Does a sequence of random variables constructed in a certain manner converge in distribution to a Gaussian?












1












$begingroup$


Let ${X_n}_{n in mathbb{N}}$ be a sequence of of IID random variables taken for simplicity with mean zero and variance one.



The Central Limit Theorem give us that
$$
frac{X_1 + dots + X_n}{sqrt{n}} xrightarrow {d} Nleft(0,1right)
$$



If one constructs a new sequence ${Y_n}_{n in mathbb{N}}$ from the first one given by the square of the sum of two consecutive terms, i.e.
$$Y_1 = (X_1 + X_2)^2, Y_2 = (X_2 + X_3)^2, dots , Y_n = (X_n + X_{n+1})^2 $$



do there exist two sequences ${mu_n}_{n in mathbb{N}}$ and ${sigma_n}_{n in mathbb{N}}$ s.t.



$$frac{1}{sigma_i^2} sum_{i=1}^n ( Y_i - mu_i) rightarrow N(0,1) $$



I was thinking of using some Lyapunov type central limit theorem to prove this but there is an obvious (weak) dependence in the sequence. Is it possible to show this or is it not true?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is the power for $Y_n$ a typo? Do you get anything from splitting up $sum_i Y_i$ in two sums (one for odd and one for even $i$)?
    $endgroup$
    – LinAlg
    Dec 8 '18 at 16:12










  • $begingroup$
    @LinAlg yes it was a typo, thanks! How would your second suggestion work?
    $endgroup$
    – Monolite
    Dec 8 '18 at 19:00










  • $begingroup$
    There are a lot of central limit theorems for sequences with different properties. Your $Y$ is m-dependent and strictly stationary. So the convergence you want to prove does indeed occur.
    $endgroup$
    – Calculon
    Dec 8 '18 at 19:04










  • $begingroup$
    @Calculon great, thanks! would you know a reference for the theorem you are citing? moreover would you know of maybe some review paper that collects the various central limit type theorems?
    $endgroup$
    – Monolite
    Dec 8 '18 at 19:56










  • $begingroup$
    Googling "central limit theorem for m-dependent random variables" yields a lot of results but I don't have the journal subscriptions required to view them.
    $endgroup$
    – Calculon
    Dec 9 '18 at 16:27
















1












$begingroup$


Let ${X_n}_{n in mathbb{N}}$ be a sequence of of IID random variables taken for simplicity with mean zero and variance one.



The Central Limit Theorem give us that
$$
frac{X_1 + dots + X_n}{sqrt{n}} xrightarrow {d} Nleft(0,1right)
$$



If one constructs a new sequence ${Y_n}_{n in mathbb{N}}$ from the first one given by the square of the sum of two consecutive terms, i.e.
$$Y_1 = (X_1 + X_2)^2, Y_2 = (X_2 + X_3)^2, dots , Y_n = (X_n + X_{n+1})^2 $$



do there exist two sequences ${mu_n}_{n in mathbb{N}}$ and ${sigma_n}_{n in mathbb{N}}$ s.t.



$$frac{1}{sigma_i^2} sum_{i=1}^n ( Y_i - mu_i) rightarrow N(0,1) $$



I was thinking of using some Lyapunov type central limit theorem to prove this but there is an obvious (weak) dependence in the sequence. Is it possible to show this or is it not true?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is the power for $Y_n$ a typo? Do you get anything from splitting up $sum_i Y_i$ in two sums (one for odd and one for even $i$)?
    $endgroup$
    – LinAlg
    Dec 8 '18 at 16:12










  • $begingroup$
    @LinAlg yes it was a typo, thanks! How would your second suggestion work?
    $endgroup$
    – Monolite
    Dec 8 '18 at 19:00










  • $begingroup$
    There are a lot of central limit theorems for sequences with different properties. Your $Y$ is m-dependent and strictly stationary. So the convergence you want to prove does indeed occur.
    $endgroup$
    – Calculon
    Dec 8 '18 at 19:04










