Convergence in square mean












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If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?



I can find a sequence that satisfies the two first conditions ($P(X_n=0)=1-1/n^2$ and $P(X_n=n)=1-1/n^2$), however I can't see how to show wether these two conditions imply no convergence in mean square.






probability







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  • $begingroup$
    I'm assuming you mean $mathbb P(X_n=n)=1/n^2$. In this case, it is true that $mathbb E[X_n^2]=1$, but this does not mean it converges in mean square to $1$. For this to be the case we must have $$mathbb E[|X_n-1|^2]stackrel{ntoinfty}longrightarrow 0 $$ which is not true.
    $endgroup$
    – Math1000
    Dec 13 '18 at 22:24
















1












$begingroup$


If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?



I can find a sequence that satisfies the two first conditions ($P(X_n=0)=1-1/n^2$ and $P(X_n=n)=1-1/n^2$), however I can't see how to show wether these two conditions imply no convergence in mean square.






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    I'm assuming you mean $mathbb P(X_n=n)=1/n^2$. In this case, it is true that $mathbb E[X_n^2]=1$, but this does not mean it converges in mean square to $1$. For this to be the case we must have $$mathbb E[|X_n-1|^2]stackrel{ntoinfty}longrightarrow 0 $$ which is not true.
    $endgroup$
    – Math1000
    Dec 13 '18 at 22:24














1












1








1





$begingroup$


If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?



I can find a sequence that satisfies the two first conditions ($P(X_n=0)=1-1/n^2$ and $P(X_n=n)=1-1/n^2$), however I can't see how to show wether these two conditions imply no convergence in mean square.






probability







share|cite|improve this question









$endgroup$




If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?



I can find a sequence that satisfies the two first conditions ($P(X_n=0)=1-1/n^2$ and $P(X_n=n)=1-1/n^2$), however I can't see how to show wether these two conditions imply no convergence in mean square.






probability




probability






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asked Dec 12 '18 at 18:39









James CaribertJames Caribert

192




192












  • $begingroup$
    I'm assuming you mean $mathbb P(X_n=n)=1/n^2$. In this case, it is true that $mathbb E[X_n^2]=1$, but this does not mean it converges in mean square to $1$. For this to be the case we must have $$mathbb E[|X_n-1|^2]stackrel{ntoinfty}longrightarrow 0 $$ which is not true.
    $endgroup$
    – Math1000
    Dec 13 '18 at 22:24


















  • $begingroup$
    I'm assuming you mean $mathbb P(X_n=n)=1/n^2$. In this case, it is true that $mathbb E[X_n^2]=1$, but this does not mean it converges in mean square to $1$. For this to be the case we must have $$mathbb E[|X_n-1|^2]stackrel{ntoinfty}longrightarrow 0 $$ which is not true.
    $endgroup$
    – Math1000
    Dec 13 '18 at 22:24
















$begingroup$
I'm assuming you mean $mathbb P(X_n=n)=1/n^2$. In this case, it is true that $mathbb E[X_n^2]=1$, but this does not mean it converges in mean square to $1$. For this to be the case we must have $$mathbb E[|X_n-1|^2]stackrel{ntoinfty}longrightarrow 0 $$ which is not true.
$endgroup$
– Math1000
Dec 13 '18 at 22:24




$begingroup$
I'm assuming you mean $mathbb P(X_n=n)=1/n^2$. In this case, it is true that $mathbb E[X_n^2]=1$, but this does not mean it converges in mean square to $1$. For this to be the case we must have $$mathbb E[|X_n-1|^2]stackrel{ntoinfty}longrightarrow 0 $$ which is not true.
$endgroup$
– Math1000
Dec 13 '18 at 22:24










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$begingroup$


If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?




If $X_n$ converges to something in square mean, it converges also in mean (see here). But $X_n$ already converges in mean to $0$, while it doesn't converge in square mean to it. Therefore $X_n$ cannot converge in square mean to anything else either.






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    $begingroup$


    If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?




    If $X_n$ converges to something in square mean, it converges also in mean (see here). But $X_n$ already converges in mean to $0$, while it doesn't converge in square mean to it. Therefore $X_n$ cannot converge in square mean to anything else either.






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      2












      $begingroup$


      If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?




      If $X_n$ converges to something in square mean, it converges also in mean (see here). But $X_n$ already converges in mean to $0$, while it doesn't converge in square mean to it. Therefore $X_n$ cannot converge in square mean to anything else either.






      share|cite|improve this answer









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        2












        2








        2





        $begingroup$


        If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?




        If $X_n$ converges to something in square mean, it converges also in mean (see here). But $X_n$ already converges in mean to $0$, while it doesn't converge in square mean to it. Therefore $X_n$ cannot converge in square mean to anything else either.






        share|cite|improve this answer









        $endgroup$




        If $X_n$ is a sequence of random variables such that $E(|X_n|)to0$ but $E(X_n^2)to 1$, does it imply that $X_n$ doesn't converge in square mean?




        If $X_n$ converges to something in square mean, it converges also in mean (see here). But $X_n$ already converges in mean to $0$, while it doesn't converge in square mean to it. Therefore $X_n$ cannot converge in square mean to anything else either.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 18:47









        FedericoFederico

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