My proof that $S_n/sqrt n$ does not converge in probability












4














I'm given a sequence $(X_n)$ of i.i.d. random variables with mean $0$ and finite variance $sigma^2$. Let $S_n=X_1 + ... + X_n$. I have to show that $S_n/sqrt n$ does not converge in probability. Here's what I did.



Since $S_n/sqrt n$ converges in distribution to a normal random variable $Z$ with mean zero, if $S_n/sqrt n$ converges in probability at all it must be to $Z$. But



$$P(|frac {S_n} {sqrt n} - Z| > epsilon) geq P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon)$$



Now I get to the main point I'm not sure of. Can I say that the random variables $S_nsqrt n$ (for any $n$) and $Z$ are independant? It seems like they might be, since in some sense we can't tell what the limit of a sequence will be from any initial segment of it.



If so, then I can continue (this all seems correct to me)



$$begin{align}
&P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon) \
=&P(|frac {S_n} {sqrt n}|<epsilon)P(|Z|>2epsilon) \
geq&(1 - frac {sigma^2} {nepsilon^2})P(|Z|>2epsilon) to P(|Z|>2epsilon) > 0
end{align}$$










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  • I would suppose that $S_n$ converges in probability for some $X_j$ (e.g., when they are normal distributed). It is only not true that it converges for all $X_j$. So it feels like there is something wrong with your proof.
    – Fabian
    Jan 2 '16 at 22:03










  • Possible duplicate of A sequence of random variables that does not converge in probability.
    – Winther
    Jan 2 '16 at 22:31










  • @Winther I don't think that really answers the question - my question is specifically about whether the approach I used is valid or can be modified to be valid.
    – Jack M
    Jan 2 '16 at 22:34












  • OK, I removed the vote! I'll keep the link above as it might be useful for others.
    – Winther
    Jan 2 '16 at 22:35






  • 2




    If (in a different question) $Y_n$ converges in probability to a non-constant $V$, and $W$ has the same distribution as $V$ but is not almost surely identical, then $Y_n$ might be said to converge in distribution to $V$ and to $W$ but cannot be said to converge in probability to $W$. So your assertion that "if $S_n/sqrt n$ converges in probability at all it must be to $Z$" looks difficult to justify
    – Henry
    Dec 5 '17 at 11:31


















4














I'm given a sequence $(X_n)$ of i.i.d. random variables with mean $0$ and finite variance $sigma^2$. Let $S_n=X_1 + ... + X_n$. I have to show that $S_n/sqrt n$ does not converge in probability. Here's what I did.



Since $S_n/sqrt n$ converges in distribution to a normal random variable $Z$ with mean zero, if $S_n/sqrt n$ converges in probability at all it must be to $Z$. But



$$P(|frac {S_n} {sqrt n} - Z| > epsilon) geq P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon)$$



Now I get to the main point I'm not sure of. Can I say that the random variables $S_nsqrt n$ (for any $n$) and $Z$ are independant? It seems like they might be, since in some sense we can't tell what the limit of a sequence will be from any initial segment of it.



If so, then I can continue (this all seems correct to me)



$$begin{align}
&P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon) \
=&P(|frac {S_n} {sqrt n}|<epsilon)P(|Z|>2epsilon) \
geq&(1 - frac {sigma^2} {nepsilon^2})P(|Z|>2epsilon) to P(|Z|>2epsilon) > 0
end{align}$$










share|cite|improve this question






















  • I would suppose that $S_n$ converges in probability for some $X_j$ (e.g., when they are normal distributed). It is only not true that it converges for all $X_j$. So it feels like there is something wrong with your proof.
    – Fabian
    Jan 2 '16 at 22:03










  • Possible duplicate of A sequence of random variables that does not converge in probability.
    – Winther
    Jan 2 '16 at 22:31










  • @Winther I don't think that really answers the question - my question is specifically about whether the approach I used is valid or can be modified to be valid.
    – Jack M
    Jan 2 '16 at 22:34












  • OK, I removed the vote! I'll keep the link above as it might be useful for others.
    – Winther
    Jan 2 '16 at 22:35






  • 2




    If (in a different question) $Y_n$ converges in probability to a non-constant $V$, and $W$ has the same distribution as $V$ but is not almost surely identical, then $Y_n$ might be said to converge in distribution to $V$ and to $W$ but cannot be said to converge in probability to $W$. So your assertion that "if $S_n/sqrt n$ converges in probability at all it must be to $Z$" looks difficult to justify
    – Henry
    Dec 5 '17 at 11:31
















4












4








4


2





I'm given a sequence $(X_n)$ of i.i.d. random variables with mean $0$ and finite variance $sigma^2$. Let $S_n=X_1 + ... + X_n$. I have to show that $S_n/sqrt n$ does not converge in probability. Here's what I did.



