather, I am required to prove a.s. convergence for the full sequence $S_n$.
          $endgroup$
          – user233650
          Dec 15 '18 at 7:13




          $begingroup$
          Hello Davide, thanks for the suggestion. Unfortunately, $L^1$ convergence is not enough for solving the exercise. Rather, I am required to prove a.s. convergence for the full sequence $S_n$.
          $endgroup$
          – user233650
          Dec 15 '18 at 7:13












          $begingroup$
          I have edited in order to address the almost sure convergence
          $endgroup$
          – Davide Giraudo
          Dec 15 '18 at 9:33




          $begingroup$
          I have edited in order to address the almost sure convergence
          $endgroup$
          – Davide Giraudo
          Dec 15 '18 at 9:33












          $begingroup$
          Monotonicity! Yes, that was the piece I was missing. Thanks so much and have a nice weekend!
          $endgroup$
          – user233650
          Dec 15 '18 at 14:54




          $begingroup$
          Monotonicity! Yes, that was the piece I was missing. Thanks so much and have a nice weekend!
          $endgroup$
          – user233650
          Dec 15 '18 at 14:54


















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