Show that $n mathbb{P}{|X_1| geq epsilon sqrt{n}} to 0$












1












$begingroup$



Let $(X_n)_n$ be a sequence of identically distributed random variables with $mathbb{E}X_1^2 < infty$. Show that
$$lim_{n to infty} nmathbb{P}{|X_1| geq epsilon sqrt{n}} = 0$$ for all $epsilon > 0$.




Here is the solution my teaching assistant provides:



We know that



$$int_0^infty mathbb{P}{X_1^2 geq tepsilon^2}dt = mathbb{E}left[frac{X_1^2}{epsilon^2}right]< infty$$



Hence $$sum_{n=1}^infty mathbb{P}{X_1^2 geq n epsilon^2} = sum_{n=1}^infty mathbb{P}{|X_1| geq sqrt{n} epsilon}< infty$$



and because $sum_{n=1}^infty frac{1}{n}= infty$, it follows that $n mathbb{P}{|X_1| geq epsilon sqrt{n}} to 0$ when $n to infty$.



Questions:



How do we deduce the last step? (I.e. $n mathbb{P}{|X_1| geq sqrt{n}epsilon} to 0$ from divergence of the harmonic series?)



I do not think this is correct. I think we can fix the argument with the Borel Cantelli lemma though.






probability-theory measure-theory probability-limit-theorems borel-cantelli-lemmas







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It is true that, if $(x_n)$ is positive and nonincreasing and if the series $sum x_n$ converges then $nx_nto0$, hence the solution in the book is formally correct... except that the divergence of the series $sum frac1n$ is off-topic hence mentioning it here is squarely misleading. One could also mention that the exercise can be completely solved by a, different, one-line argument. Which book is this taken from?
    $endgroup$
    – Did
    Dec 16 '18 at 14:29












  • $begingroup$
    @Did What short solution did you have in mind? I'd like to see a one line approach! And I wrote down that it comes from a book, but it was in fact the solution my teaching assistant wrote down.
    $endgroup$
    – Math_QED
    Dec 16 '18 at 16:13








  • 1




    $begingroup$
    Consider $A_n={|X|>epsilonsqrt n}$ and $Y_n=X^2mathbf 1_{A_n}$, then $Y_nto0$ almost surely and $|Y_n|leqslant X^2$ with $E(X^2)$ finite hence $E(Y_n)to0$ by Lebesgue dominated convergence theorem. But $E(Y_n)geqslantepsilon^2nP(A_n)$ hence $nP(A_n)to0$, qed.
    $endgroup$
    – Did
    Dec 16 '18 at 22:06










  • $begingroup$
    Very elegant indeed. Thanks.
    $endgroup$
    – Math_QED
    Dec 16 '18 at 22:10
















1












$begingroup$



Let $(X_n)_n$ be a sequence of identically distributed random variables with $mathbb{E}X_1^2 < infty$. Show that
$$lim_{n to infty} nmathbb{P}{|X_1| geq epsilon sqrt{n}} = 0$$ for all $epsilon > 0$.




Here is the solution my teaching assistant provides:



We know that



$$int_0^infty mathbb{P}{X_1^2 geq tepsilon^2}dt = mathbb{E}left[frac{X_1^2}{epsilon^2}right]< infty$$



Hence $$sum_{n=1}^infty mathbb{P}{X_1^2 geq n epsilon^2} = sum_{n=1}^infty mathbb{P}{|X_1| geq sqrt{n} epsilon}< infty$$



and because $sum_{n=1}^infty frac{1}{n}= infty$, it follows that $n mathbb{P}{|X_1| geq epsilon sqrt{n}} to 0$ when $n to infty$.



Questions:



How do we deduce the last step? (I.e. $n mathbb{P}{|X_1| geq sqrt{n}epsilon} to 0$ from divergence of the harmonic series?)



