Does sequence converge in probability?












0












$begingroup$


Let $X_1, X_2, X_3,...$ be a sequence of nonnegative random variables such that $limlimits_{ntoinfty} E[X_n]=0$.



I want to know if the sequence $Y_n=1-e^{-X_n}$ converges in probability and if so, what is the limit.



My thought is to approach this problem using the Markov inequality:



$$P[X>epsilon]leqfrac{E[X]}{epsilon}$$



Then, by plugging in $Y_n$, I get,



$$P[Y_n>epsilon]leqfrac{E[1-e^{-X_n}]}{epsilon}$$



and since $limlimits_{ntoinfty} E[X_n]=0$, then $E[1-e^{-X_n}]$ must also go to $0$ as $ntoinfty$. Therefore, $Y_n$ converges in probability to $0$.



Is this a correct proof?






probability sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the meaning of $X_n$?
    $endgroup$
    – callculus
    Jan 8 at 19:48










  • $begingroup$
    Of course, you might be asked to include a proof of the step that $E(X_n)to0$ implies $E(1-e^{-X_n})to0$. Can you do that?
    $endgroup$
    – Did
    Jan 8 at 20:58






  • 1




    $begingroup$
    @callculus There is no specific meaning other than what is stated. X_i are a sequence of rvs and the limit of that sequence has a mean of zer0.
    $endgroup$
    – Avedis
    Jan 8 at 21:02






  • 1




    $begingroup$
    @callculus It doesnt mean what you implied. It could, but the problem is specifically not commenting. You are given the facts that you have a sequence, the rv's are non-negative, and the expectation of the limit is 0.
    $endgroup$
    – Avedis
    Jan 8 at 21:08






  • 1




    $begingroup$
    You are again asserting that $E(X_n)to0$ implies $E(e^{-X_n})to1$, not proving it. Can you prove this?
    $endgroup$
    – Did
    Jan 8 at 21:31
















0












$begingroup$


Let $X_1, X_2, X_3,...$ be a sequence of nonnegative random variables such that $limlimits_{ntoinfty} E[X_n]=0$.



I want to know if the sequence $Y_n=1-e^{-X_n}$ converges in probability and if so, what is the limit.



My thought is to approach this problem using the Markov inequality:



$$P[X>epsilon]leqfrac{E[X]}{epsilon}$$



Then, by plugging in $Y_n$, I get,



$$P[Y_n>epsilon]leqfrac{E[1-e^{-X_n}]}{epsilon}$$



and since $limlimits_{ntoinfty} E[X_n]=0$, then $E[1-e^{-X_n}]$ must also go to $0$ as $ntoinfty$. Therefore, $Y_n$ converges in probability to $0$.



Is this a correct proof?






probability sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the meaning of $X_n$?
    $endgroup$
    – callculus
    Jan 8 at 19:48










  • $begingroup$
    Of course, you might be asked to include a proof of the step that $E(X_n)to0$ implies $E(1-e^{-X_n})to0$. Can you do that?
    $endgroup$
    – Did
    Jan 8 at 20:58






  • 1




    $begingroup$
    @callculus There is no specific meaning other than what is stated. X_i are a sequence of rvs and the limit of that sequence has a mean of zer0.
    $endgroup$
    – Avedis
    Jan 8 at 21:02






  • 1




    $begingroup$
    @callculus It doesnt mean what you implied. It could, but the problem is specifically not commenting. You are given the facts that you have a sequence, the rv's are non-negative, and the expectation of the limit is 0.
    $endgroup$
    – Avedis
    Jan 8 at 21:08






  • 1




    $begingroup$
    You are again asserting that $E(X_n)to0$ implies $E(e^{-X_n})to1$, not proving it. Can you prove this?
    $endgroup$
    – Did
    Jan 8 at 21:31














0












0








0


0



$begingroup$


Let $X_1, X_2, X_3,...$ be a sequence of nonnegative random variables such that $limlimits_{ntoinfty} E[X_n]=0$.



I want to know if the sequence $Y_n=1-e^{-X_n}$ converges in probability and if so, what is the limit.



