Existence of a function $f$ with $mathbb E[X^2f(X)]$ being finite for some $X$ with finite second moment
$begingroup$
$newcommand{E}{mathbb E}newcommand{PM}{mathbb P}$This question is inspired by this question which is unfortunately closed. Anyways, I found it interesting, so I tried to solve it.
The problem was as follows:
Problem. Let $X$ be a random variable which has a finite second moment, i.e. $E[X^2]<infty$. Show that there exists a non-negative function $f:mathbb Rto[0,infty)$ such that the following holds:
$f$ is non-decreasing on $(0,infty)$
$f(x)toinfty$ as $xtoinfty$
- $E[X^2f(X)]<infty$
Attempt.
Firstly, I thought that $f(x)=log(1+|x|)$ should do the work, but after some thoughts I have discovered that it is not guaranteed to work. Secondly, I have considered an Ansatz method which basically ended up with
begin{align*}
f(x) = sum_{k=1}^infty frac{1}{k} mathbf{1}_{{a_k leq x}}
end{align*}
which has all the properties if $a_k$ is strictly increasing to $infty$. Now we have
begin{align*}
Eleft[ X^2 f(X)right]&=Eleft[X^2sum_{k=1}^infty frac{1}{k}mathbf{1}_{{a_kleq X }}right]\
&stackrel{text{Fatou}}{leq} sum_{k=1}^infty frac 1 k E[X^2mathbf{1}_{{a_kleq X }}]
end{align*}
I really would like to have
begin{align*}tag{$*$}
E[X^2mathbf{1}_{{a_kleq X }}]leq frac 1 {k^alpha}
end{align*}
for some $alpha>0$. We know that
begin{align*}
lim_{ytoinfty} E[X^2 mathbf{1}_{{yleq X}}]stackrel{text{DCT}}{=}0
end{align*}
which means that we can find a sequence $a_k$ stricly increasing to infinity satisfying $(*)$. We have constructed $f$ with all the properties.
Question. Are there other methods for constructing such function $f$? Can it be done easier?
To be honest, it costed me a couple of trials to actually come up with an example.
probability-theory measure-theory expected-value
asked Dec 8 '18 at 13:16
ShashiShashi
7,1731528
$endgroup$
add a comment |
$begingroup$
$newcommand{E}{mathbb E}newcommand{PM}{mathbb P}$This question is inspired by this question which is unfortunately closed. Anyways, I found it interesting, so I tried to solve it.
The problem was as follows:
Problem. Let $X$ be a random variable which has a finite second moment, i.e. $E[X^2]<infty$. Show that there exists a non-negative function $f:mathbb Rto[0,infty)$ such that the following holds:
$f$ is non-decreasing on $(0,infty)$
$f(x)toinfty$ as $xtoinfty$
- $E[X^2f(X)]<infty$
Attempt.
Firstly, I thought that $f(x)=log(1+|x|)$ should do the work, but after some thoughts I have discovered that it is not guaranteed to work. Secondly, I have considered an Ansatz method which basically ended up with
begin{align*}
f(x) = sum_{k=1}^infty frac{1}{k} mathbf{1}_{{a_k leq x}}
end{align*}
which has all the properties if $a_k$ is strictly increasing to $infty$. Now we have
begin{align*}
Eleft[ X^2 f(X)right]&=Eleft[X^2sum_{k=1}^infty frac{1}{k}mathbf{1}_{{a_kleq X }}right]\
&stackrel{text{Fatou}}{leq} sum_{k=1}^infty frac 1 k E[X^2mathbf{1}_{{a_kleq X }}]
end{align*}
I really would like to have
begin{align*}tag{$*$}
E[X^2mathbf{1}_{{a_kleq X }}]leq frac 1 {k^alpha}
end{align*}
for some $alpha>0$. We know that
begin{align*}
lim_{ytoinfty} E[X^2 mathbf{1}_{{yleq X}}]stackrel{text{DCT}}{=}0
end{align*}
which means that we can find a sequence $a_k$ stricly increasing to infinity satisfying $(*)$. We have constructed $f$ with all the properties.
Question. Are there other methods for constructing such function $f$? Can it be done easier?
To be honest, it costed me a couple of trials to actually come up with an example.
probability-theory measure-theory expected-value
asked Dec 8 '18 at 13:16
ShashiShashi
7,1731528
$endgroup$
$begingroup$
related? canizo.org/tex/vallee-poussin.pdf
$endgroup$
– Calvin Khor
Dec 13 '18 at 9:45
1
$begingroup$
@CalvinKhor yes, thanks!!
$endgroup$
– Shashi
Dec 13 '18 at 11:19
$begingroup$
I don't have time to make an answer from that, but glad to help
$endgroup$
– Calvin Khor
Dec 13 '18 at 11:24
add a comment |
$begingroup$
$newcommand{E}{mathbb E}newcommand{PM}{mathbb P}$This question is inspired by this question which is unfortunately closed. Anyways, I found it interesting, so I tried to solve it.
The problem was as follows:
Problem. Let $X$ be a random variable which has a finite second moment, i.e. $E[X^2]<infty$. Show that there exists a non-negative function $f:mathbb Rto[0,infty)$ such that the following holds:
$f$ is non-decreasing on $(0,infty)$
$f(x)toinfty$ as $xtoinfty$
- $E[X^2f(X)]<infty$
Attempt.