Since $S_n/sqrt n$ converges in distribution to a normal random variable $Z$ with mean zero, if $S_n/sqrt n$ converges in probability at all it must be to $Z$. But



$$P(|frac {S_n} {sqrt n} - Z| > epsilon) geq P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon)$$



Now I get to the main point I'm not sure of. Can I say that the random variables $S_nsqrt n$ (for any $n$) and $Z$ are independant? It seems like they might be, since in some sense we can't tell what the limit of a sequence will be from any initial segment of it.



If so, then I can continue (this all seems correct to me)



$$begin{align}
&P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon) \
=&P(|frac {S_n} {sqrt n}|<epsilon)P(|Z|>2epsilon) \
geq&(1 - frac {sigma^2} {nepsilon^2})P(|Z|>2epsilon) to P(|Z|>2epsilon) > 0
end{align}$$










share|cite|improve this question













I'm given a sequence $(X_n)$ of i.i.d. random variables with mean $0$ and finite variance $sigma^2$. Let $S_n=X_1 + ... + X_n$. I have to show that $S_n/sqrt n$ does not converge in probability. Here's what I did.



Since $S_n/sqrt n$ converges in distribution to a normal random variable $Z$ with mean zero, if $S_n/sqrt n$ converges in probability at all it must be to $Z$. But



$$P(|frac {S_n} {sqrt n} - Z| > epsilon) geq P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon)$$



Now I get to the main point I'm not sure of. Can I say that the random variables $S_nsqrt n$ (for any $n$) and $Z$ are independant? It seems like they might be, since in some sense we can't tell what the limit of a sequence will be from any initial segment of it.



If so, then I can continue (this all seems correct to me)



$$begin{align}
&P(|frac {S_n} {sqrt n}|<epsilon, |Z|>2epsilon) \
=&P(|frac {S_n} {sqrt n}|<epsilon)P(|Z|>2epsilon) \
geq&(1 - frac {sigma^2} {nepsilon^2})P(|Z|>2epsilon) to P(|Z|>2epsilon) > 0
end{align}$$







probability-theory proof-verification






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asked Jan 2 '16 at 21:37









Jack MJack M

18.6k33880




18.6k33880












  • I would suppose that $S_n$ converges in probability for some $X_j$ (e.g., when they are normal distributed). It is only not true that it converges for all $X_j$. So it feels like there is something wrong with your proof.
    – Fabian
    Jan 2 '16 at 22:03










  • Possible duplicate of A sequence of random variables that does not converge in probability.
    – Winther
    Jan 2 '16 at 22:31










  • @Winther I don't think that really answers the question - my question is specifically about whether the approach I used is valid or can be modified to be valid.
    – Jack M
    Jan 2 '16 at 22:34












  • OK, I removed the vote! I'll keep the link above as it might be useful for others.
    – Winther
    Jan 2 '16 at 22:35






  • 2




    If (in a different question) $Y_n$ converges in probability to a non-constant $V$, and $W$ has the same distribution as $V$ but is not almost surely identical, then $Y_n$ might be said to converge in distribution to $V$ and to $W$ but cannot be said to converge in probability to $W$. So your assertion that "if $S_n/sqrt n$ converges in probability at all it must be to $Z$" looks difficult to justify
    – Henry
    Dec 5 '17 at 11:31




















  • I would suppose that $S_n$ converges in probability for some $X_j$ (e.g., when they are normal distributed). It is only not true that it converges for all $X_j$. So it feels like there is something wrong with your proof.
    – Fabian
    Jan 2 '16 at 22:03










  • Possible duplicate of A sequence of random variables that does not converge in probability.
    – Winther
    Jan 2 '16 at 22:31










  • @Winther I don't think that really answers the question - my question is specifically about whether the approach I used is valid or can be modified to be valid.
    – Jack M
    Jan 2 '16 at 22:34












  • OK, I removed the vote! I'll keep the link above as it might be useful for others.
    – Winther
    Jan 2 '16 at 22:35






