Expectation of $XY$ bounded for all bounded $Y$ implies $X$ is $L^p$
$begingroup$
I'm trying to prove:
Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.
My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?
real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces
asked Jan 10 at 19:54
D FordD Ford
675313
$endgroup$
add a comment |
$begingroup$
I'm trying to prove:
Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.
My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?
real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces
asked Jan 10 at 19:54
D FordD Ford
675313
$endgroup$
add a comment |
$begingroup$
I'm trying to prove:
Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.
My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?
real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces
asked Jan 10 at 19:54
D FordD Ford
675313
$endgroup$
I'm trying to prove:
Let $X$ be a real random variable, $p, q in (1,infty)$, $frac 1 p + frac 1 q = 1$. If there is $C < infty$ such that $|mathbb E[XY]| leq C ||Y||_q$ for any bounded random variable $Y$, then $X$ is in $mathcal L^p$.
My idea is to use the fact that $left(L^q(mathbb P)right)' cong L^p(mathbb P)$, and to show $F : L^q(mathbb P) to mathbb R$ defined by $F(Y) = mathbb E[XY]$ is continuous. For then, the isomorphism in particular is the isometry $kappa(f) = left( Y mapsto mathbb E[fY]right)$, so we must have $f = X$. But I'm not sure if that conclusion is correct, nor am I sure how to prove $F$ is continuous over $L^q(mathbb P)$; only over bounded functions in $L^q(mathbb P)$. Any suggestions?
real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces
real-analysis functional-analysis probability-theory lebesgue-integral lp-spaces
asked Jan 10 at 19:54
D FordD Ford
675313
asked Jan 10 at 19:54
D FordD Ford
675313
asked Jan 10 at 19:54
D FordD Ford
675313
asked Jan 10 at 19:54
D FordD Ford
675313
asked Jan 10 at 19:54
D FordD Ford
675313
675313
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
$$
|X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
$$ By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all bounded $Y$. Then we can use monotone convergence theorem to conclude
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.
answered Jan 10 at 20:22
SongSong
18.6k21651
$endgroup$
$begingroup$
Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
$endgroup$
– D Ford
Jan 11 at 21:30
1
$begingroup$
In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
$endgroup$
– Song
Jan 11 at 21:37
add a comment |
$begingroup$
Hint: for each fixed $n$, apply the assumption to the random variable
$$
Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
$$
where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.
Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.
answered Jan 10 at 20:25
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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active
oldest
votes
$begingroup$
Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
$$
|X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
$$ By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all bounded $Y$. Then we can use monotone convergence theorem to conclude
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.
answered Jan 10 at 20:22
SongSong
18.6k21651
$endgroup$
$begingroup$
Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
$endgroup$
– D Ford
Jan 11 at 21:30
1
$begingroup$
In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
$endgroup$
– Song
Jan 11 at 21:37
add a comment |
$begingroup$
Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
$$
|X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
$$ By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all bounded $Y$. Then we can use monotone convergence theorem to conclude
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.
answered Jan 10 at 20:22
SongSong
18.6k21651
$endgroup$
$begingroup$
Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
$endgroup$
– D Ford
Jan 11 at 21:30
1
$begingroup$
In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
$endgroup$
– Song
Jan 11 at 21:37
add a comment |
$begingroup$
Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
$$
|X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
$$ By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all bounded $Y$. Then we can use monotone convergence theorem to conclude
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.
answered Jan 10 at 20:22
SongSong
18.6k21651
$endgroup$
Hint: This is about how you can apply Riesz representation theorem saying that $[L^q]^*=L^p$. Let $Omega$ denote the underlying space. First note that by choosing an appropriate $theta(omega)$ for each $omega$ (in a measurable way), we can make
$$
|X(omega)||Y(omega)|= X(omega)Y(omega)e^{itheta(omega)}.
$$ By letting $Y'(omega)=Y(omega)e^{itheta(omega)}$, we can improve the inequality to
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all bounded $Y$. Then we can use monotone convergence theorem to conclude
$$
E[|X||Y|]le C|Y|_{L^q}
$$ for all $Yin L^q$. Now deduce the conclusion that $Xin L^p$.
answered Jan 10 at 20:22
SongSong
18.6k21651
edited Jan 10 at 20:30
answered Jan 10 at 20:22
SongSong
18.6k21651
answered Jan 10 at 20:22
SongSong
18.6k21651
answered Jan 10 at 20:22
SongSong
18.6k21651
18.6k21651
$begingroup$
Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
$endgroup$
– D Ford
Jan 11 at 21:30
1
$begingroup$
In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
$endgroup$
– Song
Jan 11 at 21:37
add a comment |
$begingroup$
Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
$endgroup$
– D Ford
Jan 11 at 21:30
1
$begingroup$
In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
$endgroup$
– Song
Jan 11 at 21:37
$begingroup$
Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
$endgroup$
– D Ford
Jan 11 at 21:30
$begingroup$
Still working on showing that $X in L^p$ at this point. Since $F(Y) = mathbb E[XY]$ is a continuous linear functional on $L^q$, Riesz representation gives us a unique $f in L^p$ so that $mathbb E[fY] = mathbb E[XY]$ for all $Y in L^q$, but is it obvious that $X = f$? Because we don't know at this point that $X in L^p$.
$endgroup$
– D Ford
Jan 11 at 21:30
1
1
$begingroup$
In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
$endgroup$
– Song
Jan 11 at 21:37
$begingroup$
In fact, Riesz representation theorem also asserts uniqueness of $f$. Anyway, if we have $$E[(X-f)Y]=0$$ for all $Yin L^q$, then we can test it for $Y_1=1_{{X-f>0}}$ and $Y_2=1_{{X-f<0}}$. See what we can say about $X-f$.
$endgroup$
– Song
Jan 11 at 21:37
add a comment |
$begingroup$
Hint: for each fixed $n$, apply the assumption to the random variable
$$
Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
$$
where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.
Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.
answered Jan 10 at 20:25
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
Hint: for each fixed $n$, apply the assumption to the random variable
$$
Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
$$
where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.
Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.
answered Jan 10 at 20:25
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
Hint: for each fixed $n$, apply the assumption to the random variable
$$
Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
$$
where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.
Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.
answered Jan 10 at 20:25
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
Hint: for each fixed $n$, apply the assumption to the random variable
$$
Y=Y_n= operatorname{sgn}left(Xright)leftlvert Xrightrvert^{p-1}mathbf 1{leftlvert Xrightrvertleqslant n},
$$
where $operatorname{sgn}left(Xright)=1$ if $X$ is positive, $-1$ if $X$ is negative and $0$ for $X=0$.
This will give a bounded on $mathbb Eleft[leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}right]$ which does not depend on $n$.
Indeed, let $X_n:=leftlvert Xrightrvert^{p}mathbf 1{leftlvert Xrightrvertleqslant n}$ and $x_n:=mathbb Eleft[X_nright]$, Then we got that $x_nleqslant cx_n^{1/q}$.
answered Jan 10 at 20:25
Davide GiraudoDavide Giraudo
128k17156268
edited Jan 12 at 10:34
answered Jan 10 at 20:25
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 10 at 20:25
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 10 at 20:25
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
add a comment |
add a comment |
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