Firstly, I thought that $f(x)=log(1+|x|)$ should do the work, but after some thoughts I have discovered that it is not guaranteed to work. Secondly, I have considered an Ansatz method which basically ended up with
begin{align*}
f(x) = sum_{k=1}^infty frac{1}{k} mathbf{1}_{{a_k leq x}}
end{align*}
which has all the properties if $a_k$ is strictly increasing to $infty$. Now we have
begin{align*}
Eleft[ X^2 f(X)right]&=Eleft[X^2sum_{k=1}^infty frac{1}{k}mathbf{1}_{{a_kleq X }}right]\
&stackrel{text{Fatou}}{leq} sum_{k=1}^infty frac 1 k E[X^2mathbf{1}_{{a_kleq X }}]
end{align*}
I really would like to have
begin{align*}tag{$*$}
E[X^2mathbf{1}_{{a_kleq X }}]leq frac 1 {k^alpha}
end{align*}
for some $alpha>0$. We know that
begin{align*}
lim_{ytoinfty} E[X^2 mathbf{1}_{{yleq X}}]stackrel{text{DCT}}{=}0
end{align*}
which means that we can find a sequence $a_k$ stricly increasing to infinity satisfying $(*)$. We have constructed $f$ with all the properties.
Question. Are there other methods for constructing such function $f$? Can it be done easier?
To be honest, it costed me a couple of trials to actually come up with an example.
probability-theory measure-theory expected-value
asked Dec 8 '18 at 13:16
ShashiShashi
7,1731528
$endgroup$
$newcommand{E}{mathbb E}newcommand{PM}{mathbb P}$This question is inspired by this question which is unfortunately closed. Anyways, I found it interesting, so I tried to solve it.
The problem was as follows:
Problem. Let $X$ be a random variable which has a finite second moment, i.e. $E[X^2]<infty$. Show that there exists a non-negative function $f:mathbb Rto[0,infty)$ such that the following holds:
$f$ is non-decreasing on $(0,infty)$
$f(x)toinfty$ as $xtoinfty$
- $E[X^2f(X)]<infty$
Attempt.
Firstly, I thought that $f(x)=log(1+|x|)$ should do the work, but after some thoughts I have discovered that it is not guaranteed to work. Secondly, I have considered an Ansatz method which basically ended up with
begin{align*}
f(x) = sum_{k=1}^infty frac{1}{k} mathbf{1}_{{a_k leq x}}
end{align*}
which has all the properties if $a_k$ is strictly increasing to $infty$. Now we have
begin{align*}
Eleft[ X^2 f(X)right]&=Eleft[X^2sum_{k=1}^infty frac{1}{k}mathbf{1}_{{a_kleq X }}right]\
&stackrel{text{Fatou}}{leq} sum_{k=1}^infty frac 1 k E[X^2mathbf{1}_{{a_kleq X }}]
end{align*}
I really would like to have
begin{align*}tag{$*$}
E[X^2mathbf{1}_{{a_kleq X }}]leq frac 1 {k^alpha}
end{align*}
for some $alpha>0$. We know that
begin{align*}
lim_{ytoinfty} E[X^2 mathbf{1}_{{yleq X}}]stackrel{text{DCT}}{=}0
end{align*}
which means that we can find a sequence $a_k$ stricly increasing to infinity satisfying $(*)$. We have constructed $f$ with all the properties.
Question. Are there other methods for constructing such function $f$? Can it be done easier?
To be honest, it costed me a couple of trials to actually come up with an example.
probability-theory measure-theory expected-value
probability-theory measure-theory expected-value
asked Dec 8 '18 at 13:16
ShashiShashi
7,1731528
asked Dec 8 '18 at 13:16
ShashiShashi
7,1731528
edited Dec 8 '18 at 13:26
Shashi
asked Dec 8 '18 at 13:16
ShashiShashi
7,1731528
asked Dec 8 '18 at 13:16
ShashiShashi
7,1731528
asked Dec 8 '18 at 13:16
ShashiShashi
7,1731528
7,1731528
$begingroup$
related? canizo.org/tex/vallee-poussin.pdf
$endgroup$
– Calvin Khor
Dec 13 '18 at 9:45
1
$begingroup$
@CalvinKhor yes, thanks!!
$endgroup$
– Shashi
Dec 13 '18 at 11:19
$begingroup$
I don't have time to make an answer from that, but glad to help
$endgroup$
– Calvin Khor
Dec 13 '18 at 11:24
add a comment |
$begingroup$
related? canizo.org/tex/vallee-poussin.pdf
$endgroup$
– Calvin Khor
Dec 13 '18 at 9:45
1
$begingroup$
@CalvinKhor yes, thanks!!
$endgroup$
– Shashi
Dec 13 '18 at 11:19
$begingroup$
I don't have time to make an answer from that, but glad to help
$endgroup$
– Calvin Khor
Dec 13 '18 at 11:24
$begingroup$
related? canizo.org/tex/vallee-poussin.pdf
$endgroup$
– Calvin Khor
Dec 13 '18 at 9:45
$begingroup$
related? canizo.org/tex/vallee-poussin.pdf
$endgroup$
– Calvin Khor
Dec 13 '18 at 9:45
1
1
$begingroup$
@CalvinKhor yes, thanks!!
$endgroup$
– Shashi
Dec 13 '18 at 11:19
$begingroup$
@CalvinKhor yes, thanks!!
$endgroup$
– Shashi
Dec 13 '18 at 11:19
$begingroup$
I don't have time to make an answer from that, but glad to help
$endgroup$
– Calvin Khor
Dec 13 '18 at 11:24
$begingroup$
I don't have time to make an answer from that, but glad to help
$endgroup$
– Calvin Khor
Dec 13 '18 at 11:24
add a comment |
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$begingroup$
related? canizo.org/tex/vallee-poussin.pdf
$endgroup$
– Calvin Khor
Dec 13 '18 at 9:45
1
$begingroup$
@CalvinKhor yes, thanks!!
$endgroup$
– Shashi
Dec 13 '18 at 11:19
$begingroup$
I don't have time to make an answer from that, but glad to help
$endgroup$
– Calvin Khor
Dec 13 '18 at 11:24