Necessity of tower property of conditional expectation [closed]
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Suppose $X$ and $Y$ defined on the same probability space $(Omega,mathcal{F},mathbb{P})$. We know that due to tower property of conditional expectation, if $sigma(X)subseteqsigma(Y)$ then $E[E[X|Y]|X]=X$. Is it also true for the opposite direction, viz, if $E[E[X|Y]|X]=X$ everywhere, can we conclude that $sigma(X)subseteqsigma(Y)$? Thanks for your help!
probability-theory
asked Nov 23 at 21:19
lychtalent
699
closed as off-topic by amWhy, Jean-Claude Arbaut, Leucippus, max_zorn, Davide Giraudo Nov 23 at 23:52
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." – amWhy, Jean-Claude Arbaut, Leucippus, max_zorn, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
Suppose $X$ and $Y$ defined on the same probability space $(Omega,mathcal{F},mathbb{P})$. We know that due to tower property of conditional expectation, if $sigma(X)subseteqsigma(Y)$ then $E[E[X|Y]|X]=X$. Is it also true for the opposite direction, viz, if $E[E[X|Y]|X]=X$ everywhere, can we conclude that $sigma(X)subseteqsigma(Y)$? Thanks for your help!
probability-theory
asked Nov 23 at 21:19
lychtalent
699
closed as off-topic by amWhy, Jean-Claude Arbaut, Leucippus, max_zorn, Davide Giraudo Nov 23 at 23:52
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." – amWhy, Jean-Claude Arbaut, Leucippus, max_zorn, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
Did you try something?
– Dante Grevino
Nov 23 at 21:31
Intuitively, if $E[E[X|Y]|X]=X$, it seems to indicate that after projecting to $sigma(Y)$, there is no information lost, but I have no clue how to prove it or even if the statement is true.
– lychtalent
Nov 23 at 21:35
You should include your work/thoughts in your question, not as comments.
– user10354138
Nov 23 at 23:43
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $X$ and $Y$ defined on the same probability space $(Omega,mathcal{F},mathbb{P})$. We know that due to tower property of conditional expectation, if $sigma(X)subseteqsigma(Y)$ then $E[E[X|Y]|X]=X$. Is it also true for the opposite direction, viz, if $E[E[X|Y]|X]=X$ everywhere, can we conclude that $sigma(X)subseteqsigma(Y)$? Thanks for your help!
probability-theory
asked Nov 23 at 21:19
lychtalent
699
Suppose $X$ and $Y$ defined on the same probability space $(Omega,mathcal{F},mathbb{P})$. We know that due to tower property of conditional expectation, if $sigma(X)subseteqsigma(Y)$ then $E[E[X|Y]|X]=X$. Is it also true for the opposite direction, viz, if $E[E[X|Y]|X]=X$ everywhere, can we conclude that $sigma(X)subseteqsigma(Y)$? Thanks for your help!
probability-theory
probability-theory
asked Nov 23 at 21:19
lychtalent
699
asked Nov 23 at 21:19
lychtalent
699
edited Nov 27 at 15:34
asked Nov 23 at 21:19
lychtalent
699
asked Nov 23 at 21:19
lychtalent
699
asked Nov 23 at 21:19
lychtalent
699
699
closed as off-topic by amWhy, Jean-Claude Arbaut, Leucippus, max_zorn, Davide Giraudo Nov 23 at 23:52
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." – amWhy, Jean-Claude Arbaut, Leucippus, max_zorn, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Jean-Claude Arbaut, Leucippus, max_zorn, Davide Giraudo Nov 23 at 23:52
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." – amWhy, Jean-Claude Arbaut, Leucippus, max_zorn, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
Did you try something?
– Dante Grevino
Nov 23 at 21:31
Intuitively, if $E[E[X|Y]|X]=X$, it seems to indicate that after projecting to $sigma(Y)$, there is no information lost, but I have no clue how to prove it or even if the statement is true.
– lychtalent
Nov 23 at 21:35
You should include your work/thoughts in your question, not as comments.
– user10354138
Nov 23 at 23:43
add a comment |
Did you try something?
– Dante Grevino
Nov 23 at 21:31
Intuitively, if $E[E[X|Y]|X]=X$, it seems to indicate that after projecting to $sigma(Y)$, there is no information lost, but I have no clue how to prove it or even if the statement is true.
– lychtalent
Nov 23 at 21:35
You should include your work/thoughts in your question, not as comments.
– user10354138
Nov 23 at 23:43
Did you try something?
– Dante Grevino
Nov 23 at 21:31
Did you try something?
– Dante Grevino
Nov 23 at 21:31
Intuitively, if $E[E[X|Y]|X]=X$, it seems to indicate that after projecting to $sigma(Y)$, there is no information lost, but I have no clue how to prove it or even if the statement is true.
– lychtalent
Nov 23 at 21:35
Intuitively, if $E[E[X|Y]|X]=X$, it seems to indicate that after projecting to $sigma(Y)$, there is no information lost, but I have no clue how to prove it or even if the statement is true.
– lychtalent
Nov 23 at 21:35
You should include your work/thoughts in your question, not as comments.
– user10354138
Nov 23 at 23:43
You should include your work/thoughts in your question, not as comments.
– user10354138
Nov 23 at 23:43
add a comment |
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Did you try something?
– Dante Grevino
Nov 23 at 21:31
Intuitively, if $E[E[X|Y]|X]=X$, it seems to indicate that after projecting to $sigma(Y)$, there is no information lost, but I have no clue how to prove it or even if the statement is true.
– lychtalent
Nov 23 at 21:35
You should include your work/thoughts in your question, not as comments.
– user10354138
Nov 23 at 23:43