Show that if $p > 1$ and $sup_{n geq 1} |X_n| in L^p$ then ${X_n, n geq 1}$ is uniformly integrable.
$begingroup$
Let $(Omega,mathcal{Sigma,mathbb{P}})$ be a complete probability space.
I have to show that $$lim_{alpha to infty} sup_{n geq 1}int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} = 0$$
assuming title's assumptions and let $alpha > 0$
my attempt :
$| X_n| leq sup_{ngeq 1}|X_n| leq [sup_{ngeq 1}|X_n|]^p$
therefore
$$int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} leq int_{{|X_n| geq alpha}} [sup_{ngeq 1}|X_n|]^p, dmathbb{P} leq mathbb{E}[ [sup_{ngeq 1}|X_n|]^p] < infty $$
so for every $n geq 1$
$$int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} < infty $$
however I'm not sure this completes the proof because although it's finite it could possibly depend on $n$ and taking the supremum with $n$'s in an expression can cause a blow up.
any hints or help will be greatly appreciated, thanks !
probability-theory lp-spaces uniform-integrability
edited Jan 21 at 22:27
Davide Giraudo
128k17156268
asked Jan 18 at 17:23
rapidracimrapidracim
1,7541419
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal{Sigma,mathbb{P}})$ be a complete probability space.
I have to show that $$lim_{alpha to infty} sup_{n geq 1}int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} = 0$$
assuming title's assumptions and let $alpha > 0$
my attempt :
$| X_n| leq sup_{ngeq 1}|X_n| leq [sup_{ngeq 1}|X_n|]^p$
therefore
$$int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} leq int_{{|X_n| geq alpha}} [sup_{ngeq 1}|X_n|]^p, dmathbb{P} leq mathbb{E}[ [sup_{ngeq 1}|X_n|]^p] < infty $$
so for every $n geq 1$
$$int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} < infty $$
however I'm not sure this completes the proof because although it's finite it could possibly depend on $n$ and taking the supremum with $n$'s in an expression can cause a blow up.
any hints or help will be greatly appreciated, thanks !
probability-theory lp-spaces uniform-integrability
edited Jan 21 at 22:27
Davide Giraudo
128k17156268
asked Jan 18 at 17:23
rapidracimrapidracim
1,7541419
$endgroup$
1
$begingroup$
I think there is a typo in your title since this is untrue in the case $p=1$ (see here). I've edited your title to remove this case. Your attempt doesn't work. The first line is wrong since if $x < 1$, $p>1$ then $x$ is not smaller than $x^p$.
$endgroup$
– Rhys Steele
Jan 18 at 17:30
add a comment |
$begingroup$
Let $(Omega,mathcal{Sigma,mathbb{P}})$ be a complete probability space.
I have to show that $$lim_{alpha to infty} sup_{n geq 1}int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} = 0$$
assuming title's assumptions and let $alpha > 0$
my attempt :
$| X_n| leq sup_{ngeq 1}|X_n| leq [sup_{ngeq 1}|X_n|]^p$
therefore
$$int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} leq int_{{|X_n| geq alpha}} [sup_{ngeq 1}|X_n|]^p, dmathbb{P} leq mathbb{E}[ [sup_{ngeq 1}|X_n|]^p] < infty $$
so for every $n geq 1$
$$int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} < infty $$
however I'm not sure this completes the proof because although it's finite it could possibly depend on $n$ and taking the supremum with $n$'s in an expression can cause a blow up.
any hints or help will be greatly appreciated, thanks !
probability-theory lp-spaces uniform-integrability
edited Jan 21 at 22:27
Davide Giraudo
128k17156268
asked Jan 18 at 17:23
rapidracimrapidracim
1,7541419
$endgroup$
Let $(Omega,mathcal{Sigma,mathbb{P}})$ be a complete probability space.
I have to show that $$lim_{alpha to infty} sup_{n geq 1}int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} = 0$$
assuming title's assumptions and let $alpha > 0$
my attempt :
$| X_n| leq sup_{ngeq 1}|X_n| leq [sup_{ngeq 1}|X_n|]^p$
therefore
$$int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} leq int_{{|X_n| geq alpha}} [sup_{ngeq 1}|X_n|]^p, dmathbb{P} leq mathbb{E}[ [sup_{ngeq 1}|X_n|]^p] < infty $$
so for every $n geq 1$
$$int_{{|X_n| geq alpha}}|X_n|, dmathbb{P} < infty $$
however I'm not sure this completes the proof because although it's finite it could possibly depend on $n$ and taking the supremum with $n$'s in an expression can cause a blow up.
any hints or help will be greatly appreciated, thanks !
