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Example: $X_n $ is uniformly integrable but $E[ f(X_n) ]=infty$ with $f(x)=omega(x)$ [closed]
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I am looking for a counterexample where the sequence of random variables is $X_n $ is uniformly integrable, that is
beginalign
lim_ k to infty sup_n ge 0 E[ |X_n| 1_ ge k ]=0
endalign
and yet for every non-negative and non-decreasing function $f(x)$ and $f(x)=omega(x)$ we have that
beginalign
sup_n ge 0 E[f(|X_n|)]=infty.
endalign
Where $f(x)=omega(x)$ means that $ lim_x to infty |f(x)|/|x|=0$.
probability-theory expectation examples-counterexamples uniform-integrability
edited Jul 17 at 7:38
zhoraster
15.2k21752
asked Jul 16 at 17:22
Lisa
629213
closed as off-topic by Did, amWhy, José Carlos Santos, Davide Giraudo, Parcly Taxel Jul 17 at 15:51
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." РDid, amWhy, Jos̩ Carlos Santos, Davide Giraudo, Parcly Taxel
add a comment |Â
up vote
0
down vote
favorite
I am looking for a counterexample where the sequence of random variables is $X_n $ is uniformly integrable, that is
beginalign
lim_ k to infty sup_n ge 0 E[ |X_n| 1_ ge k ]=0
endalign
and yet for every non-negative and non-decreasing function $f(x)$ and $f(x)=omega(x)$ we have that
beginalign
sup_n ge 0 E[f(|X_n|)]=infty.
endalign
Where $f(x)=omega(x)$ means that $ lim_x to infty |f(x)|/|x|=0$.
probability-theory expectation examples-counterexamples uniform-integrability
edited Jul 17 at 7:38
zhoraster
15.2k21752
asked Jul 16 at 17:22
Lisa
629213
closed as off-topic by Did, amWhy, José Carlos Santos, Davide Giraudo, Parcly Taxel Jul 17 at 15:51
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." РDid, amWhy, Jos̩ Carlos Santos, Davide Giraudo, Parcly Taxel
You might want to add some thoughts on the problem, e.g. why you think that such a (counter)example exists.
– saz
Jul 16 at 17:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am looking for a counterexample where the sequence of random variables is $X_n $ is uniformly integrable, that is
beginalign
lim_ k to infty sup_n ge 0 E[ |X_n| 1_ ge k ]=0
endalign
and yet for every non-negative and non-decreasing function $f(x)$ and $f(x)=omega(x)$ we have that
beginalign
sup_n ge 0 E[f(|X_n|)]=infty.
endalign
Where $f(x)=omega(x)$ means that $ lim_x to infty |f(x)|/|x|=0$.
probability-theory expectation examples-counterexamples uniform-integrability
edited Jul 17 at 7:38
zhoraster
15.2k21752
asked Jul 16 at 17:22
Lisa
629213
I am looking for a counterexample where the sequence of random variables is $X_n $ is uniformly integrable, that is
beginalign
lim_ k to infty sup_n ge 0 E[ |X_n| 1_ ge k ]=0
endalign
and yet for every non-negative and non-decreasing function $f(x)$ and $f(x)=omega(x)$ we have that
beginalign
sup_n ge 0 E[f(|X_n|)]=infty.
endalign
Where $f(x)=omega(x)$ means that $ lim_x to infty |f(x)|/|x|=0$.
probability-theory expectation examples-counterexamples uniform-integrability
edited Jul 17 at 7:38
zhoraster
15.2k21752
asked Jul 16 at 17:22
Lisa
629213
edited Jul 17 at 7:38
zhoraster
15.2k21752
edited Jul 17 at 7:38
zhoraster
15.2k21752
edited Jul 17 at 7:38
zhoraster
15.2k21752
15.2k21752
asked Jul 16 at 17:22
Lisa
629213
asked Jul 16 at 17:22
Lisa
629213
asked Jul 16 at 17:22
Lisa
629213
629213
closed as off-topic by Did, amWhy, José Carlos Santos, Davide Giraudo, Parcly Taxel Jul 17 at 15:51
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." РDid, amWhy, Jos̩ Carlos Santos, Davide Giraudo, Parcly Taxel
closed as off-topic by Did, amWhy, José Carlos Santos, Davide Giraudo, Parcly Taxel Jul 17 at 15:51
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." РDid, amWhy, Jos̩ Carlos Santos, Davide Giraudo, Parcly Taxel
You might want to add some thoughts on the problem, e.g. why you think that such a (counter)example exists.
