Find $limlimits_{n to infty}mathbb Pleft(frac1{sqrt n}sumlimits_{i=1}^nX_ile xright)$ if...












1












$begingroup$



Let $(X_{n})_{n}$ be independent random variables such that $mathbb P(X_{n}=n)=mathbb P(X_{n}=-n)=frac{1}{2n^2}$ and $P(X_{n}=0)=1-frac{1}{n^2}$.
Find $lim_{n to infty}mathbb P(frac{1}{sqrt{n}}sum_{i=1}^{n}X_{i}leq x)$.




Ideas:



I think it would be appropriate to use Borel-Cantelli.



Set $A:={X neq 0}:={omegainOmega:X_{n}(omega)neq0, forall n in mathbb N}$ and $A_{n}:={X_{n}neq0}$



Note $sum_{n=1}^{infty}P(A_{n})=sum_{n=1}^{infty}frac{1}{n^2}<infty$



Therefore $P(limsup A_{n})=P(A)=0$



This means that $P(A^{c})=1-P(A)=1$



So $A^c to 0$ a.s.



This then implies stochastic convergence. I am unsure on my selection of $A$ and $A_{n}$ though. Any corrections, guidance or recommendations?






probability-theory borel-cantelli-lemmas







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  • $begingroup$
    Your choice of $A$ is absurd. Next, the conclusion that $P(limsup A_n)=0$ holds. Later on, the assertion that $A^c to 0$ a.s., is absurd. Can you revise these steps?
    $endgroup$
    – Did
    Jan 10 at 15:53


















1












$begingroup$



Let $(X_{n})_{n}$ be independent random variables such that $mathbb P(X_{n}=n)=mathbb P(X_{n}=-n)=frac{1}{2n^2}$ and $P(X_{n}=0)=1-frac{1}{n^2}$.
Find $lim_{n to infty}mathbb P(frac{1}{sqrt{n}}sum_{i=1}^{n}X_{i}leq x)$.




Ideas:



I think it would be appropriate to use Borel-Cantelli.



Set $A:={X neq 0}:={omegainOmega:X_{n}(omega)neq0, forall n in mathbb N}$ and $A_{n}:={X_{n}neq0}$



Note $sum_{n=1}^{infty}P(A_{n})=sum_{n=1}^{infty}frac{1}{n^2}<infty$



Therefore $P(limsup A_{n})=P(A)=0$



This means that $P(A^{c})=1-P(A)=1$



So $A^c to 0$ a.s.



This then implies stochastic convergence. I am unsure on my selection of $A$ and $A_{n}$ though. Any corrections, guidance or recommendations?






probability-theory borel-cantelli-lemmas







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your choice of $A$ is absurd. Next, the conclusion that $P(limsup A_n)=0$ holds. Later on, the assertion that $A^c to 0$ a.s., is absurd. Can you revise these steps?
    $endgroup$
    – Did
    Jan 10 at 15:53
















1












1








1





$begingroup$



Let $(X_{n})_{n}$ be independent random variables such that $mathbb P(X_{n}=n)=mathbb P(X_{n}=-n)=frac{1}{2n^2}$ and $P(X_{n}=0)=1-frac{1}{n^2}$.
Find $lim_{n to infty}mathbb P(frac{1}{sqrt{n}}sum_{i=1}^{n}X_{i}leq x)$.




Ideas:



I think it would be appropriate to use Borel-Cantelli.



Set $A:={X neq 0}:={omegainOmega:X_{n}(omega)neq0, forall n in mathbb N}$ and $A_{n}:={X_{n}neq0}$



Note $sum_{n=1}^{infty}P(A_{n})=sum_{n=1}^{infty}frac{1}{n^2}<infty$



Therefore $P(limsup A_{n})=P(A)=0$



This means that $P(A^{c})=1-P(A)=1$



So $A^c to 0$ a.s.



This then implies stochastic convergence. I am unsure on my selection of $A$ and $A_{n}$ though. Any corrections, guidance or recommendations?






probability-theory borel-cantelli-lemmas







share|cite|improve this question











$endgroup$





Let $(X_{n})_{n}$ be independent random variables such that $mathbb P(X_{n}=n)=mathbb P(X_{n}=-n)=frac{1}{2n^2}$ and $P(X_{n}=0)=1-frac{1}{n^2}$.
Find $lim_{n to infty}mathbb P(frac{1}{sqrt{n}}sum_{i=1}^{n}X_{i}leq x)$.