My thought is to approach this problem using the Markov inequality:



$$P[X>epsilon]leqfrac{E[X]}{epsilon}$$



Then, by plugging in $Y_n$, I get,



$$P[Y_n>epsilon]leqfrac{E[1-e^{-X_n}]}{epsilon}$$



and since $limlimits_{ntoinfty} E[X_n]=0$, then $E[1-e^{-X_n}]$ must also go to $0$ as $ntoinfty$. Therefore, $Y_n$ converges in probability to $0$.



Is this a correct proof?






probability sequences-and-series convergence







share|cite|improve this question











$endgroup$




Let $X_1, X_2, X_3,...$ be a sequence of nonnegative random variables such that $limlimits_{ntoinfty} E[X_n]=0$.



I want to know if the sequence $Y_n=1-e^{-X_n}$ converges in probability and if so, what is the limit.



My thought is to approach this problem using the Markov inequality:



$$P[X>epsilon]leqfrac{E[X]}{epsilon}$$



Then, by plugging in $Y_n$, I get,



$$P[Y_n>epsilon]leqfrac{E[1-e^{-X_n}]}{epsilon}$$



and since $limlimits_{ntoinfty} E[X_n]=0$, then $E[1-e^{-X_n}]$ must also go to $0$ as $ntoinfty$. Therefore, $Y_n$ converges in probability to $0$.



Is this a correct proof?






probability sequences-and-series convergence




probability sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 21:54









Shashi

7,1281528




7,1281528










asked Jan 8 at 19:44









AvedisAvedis

476




476












  • $begingroup$
    What is the meaning of $X_n$?
    $endgroup$
    – callculus
    Jan 8 at 19:48










  • $begingroup$
    Of course, you might be asked to include a proof of the step that $E(X_n)to0$ implies $E(1-e^{-X_n})to0$. Can you do that?
    $endgroup$
    – Did
    Jan 8 at 20:58






  • 1




    $begingroup$
    @callculus There is no specific meaning other than what is stated. X_i are a sequence of rvs and the limit of that sequence has a mean of zer0.
    $endgroup$
    – Avedis
    Jan 8 at 21:02






  • 1




    $begingroup$
    @callculus It doesnt mean what you implied. It could, but the problem is specifically not commenting. You are given the facts that you have a sequence, the rv's are non-negative, and the expectation of the limit is 0.
    $endgroup$
    – Avedis
    Jan 8 at 21:08






  • 1




    $begingroup$
    You are again asserting that $E(X_n)to0$ implies $E(e^{-X_n})to1$, not proving it. Can you prove this?
    $endgroup$
    – Did
    Jan 8 at 21:31


















  • $begingroup$
    What is the meaning of $X_n$?
    $endgroup$
    – callculus
    Jan 8 at 19:48










  • $begingroup$
    Of course, you might be asked to include a proof of the step that $E(X_n)to0$ implies $E(1-e^{-X_n})to0$. Can you do that?
    $endgroup$
    – Did
    Jan 8 at 20:58






  • 1




    $begingroup$
    @callculus There is no specific meaning other than what is stated. X_i are a sequence of rvs and the limit of that sequence has a mean of zer0.
    $endgroup$
    – Avedis
    Jan 8 at 21:02






  • 1




    $begingroup$
    @callculus It doesnt mean what you implied. It could, but the problem is specifically not commenting. You are given the facts that you have a sequence, the rv's are non-negative, and the expectation of the limit is 0.
    $endgroup$
    – Avedis
    Jan 8 at 21:08






  • 1




    $begingroup$
    You are again asserting that $E(X_n)to0$ implies $E(e^{-X_n})to1$, not proving it. Can you prove this?
    $endgroup$
    – Did
    Jan 8 at 21:31
















$begingroup$
What is the meaning of $X_n$?
$endgroup$
– callculus
Jan 8 at 19:48




$begingroup$
What is the meaning of $X_n$?
$endgroup$
– callculus
Jan 8 at 19:48












$begingroup$
Of course, you might be asked to include a proof of the step that $E(X_n)to0$ implies $E(1-e^{-X_n})to0$. Can you do that?
$endgroup$
– Did
Jan 8 at 20:58




$begingroup$
Of course, you might be asked to include a proof of the step that $E(X_n)to0$ implies $E(1-e^{-X_n})to0$. Can you do that?
$endgroup$
– Did
Jan 8 at 20:58