  • 2




    If (in a different question) $Y_n$ converges in probability to a non-constant $V$, and $W$ has the same distribution as $V$ but is not almost surely identical, then $Y_n$ might be said to converge in distribution to $V$ and to $W$ but cannot be said to converge in probability to $W$. So your assertion that "if $S_n/sqrt n$ converges in probability at all it must be to $Z$" looks difficult to justify
    – Henry
    Dec 5 '17 at 11:31


















I would suppose that $S_n$ converges in probability for some $X_j$ (e.g., when they are normal distributed). It is only not true that it converges for all $X_j$. So it feels like there is something wrong with your proof.
– Fabian
Jan 2 '16 at 22:03




I would suppose that $S_n$ converges in probability for some $X_j$ (e.g., when they are normal distributed). It is only not true that it converges for all $X_j$. So it feels like there is something wrong with your proof.
– Fabian
Jan 2 '16 at 22:03












Possible duplicate of A sequence of random variables that does not converge in probability.
– Winther
Jan 2 '16 at 22:31




Possible duplicate of A sequence of random variables that does not converge in probability.
– Winther
Jan 2 '16 at 22:31












@Winther I don't think that really answers the question - my question is specifically about whether the approach I used is valid or can be modified to be valid.
– Jack M
Jan 2 '16 at 22:34






@Winther I don't think that really answers the question - my question is specifically about whether the approach I used is valid or can be modified to be valid.
– Jack M
Jan 2 '16 at 22:34














OK, I removed the vote! I'll keep the link above as it might be useful for others.
– Winther
Jan 2 '16 at 22:35




OK, I removed the vote! I'll keep the link above as it might be useful for others.
– Winther
Jan 2 '16 at 22:35




2




2




If (in a different question) $Y_n$ converges in probability to a non-constant $V$, and $W$ has the same distribution as $V$ but is not almost surely identical, then $Y_n$ might be said to converge in distribution to $V$ and to $W$ but cannot be said to converge in probability to $W$. So your assertion that "if $S_n/sqrt n$ converges in probability at all it must be to $Z$" looks difficult to justify
– Henry
Dec 5 '17 at 11:31






If (in a different question) $Y_n$ converges in probability to a non-constant $V$, and $W$ has the same distribution as $V$ but is not almost surely identical, then $Y_n$ might be said to converge in distribution to $V$ and to $W$ but cannot be said to converge in probability to $W$. So your assertion that "if $S_n/sqrt n$ converges in probability at all it must be to $Z$" looks difficult to justify
– Henry
Dec 5 '17 at 11:31












3 Answers
3






active

oldest

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0














Let see if this works.



Set $sigma^2=1$ (if sigma is 0 the statment is true) and assume $V_n=S_n/sqrt{n}$ converges in probability to a r.v. Z.
By the CLT, Z has standard normal distribution.



Define $W_n=frac{S_{2n}}{sqrt{2n}}$. These variables converge to Z in probability too.



Finally, take $T_n=frac{S_{2n}-S_n}{sqrt{n}}$.
T converges in distribution to a standard normal, but in probability to $(sqrt{2}-1) Z$, because $T_n=sqrt{2}W_n-V_n$, that has not standard normal distribution.






share|cite|improve this answer





















  • $S_{2n}$ and $S_n$ are not independent so I do not see why you can say that $T_n=frac{S_{2n}-S_n}{sqrt{n}}$ converges in probability to $(sqrt{2}-1) Z$
    – Henry
    Dec 5 '17 at 11:42










  • @Henry: if $X_n$ converges to X in probability and $Y_n$ converges to Y in probability too, the sum converges in probability to X+Y regerdless of the relation between the variables
    – Kolmo
    Dec 5 '17 at 21:23





















0














I have a partial answer. I think what you need is the Cauchy criterion for convergence in probability, which says:



The sequence $left(frac{S_n}{sqrt{n}}right)_{ngeq 1}$ converges in probability if and only if
begin{align}
Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)stackrel{n,mtoinfty}{longrightarrow} 0quad text{for every }epsilon>0.
end{align}



However, notice that
begin{align}
&Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)=Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}>epsilonright)+Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}<-epsilonright)\
&geq Pleft(frac{S_n}{sqrt{n}}>2epsilon,frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilon,frac{S_m}{sqrt{m}}>-epsilonright)\
&geq Pleft(frac{S_n}{sqrt{n}}>2epsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilonright)+Pleft(frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_m}{sqrt{m}}>-epsilonright)-2\
&stackrel{n,mtoinfty}{longrightarrow}2left(Q(2epsilon)+Q(-epsilon)-1right),
end{align}

where $Q(x)=intlimits_{x}^{infty} mathcal{N}(0,1),dx$.