I do not think this is correct. I think we can fix the argument with the Borel Cantelli lemma though.






probability-theory measure-theory probability-limit-theorems borel-cantelli-lemmas







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It is true that, if $(x_n)$ is positive and nonincreasing and if the series $sum x_n$ converges then $nx_nto0$, hence the solution in the book is formally correct... except that the divergence of the series $sum frac1n$ is off-topic hence mentioning it here is squarely misleading. One could also mention that the exercise can be completely solved by a, different, one-line argument. Which book is this taken from?
    $endgroup$
    – Did
    Dec 16 '18 at 14:29












  • $begingroup$
    @Did What short solution did you have in mind? I'd like to see a one line approach! And I wrote down that it comes from a book, but it was in fact the solution my teaching assistant wrote down.
    $endgroup$
    – Math_QED
    Dec 16 '18 at 16:13








  • 1




    $begingroup$
    Consider $A_n={|X|>epsilonsqrt n}$ and $Y_n=X^2mathbf 1_{A_n}$, then $Y_nto0$ almost surely and $|Y_n|leqslant X^2$ with $E(X^2)$ finite hence $E(Y_n)to0$ by Lebesgue dominated convergence theorem. But $E(Y_n)geqslantepsilon^2nP(A_n)$ hence $nP(A_n)to0$, qed.
    $endgroup$
    – Did
    Dec 16 '18 at 22:06










  • $begingroup$
    Very elegant indeed. Thanks.
    $endgroup$
    – Math_QED
    Dec 16 '18 at 22:10














1












1








1


1



$begingroup$



Let $(X_n)_n$ be a sequence of identically distributed random variables with $mathbb{E}X_1^2 < infty$. Show that
$$lim_{n to infty} nmathbb{P}{|X_1| geq epsilon sqrt{n}} = 0$$ for all $epsilon > 0$.




Here is the solution my teaching assistant provides:



We know that



$$int_0^infty mathbb{P}{X_1^2 geq tepsilon^2}dt = mathbb{E}left[frac{X_1^2}{epsilon^2}right]< infty$$



Hence $$sum_{n=1}^infty mathbb{P}{X_1^2 geq n epsilon^2} = sum_{n=1}^infty mathbb{P}{|X_1| geq sqrt{n} epsilon}< infty$$



and because $sum_{n=1}^infty frac{1}{n}= infty$, it follows that $n mathbb{P}{|X_1| geq epsilon sqrt{n}} to 0$ when $n to infty$.



Questions:



How do we deduce the last step? (I.e. $n mathbb{P}{|X_1| geq sqrt{n}epsilon} to 0$ from divergence of the harmonic series?)



I do not think this is correct. I think we can fix the argument with the Borel Cantelli lemma though.






probability-theory measure-theory probability-limit-theorems borel-cantelli-lemmas







share|cite|improve this question











$endgroup$





Let $(X_n)_n$ be a sequence of identically distributed random variables with $mathbb{E}X_1^2 < infty$. Show that
$$lim_{n to infty} nmathbb{P}{|X_1| geq epsilon sqrt{n}} = 0$$ for all $epsilon > 0$.




Here is the solution my teaching assistant provides:



We know that



$$int_0^infty mathbb{P}{X_1^2 geq tepsilon^2}dt = mathbb{E}left[frac{X_1^2}{epsilon^2}right]< infty$$



Hence $$sum_{n=1}^infty mathbb{P}{X_1^2 geq n epsilon^2} = sum_{n=1}^infty mathbb{P}{|X_1| geq sqrt{n} epsilon}< infty$$



and because $sum_{n=1}^infty frac{1}{n}= infty$, it follows that $n mathbb{P}{|X_1| geq epsilon sqrt{n}} to 0$ when $n to infty$.



Questions:



How do we deduce the last step? (I.e. $n mathbb{P}{|X_1| geq sqrt{n}epsilon} to 0$ from divergence of the harmonic series?)



I do not think this is correct. I think we can fix the argument with the Borel Cantelli lemma though.






probability-theory measure-theory probability-limit-theorems borel-cantelli-lemmas




probability-theory measure-theory probability-limit-theorems borel-cantelli-lemmas






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 22:08









Did

248k23224463




248k23224463










asked Dec 16 '18 at 13:52









Math_QEDMath_QED

7,62931452




7,62931452








  • 2




    $begingroup$
    It is true that, if $(x_n)$ is positive and nonincreasing and if the series $sum x_n$ converges then $nx_nto0$, hence the solution in the book is formally correct... except that the divergence of the series $sum frac1n$ is off-topic hence mentioning it here is squarely misleading. One could also mention that the exercise can be completely solved by a, different, one-line argument. Which book is this taken from?
    $endgroup$
    – Did
    Dec 16 '18 at 14:29