  • $begingroup$
    @Calculon great, thanks! would you know a reference for the theorem you are citing? moreover would you know of maybe some review paper that collects the various central limit type theorems?
    $endgroup$
    – Monolite
    Dec 8 '18 at 19:56










  • $begingroup$
    Googling "central limit theorem for m-dependent random variables" yields a lot of results but I don't have the journal subscriptions required to view them.
    $endgroup$
    – Calculon
    Dec 9 '18 at 16:27














1












1








1





$begingroup$


Let ${X_n}_{n in mathbb{N}}$ be a sequence of of IID random variables taken for simplicity with mean zero and variance one.



The Central Limit Theorem give us that
$$
frac{X_1 + dots + X_n}{sqrt{n}} xrightarrow {d} Nleft(0,1right)
$$



If one constructs a new sequence ${Y_n}_{n in mathbb{N}}$ from the first one given by the square of the sum of two consecutive terms, i.e.
$$Y_1 = (X_1 + X_2)^2, Y_2 = (X_2 + X_3)^2, dots , Y_n = (X_n + X_{n+1})^2 $$



do there exist two sequences ${mu_n}_{n in mathbb{N}}$ and ${sigma_n}_{n in mathbb{N}}$ s.t.



$$frac{1}{sigma_i^2} sum_{i=1}^n ( Y_i - mu_i) rightarrow N(0,1) $$



I was thinking of using some Lyapunov type central limit theorem to prove this but there is an obvious (weak) dependence in the sequence. Is it possible to show this or is it not true?










share|cite|improve this question











$endgroup$




Let ${X_n}_{n in mathbb{N}}$ be a sequence of of IID random variables taken for simplicity with mean zero and variance one.



The Central Limit Theorem give us that
$$
frac{X_1 + dots + X_n}{sqrt{n}} xrightarrow {d} Nleft(0,1right)
$$



If one constructs a new sequence ${Y_n}_{n in mathbb{N}}$ from the first one given by the square of the sum of two consecutive terms, i.e.
$$Y_1 = (X_1 + X_2)^2, Y_2 = (X_2 + X_3)^2, dots , Y_n = (X_n + X_{n+1})^2 $$



do there exist two sequences ${mu_n}_{n in mathbb{N}}$ and ${sigma_n}_{n in mathbb{N}}$ s.t.



$$frac{1}{sigma_i^2} sum_{i=1}^n ( Y_i - mu_i) rightarrow N(0,1) $$



I was thinking of using some Lyapunov type central limit theorem to prove this but there is an obvious (weak) dependence in the sequence. Is it possible to show this or is it not true?







probability 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 Dec 8 '18 at 18:59







Monolite

















asked Dec 8 '18 at 16:06









MonoliteMonolite

1,5502926




1,5502926








  • 1




    $begingroup$
    Is the power for $Y_n$ a typo? Do you get anything from splitting up $sum_i Y_i$ in two sums (one for odd and one for even $i$)?
    $endgroup$
    – LinAlg
    Dec 8 '18 at 16:12










  • $begingroup$
    @LinAlg yes it was a typo, thanks! How would your second suggestion work?
    $endgroup$
    – Monolite
    Dec 8 '18 at 19:00










  • $begingroup$
    There are a lot of central limit theorems for sequences with different properties. Your $Y$ is m-dependent and strictly stationary. So the convergence you want to prove does indeed occur.
    $endgroup$
    – Calculon
    Dec 8 '18 at 19:04










  • $begingroup$
    @Calculon great, thanks! would you know a reference for the theorem you are citing? moreover would you know of maybe some review paper that collects the various central limit type theorems?
    $endgroup$
    – Monolite
    Dec 8 '18 at 19:56










  • $begingroup$
    Googling "central limit theorem for m-dependent random variables" yields a lot of results but I don't have the journal subscriptions required to view them.
    $endgroup$
    – Calculon
    Dec 9 '18 at 16:27