I do not know how to find an $epsilon>0$ such that the term inside the brackets is strictly positive. I am not quite sure if we can find one at all. As I said, this is just a partial solution that I have worked out.



Please feel free to suggest any more additions, or strengthening of arguments.






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    -1














    One can easily construct an entity to which we can say that "$S_n/sqrt n$ converges in probability".



    Take for example $W_n = -X_1+X_2+X_3+...+X_n$



    Then



    $$lim_{nrightarrow infty} Pleft(left|frac {S_n}{sqrt n} - frac {W_n}{sqrt n}right|> epsilonright) = lim_{nrightarrow infty} Pleft(left|frac {2X_1}{sqrt n} right|> epsilonright) = 0$$



    and the criterion for convergence in probability is satisfied.



    So I suspect that "$S_n/sqrt n$ does not converge in probability" must have a more specific and narrow sense in the OP's case.





    On another front, the established phrase "$S_n/sqrt n$ converges in distribution to a random variable Z" sometimes makes us forget that the phenomenon described by "convergence in distribution" is that the sequence of distribution functions $F_n$ of $S_n/sqrt n$ converges to a certain distribution function $F$. There is really no $Z$ "at the end of the journey" waiting to "become one" with $S_n/sqrt n$ . $Z$ is a random variable, a separate entity from the distribution that characterizes it (which characterizes also an infinite number of other such $Z$'s). If there is no random variable, the question "is $S_n/sqrt n$ independent of $Z$?" cannot even be posed.






    share|cite|improve this answer























    • In response to your last paragraph: Yes, I think for this reason this proof (were it valid) would necessarily have to be posed as a proof by contradiction (unlike many proofs by contradiction which can be rephrased as direct proofs of contrapositives). You have to assume the sequence converges in probability to $Z$ in order to even be able to say what $Z$ is.
      – Jack M
      Jan 3 '16 at 7:47






    • 1




      I don't understand your construction though. To what are you claiming your $S_n/sqrt n$ converges in probability?
      – Jack M
      Jan 3 '16 at 7:50











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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

    oldest

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    active

    oldest

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    0














    Let see if this works.



    Set $sigma^2=1$ (if sigma is 0 the statment is true) and assume $V_n=S_n/sqrt{n}$ converges in probability to a r.v. Z.
    By the CLT, Z has standard normal distribution.



    Define $W_n=frac{S_{2n}}{sqrt{2n}}$. These variables converge to Z in probability too.



    Finally, take $T_n=frac{S_{2n}-S_n}{sqrt{n}}$.
    T converges in distribution to a standard normal, but in probability to $(sqrt{2}-1) Z$, because $T_n=sqrt{2}W_n-V_n$, that has not standard normal distribution.






    share|cite|improve this answer





















    • $S_{2n}$ and $S_n$ are not independent so I do not see why you can say that $T_n=frac{S_{2n}-S_n}{sqrt{n}}$ converges in probability to $(sqrt{2}-1) Z$
      – Henry
      Dec 5 '17 at 11:42










    • @Henry: if $X_n$ converges to X in probability and $Y_n$ converges to Y in probability too, the sum converges in probability to X+Y regerdless of the relation between the variables
      – Kolmo
      Dec 5 '17 at 21:23


















    0














    Let see if this works.



    Set $sigma^2=1$ (if sigma is 0 the statment is true) and assume $V_n=S_n/sqrt{n}$ converges in probability to a r.v. Z.
    By the CLT, Z has standard normal distribution.



    Define $W_n=frac{S_{2n}}{sqrt{2n}}$. These variables converge to Z in probability too.



    Finally, take $T_n=frac{S_{2n}-S_n}{sqrt{n}}$.
    T converges in distribution to a standard normal, but in probability to $(sqrt{2}-1) Z$, because $T_n=sqrt{2}W_n-V_n$, that has not standard normal distribution.






    share|cite|improve this answer





















    • $S_{2n}$ and $S_n$ are not independent so I do not see why you can say that $T_n=frac{S_{2n}-S_n}{sqrt{n}}$ converges in probability to $(sqrt{2}-1) Z$
      – Henry
      Dec 5 '17 at 11:42










    • @Henry: if $X_n$ converges to X in probability and $Y_n$ converges to Y in probability too, the sum converges in probability to X+Y regerdless of the relation between the variables
      – Kolmo
      Dec 5 '17 at 21:23
















    0












    0








    0






    Let see if this works.