– saz
Jul 16 at 17:47
add a comment |Â
You might want to add some thoughts on the problem, e.g. why you think that such a (counter)example exists.
– saz
Jul 16 at 17:47
You might want to add some thoughts on the problem, e.g. why you think that such a (counter)example exists.
– saz
Jul 16 at 17:47
You might want to add some thoughts on the problem, e.g. why you think that such a (counter)example exists.
– saz
Jul 16 at 17:47
add a comment |Â
2 Answers
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up vote
1
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I believe that you actually meant $lim_xto+infty |f(x)|/x = +infty$.
Still, there is no such counterexample as, in fact, de la Vallée Poussin condition for uniform integrability is a criterion. Here is the proof of the inverse conclusion:
Let $k_m, mge 1$ be such that $sup_nge 1mathbb Eleft[|X_n|mathbf1_X_nright]le 2^-m$; without loss of generality we can assume that $k_m+1> k_m$ for all $mge 1$. Setting $f(x) = xsum_m=1^infty mmathbf1_[k_m,k_m+1)(x)$, we have for any $nge 1$,
begingather*
mathbb E[f(|X_n|)] = sum_m=1^infty m,mathbb Eleft[|X_n|mathbf1_[k_m,k_m+1)(|X_n|)right]le sum_m=1^infty m 2^-m,
endgather*
so $sup_nge 1 mathbb E[f(|X_n|)]le sum_m=1^infty m2^-m<infty$, as required.
answered Jul 17 at 7:16
zhoraster
15.2k21752
add a comment |Â
up vote
0
down vote
Actually, the conditions on $X_n$ and $f$ imply that $sup Ef(|X_n|) <infty $: Since $f(x)<x$ for $x$ exceeding some number $x_0$ and $f(x) leq f(x_0)$ for $x leq x_0$ we see that $f(x) leq x+c$ for all $x$ where $c=f(x_0)$. Thus $Ef(|X_n|) leq c+E|X_n|$. Uniform integrability implies that $$ is bounded.
answered Jul 17 at 6:06
Kavi Rama Murthy
21k2829
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I believe that you actually meant $lim_xto+infty |f(x)|/x = +infty$.
Still, there is no such counterexample as, in fact, de la Vallée Poussin condition for uniform integrability is a criterion. Here is the proof of the inverse conclusion:
Let $k_m, mge 1$ be such that $sup_nge 1mathbb Eleft[|X_n|mathbf1_X_nright]le 2^-m$; without loss of generality we can assume that $k_m+1> k_m$ for all $mge 1$. Setting $f(x) = xsum_m=1^infty mmathbf1_[k_m,k_m+1)(x)$, we have for any $nge 1$,
begingather*
mathbb E[f(|X_n|)] = sum_m=1^infty m,mathbb Eleft[|X_n|mathbf1_[k_m,k_m+1)(|X_n|)right]le sum_m=1^infty m 2^-m,
endgather*
so $sup_nge 1 mathbb E[f(|X_n|)]le sum_m=1^infty m2^-m<infty$, as required.
answered Jul 17 at 7:16
zhoraster
15.2k21752
add a comment |Â
up vote
1
down vote
accepted
I believe that you actually meant $lim_xto+infty |f(x)|/x = +infty$.
Still, there is no such counterexample as, in fact, de la Vallée Poussin condition for uniform integrability is a criterion. Here is the proof of the inverse conclusion:
Let $k_m, mge 1$ be such that $sup_nge 1mathbb Eleft[|X_n|mathbf1_X_nright]le 2^-m$; without loss of generality we can assume that $k_m+1> k_m$ for all $mge 1$. Setting $f(x) = xsum_m=1^infty mmathbf1_[k_m,k_m+1)(x)$, we have for any $nge 1$,
begingather*
mathbb E[f(|X_n|)] = sum_m=1^infty m,mathbb Eleft[|X_n|mathbf1_[k_m,k_m+1)(|X_n|)right]le sum_m=1^infty m 2^-m,
endgather*
so $sup_nge 1 mathbb E[f(|X_n|)]le sum_m=1^infty m2^-m<infty$, as required.
answered Jul 17 at 7:16
zhoraster
15.2k21752
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I believe that you actually meant $lim_xto+infty |f(x)|/x = +infty$.
