How can I prove that $\mathbb{E}[Y|X] = a$, if $Y$ and $g(x)$ are uncorrelated with any borel measurable function $g$? Can I conclude the same for $\mathbb{E}[Y|X] = a$ where $a$ is constant?
How to prove that $\mathbb{E}[Y|X]=a$ some constant when Y and any Borel measurable function of X are uncorrelated?
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probability
statistics
probability-theory
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4The title and question are rather different – 2012-09-19
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0Oh! sorry about that! I corrected the mistake – 2012-09-19
1 Answers
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One knows that $\mathrm E(Y\mid X)=h(X)$ for some suitable measurable function $h$. If $Y$ and $h(X)$ are uncorrelated, then $\mathrm E(Yh(X))-\mathrm E(Y)\mathrm E(h(X))=0$. Since $\mathrm E(Yh(X))=\mathrm E(h(X)^2)$ and $\mathrm E(Y)=\mathrm E(h(X))$, one sees that $\mathrm{var}(h(X))=0$, hence $h(X)=c$ almost surely, for some $c$. That is, $\mathrm E(Y\mid X)=c$ almost surely. Finally, $\mathrm E(\mathrm E(Y\mid X))=\mathrm E(Y)$ hence $c=\mathrm E(Y)$, that is, $\mathrm E(Y\mid X)=\mathrm E(Y)$ almost surely.
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0thanks for the help. Do you mean that $h(X)$ is sigma-field constructed from $X$? – 2012-09-19
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1No, h(X) is a measurable function of a random variable hence is a random variable itself. If I wanted to mention the sigma-algebra generated by X, I would use the notation $\sigma$(X), like everybody else. – 2012-09-19
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01. How do you know that E[Y|X] is a function of X alone? I think that needs to be an assumption. 2. How did you go from having E(Yh(X))−E(Y)E(h(X))=0, to var(h(X))=0? I don't follow your proof at all. – 2012-09-19
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0@MichaelChernick These are standard conditional expectations definitions. First, E(Y|X) is **always** a random variable h(X) for some measurable function h (hence your 1. is wrong). Second, for every measurable function g, one knows that E(Yg(X))=E(E(Y|X)g(X))=E(h(X)g(X)) by the very definition of the conditional expectation. And here one assumes that E(Yg(X))=E(Y)E(g(X)) for every function g. Take g=h and use E(Y)=E(h(X)). – 2012-09-19
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1@did Sorry about 1 I was thinking of Y as being a function of X and another variable U but upon taking conditional expectations If U is independent of X the part involving U will just be constant and if U is dependent on X the part involving U will also be a function of X. For step 2 you showed that you gave a very poor answer by leaving out so many steps. Now Var(h(X))=E(h$^2$(X))-E$^2$(h(X)). If I foolow you correctly I take g=h so E(Yg(X))=E(Y)E(g(X)) becomes E(Yh(x))-E(h(X))E(h(X)). Then using the little trick of taking expectation of conditional expectation – 2012-09-19
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0I get E(Yh(X))=E(h$^2$(x)) and from that it becomes clear that Var(h(X))=0. Again you left out a lot of steps that were not obvious to me and probably not to the OP either. – 2012-09-19
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4@MichaelChernick How do you know? Who asked for your advice and why should we care about it? If you are lost, learn the subject (there exist some excellent books). You just exposed your poor mathematical background (and your bad manners) in plain view once again. – 2012-09-19
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1I have a masters degree in mathematics and a PhD in statistics. You admonished me for answers that were not specific. This site is a teaching site. The intent is for readers to be able to read answers and understand them even if they are not as well trained in mathematics as you or I. An answer is not a good answer if it is hard to follow. I have used bad manners in the past and i am toning it down. Your response seems to show that you don't care about the community. I filled in the blanks because I do care that the community understands. I believe the OP gave your answer a check mark without – 2012-09-19
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2@MichaelChernick Who said you represent the community or that you know what the community thinks or wants? Who asked you to insult me (and others, as the OP here) and to belittle answers posted on the site, as you repeatedly did and are still doing? Once again, if you lack some basic mathematical expertise, this is YOUR problem, not ours. Once again, if you post some wrong answers on the site, as you did in the past, I will say so, as I did in the past. At the moment, from what I see, your behaviour is a nuisance to this community. – 2012-09-20
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0@did I guess you are closed minded and we are not going to get along. But I will drop it. I probab;y will stop commenting on your answers form mow on. – 2012-09-20
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0@MichaelChernick Your last comment could have been perfect, but you could not resist one last insult, could you. – 2012-09-20
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0@did I was not trying to insult. Your previous comment was a rather vicious unwarranted attack on me. I made a point about making answers clear being more important than terse elegant answers. Instead of appreciating that comment you chose to attack it. That seems closed minded to me. These types of discussions really should not be held in comments here. A moderator should delete most of these comments. Yours and mins. – 2012-09-20
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1@MichaelChernick I fully understand why you want to hide your tracks. For me, I stand behind everything I wrote. – 2012-09-20
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0@did I stand behind my comments and accept my mistakes which I acknowledge and I do change my behavior when it is pointed out that I misbehaved. My request that a moderator remove the comments is because rules or guidelines for the site say that comments to questions should be limited to issues relevant to the question such as requests for clarifications or interesting side points that are not direct answers to the questions. Lengthy discussions related or un related to the topic should be left for chat. – 2012-09-20