    Set $sigma^2=1$ (if sigma is 0 the statment is true) and assume $V_n=S_n/sqrt{n}$ converges in probability to a r.v. Z.
    By the CLT, Z has standard normal distribution.



    Define $W_n=frac{S_{2n}}{sqrt{2n}}$. These variables converge to Z in probability too.



    Finally, take $T_n=frac{S_{2n}-S_n}{sqrt{n}}$.
    T converges in distribution to a standard normal, but in probability to $(sqrt{2}-1) Z$, because $T_n=sqrt{2}W_n-V_n$, that has not standard normal distribution.






    share|cite|improve this answer












    Let see if this works.



    Set $sigma^2=1$ (if sigma is 0 the statment is true) and assume $V_n=S_n/sqrt{n}$ converges in probability to a r.v. Z.
    By the CLT, Z has standard normal distribution.



    Define $W_n=frac{S_{2n}}{sqrt{2n}}$. These variables converge to Z in probability too.



    Finally, take $T_n=frac{S_{2n}-S_n}{sqrt{n}}$.
    T converges in distribution to a standard normal, but in probability to $(sqrt{2}-1) Z$, because $T_n=sqrt{2}W_n-V_n$, that has not standard normal distribution.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 5 '16 at 18:56









    KolmoKolmo

    909614




    909614












    • $S_{2n}$ and $S_n$ are not independent so I do not see why you can say that $T_n=frac{S_{2n}-S_n}{sqrt{n}}$ converges in probability to $(sqrt{2}-1) Z$
      – Henry
      Dec 5 '17 at 11:42










    • @Henry: if $X_n$ converges to X in probability and $Y_n$ converges to Y in probability too, the sum converges in probability to X+Y regerdless of the relation between the variables
      – Kolmo
      Dec 5 '17 at 21:23




















    • $S_{2n}$ and $S_n$ are not independent so I do not see why you can say that $T_n=frac{S_{2n}-S_n}{sqrt{n}}$ converges in probability to $(sqrt{2}-1) Z$
      – Henry
      Dec 5 '17 at 11:42










    • @Henry: if $X_n$ converges to X in probability and $Y_n$ converges to Y in probability too, the sum converges in probability to X+Y regerdless of the relation between the variables
      – Kolmo
      Dec 5 '17 at 21:23


















    $S_{2n}$ and $S_n$ are not independent so I do not see why you can say that $T_n=frac{S_{2n}-S_n}{sqrt{n}}$ converges in probability to $(sqrt{2}-1) Z$
    – Henry
    Dec 5 '17 at 11:42




    $S_{2n}$ and $S_n$ are not independent so I do not see why you can say that $T_n=frac{S_{2n}-S_n}{sqrt{n}}$ converges in probability to $(sqrt{2}-1) Z$
    – Henry
    Dec 5 '17 at 11:42












    @Henry: if $X_n$ converges to X in probability and $Y_n$ converges to Y in probability too, the sum converges in probability to X+Y regerdless of the relation between the variables
    – Kolmo
    Dec 5 '17 at 21:23






    @Henry: if $X_n$ converges to X in probability and $Y_n$ converges to Y in probability too, the sum converges in probability to X+Y regerdless of the relation between the variables
    – Kolmo
    Dec 5 '17 at 21:23













    0














    I have a partial answer. I think what you need is the Cauchy criterion for convergence in probability, which says:



    The sequence $left(frac{S_n}{sqrt{n}}right)_{ngeq 1}$ converges in probability if and only if
    begin{align}
    Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)stackrel{n,mtoinfty}{longrightarrow} 0quad text{for every }epsilon>0.
    end{align}



    However, notice that
    begin{align}
    &Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)=Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}>epsilonright)+Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}<-epsilonright)\
    &geq Pleft(frac{S_n}{sqrt{n}}>2epsilon,frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilon,frac{S_m}{sqrt{m}}>-epsilonright)\
    &geq Pleft(frac{S_n}{sqrt{n}}>2epsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilonright)+Pleft(frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_m}{sqrt{m}}>-epsilonright)-2\
    &stackrel{n,mtoinfty}{longrightarrow}2left(Q(2epsilon)+Q(-epsilon)-1right),
    end{align}

    where $Q(x)=intlimits_{x}^{infty} mathcal{N}(0,1),dx$.