  • $begingroup$
    @Did What short solution did you have in mind? I'd like to see a one line approach! And I wrote down that it comes from a book, but it was in fact the solution my teaching assistant wrote down.
    $endgroup$
    – Math_QED
    Dec 16 '18 at 16:13








  • 1




    $begingroup$
    Consider $A_n={|X|>epsilonsqrt n}$ and $Y_n=X^2mathbf 1_{A_n}$, then $Y_nto0$ almost surely and $|Y_n|leqslant X^2$ with $E(X^2)$ finite hence $E(Y_n)to0$ by Lebesgue dominated convergence theorem. But $E(Y_n)geqslantepsilon^2nP(A_n)$ hence $nP(A_n)to0$, qed.
    $endgroup$
    – Did
    Dec 16 '18 at 22:06










  • $begingroup$
    Very elegant indeed. Thanks.
    $endgroup$
    – Math_QED
    Dec 16 '18 at 22:10














  • 2




    $begingroup$
    It is true that, if $(x_n)$ is positive and nonincreasing and if the series $sum x_n$ converges then $nx_nto0$, hence the solution in the book is formally correct... except that the divergence of the series $sum frac1n$ is off-topic hence mentioning it here is squarely misleading. One could also mention that the exercise can be completely solved by a, different, one-line argument. Which book is this taken from?
    $endgroup$
    – Did
    Dec 16 '18 at 14:29












  • $begingroup$
    @Did What short solution did you have in mind? I'd like to see a one line approach! And I wrote down that it comes from a book, but it was in fact the solution my teaching assistant wrote down.
    $endgroup$
    – Math_QED
    Dec 16 '18 at 16:13








  • 1




    $begingroup$
    Consider $A_n={|X|>epsilonsqrt n}$ and $Y_n=X^2mathbf 1_{A_n}$, then $Y_nto0$ almost surely and $|Y_n|leqslant X^2$ with $E(X^2)$ finite hence $E(Y_n)to0$ by Lebesgue dominated convergence theorem. But $E(Y_n)geqslantepsilon^2nP(A_n)$ hence $nP(A_n)to0$, qed.
    $endgroup$
    – Did
    Dec 16 '18 at 22:06










  • $begingroup$
    Very elegant indeed. Thanks.
    $endgroup$
    – Math_QED
    Dec 16 '18 at 22:10








2




2




$begingroup$
It is true that, if $(x_n)$ is positive and nonincreasing and if the series $sum x_n$ converges then $nx_nto0$, hence the solution in the book is formally correct... except that the divergence of the series $sum frac1n$ is off-topic hence mentioning it here is squarely misleading. One could also mention that the exercise can be completely solved by a, different, one-line argument. Which book is this taken from?
$endgroup$
– Did
Dec 16 '18 at 14:29






$begingroup$
It is true that, if $(x_n)$ is positive and nonincreasing and if the series $sum x_n$ converges then $nx_nto0$, hence the solution in the book is formally correct... except that the divergence of the series $sum frac1n$ is off-topic hence mentioning it here is squarely misleading. One could also mention that the exercise can be completely solved by a, different, one-line argument. Which book is this taken from?
$endgroup$
– Did
Dec 16 '18 at 14:29














$begingroup$
@Did What short solution did you have in mind? I'd like to see a one line approach! And I wrote down that it comes from a book, but it was in fact the solution my teaching assistant wrote down.
$endgroup$
– Math_QED
Dec 16 '18 at 16:13






$begingroup$
@Did What short solution did you have in mind? I'd like to see a one line approach! And I wrote down that it comes from a book, but it was in fact the solution my teaching assistant wrote down.
$endgroup$
– Math_QED
Dec 16 '18 at 16:13






1




1




$begingroup$
Consider $A_n={|X|>epsilonsqrt n}$ and $Y_n=X^2mathbf 1_{A_n}$, then $Y_nto0$ almost surely and $|Y_n|leqslant X^2$ with $E(X^2)$ finite hence $E(Y_n)to0$ by Lebesgue dominated convergence theorem. But $E(Y_n)geqslantepsilon^2nP(A_n)$ hence $nP(A_n)to0$, qed.
$endgroup$
– Did
Dec 16 '18 at 22:06




