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Im currently reading mostly harmless econometrics. The three relevant pages are here: http://econ.lse.ac.uk/staff/spischke/mhe/ex_ch3.pdf

My issue is the following. They claim that epsilon is uncorrelated with any function of $h(X)$ in theorem 3.1.1 and later they claim that it is uncorrelated with $E(Y | X)$ in theorem 3.1.3. So far I always thought that its just an assumption that epsilon is uncorrelated with $X$ in a data analysis. How am I to understand their proof?

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by conditional expectation: $$E(h(X)\varepsilon) = E(h(X) E(\varepsilon|X)) = E(h(X)0) = 0,$$ i.e. $h(X)$ and $\varepsilon$ are uncorrelated.

To shed some light on the first part of the theorem in the satement, note that one can always write $Y=E(Y|X) + \varepsilon$, where $\varepsilon = Y-E(Y|X)$ yielding $E(\varepsilon|X)=0$, since $$E(Y|X) = E(Y|X) + E(\varepsilon|X) \quad \Rightarrow \quad E(\varepsilon|X)=0.$$

However, this is pretty boring since it doesn't give you any helpful information about the relationship of $Y$ and $X$. In order to model the relationship you will assume $Y=g(X) + \varepsilon$ for some specific $g(X)$. For this specific model to be meaningful (as in $g(X) = E(Y|X)$) you then need to assume $E(\varepsilon|X)=0$:

For a model of the form $Y=g(X)+\varepsilon$ we have $$E(\varepsilon|X) = 0 \Leftrightarrow g(X)=E(Y|X).$$ Indeed, If $E(\varepsilon|X)=0$ then $$E(Y|X)=E(g(X)|X)+E(\varepsilon|X)= g(X).$$ On the other hand, if $g(X)=E(Y|X)$ then $Y=E(Y|X) + \varepsilon$ and we have $$E(Y|X) = E(Y|X) + E(\varepsilon|X) \quad \Rightarrow \quad E(\varepsilon|X)=0.$$

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  • $\begingroup$ yes, that proof makes sense to me. but does that mean that E (Y I X) is also a function of X? and i thought that the indpedence of X and epsilon was a statistical assumption, not something you could prove $\endgroup$
    – leo
    Commented Oct 23, 2017 at 9:43
  • $\begingroup$ yes, the conditional expectation $E(Y|X) $ can be seen as a function of $X$. $E(\varepsilon|X)=0$ is indeed an assumption and quite a strong one in practice. while it may indeed be hard to check, ther are simple examples where you know that the assumption can not hold. for example, In a standard regression model this occurs whenever you omit relevant variables in the regression and the omited variable is correlated with $X$. it's easy to imagine that this happens quiet often in practice. $\endgroup$
    – BloXX
    Commented Oct 23, 2017 at 10:36
  • $\begingroup$ but they say they prove E(ε|X)=0 in theorem 3.1.1 i. so why is it an assumption? $\endgroup$
    – leo
    Commented Oct 23, 2017 at 10:44
  • $\begingroup$ i expanded my answer. they already modelled $Y$ in an equivalent way as assuming $E(\varepsilon|X)=0$. $\endgroup$
    – BloXX
    Commented Oct 23, 2017 at 10:53
  • $\begingroup$ thanks so much, this has made it clearer to me. So assuming the independence and choosing E(Y|X) as the function of X is equivalent. Thats why it remains an assumption. And then i suppose even the proof in theorem 3.1.1 ii (where they show the independence of epsilon from any function h(X)) rests on this assumption as it still uses E(ε|X)=0? $\endgroup$
    – leo
    Commented Oct 23, 2017 at 12:20

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