    I do not know how to find an $epsilon>0$ such that the term inside the brackets is strictly positive. I am not quite sure if we can find one at all. As I said, this is just a partial solution that I have worked out.



    Please feel free to suggest any more additions, or strengthening of arguments.






    share|cite|improve this answer


























      0














      I have a partial answer. I think what you need is the Cauchy criterion for convergence in probability, which says:



      The sequence $left(frac{S_n}{sqrt{n}}right)_{ngeq 1}$ converges in probability if and only if
      begin{align}
      Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)stackrel{n,mtoinfty}{longrightarrow} 0quad text{for every }epsilon>0.
      end{align}



      However, notice that
      begin{align}
      &Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)=Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}>epsilonright)+Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}<-epsilonright)\
      &geq Pleft(frac{S_n}{sqrt{n}}>2epsilon,frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilon,frac{S_m}{sqrt{m}}>-epsilonright)\
      &geq Pleft(frac{S_n}{sqrt{n}}>2epsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilonright)+Pleft(frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_m}{sqrt{m}}>-epsilonright)-2\
      &stackrel{n,mtoinfty}{longrightarrow}2left(Q(2epsilon)+Q(-epsilon)-1right),
      end{align}

      where $Q(x)=intlimits_{x}^{infty} mathcal{N}(0,1),dx$.



      I do not know how to find an $epsilon>0$ such that the term inside the brackets is strictly positive. I am not quite sure if we can find one at all. As I said, this is just a partial solution that I have worked out.



      Please feel free to suggest any more additions, or strengthening of arguments.






      share|cite|improve this answer
























        0












        0








        0






        I have a partial answer. I think what you need is the Cauchy criterion for convergence in probability, which says:



        The sequence $left(frac{S_n}{sqrt{n}}right)_{ngeq 1}$ converges in probability if and only if
        begin{align}
        Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)stackrel{n,mtoinfty}{longrightarrow} 0quad text{for every }epsilon>0.
        end{align}



        However, notice that
        begin{align}
        &Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)=Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}>epsilonright)+Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}<-epsilonright)\
        &geq Pleft(frac{S_n}{sqrt{n}}>2epsilon,frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilon,frac{S_m}{sqrt{m}}>-epsilonright)\
        &geq Pleft(frac{S_n}{sqrt{n}}>2epsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilonright)+Pleft(frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_m}{sqrt{m}}>-epsilonright)-2\
        &stackrel{n,mtoinfty}{longrightarrow}2left(Q(2epsilon)+Q(-epsilon)-1right),
        end{align}

        where $Q(x)=intlimits_{x}^{infty} mathcal{N}(0,1),dx$.



        I do not know how to find an $epsilon>0$ such that the term inside the brackets is strictly positive. I am not quite sure if we can find one at all. As I said, this is just a partial solution that I have worked out.



        Please feel free to suggest any more additions, or strengthening of arguments.






        share|cite|improve this answer












        I have a partial answer. I think what you need is the Cauchy criterion for convergence in probability, which says:



        The sequence $left(frac{S_n}{sqrt{n}}right)_{ngeq 1}$ converges in probability if and only if
        begin{align}
        Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)stackrel{n,mtoinfty}{longrightarrow} 0quad text{for every }epsilon>0.
        end{align}



        However, notice that
        begin{align}
        &Pleft(biggvertfrac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}biggvert>epsilonright)=Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}>epsilonright)+Pleft(frac{S_n}{sqrt{n}}-frac{S_m}{sqrt{m}}<-epsilonright)\
        &geq Pleft(frac{S_n}{sqrt{n}}>2epsilon,frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilon,frac{S_m}{sqrt{m}}>-epsilonright)\
        &geq Pleft(frac{S_n}{sqrt{n}}>2epsilonright)+Pleft(frac{S_n}{sqrt{n}}le-2epsilonright)+Pleft(frac{S_m}{sqrt{m}}leepsilonright)+Pleft(frac{S_m}{sqrt{m}}>-epsilonright)-2\
        &stackrel{n,mtoinfty}{longrightarrow}2left(Q(2epsilon)+Q(-epsilon)-1right),
        end{align}

        where $Q(x)=intlimits_{x}^{infty} mathcal{N}(0,1),dx$.



