integrable random variables $X,Y$ such that $E(X|Y)=X$ a.s. and $E(Y|X)=Y$ a.s.
0
$begingroup$
Let $X,Y$ are random variables such that $E(|X|)+E(|Y|)<infty$, and the random variable $E(X|Y)=X$ a.s. and $E(Y|X)=Y$ a.s .
Then is it true that $X=Y$ a.s. ?
probability-theory measure-theory random-variables conditional-expectation expected-value
asked Dec 20 '18 at 3:59
user521337user521337
1,2041417
$endgroup$
add a comment |
0
$begingroup$
Let $X,Y$ are random variables such that $E(|X|)+E(|Y|)<infty$, and the random variable $E(X|Y)=X$ a.s. and $E(Y|X)=Y$ a.s .
Then is it true that $X=Y$ a.s. ?
probability-theory measure-theory random-variables conditional-expectation expected-value
asked Dec 20 '18 at 3:59
user521337user521337
1,2041417
$endgroup$
$begingroup$
What do you think?
$endgroup$
– zoidberg
Dec 20 '18 at 4:43
$begingroup$
@norfair: I'm not sure ... the converse is definitely true ... I also know that $X,Y$ integrable and $E(X|Y)=Y$ and $E(Y|X)=X$ a.s. implies $X=Y$ a.s. but I haven't been able to figure out the one I ask ...
$endgroup$
– user521337
Dec 20 '18 at 4:47
$begingroup$
@norfair: do you have any ideas ?
$endgroup$
– user521337
Dec 20 '18 at 4:52
add a comment |
0
0
0
1
$begingroup$
Let $X,Y$ are random variables such that $E(|X|)+E(|Y|)<infty$, and the random variable $E(X|Y)=X$ a.s. and $E(Y|X)=Y$ a.s .
Then is it true that $X=Y$ a.s. ?
probability-theory measure-theory random-variables conditional-expectation expected-value
asked Dec 20 '18 at 3:59
user521337user521337
1,2041417
$endgroup$
Let $X,Y$ are random variables such that $E(|X|)+E(|Y|)<infty$, and the random variable $E(X|Y)=X$ a.s. and $E(Y|X)=Y$ a.s .
Then is it true that $X=Y$ a.s. ?
probability-theory measure-theory random-variables conditional-expectation expected-value
probability-theory measure-theory random-variables conditional-expectation expected-value
asked Dec 20 '18 at 3:59
user521337user521337
1,2041417
asked Dec 20 '18 at 3:59
user521337user521337
1,2041417
edited Dec 20 '18 at 4:35
user521337
asked Dec 20 '18 at 3:59
user521337user521337
1,2041417
asked Dec 20 '18 at 3:59
user521337user521337
1,2041417
asked Dec 20 '18 at 3:59
user521337user521337
1,2041417
1,2041417
$begingroup$
What do you think?
$endgroup$
– zoidberg
Dec 20 '18 at 4:43
$begingroup$
@norfair: I'm not sure ... the converse is definitely true ... I also know that $X,Y$ integrable and $E(X|Y)=Y$ and $E(Y|X)=X$ a.s. implies $X=Y$ a.s. but I haven't been able to figure out the one I ask ...
$endgroup$
– user521337
Dec 20 '18 at 4:47
$begingroup$
@norfair: do you have any ideas ?
$endgroup$
– user521337
Dec 20 '18 at 4:52
add a comment |
$begingroup$
What do you think?
$endgroup$
– zoidberg
Dec 20 '18 at 4:43
$begingroup$
@norfair: I'm not sure ... the converse is definitely true ... I also know that $X,Y$ integrable and $E(X|Y)=Y$ and $E(Y|X)=X$ a.s. implies $X=Y$ a.s. but I haven't been able to figure out the one I ask ...
$endgroup$
– user521337
Dec 20 '18 at 4:47
$begingroup$
@norfair: do you have any ideas ?
$endgroup$
– user521337
Dec 20 '18 at 4:52
$begingroup$
What do you think?
