How to prove that $\mathbb{E}[Y|X]=a$ some constant when Y and any Borel measurable function of X are uncorrelated?

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?

• The title and question are rather different Sep 19, 2012 at 6:34
• Oh! sorry about that! I corrected the mistake Sep 19, 2012 at 15:26

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.

• 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, 2012 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 Sep 19, 2012 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, 2012 at 20:52