probability-theory lp-spaces uniform-integrability
probability-theory lp-spaces uniform-integrability
edited Jan 21 at 22:27
Davide Giraudo
128k17156268
asked Jan 18 at 17:23
rapidracimrapidracim
1,7541419
edited Jan 21 at 22:27
Davide Giraudo
128k17156268
asked Jan 18 at 17:23
rapidracimrapidracim
1,7541419
edited Jan 21 at 22:27
Davide Giraudo
128k17156268
edited Jan 21 at 22:27
Davide Giraudo
128k17156268
edited Jan 21 at 22:27
Davide Giraudo
128k17156268
128k17156268
asked Jan 18 at 17:23
rapidracimrapidracim
1,7541419
asked Jan 18 at 17:23
rapidracimrapidracim
1,7541419
asked Jan 18 at 17:23
rapidracimrapidracim
1,7541419
1,7541419
1
$begingroup$
I think there is a typo in your title since this is untrue in the case $p=1$ (see here). I've edited your title to remove this case. Your attempt doesn't work. The first line is wrong since if $x < 1$, $p>1$ then $x$ is not smaller than $x^p$.
$endgroup$
– Rhys Steele
Jan 18 at 17:30
add a comment |
1
$begingroup$
I think there is a typo in your title since this is untrue in the case $p=1$ (see here). I've edited your title to remove this case. Your attempt doesn't work. The first line is wrong since if $x < 1$, $p>1$ then $x$ is not smaller than $x^p$.
$endgroup$
– Rhys Steele
Jan 18 at 17:30
1
1
$begingroup$
I think there is a typo in your title since this is untrue in the case $p=1$ (see here). I've edited your title to remove this case. Your attempt doesn't work. The first line is wrong since if $x < 1$, $p>1$ then $x$ is not smaller than $x^p$.
$endgroup$
– Rhys Steele
Jan 18 at 17:30
$begingroup$
I think there is a typo in your title since this is untrue in the case $p=1$ (see here). I've edited your title to remove this case. Your attempt doesn't work. The first line is wrong since if $x < 1$, $p>1$ then $x$ is not smaller than $x^p$.
$endgroup$
– Rhys Steele
Jan 18 at 17:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can use the fact that
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}int_{|X_n|ge alpha}|X_n|^pdBbb P.
$$ If $suplimits_{ninBbb N}|X_n|_p=M<infty$ (our case is a special case of this), then we have
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p,quadforall ninBbb N.
$$ This gives
$$
sup_{ninBbb N}int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p
$$ and we get the desired conclusion by taking $alphatoinfty.$
answered Jan 18 at 17:33
SongSong
18.6k21651
$endgroup$
$begingroup$
thanks ! , I actually know this technique to prove that the boundedness in $L^p$ implies uniform integrability, what I fail to see is that $|sup_{n geq 1} X_n|_p < infty implies sup_{n geq 1}|X_n|_p < infty$, I'm gonna think about this
$endgroup$
– rapidracim
Jan 18 at 17:42
1
$begingroup$
@rapidracim Note that $int |X_n|^p leq int (sup |X_n|)^p$. This gives the inequality you want since the right hand side is uniform in $n$.
$endgroup$
– Rhys Steele
Jan 18 at 17:45
$begingroup$
@RhysSteele I see it now, thanks !
$endgroup$
– rapidracim
Jan 18 at 17:47
add a comment |
Your Answer
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1 Answer
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$begingroup$
We can use the fact that
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}int_{|X_n|ge alpha}|X_n|^pdBbb P.
$$ If $suplimits_{ninBbb N}|X_n|_p=M<infty$ (our case is a special case of this), then we have
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p,quadforall ninBbb N.
$$ This gives
$$
sup_{ninBbb N}int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p
$$ and we get the desired conclusion by taking $alphatoinfty.$
answered Jan 18 at 17:33
SongSong
18.6k21651
$endgroup$
$begingroup$
thanks ! , I actually know this technique to prove that the boundedness in $L^p$ implies uniform integrability, what I fail to see is that $|sup_{n geq 1} X_n|_p < infty implies sup_{n geq 1}|X_n|_p < infty$, I'm gonna think about this
$endgroup$
– rapidracim
Jan 18 at 17:42
1
$begingroup$
@rapidracim Note that $int |X_n|^p leq int (sup |X_n|)^p$. This gives the inequality you want since the right hand side is uniform in $n$.
$endgroup$
– Rhys Steele
Jan 18 at 17:45
$begingroup$
@RhysSteele I see it now, thanks !
$endgroup$
– rapidracim
Jan 18 at 17:47
add a comment |
$begingroup$
We can use the fact that
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}int_{|X_n|ge alpha}|X_n|^pdBbb P.
$$ If $suplimits_{ninBbb N}|X_n|_p=M<infty$ (our case is a special case of this), then we have
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p,quadforall ninBbb N.
$$ This gives
$$
sup_{ninBbb N}int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p
$$ and we get the desired conclusion by taking $alphatoinfty.$
answered Jan 18 at 17:33
SongSong
18.6k21651
$endgroup$
$begingroup$
thanks ! , I actually know this technique to prove that the boundedness in $L^p$ implies uniform integrability, what I fail to see is that $|sup_{n geq 1} X_n|_p < infty implies sup_{n geq 1}|X_n|_p < infty$, I'm gonna think about this
$endgroup$
– rapidracim
Jan 18 at 17:42
1
$begingroup$
@rapidracim Note that $int |X_n|^p leq int (sup |X_n|)^p$. This gives the inequality you want since the right hand side is uniform in $n$.
