Big-O and small-o in probability notation: if $X_n=O_P(1/n)$ then $X_n=o_P(1)$? [closed]

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Does $ X_n = O_p(1/n) $ imply $ X_n = o_p(1) $?




probability probability-theory asymptotics




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closed as off-topic by Shailesh, John Ma, Adrian Keister, Xander Henderson, Taroccoesbrocco Aug 14 at 1:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, John Ma, Adrian Keister, Xander Henderson, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Please provide more context/details to your question. Also, what are your thoughts on this?
    – an4s
    Aug 13 at 6:33






  • 3




    What do $O_p,o_p$ mean? I mean the $ _p$ part.
    – copper.hat
    Aug 13 at 6:34






  • 1




    @copper.hat see here
    – pointguard0
    Aug 13 at 7:07















up vote
1
down vote

favorite
1












Does $ X_n = O_p(1/n) $ imply $ X_n = o_p(1) $?




probability probability-theory asymptotics




share|cite|improve this question














closed as off-topic by Shailesh, John Ma, Adrian Keister, Xander Henderson, Taroccoesbrocco Aug 14 at 1:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, John Ma, Adrian Keister, Xander Henderson, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Please provide more context/details to your question. Also, what are your thoughts on this?
    – an4s
    Aug 13 at 6:33






  • 3




    What do $O_p,o_p$ mean? I mean the $ _p$ part.
    – copper.hat
    Aug 13 at 6:34






  • 1




    @copper.hat see here
    – pointguard0
    Aug 13 at 7:07













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Does $ X_n = O_p(1/n) $ imply $ X_n = o_p(1) $?




probability probability-theory asymptotics




share|cite|improve this question














Does $ X_n = O_p(1/n) $ imply $ X_n = o_p(1) $?





probability probability-theory asymptotics





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 13 at 9:27









pointguard0

1,271821




1,271821










asked Aug 13 at 6:29









Dorro D.

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264




closed as off-topic by Shailesh, John Ma, Adrian Keister, Xander Henderson, Taroccoesbrocco Aug 14 at 1:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, John Ma, Adrian Keister, Xander Henderson, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Shailesh, John Ma, Adrian Keister, Xander Henderson, Taroccoesbrocco Aug 14 at 1:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, John Ma, Adrian Keister, Xander Henderson, Taroccoesbrocco
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    Please provide more context/details to your question. Also, what are your thoughts on this?
    – an4s
    Aug 13 at 6:33






  • 3




    What do $O_p,o_p$ mean? I mean the $ _p$ part.
    – copper.hat
    Aug 13 at 6:34






  • 1




    @copper.hat see here
    – pointguard0
    Aug 13 at 7:07













  • 1




    Please provide more context/details to your question. Also, what are your thoughts on this?
    – an4s
    Aug 13 at 6:33






  • 3




    What do $O_p,o_p$ mean? I mean the $ _p$ part.
    – copper.hat
    Aug 13 at 6:34






  • 1




    @copper.hat see here
    – pointguard0
    Aug 13 at 7:07








1




1




Please provide more context/details to your question. Also, what are your thoughts on this?
– an4s
Aug 13 at 6:33




Please provide more context/details to your question. Also, what are your thoughts on this?
– an4s
Aug 13 at 6:33




3




3




What do $O_p,o_p$ mean? I mean the $ _p$ part.
– copper.hat
Aug 13 at 6:34




What do $O_p,o_p$ mean? I mean the $ _p$ part.
– copper.hat
Aug 13 at 6:34




1




1




@copper.hat see here
– pointguard0
Aug 13 at 7:07





@copper.hat see here
– pointguard0
Aug 13 at 7:07











2 Answers
2






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up vote
1
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accepted










Yes, it does.



$X_n = mathcalO_pleft(frac 1 n right)$ means $forall ~ varepsilon > 0, ~ exists delta, N_varepsilon: ~ mathbbP(|n cdot X_n| > delta(varepsilon)) < varepsilon, ~ forall n > N_varepsilon$. Hence, $mathbbP(|X_n| > delta(varepsilon)/n) < varepsilon$.



We need to show that $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$. From $X_n = mathcalO_pleft(frac 1 n right)$ we have $delta$ of the form $delta/n$, so it can be made arbitrary through $n$. So taking $delta_1 = delta(varepsilon)/n$ we arrive at the desired result $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$.






share|cite|improve this answer






















  • I was't aware about that specific definition in probabilistic notation! That's nice to know.
    – gimusi
    Aug 13 at 7:32

















up vote
1
down vote













$X_n = o_p(1)$ means that $X_n rightarrow_p 0$



and



$X_n = O_p(1/n)$ means for all $ delta > 0 $ there is a $ C > 0 $ such that $ mathbbP left(leftvert dfracX_n1/n rightvert leq C right) geq 1- delta $






share|cite|improve this answer





























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Yes, it does.



    $X_n = mathcalO_pleft(frac 1 n right)$ means $forall ~ varepsilon > 0, ~ exists delta, N_varepsilon: ~ mathbbP(|n cdot X_n| > delta(varepsilon)) < varepsilon, ~ forall n > N_varepsilon$. Hence, $mathbbP(|X_n| > delta(varepsilon)/n) < varepsilon$.