$begingroup$
Consider $A_n={|X|>epsilonsqrt n}$ and $Y_n=X^2mathbf 1_{A_n}$, then $Y_nto0$ almost surely and $|Y_n|leqslant X^2$ with $E(X^2)$ finite hence $E(Y_n)to0$ by Lebesgue dominated convergence theorem. But $E(Y_n)geqslantepsilon^2nP(A_n)$ hence $nP(A_n)to0$, qed.
$endgroup$
– Did
Dec 16 '18 at 22:06












$begingroup$
Very elegant indeed. Thanks.
$endgroup$
– Math_QED
Dec 16 '18 at 22:10




$begingroup$
Very elegant indeed. Thanks.
$endgroup$
– Math_QED
Dec 16 '18 at 22:10










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$begingroup$

Indeed, it seems that what is written is that "if $a_n$, is a sequence of positive numbers such that $sum_n a_n/n$ is convergent, then $a_nto 0$. But this may not be true, for example if $a_n=1$ when $n$ has the for $2^k$ for some $kgeqslant 1$ and zero otherwise. However, it is true if there exists positive constants $c$ and $C$ such that for all $Ngeqslant 1$,
$$
ca_{2^{N+1}}leqslant min_{2^Nleqslant nleqslant 2^{N+1}}a_nmbox{ and } max_{2^Nleqslant nleqslant 2^{N+1}}leqslant Ca_{2^{N}} .
$$






share|cite|improve this answer









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    $begingroup$

    Indeed, it seems that what is written is that "if $a_n$, is a sequence of positive numbers such that $sum_n a_n/n$ is convergent, then $a_nto 0$. But this may not be true, for example if $a_n=1$ when $n$ has the for $2^k$ for some $kgeqslant 1$ and zero otherwise. However, it is true if there exists positive constants $c$ and $C$ such that for all $Ngeqslant 1$,
    $$
    ca_{2^{N+1}}leqslant min_{2^Nleqslant nleqslant 2^{N+1}}a_nmbox{ and } max_{2^Nleqslant nleqslant 2^{N+1}}leqslant Ca_{2^{N}} .
    $$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Indeed, it seems that what is written is that "if $a_n$, is a sequence of positive numbers such that $sum_n a_n/n$ is convergent, then $a_nto 0$. But this may not be true, for example if $a_n=1$ when $n$ has the for $2^k$ for some $kgeqslant 1$ and zero otherwise. However, it is true if there exists positive constants $c$ and $C$ such that for all $Ngeqslant 1$,
      $$
      ca_{2^{N+1}}leqslant min_{2^Nleqslant nleqslant 2^{N+1}}a_nmbox{ and } max_{2^Nleqslant nleqslant 2^{N+1}}leqslant Ca_{2^{N}} .
      $$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Indeed, it seems that what is written is that "if $a_n$, is a sequence of positive numbers such that $sum_n a_n/n$ is convergent, then $a_nto 0$. But this may not be true, for example if $a_n=1$ when $n$ has the for $2^k$ for some $kgeqslant 1$ and zero otherwise. However, it is true if there exists positive constants $c$ and $C$ such that for all $Ngeqslant 1$,
        $$
        ca_{2^{N+1}}leqslant min_{2^Nleqslant nleqslant 2^{N+1}}a_nmbox{ and } max_{2^Nleqslant nleqslant 2^{N+1}}leqslant Ca_{2^{N}} .
        $$






        share|cite|improve this answer









        $endgroup$



        Indeed, it seems that what is written is that "if $a_n$, is a sequence of positive numbers such that $sum_n a_n/n$ is convergent, then $a_nto 0$. But this may not be true, for example if $a_n=1$ when $n$ has the for $2^k$ for some $kgeqslant 1$ and zero otherwise. However, it is true if there exists positive constants $c$ and $C$ such that for all $Ngeqslant 1$,
        $$
        ca_{2^{N+1}}leqslant min_{2^Nleqslant nleqslant 2^{N+1}}a_nmbox{ and } max_{2^Nleqslant nleqslant 2^{N+1}}leqslant Ca_{2^{N}} .
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 14:14









        Davide GiraudoDavide Giraudo

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        127k16151264






























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