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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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    Oh! sorry about that! I corrected the mistake2012-09-19

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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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    @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