1




1




$begingroup$
@callculus There is no specific meaning other than what is stated. X_i are a sequence of rvs and the limit of that sequence has a mean of zer0.
$endgroup$
– Avedis
Jan 8 at 21:02




$begingroup$
@callculus There is no specific meaning other than what is stated. X_i are a sequence of rvs and the limit of that sequence has a mean of zer0.
$endgroup$
– Avedis
Jan 8 at 21:02




1




1




$begingroup$
@callculus It doesnt mean what you implied. It could, but the problem is specifically not commenting. You are given the facts that you have a sequence, the rv's are non-negative, and the expectation of the limit is 0.
$endgroup$
– Avedis
Jan 8 at 21:08




$begingroup$
@callculus It doesnt mean what you implied. It could, but the problem is specifically not commenting. You are given the facts that you have a sequence, the rv's are non-negative, and the expectation of the limit is 0.
$endgroup$
– Avedis
Jan 8 at 21:08




1




1




$begingroup$
You are again asserting that $E(X_n)to0$ implies $E(e^{-X_n})to1$, not proving it. Can you prove this?
$endgroup$
– Did
Jan 8 at 21:31




$begingroup$
You are again asserting that $E(X_n)to0$ implies $E(e^{-X_n})to1$, not proving it. Can you prove this?
$endgroup$
– Did
Jan 8 at 21:31










1 Answer
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oldest

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$begingroup$

It is well known that
$$0leq 1-e^{-x}leq x$$
for all $xgeq 0$ (this fact can be easily proven using Mean Value Theorem for instance). In the particular problem you are considering the inequality translates to
$$0leq Y_n leq X_n$$
That gives the result
$$lim_{ntoinfty} mathbb E|Y_n|=0$$
Hence $Y_n$ converges to $0$ in $L^1$ so it converges in probability as well (by Markov's inequality). You indeed said the result; here is a way to prove it.






share|cite|improve this answer











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    $begingroup$

    It is well known that
    $$0leq 1-e^{-x}leq x$$
    for all $xgeq 0$ (this fact can be easily proven using Mean Value Theorem for instance). In the particular problem you are considering the inequality translates to
    $$0leq Y_n leq X_n$$
    That gives the result
    $$lim_{ntoinfty} mathbb E|Y_n|=0$$
    Hence $Y_n$ converges to $0$ in $L^1$ so it converges in probability as well (by Markov's inequality). You indeed said the result; here is a way to prove it.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      It is well known that
      $$0leq 1-e^{-x}leq x$$
      for all $xgeq 0$ (this fact can be easily proven using Mean Value Theorem for instance). In the particular problem you are considering the inequality translates to
      $$0leq Y_n leq X_n$$
      That gives the result
      $$lim_{ntoinfty} mathbb E|Y_n|=0$$
      Hence $Y_n$ converges to $0$ in $L^1$ so it converges in probability as well (by Markov's inequality). You indeed said the result; here is a way to prove it.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        It is well known that
        $$0leq 1-e^{-x}leq x$$
        for all $xgeq 0$ (this fact can be easily proven using Mean Value Theorem for instance). In the particular problem you are considering the inequality translates to
        $$0leq Y_n leq X_n$$
        That gives the result
        $$lim_{ntoinfty} mathbb E|Y_n|=0$$
        Hence $Y_n$ converges to $0$ in $L^1$ so it converges in probability as well (by Markov's inequality). You indeed said the result; here is a way to prove it.






        share|cite|improve this answer











        $endgroup$



        It is well known that
        $$0leq 1-e^{-x}leq x$$
        for all $xgeq 0$ (this fact can be easily proven using Mean Value Theorem for instance). In the particular problem you are considering the inequality translates to
        $$0leq Y_n leq X_n$$
        That gives the result
        $$lim_{ntoinfty} mathbb E|Y_n|=0$$
        Hence $Y_n$ converges to $0$ in $L^1$ so it converges in probability as well (by Markov's inequality). You indeed said the result; here is a way to prove it.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 21:56

























        answered Jan 8 at 21:50









        ShashiShashi

        7,1281528




        7,1281528






























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