  • 1




    $begingroup$
    Is the power for $Y_n$ a typo? Do you get anything from splitting up $sum_i Y_i$ in two sums (one for odd and one for even $i$)?
    $endgroup$
    – LinAlg
    Dec 8 '18 at 16:12










  • $begingroup$
    @LinAlg yes it was a typo, thanks! How would your second suggestion work?
    $endgroup$
    – Monolite
    Dec 8 '18 at 19:00










  • $begingroup$
    There are a lot of central limit theorems for sequences with different properties. Your $Y$ is m-dependent and strictly stationary. So the convergence you want to prove does indeed occur.
    $endgroup$
    – Calculon
    Dec 8 '18 at 19:04










  • $begingroup$
    @Calculon great, thanks! would you know a reference for the theorem you are citing? moreover would you know of maybe some review paper that collects the various central limit type theorems?
    $endgroup$
    – Monolite
    Dec 8 '18 at 19:56










  • $begingroup$
    Googling "central limit theorem for m-dependent random variables" yields a lot of results but I don't have the journal subscriptions required to view them.
    $endgroup$
    – Calculon
    Dec 9 '18 at 16:27








1




1




$begingroup$
Is the power for $Y_n$ a typo? Do you get anything from splitting up $sum_i Y_i$ in two sums (one for odd and one for even $i$)?
$endgroup$
– LinAlg
Dec 8 '18 at 16:12




$begingroup$
Is the power for $Y_n$ a typo? Do you get anything from splitting up $sum_i Y_i$ in two sums (one for odd and one for even $i$)?
$endgroup$
– LinAlg
Dec 8 '18 at 16:12












$begingroup$
@LinAlg yes it was a typo, thanks! How would your second suggestion work?
$endgroup$
– Monolite
Dec 8 '18 at 19:00




$begingroup$
@LinAlg yes it was a typo, thanks! How would your second suggestion work?
$endgroup$
– Monolite
Dec 8 '18 at 19:00












$begingroup$
There are a lot of central limit theorems for sequences with different properties. Your $Y$ is m-dependent and strictly stationary. So the convergence you want to prove does indeed occur.
$endgroup$
– Calculon
Dec 8 '18 at 19:04




$begingroup$
There are a lot of central limit theorems for sequences with different properties. Your $Y$ is m-dependent and strictly stationary. So the convergence you want to prove does indeed occur.
$endgroup$
– Calculon
Dec 8 '18 at 19:04












$begingroup$
@Calculon great, thanks! would you know a reference for the theorem you are citing? moreover would you know of maybe some review paper that collects the various central limit type theorems?
$endgroup$
– Monolite
Dec 8 '18 at 19:56




$begingroup$
@Calculon great, thanks! would you know a reference for the theorem you are citing? moreover would you know of maybe some review paper that collects the various central limit type theorems?
$endgroup$
– Monolite
Dec 8 '18 at 19:56












$begingroup$
Googling "central limit theorem for m-dependent random variables" yields a lot of results but I don't have the journal subscriptions required to view them.
$endgroup$
– Calculon
Dec 9 '18 at 16:27




$begingroup$
Googling "central limit theorem for m-dependent random variables" yields a lot of results but I don't have the journal subscriptions required to view them.
$endgroup$
– Calculon
Dec 9 '18 at 16:27










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $left(X_iright)_{igeqslant 1}$ be an i.i.d. sequence and let $fcolon mathbb R^2to mathbb R$ be a function such that such that the random variable $Y_i:=f(X_i,X_{i+1})$ is centered and square integrable.