Ideas:



I think it would be appropriate to use Borel-Cantelli.



Set $A:={X neq 0}:={omegainOmega:X_{n}(omega)neq0, forall n in mathbb N}$ and $A_{n}:={X_{n}neq0}$



Note $sum_{n=1}^{infty}P(A_{n})=sum_{n=1}^{infty}frac{1}{n^2}<infty$



Therefore $P(limsup A_{n})=P(A)=0$



This means that $P(A^{c})=1-P(A)=1$



So $A^c to 0$ a.s.



This then implies stochastic convergence. I am unsure on my selection of $A$ and $A_{n}$ though. Any corrections, guidance or recommendations?






probability-theory borel-cantelli-lemmas




probability-theory borel-cantelli-lemmas






share|cite|improve this question















share|cite|improve this question













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edited Jan 10 at 15:54









Did

249k23228466




249k23228466










asked Jan 10 at 15:50









SABOYSABOY

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  • $begingroup$
    Your choice of $A$ is absurd. Next, the conclusion that $P(limsup A_n)=0$ holds. Later on, the assertion that $A^c to 0$ a.s., is absurd. Can you revise these steps?
    $endgroup$
    – Did
    Jan 10 at 15:53




















  • $begingroup$
    Your choice of $A$ is absurd. Next, the conclusion that $P(limsup A_n)=0$ holds. Later on, the assertion that $A^c to 0$ a.s., is absurd. Can you revise these steps?
    $endgroup$
    – Did
    Jan 10 at 15:53


















$begingroup$
Your choice of $A$ is absurd. Next, the conclusion that $P(limsup A_n)=0$ holds. Later on, the assertion that $A^c to 0$ a.s., is absurd. Can you revise these steps?
$endgroup$
– Did
Jan 10 at 15:53






$begingroup$
Your choice of $A$ is absurd. Next, the conclusion that $P(limsup A_n)=0$ holds. Later on, the assertion that $A^c to 0$ a.s., is absurd. Can you revise these steps?
$endgroup$
– Did
Jan 10 at 15:53












1 Answer
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oldest

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1












$begingroup$

Your idea of using Borel-Cantelli lemma is basically right. Let $$
A={X_n ne 0 text{ for infinitely many }ninmathbb{N}}.
$$ Since $sum_{n=1}^infty P(X_n ne 0)=sum_{n=1}^infty frac{1}{n^2}<infty$, Borel-Cantelli Lemma implies that $P(A) = 0$. So, for almost every $omega inOmega$, there exists $N(omega)inmathbb{N}$ such that$$
nge N(omega) Rightarrow X_n(omega) = 0.
$$ This shows
$$
lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^nX_i(omega) = lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^{N(omega)-1}X_i(omega)=0
$$ for almost every $omega$, hence giving the result
$$
lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile xright) =begin{cases} 1,quad x>0\0,quad x<0end{cases}.
$$ The case $x=0$ is more subtle. Note that by the symmetry of the distribution, we have
$$
2Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ige 0right)=1+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_i=0right).
$$ Therefore,
$$
lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=frac{1}{2}+frac{1}{2}lim_{nto infty} Pleft(sum_{i=1}^nX_i =0right).
$$ My guess is that $lim_{ntoinfty} Pleft(sum_{i=1}^nX_i =0right)=0$, but as of now, I'm not sure.






