# 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?

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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

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 '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