Let $n$ be a fixed integer and $qinleft{1,dots,nright}$. We write
begin{align}
sum_{i=1}^nY_i&= sum_{i=1}^{qleftlfloor frac nqrightrfloor}Y_i+
sum_{qleftlfloor frac nqrightrfloor}^nY_i\
&= sum_{k=1}^{leftlfloor frac nqrightrfloor}
sum_{i=(k-1)q+1}^{kq}Y_i+sum_{qleftlfloor frac nqrightrfloor}^nY_i\
&= sum_{k=1}^{leftlfloor frac nqrightrfloor}
sum_{i=(k-1)q+2}^{kq}Y_i+sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}+sum_{qleftlfloor frac nqrightrfloor}^nY_i.
end{align}

Denoting
$Z^q_k:= sum_{i=(k-1)q+2}^{kq}Y_i$, the sequence $left(Z^q_kright)_{kgeqslant 1}$ is i.i.d. hence we could apply the central limit theorem but the problem is that in order to make the contribution of $n^{-1/2}sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}$ small but we can choose $q$ depending on $n$ and apply the central limit theorem for arrays.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Your specification corresponds to $Y_i = f(X_{i-1},X_i)$ for $i=1,2,3,ldots$, but CLT can be extended to more general form $Y_i = f(cdots,X_{i-2},X_{i-1},widehat{X_i},X_{i+1},ldots)$ ($widehat{X_i}$ denotes the center of the variables) given that $ldots,Y_{i-1},Y_i$ and $Y_{i+N},Y_{i+N+1},ldots$ are "asymptotically independent" as $Nto infty$, which can be rigorously defined in terms of mixing conditions. The given case is among the simplest since $ldots,Y_{i-1}, Y_i$ and $Y_{i+2},Y_{i+3},ldots$ are independent. Proof of this fact requires a bit of knowledge in ergodic and martingale theory. If you are interested, see http://www.stat.yale.edu/~mjk56/MartingaleLimitTheoryAndItsApplication.pdf for martingale limit theory and C. C Heyde, On the central limit theorem for stationary processes, 1974 for ergodic CLT.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Let $left(X_iright)_{igeqslant 1}$ be an i.i.d. sequence and let $fcolon mathbb R^2to mathbb R$ be a function such that such that the random variable $Y_i:=f(X_i,X_{i+1})$ is centered and square integrable.



      Let $n$ be a fixed integer and $qinleft{1,dots,nright}$. We write
      begin{align}
      sum_{i=1}^nY_i&= sum_{i=1}^{qleftlfloor frac nqrightrfloor}Y_i+
      sum_{qleftlfloor frac nqrightrfloor}^nY_i\
      &= sum_{k=1}^{leftlfloor frac nqrightrfloor}
      sum_{i=(k-1)q+1}^{kq}Y_i+sum_{qleftlfloor frac nqrightrfloor}^nY_i\
      &= sum_{k=1}^{leftlfloor frac nqrightrfloor}
      sum_{i=(k-1)q+2}^{kq}Y_i+sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}+sum_{qleftlfloor frac nqrightrfloor}^nY_i.
      end{align}

      Denoting
      $Z^q_k:= sum_{i=(k-1)q+2}^{kq}Y_i$, the sequence $left(Z^q_kright)_{kgeqslant 1}$ is i.i.d. hence we could apply the central limit theorem but the problem is that in order to make the contribution of $n^{-1/2}sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}$ small but we can choose $q$ depending on $n$ and apply the central limit theorem for arrays.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $left(X_iright)_{igeqslant 1}$ be an i.i.d. sequence and let $fcolon mathbb R^2to mathbb R$ be a function such that such that the random variable $Y_i:=f(X_i,X_{i+1})$ is centered and square integrable.



