Second lemma of Borel-Cantelli: Normal distribution












3












$begingroup$


In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.




Let $X_n$, $ngeq 1$, be a sequence of independent $N(0, sigma^2)$- distributed random variables, with $sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $limsup_n X_n = infty$.




For sake of completeness: $N(0, sigma^2$) denotes the normal distribution with mean $=0$ and variance $sigma$. Our 2nd lemma of Borel Cantelli says:




Let $A_n$, $n geq 1$, be a sequence of independent events on a probability space. Then: $$sum_n P(A_n) = infty Rightarrow Pleft[limsup A_nright]=1.$$




What I don't see is why the normal distribution of $X_n$ implies that $sum_nP({X_n leq x}) = infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.






probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
    $endgroup$
    – saz
    Sep 30 '17 at 11:00










  • $begingroup$
    @saz You are right. I have corrected it.
    $endgroup$
    – Quasar
    Oct 1 '17 at 8:33










  • $begingroup$
    You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
    $endgroup$
    – saz
    Oct 1 '17 at 8:43


















3












$begingroup$


In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.




Let $X_n$, $ngeq 1$, be a sequence of independent $N(0, sigma^2)$- distributed random variables, with $sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $limsup_n X_n = infty$.




For sake of completeness: $N(0, sigma^2$) denotes the normal distribution with mean $=0$ and variance $sigma$. Our 2nd lemma of Borel Cantelli says:




Let $A_n$, $n geq 1$, be a sequence of independent events on a probability space. Then: $$sum_n P(A_n) = infty Rightarrow Pleft[limsup A_nright]=1.$$




What I don't see is why the normal distribution of $X_n$ implies that $sum_nP({X_n leq x}) = infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.






probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
    $endgroup$
    – saz
    Sep 30 '17 at 11:00










  • $begingroup$
    @saz You are right. I have corrected it.
    $endgroup$
    – Quasar
    Oct 1 '17 at 8:33










  • $begingroup$
    You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
    $endgroup$
    – saz
    Oct 1 '17 at 8:43
















3












3








3





$begingroup$


In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.




Let $X_n$, $ngeq 1$, be a sequence of independent $N(0, sigma^2)$- distributed random variables, with $sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $limsup_n X_n = infty$.




For sake of completeness: $N(0, sigma^2$) denotes the normal distribution with mean $=0$ and variance $sigma$. Our 2nd lemma of Borel Cantelli says:




Let $A_n$, $n geq 1$, be a sequence of independent events on a probability space. Then: $$sum_n P(A_n) = infty Rightarrow Pleft[limsup A_nright]=1.$$




What I don't see is why the normal distribution of $X_n$ implies that $sum_nP({X_n leq x}) = infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.






probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas







share|cite|improve this question











$endgroup$




In my textbook is an example of an application of Borel-Cantelli's lemma which I don't understand.




Let $X_n$, $ngeq 1$, be a sequence of independent $N(0, sigma^2)$- distributed random variables, with $sigma > 0$. From the second lemma of Borel-Cantelli it follows that: P- almost surely (P-a.s.) $limsup_n X_n = infty$.




For sake of completeness: $N(0, sigma^2$) denotes the normal distribution with mean $=0$ and variance $sigma$. Our 2nd lemma of Borel Cantelli says:




Let $A_n$, $n geq 1$, be a sequence of independent events on a probability space. Then: $$sum_n P(A_n) = infty Rightarrow Pleft[limsup A_nright]=1.$$




What I don't see is why the normal distribution of $X_n$ implies that $sum_nP({X_n leq x}) = infty$ (or does it not?) and how from this we then can apply Borel Cantelli's lemma.






probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas




probability-theory random-variables limsup-and-liminf borel-cantelli-lemmas






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 9:44









Davide Giraudo

127k16151264




127k16151264










asked Sep 30 '17 at 10:45









QuasarQuasar

1278




1278








  • 1




    $begingroup$
    Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
    $endgroup$
    – saz
    Sep 30 '17 at 11:00










  • $begingroup$
    @saz You are right. I have corrected it.
    $endgroup$
    – Quasar
    Oct 1 '17 at 8:33










  • $begingroup$
    You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
    $endgroup$
    – saz
    Oct 1 '17 at 8:43
















  • 1




    $begingroup$
    Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
    $endgroup$
    – saz
    Sep 30 '17 at 11:00










  • $begingroup$
    @saz You are right. I have corrected it.
    $endgroup$
    – Quasar
    Oct 1 '17 at 8:33










  • $begingroup$
    You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
    $endgroup$
    – saz
    Oct 1 '17 at 8:43










1




1




$begingroup$
Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
$endgroup$
– saz
Sep 30 '17 at 11:00




$begingroup$
Note that $P(X_n)$ doesn't make any sense. $X_n$ is a random variable; $P(A)$ is only defined for events $A$.
$endgroup$
– saz
Sep 30 '17 at 11:00












$begingroup$
@saz You are right. I have corrected it.
$endgroup$
– Quasar
Oct 1 '17 at 8:33




$begingroup$
@saz You are right. I have corrected it.
$endgroup$
– Quasar
Oct 1 '17 at 8:33












$begingroup$
You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
$endgroup$
– saz
Oct 1 '17 at 8:43






$begingroup$
You are interested in $X_n$ having large values, right? So it would make more sense to study $sum_n P({X_n color{red}{geq} x})$ ... Use that the random variables are identically distributed!
$endgroup$
– saz
Oct 1 '17 at 8:43












1 Answer
1






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0












$begingroup$

Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
$$
forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
$$
Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.



Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.






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    $begingroup$

    Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
    $$
    forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
    $$
    Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.



    Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
      $$
      forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
      $$
      Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.



      Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
        $$
        forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
        $$
        Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.



        Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.






        share|cite|improve this answer











        $endgroup$



        Let us fix a positive integer $N$ and define the (independent) events $A_n:=left{X_ngeqslant Nright}$. Since the random variable $X_n$ has the same distribution as $X_1$, $mathbb Pleft(A_nright)=mathbb Pleft(A_1right)gt 0$, which implies that the series $sum_{ngeqslant 1}mathbb Pleft(A_nright)$ is divergent. By the second Borel-Cantelli lemma, $mathbb Pleft(limsup_{nto +infty}A_nright)=1$, which means that there exists a set $Omega_N$ of probability one such that for all $omegainOmega_N$, the set ${ninmathbb N, X_n(omega)geqslant N}$ is finite. This implies that
        $$
        forall omegainOmega_N, limsup_{nto +infty}X_n(omega)geqslant N.
        $$
        Let $Omega':=bigcap_{Ngeqslant 1}Omega_N$. Then $Omega'$ has probability one and $limsup_{nto +infty}X_n(omega)=+infty$ for all $omegainOmega'$.



        Remark that we do not really need the $X_n$ to have a normal distribution: it suffices that the $X_n$ have the same distribution and that $mathbb Pleft(left{X_1geqslant Nright}right)gt 0$ for all $N$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 22 at 9:46

























        answered Jan 21 at 23:07









        Davide GiraudoDavide Giraudo

        127k16151264




        127k16151264






























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