    We need to show that $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$. From $X_n = mathcalO_pleft(frac 1 n right)$ we have $delta$ of the form $delta/n$, so it can be made arbitrary through $n$. So taking $delta_1 = delta(varepsilon)/n$ we arrive at the desired result $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$.






    share|cite|improve this answer






















    • I was't aware about that specific definition in probabilistic notation! That's nice to know.
      – gimusi
      Aug 13 at 7:32














    up vote
    1
    down vote



    accepted










    Yes, it does.



    $X_n = mathcalO_pleft(frac 1 n right)$ means $forall ~ varepsilon > 0, ~ exists delta, N_varepsilon: ~ mathbbP(|n cdot X_n| > delta(varepsilon)) < varepsilon, ~ forall n > N_varepsilon$. Hence, $mathbbP(|X_n| > delta(varepsilon)/n) < varepsilon$.



    We need to show that $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$. From $X_n = mathcalO_pleft(frac 1 n right)$ we have $delta$ of the form $delta/n$, so it can be made arbitrary through $n$. So taking $delta_1 = delta(varepsilon)/n$ we arrive at the desired result $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$.






    share|cite|improve this answer






















    • I was't aware about that specific definition in probabilistic notation! That's nice to know.
      – gimusi
      Aug 13 at 7:32












    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    Yes, it does.



    $X_n = mathcalO_pleft(frac 1 n right)$ means $forall ~ varepsilon > 0, ~ exists delta, N_varepsilon: ~ mathbbP(|n cdot X_n| > delta(varepsilon)) < varepsilon, ~ forall n > N_varepsilon$. Hence, $mathbbP(|X_n| > delta(varepsilon)/n) < varepsilon$.



    We need to show that $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$. From $X_n = mathcalO_pleft(frac 1 n right)$ we have $delta$ of the form $delta/n$, so it can be made arbitrary through $n$. So taking $delta_1 = delta(varepsilon)/n$ we arrive at the desired result $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$.






    share|cite|improve this answer














    Yes, it does.



    $X_n = mathcalO_pleft(frac 1 n right)$ means $forall ~ varepsilon > 0, ~ exists delta, N_varepsilon: ~ mathbbP(|n cdot X_n| > delta(varepsilon)) < varepsilon, ~ forall n > N_varepsilon$. Hence, $mathbbP(|X_n| > delta(varepsilon)/n) < varepsilon$.



    We need to show that $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$. From $X_n = mathcalO_pleft(frac 1 n right)$ we have $delta$ of the form $delta/n$, so it can be made arbitrary through $n$. So taking $delta_1 = delta(varepsilon)/n$ we arrive at the desired result $forall varepsilon_1, delta_1 ~ ~ exists N_varepsilon, delta: mathbbP(|X_n| > delta_1) < varepsilon_1$ for all $n > N_varepsilon, delta$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 13 at 7:12

























    answered Aug 13 at 7:05









    pointguard0

    1,271821




    1,271821











    • I was't aware about that specific definition in probabilistic notation! That's nice to know.
      – gimusi
      Aug 13 at 7:32
















    • I was't aware about that specific definition in probabilistic notation! That's nice to know.
      – gimusi
      Aug 13 at 7:32















    I was't aware about that specific definition in probabilistic notation! That's nice to know.
    – gimusi
    Aug 13 at 7:32




    I was't aware about that specific definition in probabilistic notation! That's nice to know.
    – gimusi
    Aug 13 at 7:32










    up vote
    1
    down vote













    $X_n = o_p(1)$ means that $X_n rightarrow_p 0$



    and



    $X_n = O_p(1/n)$ means for all $ delta > 0 $ there is a $ C > 0 $ such that $ mathbbP left(leftvert dfracX_n1/n rightvert leq C right) geq 1- delta $






    share|cite|improve this answer


























      up vote
      1
      down vote













      $X_n = o_p(1)$ means that $X_n rightarrow_p 0$



      and



      $X_n = O_p(1/n)$ means for all $ delta > 0 $ there is a $ C > 0 $ such that $ mathbbP left(leftvert dfracX_n1/n rightvert leq C right) geq 1- delta $






      share|cite|improve this answer
























        up vote
        1
        down vote










        up vote
        1
        down vote









        $X_n = o_p(1)$ means that $X_n rightarrow_p 0$



        and



        $X_n = O_p(1/n)$ means for all $ delta > 0 $ there is a $ C > 0 $ such that $ mathbbP left(leftvert dfracX_n1/n rightvert leq C right) geq 1- delta $






        share|cite|improve this answer














        $X_n = o_p(1)$ means that $X_n rightarrow_p 0$



        and



        $X_n = O_p(1/n)$ means for all $ delta > 0 $ there is a $ C > 0 $ such that $ mathbbP left(leftvert dfracX_n1/n rightvert leq C right) geq 1- delta $







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 13 at 7:28









        pointguard0

        1,271821




        1,271821










        answered Aug 13 at 7:01









        Dorro D.

        264




        264












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