        Let $n$ be a fixed integer and $qinleft{1,dots,nright}$. We write
        begin{align}
        sum_{i=1}^nY_i&= sum_{i=1}^{qleftlfloor frac nqrightrfloor}Y_i+
        sum_{qleftlfloor frac nqrightrfloor}^nY_i\
        &= sum_{k=1}^{leftlfloor frac nqrightrfloor}
        sum_{i=(k-1)q+1}^{kq}Y_i+sum_{qleftlfloor frac nqrightrfloor}^nY_i\
        &= sum_{k=1}^{leftlfloor frac nqrightrfloor}
        sum_{i=(k-1)q+2}^{kq}Y_i+sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}+sum_{qleftlfloor frac nqrightrfloor}^nY_i.
        end{align}

        Denoting
        $Z^q_k:= sum_{i=(k-1)q+2}^{kq}Y_i$, the sequence $left(Z^q_kright)_{kgeqslant 1}$ is i.i.d. hence we could apply the central limit theorem but the problem is that in order to make the contribution of $n^{-1/2}sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}$ small but we can choose $q$ depending on $n$ and apply the central limit theorem for arrays.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $left(X_iright)_{igeqslant 1}$ be an i.i.d. sequence and let $fcolon mathbb R^2to mathbb R$ be a function such that such that the random variable $Y_i:=f(X_i,X_{i+1})$ is centered and square integrable.



          Let $n$ be a fixed integer and $qinleft{1,dots,nright}$. We write
          begin{align}
          sum_{i=1}^nY_i&= sum_{i=1}^{qleftlfloor frac nqrightrfloor}Y_i+
          sum_{qleftlfloor frac nqrightrfloor}^nY_i\
          &= sum_{k=1}^{leftlfloor frac nqrightrfloor}
          sum_{i=(k-1)q+1}^{kq}Y_i+sum_{qleftlfloor frac nqrightrfloor}^nY_i\
          &= sum_{k=1}^{leftlfloor frac nqrightrfloor}
          sum_{i=(k-1)q+2}^{kq}Y_i+sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}+sum_{qleftlfloor frac nqrightrfloor}^nY_i.
          end{align}

          Denoting
          $Z^q_k:= sum_{i=(k-1)q+2}^{kq}Y_i$, the sequence $left(Z^q_kright)_{kgeqslant 1}$ is i.i.d. hence we could apply the central limit theorem but the problem is that in order to make the contribution of $n^{-1/2}sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}$ small but we can choose $q$ depending on $n$ and apply the central limit theorem for arrays.






          share|cite|improve this answer









          $endgroup$



          Let $left(X_iright)_{igeqslant 1}$ be an i.i.d. sequence and let $fcolon mathbb R^2to mathbb R$ be a function such that such that the random variable $Y_i:=f(X_i,X_{i+1})$ is centered and square integrable.



          Let $n$ be a fixed integer and $qinleft{1,dots,nright}$. We write
          begin{align}
          sum_{i=1}^nY_i&= sum_{i=1}^{qleftlfloor frac nqrightrfloor}Y_i+
          sum_{qleftlfloor frac nqrightrfloor}^nY_i\
          &= sum_{k=1}^{leftlfloor frac nqrightrfloor}
          sum_{i=(k-1)q+1}^{kq}Y_i+sum_{qleftlfloor frac nqrightrfloor}^nY_i\
          &= sum_{k=1}^{leftlfloor frac nqrightrfloor}
          sum_{i=(k-1)q+2}^{kq}Y_i+sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}+sum_{qleftlfloor frac nqrightrfloor}^nY_i.
          end{align}

          Denoting
          $Z^q_k:= sum_{i=(k-1)q+2}^{kq}Y_i$, the sequence $left(Z^q_kright)_{kgeqslant 1}$ is i.i.d. hence we could apply the central limit theorem but the problem is that in order to make the contribution of $n^{-1/2}sum_{k=1}^{leftlfloor frac nqrightrfloor}Y_{(k-1)q+1}$ small but we can choose $q$ depending on $n$ and apply the central limit theorem for arrays.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 11:03