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    $begingroup$

    Your idea of using Borel-Cantelli lemma is basically right. Let $$
    A={X_n ne 0 text{ for infinitely many }ninmathbb{N}}.
    $$     Since $sum_{n=1}^infty P(X_n ne 0)=sum_{n=1}^infty frac{1}{n^2}<infty$, Borel-Cantelli Lemma implies that $P(A) = 0$. So, for almost every $omega inOmega$, there exists $N(omega)inmathbb{N}$ such that$$
    nge N(omega) Rightarrow X_n(omega) = 0.
    $$     This shows
    $$
    lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^nX_i(omega) = lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^{N(omega)-1}X_i(omega)=0
    $$     for almost every $omega$, hence giving the result
    $$
    lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile xright) =begin{cases} 1,quad x>0\0,quad x<0end{cases}.
    $$     The case $x=0$ is more subtle. Note that by the symmetry of the distribution, we have
    $$
    2Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ige 0right)=1+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_i=0right).
    $$     Therefore,
    $$
    lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=frac{1}{2}+frac{1}{2}lim_{nto infty} Pleft(sum_{i=1}^nX_i =0right).
    $$     My guess is that $lim_{ntoinfty} Pleft(sum_{i=1}^nX_i =0right)=0$, but as of now, I'm not sure.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Your idea of using Borel-Cantelli lemma is basically right. Let $$
      A={X_n ne 0 text{ for infinitely many }ninmathbb{N}}.
      $$       Since $sum_{n=1}^infty P(X_n ne 0)=sum_{n=1}^infty frac{1}{n^2}<infty$, Borel-Cantelli Lemma implies that $P(A) = 0$. So, for almost every $omega inOmega$, there exists $N(omega)inmathbb{N}$ such that$$
      nge N(omega) Rightarrow X_n(omega) = 0.
      $$       This shows
      $$
      lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^nX_i(omega) = lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^{N(omega)-1}X_i(omega)=0
      $$       for almost every $omega$, hence giving the result
      $$
      lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile xright) =begin{cases} 1,quad x>0\0,quad x<0end{cases}.
      $$       The case $x=0$ is more subtle. Note that by the symmetry of the distribution, we have
      $$
      2Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ige 0right)=1+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_i=0right).
      $$       Therefore,
      $$
      lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=frac{1}{2}+frac{1}{2}lim_{nto infty} Pleft(sum_{i=1}^nX_i =0right).
      $$       My guess is that $lim_{ntoinfty} Pleft(sum_{i=1}^nX_i =0right)=0$, but as of now, I'm not sure.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Your idea of using Borel-Cantelli lemma is basically right. Let $$
        A={X_n ne 0 text{ for infinitely many }ninmathbb{N}}.
        $$         Since $sum_{n=1}^infty P(X_n ne 0)=sum_{n=1}^infty frac{1}{n^2}<infty$, Borel-Cantelli Lemma implies that $P(A) = 0$. So, for almost every $omega inOmega$, there exists $N(omega)inmathbb{N}$ such that$$
        nge N(omega) Rightarrow X_n(omega) = 0.
        $$         This shows
        $$
        lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^nX_i(omega) = lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^{N(omega)-1}X_i(omega)=0
        $$         for almost every $omega$, hence giving the result
        $$
        lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile xright) =begin{cases} 1,quad x>0\0,quad x<0end{cases}.
        $$         The case $x=0$ is more subtle. Note that by the symmetry of the distribution, we have
        $$
        2Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ige 0right)=1+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_i=0right).
        $$         Therefore,
        $$
        lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=frac{1}{2}+frac{1}{2}lim_{nto infty} Pleft(sum_{i=1}^nX_i =0right).
        $$         My guess is that $lim_{ntoinfty} Pleft(sum_{i=1}^nX_i =0right)=0$, but as of now, I'm not sure.






        share|cite|improve this answer











        $endgroup$



        Your idea of using Borel-Cantelli lemma is basically right. Let $$
        A={X_n ne 0 text{ for infinitely many }ninmathbb{N}}.
        $$         Since $sum_{n=1}^infty P(X_n ne 0)=sum_{n=1}^infty frac{1}{n^2}<infty$, Borel-Cantelli Lemma implies that $P(A) = 0$. So, for almost every $omega inOmega$, there exists $N(omega)inmathbb{N}$ such that$$
        nge N(omega) Rightarrow X_n(omega) = 0.
        $$         This shows
        $$
        lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^nX_i(omega) = lim_{ntoinfty}frac{1}{sqrt{n}}sum_{i=1}^{N(omega)-1}X_i(omega)=0
        $$         for almost every $omega$, hence giving the result
        $$
        lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile xright) =begin{cases} 1,quad x>0\0,quad x<0end{cases}.
        $$         The case $x=0$ is more subtle. Note that by the symmetry of the distribution, we have
        $$
        2Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ige 0right)=1+Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_i=0right).
        $$         Therefore,
        $$
        lim_{nto infty} Pleft(frac{1}{sqrt{n}}sum_{i=1}^nX_ile 0right)=frac{1}{2}+frac{1}{2}lim_{nto infty} Pleft(sum_{i=1}^nX_i =0right).
        $$         My guess is that $lim_{ntoinfty} Pleft(sum_{i=1}^nX_i =0right)=0$, but as of now, I'm not sure.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 10 at 19:03

























        answered Jan 10 at 17:51









        SongSong

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