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
asked Dec 12 '18 at 18:39
James CaribertJames Caribert
162
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add a comment |
$begingroup$
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
asked Dec 12 '18 at 18:39
James CaribertJames Caribert
162
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 12 '18 at 18:39
James CaribertJames Caribert
162
$endgroup$
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
asked Dec 12 '18 at 18:39
James CaribertJames Caribert
162
asked Dec 12 '18 at 18:39
James CaribertJames Caribert
162
asked Dec 12 '18 at 18:39
James CaribertJames Caribert
162
asked Dec 12 '18 at 18:39
James CaribertJames Caribert
162
asked Dec 12 '18 at 18:39
James CaribertJames Caribert
162
162
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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.
answered Dec 12 '18 at 18:47
FedericoFederico
5,124514
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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.
answered Dec 12 '18 at 18:47
FedericoFederico
5,124514
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 18:47
FedericoFederico
5,124514
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 18:47
FedericoFederico
5,124514
$endgroup$
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.
answered Dec 12 '18 at 18:47
FedericoFederico
5,124514
answered Dec 12 '18 at 18:47
FedericoFederico
5,124514
answered Dec 12 '18 at 18:47
FedericoFederico
5,124514
answered Dec 12 '18 at 18:47
FedericoFederico
5,124514
5,124514
add a comment |
add a comment |
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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