        I do not know how to find an $epsilon>0$ such that the term inside the brackets is strictly positive. I am not quite sure if we can find one at all. As I said, this is just a partial solution that I have worked out.



        Please feel free to suggest any more additions, or strengthening of arguments.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 5:01









        KarthikKarthik

        977218




        977218























            -1














            One can easily construct an entity to which we can say that "$S_n/sqrt n$ converges in probability".



            Take for example $W_n = -X_1+X_2+X_3+...+X_n$



            Then



            $$lim_{nrightarrow infty} Pleft(left|frac {S_n}{sqrt n} - frac {W_n}{sqrt n}right|> epsilonright) = lim_{nrightarrow infty} Pleft(left|frac {2X_1}{sqrt n} right|> epsilonright) = 0$$



            and the criterion for convergence in probability is satisfied.



            So I suspect that "$S_n/sqrt n$ does not converge in probability" must have a more specific and narrow sense in the OP's case.





            On another front, the established phrase "$S_n/sqrt n$ converges in distribution to a random variable Z" sometimes makes us forget that the phenomenon described by "convergence in distribution" is that the sequence of distribution functions $F_n$ of $S_n/sqrt n$ converges to a certain distribution function $F$. There is really no $Z$ "at the end of the journey" waiting to "become one" with $S_n/sqrt n$ . $Z$ is a random variable, a separate entity from the distribution that characterizes it (which characterizes also an infinite number of other such $Z$'s). If there is no random variable, the question "is $S_n/sqrt n$ independent of $Z$?" cannot even be posed.






            share|cite|improve this answer























            • In response to your last paragraph: Yes, I think for this reason this proof (were it valid) would necessarily have to be posed as a proof by contradiction (unlike many proofs by contradiction which can be rephrased as direct proofs of contrapositives). You have to assume the sequence converges in probability to $Z$ in order to even be able to say what $Z$ is.
              – Jack M
              Jan 3 '16 at 7:47






            • 1




              I don't understand your construction though. To what are you claiming your $S_n/sqrt n$ converges in probability?
              – Jack M
              Jan 3 '16 at 7:50
















            -1














            One can easily construct an entity to which we can say that "$S_n/sqrt n$ converges in probability".



            Take for example $W_n = -X_1+X_2+X_3+...+X_n$



            Then



            $$lim_{nrightarrow infty} Pleft(left|frac {S_n}{sqrt n} - frac {W_n}{sqrt n}right|> epsilonright) = lim_{nrightarrow infty} Pleft(left|frac {2X_1}{sqrt n} right|> epsilonright) = 0$$



            and the criterion for convergence in probability is satisfied.



            So I suspect that "$S_n/sqrt n$ does not converge in probability" must have a more specific and narrow sense in the OP's case.





            On another front, the established phrase "$S_n/sqrt n$ converges in distribution to a random variable Z" sometimes makes us forget that the phenomenon described by "convergence in distribution" is that the sequence of distribution functions $F_n$ of $S_n/sqrt n$ converges to a certain distribution function $F$. There is really no $Z$ "at the end of the journey" waiting to "become one" with $S_n/sqrt n$ . $Z$ is a random variable, a separate entity from the distribution that characterizes it (which characterizes also an infinite number of other such $Z$'s). If there is no random variable, the question "is $S_n/sqrt n$ independent of $Z$?" cannot even be posed.






            share|cite|improve this answer























            • In response to your last paragraph: Yes, I think for this reason this proof (were it valid) would necessarily have to be posed as a proof by contradiction (unlike many proofs by contradiction which can be rephrased as direct proofs of contrapositives). You have to assume the sequence converges in probability to $Z$ in order to even be able to say what $Z$ is.
              – Jack M
              Jan 3 '16 at 7:47






            • 1




              I don't understand your construction though. To what are you claiming your $S_n/sqrt n$ converges in probability?
              – Jack M
              Jan 3 '16 at 7:50














            -1












            -1








            -1






            One can easily construct an entity to which we can say that "$S_n/sqrt n$ converges in probability".



            Take for example $W_n = -X_1+X_2+X_3+...+X_n$



            Then



            $$lim_{nrightarrow infty} Pleft(left|frac {S_n}{sqrt n} - frac {W_n}{sqrt n}right|> epsilonright) = lim_{nrightarrow infty} Pleft(left|frac {2X_1}{sqrt n} right|> epsilonright) = 0$$



            and the criterion for convergence in probability is satisfied.