Still, there is no such counterexample as, in fact, de la Vallée Poussin condition for uniform integrability is a criterion. Here is the proof of the inverse conclusion:
Let $k_m, mge 1$ be such that $sup_nge 1mathbb Eleft[|X_n|mathbf1_X_nright]le 2^-m$; without loss of generality we can assume that $k_m+1> k_m$ for all $mge 1$. Setting $f(x) = xsum_m=1^infty mmathbf1_[k_m,k_m+1)(x)$, we have for any $nge 1$,
begingather*
mathbb E[f(|X_n|)] = sum_m=1^infty m,mathbb Eleft[|X_n|mathbf1_[k_m,k_m+1)(|X_n|)right]le sum_m=1^infty m 2^-m,
endgather*
so $sup_nge 1 mathbb E[f(|X_n|)]le sum_m=1^infty m2^-m<infty$, as required.
answered Jul 17 at 7:16
zhoraster
15.2k21752
I believe that you actually meant $lim_xto+infty |f(x)|/x = +infty$.
Still, there is no such counterexample as, in fact, de la Vallée Poussin condition for uniform integrability is a criterion. Here is the proof of the inverse conclusion:
Let $k_m, mge 1$ be such that $sup_nge 1mathbb Eleft[|X_n|mathbf1_X_nright]le 2^-m$; without loss of generality we can assume that $k_m+1> k_m$ for all $mge 1$. Setting $f(x) = xsum_m=1^infty mmathbf1_[k_m,k_m+1)(x)$, we have for any $nge 1$,
begingather*
mathbb E[f(|X_n|)] = sum_m=1^infty m,mathbb Eleft[|X_n|mathbf1_[k_m,k_m+1)(|X_n|)right]le sum_m=1^infty m 2^-m,
endgather*
so $sup_nge 1 mathbb E[f(|X_n|)]le sum_m=1^infty m2^-m<infty$, as required.
answered Jul 17 at 7:16
zhoraster
15.2k21752
edited Jul 17 at 7:22
answered Jul 17 at 7:16
zhoraster
15.2k21752
answered Jul 17 at 7:16
zhoraster
15.2k21752
answered Jul 17 at 7:16
zhoraster
15.2k21752
15.2k21752
add a comment |Â
add a comment |Â
up vote
0
down vote
Actually, the conditions on $X_n$ and $f$ imply that $sup Ef(|X_n|) <infty $: Since $f(x)<x$ for $x$ exceeding some number $x_0$ and $f(x) leq f(x_0)$ for $x leq x_0$ we see that $f(x) leq x+c$ for all $x$ where $c=f(x_0)$. Thus $Ef(|X_n|) leq c+E|X_n|$. Uniform integrability implies that $$ is bounded.
answered Jul 17 at 6:06
Kavi Rama Murthy
21k2829
add a comment |Â
up vote
0
down vote
Actually, the conditions on $X_n$ and $f$ imply that $sup Ef(|X_n|) <infty $: Since $f(x)<x$ for $x$ exceeding some number $x_0$ and $f(x) leq f(x_0)$ for $x leq x_0$ we see that $f(x) leq x+c$ for all $x$ where $c=f(x_0)$. Thus $Ef(|X_n|) leq c+E|X_n|$. Uniform integrability implies that $$ is bounded.
answered Jul 17 at 6:06
Kavi Rama Murthy
21k2829
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Actually, the conditions on $X_n$ and $f$ imply that $sup Ef(|X_n|) <infty $: Since $f(x)<x$ for $x$ exceeding some number $x_0$ and $f(x) leq f(x_0)$ for $x leq x_0$ we see that $f(x) leq x+c$ for all $x$ where $c=f(x_0)$. Thus $Ef(|X_n|) leq c+E|X_n|$. Uniform integrability implies that $$ is bounded.
answered Jul 17 at 6:06
Kavi Rama Murthy
21k2829
Actually, the conditions on $X_n$ and $f$ imply that $sup Ef(|X_n|) <infty $: Since $f(x)<x$ for $x$ exceeding some number $x_0$ and $f(x) leq f(x_0)$ for $x leq x_0$ we see that $f(x) leq x+c$ for all $x$ where $c=f(x_0)$. Thus $Ef(|X_n|) leq c+E|X_n|$. Uniform integrability implies that $$ is bounded.
answered Jul 17 at 6:06
Kavi Rama Murthy
21k2829
answered Jul 17 at 6:06
Kavi Rama Murthy
21k2829
answered Jul 17 at 6:06
Kavi Rama Murthy
21k2829
answered Jul 17 at 6:06
Kavi Rama Murthy
21k2829
21k2829
add a comment |Â
add a comment |Â
You might want to add some thoughts on the problem, e.g. why you think that such a (counter)example exists.
– saz
Jul 16 at 17:47