          Davide GiraudoDavide Giraudo

          127k17154268




          127k17154268























              1












              $begingroup$

              Your specification corresponds to $Y_i = f(X_{i-1},X_i)$ for $i=1,2,3,ldots$, but CLT can be extended to more general form $Y_i = f(cdots,X_{i-2},X_{i-1},widehat{X_i},X_{i+1},ldots)$ ($widehat{X_i}$ denotes the center of the variables) given that $ldots,Y_{i-1},Y_i$ and $Y_{i+N},Y_{i+N+1},ldots$ are "asymptotically independent" as $Nto infty$, which can be rigorously defined in terms of mixing conditions. The given case is among the simplest since $ldots,Y_{i-1}, Y_i$ and $Y_{i+2},Y_{i+3},ldots$ are independent. Proof of this fact requires a bit of knowledge in ergodic and martingale theory. If you are interested, see http://www.stat.yale.edu/~mjk56/MartingaleLimitTheoryAndItsApplication.pdf for martingale limit theory and C. C Heyde, On the central limit theorem for stationary processes, 1974 for ergodic CLT.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Your specification corresponds to $Y_i = f(X_{i-1},X_i)$ for $i=1,2,3,ldots$, but CLT can be extended to more general form $Y_i = f(cdots,X_{i-2},X_{i-1},widehat{X_i},X_{i+1},ldots)$ ($widehat{X_i}$ denotes the center of the variables) given that $ldots,Y_{i-1},Y_i$ and $Y_{i+N},Y_{i+N+1},ldots$ are "asymptotically independent" as $Nto infty$, which can be rigorously defined in terms of mixing conditions. The given case is among the simplest since $ldots,Y_{i-1}, Y_i$ and $Y_{i+2},Y_{i+3},ldots$ are independent. Proof of this fact requires a bit of knowledge in ergodic and martingale theory. If you are interested, see http://www.stat.yale.edu/~mjk56/MartingaleLimitTheoryAndItsApplication.pdf for martingale limit theory and C. C Heyde, On the central limit theorem for stationary processes, 1974 for ergodic CLT.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Your specification corresponds to $Y_i = f(X_{i-1},X_i)$ for $i=1,2,3,ldots$, but CLT can be extended to more general form $Y_i = f(cdots,X_{i-2},X_{i-1},widehat{X_i},X_{i+1},ldots)$ ($widehat{X_i}$ denotes the center of the variables) given that $ldots,Y_{i-1},Y_i$ and $Y_{i+N},Y_{i+N+1},ldots$ are "asymptotically independent" as $Nto infty$, which can be rigorously defined in terms of mixing conditions. The given case is among the simplest since $ldots,Y_{i-1}, Y_i$ and $Y_{i+2},Y_{i+3},ldots$ are independent. Proof of this fact requires a bit of knowledge in ergodic and martingale theory. If you are interested, see http://www.stat.yale.edu/~mjk56/MartingaleLimitTheoryAndItsApplication.pdf for martingale limit theory and C. C Heyde, On the central limit theorem for stationary processes, 1974 for ergodic CLT.






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                  $endgroup$



                  Your specification corresponds to $Y_i = f(X_{i-1},X_i)$ for $i=1,2,3,ldots$, but CLT can be extended to more general form $Y_i = f(cdots,X_{i-2},X_{i-1},widehat{X_i},X_{i+1},ldots)$ ($widehat{X_i}$ denotes the center of the variables) given that $ldots,Y_{i-1},Y_i$ and $Y_{i+N},Y_{i+N+1},ldots$ are "asymptotically independent" as $Nto infty$, which can be rigorously defined in terms of mixing conditions. The given case is among the simplest since $ldots,Y_{i-1}, Y_i$ and $Y_{i+2},Y_{i+3},ldots$ are independent. Proof of this fact requires a bit of knowledge in ergodic and martingale theory. If you are interested, see http://www.stat.yale.edu/~mjk56/MartingaleLimitTheoryAndItsApplication.pdf for martingale limit theory and C. C Heyde, On the central limit theorem for stationary processes, 1974 for ergodic CLT.







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                  answered Dec 11 '18 at 13:11









                  SongSong

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