Prove that if $E[X|\sigma(Y)] = Y$ and $E[Y|\sigma(X)] = X$ then $X = Y$ almost surely.

Asked Aug 15 '15 at 16:41

Active Aug 15 '15 at 17:18

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Prove that if $E[X|\sigma(Y)] = Y$ and $E[Y|\sigma(X)] = X$ then $X = Y$ almost surely. This is my idea:

By assumption, $Y = E[X|\sigma(Y)] = E\left\lbrack E[Y|\sigma(X)]|\sigma(Y)\right\rbrack$, and I would like to show that this equals $E[Y|\sigma(X)]$ which equals $X$. If I can show that $E[Y|\sigma(X)]$ is $\sigma(Y)$ measurable, then the proof is finished. But I am not sure how to do this.

Any help would be greatly appreciated.

edited Aug 15 '15 at 16:56

Augustin

8,446

asked Aug 15 '15 at 16:41

Ann Marie

1

As a hint, you can try computing $E[(X-Y)^2]$. – Michael Aug 15 '15 at 17:11
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Also, I find it easier to just use $E[X|Y]=Y$ and $E[Y|X]=X$. – Michael Aug 15 '15 at 17:12
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Well, I guess my approach also assumes second moments are finite. – Michael Aug 15 '15 at 17:17
@ByronSchmuland: Yes, it is a duplicate, but I don't think that the answer given there is correct. – PhoemueX Aug 15 '15 at 17:58
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@PhoemueX Try this one: http://math.stackexchange.com/questions/385109/question-about-conditional-expectation – Aug 15 '15 at 18:06
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Thanks Augustin for helping edit my question. Thanks Michael and Byron for the help. I think I understand it mostly. – Ann Marie Aug 15 '15 at 18:27
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Also @Michael your first hint about computing E[(X−Y)2] is much shorter and simpler. Thanks a lot :) – Ann Marie Aug 15 '15 at 18:35
@Michael That is much better than the answers given in the other questions! – BCLC Sep 08 '15 at 01:21
@Michael Oh right, second moments aren't necessarily finite. Oh well. Can't use that, Ann Marie. – BCLC Sep 08 '15 at 01:23

Prove that if $E[X|\sigma(Y)] = Y$ and $E[Y|\sigma(X)] = X$ then $X = Y$ almost surely.

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