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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?

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The title and question are rather different – Henry Sep 19 '12 at 6:34
Oh! sorry about that! I corrected the mistake – ChuckM Sep 19 '12 at 15:26
up vote 2 down vote accepted

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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No, 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. – Did Sep 19 '12 at 17:24
@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 – Michael Chernick Sep 19 '12 at 20:39
@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. – Did Sep 19 '12 at 20:52
I 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 – Michael Chernick Sep 19 '12 at 21:22
@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. – Did Sep 20 '12 at 5:06

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