$endgroup$
– zoidberg
Dec 20 '18 at 4:43
$begingroup$
What do you think?
$endgroup$
– zoidberg
Dec 20 '18 at 4:43
$begingroup$
@norfair: I'm not sure ... the converse is definitely true ... I also know that $X,Y$ integrable and $E(X|Y)=Y$ and $E(Y|X)=X$ a.s. implies $X=Y$ a.s. but I haven't been able to figure out the one I ask ...
$endgroup$
– user521337
Dec 20 '18 at 4:47
$begingroup$
@norfair: I'm not sure ... the converse is definitely true ... I also know that $X,Y$ integrable and $E(X|Y)=Y$ and $E(Y|X)=X$ a.s. implies $X=Y$ a.s. but I haven't been able to figure out the one I ask ...
$endgroup$
– user521337
Dec 20 '18 at 4:47
$begingroup$
@norfair: do you have any ideas ?
$endgroup$
– user521337
Dec 20 '18 at 4:52
$begingroup$
@norfair: do you have any ideas ?
$endgroup$
– user521337
Dec 20 '18 at 4:52
add a comment |
2 Answers
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If $Y=X^{3}$ then the two equations are satisfied but $X =Y$ may not hold.
answered Dec 20 '18 at 5:28
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
0
$begingroup$
Let $Y=2X$ for any nondegenerate and integrable RV $X.$
answered Dec 20 '18 at 5:27
spaceisdarkgreenspaceisdarkgreen
33.6k21753
$endgroup$
add a comment |
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2 Answers
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$begingroup$
If $Y=X^{3}$ then the two equations are satisfied but $X =Y$ may not hold.
answered Dec 20 '18 at 5:28
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
0
$begingroup$
If $Y=X^{3}$ then the two equations are satisfied but $X =Y$ may not hold.
answered Dec 20 '18 at 5:28
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
0
0
0
$begingroup$
If $Y=X^{3}$ then the two equations are satisfied but $X =Y$ may not hold.
answered Dec 20 '18 at 5:28
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
If $Y=X^{3}$ then the two equations are satisfied but $X =Y$ may not hold.
answered Dec 20 '18 at 5:28
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Dec 20 '18 at 5:28
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Dec 20 '18 at 5:28
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Dec 20 '18 at 5:28
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
69.1k53169
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add a comment |
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Let $Y=2X$ for any nondegenerate and integrable RV $X.$
answered Dec 20 '18 at 5:27
spaceisdarkgreenspaceisdarkgreen
33.6k21753
$endgroup$
add a comment |
0
$begingroup$
Let $Y=2X$ for any nondegenerate and integrable RV $X.$
answered Dec 20 '18 at 5:27
spaceisdarkgreenspaceisdarkgreen
33.6k21753
$endgroup$
add a comment |
0
0
0
$begingroup$
Let $Y=2X$ for any nondegenerate and integrable RV $X.$
answered Dec 20 '18 at 5:27
spaceisdarkgreenspaceisdarkgreen
33.6k21753
$endgroup$
Let $Y=2X$ for any nondegenerate and integrable RV $X.$
answered Dec 20 '18 at 5:27
spaceisdarkgreenspaceisdarkgreen
33.6k21753
answered Dec 20 '18 at 5:27
spaceisdarkgreenspaceisdarkgreen
33.6k21753
answered Dec 20 '18 at 5:27
spaceisdarkgreenspaceisdarkgreen
33.6k21753
answered Dec 20 '18 at 5:27
spaceisdarkgreenspaceisdarkgreen
33.6k21753
33.6k21753
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What do you think?
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– zoidberg
Dec 20 '18 at 4:43
$begingroup$
@norfair: I'm not sure ... the converse is definitely true ... I also know that $X,Y$ integrable and $E(X|Y)=Y$ and $E(Y|X)=X$ a.s. implies $X=Y$ a.s. but I haven't been able to figure out the one I ask ...
$endgroup$
– user521337
Dec 20 '18 at 4:47
$begingroup$
@norfair: do you have any ideas ?
$endgroup$
– user521337
Dec 20 '18 at 4:52