            So I suspect that "$S_n/sqrt n$ does not converge in probability" must have a more specific and narrow sense in the OP's case.





            On another front, the established phrase "$S_n/sqrt n$ converges in distribution to a random variable Z" sometimes makes us forget that the phenomenon described by "convergence in distribution" is that the sequence of distribution functions $F_n$ of $S_n/sqrt n$ converges to a certain distribution function $F$. There is really no $Z$ "at the end of the journey" waiting to "become one" with $S_n/sqrt n$ . $Z$ is a random variable, a separate entity from the distribution that characterizes it (which characterizes also an infinite number of other such $Z$'s). If there is no random variable, the question "is $S_n/sqrt n$ independent of $Z$?" cannot even be posed.






            share|cite|improve this answer














            One can easily construct an entity to which we can say that "$S_n/sqrt n$ converges in probability".



            Take for example $W_n = -X_1+X_2+X_3+...+X_n$



            Then



            $$lim_{nrightarrow infty} Pleft(left|frac {S_n}{sqrt n} - frac {W_n}{sqrt n}right|> epsilonright) = lim_{nrightarrow infty} Pleft(left|frac {2X_1}{sqrt n} right|> epsilonright) = 0$$



            and the criterion for convergence in probability is satisfied.



            So I suspect that "$S_n/sqrt n$ does not converge in probability" must have a more specific and narrow sense in the OP's case.





            On another front, the established phrase "$S_n/sqrt n$ converges in distribution to a random variable Z" sometimes makes us forget that the phenomenon described by "convergence in distribution" is that the sequence of distribution functions $F_n$ of $S_n/sqrt n$ converges to a certain distribution function $F$. There is really no $Z$ "at the end of the journey" waiting to "become one" with $S_n/sqrt n$ . $Z$ is a random variable, a separate entity from the distribution that characterizes it (which characterizes also an infinite number of other such $Z$'s). If there is no random variable, the question "is $S_n/sqrt n$ independent of $Z$?" cannot even be posed.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 11 '18 at 9:22

























            answered Jan 3 '16 at 6:15









            Alecos PapadopoulosAlecos Papadopoulos

            8,20811535




            8,20811535












            • In response to your last paragraph: Yes, I think for this reason this proof (were it valid) would necessarily have to be posed as a proof by contradiction (unlike many proofs by contradiction which can be rephrased as direct proofs of contrapositives). You have to assume the sequence converges in probability to $Z$ in order to even be able to say what $Z$ is.
              – Jack M
              Jan 3 '16 at 7:47






            • 1




              I don't understand your construction though. To what are you claiming your $S_n/sqrt n$ converges in probability?
              – Jack M
              Jan 3 '16 at 7:50


















            • In response to your last paragraph: Yes, I think for this reason this proof (were it valid) would necessarily have to be posed as a proof by contradiction (unlike many proofs by contradiction which can be rephrased as direct proofs of contrapositives). You have to assume the sequence converges in probability to $Z$ in order to even be able to say what $Z$ is.
              – Jack M
              Jan 3 '16 at 7:47






            • 1




              I don't understand your construction though. To what are you claiming your $S_n/sqrt n$ converges in probability?
              – Jack M
              Jan 3 '16 at 7:50
















            In response to your last paragraph: Yes, I think for this reason this proof (were it valid) would necessarily have to be posed as a proof by contradiction (unlike many proofs by contradiction which can be rephrased as direct proofs of contrapositives). You have to assume the sequence converges in probability to $Z$ in order to even be able to say what $Z$ is.
            – Jack M
            Jan 3 '16 at 7:47




            In response to your last paragraph: Yes, I think for this reason this proof (were it valid) would necessarily have to be posed as a proof by contradiction (unlike many proofs by contradiction which can be rephrased as direct proofs of contrapositives). You have to assume the sequence converges in probability to $Z$ in order to even be able to say what $Z$ is.
            – Jack M
            Jan 3 '16 at 7:47




            1




            1




            I don't understand your construction though. To what are you claiming your $S_n/sqrt n$ converges in probability?
            – Jack M
            Jan 3 '16 at 7:50




            I don't understand your construction though. To what are you claiming your $S_n/sqrt n$ converges in probability?
            – Jack M
            Jan 3 '16 at 7:50


















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