$endgroup$
– Rhys Steele
Jan 18 at 17:45
$begingroup$
@RhysSteele I see it now, thanks !
$endgroup$
– rapidracim
Jan 18 at 17:47
add a comment |
$begingroup$
We can use the fact that
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}int_{|X_n|ge alpha}|X_n|^pdBbb P.
$$ If $suplimits_{ninBbb N}|X_n|_p=M<infty$ (our case is a special case of this), then we have
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p,quadforall ninBbb N.
$$ This gives
$$
sup_{ninBbb N}int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p
$$ and we get the desired conclusion by taking $alphatoinfty.$
answered Jan 18 at 17:33
SongSong
18.6k21651
$endgroup$
We can use the fact that
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}int_{|X_n|ge alpha}|X_n|^pdBbb P.
$$ If $suplimits_{ninBbb N}|X_n|_p=M<infty$ (our case is a special case of this), then we have
$$
int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p,quadforall ninBbb N.
$$ This gives
$$
sup_{ninBbb N}int_{|X_n|ge alpha}|X_n|dBbb Plefrac{1}{alpha^{p-1}}M^p
$$ and we get the desired conclusion by taking $alphatoinfty.$
answered Jan 18 at 17:33
SongSong
18.6k21651
edited Jan 18 at 17:39
answered Jan 18 at 17:33
SongSong
18.6k21651
answered Jan 18 at 17:33
SongSong
18.6k21651
answered Jan 18 at 17:33
SongSong
18.6k21651
18.6k21651
$begingroup$
thanks ! , I actually know this technique to prove that the boundedness in $L^p$ implies uniform integrability, what I fail to see is that $|sup_{n geq 1} X_n|_p < infty implies sup_{n geq 1}|X_n|_p < infty$, I'm gonna think about this
$endgroup$
– rapidracim
Jan 18 at 17:42
1
$begingroup$
@rapidracim Note that $int |X_n|^p leq int (sup |X_n|)^p$. This gives the inequality you want since the right hand side is uniform in $n$.
$endgroup$
– Rhys Steele
Jan 18 at 17:45
$begingroup$
@RhysSteele I see it now, thanks !
$endgroup$
– rapidracim
Jan 18 at 17:47
add a comment |
$begingroup$
thanks ! , I actually know this technique to prove that the boundedness in $L^p$ implies uniform integrability, what I fail to see is that $|sup_{n geq 1} X_n|_p < infty implies sup_{n geq 1}|X_n|_p < infty$, I'm gonna think about this
$endgroup$
– rapidracim
Jan 18 at 17:42
1
$begingroup$
@rapidracim Note that $int |X_n|^p leq int (sup |X_n|)^p$. This gives the inequality you want since the right hand side is uniform in $n$.
$endgroup$
– Rhys Steele
Jan 18 at 17:45
$begingroup$
@RhysSteele I see it now, thanks !
$endgroup$
– rapidracim
Jan 18 at 17:47
$begingroup$
thanks ! , I actually know this technique to prove that the boundedness in $L^p$ implies uniform integrability, what I fail to see is that $|sup_{n geq 1} X_n|_p < infty implies sup_{n geq 1}|X_n|_p < infty$, I'm gonna think about this
$endgroup$
– rapidracim
Jan 18 at 17:42
$begingroup$
thanks ! , I actually know this technique to prove that the boundedness in $L^p$ implies uniform integrability, what I fail to see is that $|sup_{n geq 1} X_n|_p < infty implies sup_{n geq 1}|X_n|_p < infty$, I'm gonna think about this
$endgroup$
– rapidracim
Jan 18 at 17:42
1
1
$begingroup$
@rapidracim Note that $int |X_n|^p leq int (sup |X_n|)^p$. This gives the inequality you want since the right hand side is uniform in $n$.
$endgroup$
– Rhys Steele
Jan 18 at 17:45
$begingroup$
@rapidracim Note that $int |X_n|^p leq int (sup |X_n|)^p$. This gives the inequality you want since the right hand side is uniform in $n$.
$endgroup$
– Rhys Steele
Jan 18 at 17:45
$begingroup$
@RhysSteele I see it now, thanks !
$endgroup$
– rapidracim
Jan 18 at 17:47
$begingroup$
@RhysSteele I see it now, thanks !
$endgroup$
– rapidracim
Jan 18 at 17:47
add a comment |
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$begingroup$
I think there is a typo in your title since this is untrue in the case $p=1$ (see here). I've edited your title to remove this case. Your attempt doesn't work. The first line is wrong since if $x < 1$, $p>1$ then $x$ is not smaller than $x^p$.
$endgroup$
– Rhys Steele
Jan 18 at 17:30